From Dr. Roy Spencer’s Global Warming Blog
by Roy W. Spencer, Ph. D.
This month I’m adding plots for USA48 and Canada, too.
The Version 6.1 global average lower tropospheric temperature (LT) anomaly for April, 2026 was +0.39 deg. C departure from the 1991-2020 mean, which remains statistically unchanged for 4 months now.
The Version 6.1 global area-averaged linear temperature trend (January 1979 through April 2026) remains at +0.16 deg/ C/decade (+0.22 C/decade over land, +0.13 C/decade over oceans).
The following table lists various regional Version 6.1 LT departures from the 30-year (1991-2020) average for the last 28 months (record highs are in red). Note I’ve added Canada to the table this month, by request (although WordPress won’t allow me to add September 2024 for some reason). The warmest April in Canada was in 2010 (+2.61 deg. C), while the warmest anomaly out of all months was in January 1981 (+3.75 deg. C).
| Year | Mon | Globe | NHem | SHem | Tropic | US48 | Arctic | Aust. | Can. |
| 2024 | Jan | +0.80 | +1.02 | +0.57 | +1.20 | -0.19 | +0.40 | +1.12 | +0.97 |
| 2024 | Feb | +0.88 | +0.94 | +0.81 | +1.16 | +1.31 | +0.85 | +1.16 | +2.45 |
| 2024 | Mar | +0.88 | +0.96 | +0.80 | +1.25 | +0.22 | +1.05 | +1.34 | +1.12 |
| 2024 | Apr | +0.94 | +1.12 | +0.76 | +1.15 | +0.86 | +0.88 | +0.54 | +1.39 |
| 2024 | May | +0.77 | +0.77 | +0.78 | +1.20 | +0.04 | +0.20 | +0.52 | +0.67 |
| 2024 | June | +0.69 | +0.78 | +0.60 | +0.85 | +1.36 | +0.63 | +0.91 | +0.19 |
| 2024 | July | +0.73 | +0.86 | +0.61 | +0.96 | +0.44 | +0.56 | -0.07 | +1.15 |
| 2024 | Aug | +0.75 | +0.81 | +0.69 | +0.74 | +0.40 | +0.88 | +1.75 | +1.36 |
| 2024 | Sep | +0.81 | +1.04 | +0.58 | +0.82 | +1.31 | +1.48 | +0.98 | |
| 2024 | Oct | +0.75 | +0.89 | +0.60 | +0.63 | +1.89 | +0.81 | +1.09 | +0.89 |
| 2024 | Nov | +0.64 | +0.87 | +0.40 | +0.53 | +1.11 | +0.79 | +1.00 | +1.61 |
| 2024 | Dec | +0.61 | +0.75 | +0.47 | +0.52 | +1.41 | +1.12 | +1.54 | +1.65 |
| 2025 | Jan | +0.45 | +0.70 | +0.21 | +0.24 | -1.07 | +0.74 | +0.48 | +1.04 |
| 2025 | Feb | +0.50 | +0.55 | +0.45 | +0.26 | +1.03 | +2.10 | +0.87 | -0.35 |
| 2025 | Mar | +0.57 | +0.73 | +0.41 | +0.40 | +1.24 | +1.23 | +1.20 | +0.80 |
| 2025 | Apr | +0.61 | +0.76 | +0.46 | +0.36 | +0.81 | +0.85 | +1.21 | +0.45 |
| 2025 | May | +0.50 | +0.45 | +0.55 | +0.30 | +0.15 | +0.75 | +0.98 | +0.81 |
| 2025 | June | +0.48 | +0.48 | +0.47 | +0.30 | +0.80 | +0.05 | +0.39 | -0.22 |
| 2025 | July | +0.36 | +0.49 | +0.23 | +0.45 | +0.32 | +0.40 | +0.53 | -0.23 |
| 2025 | Aug | +0.39 | +0.39 | +0.39 | +0.16 | -0.06 | +0.82 | +0.11 | +0.62 |
| 2025 | Sep | +0.53 | +0.56 | +0.49 | +0.35 | +0.38 | +0.77 | +0.30 | +2.44 |
| 2025 | Oct | +0.53 | +0.52 | +0.55 | +0.24 | +1.12 | +1.42 | +1.67 | +2.59 |
| 2025 | Nov | +0.43 | +0.59 | +0.27 | +0.24 | +1.32 | +0.78 | +0.36 | +1.47 |
| 2025 | Dec | +0.30 | +0.45 | +0.15 | +0.19 | +2.10 | +0.32 | +0.37 | -1.86 |
| 2026 | Jan | +0.35 | +0.51 | +0.19 | +0.09 | +0.30 | +1.40 | +0.95 | +1.17 |
| 2026 | Feb | +0.39 | +0.54 | +0.23 | +0.03 | +1.91 | -0.48 | +0.73 | +0.32 |
| 2026 | Mar | +0.38 | +0.33 | +0.42 | +0.07 | +3.74 | -0.48 | +1.14 | -3.17 |
| 2026 | Apr | +0.39 | +0.43 | +0.34 | +0.23 | +1.20 | +0.30 | +0.70 | -0.89 |
| Year | Mon | Globe | NHem | SHem | Tropic | US48 | Arctic | Aust. | Can. |
Time Series Plots for USA48 and Canada
Starting this month I will include time series graphs for USA48 and Canada, in addition to the usual global plot. Note that for the previous month (March) the record warmth in USA48 (+3.74 deg. C) was in stark contrast to the coldest March in Canada in the 48-year satellite record (-3.17 deg. C).
The full UAH Global Temperature Report, along with the LT global gridpoint anomaly map for April, 2026 and a more detailed analysis by John Christy, should be available within the next several days here.
The monthly anomalies for various regions for the four deep layers we monitor from satellites will be available in the next several days at the following locations:
Why are we still looking at average global temperatures?
Because it allows us to determine how quickly our planet is recovering from the ice age 10,000 years ago…now that only vestigial masses of ice remain, mostly in polar regions.
It is so that we can test the hypothesis that the global average temperature will increase/decrease.
This should answer your question: ... https://www.rgkernodle.com/the-intellectual-fraud-of-global-temperature-a-crime-against-reason
Good summary.
“global”
Definition:
Pertaining to the whole of something
Which means that the “global temperature” must pertain to my street, and all the other streets around the globe.
So it’s currently “15C and Sunny” everywhere around the planet, just as it is at my location?
Here’s an idea – why don’t we start using more precise, accurate descriptors for such numeric constructs as this, say something like –
“formulaic constructs of selected temps readings at different times from around the world”?
Make what you will of them.
The measured surface temperature has been rather stagnant too. The anomaly average (base 1961-90) for April was 1.01°C, only slightly down from March 1.025°C. Here is the graph of the last four years (different base years):
Really not much down from the 2024 record warm.
These are not measurements. They are the output of questionable statistical algorithms using local heat contaminated convenience samples as input. Junk science personified.
Yep.
It’s real X-Files inspired stuff –
“I want to believe!”
I showed above a graph of the major surface indices, along with satellite UAH and RSS, over the last 4 years on a common anomaly base, 1981-2010. Satellite and surface are all very close. UAH is a little lower, but then, RSS is higher.
And that means what? Show it for the period beginning in 1900.
Show satellite data starting 1900?
Sure, why not?
Just make some numbers up.
Reading difficulties? “…major surface indices,…”
Perhaps you are unaware that history extends to before 1981.
Not for satellite data. Nick’s graph is a comparison of surface and satellite datasets. It’s not possible to extend that comparison back to 1900.
Yes, but that’s not my point. I asked what significance the (short) time series has, meaning in the context of the subject of climate, which is, after all, the whole purpose of the discussion here.
It is possible to include the entire temperature history instead of choosing one small bit, ostensibly for the purpose of supporting claims of human interference in the system and Nick’s claim of 2024 being a “record” in terms of warmth.
Nick isn’t claiming human interference here. He is claiming that the surface and satellite datasets are similar. He is essentially testing the hypothesis that the surface datasets will show something substantially different than satellite datasets. As you can see the hypothesis is false.
I’m not suggesting we shouldn’t discuss temperatures back to 1900. We should. But that’s a different conversation and one in which we’ll have to forgo consideration of satellite data since it didn’t exist that far back.
[JCGM GUM-6:2020[ on Developing and using measurement models says it is.
Only for repeatability conditions — the same quantity measured with the same apparatus!
The same quantity here is the regional weather, which several independent thermometers are measuring.
To which region do you refer?
Air temperature measurements are a time series, there is exactly one opportunity to measure a single temperature at a single location when it is gone forever.
Most regions. Exceptions are remote areas of the planet (e.g., the High Arctic and Antarctic, the Indian Ocean, etc.)
“Air temperature measurements are a time series, there is exactly one opportunity to measure a single temperature at a single location when it is gone forever.”
But even at a single location, multiple thermometers can measure temperature simultaneously, giving you repeated observations rather than just one opportunity.
Two thermometers, maybe, doubling the cost of the instrumentation for very little gain (other than failure redundancy, perhaps).
And still, the next instant of time, that temperature is gone forever.
Usually its just the opposite — Fake Data inserted into nonexistent sites.
The measurement uncertainty of each thermometer STILL ADDS. Uncertainty *always* adds for multiple measurement devices. You simply cannot reduce uncertainty by using different measurement devices.
Uncertainty only cancels when measuring the same thing multiple times under the same environment using the same device – and even then certain conditions have to be confirmed before assuming cancellation, such as a Gaussian distribution of measurement values.
Saying it doesn’t make it true. The irony is that the majority of the examples in the GUM are of measurements that are themselves dependent upon measurements of different things. And the majority of those require the use of a different apparatus. Literally the first example (pg. 3) in the GUM is combining measurements of two different things to compute a third measurement. And the second example (pg. 4) combines measurements requiring the use of a different apparatus. The examples expand from there and get arbitrarily more complex. So your statement here is patently wrong.
This is just your usual sophistry and abuse of the GUM to justify ignoring variance.
And you continue to refuse to understand or acknowledge that an air temperature can only be measured once, averaging multiple measurements of the same quantity is impossible.
And again, the average formula is NOT a “measurement model”, your need for it to be so notwithstanding.
It’s me pointing out that your statements are contradictory to what the GUM actually says.
The reason I refuse to acknowledge your position here is because its demonstrably wrong according to the GUM.
Which you keep saying. Show me exactly where the GUM prohibits y = f(x1, x2, …, xN) = Σ[x: {1 to N}] / N.
Don’t defect. Don’t divert. Don’t post irrelevant content from the GUM that has nothing to do with the topic.
Post the exact text that says that equation or the operations used in that equations are prohibited when defining y.
a through i, all ignored by climatology, and especially:
You ignore variations under differing conditions!
You have never explained how you get a Type A uncertainty from measurements of differing conditions!
You have never shown why it is acceptable to ignore the uncertainty contributions of your fav “measurement model”:
4.1.3 is not your friend, either:
You just skip ahead to 4.1.4 and hope it applies to your abuse of the GUM and glomming together a bazillion different thermometers, then ignore 4.1.5.
None of 4.1.3, 4.1.4, or 4.1.5 say that you cannot define y = f(x1, x2, …, xN) = Σ[x: {1 to N}] / N. Nor does it insinuate, imply, or even hint that it is prohibited.
The rest of your comment has no bearing whatsoever to anything I said. You’re just randomly picking sections of GUM to defect and divert away from the fact that no where in any of the GUM documents does it prohibit a measurement model from computing an average of the input quantities.
The GUM defines that this ONLY APPLIES TO A SINGLE MEASURAND!
“4.1.1 In most cases, a measurand Y”
Measurand – singular!
4.1.2 “The input quantities X1, X2, …, XN upon which the output quantity Y”
Y – singular measurand!
“4.1.4 An estimate of the measurand Y, denoted by y, is obtained from Equation (1) using input estimates x1, x2, …, xN for the values of the N quantities X1, X2, …, XN.”
measurand – SINGULAR
x1, x2, etc are MEASUREMENT estimates for the Singular measurand.
The average is *NOT* a measurement. It is *NOT* an input quantity for anything. It is *NOT* a measurement of multiple different things.
You have *NEVER* understood what the values X/x and Y/y stand for in the GUM. It’s because you keep cherry picking from the GUM instead of actually reading it for meaning and context!
Again, x1/x2/etc are input quantiity ESTIMATES for a singular measurand. You are trying to extend it to mean a Y value estimate for *multiple* measurands.
READ THE GUM. STOP CHERRY PICKING!
Projection mode?
The GUM is a standard for EXPRESSING uncertainty, it is not immune to GIGO abuse — Garbage In, Garbage Out.
Actually it does.
From JCGM 200-2012
2.48
A mathematical relation among all quantities involved in a measurement.
What is the mathematical relation between the data points in a mean? There is none. Each data point is independent from the others. If this wasn’t true then one would need to include a covariance factor in the uncertainty. I have never seen climate science do that.
“What is the mathematical relation between the data points in a mean?”
Y = (X1 + X2 + … + Xn) / n
What is the physical relationship between X1 and X2 and Xn that allows you divide each by “n”, the same counting number (not a constant).
Its the Magic Number!
Who says there has to be any relationship between the different Xs?
Most likely though they are telated by comming from a particular population, but that has no relevance to the question of propagating measurememt uncertainties.
The divide by n is simply what you do to get an average. I’ve no idea why you think you need permission to do it.
“Its the Magic Number!”
Sure is. Any other number would be wrong.
Because if Y = f(X1,X2,…)
Then f(X1,X2,…)/n = Y/n
Y ≠ Y/n
You defined two different measurement models that have different values.
I can only repeat you don’t know how functions work.
The function of an average involves dividing by n. You don’t then divide the average by n.
Either,
f(X1,X2,…) = (X1 + X2 + …) / n
In which case Y = f(X1,X2,…), means Y is the average. Or
f(X1,X2,…) = (X1 + X2 + …)
In which case Y = f(X1,X2,…) is the sum, and Y / n is the average.
It almost defies credulity to think that to this day you still cannot consistently distinguish between a function that computes an average and one that computes a sum.
An average is *NOT* a function – it is a statistical descriptor of a set of values. Look at that term SET carefully!
The values in a data set are *NOT* independent variable values being input into a function to get an output value. There is no dependent/independent relationship between AN INPUT VARIABLE SET OF DATA and an output value.
A statistical descriptor is *NOT* a mapping from an input variable to a dependent output variable.
A statistical descriptor is a mapping from a dataset to a summary value.
The input of a statistical descriptor is *NOT* an independent input variable.
A statistical descriptor does *NOT* result in a physical relationship. A FUNCTION describes a causal or deterministic relationship.
You are trying to say that if you can write an equation then it is a function. It is *NOT* a function unless it meets the mapping requirement from an input independent variable resulting in a dependent output.
Whoever taught you that a statistical descriptor equation is a “function” taught you wrong.
Think of it this way: a function can be used in a differential equation. Can you use a statistical descriptor in a differential equation?
Think of it this way: can you use a statistical descriptor to infer causal physical relationships? A function can, a statistical descriptor cannot.
An average does not tell you anything about a physical relationship. It only tells you a partial description of a set of data.
You don’t think f(a,b) = (a+b)/2 is a function?
Do you think f(a,b) = a+b is a function?
Do you think NIST is wrong to allow me to enter both into the uncertainty machine?
The use of the letter “f” does *NOT* mean you have a function.
A function maps single inputs to one output. It represents a physically meaningful mapping.
If (a+b)/2 is meant to be an average that is *NOT* a function. The base equation for an average is (Σx_i)/n and in this case x_i = (a+b). The equation (Σx_i)/n describes the collection of possible x_i values.
It is a STATISTICAL DESCRIPTOR for the set of x_i values. It describes the collection of possible input values, it does *NOT* map any individual x_i to an output y.
Are saying that neither f(a,b) = (a+b)/2 nor f(a,b) = a+b are functions?
Why do you keep asking loaded questions?]
They are mathematical functions. Simple equations. They are not functional relationships if there is no physical connection between the variables.
The functional relationship of A = 1/2(a x b) relates a to b to define a physical value.
f(a,b) = 1/2(a+b) is not a physical functional relationship unless you provide the physical connection between a and b and the 1/2.
In the case of a midsegment length in an isosceles trapezoid there is a physical connection between f(a,b) = m = 1/2(a + b). The functional relationship maps to a definite measurement. f(a,b) = 1/2(a + b) does not map temperature to a measured value for temperature. For temperature it is a statistical descriptor to find a central value between to temperatures, not a functional relationship.
Same for f(a,b) = a + b. It is not a physical functional relationship unless you provide the physical connection between a and b.
A functional relationship maps the two variables onto a physical value in the measurement of length, volume, any extensive property.
f(a,b) = a + b, does not map temperatures onto a single physical value. It is not a functional relationship for temperature.
If you want to continue to argue about averaging temperature, why don’t you find a source that says temperatures CAN be averaged and then list the requirements that allows one to do that.
If you are using “f” to denote a function then using it to describe the mathematical operation of calculation of an average, a statistical description, is confusing you.
If you want an average, you should use a different symbol for it. E.g. d(a,b) = (a+b)/2, where “d” means description.
Using the symbol “f” doesn’t mean that what follows is a function.
If that is confusing for you then I suggest that you *do* start using the symbol “d” instead of “f” when describing the calculation of an average. I would suggest using “s” but that is likely to confuse you even more into thinking the calculation is for standard deviation.
A function maps an input to an output. A function does not describe the “shape” of a collection of inputs.
A statistical descriptor describes the “shape” of a collection of inputs. A statistical descriptor does not map an input to an output.
A function is not a statistical descriptor. A statistical descriptor is not a function.
The “different” things are actually different properties of a SINGULAR measurand. bdgwx simply can’t seem to get that into his head.
He has *never* taken the time to understand that Y/y and X/x are defined as in the GUM. Cherry picking champion!
“Saying it doesn’t make it true. The irony is that the majority of the examples in the GUM are of measurements that are themselves dependent upon measurements of different things.”
Malarky! You are equivocating again! “different things” are PROPERTIES OF THE SAME MEASURAND. Such as the length and width of a table top! Two different properties used to calculate the area OF A SINGULAR measurand.
If you measure the length and width of two table tops and get two values, Y1 and Y2, for the areas of each, then if you average those two values YOU ADD THE UNCERTAINTIES OF EACH together to get a total uncertainty. You do *NOT* average the uncertainties – they ADD!
The uncertainties of Y1 and Y2 will either be Type A uncertainties determined by multiple measurements of each individual property added in root-sum-square or a Type B uncertainty.
AGAIN, read the GUM for meaning and context. STOP CHERRYPICKING!
The record caused by a very strong El Nino.
…. with zero evidence of any human causation expect for the highly suspect urban surface sites the measurements were made at.
Yet USCRN continues to warm faster than the adjusted nClimDiv…
It is mathematical idiocy to compare trends between real and fake data. !
Oh yeah, forgot. NOAA are faking nClimDiv in order to make it warm as fast as USCRN.
You would know all about idiocy.
The Hunga Tonga eruption likely contributed as well. But in Nick’s world, C02 is the driver of everything.
And the evidence for this is….?
Let’s see it from 1930.
OK, but it needs a 12-month running average to reduce the noise:
Reverting to the long term warming average. It’ll be interesting to see if it continues to fall below that long term average for a while.
The global anomalies for 2026 have started off remarkably consistent. Only 0.04C separating all four months. Each month has been either 6th or 5th warmest.
Ten warmest Aprils since 1978:
Here’s my unofficial version of the UAH map.
Still a work in progress, but here is my estimate of anomalies for each country, starting with the 10 warmest anomalies.
and the 10 coldest anomalies:
Of course. smaller countries are more likely to have more extreme anomalies. so this ranking don’t mean much.
In contrast to last month, no individual country had a record hot or cold April.
France and Italy both had their 2nd warmest April in the UAH data set, and several countries had their 3rd warmest – including the USA (that’s the entire states, not just USA48).
Puerto Rico had its equal 5th coldest April, and Belarus its 6th coldest.
But, say Canada at 50-60 degree latitude, covers nearly 1/4 of the planet circumferentially and it is -0.89….so probably more than balances Luxembourg at 1.98….just kidding…
As nobody asked, here’s a trend per country breakdown.
The 20 fastest warming countries since December 1978:
Uncertainties are for a 2σ, and are not corrected for auto-correlation.
It’s not too surprising when you look at the global trend map, that this list is dominated by Central European countries.
Here are the slowest warming countries. (Countries are using the names from rnaturalearth. Some of these are not actually countries.)
No country has a negative trend, and all but the French Southern and Antarctic Lands have a significant trend.
Antarctica is the main slow warming land mass – again not surprising. Much of the rest of the list come from small countries and islands. This may be a result of the satellite data mixing land and sea over each grid point.
Here’s the top 20 for just the month of April
and the lowest trends
Most of these are not significant, and again are dominates by islands. But it’s interesting to see Canada as the only country with a non-positive trend for April. This is similar in February and March. Not sure why this should be, but I wonder if it’s connected with a warming Arctic
Here is my corresponding map of the surface temperatures:
Nick using a reference period that was the COLDEST since 1900..
The period of the GREAT GLOBAL COOLING SCARE
Not fooling anyone.
And GHCN is a totally bogus and unscientific FANTASY fabrication anyway.
AAwww, that’s so pretty, Nick.
(you could lose the black splotches though, I reckon)
How about just a kinda mauve & teal shaded combo?
Do you like this better?
I prefer our blue planet.
https://imgs.search.brave.com/j0BaOVMOYzaIBlOLqiG_XgFa4AF2POGmlTjiTw1sm3U/rs:fit:500:0:1:0/g:ce/aHR0cHM6Ly9pbWFn/ZXMucGV4ZWxzLmNv/bS9waG90b3MvMzA1/OTYyNDUvcGV4ZWxz/LXBob3RvLTMwNTk2/MjQ1L2ZyZWUtcGhv/dG8tb2Ytc3R1bm5p/bmctdmlldy1vZi1l/YXJ0aC1mcm9tLXNw/YWNlLmpwZWc_YXV0/bz1jb21wcmVzcyZj/cz10aW55c3JnYiZk/cHI9MSZ3PTUwMA
Nick ate all the blue and green crayons.
We are doomed !
The eastern part of the Pacific has shown cooling over the last 30 years. Your graph indicate warming. I see no white that shows any of the Pacific is cooling.
You do understand that a single month is not a trend?
Here’s the UAH trends over the last 30 years. There’s only a small patch of the pacific that’s shown a negative trend.
And here’s the same over the last 45 years.
1979 was the COLDEST year since the much warmer 1930,40.
Thank goodness the planet started to warm out of the Great Global Cooling Scare.
Nope. In the 1970s only 1977 was warmer, globally, than 1979; 1973 was a tie. The only years warmer that 1979 globally since 1930 were 1941 and 1944.
No year has been colder than 1979 globally since 1985.
Pretty non-descript isn’t it. 🙂
All that white are must be caused by CO2 !! 😉
You sure it’s 0.04 degrees, not 0.0324 degrees?
Could be the difference between humanity surviving or disappearing forever.
Could Dr Spencer, whose corpus is authentic data and not propaganda, please present measurements in absolute units as part of his output.
This “anomaly” nonsense obscures any of our attempts to talk facts, figures and trends to our nearest and dearest and the wider community.
Specifically, what rate of change and the direction of change of the surface temperature of the northern hemisphere and southern hemisphere is becoming apparent. I am looking for data which confirms or refutes the current GSM cycle as “cooling” and whether this one is really going to be a minimum comparable to previous quite cold spells like the Maunder Minimum.
The trend in the absolute data is identical to the trend in the anomaly data. Do you understand what anomalies are?
You seem to be knowledgeable enough to criticize others. Why don’t you post the experimental standard deviation of the means, the k factor used, and the degrees of freedom for the measurement uncertainty in these temperatures? Include both the anomalies and the absolute temperature uncertainties.
Why doesn’t Roy Spencer post the standard deviation of the means and the monthly measurement uncertainties? It’s UAH’s data, not mine.
To Roy W. Spencer, Ph. D., I do appreciate you releasing these monthly UAH data plots and data tables to the public . . . but PLEASE, PLEASE, PLEASE insert a disclaimer to the effect that “reporting a satellite-derived temperature anomaly average to a precision of +/- 0.01 C for any specified geographic area and/or any averaged time interval is a MATHEMATICAL artifact only and is not scientifically justified.”
Funny how UAH data becomes “scientifically unjustified” when it confirms continued warming. For years here it was championed as the ‘gold standard’. See Monckton et al. Not a word about ‘mathematical artifacts’ when a minute several-year cooling trend could be scratched out if it.
UAH does not confirm continued warming
It confirms warming ONLY at strong El Nino events.
There is no “CO2 warming” signature at all in the UAH data.
Apparently you missed this statement by Dr. Spencer that is immediately below the first graph, UAH GLAT temperature history, in his above article:
“The Version 6.1 global area-averaged linear temperature trend (January 1979 through April 2026) remains at +0.16 deg/ C/decade (+0.22 C/decade over land, +0.13 C/decade over oceans).“
(my bold emphasis added)
That would be . . . let’s see . . . yeah, 47+ years of “continued warming”.
Reading comprehension 101.
What is the linear temperature trend from January 2024 to the present?
Is it a new temperature trend from January 2024?
Yeah, let’s start the data at the peak of the last El Nino warming and then claim there has been ‘global cooling’!
I think even the regulars here have cottoned on to the silliness of this – eventually!
How is the Earth’s decreasing average temperature from 2024 Apr +0.94 to 2026 Apr +0.39 explained in the Earth’s energy balance diagram?
0.39-0.94=-0.55 degrees Celsius.
Where has the temperature decrease of -0.55 degrees Celsius gone in the Earth’s energy balance diagram?
When the Earth’s average temperature is decreasing, there is less incoming energy and more outgoing energy in the Earth’s energy balance diagram.
https://wattsupwiththat.com/2026/04/24/earth-energy-imbalance-the-sun-versus-co2/
It isn’t captured in most of the energy balance models per se because they typically use long averaging periods. However, if were to apply to the same 3 layer model concept to only the period from 2024/04 to 2026/04 it would be captured as a decrease of the components going into layer 2 (atmosphere) and an increase of the components of coming or a combination thereof.
Yes and no. The incoming and outgoing energy at layer 3 (TOA) could be unchanged while layer 1 (surface) and layer 2 (atmosphere) adjust to reduce the net flow of energy in layer 2. I will say typically layer 3 does see a net positive gain/loss when responding to El Nino/La Nino.
Do you have any examples of El Nino/La Nino causing similar global temperature decreases in the past?
There is a weak correlation between the SC25 MGII index and the Earth’s mean temperature. Is SC25 the cause of the Earth’s decreasing mean temperature?
1998, 2010, and 2016 have magnitudes similar to 2024.
Yes; partly anyway. My model says the SC25 contribution peaked at +0.09 C in 2025/04 and is now at +0.03 C in 2026/04 for an influence of -0.06 C.
Perhaps you don’t understand that the warming has ONLY occurred at very specific times.. ie .. at El Nino events.
You can put an “linear trend” through anything.. doesn’t mean the data is actually linear.
Mathematical comprehend 101. !
Perhaps you don’t understand that the ENSO system is an ‘oscillation’ (what the ‘O’ in the acronym stands for).
Oscillations do not produce heat; they just move heat around within the system. Otherwise it would be a ‘forcing’, not an oscillation… (ENSF??)
Funny how you attribute the opinion of one person to everyone who visits WUWT.
Stereotype much?
Scientifically justified measurements include things like uncertainty budgets, propagated uncertainty, standard deviations, k factors, degrees of freedom. Would you take a medicine or allow a treatment that did not disclose these items in the studies that were done for their approval? Same thing, different verse.
You mean something like what happened with the mRNA vaccines against COVID-19 administered to 250+ million American citizens (on advice of the FDA, CDC, NIAID (Dr. Fauci), and White House)?
ROTFLMAO!
You didn’t answer the question.
Answer to your (rhetorical?) question: Hell, no.
But then again, things like uncertainty budgets, propagated uncertainty, standard deviations, k factors, and degrees of freedom were never obtained, let alone disclosed, from “studies” done in advance of the rushed-to-implementation mRNA vaccines for COVID-19.
Now, please answer my direct question to you asked in my previous response above.
I did not comment that UAH GLAT data trending is “scientifically unjustified”.
I specifically pointed out that reporting UAH temperature anomaly data to two decimal places is scientifically unjustified.
Reading comprehension 101.
That’s actually not true.
According to [JCGM 100:2008] you report the measurement with the same number of digits as the uncertainty. UAH reports their uncertainty with 2 significant digts. [Christy et al. 2006]. And their use of 2 significant digits for the uncertainty is consistent with rules set forth in the GUM.
From your second paragraph:
“. . . you report the measurement . . .”
You obviously have confused a “measurement” with the clearly-stated-by-UAH-reporting of averages of measurements. The two are not considered to be the same by most people familiar with data handling.
[JCGM GUM-6:2020] considers averages of measurements to be a measurement itself.
See above, please stop abusing the GUM.
Which is an excellent reason to not put too much credibility toward that document!
For those involved with practical data analysis and reporting, it is common knowledge that AVERAGING a set of measurements of a physical parameter obtained over a range of time can—indeed almost always does—increase the uncertainty of the average value compared to the uncertainty of any single measurement value within that data set. This is caused by unavoidable factors such as instrumentation (calibration) drift, irreproducibility of consecutive measurements made at high precision, and inherently stochastic temporal and/or spatial and/or amplitude variations in the parameter that is being measured empirically.
There are good reasons to not blindly follow JCGM GUM, especially since GUM is self-defined as a “Guide”, not a rule book.
You don’t think the Joint Committee for Guides in Metrology chaired by the International Bureau of Weights of Measures with partners including NIST, UKAS, ISO, etc. is a credible source?
That’s just patently false. According to the law of propagation of uncertainty it is mathematically impossible for averaging to yield an uncertainty higher than that of the individual values. And in almost all cases it will be lower.
If you want I can walk you through the mathematical proof.
You can also verify this with the NIST uncertainty machine.
Something you desperately need to be true.
It isn’t.
“According to the law of propagation of uncertainty it is mathematically impossible for averaging to yield an uncertainty higher than that of the individual values. And in almost all cases it will be lower.”
That statement is visually falsified by the first UAH plot of GLAT at the top of the above article. Look at the apparently random data scatter (approximating “uncertainty”) in the red curve designated as the “running, centered 13-month average” . . . it’s on the order of +/- 0.05 C, whereas UAH asserts uncertainty of any individual measurement by them (even acknowledging that such “measurement” is an average obtained over one month) is +/- 0.01 C.
“Test all things; hold fast what is good” — The Bible, 1 Thessalonians 5:21
You think UAH falsifies the law of proposition of uncertainty?
Oh, and UAH asserts an uncertainty of ±0.20 C. [Christy et al. 2006]
Don’t just quote a figure. Quote the uncertainty budget so all the uncertainties and their values are accounted for and all can see them.
I have mentioned this on most threads about UAH. NOAA has the same problem in that they never publish an uncertainty budget.
I don’t think UAH has ever done an uncertainy budget to obtain the components used in the propagation of uncertainty. If they have I can not find it published.
I suspect UAH finds a standard deviation and divides that by √n where n is a large number. That is only one component in an uncertainty budget.
No it considers the average of a number of observations of the same input quantity to be an estimate of a measurand. It explicitly defines the series of observations of the input quantity as a set of data in a random variable. The random variable has both a mean and variance.
The observations of a single measurand are not scaled, they exist as an actual estimate of the measurand’s physical property. Remember, you are defining x1, x2, xn as unique values divided by a constant. That requires each observation of a unique input quantity to be scaled by a constant.
The GUM, all of them, define X1 to be the “true value” of an input quantity. Then they define x1 to be the estimate of the value of that input quantity. If you want x1 to actually be x1/n, then you must scale each and every observation by n. Finding the mean of x1 as described in JCGM 100-2008 Section 4.2 requires the use of the actual measurement of the input quantity, not a scaled value.
X1 can not simultaneously be x1 and x1/n. I would love to see that done in a publ9shed paper.
Different input quantities (each being a separate measurand) may be combined using a functional relationship that controls the output value using those inputs. This is your stumbling block, you must define the input quantity’s value and the process for determining it prior to making measurements. If you want them scaled you must define why and how they are to be scaled before the first measurement is taken.
