Double secret probation at tallblokes again!

Another post I should not have to make!!!!

Tried posting 2 times on tallblokes using different browsers but nothing got through including a 3rd post with no links

the post was something like:

Your electricity prices differ from uk government figures:

https://www.gov.uk/government/uploads/system/uploads/attachment_data/file/511952/table_531.xls

 

another post at tallblokes

thefordprefect says: Your comment is awaiting moderation.

Thank you Tom. The truth of your comments is obvious. For example:
Once upon a time Steve McIntyre usefully analysed climate papers. His website over the last year has degenerated into a defamation fest.
WUWT has turned into series of very opinionated posts which even watts has distanced himself.
And the slayers sites are just unbelievable!

Climate science will never be able to show 100% proof that AGW is true or for that matter false. There are too many variables and too much noise in the data. Predicting the future is just about impossible. reaching backwards in time before the thermometers were correctly sited is just about impossible – tree-mometers clam-mometers,ice-mometers all have their problems!. Proving the existence or not of MWP is not possible.

The science of GHGs is understood. But just how much warming will result if CO2 increases – are the feedbacks negative or positive?

We need scientific debate from the best minds and we need communicators to educate us masses about the science.

Many blogs even limit debate by banning posters who seem to me to be talking calmly and sensibly but are against the blog “theme”.

A warming world is really a problem even if some say 1°C increase would improve their environment. They seem to forget about the extra energy powering “weather” and forget that those living on the edge of thermal disaster may be pushed over that edge.

There is only the earth on which we live, and running uncontrolled experiments on its climate does not seem sensible at all.

Free and open debate is needed!!!!!!!!!!

talkshop held post why – its just a comment

Result! Coalition slashes funds to UN climate scroungers

Posted: December 1, 2014 by tallbloke in solar system dynamics     

thefordprefect says: Your comment is awaiting moderation.

so it seems that you are saying in one breath that climate research is inconclusive but you are happy that it is being defunded so more research is unlikely.

This sounds as if you do not want more research because you may not like the consequences.

If AGW is valid then we MUST know when and how much and what to do. If AGW is not occurring then research will eventually show this and researchers can look to increase knowledge elsewhere.

In either case more research is needed not less!

Hiding your head in the sand will do no good!

talkshop post

http://www.newscientist.com/article/mg21628850.200-the-sickening-truth-about-wind-farm-syndrome.html#.U-i1JXhwa9I

“The list includes “deaths, yes, many deaths”, none of which have ever come to the attention of a coroner, cancers, congenital malformations, and every manner of psychiatric problem. But mostly, it includes common health problems found in all communities, with wind turbines or not. These include greying hair, energy loss, concentration lapses, weight gain and all the problems of ageing. Sleep problems are mentioned most, but insomnia is incredibly common. Animals get a look in. Chickens won’t lay; earthworms vanish; hundreds of cattle and goats die horrible deaths from “stray electricity”.

Bird deaths are regrettable but should be considered in respect of other causes
http://www.climateandstuff.blogspot.co.uk/2013/06/bird-deaths-by-wind-turbines.html

a simple post at wuwt

more believable if they had not have wiped out Canada Greenland and the odd island or 2 in their desire to show (incorrectly) what most climate scientists predict

ref: heartland video showing Greenland gone with the sea ice!!!!!!!!!!!!!!!

tall bloke backup comments

  1. thefordprefect says: Your comment is awaiting moderation.
  2. BNP leader explains his defeat by saying people voted for “Ukip’s racist policies instead”.
  3.  

  1. Your comment is awaiting moderation.

    ps ref:
    http://news.uk.msn.com/nick-griffin-calls-ukip%E2%80%99s-policies-%E2%80%9Cracist%E2%80%9D

     

    plus this from reference:

    “I’ve lost count of the number I’ve spoken to who say, ‘We really like the BNP but we are voting Ukip because there is more chance they will stop immigration and send them all home’. As there is not a hope in hell of that, people are going to be very disappointed when they find out what Ukip really stands for and that huge vote is going to come back to us.”

    ….

    “While Front National is nearly as soft as Ukip, such votes show the people are waking up. Real change will follow.”

yet more backup from tallblokes

thefordprefect says:
Your comment is awaiting moderation.

