# Argo, Latitude, Day, and Reynolds Interpolation

Guest post by Willis Eschenbach

This is another of my occasional reports from my peripatetic travels through the Argo data (see the Appendix for my other dispatches from the front lines). In the comments to my previous post, I had put up a graphic showing how the January/February/March data for one gridcell varied by latitude and day of the year. Figure 1 shows that graphic:

Figure 1. Argo surface temperatures for three months (JFM) in the gridcell 25°-30°N, 30°-40°W. Color of the data points shows the year of the observation.

As you can see, in this gridcell the ocean gets cooler as you go north. It also gets cooler from the first of the year to the end of March, although obviously that will change during the year as it warms and cools.

I decided to take a look at every bit of the available data for that gridcell, not just three months of each year. Figure 2 shows that result.

Figure 2. All Argo temperature measurements for the gridcell, from 2002 to 2012. (a) Upper panel shows measurements only (blue diamonds). (b) Lower panel shows measurements (blue) and the trend (yellow/orange circles) as estimated by a linear model involving the day of year and the latitude of each measurement.

Note that my linear model does a pretty good job of resolving the variation by day and by latitude. For example, in the coldest and warmest parts of 2010 and 2011, there were a variety of measurements. The model does a good job of replicating that.

Curiously, this linear model does a better job than another linear model I tried using a single variable, the insolation by day and latitude. You’d think the insolation data by day and latitude would encapsulate the day and latitude data I used in Figure 1 (b). But that turned out not to be the case. It wasn’t nearly as successful at separating out the “by latitude” variation. Which was odd, because I hadn’t even bothered to use cosine(latitude) … hang on … OK, I just went and checked it with cos(latitude), and the results are indistinguishable. Not too surprising, I guess, it’s only a 5° slice so linear and cosine are not terribly different. It is surprising that it does so much better than the insolation data, though.

My model is fitted, of course. It’s kind of a linear model. For the day variable, I iteratively fit a function of the form

Temperature = (2 + Sin(2 * Pi * (JulianDay+Lag)/365.254)) ^ Somepower

JulianDay is the day count from some fixed starting point. Lag is an adjustment for the phase lag due to the delayed heating and cooling of the ocean mass. Somepower is the power to which the function is raised. The number 365.254 is days per year. The 2*PI is to convert to radians for the Sin calculation.

This (2+Sin(time))^Somepower type function is a reasonable approximation of the form of a natural system where the heat input is cyclical and the heat loss is a power function of the temperature. See, for example, the ocean temperatures shown in cyan in Figure 1 here, which spike upwards to a point and then drop quickly in the summer, but have a rounded base in the winter.

In the case of the ocean, heat loss varies linearly with wind, it varies as T^2 with Clausius-Clapeyron for evaporation, and it varies as T^4 for radiation. Of course, all of that is mitigated by a host of factors. In this case, Somepower had a best fit at 1.4.

Moving forwards, I decided to compare the Argo data with the Reynolds sea surface temperature data. Figure 3 shows a comparison of the Argo float data with the Reynolds Optimally Interpolated satellite plus observational based temperatures. The Reynolds data is from KNMI.

Figure 3. Argo temperatures (blue diamonds) and Reynolds Optimally Interpolated surface temperatures (yellow/red circles)

Generally, this is good news, as the agreement between the two is quite surprising. I used a cubic spline to interpolate the Reynolds OI data to match the dates of the Argo data, and it shows a surprisingly close match.

Now here’s the oddity. The Reynolds data has a trend. The Argo data has no trend at all. Figure 4 shows the residuals, what is left after removing the cyclical portions of the signal. It also shows the linear trends of the residuals.

Figure 4. Residuals after the removal of confounding variables. Upper panel shows the Argo data after the removal of the day-of-year and latitude effects. The lower panel shows the Reynolds data after removal of the monthly averages.

I’m not entirely sure what to make of all of this. I am encouraged that the residuals of the Reynolds data are well correlated with those of the Argo data. Since the datasets are produced entirely independently and with no common procedural steps, this is good news. But the difference in the trends is quite large.

My guess about the reason for the difference? Bear in mind that of these two, only the Argo measurements are actual observational data. The Argo floats measure one point in time, to an accuracy of 0.005°C. The Reynolds dataset, on the other hand, is interpolated. You can find a description of the method and other data here. They say the interpolation is “optimal”, which always makes me nervous, but never mind that.