“The GUM, all of them, define X1 to be the “true value” of an input quantity. Then they define x1 to be the estimate of the value of that input quantity. If you want x1 to actually be x1/n, then you must scale each and every observation by n.
…
X1 can not simultaneously be x1 and x1/n.
”
You really don’t understand how functions work, do you.
And you don’t understand what X and x are. They are not measurements of different things. The are measurements of THE SAME THING.
X is a random variable representing the probability of all possible measurements, x is a realization from the variable, that is it’s a measurement. They also use X to be a measurand. Or as Jim puts it, a “true value” of the measurand, though I doubt the GUM says that.
You need to reread the GUM. “X1, X2, Xn” are input quantities defined in the measurement model. They are not the estimated values obtained by observations.
The estimated values of each input quantity obtained by actual physical measurements are designated as x1, x2, xn.
From the GUM.
In other words,
Y = f(X1, X2, …, XN). ( The model)
y = f(x1, x2, …, xN). (The estimated)
XN and xn are each independent input quantities having independent evaluations.
Since they are independent, each must be scaled independently. In other words,
Y = f(X1/N, X2/N …, XN/N). (model)
y = f(x1/N, x2/N, …, xN/N). (estimated)
Please EXPLAIN why each physically observed measurement is scaled in the measurement model by the number of input quantities..
I can not remember any measurement model in physics, chemistry, electronics, etc. of a physical measurand ever being scaled by the count of the number of input quantities. The GUM does not show any example of where individual input quantities are scaled using the number of input quantities.
Perhaps you can find one.
“You need to reread the GUM.”
I don’t because I understand it, as eell as how functions work. Your claim wad that a function that scaled an input value meant tgatvthe input value had to exist in two states. That’s not how functions work. The inputs are invariable. A function such as y = x/2 does not mean you make x half the size, and therefore x is sumultaniously two values. It means that the value of the function is equal to half of x.
X1, X2, etc. are input quantities and the function defines the relationship between those and the output.
x1, x2, etc. are the actual “estimates” obtained that are observed while making actual physical measurements.
y = x1 divided by 2 requires the measurement model to be Y = X1/2. You can’t say Y = X1 = x1/2
That means you must have a physical reason when creating the measurement model that defines why you are dividing a physical thermometer reading, a length, a mass, ectc., by 2 (or any counting number). Tell us what physical reason exists for dividing each measurement by “n”.
A physical functional relationship is defined through experimentation, not by just saying will I can create any old function that I please to give me what I want. Your measurement model must have a statement that you have experimented and determined that your function provides an output that is found in the real physical world and reproducible.
Lastly, you and bdgwx think you can reduce uncertainty by a factor of “n”. You can not. x1 has an uncertainty based upon the GUM.
You would like to say that (1/n)(u(xₙ)/n) is the uncertainty. It is not. Since you are dividing, you must use relative uncertainties. That makes each term:
(1/n)(u(xₙ)/(xₙ/n) = (1/n)(n(u(xₙ)/xₙ) = (1/
n)(n(u(xn))/xn)and your counting number disappears. All for nothing.
“y = x1 divided by 2 requires the measurement model to be Y = X1/2. You can’t say Y = X1 = x1/2”
Yes. I don’t know why you think that’s a problem.
“That means you must have a physical reason when creating the measurement model that defines why you are dividing a physical thermometer reading, a length, a mass, ectc., by 2 (or any counting number).“
In this case it’s becasue we are averaging. In another case it might be becasue we measured a diameter but want the radius.
“Tell us what physical reason exists for dividing each measurement by “n”. ”
Because that’s how you make an average.
“A physical functional relationship is defined through experimentation, not by just saying will I can create any old function that I please to give me what I want. “
So you won’t accept that the volume of a cylinder is πR²H, unless it’s been determined through experimentation? Using a function to give you what you want is what the measurement model is all about.
“Lastly, you and bdgwx think you can reduce uncertainty by a factor of “n”. You can not. x1 has an uncertainty based upon the GUM.”
When all else fails, just stamp your feet and cry “you can’t do that”.
“You would like to say that (1/n)(u(xₙ)/n) is the uncertainty.”
No it isn’t. The uncertainty of a single input scaled by 1/n is (1/n)u(xₙ). Your obsession with getting a simple formula wrong in so many different ways is baffling.
“That makes each term:
(1/n)(u(xₙ)/(xₙ/n) = (1/n)(n(u(xₙ)/xₙ) = (1/
n)(n(u(xn))/xn)”As I said, you just don’t understand how functions work. At this point you just seem to be typing randomly. I’m guessing you are making the same mistake as Tim last time, and thinking equation 10 is using relative uncertainties.
“Yes. I don’t know why you think that’s a problem.”
Because “y” is an estimate of “Y”. You can’t say Y = X and y = x and then say y = X/n or y = x/n. That is saying “y” is not an estimate of “Y”.
The two functions Y = X and Y/n = X/n gives two different values for results. Y ≠ Y/n and X ≠ X/n!!
Y and Y/n ARE TWO DIFFERENT MEASUREMENT MODELS. You can’t say they are measurement models for the same thing!
“In this case it’s becasue we are averaging.”
If you divide one side of an equation by a constant then you must also divide the other side by the same constant.
Y = X –> Y/2=X/2
But Y and Y/n ARE NOT THE SAME VALUE.
You and bdgwx keep wanting to say that Y = X/n. It’s just wrong. There is no other way to classify it.
“If you divide one side of an equation by a constant then you must also divide the other side by the same constant.”
How simple does this need to be for you to understand?
Say the equation is Y = f(X1, X2). Where f(X1, X2) is defined as (X1 + X2)/2.
That means that Y = (X1 + X2)/2. You do not then divide Y by 2. Y is already the result of dividing the sum of the inputs by 2.
As I said, you and Jim really don’t seem to understand how functions work.
Y2 = f2(X1,x2) = (X1+X2)2
Y1 = f1(X1,X2) = (X1+X2)
Y2 = Y1/2
They are different functions, f1 and f2.
They are associated with stated values of measurements.
They are *NOT* associated with measurement uncertainty.
The GUM says the measurement uncertainty ADDS, it does *NOT* average, see Eq 11. The GUM does not define an average measurement uncertainty. There is only f1 for measurement uncertainty, there is no f2.
“They are different functions, f1 and f2.”
You don’t say!? Ones an average the others a sum.
“The GUM says the measurement uncertainty ADDS,”
In quadrature, with each term multiplied by the partial derivative for that term.
“The GUM does not define an average measurement uncertainty.”
The GUM assumes you are intelligent enough to define your own function. I’ve no idea why you are interested in average measurement uncertainty. The point is to find the measurement uncertainty of the average.
“There is only f1 for measurement uncertainty, there is no f2.”
You are really confused. The function is the function of your measurement model. It is not the function to calculate uncertainty. The equation for uncertainty uses the partial derivatives of your function, it is not replacing the uncertainty function with your model function.
Bullshite. If the international metrology community thought that an average uncertainty was meaningful they would have included it in the GUM. They didn’t do so because it is not meaningful for uncertainty.
You can average measurements if they meet the needed restriction in order to obtain a “best estimate”. You do *NOT* average uncertainties to obtain a “best uncertainty estimate”.
How many times do you have to be told that the standard deviation of the sample means is the SAMPLING UNCERTAINTY and *NOT* the measurement uncertainty? The sampling uncertainty is just ONE COMPONENT in the total measurement uncertainty of the average.
A set of highly inaccurate measurement data with large measurement uncertainties can have its mean calculated very precisely if the sample size is large enough and you have enough samples. That means that the standard deviation of the sample means will be very small. BUT THE MEASUREMENT UNCERTAINTY OF THE AVERAGE WILL BE VERY LARGE. The average simply can’t be made more accurate than the data used to calculate it. Your claim otherwise is just your continued use of the meme that “all measurement uncertainty is random, Gaussian, and cancels”. Therefore the standard error is the measurement uncertainty of the average.
One more time, THE AVERAGE IS NOT A FUNCTION. It is a STATISTICAL DESCRIPTOR. The statistical descriptor only describes the shape of the data set, it provides NO mapping of the input values to an output value via a casual relationship – a requirement to be a function.
An average is *NOT* a measurement model. Think of it this way: A function requires dimensional consistency between the input values and the output value. A statistical descriptor has no such restriction. You *can* average a data set whose components have entirely different dimensions, just ignore the dimensions. You can object that the components must have the same dimensions to be physically meaningful BUT think about what you are saying when you use that objection! It is actually *NOT* a requirement in order to find an average. The average only tells you the shape of the data set component values. There is nothing in the definition of the statistical descriptor that requires the components to have the same dimension!
The average is just a mathematical construct, it is *NOT* required to have any physical consistency. Think about it. What dimension does the scores from a history test have? Ans: NONE. But you can average the scores – the mathematical construct known as the statistical descriptor “average does *NOT* require any dimensions to be used at all. In physical science, engineering, and metrology the use of dimensions *is* a requirement in order to provide physical meaning. But that also requires that EXTENSIVE values be used to describe averages of different objects. Averaging temperatures of ten different objects has no physical meaning, it is merely a mathematical construct describing the shape of the data set consisting of the temperatures.
“…because it is not meaningful for uncertainty.”
That’s what I keep telling you. The measurement uncertainty of the events in a mean is not going to be important. You are not normally averaging a set of things because you are interested in their exact average. You are taking a sample in order to estimate the population mean. That has much larger uncertainties than could reasonably be expected from the measurements. Measurement uncertainty is only a problem if it introduces a systematic bias.
Rest of your rant ignored. It’s not worth going other your repeated misunderstandings this late in the comment period.
Huh? If you are not interested in what the exact average is then why would you calculate the standard error? If you are sampling in order to estimate the population mean then why not use the SMALLEST sample size possible in order to maximize calculation efficiency?
You are still living in a blackboard statistical world.
Malarky! Sampling error gets smaller with larger sample sizes, even with measurements of different things using different instruments under different conditions. Measurement uncertainty GROWS instead, it doesn’t reduce. It’s why measurement uncertainty will be FAR bigger than sampling error!
Sampling error tells you the interval within which the population mean lies. Measurement uncertainty tells you the accuracy of the mean.
A very inaccurate yet very precise measuring instrument may give you the exact same value for each measurement. There will be no sampling error since the SD will be zero. Yet the result will be highly inaccurate.
“If you are not interested in what the exact average is then why would you calculate the standard error?”
Because that’s part of the uncertainty of the population mean.
“If you are sampling in order to estimate the population mean then why not use the SMALLEST sample size possible in order to maximize calculation efficiency?”
Why do you keep coming out with all this nonsense just before comments will close? You question makes no sense and I have no intention on being dragged down yet another rabbit hole at this point. Simple, obvious, answer is that the larger the sample size the smaller the SEM.
“Measurement uncertainty GROWS instead, it doesn’t reduce.”
And there’s the problem. 5 years of careful attempts by multiple people to explain why that isn’t the case, and you just go back to argument by assertion. You are just unteachable.
It is *NOT* part of the accuracy of the mean, i.e. the MEASUREMENT uncertainty of the mean. You are back to using Equivocation to try and convince people that the SEM, a measure of sampling error, is actually measurement uncertainty. It is *NOT* measurement uncertainty.
If you would use the proper terminology this would be evident very quickly.
Meaning the more precisely you can locate the highly inaccurate population mean! The SEM tells you *NOTHING* about the accuracy of the mean, the propagated *measurement” uncertainty.
tpg: ““Measurement uncertainty GROWS instead, it doesn’t reduce.””
Here is your vaunted Eq 10 from the GUM:
u_c^2(y) = Σ(∂f/∂x_i))^2 u^2(x_i)
That capital sigma, Σ, represents a SUM of the measurement uncertainties. It’s not divided by sqrt(n). It’s not a subtraction. It is the sum of the variances of the measurement terms, u^2(x_i).
And if the term of interest is a multiplication or division of measurements then the equation devolves to Eq 12 (i.e. use relative uncertainty) and (∂f/∂x_i) becomes the power of the term of interest (see Possolo and his derivation of the measurement uncertainty propagation of the measurements of a barrel).
The standard deviation of the sample means, the SEM, is *NOT* measurement uncertainty no matter how many times you claim it to be. It is the SAMPLING uncertainty associated with locating the population mean.
The real problem is that this is a time series. Has seasonality been dealt with? Has auto-correlation been dealt with? Is an SARIMA been used to remove these factors?
Has the fact that diurnal ranges have changed due to night temps increasing created a change in standard deviations that result in a false trend.
Please direct those questions to the person/party responsible for analyzing and reporting the UAH satellite temperature data.
Personally, I don’t see that seasonality, auto-correlation, SARIMA (especially since it is used for forecasting), and diurnal range changes have any meaningful bearing on the plotted data points and associated LS regression curve fit over the 47+ years of UAH satellite data as presented in the first graph in the above article.
Is this an attempt to throw some out-of-left-field concerns “against the wall” to see what sticks?
So you think there is no auto-correlation involved? How about seasonal changes.
Do you believe that UAH doesn’t put daytime and nighttime temperatures together in determining an average?
You say SARIMA is used for forecasting. That what regressions in climate science are used for also.
Linear trends from regressions take nothing into account.
Remember linear regression is designed to show linearity between an independent variable and a dependable variable such that an equation of y=mx + b will be a functional relationship allowing the calculation of a value for “y” using a value for “x”.
Linear regressions for a time series do not provide a functional relationship. The linear regression will never provide when a change like ENSO will occur, nor it effects.
The other thing SARIMA does is provide a proper procedure for decaying shocks such as ENSO over a short time. Otherwise, a shock will continue being propagated into future values.
Since the UAH graph in question in a plot of monthly, global-averaged data, and since the seasons in Earth’s northern hemisphere are 180-degrees out of phase with those in the southern hemisphere, you’ll have to tell me exactly HOW and WHY seasonal changes should be “adjusted” in the plotted data points.
Hah! A least-squares linear regression does indeed take the degree of data scatter into account, by the degree-of-fit of the linear trend (y=mx + b) to the full data set, commonly known by anyone familiar with the mathematical procedure for doing such as the “coefficient of determination”, assigned the symbol R^2 or r^2, and varying in value from zero to 1.000.
R^2 is a statistical measure in least squares regression that indicates how well the regression model fits the observed data. That is important to know.
The global average temperature does have seasonality to it. Here’s is UAH’s monthly anomaly baselines.
Month 1991-2020 (K)
1 263.18
2 263.27
3 263.43
4 263.84
5 264.45
6 265.10
7 265.42
8 265.23
9 264.64
10 263.95
11 263.41
12 263.19
The reason UAH (and others) remove the seasonality component is to isolate the changes in temperature not caused by Earth’s orbital cycle. The way they remove the seasonality component is by subtracting the above baselines from the measured value for the same month pairs.
“. . . isolate the changes not caused by Earth’s orbital cycle.” Huh? Do you really think that adjusting the monthly data using a single “baseline” value for each month—each value being an average obtained over a 29-year span—really does that??? Doing that necessarily assumes those 29 years of past “seasonality” is an accurate representation of the “seasonality” of Earth for the last 6 or so years as well as going forward. Problematic, that.
Also,changes in solar insolation due to the 11-year (more or less) sunspot cycle are not caused by Earth’s orbital cycle and are not properly normalized by using anomaly baselines of 29-year averages. And, should or should not the anomaly baselines themselves be “adjusted” to remove ENSO and AMOC variations, both of which are not caused by Earth’s orbital cycle?
The whole use of “anomalies” is based on assuming that the anomalies don’t inherit the uncertainty of the component values.
If the current values have variance (and they *do* have such) then using a CONTANT to scale the current values into an ANOMALIES is a linear transformation of the current data using a constant. The variance of the current data is inherited by the anomaly, i.e. the standard deviation of the original data and the anomaly data is the *same*.
Climate science is based on the assumption that anomalies have less uncertainty (i.e. variance) than the absolute values. All they have are “smaller” absolute values – but the standard deviation stays the same. Meaning the variance stays the same and so does the uncertainty of the data.
Climate “science” is just riddled with such assumptions.
First…it’s not a single baseline. It’s 12; one for each month of the year. Second…it’s a 30 year average. Anyway, the answer to your question is yes.
Of course not. Variations caused by ENSO and/or AMOC would be something we would want to see in the data.
If it would help, here’s a comparison of monthly anomalies and absolute temperatures for UAH. They are both on the same scale, but absolute temperatures have been shifted.
I see clear seasonality in one, and not the other.
Looking at the variation in the maximums of your right-hand graph compared to the variation of maximums of your left-hand graph—and doing likewise comparisons for the minimums of both graphs—I see clear seasonality in both graphs.
“I see clear seasonality in both graphs.”
Then you must have very imaginative eyes.
Decomposing the two time series suggests for temperatures a seasonal variation of almost ±1°C, but for anomalies, less than ±0.02°C. Any seasonality in the anomalies is just noise.
Here’s the decomposition for anomalies.
And here’s the decomposition for temperatures. Compare the scales for seasonality.
They both have the same relative uncertainty.
50 +/- 1 has the same relative uncertainty as 1 +/- .02. You don’t give the actual means but I’ll bet they are close to 50 and 1.
Both have a relative uncertainty of 2%.
If seasonal variation appears in the absolute temperature, then it will also appear in the anomaly.
This can be seen by comparing the relative uncertainties. You can’t get rid of the relative uncertainty using a linear transformation by a constant.
What uncertainty? We are talking about seasonality here, not uncertainty. And I’ve no idea whrre you get values of 50 or 1 here.
“If seasonal variation appears in the absolute temperature, then it will also appear in the anomaly.”
Do you see the same seasonal variation in the graphs? If I do may I suggestbyiu get your eyes tested.
Again you seem to be confusing seasonal variation with natural variation that might change slightly on a seasonal basis.
“You can’t get rid of the relative uncertainty using a linear transformation by a constant.”
Again, we are not talking about uncrrtainty. but seasonality.
Seasonality causes the data to have different variances for different months. You can’t just throw these variances away, you have to propagate them into the total uncertainty. Either that or you have to account for them by making the data stationary.
Oh Yeah!
You are merely modifying the original value by reducing its absolute value.
The ±1°C is the variance in the data. It is determined from the data points in a random variable. Similarly, the baseline has a variance determined from the data points in another random variable.
Subtracting the means of two random variables requires the variances to add. Consequently, the value of the anomaly’s variance is > ±1°C.
You have just thrown that inherited variance away into the trash and instead found the variance of very small numbers and declared that to be be the total variance.
“You are merely modifying the original value by reducing its absolute value.”
Yes, that’s the point. Reducing the seasonal variation.
“The ±1°C is the variance in the data.”
It isn’t.
“Subtracting the means of two random variables requires the variances to add.”
Pointless discussion this unless you state what variance you are talking about. You and Tim keep switching the argument. The variance in the base value is irrelevant to the variance in the monthly anomalies because as Tim keeos pointing out you are just subtracting a constant.
If you mean uncertainty, then yes the uncertainty if the anomaly is the uncertainty of the monthly temperature plus the uncertainty of the base value added in quadrature. But the monthly variance is not uncertainty. The variance does contribute to the uncertainty if the trend, but that’s already present in the anomalies.
“Consequently, the value of the anomaly’s variance is > ±1°C.”
No idea how you get that, but as you refuse to show your workings, I doubt you understand it either. Again, talk to Tim. Even he understands that subtracting a constant doesn’t change the variance.
“…found the variance of very small numbers and declared that to be be the total variance.”
You don’t need to worry about what’s been thrown away. The variance of the data is the variance of the data. But regardless. the seasonal component is not the variance, it’s the monthly seasonal offset that can be exteacted from the data.
“Yes, that’s the point. Reducing the seasonal variation.”
You can’t get rid of the seasonal variation by just scaling the values using a linear transformation by a constant.
Nor can you get rid of the seasonal variation by just ignoring it as if it doesn’t exist. *That* is what climate science does – just ignore the variance of everything; i.e. “all measurement uncertainty is random, Gaussian, and cancels”, the exact meme you always apply even though you won’t admit it.
“You can’t get rid of the seasonal variation by just scaling the values using a linear transformation by a constant.”
You are still confused about what seasonal variation means. It’s the mean changes over the year, not the variation for individual month over the years. Anomalies do not transform monthly values by a constant, they transform monthly values by different values for each month of the year. That’s what removes the seasonal variation.
What you are describing is heteroskedasticity. Variance changes over time. But in this case only cyclically across each year, and not be a very large amount. I’ve given you a weighted linear regression using the monthly variances. It made a small, insignificant difference to the trend.
You and Jim in contrast, have never once tried to do your own analysis. All you do is rage from the sidelines.
In a time series, seasonal changes can be year‑over‑year. Seasonality refers to regular, repeating patterns that occur at fixed intervals (e.g., monthly, quarterly, annually).
With temperature, daily and monthly changes are more an auto-correlation problem. Yesterday’s temperature is a good predictor of today, and today for tomorrow.
Atmospheric temperature time series are auto‑correlated, meaning that temperature measurements at one time are statistically related to measurements at nearby times. This is well‑documented.
Here is a good analysis.
https://andymaypetrophysicist.com/2021/11/13/autocorrelation-in-co2-and-temperature-time-series/
From the discussion. (LS is least square)
“Atmospheric temperature time series are auto‑correlated, meaning that temperature measurements at one time are statistically related to measurements at nearby times.”
Yes. But you were talking about seasonality, not auto-correlation.
Using an SARIMA takes care of both. You can specify a lag of 1 for day to day or month to month. You can specify a 12 month for seasonality (annual).
You mentioned heteroscedasticity. That deals with changing variance. When creating graphs of traditional Tavg, you are in essence showing the combination of two variables, Tmax and Tmin. If the variance between them changes over time, such as warmer nights, you will not have heteroscedasticity. That is one reason I do only Tmax and Tmin separately.
“Using an SARIMA takes care of both.”
And what does your SARIMA model tell you about the seasonality in the anomaly series? Because whenever I try it it says there is no noticeable seasonality.
“When creating graphs of traditional Tavg, you are in essence showing the combination of two variables, Tmax and Tmin. If the variance between them changes over time, such as warmer nights, you will not have heteroscedasticity.”
Wrong. The variance is the variance in the data you are using, not in how it was derived. A time series of Tavg, however Tavg was calculated, doesn’t care about the diurnal range, only how much variance there is in the Tavg values.
You are still failing to address WHY the anomaly DOES NOT inherit the combined variance of the random variables used to calculate it.
The fact that scaling a value to a smaller value by subtracting a constant does reduce the absolute variance between the smaller numbers. Duh. But it does require throwing away the combined variance each number inherits from the original random variables.
BTW, your identification of these being random variables whose variance do combine, destroys your claim that they are individual input quantities each divided by “n” counts in a functional relationship. Your arguments are not coherent.
If you are declaring the monthly average as a measurand, then the variance IS part of the uncertainty. Jeez,!
JCGM 100-2008
“You are still failing to address WHY the anomaly DOES NOT inherit the combined variance of the random variables used to calculate it.”
Impossible to say until you say what variance you are talking about. And then say what you mean by inherit.
“The fact that scaling a value to a smaller value by subtracting a constant does reduce the absolute variance between the smaller numbers.”
You are not scaling a value. Subtracting the base value is translating, not scaling.
“BTW, your identification of these being random variables whose variance do combine, destroys your claim that they are individual input quantities each divided by “n” counts in a functional relationship. Your arguments are not coherent.”
What in earth are you talking about now. Trying to have multiple stupid conversation sumultaniously is too confusing. The comment you are replying to was about seasonality, but now you seem to have switched to a discussion about averages, and still doesn’t make sense.
and the same applies to subtraction.
σ₍ₓ – ᵧ₎² = σₓ² + σᵧ²
Sampling and Combination of Variables – Statistics and Probability Tutorial
Wow! A one track mind. This is in relation to your insistence that
Y = y = x1/n + x2/n + … + xn/n
This is where you dismiss GUM 4.2 process for using a random variable to characterize measurement observations and use your own definition of a “functional relationship” being an average of individual input quantities.
I can’t help it if you can’t keep track of your cherry picks. It isn’t my problem to deal with. Stay consistent with your assertions and you won’t have a problem.
Why can you never just answer the question? What variances were you talking about?
“This is consistent with the fact that variance has units of the square of the variable.”
This is what I keep telling you. It’s why the variance of an average decreases. W = 0.5X + 0.5Y. var(W) = 0.25var(X) + 0.25var(Y).
“Variances must increase when two variables are combined“
What a shame. You almost had it.
Look at the poorly written article you reference. Under variance of sample means. What happens to the variance when you combine variables by taking an average? Does dividing by n increase the variance?
“Wow! A one track mind. This is in relation to your insistence that”
It’s because I’m trying to run down multiple tracks at the same time that your sudden jumping to a completely different conversation creates confusion. Try to stick to the subject under discussion.
“This is in relation to your insistence that
Y = y = x1/n + x2/n + … + xn/n”
You mean if this the function for the mean. Yes I do insist that you divide by n to get the average. I’m not sure how that relates to your comment
What’s not coherent about dividing values by n to get an average?
“This is where you dismiss GUM 4.2 process for using a random variable to characterize measurement observations and use your own definition of a “functional relationship” being an average of individual input quantities.”
You are so far gone in misunderstanding how this works, you can’t even make a coherent argument.
You use a random variable to represent measurement uncertainty. Each input quantity is estimated by a measurement which is assumed to come from that random variable. You apply a function to these estimates to obtain a result. E.g. when the function is an average the function is y = x1/n + x2/n + … + xn/n. You use equation 10 on the standard deviations of the random variables representing the uncertainty to get the standard deviation of the random variable representing y, and say that is your estimate of the uncertainty of y.
I’ve no idea why you think I’m dismissing 4.2 using random variables. It’s all just the use of probability I keep trying to explain to you.
And where did the 0.5 come from?
Did you miss this from the resource.
Because you are cherry picking equations with no idea what you are doing.
You are declaring that Tmonth_avg is a measurement made up of “n” input quantities, one for each day, that are averaged .
Each input quantity is a unique and independent measurement with its own uncertainty.
With your “functional relationship” there is no random variable with qₖ observations. Therefore Section 4.2 has no purpose.
You can’t have it both ways. You either treat the “n” values as independent measures in a functional relationship or you treat them as individual observations in a random variable.
You still haven’t given an explanation WHY your measurement model for Tmonth_avg divides each actual physical measurement by “n”. A functional relationship requires a reason for using each variable. In essence you need “n” explanations defining how and why each input quantity is used, including “n”.
“And where did the 0.5 come from?”
It’s an average.
“X and Y are simply added together”
We are not simply adding them together, it’s an average.
“You are declaring that Tmonth_avg is a measurement made up of “n” input quantities, one for each day, that are averaged .”
If you like, but it works for any average.
“Each input quantity is a unique and independent measurement with its own uncertainty.”
Independent is unlikely for your monthly average.
“With your “functional relationship” there is no random variable with qₖ observations.”
What do you think the uncertainty is?
“You either treat the “n” values as independent measures in a functional relationship or you treat them as individual observations in a random variable.”
There’s no two ways about it. Both are true. Each independent observation is an individual observation from a random variable.
“You still haven’t given an explanation WHY your measurement model for Tmonth_avg divides each actual physical measurement by “n”.”
Because that’s how you take an average. I really think I’ve been giving you too much credit. I assume you understand at least the basics of how this works, but you keep indicating you are completely bewildered by it.
And here we go again. A constant, calculated from component values with uncertainty, can’t have uncertainty itself.
The meme: “all measurement uncertainty is random, Gaussian, and cancels”.
Unfreakingbelievable!
Of course it is!
GUM: “3.3.5 The estimated variance u^2 characterizing an uncertainty component obtained from a Type A evaluation is calculated from series of repeated observations and is the familiar statistically estimated variance s^2 (see 4.2). The estimated standard deviation (C.2.12, C.2.21, C.3.3) u, the positive square root of u^2, is thus u = s and for convenience is sometimes called a Type A standard uncertainty. For an uncertainty component obtained from a Type B evaluation, the estimated variance u^2 is evaluated using available knowledge (see 4.3), and the estimated standard deviation u is sometimes called a Type B standard uncertainty.”
Couple this with GUM, Eq 11: u_c(y)^2 = Σu(y_i)^2
and the monthly variance, u_c(y)^2 is the sum of the individual variances u(y_i)^2.
Someday you REALLY NEED to actually read the GUM for meaning and context.
The uncertainty of the trend is determined by the uncertainty of the individual components. The uncertainty of the trend is *NOT* just the “best-fit” calculation of the stated values but also of the uncertainties of individual components.
Variance_total = Variance1 + Variance2
Monthly_Variance = Variance_currentdata + Variance_monthlyaverage
“Again, talk to Tim. Even he understands that subtracting a constant doesn’t change the variance.”
You *still* can’t read! I said it doesn’t SCALE OR REDUCE the variance.
tpg: “This can be seen by comparing the relative uncertainties. You can’t get rid of the relative uncertainty using a linear transformation by a constant.”
Here we go again. “all measurement uncertainty is random, Gaussian, and cancels”.
“And here we go again. A constant, calculated from component values with uncertainty, can’t have uncertainty itself.”
Could you please try to understand what you are saying. Whatever you type never makes any sense becasue you keep changing the frame of reference. You said, that subtracting a constant from a distribution does not change the variance of the distribution. Correct. But now you change the argument to uncertainty and say that a constant can have no uncertainty. That’s something you keep claiming, and it’s wrong. A constant can have uncertainty. It’s epistemic rather than aleatoric uncertainty, but it’s still uncertainty.
So when you subtract a constant base value from a monthly average, it won’t affect the variance, but it does add, slightly to the overall uncertainty.
Your problem I suspect is you keep using variance and uncertainty as if they were the same thing. You are confusing the use of variance when measuring the same thing in order to determine the measurement uncertainty with variance caused by things actually varying randomly.
“Of course it is!”
No it isn’t. Variance in monthly values is not a measure of how uncertain you are about that monthly values. You expect each year to have a different value, that’s what the variance is measuring.
“The uncertainty of the trend is determined by the uncertainty of the individual components.”
By the variance of the individual components, not necessarily uncertainty about them.
In response to me asking how Jim could claim “Consequently, the value of the anomaly’s variance is > ±1°C”, you reply
“Monthly_Variance = Variance_currentdata + Variance_monthlyaverage”
I suspect what you are trying to say is the uncertainty of the base value has to be added to the uncertainty of a specific monthly value, when taking an anomaly. But that is not the same as the monthly variance, and regardless, neither is > 1°C.
“You *still* can’t read! I said it doesn’t SCALE OR REDUCE the variance.”
Your exact words were
“Here we go again. “all measurement uncertainty is random, Gaussian, and cancels”.”
This in response to me saying “But regardless. the seasonal component is not the variance, it’s the monthly seasonal offset that can be extracted from the data.” Which should make it clear that you just writes this mantra as some sort of nervous tick, regardless of relevance.
If you can parse my comment and explain exactly what it has to do with Gaussian distributions or randomness or cancellation, go ahead.
*I* didn’t say that at all. Your lack of reading comprehension skills is showing again. *YOU AND CLIMATE SCIENCE* assume the base average is a constant with no uncertainty!
The uncertainty of an anomaly is the SUM of the uncertainty of the current data and of the baseline average!
*YOU AND CLIMATE SCIENCE* just assume that the anomaly has no uncertainty – i.e. the components making up the anomaly have no uncertainty!
It’s at the base of assuming you can trend the anomalies and use the residuals as a best-fit metric without covering the entire range of possible trends that exist inside the variance associated with the random variable that is the anomaly! For if the anomaly has an uncertainty, then by definition it has a variance!
Then if the current data has a variance (read as uncertainty) and the baseline average has a variance (read as uncertainty), how can the anomaly variance not be the sum of the two variances? Even if the baseline average is 100% accurate it becomes a constant and you *still* wind up doing nothing but a linear transformation by a constant – leaving the variance of the current data as the variance of the anomaly!
And the variance of the current data is *NOT* the standard deviation of the sample divided by the square root of the sample size => IT IS THE SQUARE OF THE STANDARD DEVIATION OF THE SAMPLE.
“Has seasonality been dealt with?”
What seasonality? You keep saying this every month, but never give any indication you understand what it means. If you disagree with Spencer’s linear regression, explain how you would do it differently.