April 8, 2014 at 12:27 pm
Evolution
We need to stop the cult of dawinism from being taught this instant. After all there is no half dinosaur/half man skeleton. What proof is there of evolution.

You just need to look at the creation museum for proof of which I speak.
http://creationmuseum.org/whats-here/photo-preview/

Then look at the names subscribing to the Cornwall Alliance
http://www.cornwallalliance.org/articles/read/an-evangelical-declaration-on-global-warming/

to understand how wrong the greens are

WHAT WE DENY

1.We deny that Earth and its ecosystems are the fragile and unstable products of chance, and particularly that Earth’s climate system is vulnerable to dangerous alteration because of minuscule changes in atmospheric chemistry. Recent warming was neither abnormally large nor abnormally rapid. There is no convincing scientific evidence that human contribution to greenhouse gases is causing dangerous global warming.

and these people support
Dr. Roy W. Spencer (Principal Research Scientist in Climatology, University of Alabama, Huntsville,
Dr. Joseph D’Aleo (Executive Director and Certified Meteorologist, Icecap
Dr. David Legates (Associate Professor of Climatology, University of Delaware
Dr. Ross McKitrick (Associate Professor of Economics, University of Guelph, Ontario, Canada,
Dr. Cornelis van Kooten (Professor of Economics and Research Chair in Environmental Studies and Climate, University of Victoria, British Columbia, Canada, Expert Reviewer, Intergovernmental Panel on Climate Change)
Dr. Kenneth W. Chilton (Founder and Emeritus Director, Institute for the Study of Economics and the Environment, Lindenwood College);

so it cannot be wrong

more tallbloke backup

  1. thefordprefect says: Your comment is awaiting moderation.
  2. Effects of elevated CO2 and nitrogen on wheat growth and photosynthesis
    M. PAL1, L.S. RAO, V. JAIN, A.C. SRIVASTAVA*, R. PANDEY, A. RAJ and K.P. SINGH
    Division of Plant Physiology, Indian Agricultural Research Institute, New Delhi, India
    Abstract
    The effects of nitrogen [75 and 150 kg (N) ha-1] and elevated CO2 on growth, photosynthetic rate, contents of soluble leaf proteins and activities of ribulose-1,5-bisphosphate carboxylase/oxygenase (Rubisco) and nitrate reductase (NR) were studied on wheat (Triticum aestivum L. cv. HD-2285) grown in open top chambers under either ambient (AC) or elevated (EC) CO2 concentration (350 ± 50, 600 ± 50 μmol mol-1) and analyzed at 40, 60 and 90 d after sowing. Plants grown under EC showed greater photosynthetic rate and were taller and attained greater leaf area along with higher total plant dry mass at all growth stages than those grown under AC. Total soluble and Rubisco protein contents decreased under EC but the activation of Rubisco was higher at EC with higher N supply. Nitrogen increased the NR activity whereas EC reduced it. Thus, EC causes increased growth and PN ability per unit uptake of N in wheat plants, even if N is limiting.
  3.  

  1. Your comment is awaiting moderation.

    Effects of elevated CO2 on the protein concentration of food crops: a meta-analysis
    DANIEL R. TAUB1,2, BRIAN MILLER1 and HOLLY ALLEN2
    Abstract
    Meta-analysis techniques were used to examine the effect of elevated atmospheric carbon dioxide [CO2] on the protein concentrations of major food crops, incorporating 228 experimental observations on barley, rice, wheat, soybean and potato. Each crop had lower protein concentrations when grown at elevated (540–958 μmol mol−1) compared with ambient (315–400 μmol mol−1) CO2. For wheat, barley and rice, the reduction in grain protein concentration was ∼10–15% of the value at ambient CO2. For potato, the reduction in tuber protein concentration was 14%. For soybean, there was a much smaller, although statistically significant reduction of protein concentration of 1.4%. The magnitude of the CO2 effect on wheat grains was smaller under high soil N conditions than under low soil N….

