How to Fool Yourself with a Pyrgeometer (video)
CO2 alarmism feeds on an idea of massive backradiation or Downwelling Longwave Radiation DLR from the atmosphere to the Earth surface, about 330 W/m2 to be compared with 170 W/m2 absorbed shortwave radiation from the Sun.
DLR thus triples the radiation from the Sun to an alarming 500 W/m2 hitting the Earth surface. This should make it possible to boil eggs on the bare ground, but since this does not work out, we ask: What is the evidence that there is massive DLR?
The answer by a CO2 alarmist [and lukewarmer] is: DLR exists because you can measure it, e.g. it by a pyrgeometer:
- a device that measures the atmospheric infra-red radiation spectrum that extends approximately from 4.5 µm to 100 µm.
**watch Prof Johnson’s video presentation:
Here is how it works according to Wikipedia:
The atmosphere and the pyrgeometer (in effect the earth surface) exchange long wave IR radiation. This results in a net radiation balance according to:
Where:
Enet – net radiation at sensor surface [W/m²]
Ein – Long-wave radiation received from the atmosphere [W/m²]
Eout – Long-wave radiation emitted by the earth surface [W/m²]
Enet – net radiation at sensor surface [W/m²]
Ein – Long-wave radiation received from the atmosphere [W/m²]
Eout – Long-wave radiation emitted by the earth surface [W/m²]
The pyrgeometer’s thermopile detects the net radiation balance between the incoming and outgoing long wave radiation flux and converts it to a voltage according to the equation below.
Where:
Enet – net radiation at sensor surface [W/m²]
Uemf – thermopile output voltage [V]
S – sensitivity/calibration factor of instrument [V/W/m²]
Enet – net radiation at sensor surface [W/m²]
Uemf – thermopile output voltage [V]
S – sensitivity/calibration factor of instrument [V/W/m²]
The value for S is determined during calibration of the instrument. The calibration is performed at the production factory with a reference instrument traceable to a regional calibration center.[1]
To derive the absolute downward long wave flux, the temperature of the pyrgeometer has to be taken into account. It is measured using a temperature sensor inside the instrument, near the cold junctions of the thermopile. The pyrgeometer is considered to approximate a black body. Due to this it emits long wave radiation according to:
Where:
Eout – Long-wave radiation emitted by the earth surface [W/m²]
σ – Stefan-Boltzmann constant [W/(m²·K4)]
T – Absolute temperature of pyrgeometer detector [kelvins]
From the calculations above the incoming long wave radiation can be derived. This is usually done by rearranging the equations above to yield the so called pyrgeometer equation by Albrecht and Cox.
Where all the variables have the same meaning as before.
As a result, the detected voltage and instrument temperature yield the total global long wave downward radiation.
So now we now how DLR is measured. Does this mean that DLR exists as a physical transfer of energy from atmosphere to Earth surface? No, it does not as explained as myth of backradiation or DLR. We recall:
A pyrgeometer measures a net transfer and then invents DLR by adding the net to outgoing radiation according to Stefan-Boltzmann for a blackbody emitting into a void at 0 K.
We see that a pyrgeometer does not measure DLR directly but invents it from the formula
- E_in = E_net + E_out,
which is supposed to result from E_net = E_in – E_out expressing a Stefan-Boltzmann law of the form
- E_net = sigma Ta^4 – sigma Te^4,
where Ta and Te are the temperatures of atmosphere and Earth surface. But Stefan-Boltzmann’s law is not described this way in physics literature, where it instead takes the form
- E_net = sigma (Ta^4 – Te^4),
which does not allow extracting DLR as sigma Ta^4.
DLR and backradiation is thus fiction invented from an ad hoc formula without physical reality, which is not described in the physics literature. Nevertheless there are companies selling pyrgeometers at price of 4.000 Euro, but of course selling fiction can also serve as a business idea. But is it legal to sell fiction as science? As science fiction?
To sum up: Working with fictional differences of massive gross flows feeds alarm, while physically correct net flow does not.