And why did you never raise any of these objections all the time you were praising Monckton’s pause “analysis”? All you ever did was attack me for pointing out the uncertainty in the trends.
Seasonality can introduce spurious trends due to variance changes, that is, difference in winter vs summer in different hemispheres. Trends reflect a change in variance rather than an actual increase.
It is why ag science was the group to show that nighttime temperatures were increasing rather than climate science. Climate science ignored the actual reason by concentrating on a daily average and regressions that showed increasing temperature. CAGW!
A linear regression is used to develop a functional relationship between an independent variable (x) and a dependent variable (y). This functional relationship can then be used to predict values not collected in the original data set. As it stands right now, any linear regression of temperature has two problems.
One, it does not predict when step changes will occur, nor why nor does it predict when pauses will occur.
Two, extending the linear regression into the future results in temperatures that are unrealistic. Remember, CO2 is not the independent variable, time is. So as time increases far into the future so will temperature.
Why do you think nowhere in climate science does anyone map CO2 vs temperature and do a linear regression? You have the data, why don’t you do the regression?
I have done this on a number of stations while teaching myself python. SARIMA stands for Seasonal Auto Regression Moving Average. It has been developed for time series analysis. It allows one to have a statistically constant baseline for a model. One can then develop variables that provide the actual values and apply them to the base. I assure you that CO2 concentrations alone do not give adequate projections throughout the complete time line.
This started me down the path of determining how the intermediate land/oceans warm from the sun, and then release their energy into the atmosphere. Using averages quickly lead you astray when doing the thermodynamic analysis. For land, the soil temperature distribution is something that is not amenable to arithmetic averaging and I’m sure the atmosphere is no different.
I seldom look at temperature trends nymore. The measurement uncertainty is never given appropriately, UHI contamination is rife along with poorstation siting. Data modification in station data, while well intentioned still poisons the well, and daily averages have already been provenusless in determining the reason for increasing values.
The very thing that anomalies remove.
“The very thing that anomalies remove.”
Anomalies remove NOTHING. They only scale the absolute values which is meaningless.
Creating an anomaly is nothing more than a linear transformation using a constant. The standard deviation (variance) stays exactly the same even if the absolute values are scaled. The variance (standard deviation squared) is the metric used for uncertainty. If the variance doesn’t change then neither does the uncertainty of the data being scaled.
If the variance is the metric for uncertainty then the mean needs to be a *weighted* mean reflecting the differing uncertainty associated with each. Since winter temperatures typically have a higher variance they also have a larger uncertainty. Their percentage contribution to the mean should be less than the percentage contribution from temps (i.e. summer) with smaller variance.
This is just BASIC statistical analysis – which climate science *always* ignores.
If you KNOW that, then you should be able to show how anomalies remove seasonality using mathematics.
You show no references to support your assertion. That makes your assertion worthless.
If you knew what you are talking about, you would mention the variances associated with the mean values that make up anomalies. Do those variances between seasons result in spurious trends? Show how you know if that is true or not. Remember, seasons are not correlated between the NH and SH.
From: TSCS_Week5_Trends.pdf
E(∆yt) = E(a0 +εt) = a0
var(∆yt) = E(∆yt −a0)2 = E(εt)2 = σ2
cov(∆yt,∆yt−s) = E(εt,εt−s) = 0
From: Linear Regression in Time Series: Sources of Spurious Regression | Towards Data Science
I hate to be the bearer of bad news, but temperatures are auto-correlated. The temperature of yesterday is a good predictor of todays temperature, which is a good predictor of tomorrows temperature.
“Seasonality can introduce spurious trends”
I asked what seasonality are you talking about, not what effect seasonality can have. I know it can introduce spurious trends – if you remember I tried to explain that to Monckton when he used CET monthly temperatures.
But the point is we are talking about anomalies, not temperatures. There is no noticeable seasonality.
“Trends reflect a change in variance rather than an actual increase.”
How do you think that works? How does a change in variance change a trend?
“It is why ag science was the group to show that nighttime temperatures were increasing rather than climate science.”
You’re just making stuff up now aren’t you? Regardless, what has that got to do with seasonality?
“A linear regression is used to develop a functional relationship between an independent variable (x) and a dependent variable (y).”
Linear regression does not give you a functional relationship. There is always an epsilon term that is not predicted by the independent variables.
And you are again avoiding the question. I asked you explain what analysis you would do on the data.
“One, it does not predict when step changes will occur, nor why nor does it predict when pauses will occur.”
Why would you expect it to? If you have change points, you don’t have a linear regression over the entire data set. You need to use other techniques to estimate where such a change has happened.
“Two, extending the linear regression into the future results in temperatures that are unrealistic.”
Which is why you should not extend a linear trend into the future. It’s linear regression 101, that you should not extend the regression outside the data range.
“Remember, CO2 is not the independent variable, time is.”
CO2 can easily be an independent variable. I’ve shown this to you many times.
“Why do you think nowhere in climate science does anyone map CO2 vs temperature and do a linear regression?”
I keep showing you my graphs doing just that. E.g.
“I assure you that CO2 concentrations alone do not give adequate projections throughout the complete time line.”
You don’t need to “assure” me, just show your workings.
You are the one with the trends, show us your predictions. You do know that is the purpose of a regression model for a physical phenomenon. It shouldn’t take much to plug in a doubling of CO2 from 330 ppm (log₂ (330) = 8.4) to 660 ppm (log₂ (660) = 9.4). What is the resultant prediction from your regression equation?
“You are the one with the trends, show us your predictions.”
Spencer’s the one gibing the linear trends, and Monckton used to when he was posting. The only linear regression I’ve made in these comments, was my global map of trends, produced to test your claim about cooling in the Eastern Pacific.
“You do know that is the purpose of a regression model for a physical phenomenon.”
Nope, I do not know that. In that I think the opposite. You should not project a linear regression too far outside it’s data range.
“Two, extending the linear regression into the future results in temperatures that are unrealistic.”
I think your problem is you only see the role of linear regression as predicting what will happen in the future, which is not something you should really be doing.
The main point of linear regression is to identify if correlations exist between different variables, and in the case of a time series, to identify if something has been changing over time, and if so how quickly.
You might be able to use a regression to make a rough estimate of what will happen in the future – but that’s only valid if you can assume that the trend will continue, and that depends on what the reasons are for the change. If you want to estimate how the climate will change over the next century, you have to look at climate models, not just assume a linear trend will continue indefinitely.
From: Linear regression | Definition, Formula, & Facts | Britannica
In physical science, experiments are run to gather data that occurs in the dependent variable when changes in the independent variable are made.
Here you are equating CO2 ppm concentration to the independent variable and the anomaly to the dependent variable. This isn’t a statistical analysis, it is determining the physical functional relationship between the two variables. If the relationship is a viable one, then predictions can be made of what the anomaly will be at various CO2 concentrations.
Finally, to be honest, this whole focus on anomalies is worthless. It tells no one what is actually happening at any point on the globe or in time. The whole process is being done to attempt to convince folks that there are places on the globe warming to such an extent that cooler places do not offset the warming. If you want to do something worthwhile, show us WHERE this warming is occurring on a constant basis. Greenland? Antarctica? Africa? Or is well mixed CO2 just wandering around warming here and there until the entire globe is warmer. I’m very interested in where here and there actually are.
“A primary use of the estimated regression equation is to predict the value of the dependent variable when values for the independent variables are given. ”
You still don’t get that prediction does not mean predicting the future. It’s always dangerous to predict outside the data range. Look at the example in your source
That’s in relation to a graph giving a stress test range of 50 – 100, i.e. within the range of the data.
I also think you misunderstanding the word “prediction” in this context. The predicted value is not the actual value you will get. It’s what you would expect to get on average. Using this you can compare the actual result against the predicted result. You can use the prediction to see if someones blood pressure is unusually high or low.
“In physical science, experiments are run to gather data that occurs in the dependent variable when changes in the independent variable are made.”
But in this case we are using time as an independent variable, simply to see if temperatures are increasing or decreasing significantly, and if so by how much. Once again, you are hung up on one use of a regression and dismiss any other use as “not physical science”. You have a very blinkered view of science.
“This isn’t a statistical analysis, it is determining the physical functional relationship between the two variables.”
I’ve no idea how you think you can do that without a statistical analysis. And again, there is not a functional relationship between CO2 and temperature.
“If the relationship is a viable one, then predictions can be made of what the anomaly will be at various CO2 concentrations.”
How? You keep demanding the impossible then complaining when nobody does it for you? There are any number of factors that will determine the global temperature in a given year. And the relationships between them are not likely to be linear. If you think it’s possible to predict the exact global temperature for a given year, then why don’t you demonstrate how?
“The whole process is being done to attempt to convince folks that there are places on the globe warming to such an extent that cooler places do not offset the warming.”
What are you on now? Again, if you have a problem with the UAH data set, take your complaints to Spencer and Christie, before they retire.
“If you want to do something worthwhile, show us WHERE this warming is occurring on a constant basis.”
Did you notice my global anomaly maps, or the trend maps?
“But the point is we are talking about anomalies, not temperatures. There is no noticeable seasonality.”
This has been pointed out to you multiple times and you just ignore it: Anomalies are linear transformations by a constant. The anomalies inherit the EXACT VARIANCE of the parent.
If temperatures have variances based on seasonality then so will the anomalies generated by linear transformation using a constant.
I don’t ignore what you “point out”, I explain why it’s wrong or irrelevant.
First. we are talking about seasonality, not variance.
Second, you need to state what variance you are talking about. If you mean the variance in monthly values across the year, then anomalies definitly have less variance than temperatutes. That’s because you are not transforming by a constant. The value you subtract in January will be different to the value you subtract in July. The variance you see in temperatures across the year is caused by seasonality. Using anomalies removes that seasonality.
If you mean the variance for a specific time of year across the years, then yes, there will be the same variance for all Decembers whether you use anomalies or temperatutes. This variance may be different for different months, although the difference isn’t large on a global scale. But the question is, so what? What specific problem do you think that causes, and how could it result in a spurious trend?
Please give a specific example, rather than your usual hand waving.
“ If you mean the variance in monthly values across the year, then anomalies definitly have less variance than temperatutes.”
The variance of each monthly value *should* inherit the added variance of the component daily values. Var_total = Var1 + Var2 +…. VarN
Anomalies obtained from the monthly values and their inherited variances will have the exact same variance.
For the umpteenth time, variance is the uncertainty metric. Uncertainty adds, especially when different things are being measured. You, and climate science as well *ALWAYS* assume that uncertainty is random, Gaussian, and cancels so that monthly values don’t inherit the summed uncertainties of the component daily values! The variance (i.e. the uncertainty) is *NOT* the standard deviation of the monthly values, it is the added variances of the monthly values.
YOU CAN’T JUST IGNORE THE VARIANCE OF THE BASE MEASUREMENTS. You continually say that you don’t assume that measurement uncertainty is random, Gaussian, and cancels but you just proved that you *DO* assume just that! EVERY SINGLE TIME!
“That’s because you are not transforming by a constant.”
You *are* transforming by a constant. The mean value used to form the anomaly *is* a constant. It is used in forming every single anomaly. You may change that constant from decade-to-decade or from 30year mean to a different 30 year mean, but the *same* value is used to scale every single measurement over the time period covered. It is *not* a variable. If it was then it would have to be quoted with an uncertainty and the variance of the anomaly would be the sum of the uncertainty of the mean value plus the uncertainty of the current temperature. You are caught between the rock and the hard place you always wind up at when trying to justify the belief that anomalies have less uncertianty than the components used to calculate the anomaly.
“The variance of each monthly value *should* inherit the added variance of the component daily values.. Var_total = Var1 + Var2 +…. VarN”
OK, so you don’t understand what a variance is. You’ve demonstrated this enough times, you don’t have to keep repeating it.
What you are describing is what happens when you add multiple random variables. The average of multiple random variables is not a sum. The Var_average will be Var_total / N².
“Anomalies obtained from the monthly values and their inherited variances will have the exact same variance.”
Did you read my comment? That’s what I said, if you are talking about the variance of values for a specific month across the years. But that’s not the same as the variance of all monthly values across the years.
“For the umpteenth time, variance is the uncertainty metric.”
It isn’t really, it’s standard deviation that’s the metric for uncertainty. But this has nothing to do with the discussion, which was about accounting for seasonality in a linear regression.
“random, Gaussian, and cancels”
You’ve lost the argument, and the plot, again.
“You *are* transforming by a constant.”
You think you subtract the same value for January as you do for July?
You *have* to add the values of the random variables in order to calculate the average!
Each month is linearly transformed by a constant. That constant does not have to be the same for each different month. That does *NOT* change the fact that the anomaly created from the linear transformation inherits the uncertainty of the constant *and* the current data. The standard deviation remains the same no matter how much you scale the absolute value with a constant value.
Malarky! Some day you and bdgwx NEED TO READ THE GUM FOR MEANING AND CONTEXT.
From the GUM:
——————————————
3.3.5 The estimated variance u2 characterizing an uncertainty component obtained from a Type A evaluation is calculated from series of repeated observations and is the familiar statistically estimated variance s2 (see 4.2).”
—————————————-(bolding mine, tpg)
The variance of a Gaussian distribution is related to the peakedness index. pi = max-value/sqrt(variance). The larger the variance the less the peakedness index. The “hump” around the average value gets flattened as the variance increases. That means that values surrounding the mean get closer to the mean value which raises the uncertainty of the mean.
This has been explained to you multiple times. Yet you just continue to ignore it. Why?
READ THE GUM FOR MEANING AND CONTEXT. STOP CHERRY PICKING.
As usual, you just refuse to actually learn the basics of metrology.
Linear transformation of a distribution using a constant does *NOT* change the standard deviation of the distributionVariance *is* the metric for uncertaintyMeasurement uncertainty ADDS when multiple measurands are involved, it does *not* cancel.You can’t create a larger system by combining the intensive properties of multiple measurandsREAD THE GUM FOR MEANING AND CONTEXT. STOP CHERRY PICKING.
“You *have* to add the values of the random variables in order to calculate the average!”
And thrn you divide by N, and tgat neans dividing the variance by N². Your inability to remember or understand this basic fact is your problem. All I can do is keep pointing it out.
“Each month is linearly transformed by a constant. That constant does not have to be the same for each different month.”
Yes, that’s what I keep telling you. As I keep saying, these conversations woukd be far les painful if you just explained what variance you are talking about.
“The standard deviation remains the same no matter how much you scale the absolute value with a constant value.”
Again, what standard deviation? We were talking about removing seasonality by taking anomalies. That implies that the variance of the monthly anomalies is less than that gor temperatures because you’ve removed the variance caused by seasonality.
That’s complety different from the variance you see when looking at a specific calandar month and how much it varies year to year.
You can cherry pick parts of the GUM, and sometimes they suggest variance as a measure of uncertainty. But the primary unit is the standard uncertainty. which is a standard deviation. IMHO, variance is a poor value to use for incertainty giving it’s expressed in physically meaningless units and gives no real indication of the size of the uncertainty.
“The variance of a Gaussian distribution is related to the peakedness index. pi = max-value/sqrt(variance).
…
This has been explained to you multiple tomes. Yet you just continue to ignore it. Why?”
This is the first time I can remember you talking about a peakedness index. Your equation makes no sense. Could you privide a reference? As far as I can see the index was only proposed last year.
https://www.preprints.org/manuscript/202509.2604
“And thrn you divide by N, and tgat neans dividing the variance by N². Your inability to remember or understand this basic fact is your problem. All I can do is keep pointing it out.”
If you can’t add the data so that it represents a larger system then any average you calculate from the data has no physical meaning. Trending the average over time only tracks the movement of the non-physical average – meaning the trend is also useless in describing the real world.
“Yes, that’s what I keep telling you. As I keep saying, these conversations woukd be far les painful if you just explained what variance you are talking about.”
THE VARIANCE OF THE MEASUREMENT DATA!
Jeesh!
It is *still* using a CONSTANT to perform a linear transformation, even if the constant for March is different than the constant for December. That means that the variance of March data is the same for both the absolute values and for the anomaly.
Anomalies do *NOT* reduce measurement uncertainty – i.e. variance of the measurement data.
I’m not surprised you can’t accept that simple fact.
“Again, what standard deviation? We were talking about removing seasonality by taking anomalies.”
If the variance of the temperature data is impacted by the seasonality – AND IT *IS* AFFECTED BY SEASONALITY – then anomalies do *not* remove seasonality because the anomaly has the same variance as the absolute values.
For the umpteenth time, linear transformation by a constant does *NOT* reduce variance. The linearly transformed data inherits the exact same variance of the original data.
Then you are not talking about seasonality. That’s a change in the mean not in variance. You are talking about heteroscedasticity. You can fix this by doing a weighted regression, but it makes no significant difference.
When I tried it in a back of the envelope way, it just knocked about 0.001°C / decade from the trend, down to 0.155°C / decade compared to the unweighted 0.156°C / decade.
“That’s complety different from the variance you see when looking at a specific calandar month and how much it varies year to year.”
The variance of the month’s data *IS* related to seasonality. Colder temps (e.g. winter) have higher variance than warmer temps (e.g. summer). How many times has this been pointed out to you? Why do you continue to ignore that simple fact?
The variance of the year-to-year average data is separate from the variance of the individual average data element. All you are doing is ignoring the variance the monthly average inherits from the individual data elements.
It’s typical of YOU and climate science to just throw variance away – it’s part and parcel of the meme of “all measurement uncertainty is random, Gaussian, and cancels” – you know, the meme you continuously profess to not follow? But which colors each and every assertion you make!
“SOMETIMES”?????
The definition of measurement uncertainty is DEFINED as the variance! It’s consistent throughout the entire document!
JUDAS H. PRIEST!
SD = √Variance!!!!
GUM:
—————————-
3.3.5 The estimated variance u^2 characterizing an uncertainty component obtained from a Type A evaluation is calculated from series of repeated observations and is the familiar statistically estimated variance s^2 (see 4.2). The estimated standard deviation (C.2.12, C.2.21, C.3.3) u, the positive square root of u^2, is thus u = s and for convenience is sometimes called a Type A standard uncertainty.
—————————–(bolding mine, tpg)
FIRST you find the variance. THEN you find the standard deviation!
“The definition of measurement uncertainty is DEFINED as the variance! It’s consistent throughout the entire document!“
The GUM definition of uncertainty
“FIRST you find the variance. THEN you find the standard deviation!”
Why do you think you “then find the standard deviation”? You claim, standard deviation is not the measure of uncertainty, so why go to the trouble of taking a square root for something you are never going to use?
YOU CAN’T KNOW THE SD WITHOUT FIRST KNOWING THE VARIANCE!
You are simply playing word games. Section 3 of the GUM is the BASIC CONCEPTS section. I gave you the definition of the basic concept of uncertainty.
——————————
3.3.5 The estimated variance u^2 characterizing an uncertainty component obtained from a Type A evaluation is calculated from series of repeated observations and is the familiar statistically estimated variance s^2 (see 4.2). The estimated standard deviation (C.2.12, C.2.21, C.3.3) u, the positive square root of u^2, is thus u = s and for convenience is sometimes called a Type A standard uncertainty.
—————————–(bolding mine, tpg)
Variance *is* the metric characterizing the uncertainty from a Type A evaluation.
If you go to Appx C, the section on Basic Statistical Terms and Concepts you’ll find:
————————
C.2.20
variance
a measure of dispersion, which is the sum of the squared deviations of observations from their average divided by one less than the number of observations”
——————-(bolding mine, tpg)
Nor have I ever claimed that the standard deviation is *NOT* a measure of uncertainty. Variance, however, IS *THE* metric for uncertainty. It is a basic statistical rule. When combining random variables you do *NOT* add standard deviations, you add variances. The GUM specifies this for adding measurement uncertainty in Eq 11 where you calculate the sum of u^2(y), not the sum of u(y).
One of the main reasons for using standard deviation in expressing uncertainty is that it shows a plus/minus interval. Using a plus/minus interval allows the specification of a non-symmetric interval which variance doesn’t allow. So if you want a standard expression that is general, the standard deviation works. That does *NOT* mean that variance is not the overall metric for uncertainty.
“YOU CAN’T KNOW THE SD WITHOUT FIRST KNOWING THE VARIANCE!”
This is such a childish argument. You saidvthat cariance was the measure of uncertainty. I said standard deviation is the prefered measure. And now you realise you were wrong you rant and twist the argumrnt as being about how standard deviation is calculated.
Of course standard deviation is the root of variance. Variance is an intermediate step, it exists because you square to get rid of negative differencies. But in itself it isn’t a useful measurvof deviation because it’s the average of the square difference and has square units. That is why you take the square root, to get a value that is a useful metric of deviation, or uncertainty.
“Variance, however, IS *THE* metric for uncertainty.”
Nothing you quote says that.
“where you calculate the sum of u^2(y), not the sum of u(y).”
Why do you think it’s written u² and not varience? Because standard uncertainty is a standard deviation. Yes the convinience if varience is that when adding random variables you have to add the squares of the standard deviations and take the square root. which can easily be written as adding variances, but that doesn’t make the variance a useful value. It’s exactly the same as using the Pythagorean theorem. You are adding the square areas of the sides of a triangle to get the area
of the side of the hypotenuse, but the practical use is to take the square root to get the length if the hypotenuse.
“One of the main reasons for using standard deviation in expressing uncertainty is that it shows a plus/minus interval.”
Huh. Any number can be turned into a ± interval. It’s just that such an interval is meaningful for a standard deviation as it reflects the actual dispersion of values.
“Using a plus/minus interval allows the specification of a non-symmetric interval which variance doesn’t allow.’
Not this insanity again. Sorry. you just don’t understand how this works. I’m not rehashing your delusions about negative standard deviations again.
I told you why standard deviation is used. As usual, your lack of reading comprehension skills are being demonstrated.
Then why is the uncertainty always given as +/- s? You’ve always denied that sqrt(s^2) = +/- s. That the square root has a negative and positive component. “s” by itself is *NOT* a useful metric of deviation, “+/- s” *is*. The use of an interval allows for an asymmetric interval, “s” by itself does not!
GUM: “3.3.5 The estimated variance u^2 characterizing an uncertainty component obtained from a Type A evaluation”
does not say that variance characterizes an UNCERTAINTY COMPONENT obtained from a Type A evaluation?
And you wonder why people tell you that you simply can’t comprehend simple English?
Because u =s and s^2 = variance! so u^2 IS THE VARIANCE. Why do you think they use u^2 and not just “u”?
And you wonder why people tell you that you simply can’t comprehend simple English?
It is *NOT* a “convenience”. It is a mathematical consequence.
You can’t add standard deviations and get an appropriate value.
let SD = 3
(3+3) = 6, sd of combined = 6
(9 + 9) =18, sd of combined = sqrt(18) = 4
6 ≠ 4
If you have to use variance to get the right standard deviation, then it is not just useful – IT’S USE IS A REQUIREMENT!
And here we see, once again, the statistician’s meme that “numbers is just numbers”.
And here we go again. Reality doesn’t apply in statistics world. Asymmetric uncertainty intervals can’t exist in statistics world!
“I told you why standard deviation is used.”
What you said was:
“Then why is the uncertainty always given as +/- s”
It isn’t. Uncertainty is expressed as a standard uncertainty. The ± interval is expanded uncertainty. Not ± the standard deviation.
“That the square root has a negative and positive component.”
Are you really trying to restart this nonsense? You really like demonstrating your ignorance. Any positive number has two square roots, one positive one negative. The standard deviation is defined as the positive square root, as is the standard uncertainty. There is no such thing as a negative standard deviation.
If you ever actually read the GUM you would see they say that explicitly.
“does not say that variance characterizes an UNCERTAINTY COMPONENT obtained from a Type A evaluation?”
It does not say ““Variance, however, IS *THE* metric for uncertainty.”. And you wonder why I think you cannot understand simple English.
“Why do you think they use u^2 and not just “u”?”
Because for that equation you need the square of the standard uncertainty. If variance was THE metric for uncertainty they could just say u as u would already be the variance. Really try to learn basic English. It’s tedious the way you twist everything just to avoid admitting a mistake.
“It is a mathematical consequence. ”
A mathematical consequence of needing the squares of the standard deviation. Yes, that’s why I called it a convinience.
“Asymmetric uncertainty intervals can’t exist in statistics world!”
And more lies.
“NOTE 1 The parameter may be, for example, a standard deviation”
But FIRST you must find the variance. You don’t calculate standard deviation directly!
Much like an average. You usually calculate the sum and then divide by n. It doesn’t mean the sum or the variance is a meaningful or useful quantity, it’s the end result that has meaning.
“This is the first time I can remember you talking about a peakedness index. Your equation makes no sense. Could you privide a reference? As far as I can see the index was only proposed last year.”
I have told you OVER AND OVER AND OVER AND OVER ……
As the variance grows the hump around the average gets smaller – meaning the values surrounding the average get closer and closer to the average value – a metric for the measurement uncertainty surrounding the average value. In other words, the PEAKEDNESS of the hump goes down. I’ve even posted images demonstrating this to you in threads on the subject.
The peakedness index is just a method for quantifying what I’ve told you OVER AND OVER AND OVER AND OVER ….. which you have stubbornly refused to accept, just saying that the variance is *NOT* a metric for uncertainty.
The peakedness index is nothing more than quantifying what has been discussed in the GUM for YEARS! It is a concept that you just can’t seem to grasp or accept!
So you don’t know what “peakedness” means.
What you say about sd is just wrong. As usual you are assuming that all distributions a Gaussian. Try applying your logic to a rectangular distribution. Does peakedness increase as the standard deviation decreases?
I do *NOT* assume all distributions are Gaussian! Why do you think I keep saying that the 5-number statistical descriptor is what climate science should use?
Variance does *NOT* require a Gaussian distribution to be valid.
A rectangular distribution with a constant value has a variance of 0. Thus the peakedness index of a rectangular distribution would be undefined (division by 0).
That is why the GUM defines the standard uncertainty to be
(b – a) / √3.
That should be
(b – a) / 2√3
It’s just the standard equation for the SD of a rectangular distribution. Not sure what that has to do with peakedness.
What happens to the peakedness when (b-a) gets larger, i.e. the variance goes up?
Nothing.
In fact the Peakedness Index for a rectangular distribution is always 1.
Moreover, the PI for most distributions does not depend on variance. The Peakedness of a normal distribution is always 1.414…
The peakedness index is meant to compare distributions, not to classify different distributions of the same type. A Gaussian curve *always* has the same shape, regardless of x-axis scaling. The variance, however, is *not* the same. And the uncertainty of the distribution *is* related to the scaling, i.e. the variance.
Larger variance-higher uncertainty. Lower variance-lower uncertainty
“The peakedness index is meant to compare distributions, not to classify different distributions of the same type.”
Just point to your source. I know you won’t because like everything else you heard about, made up your own version if what it meant. And will refuse to accept that you might be wrong for all of eternity. But please try and prove me wrong.
“A Gaussian curve *always* has the same shape, regardless of x-axis scaling. The variance, however, is *not* the same. And the uncertainty of the distribution *is* related to the scaling, i.e. the variance.”
All correct. And nothing to do with peakedness.
“I do *NOT* assume all distributions are Gaussian!”
\sarc You claim that, but you kerp making that assumption all the time.
“Variance does *NOT* require a Gaussian distribution to be valid.”
I never said it did. But your assumption that you can tell something about “the hump” from variance does. Deviation tells you nothing about the shape of the distribution.
“A rectangular distribution with a constant value has a variance of 0.”
That’s just a point. What about a distribution with a greater than zero variance?
Really? Is that why I say you can have an asymmetric uncertainty interval while you say it can’t happen?
Even skewed distribution will have a hump somewhere, so will a multi-model distribution. The “peakedness” of that hump *is* related to the variance of the distribution.
Unfreakingbelievable!
“That’s just a point. What about a distribution with a greater than zero variance?”
The peakedness index is sqrt(12)/R^2 where R is the range. As the interval gets wider the variance goes up and the peakedness goes down.
Why would you expect anything different?
“Really?”
He can give it out, but can’t take it.
“Is that why I say you can have an asymmetric uncertainty interval while you say it can’t happen?”
And there’s yet another lie.
“The peakedness index is sqrt(12)/R^2 where R is the range.”
I asked you for a reference – no reference came. I’ll ask again, where do you get your definition of the peakedness index from? It’s obviously not from the source I gave you, and I can find no other reference to a PI online.
But wherever your definition comes from it doesn’t make sense. How can the range of a uniform distribution affect the peakedness. By definition it is always flat, it has no peak.
tpg: “Is that why I say you can have an asymmetric uncertainty interval while you say it can’t happen?”
Who is lying here? I said you can have an asymmentric uncertainty interval which is hardly Gaussian, and you claimed I think all distributions are Gaussian.
You can’t keep what you say straight, can you? It’s whatever you need to say in the moment, right?
I’m not your research assistant. And *YOU* are supposed to be the statistics expert lecturing us all on how you can average intensive values and get a physically meaningful mean. You wouldn’t even accept, and you still don’t, that variance is a metric for the uncertainty that goes with the average. And you don’t know about the peakedness index? What kind of a statistics expert are you anyhow?
As usual, you never bother to read anything for meaning and/or context.
The PI of a uniform distribution is 1. It *does* exist.
For the uniform distribution the uncertainty grows and shrinks with the range (b-a). The peakedness does not. The peakedness defines the shape of the distribution. The peakedness of a Gaussian is, and always will be, greater than the peakedness of the uniform.
If I were King of Science, I would be telling climate science that they *MUST* provide the 5-number statistical descriptor for *any and all* data sets they formulate and use along with the mean, the median (part of the 5-number), and the mode. It provides a standard and robust measure of where the bulk of the data lies for almost all distributions of data. It would put the lie to the climate science assumption that all measurement uncertainty is random, Gaussian, and cancels”. And it would eliminate the idiocy that the mis-named “standard error” is the measurement uncertainty of the average.
I don’t think there’s any point continuing this argument after this. It’s entirely obvious you are either arguing in bad faith or are clueless about what you actually said.
I asked for a reference to your definition of the Peakedness Index because it clearly didn’t agree with one I gave you. You refuse to give a reference, insult me, and then say:
“The PI of a uniform distribution is 1. It *does* exist.
For the uniform distribution the uncertainty grows and shrinks with the range (b-a). The peakedness does not. The peakedness defines the shape of the distribution. The peakedness of a Gaussian is, and always will be, greater than the peakedness of the uniform.”
Most of which is correct, except that’s not what you were saying when I was replying to you. You said two comments ago:
So you are now demonstrating you now accept I was correct and you were wrong, but as always you will never admit you made a mistake and will just pretend you never said what you said. I expect next time you’ll be claiming it was you who had to explain it to me. You’ve done this several times in the past, and I have no interest at this point in arguing with such a disingenuous childish troll.
You’ve been correct about NOTHING. You didn’t even know that the peakedness index existed! And you *still* haven’t accepted that variance *is* the uncertainty metric – even though the GUM states such explicitly. It was *I* that had to point out to you that the PI is for comparing distribution types. You didn’t even know that the PI of a uniform distribution is 1 because it is basically the range divided by the range.
I’ve made no mistake in anything. I’ve spent the last week TEACHING you about peakedness! And, like measurement uncertainty, you *still* haven’t figured it out!
Your inability to ever admit when you are wrong about anything, is what causes you to be wrong about so much.
“You didn’t even know that the peakedness index existed!”
You see, the difference between you and me, is that when I’m presented with an unfamiliar term, I try to find out what it means. You just guess and then insist your wrong guess must be correct. You still won’t admit you don’t understand it.
“It was *I* that had to point out to you that the PI is for comparing distribution types.”
What you said was it depended on the variance of the distribution. You only have to look up through the comments to see your claim that “The peakedness index is sqrt(12)/R^2 where R is the range.”, and I pointed out to you that it doesn’t change as the range changes. I also predicted that you would next try to claim that you were the one who explained that to me. I didn’t expect you to be so dumb as to do it in this very thread, where the evidence is plain to see.
“You didn’t even know that the PI of a uniform distribution is 1 because it is basically the range divided by the range.”
As always you insist on looking at these things in a strange way. The PI is the maximum probability divided by the mean probability (continuous informity). In a uniform distribution these are the same, hence the PI of 1, which means it has no peak. Range only comes into it, in as far as p(x) in a uniform distribution is equal to 1/r, where r is the range. PI = (1/r)/(1/r). You could rewrite that as r/r if you like, but it doesn’t give much of an insight.
