By S. Fred Singer
Gallia omnia est divisa in partes tres. This phrase from Julius Gaius Caesar about the division of Gaul nicely illustrates the universe of climate scientists — also divided into three parts. On the one side are the “warmistas,” with fixed views about apocalyptic man-made global warming; at the other extreme are the “deniers.” Somewhere in the middle are climate skeptics.
In principle, every true scientist must be a skeptic. That’s how we’re trained; we question experiments, and we question theories. We try to repeat or independently derive what we read in publications — just to make sure that no mistakes have been made.
In my view, warmistas and deniers are very similar in some respects — at least their extremists are.
They have fixed ideas about climate, its change, and its cause. They both ignore “inconvenient truths” and select data and facts that support their preconceived views. Many of them are also quite intolerant and unwilling to discuss or debate these views — and quite willing to think the worst of their opponents.
Of course, these three categories do not have sharp boundaries; there are gradations. For example, many skeptics go along with the general conclusion of the warmistas but simply claim that the human contribution is not as large as indicated by climate models. But at the same time, they join with deniers in opposing drastic efforts to mitigate greenhouse (GH) gas emissions.
I am going to resist the temptation to name names. But everyone working in the field knows who is a warmista, skeptic, or denier. The warmistas, generally speaking, populate the U.N.’s IPCC (Intergovernmental Panel on Climate Change) and subscribe to its conclusion that most of the temperature increase of the last century is due to carbon-dioxide emissions produced by the use of fossil fuels. At any rate, this is the conclusion of the most recent IPCC report, the fourth in a series, published in 2007. Since I am an Expert Reviewer of IPCC, I’ve had an opportunity to review part of the 5th Assessment Report, due in 2013. Without revealing deep secrets, I can say that the AR5 uses essentially the same argument and evidence as AR4 — so let me discuss this “evidence” in some detail.
Read the full essay here: http://www.americanthinker.com/2012/02/climate_deniers_are_giving_us_skeptics_a_bad_name.html#ixzz1nn0SciyO
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Henry@bart
Hi Bart, I was just wondering: do you know the principle of carbon dating?
Not that I disagree with anything you have said,
I also believe the ratio of C12/13 cannot possibly tell us how much of the CO2 in the air is human induced or natural – i.e coming from the oceans : heat + HCO3=> CO2 + H2O.
I just cannot remember how they do carbon dating on bones and articles and stuff from the past.
HenryP says:
March 6, 2012 at 11:05 am
Henry@bart
Hi Bart, I was just wondering: do you know the principle of carbon dating?
Not that I disagree with anything you have said,
I also believe the ratio of C12/13 cannot possibly tell us how much of the CO2 in the air is human induced or natural – i.e coming from the oceans : heat + HCO3=> CO2 + H2O.
I just cannot remember how they do carbon dating on bones and articles and stuff from the past.
C14 is a radioactive isotope of carbon, which has a halflife of about 5000 years. The age of an artifact is based on the residual activity due to C14. Doesn’t work in some situations, near volcanic vents for example.
Carbon-14 is an unstable isotope of carbon containing 6 protons and 8 neutrons in its nucleus. It is generated constantly in upper atmosphere, by absorption of thermal neutrons (coming from space) by Nitrogen-14. The half-life of 14C is 5,730 ± 40 years (it decays back into Nitrogen-14), so the generation and decay keep its atmospheric content in balance at 0.0000000001% of the carbon in the atmosphere.
Green plants absorb atmospheric CO2 and incorporate it in their structures, including 14C. So as long as they live and grow, the ratio of 14C to other C in their structures will be (app.) the same as in the air. Ditto for animal life (including us) which in the final instance live on plant produce. But when a plant or animal dies, it ceases to incorporate new 14C and what is in its body just decays. So the age of a dead biological structure can be determined by measuring the ratio of 14C v. other C in it.
The reason this does not work “near volcanic vents” (actually near any sources of subterranean CO2 such as springs of naturally carbonated mineral water) is that “underground” CO2 has typically been sequestered for so long that all its 14C has decayed. Plants growing near such vents therefore incorporate 14C-free CO2 – if a fresh leaf of such a plant is analyzed by radiocarbon dating, it will appear to be over 60.000 years old.
