## Weather Sensors Cause Climate Errors

ClimatePolice.com – May 11, 2007

I think this blog entry is worthy of your attention. The entry discusses the differences in measuring temperature using the ‘modern’ computer based methods and the traditional baseline methods of thermometers. The comments also are quite interesting and could be argued are more educational than the original post in the actual use of these two measuring devices and their individual bias.

I have real concerns about analyzing temperature. I do not find it to be mathematically valid to calculate ‘average’ temperatures. While the argument can be made that it is valid to average the temperature in a short period of time (probably less than an hour) in one single location, it is likely not valid to average it over longer periods (such as the entire day) and definitely not for multiple locations.

Temperature is the amalgamation of several physical properties of the body that is being measured. In the case of atmospheric temperature it includes radiation energy from surrounding entities, energy directly from the sun, and the energy level of the atmosphere surrounding the device. It also includes how much energy can be absorbed in that atmosphere which is largely influenced by the amount of water that is in the air, the pressure of that air, and the amount of movement of that air.

We tend to be confused in global warming discussions between temperature and the amount of heat energy that is stored in the atmosphere. If we would measure the amount of K-calories or Joules that are in the atmosphere at any location and at any time, we could accurately average these amounts over time at that location or over multiple locations. But to do mathematical operations on temperature is simply not mathematically valid since we are adding different things together. In reality, the high temperature yesterday at my house was not 100 deg F (we set a record) but rather was 100 deg F with the wind blowing a certain direction at a particular speed and with a pressure of X and humidity of Y resulting in Z Joules/cubic feet of air. Since today is supposed to be warmer than yesterday we could do the same measurement today and find out if there was more heat today or yesterday – discussing the temperature is actually quite irrelevant.

Thermometers evolved from noticing that certain liquids expanded at a consistent rate based on the amount of heat they were exposed to. Scientists at the time created several scales that correlated this physical expansion to a number that we now call ‘degrees’. With the advent of current electronics, we now map the movement of electrons in a particular metal to that same degree. We must not confuse these indirect measurements with the core of what we are trying to measure – the amount of heat in the atmosphere at that location.

Weather observations have been taken around the world for centuries. Up until the early 1980s, a majority of the temperature observations were taken with a Liquid in Glass (LIG) mercury thermometer. Special LIG thermometers, known as minimum and maximum thermometers, were used to record the daily high and low temperature.

In the 1980s, technology allowed for sensors to become automated and computerized. Instead of using the absolute maximum or minimum, the new digital thermometers utilize algorithms to calculate the high and low temperature. The algorithm uses a 60 second sampling rate and calculates a running 5 minute average.

A joint study conducted by the University of Nebraska and Nation Climate Data Center demonstrated that the difference between the automated sensors and LIG thermometers was between .15 and .5 degrees C.

Click through here to read the entire posting and the relevant comments.

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Filed under: Weather science

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Mike Alexander, on August 25th, 2007 at 4:05 pm Said:I think the author is confused. Temperature is simply the average kinetic energy of an ensemble of particles. It is an intensive property, that is it is independent of the mass or volume of the system, unlike energy. It can be measured in a straightforward manner by a wide variety of methods. Since it itself is an average it can be averaged,

admin, on August 26th, 2007 at 9:45 am Said:You are correct that temperature is the measurement of kinetic energy but the body that is being measured is the medium of the sensor. In the case of old thermometers this is mercury. In the case of modern thermometers this is typically one or two metals with known electrical properties. We are not measuring the kinetic energy of the surrounding mass but rather its effect on the measuring device.

So while you can say that my thermometer reads 92 degrees outside and 74 degrees inside today and average the two of them to 83 degrees, you cannot say that the average indoor/outdoor temperature is 83 degrees. You could also not measure four additional points around the house that varied in shade so resulted in different temperatures (88, 94, 93, and 92) and then 2 other locations in the house that were also 74 degrees and say that the average of inside and outside is 85 degrees. (681/8).

You could take the average of the 5 outside measurements and say that the average outside temperature around my house is 91.8 but combining that with the measurement of another physical entity that is under different physical constraints is not valid. Averaging the outside numbers in this case is valid since they are located in the same general micro-climate and you are referring to the average as being around my house. Also note that I didn’t include the measurement of the air in the exhaust of my air conditioner which has radically different physical properties than the surrounding air and measures 122 degrees. If all temperature was the same, then I should rightfully include that measurement as equally weighted with the rest.

