Worst Frosts Hit Brazil Adjusted Away!

“Averaging discrete information is already a fools game, like averaging phone numbers or street addresses”

1. Temperature is continuous, not discrete.
2. You can meaningfully average discrete numbers. e.g. number of children, or the value of a die roll.
3. The reason you cannot meaningfully average phone numbers is because they are categorical data. They are just numbers being used as identifiers. You cannot meaningfully add one to a phone number, or half it, it doesn’t mean those operations are a fool’s game.

1. They are discrete measurements of a continuous function, that don’t make them telephone numbers. By this logic all averages are a fool’s game because all measurements are discrete.
2. By all means do uncertainty analysis on global data sets, but the uncertainty will have little to do with the uncertainty in individual temperature readings. And again, you are wrong to suggest averages tell you nothing.
3. You don’t add two numbers to get an average, you add them and divide by two, When you average two numbers the average will be the mid-point. Nobody says that the mean temperature derived from max and min values is going to be exactly the same as a mean derived from the integral of temperatures throughout the day. It’s just a convenient approximation given the data available. By the way, do you still believe that the correct way to get a daily mean temperature is to multiple the maximum by 0.63?
4. And yet more assertions about what you cannot do with an average. Nobody is trying to take an average of the temperature in two different locations, but that doesn’t mean such an average will tell you nothing. If I take readings from those two places in summer and in winter, and find the winter average is colder than the summer do you not think I can determine something useful from those averages. I’m not interested in the global average temperature as such, what’s interesting is how has it changed. If the average of of all temperatures readings from different locations is higher than they were 30 years ago, that tells me something irrespective of the geographical and topographical differences.

“We have had this argument before. Uncertainty *GROWS* when you combine unrelated measurements, it doesn’t disappear.”
“The uncertainty *grows* when you add numbers. Dividing by a constant doesn’t affect that uncertainty. ”

Yes we keep having this argument, because you are unable to accept you may be wrong yet are unable or unwilling to provide any type of evidence to show why I’m wrong. I’ve showen the very books you use to argue your point show the opposite, I’ve poited out the absurdities your claim would lead to, and I’ve given empirical evidence that demonstrates you are wrong. Did you see my example of estimating π using just 2 discrete values, and how the estimate becomes more certain as sample size increases?

Let’s try another example using your point about the area under a sine wave. Calculus says that the area under a sine wave between 0 and π is 2, and dividing by π gives us approximately 0.637, which is almost what you claimed. If I’m right it and averages means something then it should be possible to get an estimate for this value by taking random samples. I also claim that as sample size increases the uncertainty of the estimate will decrease. I also claim that adding uncertainty will not have much of an impact on the estimate especially as the sample size increases.

My understanding of your point is that the average will be no different than averaging telephone numbers, that the uncertainty will increase as sample size increases, and that the average will tell you “absolutly nothing”. Is that what you would expect?
Do you want to do the experiment?

“I’ve given you the evidence. It’s right there in Taylor’s book. All you have to do is read it.”

No you haven’t. You keep pointing me to Taylor and then ignoring what he says, including the parts where he directly points out why you can divide uncertainty. I want you to point me to where Taylor says – a) uncertainty grows as sample size grows, and b) dividing a measure by a constant does not reduce the uncertainty.

“What you are doing is trying to say you can sample different things and combine them as if they are samples of the same thing.”

Yes I am. That’s because they are samples of the same thing, that same thing being the population mean. You keep failing to undersatnd that when you calulate an average you are not trying to estimate one individual measurement,you are trying to estimate the mean of something. In the case of temperature it might be the mean temperature over an area, or the mean temperature over a period of time, or a combination of both. I don’t care what one specific measurement is, excpet that it is a sample of the mean. That doesn’t mean that if I am interested in a specific value at a specific time I cannot go back to the sampe, but if I want to now what the mean is that is what I am trying to estimate.

This is in principle no different to what Taylor does in the section where he shows how you can take multiple measurements of the length of a piece of metal to get a more accurate measurement. He specifically says to measure it using different instruments and to measure different parts. Why does he say that if you don’t expect different measurements to give different results? What is the final measurement of? It cannot be any one part of the metal because then you would only need to measure at that place. The final average measurement is trying to measure the average length of the sheet.

“You implied you could model a continuous function using just two values.”

I don’t think I’ve said that, but you do keep changing the discussison. What I’ve been talking about is the way uncertainty of a mean changes with sample size. You keep changing this to the mean of temperature over a day being aproximated by the mean of the max and min values. I don’t think you can model the daily temperature cycle from just two values, just that it’s the simplest way of estimating the daily mean if you only have those two values. There might be a lot more to be said about how to best estimate the daily mean given two values, but I think it’s a distraction from what we were discussing, how sample size changes the precision of a global or monthly mean.

Part two.

So what if we add some uncertainty to the values, and what if the values are more discrete? Will the uncertainty in the mean increase as sample size increases?

I now repeat the exercise but this time round the samples to 1 dp, so each value is sine(x)±0.05. As the values only go from 0 to 1, that only leaves 11 possible values for each sample, so it is pretty discrete.

N = 10

  Mean  Error
  0.740  0.103 
  0.720  0.083
  0.660  0.023
  0.600  0.037
  0.770  0.133 
  0.750  0.113 
  0.880  0.243 
  0.520  0.117 
  0.500  0.137 
  0.800  0.163 

N=100

  Mean   Error
 0.679  0.042 
 0.649  0.012 
 0.715  0.078 
 0.628  0.009
 0.691  0.054 
 0.626  0.010 
 0.645  0.008
 0.648  0.011 
 0.648  0.011
 0.643  0.006

N=10000

  Mean    Error
 0.634  0.002
 0.641  0.004 
 0.635  0.002 
 0.634  0.003 
 0.641  0.004 
 0.637  0.001
 0.637  0.001
 0.639  0.002 
 0.636  0.001
 0.638  0.001

Even using discrete values, with an uncertainty of 0.05, using a large enough sample size gives results much better than the uncertainty of any individual measurement. And of course what does not happen is the uncertainty increases as sample size increases.