However, we did increase aerial concentration of 14C in 1950-60s: during the period of atmospheric testing of nuclear weapons these contributed additional thermal neutrons which about doubled the amount of 14C in some areas (the peak of atmospheric 14C was in 1960-1965). Plants growing in that period therefore incorporated more 14C, so unmodified radiocarbon dating would classify them as younger than they actually are. So radiocarbon labs apply corrective factors for that particular period.
Which is relevant to IPCC assumption that CO2 remains in the atmosphere for over 100 years. In fact, measurements of aerial 14C have demonstrated that over 80% (of additional 14C generated by atmospheric nuclear tests) was absorbed within 25 years (given the half-life of 5,730 ± 40 years it could not have decayed).
Henry@ur momisuglyPhil.&Miso
Thanks for the info, this helps me quite a bit,
it makes sense to believe that C14 generation stops when the plant/animal is dead,
as explained.
Do you know how they measure that ratio, C12/C14
especially with that much accuracy?
Henry@ur momisugly those who followed my argument with Phil.
I am still not convinced the back radiation is only 50% of oncoming light.
I think the form of a molecule more or less approximates that of a sphere,
therefore the second the molecule is hit by radiation,
and it is filled to saturation,
I think it returns ca. 62.5% of oncoming light in a radius of 180 degrees, at the absorptive regions,
and the rest goes the other way.
http://www.letterdash.com/HenryP/the-greenhouse-effect-and-the-principle-of-re-radiation-11-Aug-2011
HenryP says:
March 7, 2012 at 6:59 am
Henry@ur momisuglyPhil.&Miso
Thanks for the info, this helps me quite a bit,
it makes sense to believe that C14 generation stops when the plant/animal is dead,
as explained.
Do you know how they measure that ratio, C12/C14
especially with that much accuracy?
For large samples the number of counts/gram of carbon measured by scintillation counter is sufficiently accurate. For very small samples mass spectroscopy is used. Because the level of C14 in the air fluctuates a calibration curve is applied which compensates for this using samples of known age. Measurements are made relative to the levels in 1950 (before present) the reference standard being sugar harvested in that year if I recall correctly, (this is to avoid the nuclear testing contamination referred to above.
Henry@ur momisugly those who followed my argument with Phil.
I am still not convinced the back radiation is only 50% of oncoming light.
I think the form of a molecule more or less approximates that of a sphere,
therefore the second the molecule is hit by radiation,
and it is filled to saturation,
I think it returns ca. 62.5% of oncoming light in a radius of 180 degrees, at the absorptive regions,
and the rest goes the other way.
Well you’re wrong, a molecule that has absorbed a photon doesn’t become a reflector it either absorbs another one or there’s no interaction and that photon keeps on going. Any emission from the excited molecule has an equal chance of going in any direction, near the surface a large fraction of the excited molecules ore collisionally deactivated anyway.
Phil. says
a molecule that has absorbed a photon doesn’t become a reflector it either absorbs another one or there’s no interaction
Henry says:
You are still not getting it. I was hoping you did see it. I am getting desperate (with you).
Obviously it must be filled with many photons at the absorptive region, before it becomes reflective,
cq. deflective
hence, how else would you explain same radiation coming back to us via the moon?>
see fig 6 top and fig 7 here:
http://www.iop.org/EJ/article/0004-637X/644/1/551/64090.web.pdf?request-id=76e1a830-4451-4c80-aa58-4728c1d646ec
HenryP says:
March 7, 2012 at 10:22 am
Phil. says
“a molecule that has absorbed a photon doesn’t become a reflector it either absorbs another one or there’s no interaction”
Henry says:
You are still not getting it. I was hoping you did see it. I am getting desperate (with you).
Obviously it must be filled with many photons at the absorptive region, before it becomes reflective,
cq. deflective
hence, how else would you explain same radiation coming back to us via the moon?>
see fig 6 top and fig 7 here:
OK Henry I’ll try to explain what you’re seeing.
Lower part of Fig 6:
This is the clear sky spectrum i.e. no cloud effects.