     

  1. Your comment is awaiting moderation.

    Cassava is an important food for millions of people in Africa, Asia, and Latin America. When grown in conditions of increased CO2, however, its cyanide levels jump.
    http://monash.edu/science/about/schools/biological-sciences/staff/gleadow/docs/gleadow-2009-cassava-online.pdf
    The effects of nitrogen [75 and 150 kg (N) ha-1] and elevated CO2 on growth, photosynthetic rate, contents of soluble leaf proteins and activities of ribulose-1,5-bisphosphate carboxylase/oxygenase (Rubisco) and nitrate reductase (NR) were studied on wheat (Triticum aestivum L. cv. HD-2285) grown in open top chambers under either ambient (AC) or elevated (EC) CO2 concentration (350 ± 50, 600 ± 50 μmol mol-1) and analyzed at 40, 60 and 90 d after sowing. Plants grown under EC showed greater photosynthetic rate and were taller and attained greater leaf area along with higher total plant dry mass at all growth stages than those grown under AC. Total soluble and Rubisco protein contents decreased under EC but the activation of Rubisco was higher at EC with higher N supply. Nitrogen increased the NR activity whereas EC reduced it. Thus, EC causes increased growth and PN ability per unit uptake of N in wheat plants, even if N is limiting.

tallbloke back up

 

David Rose: BBC boss gags ‘sceptics’ from climate change debates

thefordprefect says: Your comment is awaiting moderation.

I think it despicable that BBC Apollo coverage has not given equal time to the fact that it was all filmed in Hollywood. I think it despicable that the BBC does not give equal airtime to the fact that vaccines are causing autism. I think it despicable that the BBC does not preface all palaeontology programmes with back to back  equal time to the fact that the earth was created 6000 years ago. Why do they not preface every travel programme with warning about falling into the void at the edge of the world.

Is fairness is required to all theories who expounds them even if they are loopy – for who is to say these cannot be correct

These programmes are watched and listened to by children who could easily be lead down the wrong path into stupidity. There has to be some significance paid to the overwhelming scientific theories as these are the best we have.

tallbloke post backup

This is of course very interesting and these warming reports are significant. Just take a look at the temperature record:

http://www.woodfortrees.org/plot/hadcrut4gl/mean:60/plot/hadcrut3vgl/mean:60/plot/hadcrut4gl/trend/plot/hadcrut3vgl/trend/plot/hadcrut4gl/last:360/trend/plot/hadcrut3vgl/last:360/trend

However look beyond 1940s and you will see another 0.5 degC ADDED to the temperatures in 1940s.

So should we not now be saying exactly the same warming messages – retreating glaciers shrinking arctic etc. – only this time it really is worse (by another 0.5C).

We may now be in a pause (just as from 1940 to 1980) but at the end what stops the temperature increasing by another 0.5C.

Remember it is not just the 0.5C that matters (after all its ONLY 0.5C and a bit warmer would be good) it is the added stored energy in the earth system that will be the problem.

You suggest it is just cyclical (and I agree that the predominant cause is a 60 year cycle – peaked in 1880s, 1940, then in 2000) but the problem is the underlying trend which is upwards

I have shown these plots before – one totally cycle based (60 and 315 years main cycles) one based on cycles plus a polynomial trend. as of this year the trend is the best fit – the cycle based plot is falling too rapidly – and rising in the 1800s when it should not).

http://climateandstuff.blogspot.co.uk/2012/12/cycle-mania-anf-hadcrut3.html

Obviously it is impossible to predict the future  but CO2 is a GHG and GHGs increase the earths temperature. Warmer planet = more water vapour and (water vapour is a GHG) so science says the world is warming.

Another 16 years will tell but will it then be too late to react?

pielke post backup

” in the near term it is simply immoral to ask the poor to make energy access sacrifices while we consume massive amounts of energy, based almost entirely on fossil fuels. Climate policy should not be used to keep poor people poor”

Presumably these “poor people” will be living in isolated communities
Presumably these poor households will not have funds to purchase cookers,fridges,computers, televisions, kettles,
Presumably being isolated there will be little industry to take advantage of the electricity – you will need to improve transport first.

Who pays for the grid to keep a 100Mw powerstation running near water whilst feeding isolated communities perhaps hundreds of miles away.
Who pays for the transport and possibly import of fuel for these stations.
Who pays for the security to check the power lines?

