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.
APPENDIX: Previous Posts on Argo
Krige the Argo Probe Data, Mr. Spock! | Watts Up With That?
Hansen’s Sea Shell Game | Watts Up With That?
Where in the World is Argo? | Watts Up With That?
Jason and the Argo Notes | Watts Up With That?
Argo Notes Part 2 | Watts Up With That?
Argo and the Ocean Temperature Maximum | Watts Up With That?
Argo Notes the Third | Watts Up With That?
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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).
You had me at “peripatetic”
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
“I’m not entirely sure what to make of all of this.”
Important words, not seen often enough in the alarmist community.
John Cooper says:
March 5, 2012 at 1:02 pm
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.
Larry Geiger says:
March 5, 2012 at 1:29 pm (Edit)
What I mean is that if you run a linear trend line through the Argo residuals, the trend is 0.0° per century …
w.
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?
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
In Baden Baden asked Sigmund Freud
He said “Son,”
From your sad face remove the grouch
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”
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?
You are a better man than I am, Gunga Din.
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.
HR says:
March 5, 2012 at 2:06 pm
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.
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.
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.
how dare you say interpolation…the Mosh beast will give you eternal nightmares if you opine that interpolation is piffle for climate data
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.
Andrejs Vanags says:
March 5, 2012 at 2:14 pm
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.
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?
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’.
Me try understand.
M&Ms of empirical data show: Sea she no get hot, sea she no get cold.
This it?
Henry says:
March 5, 2012 at 3:09 pm
The model incorporates the latitude and the day of the observation. So the “day” part is indeed a seasonal adjustment
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.
Brian Eglinton says:
March 5, 2012 at 4:31 pm
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.
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?
Great post Sir. I think you took their windpipe out. 🙂
(2 + sin(day + lag)^n)
shouldn’t this be
(2 + sin(day + lag))^n
[Thanks, fixed. -w.]