Before investing in a pyrgeometer you may ask yourself what in fact such a device is measuring: fiction or reality? Or maybe it does not matter? To help to an answer you may take a look at:
- Evaluation of DLR in General Circulation Models
- DLR Measurements and Modeling
- The Amazing Case of Backradiation or DLR
- Measuring DLR Using Modified Pyrgeometers…
Read more at claesjohnson.blogspot.com
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Norman
| #
Mr. Johnson
What is the purpose of this intentionally misleading statement?
YOU: “DLR thus triples the radiation from the Sun to an alarming 500 W/m2 hitting the Earth surface. This should make it possible to boil eggs on the bare ground, but since this does not work out, we ask: What is the evidence that there is massive DLR?”
You know this is completely false. Why did you make it?
A 500 W/m^2 energy input will raise the temperature of a BB to 91 F or 33.25 C. Not even close enough to boil eggs.
The only fiction is your con job of fooling ignorant and unscientific people with a load of dung.
Why are crackpots, like yourself, inclined to make up your own physics and reality and think you are much smarter than people who actually work in the area?
What drives the insanity of a crackpot? Delusion of Grandeur?
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John O'Sullivan
| #
Norman, only a ‘crackpot’ would consider the earth to be a blackbody. PROFESSOR Johnson is Sweden’s most cited scientist in the peer-review literature.
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Norman
| #
John O’Sullivan
the Earth’s surface is close to a blackbody with an emissivity of around 095 (average). It does not matter his credentials. 500 W/m^2 hitting the surface will NOT boil eggs. What is not absorbed will be reflected (the 0.05%). He starts with a deceptive comment to see if the readers will accept it, once he hooks them he can peddle any nonsense.
Also, what are his peer-reviewed articles about? If they are math then sure, he is an expert in that field. It does not mean he knows about the other topics. I have been to his blog where experts in the fields, he brings up, slam his goofy made up ideas.
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John O'Sullivan
| #
Norman, “Close to” doesn’t cut it – it is the same old lame fudge factor time and again employed by incompetent/corrupt govt ‘scientists’ on a multiplicity of climate factors because they simply aren’t capable of performing higher level math. Experts in Applied Mathematics, such as Professor Claes Johnson, have exposed the scam by showing how fudging the data on issues such as ‘back radiation,’ omitting the water cycle (ocean currents, cloud albedo) , sun spots, cosmic rays, etc, etc, more than explains the alleged “missing 33 degrees” of heat, which govt ‘scientists’ idiotically call the “greenhouse gas effect.” Wake up and realize you’ve been scammed.
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jerry krause
| #
Hi Claes and John,
Claes, you began: ” CO2 alarmism feeds on an idea of massive backradiation or Downwelling Longwave Radiation DLR from the atmosphere to the Earth surface, about 330 W/m2 to be compared with 170 W/m2 absorbed shortwave radiation from the Sun.”
330 W/m2 for the DLR is an averaged, but sometimes actual and often measured value by the instrument whose measurements you claim cannot be valid. An average value and an actually measured value are quite similar because the DLR is not observed to have a significant diurnal oscillation because atmospheric soundings disclose that only the atmosphere’s temperature near the earth’s surface have a significant diurnal temperature oscillation. Which observation one might consider to refute the greenhouse effect because the influence of this atmospheric layer, whose base is the surface, and whose temperature can, given an apparent cloudless sky, have a significant diurnal temperature oscillation, while the DLR does not.
But relative to the pyrgeometer performance, you ignore the fact that the same instrument is being used to measure the upwelling longwave radiation (ULR) being emitted from the earth’s surface. And this measured ULR, whose values does have a significant diurnal temperature oscillation, which is consistent with the actually measured surface temperature (https://www1.ncdc.noaa.gov/pub/data/uscrn/products/hourly02/). Of course, there are some who argue, as you are about the pyrgeometer , that the instrument used to measure the surface temperature cannot measure a valid surface temperature.