Just how are the people going to pay for appliances to connect to the grid?
just how are the people going to pay for the energy consumed.?

Heating of homes does not require electricity but would be better done with local heat stores.

One useful item would be lights but these can be powered locally from solar + battery I use 20 watts to brightly light my house usuing relatively cheap LEDs. But someone still has to pay for the lights battery and solar cells even if you have disposed of the very costly power lines from generator to consumer.

Give the poor a lead acid battery, a 300W solar panel (perhaps $500) and they will be able to light their village, pump water, UV purify water, charge laptops(!) for perhaps 10 years, How much would a coal/gas station + wires + fuel cost over the same period (UK price is £0.05 CCGT Fuel only per kWh – uk DE Electricity Generation Costs 2013 – assuming solar as described give 3kWh/day then CCGT at fuel cost only would exceed the solar price in approx 4 years)

Please tell me exactly whom is going to pay for the power stations, the grid, the appliances, the roads, the security, the industry?

Please be very specific how the “cheap” electricity will help the poor and how you would fund its introduction.

The west has grown up with electricity and so has the infrastructure, the factories, the communications. to start from scratch is a whole new ball game.

Fri Jan 31, 08:07:38 AM MST

the annual co2 cycle posted to wuwt

The annual CO2 change: Co2 absorbed in the ocean removes the full CO2 molecule and so will not change O2 levels (see link) plant growth requires CO2 but on a yearly basis that which is absorbed is not emitted until decomposition – a slow process hence growth/decomposition does not create a yearly cycle. The balance between plant photo synthesis and respiration however occur according to the dark/light ratio. In NH summer is light so more co2 converted to O2 Winter is dark so more O2 converted to CO2 This seems to me to be the only process that is rapid enough to cause the annual co2 cycle.

The following post shows that CO2 and O2 levels are in antiphase and synchronised to the light dark cycle

http://bit.ly/1i0OlfL

Postma blog

Your comment is awaiting moderation.

Joseph E Postma says: 2013/05/25 at 9:18 PM
1. You’re confusing insulation and/or reduction of convection with the GHE. 2. The icecubes get heated by you…they don’t heat you up and you don’t have to warm up in order to heat them up – it wouldn’t happen by conduction, and it doesn’t happen by radiation. Touching icecubes all over your body doesn’t heat you up. Not touching them near your body doesn’t heat you up either.
—————
If a human body was in space between galaxies with a bottle of oxygen for breathing.
Are you saying it would make no difference if you were in an ice sphere (painted inside to give BB i.e. no reflection of body radiation) at -10°C or if you were floating unprotected.
i.e.
No air = no conduction or convection in both cases.
Body is warmer than ice so no radiation absorbed
Body is warmer than space so no radiation absorbed.
Body therefore cools at same rate in both cases????

Your comment is awaiting moderation.

OK.
First what I hope is an uncontrovesial graph
Temperature 0K to 10K – Quanta emitted LINEAR (because it fits the plot better but 10^4 works too) with respect to the temperature
The core A is actively stabilized from 10K to 0K and emitts 100 to 0 quanta
The shell B is actively stabilised from 0K to 10K and emits 0 to 100 quanta
Apart from the 10^4 quanta vs temperature hopefully there is nothing “wrong” with this

http://1.bp.blogspot.com/-73PK0_vgbFg/UWTfiCNpjCI/AAAAAAAABEU/kT5cBjLFOMo/s1600/quanta+vs+temp.jpg

Now the other plots. This assumes the slayer premise that quanta are not utilised when the reach a colder body than what they were emitted from. i.e. when A temperature is below B temperature A quanta have no effect on B (graph shows this as null)

There is also the plot showing the warmist view that all quanta reach and add to the energy in the target
You will note that when looking at net quanta from A to B (slayer mode) at the point A temperature = B Temperature the net quanta in B goes from 0 to -50 so it will begin cooling rapidly.
The warmist mode shows no discontinuity!