Note:
Although the rounded figures do give values that are close to the expected value, the rounding does introduce a systematic error. The average of all rounded values will give us the average of an approximation of the sine wave, but it happens to be a reasonable approximation. Using a much larger sample size (N=1000000) all the results come back as 0.638 to 3dp, out by 0.001. The mean is precise but not completely true.

“You tried to imply you could use two measurements to establish a sine wave”

No I haven’t. You’re the one who keeps wanting to talk about the daily mean values. All I’ve been trying to do is establish that the uncertainty in a mean decreases as sample size increases. Daily mean temperatures are a distraction from that you keep bringing up. You were the one who wants to treat the daily temperature cycle as a sign wave and derive the mean daytime average by multiplying the max by 0.63.

“…an uncertainty of +/- 0.5C if not higher! That is *far* greater than the uncertainty you are attributing to your sample data!”

The exact value doesn’t matter, the 1dp is with regard to a value that moves between 0 and 1.

“Averages of different things can *NOT* tell you anything.”
“The *mean* of the data set can be calculated more and more accurately with more samples IF YOU ARE MEASURING THE SAME THING!”

You really need to explain what you mean by “different things” verses the “same thing”. If the sine wave moves from 0 to 1 and back is it the same thing or different things? If the temperature changes from day to day or across the globe is it the same or different things? If I measure a sheet of metal at different points with different instruments am I measuring the same thing or different things? If I measure a stack of paper and use that to calculate the thickness of a single sheet of paper, am I measuring the same or different things? If I measure the number of days between babies being born in a hospital am I measuring the same or different things?

“You still won’t know what length each board is by looking at the average!”

No, of course you don’t. That is why an average is not a list of thousands of different values, it’s a summary of them. Not knowing what every value that makes up an average is, does not mean the average tells you nothing. Pointing to examples of averages that are not very useful does not mean that all averages are useless.

I told you several times before about cooling degree-days and heating degree days. These are integrals of the temperature curve above and below specific set points (e.g. 65F). These values give a *much* better picture of the climate at a specific location than a mid-range value.

Dawn and dusk are arbitrary points on the daily time line. Adjust them as you want. It is still the integral of the temperature curve that defines climate, not the mid-range value.

The point is (that you seem to be actively trying to convince yourself that somehow, some way the uncertainty of measuring two different things can have the uncertainty of their additive/subtractive sum somehow divided by two so the uncertainty can be lessened instead of growing. Thus leading to the conclusion that if you just have enough samples from different things you can eliminate the uncertainty associated with adding/subtracting them by dividing by the number of samples. The law of large numbers *only* work to lessen uncertainty when you are measuring the same thing, not different things. Maximum temp and minimum temp are TWO DIFFERENT THINGS. Each has an uncertainty and when you add the two together the uncertainty grows by at least sqrt(2) if not a direct addition. You don’t divide that uncertainty by 2 because you have two samples!

Stop trying to convince yourself that mid-range values hold some magic meaning. They don’t. They aren’t even a good representation of the climate at even one location because you lose the data telling you the min and max temps which is a much better representation of the climate at a location.

*IF* day and night are equal in length then .63 * Tmax gives a much better representation of the daytime climate than a mid-range value. Same for night – .63 * Tmin gives you a much better representation of the nighttime climate. If day/night intervals are not the same then it just complicates the calculation but it doesn’t invalidate it. It remains a much better representation of climate at a location than a mid-range value. And it doesn’t lose data, you can still find Tmax and Tmin which you can *not* do with a mid-range value.

We have had the ability to collect 1-minute temperature data for at least two decades, if not longer. That would allow a much better representation of the actual temperature curve and would allow a numerical integration of the curve at least. There is no doubt that it would complicate the models and the modelers tasks but that should not be used as an excuse for not moving to a better representation of climate from the models. My guess is that the reason this isn’t being done is because it would also show just how badly the models are at actually predicting future temps and climate!

“Yes, and this has nothing to do with multiply the max temperature by 0.637.”

Do you understand what you are saying? The average of a sine wave *is* .637 * Amplitude. It is the area under the sine wave from 0deg-180deg. What do you think cooling degree-days and heating degree-days *are*?

Cooling and heating degree-days were developed for use by engineers trying to size HVAC systems. If the temperature never goes above the set point then that tells the engineer that air conditioning isn’t a real need at that location. I.e. the cooling degree-day value is ZERO. Same for heating degree-day values.

“But if you can assume the day follows the sine wave, then the mean derived by finding the mid-point between max and min will also be just as good an estimate of the true mean”

Not for evaluating climate! Two different locations can have the same mid-range value while having vastly different climates! It is the average, i.e. the cooling degree-day value, that tells you what the daytime climate is. Same for nighttime. The mid-range value is useless – as is the GAT!

“I’m talking about the mean of global anomalies or the means of monthly or annual anomalies, not using two samples, but thousands.”

Anomalies derived from mid-range values are just as useless as the mid-range values the anomalies are calculated from. And, once again, you have thousands of samples THAT AREN’T MEASURING THE SAME THING! The uncertainty grows with root-mean-square and is *NOT* divided by the number of samples! Why is that so hard to understand? Taylor explains it magnificiently!

“Rather than trying to convince me, you should try to convince the authors of every text book on statistics, who all say it does.”

Uncertainty is *NOT* a probability distribution and is, therefore, not subject to statistical analysis. If you read the statistics books they speak to a data set that is a random, *dependent” data set – i.e. measurements of the same thing. This is subject to statistical analysis. The uncertainty of random, independent measurements is simply not the same thing.

“A mid-range daily mean is not meant to be representing the daytime temperatures, it’s representing the entire 24 hour period – day and night.”

Then how does it tell you anything about climate? Two vastly different locations can have the same mid-range value. How do you differentiate the difference in the climate at each location? The mid-range value certainly won’t tell you!

“Which is great as long as you don’t want to compare them with temperatures from the last century.”

Why is that necessary? Why not just track the values over the past twenty years? The past is the past. You can’t change it. Knowing it won’t actually tell you what is happening *now*. I would much rather have a more accurate picture of what has happened over the past twenty years than an inaccurate picture of how the past compares to today!