The problem is, nature doesn’t do interpolation. There isn’t any gradual transition between say the cool waters off of New York and the Gulf Stream, it is a sudden jump. And when I used to fish albacore off of the California coast, I learned that in the fall you drive the fishing boat offshore through green cold water covered with fog, miles and miles of fog and green water, and then suddenly you emerge from the fog bank to see blue warm water and clear skies, and a clear dividing line between the two … there’s no way to “interpolate” that.

So if you are interpolating between two areas that are warming, the intervening area will interpolate as warming as well … and that may or may not be the case.

As before, I am making no great over-arching claims for these results. It is one gridcell in a large ocean, and I’m using these posts to list and discuss some of the things I’m learning about it as I go. I’m working to assess the validity of the Argo data, compare it to other datasets, and take some guesses as to its accuracy. I do note that in this gridcell, over the full decade, both Reynolds and Argo show a trend that has a statistical error of plus/minus a tenth of a degree per decade, and yet the trends are half a degree per decade different from each other, and that’s a lot … so one of them has to be pretty wrong. My money’s on the Argo data as being the better of the two … but to what kind of accuracy?

There’s one final issue I want to discuss. The assumption is often made, by Hansen and others, that correlation between two datasets implies similar trends. Based on good correlation between stations, the GISS folks believe that you can use one dataset to extrapolate temperatures out 250 to 1200 kilometres away from the nearest stations. I have argued against this misconception in a post called “GISScapades“.

But look at these two datasets, Argo and Reynolds. The correlation between the Argo and Reynolds residuals is 0.66, and the correlation between the first differences of the residuals is 0.82, both quite good. There’s a reasonable amount of data, 953 data points. For climate science where small dissimilar datasets are the norm, those are impressive correlations, particularly given that the two analyses are totally independent.

Yet the Argo data shows no trend at all, 0.0°C per century … while the Reynolds data, which claims to be measuring the same thing at the same time, shows a trend of 5.1°C per century! (Yes, I know, it’s only a decade of data, I know you can’t extrapolate that out a century, and I’m not doing that. This is not a forecast of any kind. I mention the difference over a century purely because I want people to be clear about the size of the difference in trends between these two very well correlated datasets.)

Go figure …

My best regards to everyone,

w.

## 52 thoughts on “Argo, Latitude, Day, and Reynolds Interpolation”

1. John Cooper says:

It took me a moment to visualize the three-dimensional scatter graph, because the two lines representing the hidden top-back edge of the cube should have been dashed. I learned this in drafting class in 1962 (grin).

2. Dave N says:

3. Larry Geiger says:

Hi Willis
I confuse easily. Twice I noticed that you said “Argo data shows no trend at all”. Is this just statistical shorthand for: The trend line in the data is flat? It appears that you are graphing a trend, yet you say there is none. Is it only “a trend” if it goes up? Not trying to be argumenative. Just trying to understand.
Thanks

4. JEM says:

“I’m not entirely sure what to make of all of this.”

Important words, not seen often enough in the alarmist community.

5. Willis Eschenbach says:

John Cooper says:
March 5, 2012 at 1:02 pm

It took me a moment to visualize the three-dimensional scatter graph, because the two lines representing the hidden top-back edge of the cube should have been dashed. I learned this in drafting class in 1962 (grin).

Yeah, I know, but I couldn’t figure out how to make the software ( R ) do it so I said bad words and published it as is …

w.

6. Willis Eschenbach says:

Larry Geiger says:
March 5, 2012 at 1:29 pm (Edit)

Hi Willis
I confuse easily. Twice I noticed that you said “Argo data shows no trend at all”. Is this just statistical shorthand for: The trend line in the data is flat? It appears that you are graphing a trend, yet you say there is none. Is it only “a trend” if it goes up? Not trying to be argumenative. Just trying to understand.
Thanks

What I mean is that if you run a linear trend line through the Argo residuals, the trend is 0.0° per century

w.

7. More interesting stuff Willis.

Upper Figure 4 went click. I was just looking at SORCE TSI from 2003 but that is humans for you.

You notice a trend. What happens to r2 if you remove the trend from Reynolds?