Another relevant example. In a closed room with no circulation and perfect insulation, you fill the room with standard atmosphere that is 78 degrees. You also insert into the room a gallon of boiling water (212). In addition, you put a man (98), woman (99), child (100), and their pet beagle (102) into the room. What is the average temperature of everything in the room (689/6): 114.8. However, this number is irrelevant and means nothing. (Since the room is a perfect insulator, there is also a macabre technique that would allow one to measure the average temperature but lets not get into that here).

If you want to know how much “heat” is in the room, then you need to multiply each element by its physical characteristics to ascertain the amount of heat each entity possesses. This would be in Joules or Calories and could be added together and the statement could easily be that the average amount of heat in the room is X Joules.

What we tend to forget in the global warming discussion is not temperature but rather heat. Averaging temperature of 5000 separate entities is not the point, but rather how much heat is located at each of those separate points.

Mike Alexander, on August 26th, 2007 at 5:42 pm Said:I think you are confused. Do you know how to analyze an aqueous solution for ethanol content using gas chromatography? If not then you wouldn’t try to measure a blood alcohol level would you? From you post is is clear that you do not know how to measure air temperature using any of the common techniques: RTD, a thermocouple, thermometer etc.

You might think, hey I use a thermometer when my kid is sick (which is important that it be correct) so whats the big deal? The big deal is air has less than 0.1% of the heat capacity of mouth tissue and so air temperature measures are much more sensitive to disruption. S0 you don’t measure air temperature under varies degrees of sunlight. You have to measure it in the shade. Solar radiation will disrupt your measurement of air temperature whereas it has essentially no disruptive effect on the measurement of mouth temperature. The same thing is readily apparent in a backyard pool. Assuming the pool is regularly used and so the water is mixed, if you immerse the thermometer in sunny and shaded places in the pool you will get about the same reading (unless it is a very large pool). Do the same in the yard and your will get different readings.

Now the air in your backyard is at least as well mixed as the water in the pool so it should all be at the same temperature. But if won’t *appear* to be so unless you take carefully pains to make sure your temperature reading is not affected by radiation (not just from the sun) but also from surrounding structures.

SO measuring air temperature isn’t as straightforward as sticking a thermometer into the pool or you mouth. Official temperature measures are made properly by people you know how to measure air temperature. SO this issue isn’t a problem.

For construction and *average* temperature over a region, the average much be weighted by the amount of air covered by a temperature measurement. Think of this as the problem of the average temperature of a one gallon pot of boiling water (212 F)and a 20 gallon tub of room temperature (70 F) bathwater. It’s not 121 F. You have to apply weightings: 30 gallon x 70 + 1 gallon 212 divided by 21 total gallons equals 76.8 degree F. Intuitively we know that if we pour one gallon of boiling water to a bathtub of room temperature water we are not going bathwater (we will need several pots of boiling water).

The same is true for constructing a global average temperature. It is analogous to the tub problem. Each temperature measurement is associated with a different sized pot. Since the atmosphere is more or less uniformly thick over the globe the size of the cup is the amount of surface area covered by a particular measurement. In regions where there are many measurements the cup size is small. Where there are few measurements, the cup size is large. This is why the discovery of errors in the US temperature record has such a trivial effect on the global temperature. Although a large fraction of the total measurements come form the US, they are associated with tiny cups and so the accuracy of the US measurements hardly matters.

admin, on August 26th, 2007 at 8:25 pm Said:Thank you Mike, I think you have made my point!

For the record, I fully understand how to use electronic measuring sensors. While I don’t hold as advanced degree as your own, I do have a BSME from a top engineering school and did hundreds of experiments in the attainment of that degree and then the few short years that I was a practicing engineer. My age may get in my way at times but I still remember the physics involved. I also fully understand the challenges of measuring air temperature and my simplistic example should not imply otherwise.

My example was meant to explain why it was not valid to add temperatures of different things without taking into account the physical properties of what was measured. Your example with water was exactly my point.

I do not believe that you are correct that the physical properties are the same for all places on the globe. There is less heat in the air in the desert than there is in the rainforest, if they are at the same temperature. Among other variabilities, the amount of water vapor in the air in the two locations is not the same therefore it cannot be valid to average two such locations. The barometric pressure of the air also has a small impact on the amount of energy that is stored in these gases.

As you point out, multiple temperatures of different types of bodies cannot be added together. Rather, we must take into account their physical properties which allow us to correlate to the amount of heat the body can hold.

Mike Alexander, on August 27th, 2007 at 3:17 pm Said:The properties are not exactly the same but this shouldn’t be a problem because what climatologists actually use is not a global average temperature. It’s average temperature *deviations* from their *local* norms. The temperature of a given location is compared to the average temperature of that location over the 1961-1990 period. By using deviations the individual temperatures are normalized to themselves; the effect of properties on the deviations cancel out.