The sunlight passes through the atmosphere and some weak absorptions occur (red, green, orange lines). That sunlight minus the absorptions reflects from the Earth and passes through the atmosphere again and more absorption occurs in the same lines. It then reflects back from the moon and executes a third pass through the atmosphere and then is detected by the telescope. So what you’re seeing is the spectrum resulting from a triple pass through the atmosphere. If any of those features were strong absorption features you’d detect zero signal at the line. The upper part of the figure includes the effect of clouds and ground reflectivity, you can see the dark bands from the molecular species there too.
Phil. says:
So what you’re seeing is the spectrum resulting from a triple pass through the atmosphere
Henry says:
I think you do not understand what they are doing. I quote from the paper:
the current bestway to study the spectral characteristics of planet Earth is to
observe earthshine, the faint light seen on the night side of the
crescent Moon. Earthshine is light reflected from the sunlightilluminated
(day side) Earth onto the night side of theMoon and
back again into our telescopes on the night-side Earth.
So here is what really happens:
Sunshine falls on the earth, a number of substances (mostly GHG’s) absorb this sunlight, namely, especially water and CO2, leading to re-radiation or back radiation.
The earthshine (FROM THE SUNLIGHT ILLUMINATED) dayside of earth falls on the night side of the moon. From the moon it travels back, now to the night side of earth where the absorption spectra are picked up. Note that we are only looking at near infra red spectra and earth does emit this at night.
So, there is no “triple” by-pass or anything like that.
You are seeing happening what I have been saying from the start: In the absorptive region of the substances sun light is re-radiated or back radiated or deflected or reflected, whichever term you prefer, leading to cooling, i.e. if the CO2 were not there for us, more light of 2 and 4 um would hit us on our heads.
Please tell me that you understand it now.
Sorry,
Note that we are only looking at near infra red spectra and earth does emit this at night.
should be
Note that we are only looking at near infra red spectra and earth does NOT emit this at night.
HenryP says:
March 7, 2012 at 11:13 pm
Phil. says:
So what you’re seeing is the spectrum resulting from a triple pass through the atmosphere
Henry says:
I think you do not understand what they are doing. I quote from the paper:
the current bestway to study the spectral characteristics of planet Earth is to
observe earthshine, the faint light seen on the night side of the
crescent Moon. Earthshine is light reflected from the sunlightilluminated
(day side) Earth onto the night side of theMoon and
back again into our telescopes on the night-side Earth.
I understand it very well, it’s as I outlined above.
So here is what really happens:
Sunshine falls on the earth, a number of substances (mostly GHG’s) absorb this sunlight, namely, especially water and CO2, leading to re-radiation or back radiation.
There is no re-radiation at those frequencies, look at fig 6, you’ll see the absorption in those bands compared with the clouds.
The earthshine (FROM THE SUNLIGHT ILLUMINATED) dayside of earth falls on the night side of the moon. From the moon it travels back, now to the night side of earth where the absorption spectra are picked up. Note that we are only looking at near infra red spectra and earth does emit this at night.
The earth emits IR at both day and night, but not in the 0.7-2.4 μm range!
So, there is no “triple” by-pass or anything like that., it’s triple pass not bypass!
You are seeing happening what I have been saying from the start: In the absorptive region of the substances sun light is re-radiated or back radiated or deflected or reflected, whichever term you prefer, leading to cooling, i.e. if the CO2 were not there for us, more light of 2 and 4 um would hit us on our heads.
The CO2 is absorbing which is why there is less reflected light in the CO2 bands of the light reflected from the moon, read the paper carefully.
Please tell me that you understand it now.
I do!
Phil.
You don’t understand. You are still being confused by the term “absorbs”;
unfortunately many people are. I wished they had stayed with the term “extinction”.
let’s try this:
in the solar spectrum below you see the yellow (TSI) measured on top of the atmosphere.
The red is what actually lands on earth at sea level, on a cloudless day.
Now, according to your thinking, where did the difference go, exactly?
http://en.wikipedia.org/wiki/File:Solar_Spectrum.png
Well, Phil,
Why so quiet? You are the “expert”. But I know your thinking, because I heard it many times before,
and I know you were going to say that the odd 30% of sunlight not reaching earth is somehow “absorbed” in the atmosphere and mostly passed on as “heat” to neighbouring molecules….
But now how can that be,
if earth’s albedo has also been calculated as 30%?