Climate is determined by the daytime heating and the nighttime cooling. It is *NOT* determined by the mid-range temperature. It truly is that simple. Two different locations with vastly different climates can have the same mid-range value, the mid-range value tells you nothing about the climate at each location.

The daytime temp is pretty much a sine wave. So is the nighttime temp. Depending on things like wind, humidity, and clouds each may not be a *perfect* sine wave but they are pretty close. They are certainly not triangle waves or square waves or any other commonly understood types of waves. The daytime temps and nighttime temps are close enough to sine waves that .63 * Tmax or .63 * Tmin will get you pretty close to the *average* daytime or nighttime heating/cooling as measured using temperature as a proxy.

“ what do you think it would mean to say that the average day time temperature was around 9.6°C and the average nighttime temperature was 3.2°C? What would be the difference if you measured the max and min in Fahrenheit or Kelvin?”

What do you think the mid-range value would tell you? If the daytime/nighttime temps are close to a sine wave then I can tell you the Tmax and Tmin values. You can’t do that with a mid-range value or at least I know of no way to do so.

9.6C is about 50F. Divide by .63 and you get a Tmax value of about 79F. A nice daytime temp. 3.2C is about 40F, or a Tmin of about 65F. A nice nighttime min temp. About what we are seeing here in Kansas right now. It’s a nice climate to be in (except for the humidity).

Do that from a mid-range value.

“Yes, as can two locations having the same CDD”

You continue to demonstrate that you don’t understand what an integral is. How can two sine waves of different amplitudes have the same area under the curve? As you pointed out earlier the integral of sin(x) evaluated from 0 to pi equals 2. But the function we need to evaluate is Asin(x), not just sin(x). So the integral becomes 2A. So how can two locations have the same area under the curve unless A_1 and A_2 are the same? Since the average value is .63 * A how can the average value of two locations be the same unless they have the same value for A?

” But if I’m not interested in specific climates but just in the question is the world getting warmer then can be quite useful.”

What do you mean by “is the world getting warmer”? Every alarmist says it means that maximum temps are going up which is what causes the mid-range value to go up. But it can just as easily mean that the minimum temps are going up. Do you *really* care if minimum temps are going up? How many alarmists are going to say that droughts are increasing because minimum temps are going up? Who would listen to them? Higher minimum temps have all kinds of benefits such as longer growing seasons, higher plant growth at night, more food for humans and livestock, fewer homeless people in San Francisco and Seattle expiring on the street from hypothermia, etc. What *bad* impacts do *you* see from higher minimum temps?

How do you distinguish what is actually happening from the use of mid-range values? How do you tell exactly what is warming? Cooling/heating degree-days *will* tell you, mid-range values will not.

“I’ve shown empirical evidence that this is not true. I’m still waiting for your evidence that it is true.”

No, you haven’t given us any evidence. You would have us believe that you can take two independent, random boards whose length has been measured by two different device with possibly different uncertainties, lay them end to end and have the uncertainty of overall length go DOWN by a factor of two!

Such a belief flies in the face of rational thinking. Look at Taylor on page 57: “When measured quantities ae added or subtracted, the uncertainties add; when measured quantities are multiplied or divided, the frational uncertaies add. In this and the next section, I discuss how, uncer certain conditions, the uncertainties calculated by useing these rules may be unessarily large. Specifically, you will see that if the original uncertainties ae INDEPENDENT and RANDOM (caps are mine, tpg), a more realistic(and smaller) esitimate of the final uncertainty is given by similar rules i which the uncertainties (or fractional uncertainties) are DDED IN QUADRATURE (caps are mine, tpg) (a procedure defined shortly).

You want us to believe that you can take 1000 measurements of 1000 independent, random boards, each measurement with its own uncertainty (e.g. +/- 1″), lay them end to end and that the final length will have an uncertainty of 1/1000″. Simply unfreaking believable.

Remember., this is what you are doing when you calculate an average, you are laying boards end-to-end to get a final result and then dividing by the number of boards. That is no different than laying temperatures end-to-end to get a final result that is then divided by the number of temperatures you use. In both cases that average tells you nothing about the boards, you may have a bunch of short ones (e.g. nighttime minimum temps) and a bunch of long ones (e.g. daytime maximum temps). The average is meaningless and useless for trying to describe anything about the boards! Just as it is for trying to describe something like a “global average temperature).

“As I keep trying to tell you, the uncertainty in the mean of different things is mostly due to the sampling.”

Uncertainty of the calculation of the means is *NOT* the same thing as the uncertainty associated with the combination of the data. You *can* calculate the mean ever more accurately but it is only meaningful if you that data is associated with the SAME THING. Those measurements are then many times considered to be a probability distribution associated with the same thing. When you have DIFFERENT THINGS, no probability distribution is associated with the data, therefore calculating the mean more and more accurately is meaningless once you go past the uncertainty interval which is the root-sum-square of the uncertainties of the independent, random multiplicity of the things you have stuck into the data set. You can *NOT* decrease that uncertainty by adding more data, that just increases the uncertainty of the final result! Just like laying random, independent boards end-to-end. The more boards you add the more uncertain the final length becomes!

“And you still don’t seem to grasp that just because something doesn’t tell you everything, it does not mean it tells you nothing”

Mid-range temperatures tell you nothing about the climate. You *still* haven’t show how you can determine minimum and maximum temps from a mid-range value. If you can’t do that then you know nothing of the climate associated with that mid-range value! In other words, it means nothing. And if the data set you are using is composed of values that mean nothing then the end result means nothing either!

“The uncertainty in the measurements is usually of little importance”

Have you *ever* framed a house? Have you *ever* had to order I-beams for a bridge construction? Have you *ever* designed an audio amplifier for commercial use using passive parts? Have you *ever* turned the crankshaft journals in a racing engine on a lathe? I have.

The uncertainties in the measurements of all of these is of HIGH IMPORTANCE if you are going to do a professional job. These are just a sample, the number of situations in the real world where the uncertainty of measurements is of high importance is legion!

You are the typical mathematician or statistician who’s work product has never actually carried some real world liability for you if it isn’t done properly. You would *never* make it as a professional engineer!