And it doesn’t matter which distribution you have, variance is irrelevant. The PI will always be the same regardless of variance, which is what you claim to be teaching me, but is the opposite of what you said a week ago:
https://wattsupwiththat.com/2026/05/07/uah-v6-1-global-temperature-update-for-april-2026-0-39-deg-c/#comment-4193484
“I’ve made no mistake in anything. I’ve spent the last week TEACHING you about peakedness!”
See my opening comment.
Even after looking it up *YOU* didn’t know its used to compare types of distribution! *I* had to teach you that!
And different distributions have different variances! Some don’t even have a variance, it’s undefined. And *I* had to teach you that!
Like I keep saying: You can’t even do simple algebra!
1/(1/r) = r
(1/r)/1 = 1/r
r/r is *MORE* insightful because it is simpler, it’s obvious that the result is 1. It’s the same issue with you and Possolo’s uncertainty formula for a barrel volume. You couldn’t figure out how V/V^2 = 1/V so I had to be doing it wrong. Here you couldn’t figure out r/r = 1 so I have to be doing it wrong. Unfreakingbelievable!
Like *I* told you, that’s true for the same distribution. It is *NOT* true for different distributions! Different distributions have different variances! You can’t seem to even keep that straight for consecutive posts!
Since you obviously won’t take my word for it. From here: TSCS_Week5_Trends.pdf
When the variance of a time series changes over time — for example, due to structural breaks, regime shifts, or non-stationarity — it can distort statistical relationships and lead to spurious trends that appear meaningful but are not causally linked.
Including ENSO which is a dramatic shock to the trend can propagate indefinitely if not properly accounted for. Do your regressions take this into account?
Now you are talking statistics instead of physical science. That is one of the main problems in climate science. The primary use of the regression is to predict the value of the dependent variable when values for the independent variables are given.
Your graph is a mess. The x-axis is not log CO2. It is log₂ CO2. This gives values of concentration on the x-axis from about 330 to 415. You also didn’t include the regression formula. The graph is useless as a forecasting tool. Show us the regression equation you came up with and show what the predicted anomaly will be at 660 ppm, a doubling of CO2.
“Since you obviously won’t take my word for it.”
Nullius in verba.
You need to explain what you mean and preferably demonstrate it. I’m not even saying you are wrong, but unless you can show your workings as far as the UAH trends Spencer states, then how can you say there is a serious problem with them?
But all you do is quote more stuff that has nothing to do with the claim. Your slide show says nothing about seasonality, let alone how different variances over the year can effect liner regression.
If I have time I’ll try to do some analysis on this, but it’s really up to you to demonstrate what you think the correct trend should be. You keep claiming to be an expert on this, yet never do anything than say that Dr Spencer doesn’t understand what he’s doing.
“Now you are talking statistics instead of physical science.”
Of course we are talking about statistics. What do you think linear regression is? I take it you think you should have a deterministic rather than a stochastic model. In which case I can only echo the charge leveled at me, you don’t live in the real world.
“Your graph is a mess.”
Sorry. I produced it in a hurry, to counter your lies. I’s sooner have a messy graph than your non-existent ones.
“The graph is useless as a forecasting tool.”
That’s not it’s purpose.
“Show us the regression equation you came up with and show what the predicted anomaly will be at 660 ppm, a doubling of CO2.”
And again, you should not use a regression outside the data range.
For the record the regression is 2.2 ± 0.47 °C / log2(CO2). With a 2σ uncertainty interval. Though note this is not corrected for auto-regression so it should probably be a bit larger.
The prediction for 660ppm of CO2 would be 1.8 ± 0.5 °C. That’s a prediction interval, not a confidence interval. But, as I said, it’s a mistake to take this as an actual prediction.
I gave you resources to read. If you won’t take the time to read them and attempt to understand them, then it isn’t worth my time to try and explain it to you.
Benice has made it plain to you that most of the increase in UAH is due to ENSO. It does exist.
UAH’s lower‑troposphere (TLT) product is known to retain ENSO‑related shocks unusually strongly, producing step‑like increases that decay more slowly than in other satellite or reanalysis datasets. This behavior is linked to how UAH performs diurnal drift correction, synthetic channel construction, and satellite merging — not to a physical requirement of the climate system.
If you wish to know more, I suggest you delve into the UAH processing.
So your graph is useless in determining what the future holds. That is what many of us have been trying to tell you. Cherry picking early 20th century temperatures that are closely related to the Little Ice Age to start a trend will not tell you what the future holds regardless of when it ends.
“I gave you resources to read.”
Typical deflection. Your resource is a slideshow from a lecture that makes no mention of seasonality, or anything else you claimed. And I keep telling you, if you want to be taken seriously, you need to be able to explain this in your own words, and give us your analysis. How are you interpreting the trend for UAH? How does it differ from the value Spencer gives in this article?
“Benice has made it plain to you that most of the increase in UAH is due to ENSO. It does exist.”
If you mean bnice2000, why do you think that troll has demonstrated anything? You are the one claiming linear regression is wrong for UAH, yet now you want to rely on arbitrary short term linear regression to claim that ENSO is causing multi-decade warming. This has not been demonstrated statistically, and is physically implausible.
And this is yet more deflection. The discussion was about seasonality, not ENSO.
“If you wish to know more, I suggest you delve into the UAH processing.”
Or maybe we should just stop using UAH.
“So your graph is useless in determining what the future holds.”
Not completely useless, but that’s not the purpose of a linear regression. Best you can say about future projections is that, if the linear trend is valid, and if UAH is an accurate measure of global temperature, and if it continues to be linear as CO2 increases, and if there are no confounding factors, then it might be a reasonable estimate.
“Cherry picking early 20th century temperatures…”
Huh. The data goes back to 1979. That is not early 20th century.
Sorry you are so adverse to learning new things. The resource discusses spurious trends. These are real and can’t be dismissed.
For trend analysis here are some items to consider.
You keep criticizing but seldom show references. Show some resources that refute the assertion that regressions on current temperature data sets do not include spurious trends from data.
Let me point out that regression trending average temperatures results in spurious trends because of Tmax and Tmin having unique trends of their own. It leads one to say both are increasing when that is not true.
“Sorry you are so adverse to learning new things.”
You couldn’t imagine how many new things I’ve learned during the course of these arguments. It’s the only reason it’s worth continuing.
“The resource discusses spurious trends.”
But not seasonality, which is what you were claiming.
“Remove the annual cycle to avoid aliasing.”
Done that. Anomalies, remember.
“Account for autocorrelation because ENSO introduces persistence”
Ideally yes. but this generally increases uncertainty, not the trend. Remember all the times I had to point out the uncertainty in your pauses? Factoring in auto correlation is why the uncertainties are so large, but it doesn’t stop the trend being flat.
But it’s a complicated procedure and there is no one correct method. In the past I used the figures given from the trend calculator that results in UAH uncertainty being roughly 4 times larger than the default value. But I suspect this might be too large.
Most of the time I prefer to use annual averages that remove some but not allauto correlation. It also of course removes any seasonality.
“ENSO years can create apparent “steps” in tropospheric datasets due to shocks not decaying”
That’s your assetion and I see no reason to it. You have done nothing to determine these step changes statistically. but even if you did it wouldn’t counter the real world physical reasons why this is implausible.
“You keep criticizing but seldom show references”
Strange, I could have sworn this started with you criticizing the way liner regression was used in this article, and then refusing to explain how you would do it differently.
“Show some resources that refute the assertion that regressions on current temperature data sets do not include spurious trends from data.”
You want me to prove a negative. I’m not even claiming the simple regression is correct. The more you learn the more ways of doing things there are and the more possible issues. But if you think there is something fundamentally wrong with a specific use, then you need to show your own workings. Do your own regression using whatever method you want, and then we can compare it with the trend given by Spencer, or any simple software package.
“Let me point out that regression trending average temperatures results in spurious trends because of Tmax and Tmin having unique trends of their own.”
I’m not sure if you understand what “spurious” means. A correct trend of a specific quantity is still a correct trend. The trend if average temperatures is a correct trend of average temperatures. A correct trend of lower troposphere satellite data is a correct trend of that data.
“It leads one to say both are increasing when that is not true.”
It doesn’t. The change in average temperatures does not imply the same change is happening in both max and min. But in fact both min and max are increasing, so your point is moot.
Create a linear temperature trend for each decade from January 1979 to the present. Will the temperature trend be the same for each decade?
Do the temperature trends for each decade tell more than a temperature trend for the entire period?
No. A period of ‘climatology’ is based on at least 30-years of continuous data, according to the WMO.
Every global temperature data set we have, surface or satellite, shows statistically significant warming over the past 30 years.
(‘Statistically significant’ meaning >95% probability that the observed warming trend would not have occurred in a randomly fluctuating climate system over a 30-year period. The chances of it not being a true warming trend are the same as the chances of tossing a coin 30 times and it coming up heads on ~29 of the tosses. The warming is real.)
Is the trend the same for the period 1979-1989 compared to the period 2009-2019?
When comparing 2 periods, you can see whether the trend is decreasing or increasing.
Better yet, compare it to 1915-1945.
Temperature trends are a joke. The trends are all against time, auto-correlated, and with seasonality. If you want to use time series analysis to find a stationary trend with constant statistical values for means, and variance, that is a proper method to eliminate these problems. Then one can create a model that allow different variables to be assessed as to how they affect they trend.
Bellman continues to show regressions with a linear equation that continues indefinitely. Doing so ignores cooling, as over the last few months, pauses, and step changes at El Nino occurrences.
If you really want to get down into the weeds, start using Tmax and Tmin from every day separately. Tavg = (Tmin + Tmax)/2 is a joke. They have different means, variances, functional relationships.
Here is a good graph showing the relationship between sun, soil, and atmosphere. Just looking at temperature will tell you nothing.
Ag studies have shown that daytime temperatures are well within the tolerances of most grain crops in the U.S. Night temperatures have increased such that last frost and first frosts allow for longer growing seasons. Grain harvests worldwide, confirm that global temperatures are not a problem.
Averaging temperatures, especially from far flung places is not scientific. NH and SH have reverse seasons, do you think their variances are different? Where do those disappear to in the calculation of a global temperature?
OK, as Jim won;t show his seasonal analysis, I might as well have a stab. I’ll use the method of treating each month as a factor in the linear regression.
First let’s compare the linear trend without accounting for seasonality.
For anomalies I get
lm(formula = Anomaly ~ Time, data = .) Residuals: Min 1Q Median 3Q Max -0.50432 -0.12224 -0.01685 0.10373 0.71156 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -3.116e+01 1.187e+00 -26.25 <2e-16 *** Time 1.555e-02 5.928e-04 26.23 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.1936 on 567 degrees of freedom Multiple R-squared: 0.5482, Adjusted R-squared: 0.5474 F-statistic: 687.9 on 1 and 567 DF, p-value: < 2.2e-16The trend is 0.1555°C / decade. Consistent with Spencer’s quoted value of 0.16.
(Whine about the number of decimal places all you want, I want to see any difference between the methods.)
Now for absolute temperatures.
lm(formula = Temperature ~ Time, data = .) Residuals: Min 1Q Median 3Q Max -1.3641 -0.7519 -0.2041 0.7699 1.8004 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.333e+02 5.111e+00 45.638 < 2e-16 *** Time 1.537e-02 2.552e-03 6.022 3.1e-09 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.8334 on 567 degrees of freedom Multiple R-squared: 0.06012, Adjusted R-squared: 0.05846 F-statistic: 36.27 on 1 and 567 DF, p-value: 3.097e-09The trend is slightly lower at 0.1537°C / decade. This is becasue of the seasonal variation, especially as the time series is ending during the colder months. Also note that the R² value is just 0.058, as most of the variation is in the seasonal fluctuations.
So now the seasonal model, used with anomalies.
lm(formula = Anomaly ~ Time + factor(Month), data = .) Residuals: Min 1Q Median 3Q Max -0.48861 -0.12165 -0.01403 0.10233 0.70243 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -3.117e+01 1.198e+00 -26.009 <2e-16 *** Time 1.555e-02 5.983e-04 25.985 <2e-16 *** factor(Month)2 -6.229e-03 3.987e-02 -0.156 0.876 factor(Month)3 1.102e-02 3.987e-02 0.276 0.782 factor(Month)4 1.860e-02 3.987e-02 0.466 0.641 factor(Month)5 1.626e-02 4.008e-02 0.406 0.685 factor(Month)6 4.826e-03 4.008e-02 0.120 0.904 factor(Month)7 1.338e-02 4.008e-02 0.334 0.739 factor(Month)8 1.481e-02 4.008e-02 0.369 0.712 factor(Month)9 7.668e-03 4.008e-02 0.191 0.848 factor(Month)10 1.141e-02 4.008e-02 0.285 0.776 factor(Month)11 9.947e-03 4.008e-02 0.248 0.804 factor(Month)12 1.225e-02 3.987e-02 0.307 0.759 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.1953 on 556 degrees of freedom Multiple R-squared: 0.5487, Adjusted R-squared: 0.539 F-statistic: 56.34 on 12 and 556 DF, p-value: < 2.2e-16The trend is 0.1555°C identical to 4 significant figures to the standard model.
The estimate values for each month is the offset relative to January, and they are all very small, less than 0.02°C, and show little pattern. All are insignificant.
Here’s the graph using this regression. You can see a slight wobble in the trend line, but it’s essentially the same as the standard regression.
Now for temperatures using the seasonal model.
lm(formula = Temperature ~ Time + factor(Month), data = .) Residuals: Min 1Q Median 3Q Max -0.48855 -0.12152 -0.01206 0.10228 0.70240 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.320e+02 1.198e+00 193.617 < 2e-16 *** Time 1.554e-02 5.983e-04 25.976 < 2e-16 *** factor(Month)2 8.251e-02 3.987e-02 2.069 0.039 * factor(Month)3 2.587e-01 3.987e-02 6.490 1.91e-10 *** factor(Month)4 6.821e-01 3.987e-02 17.108 < 2e-16 *** factor(Month)5 1.285e+00 4.008e-02 32.059 < 2e-16 *** factor(Month)6 1.924e+00 4.008e-02 48.003 < 2e-16 *** factor(Month)7 2.252e+00 4.008e-02 56.178 < 2e-16 *** factor(Month)8 2.069e+00 4.008e-02 51.625 < 2e-16 *** factor(Month)9 1.465e+00 4.008e-02 36.555 < 2e-16 *** factor(Month)10 7.758e-01 4.008e-02 19.354 < 2e-16 *** factor(Month)11 2.356e-01 4.008e-02 5.878 7.15e-09 *** factor(Month)12 2.384e-02 3.987e-02 0.598 0.550 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.1953 on 556 degrees of freedom Multiple R-squared: 0.9494, Adjusted R-squared: 0.9483 F-statistic: 868.8 on 12 and 556 DF, p-value: < 2.2e-16The trend is 0.1554°C / decade, virtually identical to that of the anomalies. Compare this with the original trend for temperature of 0.1537°C / decade., and you can see it’s removed the spurious change caused by seasonality.
The monthly offsets compared to January now reflect the season variation, going up to 2.3°C for July, and all but December are significant.
The R² is now 0.95, but that’s becasue the majority of the variation is caused by the seasonal cycle.
This should be clear in the graph.
I have shown you this before.
This graph was made using the ADIFF command in an Excel stats package. It reduces auto-correlation by first differences of 1-lag and removes seasonality of 12 month periodicity.
Do you see a trend?
That is the baseline trend you should be starting with.
You say a linear regression is not useful for making predictions. It only is useful for displaying a line of best fit for what has occurred in the past. Whoopee!
A time series analysis will allow one to derive multiple variables that can be used to evaluate what causes temperature to change.
It is one reason I have started researching how insolation is processed through soil and then to the atmosphere. Let’s see you do that with your regression.
“This graph was made using the ADIFF command in an Excel stats package. It reduces auto-correlation by first differences of 1-lag and removes seasonality of 12 month periodicity.
Do you see a trend?”
The main point of taking differences is to remove the trend.
No, that is not the case. It does remove linear trends that may very well be due to auto-correlation. In other words when yesterday’s temperature influences today’s temperature.
Here is a web page from Michael Mann’s u.niversity
https://online.stat.psu.edu/stat462/node/188/
As I mentioned earlier, which you obviously missed, is that this gives a baseline with constant statistical parameters. You can then use that baseline to add changes contributed by other variables. CO2 for example. It allows one to create a model that more accurately makes forecasts.
If you are confident that your linear regression accurately predicts the change in temperature, then you can easily add this to the baseline trend and obtain a forecast of future temperatures. Then you can compare it to actual values each month in the future. Good luck.
“No, that is not the case.”
https://milvus.io/ai-quick-reference/what-is-differencing-in-time-series-and-why-is-it-used
https://apxml.com/courses/time-series-analysis-forecasting/chapter-2-decomposition-stationarity/differencing-for-stationarity
Bottom line is, if you detrend a time series, don’t complain if you can’t see the trend.
Differencing is not just about removing a trend. It is make a time series more stationary by removing auto-correlation between data points. A lag-1 is usually sufficient but may need different lag values.
Removing seasonality helps further stabilize the statistical parameters of the time series.
Stationarity is a requirement for many statistical theorems to hold. This includes the Law of Large Numbers and the Central Limit Theory both of which are fundamental to the divide by √n for standard deviation of the mean. This paper has more details than your references. Read Section 2.3 of this paper.
https://www.tandfonline.com/doi/full/10.1080/00273171.2024.2436413#d1e171
I didn’t say it was the only reason. But the point is it will remove a linear trend, and so claiming your graph shows no trend tells you little.
“It is make a time series more stationary by removing auto-correlation between data points.”
Removing auto-correlation does not make the series stationary. That only happens when you have a random walk.
I didn’t say it makes the series stationary. I said “more stationary”.
You gripe about the base trend having a constant mean and standard deviation then turn around and gripe that SARIMA doesn’t make the series stationary. Get your gripes sorted out.
Here is what Penn State says.
https://online.stat.psu.edu/stat462/node/188/
I have shown you this before.
This graph was made using the ADIFF command in an Excel stats package. It reduces auto-correlation by first differences of 1-lag and removes seasonality of 12 month periodicity.
Do you see a trend?
That is the baseline trend you should be starting with.
You say a linear regression is not useful for making predictions. It only is useful for displaying a line of best fit for what has occurred in the past. Whoopee!
A time series analysis will allow one to derive multiple variables that can be used to evaluate what causes temperature to change.
It is one reason I have started researching how insolation is processed through soil and then to the atmosphere. Let’s see you do that with your regression.
“Let’s see you do that with your regression.”
You won’t see it happen.
He actually reports it to 0.001 C.
Why not 0.001 mK?
I don’t know. You’ll have to ask Dr. Spencer.
If it follows the pattern since the beginning it should cool a bit more then oscillate around a constant value a little warmer than before this super El Niño. Might take 3-4 years to know.
All the warming to date looks like residual heat from super El Niños. No GHG increase caused warming at all.
This is known as a “shock” in a time series. It is carried throughout following averages. The decay of a shock requires specific treatment of the data to remove. Time series analysis is in short supply in climate science.
Please note the red line in the topmost graph of UAH data that is labeled “Running, centered 13-month average”.
Any “shock” data in the UAH plotted time-series data that does nor fall within that moving 13-month time window DOES NOT get “carried forward” nor does it affect that “running centered average” trend line more than 14 months before or more than 14 months after the interval of that “shock”.
In this case, the “specific treatment” to remove the “shock” from “following averages” is to allow time to pass. /sarc
The average ENSO state since 1979 is 0. [1]
Anyway, do you have a prediction of when you think El Ninos (and thus the warming) will stop?
I keep seeing predictions of an imminent super El Niño which is strange since we are just coming out of one.
The climate scaremongers DESPERATELY want another El Nino.
They KNOW it is the only cause of any warming, and a totally natural solar forced occurrence….
.. but still need it so they can continue to PRETEND that human CO2 is the cause.
From the WUWT reference page, here is a plot of T at depth 55m, along the equator. Time on the y axis – present time at bottom. Something is going on:
This chart is absolute confirmation that the warming is not cause by human released CO2.
Well . . confirmation over the charted period of start-May 2025 to end-April 2026 . . . one year.
Yes, it’s called natural variations in ocean temperature, particularly along Earth’s equator on the east side of the Pacific Ocean (~70 to 100 W longitude).
Interesting that you only focused on presenting data over the relatively narrow time period of May 2025 to end-April 2026.
According to https://ggweather.com/enso/roni.htm there was a weak La Niña (decline in ocean near-surface temperatures) in 2025-2026.
“Interesting that you only focused on presenting data”
It isn’t my graph. It is what is displayed on the WUWT Reference Page
A distinction without a difference.
It is YOU that presented it in YOUR post with YOUR statement that “Something is going on”.
That statement is not on the WUWT Reference Page.
If you’re seriously interested in this question, please study this report, especially the slides #10-12, 15* & (for ‘projections’ fwd) #25*:
http://www.cpc.ncep.noaa.gov/products/analysis_monitoring/lanina/enso_evolution-status-fcsts-web.pdf
And, at your own risk, the up-to-date (labeled 6th of May) projections are presented in the graphically in the midst of too-much-more at …
https://arctic-news.blogspot.com/
Use this phrase to search for the (2) graphs:
We’re coming out of a La Nina. [1]
The most recent, moderate El Niño event occurred from summer 2023 through early spring 2024, with peak impacts around the end of 2023. This El Niño was officially declared to have ended by April 2024, with a weak La Niña period beginning in late 2024 and continuing into 2026. (see https://ggweather.com/enso/roni.htm )
These facts are in excellent agreement with the UAH graph of GLAT posted in the above article.
The Hunga-Tonga submarine volcano is claimed to have injected “massive amounts” of water vapor into the stratosphere in mid-January 2022 . . . H-T water vapor, where are you now and where were you from January 2022 through, oh, May 2023???
I have a graphic of the water vapour.
Excellent!
That contour plot (time on the x-axis) indicates that there is still a significant amount of excess water vapor, at least at the 26 hPa pressure altitude, from 75 S to 75 N latitudes, compared to pre-January 2022 levels. Despite this, the UAH plot of GLAT shows that global average lower atmospheric temperature has returned to being right at the 47+ year historic linear temperature trend line (January 1979 through April 2026).
HT water vapor . . . if you’re really up there in the stratosphere, you’re really not doing what many claimed you were capable of!
The Hunga-Tongan volcano worshippers are currently the biggest embarrassment to climate skepticism that exists. They spent decades arguing that a 200ppm increase in CO2 had nothing to do with global temperatures, only to turn around and insist that a 1ppm increase in H2O was driving the highest temperature spike in the satellite record. They need to be excised from the movement. History should not be kind to these people.
Ahhh stuff the global boiling the natives are getting hot under the collar again-
Brussels mulls scrapping methane fines amid energy crisis – leak
I have to interject here that all of this discussion is nothing more than mathematical masturbation.
These truths indicate that adding local and regional temperature systems to create a larger “global” system is decidedly non-physical. In fact, using local temperatures to create a larger regional system is also non-physical. The microclimate at PointA is different than the microclimate at PointB – different systems with different intensive properties. Adding the temperatures of the two points to create a mean tells you nothing physical, it does not create a larger system whose intensive temperature property can be determined.
Yes, you can use the data to create a MATHEMATICAL MEAN but you are *NOT* creating a physically meaningful statistical descriptor of a larger system.
Statistical descriptors are *NOT* measurements. They are mathematical attributes of a data set that is useful in understanding the data set. But if the data set is not attributable to a physical system then they tell you nothing about reality.
Tracking a non-physical mathematical mean is an attractive nuisance that climate science tries to assign some physical meaning. An attractive nuisance is a legal doctrine concerning hazardous conditions that are attractive to children. That is a perfect description of the “global average temperature”.
“Yes, you can use the data to create a MATHEMATICAL MEAN but you are *NOT* creating a physically meaningful statistical descriptor of a larger system.”
Keep thinking, you are almost getting there.
The mean is not a physical thing. That’s true for intensive or intensive properties. The average weight of a person is not the combined weight of all people. The average temperature if a person does not have to be the physical temperature of all people combined. But that doesn’t mean that the average is useless. The average tells you something about the population and it alliws you to distinguish between different populations.
If you want a phyisical meaningful description of an extensive property of an entire population, you want the sum of all the properties.
If you want a physically meaningful property of a combined population the sum will not give you that. Thevaverage may or may not give you that property, depending on exactly what and how you are aceraging, but at the least an average will be closer to that physical property than the sum.
Bull hockey. You are dealing with measurements of a physical property. The mean is the best estimate if the distribution of measurements varies randomly. See GUM 4.2.1. However, the dispersion of observations about the mean, the standard deviation, is an integral part of the measurement. See GUM 4.2.2.
If you want to pretend the mean of measurements has no physical meaning, then you are free to do so. Just don’t try to tell folks that you are discussing measurements like a global temperature. If it isn’t physical, then it isn’t a temperature. It is just a number you are playing around with that has no meaning in the real world.
“The mean is the best estimate if the distribution of measurements varies randomly.”
I think you are confusing yourself again. You are talking about the average of multiple measurements of the same thing. I was talking about the average of different things.
“If you want to pretend the mean of measurements has no physical meaning”
I said it wasn’t a physical thing, not that it didn’t have a physical meaning. I really think you are tying yourself in knots with all these phrases. The claim Tim is making is that the average isn’t a physical property of the collection of all things. The average weight of a person is not the combined weight of all people.
“If it isn’t physical, then it isn’t a temperature.”
That’s what I’m saying. The average temperature is not a temperature. It’s the average of all temperatures.
“It is just a number you are playing around with that has no meaning in the real world.”
And there’s your confusion again. Assuming that if a value isn’t a physical thing, it can have no meaning.
“I said it wasn’t a physical thing, not that it didn’t have a physical meaning. I really think you are tying yourself in knots with all these phrases. The claim Tim is making is that the average isn’t a physical property of the collection of all things. The average weight of a person is not the combined weight of all people.”
That is *NOT* what I am saying at all.
I am saying that if you have a large enough scale you can put 10 people on the scale and get a total weight. You can then find the average of that total weight.
I am saying that if you put 10 rocks on a thermometer, each at 20C, you will *NOT* get a total of 200C! If you can’t get a total value for the larger system then you cannot get a meaningful, physical mean for the larger system.
The difference is that one is an extensive property, weight, and the other is an intensive property, temperature.
A statistician or mathematician will say you can average ANYTHING and the average is meaningful. It is only meaningful in that it describes the distribution of the values you include in the data set. That doesn’t mean you can deduce anything meaningful in the real world from that statistical descriptor.
You *can* average the weight of measurand1, the length of measurand2, the temperature of measurand3, the speed of measurand4, and get a value. So what? It just tells you the average of the values you include in the data set. What use can you make of that average? If you track the average of those properties of the multiple measurands over time what can you discern from the change in the average of the values in the data set? Jeesshhh, the values of the components can change while the average stays the same!
Averaging temperatures of multiple measurands is exactly the same. Changes in the average for the phantom larger system tells you exactly nothing about the real world. It’s why climate science kept saying the world was going to burn up from higher temperatures and failed to recognize that the higher average was due to expanding growing seasons! Using mid-range values as an “average” intensive property is idiotic, averaging averages of intensive properties is even worse!
“That is *NOT* what I am saying at all.”
What you said was:
Perhaps if you calmed down and tried to define your terms, we wouldn’t keep having these disagreements. What exactly do you mean by “physically meaningless”?
My point is that there are two different concepts here – one the average as a meaningful statistical descriptor of the population (the larger system), and that is a useful value to know for a variety of reasons. But it does not necessarily describe a physical property of the larger system. In the case of extensive properties, it never does. The average mass of a person is not the mass of the population.
In the case of intensive properties, the mean is a closer approximation to a physical property of the larger system, and may be a more exact description of that property if properly weighted. But that is not generally the point of the average. The point is exactly the same as for extensive properties, to have a value that describes part of the population.
“I am saying that if you put 10 rocks on a thermometer, each at 20C, you will *NOT* get a total of 200C! If you can’t get a total value for the larger system then you cannot get a meaningful, physical mean for the larger system.”
But you don’t need to measure the larger system in one go. You can just add the values. And it doesn’t matter if that sum is not physically meaningful, the average will still be as useful as it is for the extensive property, Really, this idea that every step of an equation has to be physically meaningful for the end result to be meaningful is just nonsense.
“A statistician or mathematician will say you can average ANYTHING and the average is meaningful.”
They shouldn’t. But as usual you think that all statisticians and mathematicians are idiots who don;t understand their own research, and all scientists who use statistics are idiots. You really need to consider that maybe you are the one who is misunderstanding things.
“You *can* average the weight of measurand1, the length of measurand2, the temperature of measurand3, the speed of measurand4, and get a value.”
You cannot. An average has to be of things with the same units.
“Averaging temperatures of multiple measurands is exactly the same.”
Exactly the same, in the sense it’s completely different. Do words have no meaning for you?
“Changes in the average for the phantom larger system tells you exactly nothing about the real world.”
You think a significant change in average temperature can occur without there being any change in the larger system?
“…failed to recognize that the higher average was due to expanding growing seasons”
I’m not sure you understand cause and effect here. Growing seasons only change becasue temperatures are changing. The changing average indicates that temperatures are changing. You’ve just demonstrated that the average is not completely meaningless.
“Perhaps if you calmed down and tried to define your terms, we wouldn’t keep having these disagreements. What exactly do you mean by “physically meaningless”?”
Put the bottle away. I explained in DETAIL what “physically meaningless” is.
If you put 10 rocks, whose temperatures range consecutively from 10C to 19C, on a thermometer you do *NOT* get a 145C reading on the thermometer. Thus the average of 15C is PHYSICALLY MEANINGLESS. The collection of rocks simply can’t be combined into a larger system since they are intensive properties.
The average value obtained from the temperature at 0000GMT on Pikes Peak and the temperature at 0000GMT in Colorado Springs is PHYSICALLY MEANINGLESS. The average value obtained from the temperature at 0000GMT in Kansas City and the temperature at 0000GMT in Miami is PHYSICALLY MEANINGLESS. Neither of these examples can be combined into a larger system that can be averaged.
Even worse is the climate science belief that you can combine disparate intensive property values into a data set and it will always come out to be Gaussian – i.e. a distribution that the mean actually describes – instead of being a distribution that needs at least the 5-number statistical description to characterize the distribution.
“My point is that there are two different concepts here – one the average as a meaningful statistical descriptor of the population (the larger system), and that is a useful value to know for a variety of reasons. But it does not necessarily describe a physical property of the larger system. In the case of extensive properties, it never does. The average mass of a person is not the mass of the population.”
You have totally confused yourself!
The average mass of a GROUP of people is not the mass of each individual. SO WHAT! The mean mass value *is* a physically meaningful value for that group of people – because MASS is an EXTENSIVE property that *can* be combined into a larger system. It *DOES* describe a physical property of the larger system!
But temperature, an intensive property, can *NOT* be combined into a larger system. Thus the mean value is not physically meaningful, it does *NOT* describe a physical property of the larger system.
A physically meaningless mean *is* useless in describing physical reality. You can’t say that the mean of a set of disparate temperatures describes anything physical – its why the “global average temperature” is physically meaningless.
The entire concept based on using temperature is nothing more than the idea of TRADITION. Temperature was the only data available in the earlier centuries so it’s good enough for us! Teyve’s view in “Fiddler on the Roof”! The sad thing is that temperature is *NOT* a good metric for climate! Using a mid-range daily temperature value as the metric is even worse!
“The average mass of a GROUP of people is not the mass of each individual.”
What do you mean by the average mass of a group of people?
“The mean mass value *is* a physically meaningful value for that group of people – because MASS is an EXTENSIVE property that *can* be combined into a larger system. It *DOES* describe a physical property of the larger system!”
What are you on about? How does the average mass of a person, or a group of people, describe a physical property of a larger system? It was only a few weeks ago you were arguing the exact opposite. That the average mass of an asteroid told you NOTHING about the mass of all asteroids.
“A physically meaningless mean *is* useless in describing physical reality.”
Say what you mean by “physically meaningful”. An average of a physical property has meaning. It can be used to identify differences in populations for a start.