Relative to the “170 W/m2 absorbed shortwave radiation from the Sun”, I have to speculate, because you do not define what this value is. I am very certain it is not the midday value of the incident solar radiation, given an apparent cloudless sky, upon the earth’s surface which (https://principia-scientific.com/record-temperature-result-of-cloud-revised-updated/) evidently warms the surface to a measured 60C and the soil, 5cm beneath the surface, to 50C. It is the latter soil temperature, which I believe that you cannot claim to be not valid, that you must address and to admit that the 170 W/m2 solar radiation could never cause. So whatever reasoning produced that value must be wrong.
Have a good day, Jerry
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William Masters
| #
NASA satellite data found here:
https://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
States:
“Just as the major atmospheric gases (oxygen and nitrogen) are transparent to incoming sunlight, they are also transparent to outgoing thermal infrared. However, water vapor, carbon dioxide, methane, and other trace gases are opaque to many wavelengths of thermal infrared energy. Remember that the surface radiates the net equivalent of 17 percent of incoming solar energy as thermal infrared. However, the amount that directly escapes to space is only about 12 percent of incoming solar energy. The remaining fraction—a net 5-6 percent of incoming solar energy—is transferred to the atmosphere when greenhouse gas molecules absorb thermal infrared energy radiated by the surface.
Why doesn’t the natural greenhouse effect cause a runaway increase in surface temperature? Remember that the amount of energy a surface radiates always increases faster than its temperature rises –(the Stefan-Boltzmann Law)—outgoing energy increases with the fourth power of temperature. As solar heating and “back radiation” from the atmosphere raise the surface temperature, the surface simultaneously releases an increasing amount of heat—equivalent to about 117 percent of incoming solar energy. The net upward heat flow, then, is equivalent to 17 percent of incoming sunlight (117 percent up minus 100 percent down).”
Thus the total backscatter radiation measured by NASA is equal to 5-6% of the total solar energy reaching the earth.
In other words, whatever the temperature outside is, 5-6% of that heat is reflected heat from the GHGs in the atmosphere.
If it is 80F outside, the 4F degrees of that, is backscatter radiation, while 76F degrees are from the Sun.
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William Masters
| #
NASA satellite data found here:
https://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
States:
“Just as the major atmospheric gases (oxygen and nitrogen) are transparent to incoming sunlight, they are also transparent to outgoing thermal infrared. However, water vapor, carbon dioxide, methane, and other trace gases are opaque to many wavelengths of thermal infrared energy. Remember that the surface radiates the net equivalent of 17 percent of incoming solar energy as thermal infrared. However, the amount that directly escapes to space is only about 12 percent of incoming solar energy. The remaining fraction—a net 5-6 percent of incoming solar energy—is transferred to the atmosphere when greenhouse gas molecules absorb thermal infrared energy radiated by the surface.
Why doesn’t the natural greenhouse effect cause a runaway increase in surface temperature? Remember that the amount of energy a surface radiates always increases faster than its temperature rises –(the Stefan-Boltzmann Law)—outgoing energy increases with the fourth power of temperature. As solar heating and “back radiation” from the atmosphere raise the surface temperature, the surface simultaneously releases an increasing amount of heat—equivalent to about 117 percent of incoming solar energy. The net upward heat flow, then, is equivalent to 17 percent of incoming sunlight (117 percent up minus 100 percent down).”
Thus the total backscatter radiation measured by NASA is equal to 5-6% of the total solar energy reaching the earth.
In other words, whatever the temperature outside is, 5-6% of that heat is reflected heat from the GHGs in the atmosphere.
If it is 80F outside, the 4F degrees of that, is backscatter radiation, while 76F degrees are from the Sun.
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jerry krause
| #
Hi William,
You conclude: “In other words, whatever the temperature outside is, 5-6% of that heat is reflected heat from the GHGs in the atmosphere.” Then you conclude: “If it is 80F outside, the 4F degrees of that, is backscatter radiation, while 76F degrees are from the Sun.”