Clearly, that does not add up?
How is it possible for a tiny bit of ozone in the upper atmosphere to strike off almost 20% of all incoming sunlight, especially in the UV region?
We know that ozone absorbs in the UV region (as also evident from the previously referred-to paper)
The only way for this to be is if things happen exactly as I told you: the Ozone absorbs UV light and re-radiates it, at least 50% of all on-coming light in a radius of 180 degrees the direction where it came from…
(I still think it is 62,5%, if the molecule approximates a sphere – I was hoping for an expert to tell me, but clearly I am the only expert here).
“Absorbed” was the wrong term to have been used by the analytical chemists. They should have continued with the term “extinction”, meaning the light is lost – it went the other way.
Please tell me that you do understand it now.
BTW you might have picked up on the way (if you really studied the paper and graphs ) that O2 also absorbs (and therefore re-radiates). Last time you still argued with me that oxygen does not have any absorptions.
HenryP says:
March 9, 2012 at 8:03 am
Well, Phil,
Why so quiet? You are the “expert”.
Yes I am, but I also have a life and only just saw this post!
But I know your thinking, because I heard it many times before,
and I know you were going to say that the odd 30% of sunlight not reaching earth is somehow “absorbed” in the atmosphere and mostly passed on as “heat” to neighbouring molecules….
But now how can that be,
if earth’s albedo has also been calculated as 30%?
Clearly, that does not add up?
Sorry, your mind-reading failed you this time, the Bond Albedo is “is the fraction of power in the total electromagnetic radiation incident on an astronomical body that is scattered back out into space”, so I would have told you that it was scattered back unchanged into space.
How is it possible for a tiny bit of ozone in the upper atmosphere to strike off almost 20% of all incoming sunlight, especially in the UV region?
We know that ozone absorbs in the UV region (as also evident from the previously referred-to paper)
It’s a very strong absorber of UV. Below 240 nm O2 is a very strong absorber of UV which causes the O2 to disintegrate into two O atoms. Note that the O2 is unable to re-radiate because it no longer exists! One of those O atoms can form a molecule of O3 if it collides with an O2 molecule but it needs to collide with another molecule (e.g. N2) to remove some of the excess energy so that the O3 molecule doesn’t rapidly fall apart. The O3 is extremely effective at absorbing UV around 250nm, however that causes the O3 molecule to fall apart into O2 + O.
Yet again the O3 will not re-radiate because it no longer exists! This process is known as photodissociation, I suggest you look it up.
The only way for this to be is if things happen exactly as I told you: the Ozone absorbs UV light and re-radiates it, at least 50% of all on-coming light in a radius of 180 degrees the direction where it came from…
As shown above chemists who know what they are talking about find the Chapman mechanism more persuasive.
(I still think it is 62,5%, if the molecule approximates a sphere – I was hoping for an expert to tell me, but clearly I am the only expert here).
You have been told you just won’t listen. What exactly is meant by “in a radius of 180 degrees the direction where it came from…”, is not very clear, radii are distances, not angles. In any case you should be using solid angles. 50 % of the re-radiated light will be emitted within 90º of the direction of the incident photon. Where on earth you get 62.5% from is a mystery.
“Absorbed” was the wrong term to have been used by the analytical chemists. They should have continued with the term “extinction”, meaning the light is lost – it went the other way.
No, they knew what they were doing!
Please tell me that you do understand it now.
I always did, you on the other hand…..
BTW you might have picked up on the way (if you really studied the paper and graphs ) that O2 also absorbs
Yes, some rather weak bands in the visible and near IR which are orders of magnitude weaker than neighboring water bands, the ones that show up in the reflection spectrum happen to lie in gaps in the water spectrum.
(and therefore re-radiates). Last time you still argued with me that oxygen does not have any absorptions.
In the IR appropriate to the Earth’s energy budget it does not, the solitary band at around 6.46 microns is swamped by many orders of magnitude by CO2 (hundreds of billions of times weaker) and similarly by water. I’ve posted about this here before.
Henry@Phil.
The formation of ozone from UV and O2 is clear
UV + 3O2 => 6O => 2O3
Then same ozone turns against the UV sunlight absorbing it, until saturation, and starts deflecting it, when the UV falls on it..