“You can’t say that the mean of a set of disparate temperatures describes anything physical – its why the “global average temperature” is physically meaningless.”
All you keep stating is your personal belief as if it had any meaning. Just because you can’t understand how an average of surface temperature can be used, doesn’t make it meaningless.
“Temperature was the only data available in the earlier centuries so it’s good enough for us!”
Yes, if you want to compare temperatures with the past, you have to use temperatures. Do you not use temperatures in your modern world? Weather forecasters never tell you what the temperature will be, Doctors never take your temperature because it’s old fashioned? Do you live on Laputa?
“All you keep stating is your personal belief as if it had any meaning.”
It’s not my personal opinion. It’s a physical science fact.
Give me ONE solid reference that says you can add intensive property values for different objects and obtain a physically meaningful average.
JUST ONE. All you need is ONE!
copilot:
————————————-
can you average the intensive property values of different objects such as rocks.
Short, direct answer:
You can compute a numerical average of intensive property values from different rocks, but that average is not itself an intensive property of any rock and not a thermodynamic state variable. It is only a statistical descriptor of the collection.
————————————-
grok:
—————————————
Temperature Yes Heat capacity weighted Most common case
———————————-
tpg: (heat capacity weighted gives an extensive value)
chatgpt:
————————————
So the short answer is:
————————————
(tpg: mass and heat capacity gives an extensive value)
I could give you any number of references from scientific sites but it’s obvious that you think you know more than physical science about averaging the intensive property value of temperatures, so it would be a waste of time.
“In the case of intensive properties, the mean is a closer approximation to a physical property of the larger system, and may be a more exact description of that property if properly weighted.”
Your lack of knowledge of physical science is sad.
You cannot create a larger system from diverse intensive property values. You *can* do so with extensive property values. It’s the very definition of intensive vs extensive properties!
The only way your conclusion could be true is if a rock at 10C and a second rock at 20C, both placed on a thermometer at the same time, would cause the thermometer to indicate 30C.
I suggest you try it sometime and see what the thermometer shows. Let us know when you do the experiment what you find.
“You cannot create a larger system from diverse intensive property values. You *can* do so with extensive property values. It’s the very definition of intensive vs extensive properties!”
That’s not the definition.
“The only way your conclusion could be true is if a rock at 10C and a second rock at 20C, both placed on a thermometer at the same time, would cause the thermometer to indicate 30C.”
Could you please keep banging your head against a wall until you understand the difference between a sum and an average. And understand that temperatures do not start at freezing point.
My contention is that your two rocks would register something between 10 and 20°C, assuming you have a thermometer you could place things on. Possibly close to 15°C, if the rocks are of similar size and material.
But my further contention is that it doesn’t matter, because the average is not a physical property, but an average of physical properties.
In other words you are guessing. Show us the math behind your guess of how the energy is redistributed between the two rocks. Give us the EXACT requirements for your guess to be true.
Averaging intensive properties of distinct objects is only correct when the measurand is explicitly defined as a ratio of extensive quantities that can be aggregated. Otherwise, the average is not a property of anything.
This is the exact reason climate metrics, station networks, and heterogeneous measurements must be handled with extreme care.
“In other words you are guessing.”
Yes, because I don’t happen to have two rocks of different temperature on my person at the moment. But if you think the two rocks can have a combined temperature greater than 20°C or less than 10°C I’d like to
see your experimental evidence.
So temperatures DONT add, nor do they subtract. Who knew?
If they don’t add, how do you obtain an average that is physically meaningful?
How many more times – temperatures don’t add in the sense that the temperature of two rocks is not the sum of the two temperatures. But it’s entirely possible to add the numbers and divide by n to get a meaningful average.
If you find it too difficult to add large numbers you can use the approximation method, guess an average, subtract from each value, average these differencies, and add to your guess.
And if averaging differences of temperature is still objectionable to you, you can just find the mid point of your two rocks. Subtract one from the other divide by two and add. E.g (20 – 10) = 10, 10/2 = 5, 5 + 10 = 15. 15°C is the mid point, and average. With a small amount of effort this can be expanded to any number of rocks, using a recursive function and appropriate weighting.
But it’s a lot simpler just to add all the values and divide by n, with no assumption that the sum means anything than the sum of the values.
Meaningful to you maybe but not scientifically.
No more arguing. Find and show references that describe how averaging temperatures is meaningful AND what assumptions are necessary to perform that task. Then tell us how atmospheric temperatures meet those restrictions.
PS: Here is a link from another thread post from Robert Kernodle that is enlightening. Read it and need it.
https://www.rgkernodle.com/the-intellectual-fraud-of-global-temperature-a-crime-against-reason
NIST found meaning in averaging temperatures of different parcels of air on different days in TN 1900 E2. Not that it matters since you already know about it and have found it to be unconvincing. If that doesn’t convince you then nothing will.
I wish I knew. It just becomes a bit of an obsession.
It does force me to learn new stuff, which is something.
You don’t even know what NIST used as it’s basis for it method.
NIST declared
What does no other uncertainty provide? You don’t need to deal with the uncertainty of single measurements in addition to sampling variation.
Now read JCGM 100-2008, F.1.1.2, H.6, and specifically H.6.3.
H.6 deals with uncertainty where sampling the same thing multiple times is not possible. The conditions of measurement fall under reproducibility and not repeatability.
Daily and monthly averages deal with measuring a property that varies over time. Developing a combined uncertainty requires propagating both single measurement uncertainty for each measurement AND the sampling uncertainty into the total.
Why do you think NIST arrived at an uncertainty of ±1.8°C while discounting single measurement uncertainty.
Note – What might the uncertainty of a daily average be? It is notable that NIST used Tmax in order to skip over the sampling error associated with Tavg.
Once again you completely ignore the main point. That point is that you bitch and moan about how averaging temperature is useless and meaningless and yet NIST does. And it’s not even the same temperature measurand they are averaging. It is different parcels of air measured on a different days yielding a completely different values. That should shut down any argument you have against averaging intensive properties or averaging different things. Yet you persist.
“ temperatures don’t add in the sense that the temperature of two rocks is not the sum of the two temperatures. But it’s entirely possible to add the numbers and divide by n to get a meaningful average.”
ROFL!!!
Temperatures don’t add but you can add them and average them and get a meaningful average!
But you just can’t quite identify what that “meaningful average” actually tells you! It’s just meaningful because it is a calculated statistical descriptor and calculated statistical descriptors are always meaningful.
Statistical world personified.
Its the Magic Number. Again!
Not unlike the Magic Molecule.
“Temperatures don’t add but you can add them and average them and get a meaningful average!”
You missed the bit where I said “don’t add in the sense that the temperature of two rocks is not the sum of the two temperatures.” That doesn’t mean you can’t add them, it’s just not going to give you a result that tells you what the combined temperature is. Having added them you can work out the average, which is the purpose of the procedure.
Is it your profound lack of a mathematical education that stops you from seeing how easy this is? Numbers are abstractions, they do not have to make physical sense at every stage of an operation, but they can give you results that are meaningful in the real world. Many scientists are able to do this, but for some reason you cannot. And it’s your your nature to assume that if you can’t do it it must be wrong. Your reveal this ignorance every time you use “numbers are numbers” as an insult.
“But you just can’t quite identify what that “meaningful average” actually tells you!”
I’ve given you lots of examples, starting with the average in the headline of this very article. You can use the average to see if the world is getting hotter, colder or staying the same. Countless articles were written on this site using averages to make all sorts of claims about what global temperatures have been doing, including all those ones about pauses. The articles were nonsense, but they do illustrate the point that the only way to establish that warming has paused would be by using average temperatures.
Your posts sound condescending, until the context of the discussion becomes apparent.
If adding them and dividing by the count of them doesn’t give you a result that tells you what the combined temperature is, then from a scientific standpoint, the result is meaningless.
Can you do an average on any set of numbers mathematically. Sure, but is the result meaningful, no. Discussing a “global average” as if it is a scientific determination of the temperature of the globe is portraying a meaningless number as a real, physical measurement. It is not.
“Discussing a “global average” as if it is a scientific determination of the temperature of the globe…”
And that in a nutshell is the problem. You want the average surface temperature to be the temperature of the globe, and I’m saying it isn’t and doesn’t have to be to be meaningful. For a start nobody is promoting the average surface temperature, it’s always the anomaly. The anomaly doesn’t tell you what the temperature is, but it does illustrate the main use – to tell you how temperature is changing.
Even if you can find the average temperature it isn’t the temperature of the globe, it’s the temperature of a specific part of the globe, e.g. the air 2m above the surface, or in this case the lower troposphere.
The usefulness of these figures is not in saying what the specific global temperature is (whatever that would mean), it’s in detecting changes in part of the globe over time or different regions. I keep telling you the mean is not normally a physical property of the whole. It’s a statistically useful value that allows you to test hypothesis about the physical world. This is the case regardless of whether you are measuring an extensive or intensive property. In my view that makes the average meaningful. It is telling you something about the physical world.
You need to pick one and stick with it. You just contradicted yourself!
If the sum is not physically meaningful then the average won’t be either.
If ANY step in an equation is non-physical, then the result of the equation will be non-physical as well. For Pete’s Sake, did you actually read this before posting it?
Your claim seems to be: The number of angels multiplied by the area occupied by an angel tells you the size of the head of a pin!
Good luck with that!
“Really, this idea that every step of an equation has to be physically meaningful for the end result to be meaningful is just nonsense.”
/BLINK/ — did he really write this?!
Well folks, you heard it here first — pressure and volume do not need to be physically meaningful quantities in PV = nRT!
Only a “numbers is numbers” person would make such a claim.
It ranks up there with his claim that division with the Magic Number N magically reduces measurement uncertainty.
!!
“If the sum is not physically meaningful then the average won’t be either.
/BLINK/ — did he really write this?!”
No that’s Tim’s quote. I’m saying the opposite. ot every syep of an equation has to be “physically meaningful” for the end result to be meaningful.
“pressure and volume do not need to be physically meaningful quantities in PV = nRT!”
Learn some logic.
“It ranks up there with his claim that division with the Magic Number N magically reduces measurement uncertainty.”
To s primative person I’m guessing basic maths must seem like magic.
Oops, try again:
https://wattsupwiththat.com/2026/05/07/uah-v6-1-global-temperature-update-for-april-2026-0-39-deg-c/#comment-4193564
As always, you fly in circles.
Sarcasm fail.
You quoted Tim.
I quoted Tim, who quoted YOU. Play again?
Much easier to read Tim’s replies than your unreadable tomes.
Nope, it’s all Tim’s
https://wattsupwiththat.com/2026/05/07/uah-v6-1-global-temperature-update-for-april-2026-0-39-deg-c/#comment-4193821
What I said was
It’s Tim who said
Which I agree, is a dumb think to say.
Do I have to use the sarcasm tag again?
It is not dumb. If the sum of intrinsic values does not give a physically meaningful value, dividing that meaningless value by a counting number will not result in a magical change to a number that is physically meaningful.
Intensive properties don’t add — so their averages are usually meaningless. Temperature, pressure, density, concentration, pH, etc. are not additive.
If you take:
– Object A at 300 K
– Object B at 400 K
The arithmetic mean (350 K) is not the temperature of any physical system unless very special constraints hold.
This is the same reason you can’t average:
– Pressures of two sealed tanks
– Humidities of two air parcels
– pH values of two solutions
– Wind speeds from two stations
The arithmetic mean has no physical interpretation.
An average can be correct only if the measurand is a ratio of extensive quantities.
An intensive property can only be averaged when the average corresponds to a ratio of extensive quantities that can be meaningfully summed.
Examples are:
In these cases, the “average” is not really an average of intensives — it’s a ratio of sums of extensives.
When the two objects are distinct, averaging is almost always invalid. If objects differ in:
then the arithmetic mean of their intensive properties is not a property of any real or hypothetical combined system.
This is the situation in most real-world measurement problems.
“If the sum of intrinsic values does not give a physically meaningful value, dividing that meaningless value by a counting number will not result in a magical change to a number that is physically meaningful.”
Just keep repeating that, maybe it will become true. And could you please define “physically meaningful”.
If you don’t want to add up all values to get an average, will you accept that a median value could be meaningful?
And what about if you calculate the mean without first adding up all intrinsic values?
Do you allow that the difference between two intensive values can be meaningful? What about the sum of two differences?
“The arithmetic mean (350 K) is not the temperature of any physical system”
And I keep telling you the average does not have to be the property of a physical system to be useful. Averages are generally used for statistical analysis. This is true whether the property averaged is extensive or intensive.
Try your argument with an extensive property – two objects, one with a mass of 300g the other with a mass of 400g. The average of 350g is not the mass of any physical system.
Let’s look at what the GUM says about an average.
Regardless of averaging intensive properties, do temperature measurements meet all of these requirements for finding a best estimate using the mean of observations?
If any of the answers are “no”, the mean of a Gaussian distribution is not meaningful.
Observations that come from a distribution with occasional large deviations — e.g., Laplace, Cauchy, mixture distributions, or any contamination model — the median minimizes the expected absolute error.
If the underlying distribution is skewed (e.g., lognormal, gamma, exponential), the median is often closer to the “typical” value than the mean.
“Let’s look at what the GUM says about an average.”
Why? You keep insisting an average is not a measurement. It’s only describing averaging measurements of the same thing.
“varies randomly – Gaussian (normal)”
No requirement for the distribution to be Gaussian. Your quote doesn’t even mention Gaussian.
“independent observations – no correlation”
Yes, it’s an assumption of the CLT that the random variables are iid. But you never get that this doesn’t mean you cannot work with imperfect data. Virtually no real world data is entirely iid.
“same conditions of measurement – same thing, same device, short time”
You are describing the usage in the GUM. But in real world sampling you are not going to be measuring the same thing each time. That would be pointless.
“If any of the answers are “no”, the mean of a Gaussian distribution is not meaningful.”
Completely wrong.
“If the underlying distribution is skewed (e.g., lognormal, gamma, exponential), the median is often closer to the “typical” value than the mean.”
You haven’t said if you regard a median of temperatures are acceptable? It’s still an average but doesn’t use the verboten sum. Yet it will give you a similar result to the supposedly meaningless mean.
Regardless, if you want a typical value you want the mode. But the purpose the mean is not necessarily to find a typical value
You don’t really have a clue do you?
How many times have you seen my statement that Tavg is a bogus average?
Why do you think I no longer post or deal with “global” averages of Tavg, only local stations and only Tmax or Tmin separately. Averaging temperatures of different stations is a joke. I know you’ve seen Tim’s and my statement that if temperature was a good proxy for climate then the climates in Las Vegas and Miami would be the same. Their climates are vastly different. Why? Enthalpy, an extensive property.
An average of sample measurements used to determine a property of a substance or phenomenon under reproducible conditions is a valid process.
If one defines the measurand as a monthly average of Tmax, then the process is sampling the phenomenon. This covered in GUM F.1.1.2. You won’t like what it requires. The combined uncertainty is calculated with the uncertainty of each input quantity and the standard deviation is added to that.
Since there are no “repeated” observations, one must use a Type B uncertainty for the single sample. Then add the standard deviation among the samples.
See H.6 for an example of determining the uncertainty when reproducible conditions are used.
That the result that you get from your math is something that can exist physically. How would *YOU* define it? In the equation 2units + 3units = 6units, each value *can* exist on a standalone basis, but the equation gives a non-physical result. In the equation 10C + 12C = 22C, each value *can* exist on a standalone basis but the equation gives a non-physical result. The average of the two items is 11C but that doesn’t make the average into a physical reality because the sum is non-physical.
(non-physical sum) / (physical count number) ≠ physical result.
(non-physical sum) / (physical count number) = non-physical result
The operative words here are “could be”. How do you judge if it is meaningful as a standalone statistical descriptor? What is the median of 0C,50C,50C,100C,100C,100C? What would the median tell you? What would the mode tell you? If those temperatures are from six different water baths sitting on a table in front of you, what would the mean, the median, and the mode tell you about what you have? Would the mean = 67C, the median = 75C, or the mode = 100C tell you anything that is physically meaningful about the environment consisting of the table top?
If someone told you five minutes later that now the mean is 60C, the median is 70C, and the mode is 50C would that tell you something physical about what changed in the table top environment? Would you be able to say “we need to change the value of x water bath to yC in order to bring the table top environment back to optimum”?
How do you propose doing so? You can’t create a larger system if each component is physically separate and each value is an intensive property.
I am assuming that you mean the values of the intensive properties of two different things. Since an intensive property is not dependent on the size of the system, what could a difference tell you? That one has a different temperature than the other? You still wouldn’t be able to find an average that is physically meaningful.
If you are talking about the value of the same intensive property of the same physical object then the difference would imply a physical change in the object – but it would be one that is not based on the size of the system. In order to get a physically meaningful value for the object it would require using the intensive property difference to create an extensive property – e.g. the number of joules added/subtracted which also requires knowing the mass of the object and the specific heat capacity of the object.
That is *NOT* the “global average temperature” which is a collection of the intensive property values for multiple objects of unknown extensive properties.
It doesn’t really matter if it is a sum of differences for multiple objects or a single object. sum = positive + positive or sum = positive + negative are both sums. If a single object changes from 70F to 75F, i.e. 70F + 5F, exactly what do you think that tells you physically? If the change is from 75F to 70F, i.e. 75F – 5F, exactly what do you think that tells you physically? If you have two different objects that each exhibit the exact same changes as above, what does the change in temperature for each object tell you physically? What would the sum of the differences for each tell you physically?
So your definition of “physically meaningful” is
I’m not sure how that precludes average temperature. The value of an average temperature can exist physically.
“In the equation 2units + 3units = 6units”
You might want to check your maths here.
“each value *can* exist on a standalone basis, but the equation gives a non-physical result.”
Why? What are the units here?
“The average of the two items is 11C but that doesn’t make the average into a physical reality because the sum is non-physical. ”
You are now making the point I said you were making some time ago, and you disagreed. You are saying 11°C does not have a physical reality, not that it isn’t physically meaningful.
As a statistic, I think an average can have meaning about the physical world. That doesn’t mean it has to be a physical reality. And again this applies as much to extensive and intensive properties. By meaning, I mean for instance you can use it to test hypothesis, and detect changes in the physical world.
If you are sampling items coming of a production line and you detect a significant change in an average property, that might indicate there is a fault with the process. This can be true if you are monitoring mass or temperature.
If you think longer days might increase temperature, you can test this by taking the average temperature in July and compare it with the average temperature in January. If the average temperature was meaningless that won’t work.
“The operative words here are “could be”.”
Is that a yes or a no? You claim a mean if an intensive property isn’t meaningful simply because the sum isn’t. The median isn’t based on a sum so is it physically meaningful?
” What is the median of 0C,50C,50C,100C,100C,100C? What would the median tell you?”
Not much given it’s a very small sample.
“How do you propose doing so?”
I suggested a couple of ways somewhere. One is to use the approximation technique. Guess a reasonable average, then average the differences between each value and your guess and add it to the guess. Of course, you might say that’s still adding up temperatures, except you are not actually adding temperatures, but temperature differences. (In fact you are already doing that when you add Celsius, but you never seem to notice). The point is there’s no pretense that the sum is a physical property, it’s just the net difference.
The other way is to note that what you are really looking for is a mid point. With just two values you can get that just by halving the difference between them and adding it to the smaller value. But you could extend that to any number of values using a recursive algorithm.
“Since an intensive property is not dependent on the size of the system, what could a difference tell you?”
How much hotter one thing is from another.
“You still wouldn’t be able to find an average that is physically meaningful.”
Just keep stating this is not an argument. If the mid point temperature between two rocks is not physically meaningful, why would the mid point mass between two rocks be physically meaningful?
“If a single object changes from 70F to 75F, i.e. 70F + 5F, exactly what do you think that tells you physically?”
That it got hotter.
“If the change is from 75F to 70F, i.e. 75F – 5F, exactly what do you think that tells you physically?”
I’m sure you could work that out from my previous answer.
“If you have two different objects that each exhibit the exact same changes as above, what does the change in temperature for each object tell you physically?”
That they both got hotter.
“What would the sum of the differences for each tell you physically?”
The sum on it’s own? Not much. The average would tell me what the average change was.
“I’m not sure how that precludes average temperature. The value of an average temperature can exist physically.”
This is the statistician’s world view in statistics world – averages are always physically meaningful. The average of intrinsic property values doesn’t exist physically. You said this yourself when you admitted that statisticians shouldn’t always calculate averages.
“Why? What are the units here?”
Good lord. You just said I needed to check my math! *YOU* need to be able to read and understand basic English!
Try writing in a way that makes sense if you don’t want to be misunderstood. You gave an equation of
In the equation 2units + 3units = 6units
And said “each value *can* exist on a standalone basis, but the equation gives a non-physical result.”.
Obviously the result is not physical as your maths is wrong. But correcting for that, it’s impossible to to say if it is a non-physical unless you state what the units are.
“ You are saying 11°C does not have a physical reality, not that it isn’t physically meaningful.”
I said: “ that doesn’t make the average into a physical reality”
As usual, you can’t read and understand basic English.
“As a statistic, I think an average can have meaning about the physical world. That doesn’t mean it has to be a physical reality.”
You still refuse to understand that an average is a STATISTICAL DESCRIPTOR. It is not a measurement of a physical reality. But it has to be physically meaningful or it is just a statistical descriptor of an arbitrary data set. You even admitted that when I gave you the values of several different things and asked you to average them.
Again, the operative word in your sentence is “can”. The average of intensive property values is *NOT* physically meaningful. There is no way it *can* tell you anything about the physical world. It tells you *nothing* about the physical world. If it did then the sum of 10C and 12C rocks in your hand would be 22C in your hand.
You won’t understand this until you admit TO YOURSELF that you wouldn’t have a total of 22C in your hand.
“You still refuse to understand that an average is a STATISTICAL DESCRIPTOR.”
What do you think I meant when I said “As a statistic.”?
“It is not a measurement of a physical reality”
That’s what I’ve been telling you.
“But it has to be physically meaningful…”
And we are back to what you mean by “physically meaningful”. Refusing to explain what you mean demonstrates you just want to keep equivocating about it.
“You even admitted that when I gave you the values of several different things and asked you to average them. ”
Pathetic. What I said was you couldn’t have an average of things with different dimensions. That should be obvious even to you, but instead you treat it as some “admission” on my part, and read all sorts of things into it hat aren’t there.
“There is no way it *can* tell you anything about the physical world. ”
And that’s where I disagree. You just endlessly repeating the same mantra doesn’t make it true.
“You won’t understand this until you admit TO YOURSELF that you wouldn’t have a total of 22C in your hand. ”
And I think we’ve reached the stage we’re it’s impossible tell if you really are this dense, or are just rolling. There’s only so many ways I can explain that an average is not a sum.
No, it isn’t physically meaningful. Tell us all EXACTLY what you think it tells you that is physically meaningful! It is *NOT* a mid-point temperature because those temperatures could be taken from objects on different planets with different compositions. On the other hand, if those values were in kilograms then you *would* have something physically meaningful.
And you think the net difference in temperature makes the difference into a physically meaningful value? You *STILL* don’t understand physical science. Tell us EXACTLY what you think it tells you that is physically meaningful. Is it whether or not you will get burned if you pick one of the objects up in your hand? What does that have to do with an AVERAGE temperature of different objects?
The *average* temperature will tell you that? How does it do that?
“Tell us all EXACTLY what you think it tells you that is physically meaningful!”
In global terms it’s the temperature where half the world is warmer and half is colder. Practically it has similar uses as the mean. It can tell if temperatures are changing and if, on the whole, they are increasing or decreasing. Comparing it with the mean. This has meaning to most people. Whether you consider this as “physically” meaningful would depend on your definition, which you refuse to spell out. As far as I’m concerned, it tells you something about the physical world, and so is physically meaningful.
“On the other hand, if those values were in kilograms then you *would* have something physically meaningful. ”
Why? Your only argument is about summing intensive properties. The median doesn’t require summing. Why do you think it matters if it’s of an extensive or intensive property?
“The *average* temperature will tell you that? How does it do that?”
Check your own reading comprehension. You asked what the difference between two objects could tell you, not the average temperature.
So what does that tell you? It tells you ZERO about climate since for surface observations it’s based on mid-range temperatures which cannot distinguish between climates. In the case of UAH it’s not even consistent on time of observation around the globe so you can’t tell what is going on at any specific time with the globe. Nor does it tell you *anything* about the thermodynamic state of the globe since you simply don’t know the rest of the parameters needed to judge the thermodynamic state.
It’s like measuring the temperature inside your oven at 8AM and in the master bedroom at 6PM and saying that the midpoint or average of the two temperatures tells you what is going on in your house as far as “climate” of the house. You won’t know if there is a large pan of water in the oven at the time of measurement or if the furnace or air conditioner is running in the evening (think humidity).
And none of it includes the measurement uncertainty associated with the measuring devices or measurement processes (was the oven door opened to insert a thermometer or was one inside the oven read through a piece of glass in the door). Contrary to your adamant insistence that anomalies are highly accurate the truth is that they are not. They are just as inaccurate as the absolute data values used to determine the anomalies.
You can’t tell unless the change is greater than the measurement uncertainty – and assuming that all measurement uncertainty is random, Gaussian, and cancels so you can “calculate” a change clear down to the hundredths of a degree is just plain non-physical. And if you are using a mid-point temperature to track changes you are only fooling yourself since the max temp can go up while the min temp goes down so the mid-point stays the same! You won’t know if anything changed at all!
Not to a physical scientist that actually understands what temperature is and what climate is. It’s like saying that most people understand that the rich don’t pay their fair share of taxes when in actuality they pay much more than their fair share!
It’s been spelled out to you in excruciating detail multiple times. The fact that you simply cannot read and comprehend simple English doesn’t invalidate that fact. It’s explained easily by the fact that you think measuring the temperature of six solution baths on a table top can give you a meaningful average that explains the “climate” existing on the table top. That average may or may not tell you whether any change over time actually occurred or not. And the average tells you exactly nothing that is physically meaningful about the environment (read climate) on the table top, it may have a difference from one end of the table to the other. The difference may be large and getting larger or it may be small and getting smaller or it may even be stagnant – and you won’t be able to tell!
You simply cannot combine intensive property values like temperature into a larger system and know anything about that “larger system” that is physically meaningful. The fact that you can perform a mathematical operation of the intensive property values doesn’t mean that it has any physically meaningful value, all you learn is a statistical descriptor of the collection of data.
A statistical descriptor of intensive property values can *not* be mapped to a function output value that provides any casual relationship. The statistical descriptor calculation known as the “mean” or “average” is *NOT* a function no matter how much you and bdgwx want to insist that it is. Your claim that the average calculation is a function is just unassailable proof of your lack of knowledge of physical science. A function describes a casual relationship between the input values (the independent variables) and the output value (the dependent variable). A statistical descriptor does not provide that casual relationship. An average summarizes a set of data, it does not map input values to output values based on casual relationships.
Statistical descriptors are *NOT* model mechanisms and are, therefore, not functions.
Not indulging you cry for attention at this stage.
But one point, you are still confused about what a function means. Understandable, as it’s one of those words that mean different things in different contexts. But mathematically a function is just a mapping from one set of values to another. It does not imply causality.
It wouldn’t even make sense to talk about causality in most cases. Consider your favourite function, V = HπR². Does the height of a cylinder cause the volume?
I asked copilot to describe a function vs a statistical descriptor. Its answer is almost a carbon copy of what I’ve tried to tell youL
————————————-
me: “define a mathematical function verses a mathematical statistical descriptor”
A mathematical function and a statistical descriptor look similar on the surface—both take “inputs” and produce an “output”—but they are fundamentally different kinds of mathematical objects. Here is the clean, engineering‑precise distinction you’ve been circling.
Concise takeawayA mathematical function maps independent variables to dependent outputs.
A statistical descriptor maps an entire dataset to a summary value.
So a function describes a relationship, while a descriptor summarizes a collection.
———————————–
A function MAPS related input values to an output.
A statistical descriptor describes a collection of values that may or may not be related.
Both involve calculations but the difference between the two is what you get as a result of the calculations.
You can argue till you are blue in the face that a statistical descriptor is a “function” but that is just not the definition.
You can do this with a function: ∂(x_1*x_2)/∂(x_1)
You can *NOT* do this with a statistical descriptor: ∂(median)/∂x
The falicy of argument by AI. You deny any use of statistics then try to get a statistical program to say what you want it to say.
“you’ve been circling.”
A telling comment. How long have you been torturing your imaginary friend before it gives you the answer you want.
Here for what is’s is what the Google random word generator tells me.
Me: “Is the mean a function”
Google ai:
But, regardless, you keep dodging the central contradiction. You are the one who insist on propagating the measurement uncertainty of individual readings onto the global average. You kept insisting this has to be done using the general equation. But now you say you can’t use equation 10 because averaging isn’t really a function. So what do you use? How can you propagate measurement uncertainty onto something that in your view isn’t a measurand?
“You can *NOT* do this with a statistical descriptor: ∂(median)/∂x”
Because the median function is not analytic. That’s why you would want to use an MC method. The mean is analytic, you can estimate it’s uncertainty using the partial derivatives or by MC.
“Because the median function is not analytic.”
Which means that a statistical descriptor is *NOT* a function.
“That’s why you would want to use an MC method. “
Oh, malarky!
The ∂(x_1*x_2)/∂(x_1) doesn’t represent uncertainty. There’s no u(x_i) in that expression!
Monte Carlo merely varies the extensive input variables to get different intensive output values. If the output values represent an intensive property you *still* can’t average them. E.g. density is a function of mass/volume, the ratio of two extensive properties. You *still* can’t average densities because it is an intensive property.
The issue here isn’t the uncertainty of the output values. It’s what the output values represent.
“The falicy of argument by AI. You deny any use of statistics then try to get a statistical program to say what you want it to say.”
I’ve given you a reference from an AI. YOU HAVE YET TO PROVIDE *ANY* REFERENCE AT ALL TO BACK UP YOUR ASSERTION THAT INTENSIVE PROPERTIES CAN BE AVERAGED!!!!
How about this from wikipedia:
—————————————
According to International Union of Pure and Applied Chemistry (IUPAC), an intensive property or intensive quantity is one whose magnitude (extent) is independent of the size of the system.[3] An intensive property is not necessarily homogeneously distributed in space; it can vary from place to place in a body of matter and radiation. Examples of intensive properties include temperature, T; refractive index, n; density, ρ; and hardness, η.
By contrast, an extensive property or extensive quantity is one whose magnitude is additive for subsystems.[4] Examples include mass, volume and Gibbs energy.[5](bolding mine, tpg)
——————————————-
“Intensive and Extensive Properties: Thermodynamic balance”, Pierre-Marie Robitaille and Stephen J. Crothers, 3/2019
————————————————————–
Tolman writes: “In order to decide whether a given quantity has extensive or intensive magnitude it is sometimes helpful to see if the simultaneous presence of two systems, each having a definite quantity of the kind in question, can be regarded as giving a larger system with twice the quantity; if so that quantity has extensive magnitude.” A similar scenario is advanced by Redlich.13 Importantly, this scenario establishes that extensive properties scale linearly with system size and therefore, must be additive. Alternatively, for simplicity, one could consider two identical homogenous spheres, B1 and B2, in thermal equilibrium. The two spheres could be combined to create a larger sphere, BT. Those properties like mass, M, volume, V, internal energy, U, and entropy, S, which are doubled in forming BT are extensive, as they are additive for subsystems (e.g., MT ¼M1þM2). Mathematically, they can be viewed as homogeneous functions of degree 1. Thus, if mass M can be viewed as some function of x, y, and z, this function is considered homogenous in the first degree since when multiplying the arguments of the function by a quantity k, the following holds: Mðkx;ky;kzÞ¼kkMðx;y;zÞ, where k¼1.9 All extensive properties can be viewed in the same manner. They are homogenous functions of degree 1 and are therefore additive across subsystems.15 This is a central feature in macroscopic thermodynamics. Conversely, those properties like temperature and pressure which remain unaltered when combining the two smaller spheres are considered intensive. Intensive properties are independent of system size. Mathematically, they are Physics Essays 32, 2 (2019) 161 homogeneous functions of degree 0, as they are usually partial derivatives of homogenous functions of order 1 in the extensive coordinates. Thus, if temperature T can be viewed as some function of x, y, and z, when multiplying the arguments of the function by a value k, then the following holds: Tðkx;ky;kzÞ¼kkTðx;y;zÞ¼Tðx;y;zÞ, as k¼0.9 Intensive properties are clearly not additive and all such properties can be viewed in the same manner.