8. Willis Eschenbach says:

Dave N says:
March 5, 2012 at 1:27 pm

I got it from the Harry Belafonte song “Man Piaba”, viz …

“It was clear as mud but it covered the ground
And the confusion made the brain go ’round.
I grabbed a boat and went abroad
He said “Son,”

Put the body down up on the couch
I can see from your frustration a neurotic sublimation
Hey love and hate is psychosomatic
Your Rorschach shows that you’re a peripatetic
It all started with a broken sibling
In the words of the famous Rudyard Kipling”

9. HR says:

Willis

The comparison of the ARGO and Reynolds data residuals. Why does the Reynolds data also have a big hole in the data around 2007 just like the ARGO data? In fact across the timeline the Reyolds seems to have the same spread of data points as the ARGO data. Given the very different sources this seems like too much of a coincidence. Are you sure you haven’t made an error with your data sets there?

10. JimB says:

You are a better man than I am, Gunga Din.

11. Andrejs Vanags says:

You are comparing the trend of the residuals. For it to be zero, the ‘model’ curve must have zero bias as well. Did you check that for the time period chosen the model curve has a bias of zero? easy not to have it, if the interval is not exactly an integer of the period of the sinusoid.

12. Willis Eschenbach says:

HR says:
March 5, 2012 at 2:06 pm

Willis

The comparison of the ARGO and Reynolds data residuals. Why does the Reynolds data also have a big hole in the data around 2007 just like the ARGO data? In fact across the timeline the Reyolds seems to have the same spread of data points as the ARGO data. Given the very different sources this seems like too much of a coincidence. Are you sure you haven’t made an error with your data sets there?

Thanks, HR. I have selected and used the Reynolds data for the same dates as the Argo data, so that I am comparing apples to apples.

w.

13. P. Solar says:

RE: the mystery:

why is there a split in 2007?

To my eye the latter parts of both records are near flat trend in fig.4.

In the first half Reynolds in fairly flat but lower. So again, why the split , is fitting a trend to the ensemble reasonable?

Argo, I see as decidedly downwards in first half but about the same level.

how about fitting trend and offset to each half, it may tell you something.

BTW, someone tells you they have thousands of pieces of production equipment in a hostile environment stable to 0.005K over 10 years? Show me the spec sheet and the warranty.

then show me the field tests where they verified a few random devices.

My gut feeling is we’re being bullshitted on that one.

14. Henry says:

Just be clear: you seem to be fitting a cyclical function to the data, so your model has no room for a trend. You then look at the residuals between the data and the fitted cyclical function and find no trend in the residuals. This looks like a form of seasonal adjustment.

I am not sure whether your model is in Kelvin or Celsius: nor can I see a reason for the “2+” which I would have thought you could let the data decide.

15. how dare you say interpolation…the Mosh beast will give you eternal nightmares if you opine that interpolation is piffle for climate data

16. Greg Locke says:

Willis, I’m no wordsmith, but i think “peripetetic travels” is a redundancy.

But no matter. I’m in way over my head on this one. When I read Mr. Vanags comment, I broke down and cried.

17. Willis Eschenbach says:

Andrejs Vanags says:
March 5, 2012 at 2:14 pm

You are comparing the trend of the residuals. For it to be zero, the ‘model’ curve must have zero bias as well. Did you check that for the time period chosen the model curve has a bias of zero? easy not to have it, if the interval is not exactly an integer of the period of the sinusoid.

An interesting question, Andrejs. I don’t see why the model has to have zero trend, other than in the idealized condition. All the model has to do is accurately remove the effects of latitude and day of year. If it does so and there is a trend in the original data, the residuals will show it.

But if the known confounding variables of latitude and day of year are accurately removed, and there is no trend in the residuals, then that means there was no trend in the original data.

Bear in mind that if the data is evenly distributed by latitude and day, my model will indeed have a zero trend. But since the data is unevenly distributed in space and time, only by removing the modeled results and considering the residuals can we see if the underlying data has a trend or not.

w.

18. Brian Eglinton says:

I think P Solar has a good idea.

Can you do a trend comparison between the first 5 years and last 5 years?
You might find a much closer trend in each half.

And that might suggest something about the way the satellite data is being generated or processed.
It might also suggest that the calculated error bounds on this trend are not realistic?

19. Louis Hooffstetter says:

What!? You’re not extrapolating a decade of data out over a century and forecasting catastrophic warming of the oceans by 5.1°C?

Willis, you’ll never be a ‘real climatologist’.

20. RoHa says:

Me try understand.