“If I know the mean temperature of a place in December is -5°C, and the mean temperature in July is 20°C, can you not deduce something about the climate of the place during those two months?”

You can deduce a seasonal influence, that’s about all. If the mean temperature in July is 20C what is the maximum temp associated with that? What is the *mean* maximum temperature? If you don’t know those then how do you judge what is happening to the July climate? If next year the mean temperature in July is 21C how do you know what caused the increase? Did max temps go up? Did min temps go up? Was it a combination of both? If you can’t answer these then how do you judge anything about the local climate?

“You keep saying I don’t understand calculus, but when you first made this claim, I showed you the integral, explained why you were wrong, and suggested what you might be grasping at. Yet you still keep repeating this meaningless claim that 0.63*TMax is close to the daytime average, etc. Explain why I’m wrong, or show your workings.”

You didn’t do *any* of that. You just showed that you don’t know what an integral actually is. You can’t come up with the same area under two sine curve unless the amplitude of each curve is the same!

“Explain why I’m wrong, or show your workings.”

I’ve explained it over and over and over and over and over till I’m blue in the face. CLIMATE IS THE ENTIREITY OF THE TEMPERATURE CURVE. Climate is not defined by a mid-range value. Two different locations can have the same mid-range value while having different temperature curves and different climates. Trying to define climate using mid-range values is just a joke on the uninformed.

The average value of the daytime sine wave defines the entire daytime sine wave. From it I can calculate the maximum temp. Same for the nighttime sine wave. I can tell you if maximum temps are going up/down, I can tell you if nighttime temps are going up/down, or if it is a combination. I can tell you immediately what is happening to the climate at a location. *YOU* can’t do that with your mid-range values therefore you can’t tell what is happening to the climate. Since mid-range values contain no information on the climate then combining a bunch of mid-range values to form another average won’t tell you anything about the climate either!

“So how do you define “average” so that the average daytime temperature is less than the coolest part of the daytime (assuming daytime starts and ends at 20°C) and even more impressively how the average nighttime temperature is less than the minimum temperature for the whole day.”

You *really* don’t understand integrals at all, do you? The integral of sin(x) from pi to 0 is a -2. Divide by pi to get the average value and you get a -.63. So the average nighttime value is (-.63) * Nmax. Since you are multiplying by a decimal then how does the average nighttime temp wind up being lower than Nmax?

And how did you come up with the daytime start and end points are 20C when you said that the minimum daytime temp was 10C? An average of 19C is certainly between 10C and 30C! Slow down and check your work!

“And if you don’t mean average day or night time temperatures but CDD and HDD then explain how the colder the minimum temperature gets the lower the HDD, or how you can have a negative CDD or HDD.”

Wow! You haven’t studied up on anything I’ve given you, have you? CDD and HDD ARE NOT AVERAGE VALUES. They are the area under the curve defined by the set points you pick! You don’t divide by anything to get an average. You just get the integral value – the area under the curve.

A sine wave with an average of 9.6 will have a maximum of 15C. That’s about 60F. You are right. I shouldn’t have changed scales. But it doesn’t alter my point at all! I can calculate the max temp from the average value. *YOU* can’t do that with a mid-range value. Same for the nighttime temps.

And that is *still* the whole point. Mid-range values are useless for describing climate. They contain no information about climate.

“You still don’t get that you are integrating a sine centered on zero and that therefore results will be different for a temperature profile that is not centered on zero, that is nearly all of them.”

You’ve just hit on one of the major problems with climate models today even though you probably don’t understand it.

What does (Tmax-Tmin)/2 trend toward in the limit? It tends toward ZERO. As the daytime and nighttime temperature excursions get closer together the mid-range tends toward zero instead of the absolute temperature. And that is true no matter what scale you use, celsius, fahrenheit, or kelvin. The mid-range temperature value has an in-built bias that can’t be eliminated.

So why do so many climate scientists, mathematicians, and statisticians remain so adamant that it properly represents the climate anywhere, let alone the global climate?

“What, now you want to multiply by a negative number? What is Nmax?”

What is the integral of sin(x) from pi to zero? Do I need to work it out for you? Nmax is my shorthand for the maximum nighttime temperature excursion. Same as Tmin.

The integral of sin(x) = -cos(x). Evaluated from pi to 0 you get -cos(0) – (-cos(pi)). -cos(0) = -1. Cos(pi) = -1. So you get -(-(-1)) for the second term or -1. -1-1 = -2.

Did you *really* take calculus in school?

“But you get completely different values if you do this with Fahrenheit, Celsius or Kelvin.”

I also get different measurements when I use the different scales. You won’t get the same error that you see when converting between scales. And nothing will change the fact that mid-range values trend toward zero as temperature excursions trend toward zero – an in-built bias from using mid-range values. Think a 24 hour blizzard in Siberia where the temperature might only change 1C from daytime to nighttime. The absolute temp might be -20C to -21C. You wind up with about a 20C bias in the mid-range value. How do you overcome that?

“You keep doing this, changing the subject. In this comment I’m not interested in whether an average daytime value would be better than a mean. I’m simply trying to help you understand why you cannot multiply Tmax by 0.637 to get a daytime average.”

Of course you can! The only reason you can’t would be if the temperature curve does not approach a sine wave. Daytime temps are mostly controlled by the sun. The angle of incidence from the sun to the earth is a sine wave so the temperature naturally tends to follow that same sine wave. You get the largest contribution to temperature when the sun is overhead and the sin(90) = 1. At sunrise and sunset the angle of incidence approaches 0 and the contribution of the sun to surface heat is sin(0) = 0. Of course the surface temp lags the actual sun position because it takes time for the heat input to actually result in a temperature rise.

I’ve attached a graph of our past weeks temperatures (or at least I’ve tried. I don’t know why it isn’t showing up. I’ll do it again in a separate message) If that temperature envelope doesn’t look like a sine wave (distorted perhaps but still some kind of a sine wave) then I don’t know what it looks like. If you want to quibble about the actual value of the integral then have at it. It might be 0.5 or 0.7 or something else. But it is *still* better than the mid-range value for representing the actual climate!