————————(bolding mine, tpg)
I can give you more non-AI references if you need them. But they are all going to say the same thing. Intensive properties cannot be combined into creating a larger system where the properties add.
Temperature is no different than color. A ten inch piece of copper tubing is the exact same color as a 20 inch piece of copper tubing. You can’t average the two and get a different color that is 15 inches of copper.
You can continue to disbelieve this if you want, it will just confirm that you remain ignorant of physical science. Same with climate science.
“The sum on it’s own? Not much. The average would tell me what the average change was.”
What would the average change in the temperature of two different objects tell you that is physically meaningful?
bellman will never admit this. For him *all* averages are physically meaningful and even if they aren’t they are still meaningful. The concept that tracking the average of non-physical data is nothing more than tracking the value of the non-physical average just flys over their head. If the values of the average are non-physical it doesn’t matter, knowing the trend of the average of non-physical data is important.
“then the arithmetic mean of their intensive properties is not a property of any real or hypothetical combined system.
This is the situation in most real-world measurement problems.”
bellman, AND CLIMATE SCIENCE, believes that if you are holding two rocks in your hand, each at 50C, you are holding a total of 100C in your hand – the boiling point of water. They believe that if you are holding two rocks in your hand, one at 100C and one at 0C, that you are holding an average of 50C in your hand and you won’t get burned.
The fact that climate science didn’t discover the change in growing season length, in fact they *can’t* even show it now with their average of averages of mid-point temperatures, but ag science *did* should be a clue to climate science (and bellman) that their trendology is non-physical. But climate science (and bellman) remain trapped in their “Tradition!” meme and in the meme that averages of intensive properties *is* physically meaningful.
“For him *all* averages are physically meaningful and even if they aren’t they are still meaningful.”
Lie.
“The concept that tracking the average of non-physical data is nothing more than tracking the value of the non-physical average just flys over their head.”
Temperatures are physical data. Averages are not physical properties, but tracking them can be useful.
“knowing the trend of the average of non-physical data is important.”
What non-physical data are you talking about. An average temperature is an average of physical data. Are you claiming temperature isn’t physical?
“bellman, AND CLIMATE SCIENCE, believes that if you are holding two rocks in your hand, each at 50C, you are holding a total of 100C in your hand”
Why do you have to make up so many lies about me? Is it because it’s the only way you think you can win the argument, or is it because you genuinely don’t understand what I’m saying?
Also, why do you never understand that Celsius is not an absolute temperature scale. Adding 50°C to 50°C to get a total of 100°C is not adding two temperatures. It’s adding two offsets.
Useful enough that climate scientists and warmists using linear regressions never realized that Tmin was what was raising the “average” temperature. Left it up to the farmers to find out!
Just more deflections. What does averaging Celsius vs Kelvin have to do with the subject? They are both temperatures. Why don’t you include Fahrenheit and Rankin while you are at it.
The crux of the problem is that temperature is not an extensive property that can be used to discern the total energy contained in a given volume. An average of air temperature ignores any latent heat energy difference.
You never answer the question. Why do Las Vegas and Miami have different climates when their average temperatures are similar. Do you think an average of enthalpy (an extensive property) would be a better indicator of the climates at the two locations?
You have yet to show ANY REFERENCE that supports your assertion that an average of temperatures is a meaningful value. With all the criticism from you about temperatures being intensive and not averageable and assertions you have made about the averages being useful, there should be reams of pertinent scientific information on the web that supports you.
“Useful enough that climate scientists and warmists using linear regressions never realized that Tmin was what was raising the “average” temperature”
What on earth are you about now? This is from the IPCC AR2:
That was over 30 years ago.
“Left it up to the farmers to find out!”
Could you provide a reference to the farmers reporting an increase in average global minimum temperatures? And how do you know it’s correct if you think average temperatures have no physical meaning?
“The crux of the problem is that temperature is not an extensive property that can be used to discern the total energy contained in a given volume.”
So? That’s not the point of an average temperature.
“An average of air temperature ignores any latent heat energy difference.”
Which has nothing to do with it being intensive. If you want to know the total energy change then you will have to calculate that. Average temperature is just telling you how much hotter it’s getting. When you are told that the forecast for tomorrow will be 20°C do you say that’s a useless figure because it doesn’t tell you how much energy is in the air?
“Why do Las Vegas and Miami have different climates when their average temperatures are similar.”
Because they are in different geological regions, and have different levels of humidity – I assume. Here’s what Wikipedia says:
According to those entries the annual mean temperature for Miami is 24.5°C and for Las Vegas 21.2°C, so I’m not sure why you say they have similar average temperatures. And that’s for the annual means, the monthly averages are quite different.
Talking about averages, do you find no meaning in the fact that their monthly average temperatures are so different, along with the average min and max temperatures. Do you not think it’s possible to look at those averages and figure out something about their climates? Or are they all meaningless random numbers to you?
“You have yet to show ANY REFERENCE that supports your assertion that an average of temperatures is a meaningful value.”
I love the way you personalize this. It’s hardly my assertion. It’s the assertion of everyone who’s ever used average temperature in a meaningful way. That includes the authors of this page, and every other global or local climate report. I doubt there are many papers written explain why these averages are not meaningless. But many of the sources you site, the GUM, NIST for example happily show you who to average temperatures, and never once warn you that any such average will be meaningless.
Perhaps you should ask, checks notes, a Jim Gorman. He was insisting a few days ago that he could use average global temperatures to prove that the world was warming in step changes caused by El Niños. I doubt that result, but he obviously though there was some meaning in average temperatures.
U.S. Agro-Climate in 20th Century: Growing Degree Days, First and Last Frost, Growing Season Length, and Impacts on Crop Yields
Why do you the AG science folks began examining this? Farmers! Planting dates being earlier. Longer maturing crops.
Close enough for back of the envelope.
Here is what you missed in terms of HEAT in the two locations.
What do you think the different enthalpy’s, a measure of heat, will be at the two locations? Remember, enthalpy is an extensive property and can be averaged.
You mean this?
https://www.nature.com/articles/s41598-018-25212-2
Publication 2018. Do you really think this was the first time anyone looked at minimum temperature changes?
As I said, the 1995 IPCC report mentions that minimum temperatures were warming faster than maximums globally, not just in the USA.
You do realize that your report is using average temperatures? Specifically the average between Tmax and Tmin.
“Close enough for back of the envelope.”
A difference in annual mean temperatures of over 3°C is close enough to be the same? By that logic England has a similar temperature to Spain.
“Here is what you missed in terms of HEAT in the two locations.”
I specifically said humidity was a reason for the different climates. Aside from anything else this means the “feels like” temperature will be different, becasue of your ability to sweat. But it also has a big impact on the range of temperatures. Las Vegas has hotter day time and summer temperatures, Miami much less variation. All of this is can be seen by looking at average temperatures. I don’t know how you could tell if you ignore average temperatures.
“What do you think the different enthalpy’s, a measure of heat, will be at the two locations?”
Enthalpy is not a measure of heat. Surely you’ve learnt that by now. Heat is the transfer of energy from one body to another. Enthlapys a measure of internal energy. Ans as enthalpy is an extensive property you can’t really say how much there is at a location without knowing what mass of air you are talking about. Specific Enthalphy is going to be more relevant.
Using an online calculator I would estimate that an average summer day in Miami has a specific enthalpy of ~ 60kJ/kg, and for Las Vegas of 40kJ/kg.
“Remember, enthalpy is an extensive property and can be averaged.”
How, and what, are you averaging. If you are talking about specific enthalpy, that’s an intensive property, and if you are averaging by time, you need to sum enthalpy for each unit of time, which seems like a physically meaningless property. What you need to do is have units of enthalpy ✕ time, and integrate that over a period of time and divide by total time. Which is exactly what you can do with temperature. temperature ✕ time is extensive.
As to the IPCC, they continued to proselytize that the earth was burning up and crop failures would occur even after knowing:
The IPCC also continued the tradition of using (Tmax + Tmin)/2 knowing full well that it hides what is actually happening. Gotta keep that CAGW meme going don’t you know.
And, you have just provided, whether you know it or not, a reason for variance change affecting the time series trend of Tavg.
Don’t put words in my mouth. Quote what I said in context. My comment was about using trends with no attention paid to what shocks can do and that there are steps that can be taken to insure they are not permanently propagated. The trend could have just as easily been babies born over time with shocks introduced due to power outages.
“As to the IPCC, they continued to proselytize that the earth was burning up and crop failures would occur even after knowing”
An exact quote please. Especially about the earth “burning up”. That seems very unlikely.
“The IPCC also continued the tradition of using (Tmax + Tmin)/2”
What non-traditional data would you use instead? Limiting yourself to just max or min temperatures means you have to ignore ocean temperatures and satellite data, and reduces the data for land.
“And, you have just provided, whether you know it or not, a reason for variance change affecting the time series trend of Tavg.”
You really have no idea what you are talking about, do you?
“Don’t put words in my mouth. Quote what I said in context.”
https://wattsupwiththat.com/2026/05/07/uah-v6-1-global-temperature-update-for-april-2026-0-39-deg-c/#comment-4193530
How do you determine that without resorting to meaningless monthly average temperatures.
https://wattsupwiththat.com/2026/05/07/uah-v6-1-global-temperature-update-for-april-2026-0-39-deg-c/#comment-4193456
As the “pauses” are based on monthly UAH global average data, how is your claim meaningful if the monthly averages are meaningless?
“The trend could have just as easily been babies born over time with shocks introduced due to power outages.”
But they were actually about “meaningless” average temperatures.
Can you cite the document and page number?
The IPCC use reanalysis products like ERA5 which use more robust daily averaging methods.
It’s worth mentioning that scientists are fully aware of and analyze enthalpy based metrics like equivalent potential temperature as well despite your claims otherwise. See [Song et al. 2022] as example.
Yes some do, but many, many do not, even including NOAA. As a reference I would ask you to review many of the temperature graphs and linear regressions posted here rather than graphs of enthalpy.
“Temperatures are physical data. Averages are not physical properties, but tracking them can be useful.”
You are *STILL* trying to say that you can average intensive properties of different things.
When are you going to provide a reference from anywhere that states you can do this and get a physically meaningful average?
“You are *STILL* trying to say that you can average intensive properties of different things.”
Yes, that’s what I’m trying to tell you. Not only can you do it, but scientists do it all the time.
“Yes, that’s what I’m trying to tell you.”
And the average is physically meaningless – and that’s why scientists don’t use such.
“Not only can you do it, but scientists do it all the time.”
Climate scientists do – not actual physical scientists. You can’t give me a reference, NOT ONE, that says you can get a physically meaningful average from the intensive properties of different things. It’s why climate science is such a joke to those who actually use metrology on a daily basis and NEED to use metrology properly to avoid civil and criminal liability for what they produce.
You told me that I couldn’t average things like velocity, mass, and area values in a single data set because they have different units. But the mathematical formula for the statistical descriptor known as the average or mean doesn’t require the data set components to even have units! Things such as test scores don’t have units but you can average them. A data set has no defined meaning other than it is a set of data. There doesn’t have to be a physical relationship among the numbers.
All I need to do is tell you the numbers 70, 50, and 6 and you can average them. You don’t need to know what they represent. That’s because the statistical descriptor only describes the data set, not the physical relationship between the numbers.
And the intensive property values of different things are exactly the same. The average only describes the data set, not the physical relationship between the numbers. The average of intensive properties of different things have no physical meaning because the numbers together don’t represent a larger system – they represent NOTHING and their average represents nothing but the shape of the data set. The average is *NOT* a function that maps inputs to outputs. Statistical descriptors are *NOT* physical models laying out casual relationships.
We’ve given you many references. I think what you really mean is that we haven’t given you “NOT ONE” that you would accept. And given the fact that you’ve been presented references from NIST and the GUM it’s hard to imagine that you would accept anything at this point. If you think that is an unfair on my part then perhaps you could provide us your criteria for acceptance?
“We’ve given you many references.”
You’ve given us ONE reference, Nist 1900, Ex2 – WHICH IS NOT APPLICABLE BECAUSE OF THE ASSUMPTIONS MADE IN THE TEACHING EXAMPLE.
All the other possible examples, be it measuring the temperature of a water bath or the hardness of an object, INVOLVES MULTIPLE MEASUREMENTS OF A SINGLE MEASURAND. The actual measurement process for making the multiple measurements may be different but they are all done to establish the best possible estimate of the value of a property associated with that single measurand.
You keep trying to conflate multiple measurements of a single object with single measurements of multiple objects and all it shows is that you have no understanding of the context and meanings in the GUM or in metrology in general.
You *can* measure the value of a property of a single object multiple times. You *can* measure the value of a property of multiple objects a single time each. Even INTENSIVE properties. You *can* average the multiple measurements of a SINGLE OBJECT to get a “best estimate” of the value of an intensive property for that single object. You can *NOT* average the single measurements of multiple objects to get a “best estimate” of a non-existent larger system created by combining those single measurements.
Until you can accept that, for intensive properties at least, multiple measurements of the same thing and multiple single measurements of different things HAVE TO BE TREATED DIFFERENTLY, you’ll never understand the basic physical truth that you cannot create a larger system from the intensive properties of different objects.
You *can* average the mass of a rock at 10g and one at 12g by combining them into a larger system. You can then put two rocks, each of 11g in your hand and experience the same physical system as with one at 10g and one at 12g.
You can *NOT* average the temperature of a rock at 20C and a second rock at 100C and experience the same physical system from two rocks at 60C. TEMPERATURES DO NOT ADD. If you can’t combine two temperatures into a larger system then averaging them makes exactly zero physical sense.
What is it that you simply do not understand about this simple and obvious fact?
“and that’s why scientists don’t use such. ”
Could you list all these scientists who don’t use hypothesis testing, and what century were they born in?
“You can’t give me a reference, NOT ONE, that says you can get a physically meaningful average from the intensive properties of different things.”
Nobody need to state it, because it’s bleeding obvious you can. Here’s an example you keep giving me
https://www.nature.com/articles/s41598-018-25212-2
They use average temperatures to estimate GDD, then average those across the country to determine where GDD is increasing. If mean temperature was physically meaningless you would have to say increased GDD was physically meaningless.
“You told me that I couldn’t average things like velocity, mass, and area values in a single data set because they have different units.”
I really don’t understand why you keep arguing that. Give me one reference that says it makes sense to average things of different units. What would the average of the height of the Eiffel Tower and the weight of Statue of Liberty be? I’m not asking what it would mean because it would obviously be meaningless, but what would you state it was when combining meters and kilograms?
“But the mathematical formula for the statistical descriptor known as the average or mean doesn’t require the data set components to even have units!”
But if they do have units you can’t combine different ones.
“A data set has no defined meaning other than it is a set of data.”
It does if it’s any use.
“All I need to do is tell you the numbers 70, 50, and 6 and you can average them. You don’t need to know what they represent.”
Yes, that’s how numbers work. They are abstractions of specific quantities.
“That’s because the statistical descriptor only describes the data set, not the physical relationship between the numbers. ”
Not if your statistical descriptor is for real world properties. You can apply your logic to any real world arithmetic. You can say that 2 + 2 = 4, it doesn’t mean you can add two apples to two sheep and get a meaningful result.
“The average only describes the data set, not the physical relationship between the numbers.”
Wrong. Please think harder.
“The average of intensive properties of different things have no physical meaning because the numbers together don’t represent a larger system”
What physical meaning does the average of extensive properties of different things have? How does it represent about the larger system?
If you want to test a hypothesis, what is the difference between a hypothesis about the average weight of a human, verses one about the average body temperature?
“The average is *NOT* a function that maps inputs to outputs.”
Your education is really lacking.
I said: “And the average is physically meaningless – and that’s why scientists don’t use such.”
Is this the best you got? Misquoting me?
You only give the link but NOT ONE SINGLE QUOTE FROM THE LINK. Here’s one that is applicable:
————————————
“Specifically, this study presents research efforts to investigate the changes in growing degree days throughout the year and how these changes vary spatially and temporally across the conterminous United States (CONUS) during this period,”
———————————–(bolding mine, tpg)
As usual, you are cherry picking without one iota of the context or meaning of what you are cherry picking from. You can’t even tell the difference between “degrees” and “degree-days”. (Hint: degree-days are *NOT* an intensive value).
They did *NOT* use the average daily temperature. They used the mid-point daily temperature and subtracted the crop type base temperature to determine if a day was a growing degree day. The daily mid-point temperature and the daily average temperature ARE NOT THE SAME – something you have continued to adamantly deny by claiming they *are* the same.
from the study:
——————————
Each cropping system (maize, soybean, sorghum, cotton, and spring and winter wheat) has different growing season. Thus, the values growing degree days (GDD), or thermal units, were computed for each crop separately as the accumulation of Tavg exceeding a base temperature for each crop as
GDD = [(Tmax+Tmin)/2] – Tbase”
——————————
You can’t even tell when an average is being used and when a mid-point value is being used. In addition, the mid-point temperature is *NOT* being used to create a larger system which can be averaged but only as a trigger to establish the existence of an EXTENSIVE value – a “growing degree day”. It is the GROWING DEGREE DAYS that get added, not the mid-point temperature!
Again, STOP CHERRY PICKING and try to understand the meaning and context of what you are cherry picking from. Do that and you won’t get caught making idiotic assertions!
I didn’t say it made sense. I said that you *CAN* do it. All the statistical descriptors do is describe the shape of the collection of values. They don’t map any dimensional attribute of an input value to an output value dimension as a function does.
All the average of intensive properties like temperature tells you is a description of the values you include in the data set. That average tells you NOTHING about a non-physical larger system because you can’t add temperatures in reality. It’s not one iota different than using values like 10 m/sec, 39kg, 18m^2, and 50C. You can’t combine those values into a larger physical system so you can’t find a physically meaningful average from them – you can find an average value from the item values but it is physically meaningless. It’s the same with combining the temperatures of different things. You can’t create a larger physical system so you can’t find a physically meaningful average from them.
Why not? If you can’t combine the values into a larger system then it doesn’t matter if they have units or not!
Why? If I give you a set of values with no units can you tell if they are related or not?
Ahhhh… The old “numbers is just numbers” meme.
But the sum of a set of temperatures is *NOT* a real world property!
You don’t realize it but you just admitted that you can’t calculate an average of temperatures from different things because the sum is not a real world property!
A statistical descriptor only describes the shape of the COLLECTION OF VALUES. The statistical descriptor does *NOT* depend in any way, shape, or form on a relationship between the values in the collection.
“You can say that 2 + 2 = 4, it doesn’t mean you can add two apples to two sheep and get a meaningful result.”
You can’t even understand that you are admitting to what I am trying to tell you! You can’t add two temperatures of different things and get a meaningful result!
You can’t add apple1 to apple2 and get one larger apple. You can’t add sheep1 to sheep2 and get a larger sheep! You can’t add apple1 to sheep1 and get a physically larger hybrid.
If you have two sheep what do you get for an average? 2sheep/2 = 1 sheep. What sense does that make, 2 = 1? You have an average of 1 sheep when there are two of them right in front of you? The unit of “sheep” is an intensive value just like “temperature”.
You can’t add temperature1 to temperature2 and get a larger temperature so you can’t get a meaningful result.
You are kidding, right? What do you think we’ve been trying to teach you for three years? If I am calculating the price of a load of 2″x4″x8′ boards to a contractor I can use the average length of 8′ to get the number of board-feet to be charged. If I know the average size of a set of fenceposts +/- the uncertainty then I can estimate the size of auger I need to use while minimizing post wastage versus concrete backfill. Length is an extensive property. Volume is an extensive property. Temperature is *not* an extensive property.
Ask the pilot of a large airliner sometime if they use the “average body weight” of humans to calculate how much fuel they will need to get from PointA to PointB. Have you ever seen a weight limit sign before entering a bridge? As usual, your total lack of experience in the real world most of us operate in just shines through.
The simple answer is that bellman can not find any resource that explicitly says you can average temperature without very specific assumptions.
Other than the situation where two exactly similar masses at the exact same temperature, you just can’t do an average. In fact, in this situation the “average” is still meaningless when you really think about it.
Take color. I have two magnets with the same shade of red. I let them come together. I have a larger system in that single shade of red. Now one can say the average color is also red, and try to make the point that you can average color. However, if one magnet is red and the other blue and you let them come together, do you have the average between blue and red? No.
Temperature is no different. Unless you can create experiments that show a general rule of being able to average temperatures, you can’t just say you can do so from one experiment.
And he still maintains that averaging different quantities reduces measurement uncertainty. He will be forever stuck in the mud until he lets loose of this one.
“And he still maintains that averaging different quantities reduces measurement uncertainty”
Finally. Glad someone’s been listening.
And it remain nonsense: uncertainty INCREASES (except in bellman-world apparently).
I have you the example you gave me about ggd. Uses the mean of max and min. Saying this is just the mid point is just agreeing with me. The average is just a mid point. It doesn’t require the sum of all temperatures, that’s just a means to an end.
Here’s an example of testing mean body temperature
https://online.stat.psu.edu/statprogram/reviews/statistical-concepts/hypothesis-testing
The fact is, if it was impossible to average temperature, then every source would be staying that you shouldn’t do it. Instead you can find many examples where average temperature is used and described, with zero warning. Including examples from the simple NIST document.
Outside of that stupid Essex paper can you find any paper explaining why an average if temperature is meaningless? The only reason for this claim is to discredit evidence for global warming.
“However, if one magnet is red and the other blue and you let them come together, do you have the average between blue and red? No.”
Depends on what question you are asking. If you are treating the colours as categorical the average is meaningless but the average frequency makes sense If you are treating them as continuous values then the average can make sense. Have you never heard of scientists trying to estimate the average colour if the universe? In graphic work, the average colour can be very useful. Say you have a high resolution picture, and need to lower the resolution.
But as always, you come up with meaningless examples, then imply that means all uses are meaningless.
“Unless you can create experiments that show a general rule of being able to average temperatures, you can’t just say you can do so from one experiment.”
What do you think UAH is doing? If you wasn’t to claim any of the averages are meaningless, you need to show that? A test would be to demonstrate contradictionary results. E.g. if you could show that half the times the absolut NH summer was colder than the winter.
“The simple answer is that bellman can not find any resource that explicitly says you can average temperature without very specific assumptions.”
Seeing as you put some meaning on generative ai, this is what Google told me when I asked can you use the average of temperature in a hypothesis test.
You betray your methanation/statistician background. The whole issue here is not hypothesis testing. It is determining a measurement.
I suspect Google showed you this as it did me. Why did you leave out the need for a “hypothesized value”?
I asked Google this:
What is the typical rule for averaging intensive properties like temperature?
Response:
Funny how AI’s work best when you ask the appropriate answer.
“It is determining a measurement.”
In order to test hypothesis. Nobody apart from you cares what the exact temperature of the Earth is. Nobody is even quoting an absolute temperature, in case you haven’t noticed they use anomalies.
“Funny how AI’s work best when you ask the appropriate answer.”
Funny how you get the answer you want.
And global average temperature is a weighted average. Weighted by surface area.
But you keep labouring under the impression that a global average anomaly is intended to tell you the thermodynamic temperature of the entire planet. I keep trying to tell you that isn’t the point.
That is NOT true. People are concerned about local and regional changes, IN ABSOLUTE TEMPERATURES.
When was the last weather report you have seen that shows anomalies? How about the Farmer’s Almanac. Farmers could care less, they want to plan for when to start planting, what seed variety to purchase, when they can expect to harvest, what delivery date to put on contracts for their crop. People need to plan cooling expenses, they could care less about anomalies, they want to know temperatures.
Climate science concentration has one focus – radiation and the temperatures that cause it. Forget variation. Forget thermodynamics. It is all SB radiation as determined from space. They don’t care if in one month England is hot while Poland is not, and then reverses the next month. Just as long as the anomalies balance out.
How many times have I told you anomalies are not temperatures? They still are not. They are a rate just like miles per hour, joules per second. They are degrees per time interval. A global average anomaly supplies no information about the heat distribution around the globe. It is worthless.
A global average anomaly is an average of an average of an average of an average. You keep laboring under the misconception that the global average anomaly has no uncertainty because of the number of stations used. I can assure you the uncertainty is so large that any anomaly value has no statistical significance.
This is the simple truth that you cannot combine intensive values into a larger system. If you can’t form a larger system you can’t have an average value.
What do you think UAH is doing?
I don’t really care what UAH or NOAA or anyone else is doing with their temperature measurements. Global Average Temperature is a joke because it doesn’t take enthalpy (an extensive property) into account when determining the energy at a given point.
If you can’t average temperatures, then the very start of using (Tmax + Tmin)/2, is incorrect and any additional calculations based upon it are also incorrect. That doesn’t even cover inadequate treatment of time series forecasts.
That is why I am doing my own research into Tmax and Tmin as separate functions whith different shapes. It has also led me to plotting insolation against soil temperature and soil temperature against atmospheric temperature.
Climate science being trapped as it is in the past, simply is unable to move forward to using 1 and 5 minute data on both Tmax, Tmin, insolation, and humidity. Climate science had a wedding somewhere to radiation/SB. They simply will not allow the idea that something else is going on.
I just read an article today about a new paper in biology that is recommending that biologists begin to move away from temperature and start analyzing microclimates and how creatures react. Some like the heat, some don’t. Some can mitigate stress by moving while other can’t. In essence it said we are past the time where the simple adage that CAGW is going to eradicate life on earth is incorrect.
“I don’t really care what UAH or NOAA or anyone else is doing with their temperature measurements.”
Which is why you spend souch time discussing it.
“it doesn’t take enthalpy (an extensive property) into account”
Why do you think the purpose of a temperature measurement is to tell you about enthalpy?
“If you can’t average temperatures, then the very start of using (Tmax + Tmin)/2”
A couple of posts ago you were defending using that to determine GGD.
“That is why I am doing my own research into Tmax and Tmin as separate functions whith different shapes. ”
Good for you. Are you ever going to say what the results are?
“Climate science being trapped as it is in the past,”
Once upon a time we were told that science was stuck in the past as it didn’t use the satellite data. Now satelites are considered old fashioned by you.
“A couple of posts ago you were defending using that to determine GGD.“
They are not using this as an average! Like I said you are cherry picking again without understanding the meaning and context. They are identifying a trigger point for classifying a day as a GDD. The mid-point temp is *not* used as a value in determining the quantity of GDD.
“The mid-point temp is *not* used as a value in determining the quantity of GDD.”
That’s exactly what they do.
but the average frequency makes sense
LOL. No it doesn’t. Neither magnet is broadcasting an average frequency. IT DOESN’T EXIST!
No it doesn’t. You can’t take an rgb value at the left side of a photo and an rgb value from the right side. Compute the average and say the middle of the photo must be this average.
I know because I had a small business working with tintype and Polaroid photos using the 1990 version of photoshop. Most pixels are not the result of averaging, they are shading differences. You had to beware changing certain things or you could inadvertently remove a glint from an eye or cufflink. Most often, the pixels are a continuous change in shading or in color, not an average at all.
“but the average frequency makes sense”
Sorry that should have been average colour.
“No it doesn’t.”
You are saying noone ever uses average colour in graphic processing?
“Compute the average and say the middle of the photo must be this average.”
That’s not what I said. You can take the average of several pixels to get the required colour of a larger pixel. You can either average all the pixels to get an average colour for the image.
From the website.
When was the last time you hypothesized that the temperature in Mexico City was, say 70F? How about Calgary having 32F?
Do you reckon that body temperature is pretty much a constant? Remember what I’ve already said?
Here it is.
Body temperature is controlled. It is not a variable controlled by various inputs.
I also notice that the example showed no measurement uncertainty whatsoever. Didn’t even mention it in an upper level university level course. A great example of why statisticians never consider that in their analysis.
Do you know what kind of grade I would have received for not analyzing measurement errors and uncertainty? You don’t forget that training.
First, it isn’t “ggd”, it is GDD. Secondly, the value is a trigger point, not a measurement. Do you understand the difference? In fact, for GDD, a mid-point might be a good value to discern what the growing potential is.
I’m definitely saying you can and that scientists do it frequently.
I’ve provided you a lot of references over the years. One in particular that we’ve been discussing the last couple of days is NIST TN 1900 E2.
You STILL don’t understand what was being done in NIST TN 1900 E2.
If you would EVER list out the assumptions made in that example it would become obvious that it was a teaching example and not a real world situation. But you can’t list out the assumptions, can you? Is that because you don’t know them or you don’t want to have to admit what they do?
PHYSICAL SCIENTISTS do *NOT* average intensive properties of DIFFERENT THINGS all the time. And you have yet to give a reference stating that such an average is physically meaningful and not just mathematical masturbation.
Do whatever floats your barge.
Of all the troll-persons inhabiting WUWT, you are the most disingenuous which is why I refuse to debate anything with you.
You talk in circles, your typing is atrocious, you backpedal like crazy when your bizarre ideas are exposed, and you have a bad case of LWS.
Bizarrea ideas such as averaging temperatures? The thing all these UAH updates are doing.
If everyone was jumping off a tall bridge, would you?
“As always, you fly in circles”
His reading comprehension skills are poor.
I have yet to see either of them post a real, detailed Type A analysis of temperature data, let alone all the diverse Type B factors.
Nothing but handwaving and division by the Magic Number.
You have seen me post a type A analysis of the UAH data. And I did so using a method you approved of and even attempted yourself. You just got the wrong answer because you made math mistakes.
And look, the “egregious math errors” database rears itself up out of the troll swamp.
Feel better now?
As I’ve told you before it’s not my database. It’s WUWT’s database. If you don’t like them archiving your posts then take it up with them.
As expected, you had no answer for Jim’s exposure of your profound ignorance and cherry picking about a real Type A analysis.
No, you did not use a proper method.
JCGM 100-2008
You haven’t shown that your distribution varies randomly, that is normal or Gaussian. You haven’t shown that the same conditions applied for all measurements. You did not evaluate the uncertainty in each input quantity. You assumed that each input quantity had zero uncertainty.
All you did was evaluate a statistic that describes how closely the estimated mean may be to the real population mean. To use that as measurement uncertainty requires measuring the same thing multiple times under repeatable conditions and only applies to that one, single thing.
Paragraph 4.2.3 has the following.
See that Xi,k. That means multiple observations for each input quantity. You do not have that. There are not multiple observations of any Tmax or Tmin. That means the use of experimental standard deviation as described below is the statistical parameter to use throughout.
In general, measurement uncertainty uses the traditional standard deviation.
Exactly right. Air temperature don’t qualify.
I was going point this out, but figured it wasn’t worth the time as all they care about are Holy Averages performed with the Magic Number, plus they would just ignore it like every other time it has been brought up.
You thought the method was valid before your math mistake was discovered.
Now that the math mistake has been identified and the correct answer given you now claim the method was invalid.
If your position really isn’t based on the result you wanted to see as opposed to what you actually saw then you are doing everything you can to convince me otherwise.
The only math error is yours in trying to equate sampling error with measurement uncertainty. Measurement uncertainty is SD, it is *NOT* SD/sqrt(n). Sampling error is an additive factor to measurement uncertainty.
YOU DIDN’T CALCULATE MEASUREMENT UNCERTAINTY.
You only calculated sampling error!
Your analysis wound up using SD/√n.
That is *NOT* measurement uncertainty. That is the standard deviation of the sample means – it is how precisely you have located the population mean from the data in the population.
That is *NOT* the accuracy of the mean, it is not a measurement uncertainty. You can have a vanishingly small SD/√n from HIGHLY INACCURATE DATA. If the data is highly inaccurate then your so-precisely located mean value will be HIGHLY INACCURATE.
Somehow you and climate science can NEVER get this into your brains!
Do you think JCGM 100:2008 section 4.2 was wrong to suggest that u=s/sqrt(N)?
Do you think NIST TN 1900 E2 was wrong to assume u=s/sqrt(N)?
4.3.2: “4.2.3 The best estimate of σ^2(q_bar ) = σ^2/n , the variance of the mean, is given by”
This is the standard deviation of the sample means. It is HOW PRECISELY YOU HAVE LOCATED THE MEAN OF THE POPULATION.