M&Ms of empirical data show: Sea she no get hot, sea she no get cold.

This it?

21. Willis Eschenbach says:

Henry says:
March 5, 2012 at 3:09 pm

Just be clear: you seem to be fitting a cyclical function to the data, so your model has no room for a trend. You then look at the residuals between the data and the fitted cyclical function and find no trend in the residuals. This looks like a form of seasonal adjustment.

The model incorporates the latitude and the day of the observation. So the “day” part is indeed a seasonal adjustment

I am not sure whether your model is in Kelvin or Celsius: nor can I see a reason for the “2+” which I would have thought you could let the data decide.

All the temperatures are in °C.

All that the “day” part of the adjustment does is to fit a curve of the form

(2 + sin(day + lag))^n

to the data. The reason for the “2 +” is because the value of sin(x) varies between +1 and -1. I move that up to the interval =1 to +3 by adding 2 to the values. This is so the power function varies smoothly over the interval, and so that I avoid taking negative numbers to non-integral powers.

I’m just after the shape. That particular function yields a curve family of the form

I use it because it resembles the natural shape of the annual cycle.

w.

22. Willis Eschenbach says:

Brian Eglinton says:
March 5, 2012 at 4:31 pm

I think P Solar has a good idea.

Can you do a trend comparison between the first 5 years and last 5 years?
You might find a much closer trend in each half.

And that might suggest something about the way the satellite data is being generated or processed.
It might also suggest that the calculated error bounds on this trend are not realistic?

I’ll pass. I’m very uneasy just putting a linear trend on ten years of data, and it’s only because I have about a thousand data points that I’m willing to do it. Splitting that in half and fitting a linear trend is a bridge too far.

w.

23. Willis,
When I saw “peripatetic” in your post I assumed you meant it in the walking-around sense. But the Belafonte lyric “Your Rorschach shows that you’re a peripatetic” made me think you might have used the word in the Aristotelian philosophical sense. Which is it, please? Or is it both and I should be embarrassed at not getting the double entendre?

24. Ohmtgosh says:

Great post Sir. I think you took their windpipe out. :)

25. sheep92 says:

(2 + sin(day + lag)^n)

shouldn’t this be

(2 + sin(day + lag))^n

[Thanks, fixed. -w.]

26. Willis
There is another very important piece of information in the Argo data that you show. The 3 month difference between winter solstice and the minimum ocean temperature is the ocean storing and releasing the heat that keeps the higher latitudes warm(er) in winter. Near 45 deg latitude, the annual heat storage is about 1000 MJ in a 1 square meter column of water down to about 100 m. That is about 40 days worth of full summer sun. The rate of heat release is controlled by evaporation – ocean surface temperature and the wind speed. The ocean can only cool from the surface. This is the source of climate change. The total increase in downward LWIR flux from 100 ppm of CO2 is 50 MJ per sq meter per year (over 200 years). This can only penetrate 100 micron into the ocean and gets lost in the noise of the wind driven evaporation.
This also means that there is no such thing as a climate equilibrium, so all of the assumptions underlying the IPCC climate models are completely wrong – before any equations are derived or code is written. Take those radiative forcing constants, forcings, feedbacks and flux equations and put them where the sun don’t shine – at least 100 m below the ocean surface. The ocean does not have an equilibrium black body surface.
The California climate (5 year rolling average of the minimum weaterh station temperature) tracks the Pacific Decadal Oscillation quite closely. I have used this to estimate urban heat island effects in CA.
Sun, wind and water need no help from CO2 to change the Earth’s climate.

27. Len says:

Willis. I am happy that someone has the energy and talent to start looking at these huge data sets and interpret them for us. As to the climate models pre-set with a strong CO2 forcing, remember the old saying? Figures never lie, but liars figure.
Good luck with analyzing the data. The data are very interesting as are your analyses and interpretations.

28. In a climatology blog, there’s not much point in obsessing over the data from Argo. Though Argo may prove to be a boon for the purpose of modern day short term weather forecasting, it is virtually useless for the purpose of modern day climatological forecasting. Several hundred statistically independent observed events are about the minimum for the construction of and testing of a statistically validated model that predicts the outcomes of events with statistical significance but as the canonical period of a climatological event is 30 years, we’d need to collect data from Argo for about 6000 years before beginning to use Argo time series in the construction of climatological models. That’s 600 times as much data as we now have.