Reflection of radiation is a physical phenomenon. Scattering of radiation is a physical phenomenon. Reflection of radiation and scattering of radiation are not the same physical phenomenon. And you seem to suggest that temperature is heat (energy). Anyone reading what you have written should be confused because you are not accurately defining and using word as they should be defined and used.
But William I commend you for using what I accept as your given name; which Jimbo (Gymbo) seems not have courage to do and instead confused me a bit by using two different aliases while writing the same thing or nearly the same thing. I haven’t tried to compare every word to see if the two comments are identical. But as I compare what you wrote and what Jimbo wrote, I am not sure if it is at all the same.
As I write this comment I begin to see more and more which is obvious. I, like you, have double submitted my comment because I considered for a moment that I hadn’t submitted it the first time because a minute can be a long time and I was not patient enough. But there is a third unknown author–NASA. Which is where you got the ‘5-6 percent’. “The remaining fraction—a net 5-6 percent of incoming solar energy—is transferred to the atmosphere when greenhouse gas molecules absorb thermal infrared energy radiated by the surface.”
All three of you overlook a commonly observed fact of the atmosphere. For there is no question that clouds (condensed water vapor) are commonly seen in the atmosphere. Can you all believe that clouds might scatter longwave IR radiation just as we can see that clouds scatter the shorter wavelength visible portion of the solar radiation?
Have a good day, Jerry.
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Alan
| #
Why all the complicated discussion? No one has yet ever demonstrated how one molecule (CO2) in every 2500 molecules of air can possibly perform as a “blanket” or any other mechanism to limit the massive upward/outward transfer of heat from the earth/oceans into the upper atmosphere and be lost to space. It just does not happen – and cannot be demonstrated to happen.
There is no “glass lid” on the atmosphere – and the relative heat capacity of the atmosphere would require it to be 3000 degrees (with a lid) in order to raise the temperature of the oceans by 1 degree.
Any “rise” in temperature of the atmosphere MUST be the result of much more significant factors than an adding one molecule of CO2 to every 2500 molecules of air (which would be 800 ppm – which we don’t have).
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James McGinn
| #
Climatology came from Meteorology. Conversational sciences share many of the same traditions. For both truth is decided by consensus and authority.
If you object to this methodology I suggest you start with meteorology. Climatology is not going to change if they can see that meteorology is still getting away with it.
How is it you never noticed this method being used by meteorologists?
If it is alright for meteorologists to employ this methodology then we really can’t fault climatologists can we?
If you are going to let the gander get away with it you cant complain about the goose.
James McGinn / Solving Tornadoes
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Jimbo
| #
“But Stefan-Boltzmann’s law is not described this way in physics literature, where it instead takes the form
E_net = sigma (Ta^4 – Te^4),
which does not allow extracting DLR as sigma Ta^4.”
I am sorry but this is nonsense and I can prove it.
Planck’s law forms the entire basis for satellite remote sensing so to decry it as incorrect whilst relying on the fact that the satellite data disproves the land based alarmisim temperature data is hypocritical.
So let’s begin by assuming Planck’s law correctly describes blackbody radiant emissions for various temperatures (it does and there is a plethora of experimental data supporting this ) you can easily demonstrate that “net” radiation CAN be determined from E_net = sigma Ta^4 – sigma Te^4.
Here are 2 Planck curves demonstrating (proving) this – if Planck is right so are these:-
https://www.dropbox.com/s/4t65jx8o1amns35/P%20net%201.png?dl=0
https://www.dropbox.com/s/oo0inggltuyt38v/P%20net%202.png?dl=0
Plot 1 shows Planck curves for 254 K (Ta = atmosphere), and 303 K (Ts = surface) as always claimed in the greenhouse effect model as well as the quantity which Professor Johnson says can’t be calculated using algebra yet there it is in graphic detail:-
The P(net) value which is obvious in the graph is exactly sigma Ts^4 – sigma Ta^4 – it is sigma*303 K^4 minus 254K^4.