IT IS SAMPLING ERROR.
It is *not* measurement uncertainty. It is an additive factor to measurement uncertainty in the overall measurement uncertainty budget.
Do you see the (q_bar) in 4.3.2? Your reading comprehension skills truly need some work!
q_bar represents the mean of the sample means distribution. σ^2 of q_bar represents the variance of the sample means distribution.
It has *NOTHING* to do with propagating the measurement uncertainty of the measurement components onto the average value.
I’ve shown you this multiple times in the past.
If the measurement data is of the form m_n +/- u_n, e.g. m_2=5 and u_5 = 1), then the sample sets should contain the ENTIRE specification.
So sample1 may consist of m_1 +/- u_1, m_8 +/- u_8, m_10 +/- u_10, etc.
Sample2 2 may consist of m_2 +/- u_2, m_18 +/- u_18, m_33 _/- u_33, etc.
In this case the distribution of the stated values for sample1 becomes (m_1, m_8, m_10, ….) and for sample2 (m_2, m_18, m_33, …..)
You can then calculate the average of sample1 and of sample2. Those averages become the distribution of the sample means.
The measurement uncertainty of sample1 becomes (u_1 + u_8 + u_10 + …) and the measurement uncertainty of sample2 becomes u_2 + u_18 + u_33 + …;).
The uncertainty of the sample means becomes u_sample1 + u_sample2.
Not even the GUM does this involved of a calculation. It just uses the dispersion of the sample means as the uncertainty of the mean of the sample means.
But, again, this is ONLY how precisely you have located the population mean. If the measurement data in the population is highly inaccurate then that so-precisely-located population mean will be highly inaccurate as well. The standard deviation of the sample means will *NOT* tell you anything about how accurate the population mean is, i.e. it’s measurement uncertainty. It is ONLY an estimate of the sampling error associated with locating that population mean.
How many more times are you going to have to be run through this before it stops going in one ear and out the other?
So do you think JCGM 100:2008 section 4.2 was wrong to suggest that u=s/sqrt(N) and NIST TN 1900 E2 wrong to follow it or not?
Your insistence on quoting NIST TN 1900 blows your assertion that f(X1, X2, … , Xn) =Y = y = (X1 + X2 + … +Xn)/n out of the water doesn’t it?
Your insistence that JCGM 100:2008 4.2 is appropriate also blows your assertion out of the water.
Section 4.2 says:
Here is what the note on B.2.15 says:
Dispersion characteristics of the results means standard deviation of qk, not the standard deviation of the mean.
It appears you are ending the argument about f(X1, X2, … , Xn) = (X1 + X2 + … +Xn)/n being an appropriate method of finding the mean of temperatures.
I’ve never asserted that for NIST TN 1900 E2. NIST TN 1900 E2 uses a type A analysis; not a type B analysis.
For the record…I do think it is an appropriate method when doing a type B analysis. That’s because a type B analysis utilizes u(Xi) for each input i.
But you are deflecting and diverting again. A type B analysis is not relevant for NIST TN 1900 E2.
You said s/sqrt(N) was not measurement uncertainty. The GUM says it is when doing a type A analysis.
You are so far out in left field you have left the ball park. It is so far out it is hard to know where to start.
A Type B evaluation applies to each individual input quantity in the same way that a Type A evaluation applies to each input quantity. If you had read the GUM with any understanding you know that. You certainly don’t do a Type A or B evaluation analysis using measurement uncertainty for each input quantity. Those evaluations are used to determine the uncertainty.
You are trying to conflate the calculation of a combined uncertainty with doing a Type A or Type B evaluation. They are two entirely different things.
You say, “That’s because a type B analysis utilizes u(Xi) for each input i.“. Do you know how ignorant that sounds? Both Type A and Type B evaluations EVALUATE the value of u(Xi) for each input quantity. They do not “utilize” u(Xi) for anything, they determine its value.
Those values are used to calculate a combined measurement uncertainty by “adding” the individual input quantity uncertainties.
Here is a brief instruction.
u_c²( y), which is given by
Look at the bolded part. Each xi of an input quantity.
In a type A evaluation of a given input quantity,
q_k(X1) = {obs1, obs2, …, obs_n) with μ(X1) and σ(X1)
q_k(X2) = {obs1, obs2, …, obs_n)with μ(X2) and σ(X2)
q_k(Xn) = {obs1, obs2, …, obs_n)with μ(Xn) and σ(Xn)
and the mean is the best estimate and the uncertainty is the standard deviation for each input quantity.
In a Type B evaluation there are no observations, only a single measurement. The best estimate is the measured value and the standard deviation uses scientific judgement based on documentation hopefully.
Absolutely correct, he is still floundering around with cherry picking, completely missing the complete picture of the GUM.
And he doesn’t understand that a combined uncertainty can have elements of both types.
For some reason he and bellman cannot read GUM 4.2.3 and understand it. It is also where I part ways with NIST 1900.
For a unique Xi, to use the experimental standard deviation of the mean, one must have “n” independent repeated observations. That is, “n” = “k”.
Each Xi can have its own value of “k” depending on the process.
This is where so many mistakes are made. Climate science has listened to too many staticians. Of which, bellman and bdgwx are two.
To determine the specific uncertainty u(q̅) of a unique input quantity, the random variable should have 30 or more observations.
The take away, the number of samples is not the the value of “n”. This is why I have a problem with NIST TN 1900. What this example uses is basically the old “error” model with (τ + ε). However, this method does show that the difference between samples is an important factor, ala F.1.1.2.
The biggest point is that their example arrived at an uncertainty of 1.8F vs 4.2F which have the same magnitude. Rounded to 2F and 4F respectfully.
Someone needs to explain rationally how they can peer into those intervals and discern anomaly resolution to the one thousandths of a degree.
Oh yes, and another point they will never concede is that n is always exactly one, regardless of how much dancing and prancing they do around the issue. Divide by root-n all you like, it is still one.
“Climate science has listened to too many staticians. Of which, bellman and bdgwx are two.”
I never knew I had so much influence. Wish they would pay me for my advice.
They wouldn’t know measurement uncertainty if it bit them on the backside.
And then stung them on the neck!
Temperature is a measurement of a substance’s kinetic energy. It does not inform about the latent heat energy content. Averaging intensive values is generally not allowed.
One can always work out assumptions that in a unique example, averaging may work. That does not make the general rule that an intensive quantity must be converted incorrect.
Even with the necessary assumptions that would allow averaging, using the general rule, that is, conversion to extensive values, one will achieve the correct answer.
“You need to pick one and stick with it. You just contradicted yourself!”
It’s only a contradiction if you are incapable of reading.
Statement 1 – you don’t have to weigh multiple rocks in one go, you can weigh each one separately and add the results.
Statement 2 – A competent statistician should not say you can average anything and always get a meaningful result.
I really don’t want to get inside your head to figure out why you think those two statements are contradictory.
“If the sum is not physically meaningful then the average won’t be either.”
Argument by endless repetition is very boring. Do you make the same claim about standard deviation? If variance is meaningless than the standard deviation must also be meaningless.
“If ANY step in an equation is non-physical, then the result of the equation will be non-physical as well.”
So say you want to calculate kinetic energy. I see the equation is 1/2 mv². Velocity squared must be a meaningless concept, so you conclude that kinetic energy is non-physical. And by the same logic e = mc² must be meaningless, unless you can describe what the square of the speed of light looks like.
Your brain has exploded. You have obviously never taken a calculus based higher level physics class.
1/2 mv² is a common functional relationship that is reproducible in experiments. The 1/2 is a physical constant (not a counting number, think π) determined thru experiment. It is not an average! You would know this if you had physics lab in college. The formula accurately predicts correct values every time. It’s predictions are not meaningless. You should be happy that is the case as safety requirements for autos rely on its fundamental predictions.
Same with e = mc². It is not an average, it is a functional relationship that has been experimentally verified. https://www.nist.gov/news-events/news/2005/12/einstein-was-right-again-experiments-confirm-e-mc2
“1/2 mv² is a common functional relationship that is reproducible in experiments. The 1/2 is a physical constant (not a counting number, think π) determined thru experiment.”
Read what I said. I’m not arguing about the 1/2, I’m arguing about squaring velocity.
“The formula accurately predicts correct values every time.”
Almost as if you don’t need every part of the equation to be meaningful in order for the result to be meaningful.
“Same with e = mc². It is not an average”
Again, why would you think it’s an average. The question is what physically meaningful property is the square of the speed of light?
Read what I said. The formula is an experimentally derived and verified functional relationship. Consequently, v² is meaningful when used in this functional relationship.
You need to investigate and find a resource that declares squared variables are not meaningful in experimentally verified functional relationships.
Otherwise your assertion is meaningless!
You just keep missing the point. Your claim is that a result cannot be physically meaningful if part of the equation is not physically meaningful. Saying that the equation gives you verifiable physical results despite having physically meaningless parts, is making my point.
“You need to investigate and find a resource that declares squared variables are not meaningful in experimentally verified functional relationships.”
The question was about physical meaning. What physical meaning do you attach to v², a value in m²/s²? What is a square second and does it exist in the physical world.
And then explain why square velocity can have a physical meaning, but not the sum of temperatures.
Your lack of reading comprehension skills is showing again!
Velocity *is* a physically meaningful property. Mass is a physically meaningful property.
Two cars, one travelling at v1 and a second one at v2, doesn’t have a sum velocity of (v1 + v2), the sum is *NOT* a physically meaningful property. so (1/2) m * (v1+v2)^2 is *NOT* physically meaningful because (v1 + v2) is not physically meaningful.
“The question was about physical meaning. What physical meaning do you attach to v², a value in m²/s²? What is a square second and does it exist in the physical world.”
m/s *is* physically meaningful. When it is used to calculate an extensive property via integration it remains physically meaningful as does the result of the integration.
The integration actually becomes ∫ mvdv with the appropriate simplification. Again, “v” *is* physically meaningful. “m” is physically meaningful. So you have a physically meaningful value times a physically meaningful value – you do *NOT* have a physically meaningful value times a non-physically meaningful value.
This is no different than your inability to understand how Possolo came up with his measurement uncertainty equation for the volume of a barrel. It all stems from your continual cherry picking of things with no actual understanding of concepts and context.
Do you understand the difference between v = m/s and v² = m²/s²?
Was Bellman asking about v or v²?
He is saying v² is not physically meaningful.
It is a variable that is used in several physical functional relationships. It has been experimentally verified and proven to be meaningful and accurate..
It is like T⁴, it is a meaningful variable in a functional relationship that has been verified to be accurate.
Using T in an average of temperatures with individual input quantities has never been proven to provide a meaningful and accurate value.
Right…the problem is that your brother conflated v with v² in his response.
NIST averaged temperatures of different parcels of air with different temperatures on different days in TN 1900 E2. Why do you think they would do that if it was not meaningful and accurate?
But NIST did not end up dividing each days uncertainty by “n” as done under your functional relationship.
NIST TN 1900 is an example. It makes many assumptions such as individual measurement uncertainty being neglible. Zero, nada, zip! They assumed a Gaussian distribution. They assumed the measurements were of the same thing under repeatable conditions.
It even says:
NIST does not show an uncertainy budget itemizing all the uncertainty components, they are just assumed to be negligible in this example.
They have in essence set the problem up in order to use the standard deviation of the mean as the Type A uncertainty.
One other thing to point out. When using the standard deviation of the mean, one should use a t-factor to expand it into a 95% interval. NIST did that and ended up with an uncertainty of ±1.8°C. that is not far from the standard deviation and in fact rounds to ±2.0.
Show us how that is propagated into an anomaly where the baseline has an uncertainty of ±0.5.
You’re deflecting and diverting again. Uncertainty has nothing to do with this line of conversation. You get bent of shape about how meaningless and useless it is to average temperature so I point that NIST does just that. Clearly they don’t think it is meaningless and useless.
Argument by authority, again.
As usual you ignore the fine print in your rush to get the answer you need.
Next…
It most certainly does. The whole GUM series is about making measurements and determining their uncertainty. It is a standard internationally. The GUM allows averaging observations but it also has very specific requirements that must be met to do so. Such as, measuring the same thing under repeatable conditions. NIST is very careful to address these these requirements in their examples.
Would you say that averaging intensive properties can be meaningful and useful afterall?
Is a different parcel of air on a different day with a significantly different temperature the same thing? Does measuring different parcels of air over the span of many days with varying environmental states qualify as repeatable conditions?
No.
Nope, doesn’t qualify; you’ve been told this many, many times but refuse to acknowledge the truth.
Nope — is this what your vast knowledge reservoir of measurement uncertainty tells you?
Instead of cherry picking and studiously remembering individual post from among thousands from years ago, go read the relevant documents for comprehension and understanding.
Your questions scream out that you don’t (and won’t).
So what do you make of NIST…
1) …averaging an intensive property
2) …doing so to different things
3) …while under conditions that are not repeatable?
Prediction confirmed.
Just because NIST “did something” doesn’t increase its validity at all.
NIST “doing something” is all you got in the tank, fumes.
ITS A DAMN TEACHING EXAMPLE!
The fact that it assumes that all measurement uncertainty is negligible should tell you that. It would tell anyone with a lick of sense that!
It’s no different than 99% of statistics textbooks not covering measurement uncertainty at all. That is *NOT REAL WORLD*.
You live in statistical world, not in reality. Come join the rest of us in THE REAL WORLD SOMETIME!
You’ve held firm to your positions that 1) it is meaningless and useless to average intensive properties 2) otherwise averaging should only be done if measuring the same thing and 3) when the conditions are repeatable. NIST violated all 3 of those points in their teaching example.
High-falutin’ organizations are not immune to publishing bunk.
To include your fav NIST.
I wonder if the Gormans consider the work by NIST bunk as well
Gormans? You want to chime in here? What’s your take on the matter?
Did my post make it into the mighty b-x database?
And why is it you refuse to read? They’ve answered this “buh, buh, buh, NIST!!” garbage line countless times already.
But you always run away, why is this?
Is NIST TN 1900 bunk. No, it is an example of what might be done. Does it say anywhere that the example is considered to be the ONLY acceptable method to calculate a monthly average temperature and its total uncertainty? I sure didn’t find it if it did. In fact, the TN shows it as “Examples E2”.
The second you saw the assumption of Gaussian samples and that “that no other significant sources of uncertainty are in play”, bells should have gone off in your head about this being a simple illustrative example of how to find the variation between samples of a property.
I’m glad to hear that your position on the matter diverges from that of karlomonte. But don’t you see how NIST TN 1900 E2 significantly undermines your positions that 1) averaging intensive properties is meaningless and useless, 2) otherwise averaging should only be done measuring the same thing, and 3) done only under repeatable conditions?
Are you saying that despite averaging intensive properties being meaningless and useless that there is more than one acceptable way to do it? How does that work?
My use of the term was to call attention to your propensity to employ Appeal to Authority fallacies — just because an organization like NIST published something does not automatically make it technically correct.
And as Jim has pointed out to you many, many times, why do you continue to ignore the conditions and assumptions attached to this NIST document?
Where has NIST calculated the uncertainty of a “global” temperature?
Yet more loaded questions ignored.
But don’t you see how NIST TN 1900 E2 significantly undermines your positions that 1) averaging intensive properties is meaningless and useless, 2) otherwise averaging should only be done measuring the same thing, and 3) done only under repeatable conditions?
1) you didn’t read anything I wrote nor did you investigate the GUM references I gave you.
The GUM allows the best estimate of series of observations of the same thing and that are random in nature, Gaussian, to be the mean of the observations. That is a statistical parameter from the data that has been observed. This is done for every input quantity. Somehow you continually fail to understand that.
Your insistence that f(X1, X2, …, Xn) = (X1 + X2 + … +Xn)/2 does not specify that a random variable with random observations is being used. You use each Xi as a separate input quantity consisting of a single measurement.
2) A monthly average can be designated a measurand of the property. Since one does not measure the same thing multiple times, F.1.1.2 and H.6 are the applicable concepts for determining the value of the property.
Why don’t you address the assumptions made in TN 1900 that allow the usage that NIST shows? Show us that you are able to assess the example properly and that you understand what is minimized.
Teaching examples are *NOT* meaningless. BUT you have to understand the limitations of the teaching example. Which YOU seem to be unwilling to admit!
The *TEACHING* part of the example is how to calculate the measurement uncertainty from the data that was collected.
You have NEVER, not once, listed out the assumptions Possolo made (see above) and analyzed what they mean in the context of the teaching example. All you’ve EVER done is try to say that the method of the teaching example is the method that should be used in the real world. Garbage in – Garbage out.
“properties” – I.E. MULTIPLE DIFFERENT PROPERTIES
How many times must it be repeated to you that averaging measurements is only appropriate when
How does “properties” fit into those requirements?
It is when you define the measurand correctly – i.e. measuring the same thing (Tmax) using the same instruement (same measuring station) under similar conditions (same month/season).
THIS WAS AN EXAMPLE OF HOW TO MAKE A TYPE A ANALYSIS.
How many time must this be repeated to you before it finally sinks in? If you would *ever* bother to list out all the assumptions made in this TEACHING EXAMPLE, it would hopefully become obvious to you (but maybe not) that this TEACHING EXAMPLE has nothing to do with the real world!
“You get bent of shape about how meaningless and useless it is to average temperature so I point that NIST does just that.”
Your reading comprehension skill is as bad as bellman’s.
NIST is assuming the data is FROM A SINGLE MEASURAND and is describing the SAME PROPERTY OF THAT SINGLE MEASURAND.
It’s a Type A measurement uncertainty situation. Multiple measurements of the same thing using the same device under the same conditions.
You *can’t* do that with multiple single measurements of different things using a different instrument where conditions are vastly different!
I’ll ask you the same thing I asked your brother. Would you say that averaging intensive properties can be meaningful and useful afterall?
Are you suggesting that different parcels of air are the same thing?
Are you suggesting that different environmental factors qualify as same conditions?
Read JCGM 100-2008 H.6. It describes meaCsuring the hardness (a property) of a single measurand. Since the measurement process is a destructive one, multiple measurements must be made at different points. These measurements are done with the same instrument, same operator, etc.
JCGM 100-2008 F.1.1 .1and F.1.1.2 describes how to develop an estimated measurement and the uncertainty associated with this.
Let me repeat the text. An evaluation of the possible differences between samples must be added to the variance of the repeated observations made on the single sample.
A monthly average is a PROPERTY. It is not a measurement of an actual physical specimen.
The single sample you choose in a monthly can be any day you wish. Multiple observations of, say Tmax, can be used to develop the measurement uncertainty of a single measurement. (This is what NIST TN 1900 said was negligible)
Since temperature measurements do not have multiple observations, then you must apply a Type B evaluation. (LIG – ±2.0°C, ASOS – ±1.0°C, CRN – ±0.3°C) It should be noted that these uncertainties are only one component of a comprehensive uncertainty budget.)
Then a second component must be added. “… an evaluation of a component of variance arising from possible differences among samples must be added …”. This is what NIST TN 1900 did, added a component derived from the differences of the samples.
As you conveniently forget every time, on purpose or out of ignorance, NIST made the assumption that the observations were samples from a Gaussian distribution and thus the measurement uncertainty could be stated using an expanded experimental standard deviation of the mean. NIST still obtained an uncertainty of ±1.8°C.
I would also note that NIST’s measurements have two decimal digits, yet the final statements only include one decimal digit, not two or three.
I took the last 30 days of Tmax at my station and had CoPilot analyze them. Here are some of the things it observed.
Here is a quick 5-number presentation of the data.
Some pertinent info.
Gaussian
mean 75.6
variance 65.5
SD 8.1
SE ±2.9
Expanded at k=2 ±5.8°
5-Number
median 76.2
Q1 68.6
Q2 83.0
IQR 14.4
Interval -7.6 to + 6.8
I’m aware of H.6. We discussed it a little over a year ago in part because it takes measurements at different spots and is antithetical to your faux “same thing” requirement. And it’s not just JCGM 100:2008 H.6 and NIST TN 1900 E2 that contradict you.
Anyway…
So are different air parcels the same thing or not?
Or can they at least be treated as the same thing?
More dog-chasing-tail: asking the same loaded question over and over and over and over and…
It’s an indicator that bdgwx has *NO* real world experience with metrology in a situation with civil or criminal liability. Not even with a situation with nothing more than payment penalties for non-performance. He’s like bellman – blackboard mathematicians.
Simple answer: The temperature in Nome, AK *IS* a different measurand than the temperature in Omaha, NE.
You are *NOT* making multiple measurements of the same measurand.
Even in H.6 you are measuring the SAME MEASURAND. What the variation in measurements represent is the variation in the hardness of different areas of the same measurand. What you get from averaging the measurements is a BEST ESTIMATE of the overall hardness of the measurand. The uncertainty from a non-uniform measurand ADDS to the total uncertainty budget for the value of the property of that single measurand.
If you are using that measurand in a situation with strict liability issues, you may not even care about the BEST ESTIMATE of the overall hardness, you may only care about the measurement that shows the LEAST hardness since that is the best failure indicator. In other words, the “average” may not even be physically meaningful in such a situation. And that’s for a SINGLE measurand!
Have you had *any* experience at all with metrology in a situation with civil or criminal liability?
You’re deflecting and divertering again. It’s hard not to wonder if this is because you are finding it hard to reconcile your hard-lined position with NIST TN 1900 E2. And I still don’t know if you think different parcels of air on different days with completely different temperatures qualify as the same thing or not. It’s an important question because if you want to convince people that your “same thing” rule is as sacrosanct as you claim then you should at least be able to say whether or not a certain scenario is compliant with it.
Why do you ignore the fundamental measurement uncertainties of individual air thermometers?
And I still don’t know if you think different parcels of air on different days with completely different temperatures qualify as the same thing or not.
Tell us what you think this sentence is referring to. You must have an interpretation or you are just arguing about something you know nothing about.
I know what it says and what it is referring to. I was wanting to know what you think.
Here is what I think…I have no problem with thinking of different parcels of air as specimens of the broader material when doing a type A analysis. In this specific case UAH has 9504 specimens of the TLT layer temperature for April 2026. Or in the case of NIST TN 1900 E2 they had 22 specimens of the daily high for May 2012. My position shouldn’t be a surprise since I’ve always accepted everything the GUM says.
You have no clue what a Type A evaluation of a measurement is do you?
If you accept that there are different parcels of air with single measurements of each parcel, how do you do a Type A analysis? A Type A analysis requires repeated measurements of the same thing, which doesn’t exist.
Each “sample” or different parcel of air, as in TN 1900, would need multiple measurements under repeatable conditions to meet Type A conditions. BTW, that is where √n is supposed to derive from, multiple entries in each sample. You are instead equating it to the number of samples which is incorrect. NIST makes the same mistake, however by using an expanded uncertainty, they obtain a reasonable uncertainty.
Are you comfortable with a ±1.8°F uncertainty in ASOS measurements? If so, how do you propagate that value into an anomaly uncertainty?
READ THE DARN EXAMPLE FROM TOP TO BOTTOM.
LIST OUT EVERY SINGLE ASSUMPTION AND EXPLAIN WHAT IT DOES AND WHY IT IS MADE.
UNDERSTAND THE EXAMPLE FOR BOTH CONTEXT AND MEANING.
For some reason you just adamantly refuse to do *any* of these.
It’s all explained in the example and its assumptions. Why won’t you ever list those out and explain their consequences?
It’s all in the example. When Possolo assumes negligible measurement uncertainty, that the same measuring device is being used, and Tmax is the property being measured – WHAT DO YOU THINK HE IS DOING?
He’s setting the scenario up to be multiple measurements of the same thing!
Are you truly unable to read for comprehension? Or are you just playing dumb?
IT’S A TEACHING EXAMPLE. It’s no different than assuming no friction on an inclined plane to calculate the acceleration of a car on wheels down the plane! It’s no different than assuming no wind affecting the projectile fired downrange from a canon and determining the landing spot! It’s no different than calculating the speed of a rocket ten seconds into flight assuming a fixed mass for the rocket and ignoring that fact that burning and ejecting the fuel changes the mass!
Whether I think that in the real world measuring Tmax on different days is measuring the same thing or not is irrelevant to the example as it is stated. But for some reason you can’t seem to figure that out.
WHY?
Yet you have no clue about what it is telling you.
Anyone can make an unexplained assertion that F.1.1.1, F.1.1.2, and H6 do not apply. Only those that can explain why, really know what they are talking about. Tell us exactly what metrology training you have had either in advanced physics, chemistry, engineering or stand-alone classes.
Explain why you think this does not apply.
Explain why you think this does not apply.
Explain why you think this does not apply.
Here are some questions that might lead you to some answers.
Lastly, you didn’t address the data and statistical calculations at all. Do these apply at all to measurement uncertainty?
Remember, if the Type B uncertainty for repeatable single measurements is ±0.54°F (0.3°C) and the standard deviation between samples is±2.9°F √(0.54² + 2.9² = ±2.9°F). With ASOS, √(1.8² + 2.9² = ±3.4°F.
“Would you say that averaging intensive properties can be meaningful and useful afterall?”
It’s useful in determining the BEST ESTIMATE of the value of a property of
And the meaning of the average is, again, the BEST ESTIMATE if and ONLY IF, the distribution of the individual measurements of the same thing using the same instrument under the same environment represent a Gaussian distribution. For instance, a measuring instrument with hysteresis will *not* generate a Gaussian distribution.
Have you finally figured this out? Averaging the intensive property values of different things is *NOT* meaningful or useful. Does averaging the temperature of a rubber ball in Topeka, Kansas and the temperature of a concrete brick in Atlanta, GA tell you anything meaningful and useful?
You have the EXACT SAME SITUATION when averaging the temperature on Pike’s Peak with the temperature in Denver, CO. The average is just as meaningless and useful as averaging the temperature of a rubber ball on Pikes Peak at 10AM and the temperature in Denver at 3PM. Different things in both cases.
“v” is the measured variable. v^2 is used in the calculation, it is a calculation term!
Bellman was asking about (1/2)mv^2.
v is a velocity. (1/2)mv^2 is the area under the mass/velocity curve. BOTH are physically meaningful.
Area is an extensive property.
Can you read? Are you as unknowledgeable of calculus as bellman?
…says the guy who thinks d/dx (x/n) = 1.
“…says the guy who thinks d/dx (x/n) = 1.”
You are as bad as bellman. You can’t do simple algebra simplification!
[d(x/n)dx] * (n) = 1
IT’S CALLED RELATIVE UNCERTAINTY! You have *still* not figured out how Possolo got his measurement uncertainty formula for a barrel, have you?
Neither has bellman. You two make quite a pair trying to tell people how to do algebra and calculus.
Why do you have to keep bringing me into all your conversations? It sounds like you’re obsessed with me. I really hoped we could get through one comment section without you lying about this.
“[d(x/n)dx] * (n) = 1”
Only if d(x/n)dx = 1/n
“IT’S CALLED RELATIVE UNCERTAINTY!”
Relative uncertainty has nothing to do with equation 10.
“You have *still* not figured out how Possolo got his measurement uncertainty formula for a barrel, have you?”
I’ve explained this to you far too many times for you not to remember. You work out the uncertainty using equation 10, then divide through by V². Or you can use equation 12. It’s as simple as that.
Now you are going to rant some more and try to shove relative uncertainties into equation 10, and claim you are teaching me something. And I’m going to do my best to ignore it. Knock yourself out.
“Relative uncertainty has nothing to do with equation 10.”
YOU STILL HAVEN’T ACTUALLY READ THE GUM, HAVE YOU?
This has been pointed out to you multiple times and you SIMPLY refuse to actually read the GUM for meaning and context.
From 5.1.6: “This is of the same form as Equation (11a) but with the combined variance u_c^2( y) expressed as a relative combined variance [uc(y)/y]^2 and the estimated variance u^2(xi) associated with each input estimate expressed as an estimated relative variance [u(xi)/xi]^2.”
This is what Possolo did with his uncertainty formula. It is a *standard* practice. And it leaves the sensitivity coefficient of each term as the power of the variable being evaluated in the term.
YOU STILL CAN’T READ. YOU CAN’T DO SIMPLE ALGEBRA. YOU CAN’T DO CALCULUS.
If you could you would be able to figure out relative uncertainty and what Possolo did. Instead it just remains magic for you to use in claiming people can’t do partial derivatives correctly instead of understanding what cancels and what doesn’t!
“YOU STILL HAVEN’T ACTUALLY READ THE GUM, HAVE YOU?”
Better then you it would seem.
“From 5.1.6”
That’s equation 12, not 10. And we went through all this a couple of months ago. It’s pointless expecting me to be able to convince you of your misunderstanding, when comments will be closed down in a couple of days. If you didn’t understand then, you probably never will. So I’ll leave you to your melt-down.
Yesterday, about a dozen posts ago, you claimed to be checking out, and here you are today.
You can’t, you Last Word Syndrome won’t allow it.
Yes I did, but Tim didn’t get the hint.
Sits back to see if Karl responds, because he has to have last word.
For bellman, relative uncertainty doesn’t exist. It’s why he *still* can’t figure out what Possolo did in deriving his uncertainty formula for a barrel.
“For bellman, relative uncertainty doesn’t exist.”
Utterly pathetic. It’s your inability to understand the implications of relative uncertainty that results in your great misunderstanding. Remember you pointing out all those years ago that u(Avg)/Avg = u(Sum)/Sum, and not understanding what that said about the absolute size of the uncertainty of the average. You still don;t seem to understand it.
“It’s why he *still* can’t figure out what Possolo did in deriving his uncertainty formula for a barrel.”
When you know you have no arguments left, all you can do is lie. I’ve explained to you countless times how to get the uncertainty of the volume – you simply cannot accept that the technique does not work for an average. You never understand the assumptions that are stated for equation 12.
*YOU* are the one that didn’t recognize Possolo was using Eq 12 and kept saying I couldn’t do partial derivatives. Stop trying to foist off your misunderstanding of when and how to use relative uncertainty on me.
There are *no* assumptions for Eq. 12. It states:
——————————
“If Y is of the form Y = cX_1^P1X_2^P2…X_n^P_n and the exponents p_i are known positive or negative numbers having negligible uncertainties, the combined variance, Equation (10), can be expressed as
[u_c(y)/y]^2 = Σ [p_i u(x_i)/x_i]^2
This is of the same form as Equation (11a) but with the combined variance u_c^2 ( y) expressed as a relative combined variance [u_c(y)/y]^2 and the estimated variance u^2(xi) associated with each input estimate expressed as an estimated relative variance [u(xi)/xi]^2. [The relative combined standard uncertainty is u_c(y)/│y│ and the relative standard uncertainty of each input estimate is u(xi)/│xi│, │y│ ≠ 0 and │xi│ ≠ 0.]
————————-
There is no assumptions in that definition. If multiplication of terms (which can be division just as well) is involved, then you use relative uncertainties. X_1^P1 * X_2^P2 *is multiplication!
In case anyone is interested in these incessant claims about Tim explaining “Possolo” water tank example to me, here’s the original thread.
https://wattsupwiththat.com/2022/11/03/the-new-pause-lengthens-to-8-years-1-month/#comment-3636787
Sorry I promised I wasn’t going to reply to Tim this late in the cycle but it’s difficult to ignore this noisy.
“There are *no* assumptions for Eq. 12. It states:
——————————
“If Y is of the form Y = cX_1^P1X_2^P2…X_n^P_n and the exponents p_i are known positive or negative numbers having negligible uncertainties, the combined variance, Equation (10), can be expressed as”
If Y is of the form, is the assumption. Equation 12 only works if Y is of that form. It simply does not work if there is addition in the function.
But you forgot the most important part, the result.
See that summation symbol? It tells you that uncertainties add, just like equations 10 and 13. They don’t average, they don’t subtract, they aren’t divided by the √n.
Look really closely at the equation for the combined uncertainty. Does it contain the value of “c”?
Remember, the equation you are fond of is:
Y = (1/n)(Z1), where Z1 is the sum of n input quantities. That is the form of Y = c(Z1)¹. Again, where does “c = (1/n)” enter into this equation?
The result is invalid if the assumption is not correct. An average has pluses in it, equation 12 simply does not work. Please, try it. Work out what equation 19 gives you, and then try to simplify it to equation 12. It just cannot be done. It’s basic algebra.
“They don’t average, they don’t subtract, they aren’t divided by the √n.”