29. “The Argo floats measure one point in time, to an accuracy of 0.005°C.”

Can somebody please explain to me how this accuracy (not resolution) of 0.005°C is achived? What kind of sensors and electronics is used?

Platinum sensors are common in temperature measurements, with Pt100 probably the most common. 0.005°C accuracy requires an accuracy of 0.0015 ohm in resistance measurement.

Thermocouple sensors are typically 50 microvolts per degree C, and require a temperature reference.

I find it very difficult to believe that the absolute accuracy of temperature measurement is anywhere near 0.005°C.

30. P. Solar says:

>>
I think P Solar has a good idea.

Can you do a trend comparison between the first 5 years and last 5 years?
You might find a much closer trend in each half.

And that might suggest something about the way the satellite data is being generated or processed.
It might also suggest that the calculated error bounds on this trend are not realistic?

I’ll pass. I’m very uneasy just putting a linear trend on ten years of data, and it’s only because I have about a thousand data points that I’m willing to do it. Splitting that in half and fitting a linear trend is a bridge too far.
>>

LOL. I agree fitting linear trends to climate data is dumb (no matter how many data points you have). But since you are doing it AND are puzzled by the anomalous results it gives, you AND the data is already split in two for reasons you chose not to comment on, maybe you need to go that “bridge too far” or stop post pointless questions and meaning less linear trends.

Make your choice , you do it or you don’t. A bridge too far is not convincing statistical argument. You would very likely find you had better stats on the split period fits than you got on the combined data.

31. P. Solar says:

Ágúst in Iceland says: I find it very difficult to believe that the absolute accuracy of temperature measurement is anywhere near 0.005°C.

Exactly. I suspect they may be selectively misusing hte *sensitivity* of the platinum sensor not that stability or the accuracy of the whole instrument. I defy anyone to measure temperature that accurately in the field.

As I posted above: someone tells you they have thousands of pieces of production equipment in a hostile environment stable to 0.005K over 10 years? Show me the spec sheet and the warranty.

then show me the field tests where they verified a few random devices.

My gut feeling is we’re being bullshitted on that one.

32. Willis Eschenbach says:

P. Solar says:
March 5, 2012 at 10:32 pm

>>
I think P Solar has a good idea.

Can you do a trend comparison between the first 5 years and last 5 years?
You might find a much closer trend in each half.

And that might suggest something about the way the satellite data is being generated or processed.
It might also suggest that the calculated error bounds on this trend are not realistic?

I’ll pass. I’m very uneasy just putting a linear trend on ten years of data, and it’s only because I have about a thousand data points that I’m willing to do it. Splitting that in half and fitting a linear trend is a bridge too far.
>>

LOL. I agree fitting linear trends to climate data is dumb (no matter how many data points you have). But since you are doing it AND are puzzled by the anomalous results it gives, you AND the data is already split in two for reasons you chose not to comment on, maybe you need to go that “bridge too far” or stop post pointless questions and meaning less linear trends.

Make your choice , you do it or you don’t. A bridge too far is not convincing statistical argument. You would very likely find you had better stats on the split period fits than you got on the combined data.

Ummm … you say:

Make your choice , you do it or you don’t.

Ummm … I don’t.

Well, actually I said “I’ll pass”, but surely you can make the translation. So why are claiming I haven’t made my choice?

I have a curious kind of rule about what kind of analyses I do. I do what I want, and not what random folks on the internet want. Foolish, I know, but there it is. That’s the joy of being an amateur scientist, I can follow the paths that appeal to me, and I never have to worry about pleasing funders or a boss or anyone.

I’ve cited where I got the data, P. Solar. You think it is important to do that analysis, that there is something important to be found there. Me, I think it’s a dry hole, and I’ve got no time for digging in a dry hole.

So I heartily invite you to go get the data and do the analysis, and report back on the important things that you find there. I’m off looking at other stuff …

My best to you,

w.

PS—You say “…the data is already split in two for reasons you chose not to comment on …”. In fact, I’ve commented on the data a number of times over the last couple of posts, so your snark is badly misplaced. All your unpleasant comment shows is that you’re not paying attention. FYI, that’s all of the Argo data that exists for that gridcell, so I didn’t split the data in two. It came that way … sorry there’s no secret underlying “reasons I chose not to comment on”, but in fact there’s no reason at all, because that’s just how the data was.