Why now turn this into some kind of different process as those when a certain type of radiation bounces off the absorptive regions of water and CO2?
The reason I say 62.5 % has to do with the fact that the molecule is probably more likely to be shaped as a sphere, so that, when it has absorbed photons to capacity, it must start emitting back at least 62.5 of incident radiation in a radius of 180 degrees. It is a similar effect to putting your car lights on bright in misty/humid/mosit conditions: your light is returned to you
“Divide and Conquer” Mr. Singer?
But then again that is a tried and true method for dealing with the “Enemy” and we should not fall for it. Unfortunately it works so well that it is used very often by those craving power. The attack by other horsemen on Rodeos and Carriage horses come to mind as an example. Also the recent successful attack on small farmers in the USA when it is Corporate practices that are to blame for food poisoning.
There are many many other examples of “Divide and Conquer” so I guess it should not be a surprise that “Deniers” are now subject to this method of attack.
HenryP says:
March 9, 2012 at 10:52 am
Henry@Phil.
The formation of ozone from UV and O2 is clear
UV + 3O2 => 6O => 2O3
It is clear, but that’s not it, the Chapman mechanism is well established and tested:
O2 + hν ➔2 O k1 (s-1) (1)
O + O2 + M ➔O3 + M k2 (cm6 molecule-2 s-1) (2)
O3 + hν ➔O + O2 k3 (s-1) (3)
O + O3 ➔ 2 O k4 (cm3 molecules-1 s-1) (4)
M is any non-reactive species that can take up the energy released in reaction (2) to stabilize O3.
You’ll find a good account of this at:
http://www.columbia.edu/itc/chemistry/chem-c2407/hw/ozone_kinetics.pdf
Then same ozone turns against the UV sunlight absorbing it, until saturation, and starts deflecting it, when the UV falls on it..
No, your concept of tiny mirrors isn’t correct! Even more so in this case since the O3 falls apart once it absorbs the first photon.
Why now turn this into some kind of different process as those when a certain type of radiation bounces off the absorptive regions of water and CO2?
The reason I say 62.5 % has to do with the fact that the molecule is probably more likely to be shaped as a sphere, so that, when it has absorbed photons to capacity, it must start emitting back at least 62.5 of incident radiation in a radius of 180 degrees.
Why must it?
It is a similar effect to putting your car lights on bright in misty/humid/mosit conditions: your light is returned to you
You are confusing elastic scattering from particles of size greater than the wavelength of light (Fraunhofer scattering) with elastic scattering from molecules which are much smaller than the wavelength (Rayleigh scattering). Neither relates to the absorption and re-emission which we are discussing here. In either case the ratio of back scattering to forward scattering is not 5/3 as you assert, in the molecular scattering case it’s 1 (i.e. 50% each way). Note that under the conditions you describe if a car with its headlights on approaches you then a lot of the light is scattered towards you, as well as backwards.
I am most definately with CodeTech on this:
“Personally I find this to be one of the most offensive posts and subsequent batch of comments that I’ve yet seen on WUWT…..”
Not only that I would make the three catagories, Warmist, Skeptic and Trojan Horse.
It is the Trojan Horses and not those Singer calls Deniers who are the real threat. Unfortunately we see them here at WUWT all the time. So much so that I sometimes wonder if a different Trojan Horse is assigned a specific topic and paid to write comments that refute skepticism on that particular topic.
Henry@Phil.
Nevermind how the ozone layer is formed, the important thing is to remember that there is a thin layer of ozone on top of (most) of earth to shield us from the harmful UV; BTW ozone is also formed by electrical storms, and countless human processes, like pumps, electro motors, vacuum cleaners, etc..
Now on to the next graph:
http://www.google.co.za/imgresmgurl=http://www.barrettbellamyclimate.com/userimages/Sun2.jpg&imgrefurl=http://www.barrettbellamyclimate.com/page11.htm&h=965&w=963&sz=341&tbnid=I4bPEwmMiTNtKM:&tbnh=90&tbnw=90&zoom=1&docid=Y0dkNn0-Wh0hUM&sa=X&ei=OOtaT8WOK8LMhAeXx8yoBA&ved=0CEoQ9QEwBQ&dur=2256
On the bottom below we see the spectra and infra red spectra of the individual components. This is what you must study carefully. See how the absorptions of the spectra affects the outgoing radiation of earth and see how it affects the incoming radiation. For example, let us look at the absorption of ozone at around 10-11 um? It makes a dent in earth’s out going radiation at 10-11.Do you see that?