These deviations are then combined using a weighted average. Here properties will have a small effect. You are correct that dry air (e.g. polar) should be weighted a bit less than the same volume of wet (e.g. tropical) air. But I wouldn’t be surprised if the weighting took this into account (it wouldn’t be hard to do).

admin, on August 27th, 2007 at 6:42 pm Said:I haven’t seen any evidence that the official temperature data weights by location other than to make sure that there are not too many data points per hundred square miles. Most locations do not gather the humidity and the pressure at the same time that the temperature is recorded. Also, since none of this data was normalized for physical properties in the control period, normalizing it now is probably not valid.

I also question that standard mathematics is correct for temperature analysis. I have read that vector mathematics may be more applicable and I tried that technique in other posts (http://globalwarming-factorfiction.com/2007/06/08/global-warming-melts-andean-glaciers-toward-oblivion/ and http://globalwarming-factorfiction.com/2007/06/01/analysis-that-doesnt-matter/). I would be interested in your thoughts on those posts as well.

Mike Alexander, on August 29th, 2007 at 5:38 pm Said:I don’t understand why you would do the vector calculation. It appears you calculate 12 slopes for the 72 year period, one for each month. You then multiply each value by 1000. You then sum these values and divide by 12000.

Now the sum is really 1000 times the sum of the slopes. Division by 12000 gives the sum of the slopes divided by 12. This is just the average slope.

You then take the arctan of the average slope.

You then take the tan of this restoring the average slope, which you multiply by 264.

To check this out I multiplied the average slope by 264 and got the value -4.6167 which agrees with th value you got.

I also multiplied the other average and median slopes to get

-4.3547 and -5.9777.

As far as I can tell these values are forecasted temperature changes over 264 years. What is the significance of the 264?

Mike Alexander, on August 30th, 2007 at 4:28 pm Said:Whew!

I decided to look into your concern about average temperatures. You have a valid point. The true “average temperature” will be the arithmetic average of the enthalpies divided by the average heat capacity. I filled three pages with math and finally got the the result that an excellent approximation for this quantity is the arithmetic average of the temperature.

The approach I took was to consider air as a two component mixture of water vapor present at a level given by the absolute (not relative) humidity or m. The balance is dry air present as the level 1-m. The specific heat capacity of water vapor is about two times that of air (Cp). (The exact values aren’t necessary they fall out in the math).

The enthalpy of water vapor (Hw) and air (Ha) are given by

Hw = 2Cp(T* – To) = 2Cp(T + Tavg – To) = 2CpT

Ha = Cp(T*-To) = Cp(T + Tavg – To) = CpT

Here T* is the local temperature and To is the standard temperature for which enthalpy is defined as zero. Now the usual practice is to measure local daily temperature (T*) relative to the average daily temperature for that date over the 1961-1990 period, which I will call Tavg

I define T as T*-Tavg in which case T* = T + Tavg. I substituted this into the above equations. Now let us define the standard temperature as Tavg. IN this case Tavg = To and we get very simply expressions for enthalpy as a function of T

The local enthalpy will be the sum of the enthapy of the two components or

m Hw + (1-m) Ha = 2mCp T + (1-m)CpT = (1+m)Cp T

Now the average heat capacity will be the heat capacity of the water vapor plus that of the air or

m 2Cp + (1-m) Cp = (1+m) Cp

Now to calculate “average temperature over a period of time or over various locations we average (1+m)CpT. And we average (1+m)Cp. The Cp drops out and we get

average temperature = average ((1+m)*T) divided by average (1+m).

For large numbers the average of (1+m)*T is equal to the average of (1+m) times the average of T. Division by the average of (1+m) gives the average of T.

*************************************************************

I do not know enough math to show this this equivalence between the produce of averages and the average of a produce. But an experiment in Excel can serve.

On the spreadsheet let X = RAND() (a random number between 0 and 1). Let Y = RAND() (a different random number between 0 and 1). Now for a given N calculate the average value of X, the average value of Y and the average value of X*Y. Compare the last value with the average X multiplied by the average Y.

I ran the simulation 49 times for N=1000 and got the following results (resuts were to three places for “zero” means less than 0.0005).

0.000 7

0.001 17

0.002 11

0.003 9

0.004 4

0.005 1

The expected value is 0.25 = 0.5 x 0.5. The average error is thus around 0.7% which is pretty small. Since there are on the order of a 1000 stations and 365 days in a year, N = very large for a single annual “world average temperature” data point. I would conclude that the equivalence is valid.

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