33. Willis Eschenbach says:

Ágúst in Iceland says:
March 5, 2012 at 10:22 pm

“The Argo floats measure one point in time, to an accuracy of 0.005°C.”

Can somebody please explain to me how this accuracy (not resolution) of 0.005°C is achived? What kind of sensors and electronics is used?

Platinum sensors are common in temperature measurements, with Pt100 probably the most common. 0.005°C accuracy requires an accuracy of 0.0015 ohm in resistance measurement.

Good question, Ágúst. See this PDF for a variety of details. P. Solar, you too … turns out your “gut feeling” may not be all that reliable.

w.

34. Roy says:

“As you can see, in this gridcell the ocean gets cooler as you go north.”

I am affraid I can’t see any such thing. I’m sure your observation is correct but that graphic certainly doesn’t let me see it. Sadly it’s just a pretty picture unless you’ve already interpreted what’s going on using a different (probably less pretty) picture that really shows what’s happening. Three projections may be ugly but they’d work. Sorry to nit-pick but deceptive/useless graphics are a real hobby-horse of mine.

35. Different time, different place. A few lessons ago you showed Argo SST data that did not want to go higher than 30 or so degrees C in the tropics between 30N and 30S. This caused thoughts about correlation coefficients. Of course, they will be high if there are many input values that are close to each other in value. http://wattsupwiththat.com/2012/02/09/jason-and-the-argo-notes/
In reading the present post, when you came to correlation coefficients at the end, I wonder if you have a similar effect operating, namely an unrecognised physical secondary effect that improves the correlation. On a quick read I can’t put my finger on one, though I’d be looking at the interpolation steps. Just a vague thought. I’m still trying to explain the uptick on the top left of this BEST plot even though the Y variable is not SST. http://www.geoffstuff.com/Correl.JPG

36. Andrejs Vanags says: March 5, 2012 at 2:14 pm

I agree with Andrejs. The data is the sum of model and residuals. Trend calc is linear. So the trend of the data is the sum of the trends of model and residuals.

If you fit a line, the trend of residuals is always zero.

37. Where do you get 365.254 from? I thought the tropical year was more like 365.2422 (the calendar gives us 365.2425).

38. Willis Eschenbach says:

Roy says:
March 6, 2012 at 12:19 am

“As you can see, in this gridcell the ocean gets cooler as you go north.”

I am affraid I can’t see any such thing. I’m sure your observation is correct but that graphic certainly doesn’t let me see it. Sadly it’s just a pretty picture unless you’ve already interpreted what’s going on using a different (probably less pretty) picture that really shows what’s happening. Three projections may be ugly but they’d work. Sorry to nit-pick but deceptive/useless graphics are a real hobby-horse of mine.

Roy, I feel badly that you are unable to interpret a simple 3-D graph. No one else seems to be having that problem, so it sounds like a Roy problem to me. As a result, you’ll have to fix that one yourself.

You might start by actually, you know, reading the links at the bottom of the head post. Come back when you’ve figured out how to read graphics.

Or not, up to you.

w.

PS—Sorry to slap your face and ask you to learn to understand simple graphics, but obsequious nitpickers like yourself are a real hobby-horse of mine.

39. Willis Eschenbach says:

Nick Stokes says:
March 6, 2012 at 1:20 am

Andrejs Vanags says: March 5, 2012 at 2:14 pm

I agree with Andrejs. The data is the sum of model and residuals. Trend calc is linear. So the trend of the data is the sum of the trends of model and residuals.

If you fit a line, the trend of residuals is always zero.

Thanks, Nick. As I said above, if the data is evenly distributed in time and space, then my model for the Argo data is trendless. As a result, since (as you point out) the trend of the data is the sum of the trends of the model and the residuals, and the model is trendless, then the trend of the residual is the trend of the data.

Note that what is important is not the trend of the particular realization of the model based on the exact time and location of the Argo float measurements.

What counts is the trend of the theoretical model given a uniform random sampling in time and space … which for my model is zero.

w.

40. Willis Eschenbach says:

Anomaly UK says:
March 6, 2012 at 1:40 am

Where do you get 365.254 from? I thought the tropical year was more like 365.2422 (the calendar gives us 365.2425).

I was thinking that was the sidereal year, but I must have misremembered, that one is 365.256 days. In any case, you’re likely right, I should have been using the tropical year. Dang, a 0.003% error, I hates that …

w.