This is because the value sigmaT^4 is the exact same result obtained by integrating the Planck curve equation times pi over all wavelengths in my plot (using the frequency or wavenumber domain is irrelevant – the result is the same mathmatically)and this mathematical transformation gives the Stefan-Boltzmann equation.
Surely this mathematical equilivance verifies the validity of both equations given the plethora of experimental verification ?
Therefore, by the laws of calculus, the area under the Planck curve times Pi is sigmaT^4.
Pi is relevant to remove the streadian factor and give the result for the hemisphere total emissions.
The problem for Pyrgeometers is highlighted by comparing my 2 graphs – you could plot literally thousands of these for any array of temperatures you like.
Both graphs have the same “net” value of 239.7 W/m2 but this “net” value can be arrived at in limitless combinations of temperatures !
The established flow of “het” is easily observed – it is the shaded P(net) area of the graph.
You don’t have to be too clever to see that if these two objects were isolated the equilibrium temperature value would be a curve betwee the upper and lower – backradiation cannot possibly shift the red curve upwards.
Besides, as I shown previously, algebraic additions of sigmaT^4 values do not equal the Planck curve for the Stefan-Boltzmanncalculated temperature value for the algebraic sum.
If it fails mathematically for Blackbody radiation it cannot possibly be right in the real world.
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Gymbo
| #
“But Stefan-Boltzmann’s law is not described this way in physics literature, where it instead takes the form
E_net = sigma (Ta^4 – Te^4),
which does not allow extracting DLR as sigma Ta^4.”
I am sorry but this is nonsense and I can prove it.
Planck’s law forms the entire basis for satellite remote sensing so to decry it as incorrect whilst relying on the fact that the satellite data disproves the land based alarmisim temperature data is hypocritical.
So let’s begin by assuming Planck’s law correctly describes blackbody radiant emissions for various temperatures (it does and there is a plethora of experimental data supporting this ) you can easily demonstrate that “net” radiation CAN be determined from E_net = sigma Ta^4 – sigma Te^4.
Here are 2 Planck curves demonstrating (proving) this – if Planck is right so are these:-
https://www.dropbox.com/s/4t65jx8o1amns35/P%20net%201.png?dl=0
https://www.dropbox.com/s/oo0inggltuyt38v/P%20net%202.png?dl=0
Plot 1 shows Planck curves for 254 K (Ta = atmosphere), and 303 K (Ts = surface) as always claimed in the greenhouse effect model as well as the quantity which Professor Johnson says can’t be calculated using algebra yet there it is in graphic detail:-
The P(net) value which is obvious in the graph is exactly sigma Ts^4 – sigma Ta^4 – it is sigma*303 K^4 minus 254K^4.
This is because the value sigmaT^4 is the exact same result obtained by integrating the Planck curve equation times pi over all wavelengths in my plot (using the frequency or wavenumber domain is irrelevant – the result is the same mathmatically)and this mathematical transformation gives the Stefan-Boltzmann equation.
Surely this mathematical equilivance verifies the validity of both equations given the plethora of experimental verification ?
Therefore, by the laws of calculus, the area under the Planck curve times Pi is sigmaT^4.
Pi is relevant to remove the streadian factor and give the result for the hemisphere total emissions.
The problem for Pyrgeometers is highlighted by comparing my 2 graphs – you could plot literally thousands of these for any array of temperatures you like.
Both graphs have the same “net” value of 239.7 W/m2 but this “net” value can be arrived at in limitless combinations of temperatures !
The established flow of “het” is easily observed – it is the shaded P(net) area of the graph.
You don’t have to be too clever to see that if these two objects were isolated the equilibrium temperature value would be a curve betwee the upper and lower – backradiation cannot possibly shift the red curve upwards.
Besides, as I shown previously, algebraic additions of sigmaT^4 values do not equal the Planck curve for the Stefan-Boltzmanncalculated temperature value for the algebraic sum.
If it fails mathematically for Blackbody radiation it cannot possibly be right in the real world.
Reply