You’re right, my bad. The mid-range value is *still* useless in describing anything to do with climate. It *loses* data. You cannot reproduce the temperature curve from the mid-range data. If you can’t do that, then it is of no use.

BTW, you never addressed the fact that my graph shows that the temperature curve approaches a sine wave. Which begs the question of why climate scientists refuse to move to a metric that actually describes the climate.

Is it just for the funding and the ability to use it to scare people?

“I’m not denying that a sine wave might be a reasonable model for a daily cycle.”

Then why did you say:

“Says the person who wants to model every day by a perfect sine wave.”

“The point is irregardless of the shape of the wave you cannot ignore the displacement from zero, so you cannot simply multiply the max by 0.637 to get an average.”

As I told you before, if there is a DC component then subtract it out, calculate the average, and add the DC component back in.

Again, the mid-range value is *NOT* the average value of a sine wave.

“But if a sine wave is a good fit for a daily cycle, it also means that the mean derived from the average of max and min is a reasonable approximation of the actual daily mean.”

No, it isn’t! The amount of daytime heating is the integral of the temperature profile during the day. It is *NOT* the mid-range value between Tmax and Tmin. The mid-range value is *NOT* an average value. It is not a mean. It is a mid-range value. You keep mixing up terms. Is that on purpose?

“The only reason you can’t would be if the temperature curve does not approach a sine wave.”

Yet again, it’s not the shape of the sine wave that’s the problem, it’s the translation.

here is the graph.

“Where’s 0°F on your graph? What does an average daytime temperature of 85*0.637 ≈ 54°F look like on your graph?”

It doesn’t matter where zero is. The graph shows that the daily temperature curve. It *is* close to a sine wave albeit with some distortion.

If it has a DC component you don’t like then subtract it out!

*I* don’t have a problem at all. I’ve analyzed waveforms my entire life, pure and distorted.

You are looking for any excuse you can find to show that a mid-range temperature (an average!) has some meaning. And that an average of mid-range temperatures has some actual meaning in the real world. And none of your excuses have any bearing on the issue at all. You can’t even tell the difference between dependent and independent measurements, i.e. multiple measurements of the same thing and a set of measurements of different things!

*That* is where the problem lies. You have a hammer and see everything as a nail, refusing to admit that screws exist. They are just another nail to hammer in!

Part 2

“So how can two locations have the same area under the curve unless A_1 and A_2 are the same?”

You are talking about CDDs here, they are not the area under the sine wave, they are the area under the positive part of the sine wave minus the magic number. And of course, the daily temperature cycle is not Asin(x), its Asin(x) + m, where m is the mean temperature.

The most obvious way two different maximums can both have the same CDDs is if neither reach the magic number, then the CDDs are both zero. But for days when there is cooling, being able to play with both the amplitude and the displacement can easily result in similar CDD values for different max and min values.

“Minus what magic number?”

By Magic number simply meant whatever value you are using as the base line.

“And they *are* the area under the curve – that *IS* the definition of an integral.”

I take it you didn’t read the rest of my sentence where I said “they are the area under the positive part of the sine wave minus the magic number.”.

“Actually the CDD is the Asin(x) – Asin(0).”

What fresh nonsense is this? sin(0) = 0, for all values of 0.

“Zero is the baseline, not the mid-range value”

Come again? You’re using 0°C as a baseline for CDDs? How cold do you want your buildings?

“Remember, degree-days are *not* an average.”

They’re an average of temperatures, over the baseline, minus the baseline per day.

“They are not divided by time interval to get an average.”

I’m assuming that they would actually be calculated from readings taking at set times, every hour, half hour or minute. The shorter the interval the closer the average will get to the integral (that’s the definition of an integral), but I can’t see how you could do an actual integral as you have no way of knowing the true function, and any sampling method will be more accurate than simply assuming it is a sine wave.

“They are not divided by time interval to get an average.”

Oh yes they are. If you have 24 hourly readings you have to divide the total degrees above the baseline by 24 to get the value in degree days. If not you would have Cooling Degree Hours. Even if you take the integral you are still dividing the sum by the number of time intervals, it’s just that the number tends to infinity. (You also have to divide it by whatever scale you are using, so if for example you are modelling a sine wave from 0 to 2pi to represent the day, you have to divide the area under the curve by 2pi.)

“ I trust actual engineers whose personal liability depends on evaluating climate in the real world far more than a climate scientist, mathematician, or statistician whose connection to the real world is tenuous at best and has no personal liability at stake.”

Says the person who wants to model every day by a perfect sine wave.

“To get an average you would have to divide by pi, the total interval over which the integral is done.”

You do have a habit of repeating what I said earlier as if I don’t understand it.

All this stems from you saying CDDs are not an average. I say they are. If you calculate them using any sampling technique you have to divide the total through by the number of samples. this is true even if you use an integral, it’s just you are dividing through by an infinite number of samples. Of course, if you estimate the CDD using a sine wave function you also need to divide by 2pi, because a day is 1 day, not 2pi days.

“Perhaps the sin(x) is confusing you. x is not time, it is radians or an angle, e.g. theta.”

If you are saying the sine wave represents the daily temperature cycle, then x is representing time, it’s just measuring time in unusual units, where there are 2pi radians in a day.

I’m still interested in what point you are trying to make throughout the CDD discussions. First you were saying that we have instruments that can measure at 1 minute intervals, and we should use these to calculate CDDs and throw out any data that isn’t based on high frequency measurement, but now you are happy to estimate all CDDs based on max and min and on the assumption that daily temperature is close to a sine wave. I don’t have a problem with either, but you now seem to be suggesting the estimate is more accurate than the minute samples.

What makes you think the met office method is using a mean temperature? Did you not read what I quoted?

““The third is the Met Office Method – a set of equations that aim to approximate the Integration Method using daily max/min temperatures only.”

The method uses MIN AND MAX temperatures as input to a set of equations meant to account for the distorted sine wave shape in calculating degree-day values! Min and Max are not means!

tim: ““To get an average you would have to divide by pi, the total interval over which the integral is done.”
bell: “You do have a habit of repeating what I said earlier as if I don’t understand it.””