41. Roy says:

“Sorry to slap your face and ask you to learn to understand simple graphics, but obsequious nitpickers like yourself are a real hobby-horse of mine.”

Slap away; I’m man enough. But afterwards take another look at that 3D graph and tell me you (or anyone else) can really read it. You’re fooling yourself. It only seems to mean something because it reminds you of something you already know by some other means.

And I apologise for saying I was sorry to nit-piok. I’m not. I am actually very keen to point out self-delusion, confirmation bias, observer-effects or any other traits of pseudo-science, no matter which side of the debate I see them on.

42. Willis:
Thank you for the information regarding the accuracy of 0.005°C. Really interesting.

43. Chris D. says:

“Argo, Latitude, Day, and Reynolds Interpolation”

Quick, somebody call Mosher.

44. theBuckWheat says:

““I’m not entirely sure what to make of all of this.”
This is a delightfully honest question. Science is after all, the quest to find truthful answers to questions.

I was a guest at a meeting of LENR physicists on the eve of a meeting to kick off the efforts of a major university study of LENR. They complained that the current environment in which research is conducted is built to discount and to suppress the study of anomalies in the data and in experiments. Thus, brief episodes of unexplained phenomena are rejected because they cannot be reproduced. The problem with a lot of LENR research is not that excess heat is not being produced, it is exactly because it cannot be reliably produced by referees, peers and third parties. Something that cannot be reproduced never took place.

In climate research, it appears that bias like UHI can be reproduced, so when data shows an upward temperature trend it gets published. Data that refutes the trend (such as from a station not affected by UHI) gets excluded as an “outlier”. In the case of “cold fustion” or unexplained bursts of excess heat during a LENR experiment, it cannot be explained, so the researcher dare not mention it. In the latter case, that is exactly the phenomena we all are striving to understand. We want surprises in the LENR data! In the case of climate change, we have seen academics who were so immersed in an ideology they were willing to perjure themselves and to subvert the peer review process to sustain their overarching goals. For a long time LENR research was nearly impossible to get funded because of how those in control of the status quo disbelieved it. Up until recently, it appears that research that validated the Gore Climate Model got funded while research that invalidated it could not.

45. markx says:

Roy : March 6, 2012 at 2:45 am
said:
“…… take another look at that 3D graph and tell me you (or anyone else) can really read it. You’re fooling yourself. It only seems to mean something because it reminds you of something you already know by some other means….”

Roy, I don’t see your problem: it (to me) very clearly shows a scattering of points in an elongated cluster extending from the top front left hand corner of a ‘cube’ to the right hand rear corner of that ‘cube’. (for clarity – left and right as per the viewer).

Y axis is temperature (C), X axis is day of year (Jan Feb Mar, from figure title), Z axis is latitude (starting from the 25 North, heading north – from figure title).

Even if your eyes see the cluster wrapping around the back of the ‘cube’, the end result is still the same.

And seems to clearly demonstrate Willis’ point.

Or am I seeing this wrongly?

46. Odd, I seem to have been bounced? Moderator…am I too frequent?

It did not appear to me to be too much of a stretch to ask Willis to provide the S.D. for the two graphs in Figure 4. (I would figure …sorry on the pun, that it would be easier and faster for Willis to do that, as I suppose that he has the individual data points from which the graphs were constructed.) Easier than my tediously trying to work out the data from the graphs and do said analysis.

Also, there is no malice or devious intent in the request. I just feel that it puts even the Reynolds 0.5 degree C per decade in perspective, that is if it is less than a S.D. from the data!

Max

47. Roy says:

@markx
“am I seeing this wrongly?”

The extremes near the top-left and bottom-right corners of the cube are readable enough. But (to pick just one example), take those two dark blue dots that appear near the center of the left-most face. Where are they in space? Are they right on the temp-latitude face of the cube, half way along the latitude axis? Or on the temp-day face, half way along the day-of-year axis? Or somewhere along some line inside the volume of the cube?

If you think you “see” where any of the interior points are it’s because you’ve prepared your mind to see them that way. Self-delusion is a constantly present risk in science and here is an (admittedly inconsequential) illustration of how easily it happens.

As a form of graphical evidence for an argument a static 2D projection of a 3D graph is useless and I bet it wasn’t the way Willis first visualized what was going on, or if it was, he was able to rotate it on his screen to see different projections.

I wonder if my original mealy-mouthed way of criticising it got up Willis’ nose more than my actual criticism.