In other words what happens: Radiation from earth of 10-11 goes up, hits on the ozone, which is already absorpbed to capacity and therefore a great percentage (at least 50%, probably more) is sent back to earth, leading to entrapment of heat, leading to a warming effect.
Also look at water vapor and CO2 around 2 um and see how that makes a dent in the incoming solar radiation.
The only explanation that is possbile is that when the absorptive regions are filled, it starts re-radiating.
It starts behaving like a little mirror in the absorptive regions.
As per the definition of the GH effect (trapping of heat on earth), look here
http://wattsupwiththat.com/2012/02/29/climate-deniers-are-giving-us-skeptics-a-bad-name/#comment-911368
Similarly, you also have a cooling effect (re-radiation & subsequent deflection of radiation to space), because that definition can also be applied for substances reacting under incoming sunlight.
Now if those people who figured out there is a warming effect (i.e. earth emmission bouncing back from clouds, CO2, water vapor, ozone etc) can also figure out how much each of he GHG’s are cooling the atmosphere?
The link I gave @ur momisugly Phil does not work.
http://www.google.co.za/imgres?imgurl=http://www.barrettbellamyclimate.com/userimages/Sun2.jpg&imgrefurl=http://www.barrettbellamyclimate.com/page11.htm&h=965&w=963&sz=341&tbnid=I4bPEwmMiTNtKM:&tbnh=90&tbnw=90&zoom=1&docid=Y0dkNn0-Wh0hUM&sa=X&ei=OOtaT8WOK8LMhAeXx8yoBA&ved=0CEoQ9QEwBQ&dur=2256
Henry@Phil
I have updated my own blog a bit,
Please do let me know if there is anything there that you (still) do not agree with
http://www.letterdash.com/HenryP/the-greenhouse-effect-and-the-principle-of-re-radiation-11-Aug-2011
HenryP says:
March 9, 2012 at 10:41 pm
Henry@Phil.
Nevermind how the ozone layer is formed, the important thing is to remember that there is a thin layer of ozone on top of (most) of earth to shield us from the harmful UV;
Well it does matter in the context of your ‘explanation’ of its formation and hypothesis of ‘re-emission’ of UV radiation.
Now on to the next graph:
http://www.google.co.za/imgresmgurl=http://www.barrettbellamyclimate.com/userimages/Sun2.jpg&imgrefurl=http://www.barrettbellamyclimate.com/page11.htm&h=965&w=963&sz=341&tbnid=I4bPEwmMiTNtKM:&tbnh=90&tbnw=90&zoom=1&docid=Y0dkNn0-Wh0hUM&sa=X&ei=OOtaT8WOK8LMhAeXx8yoBA&ved=0CEoQ9QEwBQ&dur=2256
On the bottom below we see the spectra and infra red spectra of the individual components. This is what you must study carefully. See how the absorptions of the spectra affects the outgoing radiation of earth and see how it affects the incoming radiation. For example, let us look at the absorption of ozone at around 10-11 um? It makes a dent in earth’s out going radiation at 10-11.Do you see that?
Yes, O3 has absorbed IR radiation from the earth.
In other words what happens: Radiation from earth of 10-11 goes up, hits on the ozone, which is already absorpbed to capacity and therefore a great percentage (at least 50%, probably more) is sent back to earth, leading to entrapment of heat, leading to a warming effect.
The highlighted statement has no basis in fact, where did you get it from?
Also look at water vapor and CO2 around 2 um and see how that makes a dent in the incoming solar radiation.
Yes, H2O and CO2 are absorbing incoming solar radiation and rapidly transferring the energy to the surrounding air molecules.
The only explanation that is possbile is that when the absorptive regions are filled, it starts re-radiating.
It starts behaving like a little mirror in the absorptive regions.
No, this does not happen, the explanation is as I gave it above, in fact in this case there is clearly no re-emission, look at the graph! The absorptive regions do not fill.