Really? I repeated what you said? Here it is –

tim: “Remember, degree-days are *not* an average.”
bell: “They’re an average of temperatures, over the baseline, minus the baseline per day.”

The integral, i.e. the degree-day value, is *NOT* an average. Are you now changing your story?

“All this stems from you saying CDDs are not an average. I say they are.”

They are *NOT* an average. I showed you that using dimensional analysis of the integral. The integral determines the area under a curve. That is *NOT* an average. It is a total. To get the average you have to divide by the interval on the horizontal axis. Degree-days values are not divided by the interval.

“If you calculate them using any sampling technique you have to divide the total through by the number of samples.”

When you calculate the area of a tabletop, length times width, do you DIVIDE by the length of the table? Do you divide by the width of the table? Why would you *have to* divide the area under a curve by the interval being integrated?

” this is true even if you use an integral, it’s just you are dividing through by an infinite number of samples”

I truly despair of teaching you calculus. My teaching skills just aren’t great enough.

Asin(x) * dx
/\ /\
height width

This is an area. When you move dx incrementally along the horizontal axis you create an infinite number of area totals. The integral SUMS these into an overall area. THERE IS NO AVERAGING.

“If you are saying the sine wave represents the daily temperature cycle, then x is representing time, it’s just measuring time in unusual units, where there are 2pi radians in a day.”

You are not measuring in unusual units. Go look up steinmetz and phasor representations of a sine wave. Nor does it matter that time is the independent variable. You *still* aren’t dividing by the total time when you integrate the temperature curve. Nor is it correct to try and calculate an integral of a sine wave from 0 to 2pi. You always wind up with zero area yet it is obvious that the result can’t be correct since there is obviously area under the positive half of the sine wave and the same for the negative half of the sine wave. You lose that information when you try to integrate from 0 to 2pi. You integrate from 0 to pi for the positive half of the wave and from pi-0 for the negative half of the wave.

“calculate CDDs and throw out any data that isn’t based on high frequency measurement, but now you are happy to estimate” (bolding mine, tpg)

The two bolded words are the operative ones to consider. If you don’t have a perfect sine wave you can still calculate a pretty close estimate through successive approximation. If you don’t have a perfect sine wave then estimating from Tmax or Tmin usually winds up being less accurate. But either method will still tell you more about the temperature curve than a mid-range temperature which tells you exactly nothing.

If the curve is a perfect sine wave then either method will give the exact same answer for both the total area under the curve and for the average value of the curve – which is *NOT* the mid-range value.

Part 3

“You want us to believe that you can take 1000 measurements of 1000 independent, random boards, each measurement with its own uncertainty (e.g. +/- 1″), lay them end to end and that the final length will have an uncertainty of 1/1000″. ”

No, no no. Summing 1000 measurements, measured with independent uncertainty of ±1″, will increase the uncertainty by the square-root of the sample size. In this case the uncertainty is around ±32″.

What I, and everyone else including Taylor, say is that when you divide the sum by 1000 to get the mean length, you also divide the uncertainty by 1000 – hence the uncertainty of the mean is the original uncertainty divided by the square-root of 1000, or about ±0.032″.

Now, I know, you don’t believe you should divide the uncertainty of the sum by 1000 to find the uncertainty of the mean. Which if true would mean that say, if every board was 1 yard long, the average length of the boards would be 1 yard ±32″, meaning the actual average could be just 4″ or almost 2 yards, despite all boards being measured as between 35 and 37″.

You don;t think this is a contradiction and insist Taylor supports you. But you quote refutes that – he says “when measured quantities are multiplied or divided, the fractional uncertainties add”. I’ve already explained in detailed algebra why that means that if you divide a measureand by a constant you can also divide the uncertainty by the same constant.

You’ve already provided part of the answer by pointing out that a constant has no uncertainty, hence if you have two measures A and B, and multiply them (or divide, it’s the same argument) to get a measure AB, than the fractional uncertainty is derived by adding the uncertainties of A and B. But if B is a constant it’s uncertainty is 0, so this is just the uncertainty of A. The fractional uncertainty of A remains unchanged when scaled by B, but as it’s a fractional uncertainty, the absolute uncertainty must scale in order to keep the same fraction.

But as I say, Taylor explains this himself on page 54, section 3.4.

“You left off the most important thing Taylor said:

“According to rule (3.8) the fractional uncertainty in q = Bx is the sum of the fractional uncertainties in B and x. Because delta-B=0, this implies delta-q/q = delta-x/x.”
”

I don’t know why you think that makes your point. It’s describing exactly what I’m saying.

“He multiplies by the absolute value of B in order to get the FRACTIONAL uncertainty, which is dependent on the value Bx.”

No he doesn’t. The fractional uncertainty remains unchanged between q and x. Therefore the absolute uncertainty has to be scaled by B. If the FRACTIONAL uncertainty scaled with B it would mean that if say the uncertainty of something was 1%, and you divided it by a 1000, the fractional uncertainty would decrease to 0.001%, which would make the actual uncertainty much less.

“Taylor is *NOT* scaling the uncertainty itself by the constant.”

Yes he is.

He spells it out in the following examples. Measure a stack of 200 sheets of paper with an uncertainty of ±0.1″, divide by 200 to get a measure of a single sheet of paper with an uncertainty of ±0.0005″. How do you thin he gets 0.0005, excpet by dividing 0.1 by 200?

“Again delta-q/q = delta-x/x”

Yes. That’s the whole point. The fractional uncertainties are equal. That means that if q > x, then it must be that δq > δx. If q = 100x, then δ = 100δx, if q = x / N, then δq = δx / N.

“The uncertainty of 200 sheets of paper is the sum of the uncertainties of each individual sheet!”

No it isn’t. Have you read Taylor’s example? He’s making a single measurement of all the papers stacked together. The uncertainty of that one measurement is 0.1″. It would be difficult to measure the width of a single sheet of paper to an accuracy of 0.0005″, and even if you could the uncertainty of adding together 200 sheets would follow the adding by quadrature rule, it would be 0.0005 * sqrt(200).

Read it closer. Take a look at Annex F.

“Components evaluated from repeated observations: Type A evaluation of standard uncertainty”
Repeated measurements of the same thing! This creates a probability distribution whose true value is the mean.