48. Scott Covert says:

Roy, I see your point. You are correct.
The fact that I saw a diagonal line from corner to corner evenly surrounded by the dots was due to a preconceived bias that logically, higher lattitude above the equator should be colder. If those three dots in the lower right corner weren’t there, I would still see it the same, but it could be quite different. If there was a hidden curve, dog-leg, or other deviation not seen in the 3rd dimension, it would be dishonest to present the data in that form but I doubt very much that is the case here (Maybe I am showing my ignorance in this case).

Skepticism by default is very important in science. You display that admirably. Nonetheless, it is irritating to have that pointed out when you are trying to explain a complicated problem. It derails the train of thought and progress in understanding the topic pauses.

I assume Willis wants to get on with the discussion without impediment.

With that said, it’s still amusing to see someone get Willis’s goat while being correct at the same time.

Now behave before he get’s out the ruler! Oh, too late…

49. Willis Eschenbach says:

Roy says:
March 6, 2012 at 2:45 am

… And I apologise for saying I was sorry to nit-piok. I’m not. I am actually very keen to point out self-delusion, confirmation bias, observer-effects or any other traits of pseudo-science, no matter which side of the debate I see them on.

I didn’t believe you when you said you were sorry before, I suspected that you were lying about that, but I had no evidence.

Now you admit you were lying, and try to shrug it off with another meaningless apology.

I didn’t believe your apology then, and I don’t believe your apology now.

You obviously were lying before, you have admitted it. Why should I believe you now?

You think it gains you points to admit you were lying before … it doesn’t. It just proves that you are a liar.

Me, I don’t deal with liars. You’ll have to take your inability to read a graph somewhere else to get it fixed, I’m done with your deception. Heck, maybe the whole thing was a lie, maybe you do understand that graphic.

I don’t know, and I don’t care.

w.

PS—In my previous post, I showed the same data in 2-D … but I guess you were too dumb to be able to read that as well.

50. Willis Eschenbach says:

Max Hugoson says:
March 6, 2012 at 8:47 am (Edit)

Odd, I seem to have been bounced? Moderator…am I too frequent?

No clue.

It did not appear to me to be too much of a stretch to ask Willis to provide the S.D. for the two graphs in Figure 4. (I would figure …sorry on the pun, that it would be easier and faster for Willis to do that, as I suppose that he has the individual data points from which the graphs were constructed.) Easier than my tediously trying to work out the data from the graphs and do said analysis.

Argo residuals, 0.66°C. Reynolds residuals, 0.67°C.

Also, there is no malice or devious intent in the request. I just feel that it puts even the Reynolds 0.5 degree C per decade in perspective, that is if it is less than a S.D. from the data!

That’s not a valid comparison. One is a trend, the other is a standard deviation. You can’t compare them like that, that’s apples and oranges.

All the best,

w.

51. Geoff says:

Hi Willis,

You might be interested in this paper in press ( http://www.ocean-sci-discuss.net/9/25/2012/osd-9-25-2012.html ).

Abstract. The transport and storage of heat by the ocean is of crucial importance because of its effect on ocean dynamics and its impact on the atmosphere, climate and climate change. Unfortunately, limits to the amount of data that can be collected and stored means that many experimental and modelling studies of the heat budget have to make use of mean datasets where the effects of short term fluctuations are lost.

In this paper we investigate the magnitude of the resulting errors making use of data from OCCAM, a high resolution global ocean model. The model carries out a proper heat balance every timestep so any imbalances that are found in the analysis must result from the use of mean fields.

The study concentrates on two areas of the ocean affecting the El Nino. The first is the region of tropical instability waves north of the equator. The second is in the upwelling region along the equator.

It is shown that in both cases, processes with a period of less than five days can have a significant impact on the heat budget. Thus analyses using data averaged over five days or more are likely to have significant errors. It is also shown that if a series of instantaneous values is available, reasonable estimates can be made of the size of the errors. In model studies such values are available in the form of the datasets used to restart the model. In experimental studies they may be in the form of individual unaveraged observations.

52. Willis Eschenbach says:

Thanks, Geoff. I agree that averaging is generally an error, and that one should work from the smallest-grained data that one can find. However, I’d be very skeptical of OCCAM, especially if he’s packing a razor, drawing conclusions based on models is fraught with dangers and hidden pitfalls.

w.