As per the definition of the GH effect (trapping of heat on earth), look here
http://wattsupwiththat.com/2012/02/29/climate-deniers-are-giving-us-skeptics-a-bad-name/#comment-911368
Similarly, you also have a cooling effect (re-radiation & subsequent deflection of radiation to space), because that definition can also be applied for substances reacting under incoming sunlight.
Only if those substances are at ~5000K, see S-B law.
Now if those people who figured out there is a warming effect (i.e. earth emmission bouncing back from clouds, CO2, water vapor, ozone etc) can also figure out how much each of he GHG’s are cooling the atmosphere?
They can, it’s been done by Clough and Iacono, see for example http://www.agu.org/pubs/crossref/1995/95JD01386.shtml
http://www.agu.org/pubs/crossref/1995/95JD01386.shtml
Henry, it’s still wrong starting from where you argue that it should be extinction not absorption.
Phil. says
4) “Well it does matter in the context of your ‘explanation’ of its formation and hypothesis of ‘re-emission’ of UV radiation.”3) “The highlighted statement has no basis in fact, where did you get it from?”
2) Yes, H2O and CO2 are absorbing incoming solar radiation and rapidly transferring the energy to the surrounding air molecules.
1) They can, it’s been done by Clough and Iacono, see for example http://www.agu.org/pubs/crossref/1995/95JD01386.shtml
Henry@Phil.
Well, Phil. answering you backwards to forwards on your reply:
1) Quoting that paper you are now contradicting yourself. Are you saying you now believe that there is back radiation of sunshine by the GHG’s? I suggest you read the paper. The paper does not measure or calculate a cooling and warming rate for each of the GHG’s.
2) You still don’t get it. You keep thinking that “absorption” must be some kind of magical thing that continuously passes the light as energy on to neighbouring molecules. That is not happening, mostly. You still donot understand it. Why don’t you do the experiment with a spectrophotometer that will show you that this does not happen? Remember the O2 and N2 lets most radiation through. Let’s take one molecule CO2. There is a constant beam of light falling of 4 um on it, 12 hours per day. CO2 is used as fire retardant and does not conduct heat. In fact it insulates. How many substances in the air can take the re-radiated 4 um except the water? So it is re-radiated back to space. Hence we are able to measure as it bounces off from the moon back to earth.. Amazing actually, if you consider that the CO2 is only 0.04% of air.
3) Earth emits 24 hours per day 5-20 um, so it means that the (very thin) ozone layer at 10-11 microns must be filled up to capacity all of the time. Hence the 10-11 bounces back to earth, causing a delay in cooling.
4) Clearly, the graph/reprsentation that was quoted here
http://wattsupwiththat.com/2012/02/29/climate-deniers-are-giving-us-skeptics-a-bad-name/#comment-918080
shows that without the ozone and CO2 and H2O and other GHG’s you will get a lot more radiation on your head. In fact, you would probably fry.
Phil. says on
http://www.letterdash.com/HenryP/the-greenhouse-effect-and-the-principle-of-re-radiation-11-Aug-2011
Henry, it’s still wrong starting from where you argue that it should be extinction not absorption.
Henry@Phil.
Phil. I have given you many examples that you can actually do for yourself to experience how re-radiation works. You keep hanging on to a theory that does not make sense with the observations.
I feel bad that you did not teach your children well. You have a thick head.
Let us agree to disagree on this, as I think we did before.
Henry, sometimes you’s sensible, but …
Um, nope. Nothing to do with insulation. It doesn’t support combustion, already being “combusted”. It stifles the fire, by excluding O2. So would pure N2, or Argon, or …
Mods, my reply to this appears to be in limbo.
[REPLY: There is nothing in the queue. Your last post was yesterday. Try entering again. -REP]
Henry@BrianH
OK, you are right of course. I think I meant to say that air insulates, which is why you leave a strip of air between a double walled house. Phil thinks that if you put a beam of 2 or 4 um on CO2 (where CO2 absorbs), then that radiation is being continuously “absorbed” and “passed on” to neighbouring molecules as heat. Clearly this happens only up until absorption is complete (saturation). Subsequently the 2 and 4 is back radiated in all directions including back to the sun.
You agree?