They do if that’s the result of equation 10. It’s trivial the partial derivative for each term is 1/n, this squared in each term for equation 10. As it’s the same for each term it can be extracted out, so the whole sum is the sum of the squares of the uncertainties divided by n². Take the square root and you have the uncertainty of the sum divided by n. Now assume all terms have the same uncertainty, and you should be able to work it out from there.
I really don’t know if your inability to see this is down to your lack of understanding of algebra, or a determined effort on your part not to see it.
“Look really closely at the equation for the combined uncertainty. Does it contain the value of “c”?”
Again, in equation 12 any constant term cancels out when you divide by the result squared.
“Y = (1/n)(Z1), where Z1 is the sum of n input quantities. That is the form of Y = c(Z1)¹. Again, where does “c = (1/n)” enter into this equation?”
c(Z1)’ = c. Basic calculus.
Why don’t you ask your AI to explain it to you?
Sure it does, not. Where did you learn math. Show exactly how it cancels. Use the information from the NIST example of a volume. I’ll guarantee, you don’t divide through by V² in the process. At best you will MULTIPLY the right hand side by V².
Equation 12 has no derivative in it. That is your idea and I have no idea where you found it.
Y = c(Z1) = (1/n)(T1 + T2 + … + Tn) This is a valid use of Eq 12. I have simply factored the “n” out of each term which is an accurate simplification mathematically.
To obtain the uncertainty in the sum of Z1 one can use the SD of the collection of observations as is done in NIST TN 1900. You can even use the expanded standard deviation of the mean if you wish.
u(Z1) = SD(T1 + T2 + … + Tn)
[u꜀(y)/y]² = Σ[(1)[u(Z1)/Z1]² This is exactly what Eq 12 supplies. I see no value whatsoever for “c”. Show where it is at in your equation and how it is you divide by “y²”
See the original comment from a few years ago. Nothing has changed.
But as an example take the radius input. Partial differential with respect to R is 2πHR. This squared in the equation. Then divide through by V² which is πHR². So for the radius term you have (2πHR/πHR²)u(R) all squared. Cancel terms. π, H and one of the R’s cancel, leaving you with (2/R)u(R) all squared.
“Equation 12 has no derivative in it.”
You asked what the derivative of c(Z1) was, not about equation 12.
“Y = c(Z1) = (1/n)(T1 + T2 + … + Tn) This is a valid use of Eq 12.”
Pointless arguing at this late stage. If you can’t understand why it isn’t you never will.
But, if you understand that (12) is derived from (10), you need to dexllain why using (10) does not give you the result you are claiming.
“To obtain the uncertainty in the sum of Z1 one can use the SD of the collection of observations as is done in NIST TN 1900.”
No you can’t. That is not propagating individual measurement uncertainties. It’s treating the variance in daily values as a sort of measurement uncertainty. That is the same as treating it as a sample, I.e.SD/√N.
What you should do is explain the “π” value goes in the uncertainty equation.
The equation is V = πr²H. “π” is “c” in equation in the function Y = (c)(X₁)²(X₂) where”
c = π, X₁ = r, and X₂ = H
This a straight forward multiplication. Where did the “c” term disappear to?
It cancels when you divide through by V².
Me:
You:
π cancels when you divide through by V²? Where did you learn math?
You don’t divide by V², you multiply both sides by V². As a result, if “c” remained, you would have
u_c(V) = V{√πΣ [ u(x_i)/x_i]^2
“You don’t divide by V², you multiply both sides by V²”
What? I’m not sure if you know what you are arguing against at this point.
“What you should do is explain the “π” value goes in the uncertainty equation.”
See above. And any of the other countless times I’ve explained it.
You said “π” cancels when you divide by V².
Show us your math, not a simple deflection to keep from answering.
tpg from the thread: “YOU STILL CAN’T EXPLAIN HOW POSSOLO CAME UP WITH HIS FORMULA FOR THE RELATIVE UNCERTAINTY OF V = πR^2H.
YOU: π [ (2H x u(R)/R) + (R^2 x u(H)/H) ]
Possolo: (2 x u(R)/R) + (1 x u(H)/H)
They are *NOT* the same.”
You STILL haven’t figured out the use of relative uncertainty!
The GUM covers it in Eq 12. Taylor covers it. Bevington covers it. Possolo covers it. YOU JUST USE RELATIVE UNCERTAINTY WHEN THE FUNCTION IS MULTIPLICATIVE INSTEAD OF AN ADDITIVE FUNCTION.
If y = f(x,y) = x^2y^2 the uncertainty formula is EXACTLY what Eq 12 says and what Possolo did.
∂f(x,y)/∂x = 2y^2
∂f(x,y)/∂y = 2x^2y
The first uncertainty term becomes (2y^2)^2 * [u(x)/(x^2y^2)]^2
The second part, [u(x)/(x^2y^2)] is the relative uncertainty associated with x.
This simplifies to the term (2xy^2)^2 * (1/x^2y^2)^2 u(x)^2 –>
[2xy^2/(x^2y^2)] ^2 u(x)^2 –> (cancelling y^2 terms and x/x^2)
(2)^2 *u(x)^2 * (1/x)^2 ==> (2)^2 [ u(x)/x]^2
Exactly what Possolo did. Exactly what Eq 12 shows. The relative uncertainty of the “x” component.
As in Eq 12 you get p1^2 [ u(x_i)/x_i]^2
You couldn’t do relative uncertainty then and you still can’t. You still claim the derivation of Eq 12 is wrong even though you’ve been shown that it isn’t – it is CORRECT.
You really are desperate to keep this abuse going to the bitter end. Anybody can read through the convoluted thread and see who said what. Of course you think you are right because you don’t understand the mistakes you are making.
You keep trying to imply I don’t agree with equation 12, or understand how to derive it. What you kee avoiding is the fact you cannot use equation 12 with an average, or any other function involving addition.
bellman from the thread:
“Problem 2 – TG has no idea how to use the equations. Nothing he writes above makes any sense. You do not just add the partial derivatives as he does in the first part of that equation, and I’ve no idea what he’s doing in the second part.”
bellman couldn’t do simple algebraic functions in the old thread and he still can’t. He didn’t understand having to use relative uncertainty then and he still doesn’t understand it.
He couldn’t figure out that (2xy^2)/(x^2y^2) = 2/x then and he still can’t figure it out.
GUM:
——————-
“If Y is of the form cX1^p1 * X2^p2 * … * XN^pN
…
the combined variance, Equation (10), can be expressed as
[u_c(y)/y]^2 = Σ [p_i * u(x_i)/x_i ]^2 Eq 12
—————————
Eq 12 is not some crap function the GUM just threw in. It is derived directly from Eq 10 using the rule that a multiplicative function must use relative uncertainty. It’s just a different, simplified way to write the expression of Eq 10.
“It is derived directly from Eq 10 using the rule that a multiplicative function must use relative uncertainty. It’s just a different, simplified way to write the expression of Eq 10.”
Which is exactly what I said. The problem you never address is how to use equation 12 with an average, which is not of the correct form.
from the old thread:
bellman: ““He has the right elements, but want to jump straight to multiplying the partial derivatives by the relative uncertainty. But that’s not what the equation says, and it’s beyond me why he can’t just write out the equation and work from there.””
tpg: “It actually *IS* what you do. That’s why each element in the GUM Eq 10 is: ∂f/∂x_i times the uncertainty of the element divided by the value of the element (relative uncertainty).”
—————————————
Again, bellman didn’t understand relative uncertainty then and he doesn’t understand it now. Taylor covers this in detail in his Chapter 3 and Chapter 5. But bellman has *never* bothered to study any metrology text for meaning and context. Those tomes are only good for cherry picking from as far as bellman is concerned.
“It actually *IS* what you do. That’s why each element in the GUM Eq 10 is: ∂f/∂x_i times the uncertainty of the element divided by the value of the element (relative uncertainty).””
Yes, that’s an example of you not understanding the equation. Equation 10 is not using relative uncertainties.
“That’s equation 12, not 10.”
In other words: “Don’t confuse me. Eq 10 is the only equation in the GUM that I have cherry picked!”
The only misunderstanding is yours! The misunderstanding of when you need to use relative uncertainty. It’s why you couldn’t (and still can’t” understand how Possolo got his uncertainty equation for a barrel. For you relative uncertainty doesn’t exist!
“Don’t confuse me”
This should not be confusing, even to someo0ne of your limited education. Equation 10 is the “general” equation for propagating uncertainty. It works for any differentiable function, and uses absolute uncertainties. Equation 12 is a special case of Equation 10, that can be used for a specific type of function – one that only involves multiplication, division and raising to exact powers, and converts (10) into a simplified equation involving relative uncertainties. It cannot be used for an average. This really should not be hard to understand. It states the required condition right at the start.
“and still can’t” understand how Possolo got his uncertainty equation for a barrel. ”
Who are you trying to convince with this lie? I suspect the only reason you keep repeating it is to convince yourself, because deep down you know I’m right.
*I* am the one that had to teach you how Possolo did his propagation of uncertainties for a barrel. You kept trying to tell me I couldn’t do partial derivatives.
Yes, Eq 12 is a special case of Eq 10. SO FRIGGING WHAT?
And you still haven’t figured out that propagated measurement uncertainty is a SUM and not an average. You do *NOT* average measurement uncertainties in either Eq 10 or Eq 12. There is no “n” sqrt(n) in either one!
“Σ” denotes a SUM, not an average!
“Your lack of reading comprehension skills is showing again!
Velocity *is* a physically meaningful property. Mass is a physically meaningful property.”
The question was about whether velocity squared was a physically meaningful property. Stop lecturing people about their reading skills, when you keep missing what they’ve written.
“So you have a physically meaningful value times a physically meaningful value you do – *NOT* have a physically meaningful value times a non-physically meaningful value.”
You are confusing yourself as much as anything. When you added the two velocities you were adding two physically meaningful properties. Your point was that adding two physically meaningful intensive properties produced a physically meaningless value. Why do you assume that multiplying an intensive property by itself will produce a physically meaningful value? If a car is travelling at 10m/s, what physical meaning does 100m²/s² have?
“This is no different than your inability to understand how Possolo came up with his measurement uncertainty equation for the volume of a barrel.”
You really have the mind of a 5 year old. You are just incapable of admitting your mistake. But continuously lying like this when I the evidence is available for all to see is just pathetic.
You are arguing that you can’t use an intensive property to calculate an extensive property – without a single reference that agrees.
v^2 isn’t an extensive value. (1/2) * m * v^2 IS an extensive value. It really is the area under a curve and AREA IS AN EXTENSIVE PROPERTY!
Give us just ONE reference that says you can’t use an intensive property to calculate an extensive property!
“You are arguing that you can’t use an intensive property to calculate an extensive property – without a single reference that agrees.”
When did I say that? Again you need to check your own reading comprehension before criticising others.
This has nothing to do with intensive or extensive properties. I just used v² as an example of a physically meaningless value that never the less can result in a physically meaningful property.
“It really is the area under a curve and AREA IS AN EXTENSIVE PROPERTY!”
Yes, that’s a point I keep making. Think of an average surface temperature as the integral of temperature times area, and divide by total area. All your worries about summing intensive properties disappear.
“Give us just ONE reference that says you can’t use an intensive property to calculate an extensive property!”
Give me one comment where I’ve said you can’t. An intensive property multiplied an extensive one, is extensive. That’s my point about temperature multiplied by area. (Or volume, or time, or any other property you want to average by).
v^2 is *NOT* the measured value of an intensive property. “v” is the intensive property value. v^2 is used to calculate the area under the mass/velocity curve.
No one has ever asserted that you cannot use an intensive value to calculate an extensive value. What you cannot do is AVERAGE multiple intensive values of different things – you get neither an extensive value nor a physically meaningful intensive property value.
You *can* use the intensive property value of temperature to calculate the heat energy transported using Q = mc(T1-T2). Q is an extensive value, you can then average Q.
How do you integrate temperature and area? Where does the temperature value come from? Where in the area do you take the measurement of temperature? Can sub-parts of the area have different temperatures? If you are dividing by “total area” then you must be measuring the individual temperatures of different sub-areas. Are all of the sub-areas homegenous, same pressure, same volume, same specific heat? Does this represent the area represented by the counties of Shawnee and Jefferson in Kansas? Or all of these sub-areas actually different objects with different pressures, volumes, and composition?
Does the forumula pV = nRT have any meaning for you?
Where does “area” come into play in averaging temperatures of different objects?
“v^2 is used to calculate the area under the mass/velocity curve.”
It doesn’t matter how many times you repeat this – it’s still missing the point. Either you say a non-physical property can be used as part of a calculation and still give you a physical meaningful property, or it cannot.
“No one has ever asserted that you cannot use an intensive value to calculate an extensive value.”
Still missing the point. This has nothing to do with v being intensive, it’s the fact it’s squared that ensures it has no physical meaning.
“How do you integrate temperature and area?”
The point is the total temperature times area is not a sum of intensive properties, it’s an integral of extensive properties – and therefore your spurious logic doesn’t apply. The average temperature by area is not derived from anything non-physical.
How you measure it is another question. Unless you have an exact function for temperature it will always be an approximation, and that will be using the usual methods of sampling – and weighting by area, or having a systematic grid of values as you would get from a satellite.
The rest of your questions are irrelevant. The point is to determine an average temperature, not pressure, energy or humidity. The average temperature surface temperature is what it says it is.
“Where does “area” come into play in averaging temperatures of different objects?”
This is about averaging over area, not discrete objects. It’s just another one of your distracting techniques.
“It doesn’t matter how many times you repeat this – it’s still missing the point.”
IT IS *NOT* MISSING THE POINT!
A MEASURAND AND A CALCULATION TERM ARE TWO DIFFERENT THINGS.
Why is this so hard for you to understand?
A calculation term isn’t a measurand, it is not measured. A calculation term is neither an extensive nor an intensive property!
“The point is to determine an average temperature,”
The rest of my questions are right on point. If all of the properties I list are not the same then YOU ARE MEAURING AN EXTENSIVE PROPERTY OF DIFFERENT THINGS!
Do the words “DIFFERENT THINGS” mean anything to you?
If two objects have different pressures, have different humidities, are in different terrain or geographies, THEY ARE DIFFERENT THINGS.
You cannot average the intensive properties of different things and get a physically meaningful average.
You can dance around this all you want, you simply aren’t going to show where physical science is wrong with this.
“A MEASURAND AND A CALCULATION TERM ARE TWO DIFFERENT THINGS.”
So you accept that a calculation does not have to be a measurand. Good. Maybe you now understand why the sum of temperatures is just part of a calculation.
And please stop all this bold yelling. It’s so obvious when you start like this it’s because you suspect you were wrong.
“So you accept that a calculation does not have to be a measurand.”
For three freaking years people have been telling you that an average is not a measurand. And for three freaking years you’ve been trying to tell us that it is. The average is the result of a calculation, it is not a functional relationship mapping an input to an output.
Nor is a calculation TERM a measurand. Funny how you changed “calculation term” to just “calculation”. It’s just more proof that you can’t read simple English for meaning and context.
“Maybe you now understand why the sum of temperatures is just part of a calculation”
THAT HAS NO PHYSICAL MEANING if the values are from different things. It is mathematical masturbation, nothing more.
“For three freaking years people have been telling you that an average is not a measurand”
I think your memory loss is showing. We’ve been arguing this for over five fracking years.
“The average is the result of a calculation, it is not a functional relationship mapping an input to an output. ”
And I keep telling you that you don’t know what “functional relationship” means. You could have actually tried to learn something in all that time.
“Funny how you changed “calculation term” to just “calculation”.”
A distinction without a difference. You were shouting about a calculation being a sub-term, e.g. v² only being a term in the equation. I’m just pointing out, by the same logic the sum is just a term in the equation for the average.
“THAT HAS NO PHYSICAL MEANING if the values are from different things”
Changing your argument again. So is it becasue you are adding intensive properties, or because you are adding different things, that makes it physically meaningless?
More malarky! in 1 + 1 = 2, the value of 2 is the calculation result, each of the values of 1 is a calculation term. The result is *NOT* the same as the calculation terms.
Don’t you *ever* get tired of demonstrating your lack of reading comprehension skill?
“When did I say that? Again you need to check your own reading comprehension before criticising others.”
“v” is a measurand. An EXTENSIVE property.
“v^2” is a calculation term in a function. It is *NOT* a measurand.
Only measurands have intensive and/or extensive properties that can be measured. CALCULATION TERMS ARE NOT MEASURANDS.
Why is this so hard for you to understand?
““v” is a measurand. An EXTENSIVE property.”
You think velocity is an extensive property?
““v^2” is a calculation term in a function. It is *NOT* a measurand.”
And what is the sum of temperatures in the function for a mean?
v is PART of an extensive value. It *is* a measurand. v^2 is *NOT* a measurand, it is a calculation term.
Does that clear it up for you?
The mean is *NOT* a function. It cannot describe a casual relationship. It cannot map individual inputs to a single output. The mean is a statistical descriptor that describes a collection of individual inputs, it does no mapping of the individual inputs to an output.
You can keep circulating around the issue all you want. You can’t make the calculation equation for the average into a function. It only continues to show your lack of knowledge of physical science. And it does the same for climate science.
“v^2 is *NOT* a measurand, it is a calculation term.”
A sum of temperatures is not a measurand, it is a calculation term. Does that clear it up for you?
“The mean is *NOT* a function.”
Do you really think writing “not” in all caps makes it true? Provide a reference for why adding values and dividing by n is not a function.
Are you sure about that?
JCGM 100:2008 section 4.1.2 says they can be.
Statement 1 involves extensive properties values. Statement 2 involves excluding intensive property values.
And you can’t see how you invoking both is contradictory? *YOU* want to measure intensive property values of multiple things and add the results – which statement2 excludes since it doesn’t give a meaningful result!
Talk about boring! How do you think the standard deviation is calculated? Look at how you calculate the variance, it involves (x – x_bar) where x_bar is the average. If the average is non-physical then (x – x_bar) is non-physical as well! Meaning the variance is non-physical. The square root of a non-physical value is also a non-physical value!
You are *still* trying to justify the meme that “if I can calculate it then it is physically meaningful”. Which violates your Statement 2!
You demonstrate your lack of physical science knowledge with everything you assert. (1/2)mv^2 is ENERGY IN JOULES. Joules is an extensive property. This is actually the integration of force over distance, ∫ (ma)dx from x1 to x2. Since a = dv/dt and dx/dt = v this becomes ∫ m(v)dv –> (1/2)mv^2. This actually finds the area under the curve and AREA is an extensive value.
km has you pegged pretty well. You talk in circles and use the bandwagon fallacy all the time – “climate science does it so it must be right”.
“Statement 1 involves extensive properties values. Statement 2 involves excluding intensive property values.”
Nope. Read again. Then say why they are contradictory statements.
“How do you think the standard deviation is calculated? Look at how you calculate the variance, it involves (x – x_bar) where x_bar is the average. If the average is non-physical then (x – x_bar) is non-physical as well!”
Which is another reason why you keep talking about the variance of temperatures is contradictory.
“You are *still* trying to justify the meme that “if I can calculate it then it is physically meaningful””
Nope. But as usual you argue with strawmen of your own creation. Saying that average temperature is meaningful is not the same as claiming that all calculations are meaningful.
“Joules is an extensive property.”
How difficult do you find it to keep missing the point so badly. I am not arguing about intensive or extensive properties. I’m arguing that part of an equation can be physically meaningless, but the final result meaningful.
“ I’m arguing that part of an equation can be physically meaningless, but the final result meaningful.”
Unbelievable. If an equation has a physically meaningless input then the output will be physically meaningless as well because you can’t map independent non-physical input values to dependent physical output values.
Not unless you have a crystal ball or magical powers.
“If an equation has a physically meaningless input”
The sum of temperatures is not an input. The individual measurements are. Not that it’s relevant.
But where do you use the individual measurements in a function mapping the individual inputs to an output value?
Ans: *YOU* DON’T.
You don’t use the individual measurements in a function. You describe the collection, i.e. the data set, with a statistical descriptor. And the statistical descriptor does no mapping of an input variable (singular) to an output.
“But where do you use the individual measurements in a function mapping the individual inputs to an output value?”
Surely even you can figure that one out. You use them by adding them up and dividing by n, or if you prefer by dividing each by n and then adding. That’s where you use them.
“You don’t use the individual measurements in a function.”
Insane. Why not?
“You describe the collection, i.e. the data set, with a statistical descriptor.”
Yes, that’s called using the average function. I’m really not sure what you point you think you are making here.
“But as usual you think that all statisticians and mathematicians are idiots who don;t understand their own research, and all scientists who use statistics are idiots. “
That’s *NOT* what I’ve asserted at all. Statisticians and mathematicians that live in “statistical world” don’t understand their own research. It’s like statistical textbooks that have examples of averaging measurements without including the measurement uncertainty associated with the measurements in the results. It’s from living in a blackboard statistical world instead of reality.
YOU are a perfect example. Thinking that a non-physical average is physically meaningful!
Why? If you can’t create a larger system from things with different units then how is that different from trying to create a larger system from things with teh same units that can’t be combined into a larger system?
Apparently the words “intensive” and “extensive” have no meaning for you. They are completely different as well! But you seem to think that you can create a larger system from each simply by adding the values.
A perfect example of the fact that you think you can create a larger system from intensive properties of different things.
That’s what this all boils down to. The statisticians meme that you can average anything and have it be physically meaningful.
“If you can’t create a larger system from things with different units then how is that different from trying to create a larger system from things with teh same units that can’t be combined into a larger system?”
Try to read my comments. I didn’t say you cannot create a system from different units. I said you couldn’t average things with different units.
“Apparently the words “intensive” and “extensive” have no meaning for you.”
I’ve explained the difference enough times. But as is your usual distraction technique any attempt to explain why your claim is wrong results in you claiming I don’t understand the distinction.
Extensive and intensive oroperties are different. It does not mean you cannot average both.
“A perfect example of the fact that you think you can create a larger system from intensive properties of different things.”
Temperatures usually have the same units. You can average them.
“The statisticians meme that you can average anything and have it be physically meaningful.”
You really need to define “physically meaningful”. The average is not a physical tning, it is related to physical properties of induvidual elements of the population, it is meaningful. This is true regardless of properties being intensive or extensive. Do you regard the average weight of a person as physically meaningful? If you do, why would you not regard the average temperature of a person as physically meanngful?
Talk about not understanding cause and effect. A higher average caused by increasing maximum temperatures has a totally different effect than if the higher average is caused by increased minimum temperatures.
What has climate science been preaching for 25 years or longer? That the higher average will cause crop failures, declining polar bear populations, disappearance of polar ice, massive starvation and migration of humans and animals, etc.
It’s exactly the same as trying to assign an “effect” to your temperature trends. If the trend line is physically meaningless then you can’t assign a physical effect. And the trend of a physically meaningless average *is* itself meaningless.
Subtracting two dates to get a growing season is *NOT* 1. an average and, 2. is not using intensive properties.
“A higher average caused by increasing maximum temperatures has a totally different effect than if the higher average is caused by increased minimum temperatures.”
Lots of things are possible. The pont is that changing temperatures can cause changes to the average and growing season. Changes in the average or growing season do not cause changes in temperature. But it was just a joke about the way you worded your comment. Not worth going into a length argument about it.
“The pont is that changing temperatures can cause changes to the average and growing season.”
Changing temperatures *MAY* cause changes to the average and growing season. They also may *NOT* cause any change to either.
The change in the first frost and last frost dates are extensive values. They will *always* tell you about a change in the growing season.
“Changing temperatures *MAY* cause changes to the average and growing season.”
I think the word you are looking for is “can”. They don;t need permission top do it.
“The change in the first frost and last frost dates are extensive values”
Really? Are you sure you know what extensive means?
As I’ve shown you before that position is not consistent with the GUM. The GUM states that an average can be a measure of a measurand. See [JCGM GUM-6:2020] section 5.7, E.2.2, and 11.10.4 for relevancy.
You conveniently skipped 5.8:
Ranges, especially of input quantities, are routinely ignored and swept under the rug in climatology.
And you skipped 11.7 Models for time series, which has an example for a temperature bath.
11.10 Model selection and model uncertainty
You skipped an important part in 11.10.4:
This is never done in climatology.
Section E is titled: Cause-and-effect analysis.
E.2.2 says nothing averaging different quantities and different measurement systems.
Cherry picking, as usual, to get the answer you want and need.
There isn’t a single thing your response that challenges the notion that measurements can be averages of other quantities. In fact, the only relevant thing in your response was a mention of section 11.7 which…wait for it…contains an example of averaging temperature.
Each and every section speaks of measuring the properties of a SINGLULAR measurand. Not a large system consisting of multiple measurands which you are trying to characterize by using an intensive property of the multiple measurands.
Someday you REALLY need to read the GUM for meaning and context and stop cherry picking bits and pieces that you think confirm your misconceptions.
Measuring the temperature of a water bath 10 times and finding the average value is finding the best estimate of the temperature OF THAT SINGLE WATER BATH!
And I’m the one who’s reading comprehension skills are lacking?
Says the guy who thinks it is impossible to average temperature…
It’s impossible to average intensive properties of multiple measurands!
What is it that you don’t understand about that truth?
In order to create a physical average, you must be able to add the individual component’s property together to build a larger system that can be characterized. You can’t do that with intensive properties. Temperatures do not add into a larger system. 20C + 20C ≠ 40C.
What is it that you don’t understand about that?
When you are averaging the measurements of a single measurand you are averaging the MEASUREMENT values to get a better estimate of the measurand property, your are not averaging the intensive property itself.
What is it that you do not understand about that?
Thank you for confirming that you indeed ignore most everything in the GUM by cherry picking bits here-and-there.
How hard is it to understand that you can’t add two rocks at 20C each together to get a temperature of 40C?
Tracking a non-physical statistical descriptor over time doesn’t describe reality, all it does is track the statistical descriptor of the non-physical data set over time. That doesn’t mean you are tracking a physical property. If you aren’t tracking a physical property, then you can’t make any judgements about reality.
Numbers is numbers!
The truth is, he doesn’t care that the invention of the term “forcing”, denominated in the ludicrous units of Watts per meter squared is meaningless.
The truth is, he doesn’t care one bit that Spencer’s reporting of microwave radiance data averages to 1 mK is meaningless.
And the truth is, deep down he just wishes measurement uncertainty had never been developed.
It is all too inconvenient to his warmunist worldview.
“Numbers is numbers!”
Yep. Statistical World is *not* congruent with Real World unless you are a statistician instead of a physical scientist.
“And the truth is, deep down he just wishes measurement uncertainty had never been developed.”
Exactly!
You are incorrect. A measurement is a result of evaluating a measurement model that is a functional relationship.
An average (mean) is not a functional relationship. It is a statistic, not a mapping. A functional relationship is a rule y=f(x) that assigns one output to each input or group of unique inputs. A mean is a summary operator applied to a set of values. It does not map inputs to outputs in the sense required for a function describing a physical or mathematical relationship.
You are confusing statistical formula with a functional relationship that can predict output values from different inputs. An average can not define the relationship between input quantities.
A functional relationship must connect variables, not data points. A mean is a property of a dataset, not a law of nature. You can say “the mean is a function of the dataset,” but that is a set‑function, not a functional relationship between physical quantities.
When you say f(x) = (1/n) ∑(x1/n, x2/n, …, xn/n) you are declaring x1/n as a unique observation of an input quantity. The same for x2/n. You might explain why scaling an actual measurement like x1 or x2 is a functional relationship related to the other input quantities. I am sorry but you are trying to cram a simple statistical formula into a functional relationship.
Here is what the GUM says:
4.1.1 In most cases, a measurand Y is not measured directly, but is determined from N other quantities X1, X2, …, XN through a functional relationship f :
Each “input quantity”, X1, X2, Xn, is a unique and standalone physical quantity that is physically measured. X1 is estimated by the value of x1, which itself may have multiple measurements of the same thing to determine its estimated value. Its measured value is not dependent on any other variable in the functional relationship like x1/n would be.
Scaling a measured value must have a reason for doing so. Otherwise, one could make the argument that the height of a single rafter is 6 ft/10 because there are 10 of them. Or that a temperature at a given station, on a given day, at a given time is T/n because I have n different measurements from n different stations, on different days, and even at different times.
“The GUM states that an average can be a measure of a measurand”
As usual, your reading comprehension skills are sadly lacking.
The word MEASURAND is singlular – meaning ONE MEASURAND. The implication being that you are measuring the INTENSIVE VALUE of one single system, not a larger system consisting of multiple smaller systems.
This epitomizes why I rarely engage with you. I cite sections of the GUM that unequivocally say an average can be a measurand and you’re knee-jerk reaction is to then accuse me of lacking comprehension. Anyway, maybe your grievance isn’t with me, but with the GUM since you are constantly challenging what it says.
The issue isn’t averaging measurements of a single measurand to determine the best estimate of its property.
The issue is averaging measurements of an intensive property for MULTIPLE measurands to develop a characteristic intensive property for the larger system consisting of the multiple measurands.
All you are doing here is equivocation.
Merriam-Webster: equivocation – deliberate evasiveness in wording : the use of ambiguous or equivocal language
You keep trying to convince everyone that if you can average measurements of a single measurand to get a “best estimate” of the intensive property of that SINGLE measurand, then you can also average the measurements of the intensive property of MULTIPLE measurands to get an average value for the intensive property across a larger system consisting of all of the measurands.
You are being deliberately evasive in wording by trying to make the issue into averaging multiple measurements of a single measurand when the issue is multiple measurements of multiple measurands.
You can only average the properties of multiple measurands if you can combine those individual measurands and properties into a larger system. You cannot do that with intensive properties like temperature.
What is it that you can’t seem to understand about something so simple?
Do you *really* believe that if you have two rocks in your hand at 20C each that you are holding 40C in your hand?
Funny that I can physically sense temperature (that parameter receiving prominent, repeated mention in discussions under this article), whereas I simply cannot physically sense mathematics. Oh well.
/sarc
Climate science believes that you can add the temperature of a cubic meter of air on Pikes Peak with a cubic meter of air at Colorado Springs and form a larger system whose temperature can be averaged and used to describe the climate of the area encompassing both sites.
mass1 + mass2 = mass_total
temperature1 + temperature2 ≠ temperature_total
If I have you close your eyes and I put a 10g rock and a 20g rock in your right hand and two rocks of 15g in your left hand can you tell the difference in the weights you are having to hold up with each hand?
If I put a rock at 10C and a rock at 12C in your right hand and two rocks at 11C in your left hand does each hand experience a total of 22C? Can you say that the temperature of the rocks in each hand total to 22C?
You mistake me for being a scientific measuring instrument.
The new Monckton Pause extends to 39 months starting in 2023/02. The average of this pause is 0.55 C. The previous Monckton Pause started in 2014/06. It lasted 107 months and had an average of 0.21 C. That makes this pause 0.34 C higher than the previous one.
+0.156 ± 0.038 C.decade-1 k=2 is the trend from 1979/01 to 2026/04 covering 568 values.
+0.026 ± 0.010 C.decade-2 k=2 is the acceleration of the trend.
The value of the “pauses” is that they disprove that increasing CO2 during the pause has
noany effect. It indicates that natural variation has much larger effect on temperature than does a single variable of CO2.(edited to correct)
As I’ve told you before this is the reduction fallacy. Ironically it is this fallacy that provided the impetus for me to develop the graph above.
The “reduction fallacy” is when someone tries to explain a complex system entirely as a function of one of its parts – i.e. that climate is determined by temperature. Which is what your graph attempts to do. Heal thyself, physician.
Lurking in on these protracted, repetitive, monthly exchanges is one of my bad habits. So, wisdom of crowds. Do they die down on their own, or are they stopped after X days?
Comments are closed after 2 weeks. Usually just after Tim has written a mammoth comment accusing me of all sorts of nonsense.
And to prove my point, Tim is silent for a couple of days, then waits to the day before comments will close to fire half a dozen comments directed at me, full of insults and lies.
I should just ignore them, but I just don’t like the way the personal comments will be left unanswered.
For your infomation, I was with my wife at the cancer center and/or the eye surgeon for most of that time. If I wasn’t there I was working on jewelry commissions. I simply don’t have the time to continue to try and educate you on metrology. It seems to be a wasted effort in any case.
I’m really sorry to read that. Best we leave this here, but try to stop making this so personal.