This doesn’t apply to single measurements of different things!

go to: https://www.cdc.gov/niosh/docs/2014-151/pdfs/chapters/chapter-ua.pdf

“Within the field of industrial hygiene, the quantities uj are often standard deviation component estimates obtained from a single measurement-method evaluation, rather than from replicates. When the estimates are independent, a combined uncertainty uc may be computed (through the propagation of uncertainty approximation) as:

𝑢𝑢𝑐𝑐 = �𝑢𝑢1 2 + 𝑢𝑢2 2 + ⋯”
u_c = sqrt(u_1^2 + u_2^2 …..)

The manual even speaks to expanding this value by a coverage factor to get what is called the “expanded uncertainty interval”. They do *not* say you should divide the root-sum-square by the number of measurements!

Why do you insist on ignoring the title of the Annex?

“Components evaluated from repeated observations”

Yes, you *are* missing something. You are reading what you want to see and not what is being said.

“ When the estimates are independent, a combined uncertainty uc may be computed (through the propagation of uncertainty approximation) as:
𝑢𝑢𝑐𝑐 = �𝑢𝑢1 2 + 𝑢𝑢2 2 + ⋯”
u_c = sqrt(u_1^2 + u_2^2 …..)”

I don’t see any division by N or sqrt(N) in this. Where do you see it?

What do I keep trying to differentiate for you? Independent, random measurements vs random, dependent measurements.

Taylor says direct addition of the uncertainties associated with random, independent measurements may yield a value that is too high so you should use root-sum-square. This document implies that root-sum-square may yield a value that is too low so you use an expansion factor to correct it. Take your pick but do so using sound engineering judgement.

I don’t know why I continue trying to educate you on uncertainty in physical science. You are just going to continue saying that the central limit theory applies to *all* cases and you can minimize uncertainty by increasing N. It’s sad, truly sad. You think adding more independent random boards to a data set will *decrease* the uncertainty of the final length of all the boards laid end-to-end. No amount of examples seems to deter you from that view.

Part 4

“Mid-range temperatures tell you nothing about the climate. You *still* haven’t show how you can determine minimum and maximum temps from a mid-range value. If you can’t do that then you know nothing of the climate associated with that mid-range value!”

You cannot determine the max and min from the mean. That’s not the purpose of the mean. The mean is a summary statistic. That does not mean it tells you nothing.

And as I keep saying, having an average does not stop you from looking at other statistics.

Part 5

““The uncertainty in the measurements is usually of little importance”

Have you *ever* framed a house? Have you *ever* had to order I-beams for a bridge construction?…

The uncertainties in the measurements of all of these is of HIGH IMPORTANCE if you are going to do a professional job.”

Way to take my comment out of context. When I said uncertainty in the measurements is of little importance I meant when determining the uncertainty of a sampled mean.

Of course, uncertainty is important for all those things, but not if you are deriving an average from a random sample. What determines the error of the mean is the randomness of the sample, the more variance in the population the greater the uncertainty of the mean, the more samples you take the less uncertainty. If the things you are averaging vary by meters, and the measurements are uncertain ±1cm, any errors from the measurements will have little effect on the error of the mean compared to the effect from the randomness of the sample.

“I quoted you fully and in context. If you didn’t say what you meant then blame yourself, not me.”

You didn’t even quote a full sentence. My paragraph in full with the bit you quoted in italics

“As I keep trying to tell you, the uncertainty in the mean of different things is mostly due to the sampling. The uncertainty in the measurements is usually of little importance, but the formula is the same, divide the standard deviation by the square root of the sample size.”

“This does *NOT* apply to multiple measurements of different things. In this case there is *NO* true value around which you are creating a probability distribution.”

There certainly is a true value – it’s the population mean.

“Taylor explains this well. Yet you are apparently too stubborn to read his treatise for meaning.”

Why should I have to scour over a 300+ page book to see if he actually says what you claim. You are making the claim, you give me the quote and page number.

“You can only random sample the same thing!”

You still haven’t explained what you mean by “the same thing”.

“The samples must be dependent”

No. Most statistical analysis assumes samples are independent.

“Random samples of different things are *independent*, the average of the values do not get you closer to a true value.”

Search Taylor for the word independent. You’ll see he frequently says how much better it is if errors are assumed to be independent. If you errors are independent you can reduce the uncertainty compared to dependent errors. He says it on page 57 in the bit you quoted above.

“If your sample consists of two boards, one 1 foot long with an uncertainty u1 and the other 2 feet long with an uncertainty of u2, then averaging their length will *NOT* give you a true value. There IS NO TRUE VALUE.”

You haven’t defined the population in this two board sampling. Are we talking about all boards in existence, all boards of a particular type, boards from the same factory, or are we only interested in these two boards. Whatever the case, the TRUE VALUE is the mean of all those boards. Why you would want a sample of just 2 I don’t know.

“And that value tells you absolutely nothing about either of the boards.”

The point of taking a sample to derive an average is to estimate the true average of the population. It is not to tell me something directly about the specific samples. It might tell me something about the two boards in general, as part of the population. For example if I was comparing boards from one source with ones from another source, and wanted to test the hypothesis that one source was providing longer boards than the other. (This isn’t a serious suggestion, but that’s the problem. You keep bringing up silly examples where there’s little point taking the average, then claim that proves that averages can never be useful.)

“It doesn’t matter how precisely you calculate the average, it represents *NOTHING* in reality.”

And so on, ad nauseam. I don;t know how many times I can say this, but just because you think the population average is nothing in reality, doesn’t make it so.
“And the average wil have that same uncertainty.”

And I say it won’t.

“I don’t know why this is so hard for supposedly educated scientists, mathematicians, and statisticians to grasp”

Have you considered it’s because you’re wrong? I’m not any of those things you list, but even I can see that what you say is demonstrably false.

I also take it you don’t consider yourself to be an educated statistician, yet despite that you still feel you know that all of them are wrong.

“In other words, multiple measurements OF THE SAME THING.”

Yes, there’s he describing how you can use statistics to improve the accuracy of a measurement by repeatedly measuring the same thing and taking the average. What he is not saying, is that the laws of statistics change just because you are measuring different things.

I do think this is the root of your and other’s problems. The method of sampling to get an average of independent values, the method of determining the standard error of that mean, etc. have been around for a lot longer than Taylor’s book. All Taylor et al are doing is showing one application of these methods, applying it to improving the accuracy of measurements of “single things”. But that does not mean, that suddenly they can only be used for measuring single things.

By the way, I still don’t think you’ve given me your definition of a single verses different things.

“You can’t order t-shirts for a population of humans based on their average height.”

You wouldn’t want just the average, it would help to know the standard deviation of the population, or better still have an estimate of the density function. But if you don’t even know the average height of a person, how do you propose to start your t-shirt ordering business. You can’t use your engineering knowledge to measure every customer before making them a bespoke t-shirt.

“You can’t even tell if the average height increases because short people are getting taller or because taller people are getting taller! ”

But you can tell that the average height has increased. That’s one of the uses of statistics, comparing some average of two distinct populations and seeing if they are statistically different. If you cannot analyze the means it’s much more difficult to know if there has been any statistical change.

“Of course he is! Exactly what do you think he is describing when he says to use root-sum-square with random, independent measurements to propagate uncertainty?”

He’s saying that if you don’t know if your measurements are random independent measurements, you can only assume that that uncertainties add when you add measurements – e.g. if you measure something with an uncertainty of 1cm 100 times, the uncertainty of the sum could be 100cm, which means if you take the mean value of the 100 measurements the uncertainty is still 1cm.

But if you can assume that all your measurement errors are independent you can use root-sum-square, which means adding the squares of your uncertainties and taking the square root. In this case taking the sum of a hundred independent measures, each with an uncertainty of 1cm gives an uncertainty on the sum of just 10cm, and when you take the average reduces the uncertainty of the mean to 0.1cm.

I’m not sure how you think the statement means “The laws of statistics change just because you are measuring different things.”

“You keep saying random, independent measurements and then go on to describe random, dependent measurements!”

No I’m not. You don;t seem to understand what “independent measurements” means. If you measure the same thing multiple times and get a random error each time, that’s an independent measurement (that is the errors are random samples from the probability distribution of the uncertainty. Similarly if I take random samples from a population the errors, deviation from the average will be randomly distributed along the distribution of the population.

Measurements are not independent if for example they were caused by an error in the instrument, or if your random sample was taking from specific locations. Say for instance you were doing a timing test, but the person using the stopwatch had slow reaction times which always added a second to the time. These errors would not be independent if you used the same person to make each measurement.

What independent does not mean, is independent from the thing being measured, whether it’s the length of a piece of wood or the average of many pieces of wood.

“That is *NOT* the same thing as taking measurements of 100 different things…”

I was describing what Taylor was saying. He’s talking about using averaging to improve the uncertainty in measuring a single thing. But that “something” could just as easily be the mean value of a population, with uncertainties that are caused by the distribution of elements within that population.

“As we’ve already discussed, you do *NOT* divide the sum of the uncertainties by N”

That’s not a discussion, that’s you endlessly shouting it.

“delta-q/q = delta-x/x. The formula is *NOT*

delta-q/q = (delta-x/x)/N”

Finally you same something correct. Why you’d think the formula would be that I’ve no idea. delta-q/q = delta-x/x is the relevant part it means that the ratio of the uncertainty of q to q is equal to the ratio of the uncertainty of x to x. Why you cannot understand what that implies about the size of the uncertainty of q when it is scaled from x is beyond me. But why try to figure it out when you could just look in that big handy box Taylor gives you where he says

the uncertainty in q is just |B| times that in x.

“Is an uncertainty interval a probability distribution? If so, then tell me what kind of a distribution it is.”

An interval is not a probability distribution, but the errors that make up the uncertainty does.

“I can’t figure it out because it’s gibberish.”

It’s not gibberish. It’s at the root of uncertainty propagation.

Why do you think all the uncertainty texts I’ve given you say that for indpendent, random measurements the uncertainty is calculated by root-sum-square and not (root-sum-square)/sqrt(N)?

You keep saying I am wrong but you are really saying that all those people stating you use root-sum-square are wrong!

“Standard error of the mean is the uncertainty.”

NO! It is not! Not for random, independent measurements! I keep giving you the example of boards laid end-to-end. You can calculate the mean for the total length as precisely as you want but you will *not* decrease the total uncertainty of the final length in any manner whatsoever. It doesn’t matter if you lay 20 boards end-to-end or 2000 boards end-to-end. The preciseness which with you calculate the mean has nothing to with the uncertainty of the final length. And since the final length is used to calculate the mean the uncertainty of the final length propagates directly to the uncertainty associated with the mean. Laying more and more random, independent boards end-to-end makes the final length *MORE* uncertain which makes the uncertainty of the mean more uncertain as well!

“Single vs many?”

I asked for a definition, all you gave me was an example. What I’m really trying to get at is, do you consider measuring the same thing means measuring exactly the same physical entity, or can it mean measuring different entities to establish a common value.

For example, if you measure a stack of 200 sheets of paper to determine the thickness of a single sheet, is that measuring many things (i.e. the individual sheets of paper) or one thing (i.e.the thickness of a single sheet)?

Or, if I count how many babies on average are born per day in a hospital, am I measuring multiple things (i.e. individual babies), or a single thing (the birth rate)?

Part the Last

“You can deduce a seasonal influence, that’s about all.”

So finally, you admit you can tell something from a mean temperature. It’s a start.

“If next year the mean temperature in July is 21C how do you know what caused the increase? Did max temps go up? Did min temps go up? Was it a combination of both?”

You can tell by looking at the maximum and minimum values. If you don;t have them, be thankful that at least you have the mean value as otherwise you’d have no idea it had gone up at all.

You are in dire need of a course in digital signal processing.

It doesn’t matter which you use. Averaging telephone numbers or averaging temperatures from multiple locations using multiple devices (each with an inherent uncertainty) – neither tell you anything useful.

DSP question, how do you re-create a continuous function using just two values?

A myth the climastrologers cling to as if stuck with Velcro.