Perhaps you could show some plots which better represent your thoughts?

http://4.bp.blogspot.com/-7guLeJTj4U0/UWVPQPzXIkI/AAAAAAAABEs/LNqxJKnDIC0/s640/slay+flow.jpg

http://4.bp.blogspot.com/-yG2VdvcxuDE/UWVPSP4_apI/AAAAAAAABE0/A-9nt-8A330/s1600/warm+flow.jpg

Postma potty postings

shell. Obviously that has problems of the thermodynamics type.

Joseph E Postma says:
2013/03/12 at 2:49 PM
lgl: “So the energy from those ‘too low energy’ photons just magically disappears?”

No, but it doesn’t shift itself to to higher frequencies either. Don’t throw out what equilibrium actually means.

lgl: “If radiation is absorbed temperature increases, the frequency of the photon doesn’t matter.”

You’re a complete idiot. You have no science knowledge or training whatsoever. Of course you think something like this because then you can have cold things heating up hotter things. OMG so dumb.

lgl: ” Are you saying a warmer object will not absorb photons from a colder object? If so, again big scientific breakthrough.”

It is simple – a cooler object does not heat up a warmer object. Why are you such an idiot? How can you be this retarded? The scientific breakthrough would be if cold could heat hot, or if something could spontaneously heat itself with its own radiation.

Joseph E Postma says:
2013/03/12 at 2:51 PM
Yes Max, indeed, that is why I keep asking the question as to WHAT it is that drives these people? Some form of religion is a good bet, because you only see this type of behaviour anywhere else with creationist fundamentalists.

lgl says:
2013/03/12 at 3:09 PM
Joe
“You have no science knowledge or training whatsoever”
A strange statement since I’m only explaining the science known for centuries wereas you are peddling your physically totally impossible ‘homemade’ nonsense.

A cooler object will make a warmer object warmer that it would have been without the radiation from the cooler object, but this does not mean the cooler object is ‘warming up’ the warmer object.
This is the essence of your misunderstanding.

Greg House says:
2013/03/12 at 4:53 PM
lgl says, (2013/03/12 at 12:41 PM): “So a full mirror, a perfect mirror not passing any radiation to space, will not heat the planet either? Given the internal source is still there.”
=======================================================

Lgl, look, it is so simple. Please, conduct a real experiment: stand in front of a mirror (not too close to it to avoid suppressing convection significantly) and observe how it will warm you. Your face, for example. Your face has to feel the heat, right? If possible, tell us the scientific result.

sunsettommy says:
2013/03/12 at 5:29 PM
lgl,

where did the extra 235 come from?

Joseph E Postma says:
2013/03/12 at 6:31 PM
No lgl, that is the source of your fraud. Quantum mechanics is real and the behaviour of radiation trapped inside a cavity forms its basis, and it doesn’t cause runaway self-heating. The shell-game heating idea is the dumbest idea that has ever been invented. Read up on the work of the French philosopher Jean Baudrillard and the concept of hyperreality to understand the trap you’ve fallen in to.

sunsettommy says:
2013/03/12 at 6:43 PM
lgl writes this:

“A cooler object will make a warmer object warmer that it would have been without the radiation from the cooler object, but this does not mean the cooler object is ‘warming up’ the warmer object.”

Then according to you putting a small bag of ice on my tummy will get me warmer since YOU say that a cooler object (ice) will make my tummy and the body as a whole a bit warmer since it does send some waves of radiation to my body adding the energy needed.But then you say the very opposite in the ending part of your convoluted paragraph.

Charmed………………………

Joseph E Postma says:
2013/03/12 at 6:44 PM
They just say one thing and then the other back and forth as each gets refuted…mindless repetition, a product of the education system.

Greg House says:
2013/03/12 at 7:15 PM
Joseph E Postma says, (2013/03/12 at 6:31 PM): “No lgl, that is the source of your fraud. Quantum mechanics is real and the behaviour of radiation trapped inside a cavity forms its basis, and it doesn’t cause runaway self-heating.”
=======================================================

Not being an expert on quantum mechanics I have tried a different approach to the issue of behavior of trapped radiation at Tallbloke’s today:

“Dear friends of back radiation’s ,

I know from my experience on blogs that it is sometimes hard to understand, let us say, that it is at least possible that back radiation won’t warm the source. The counterargument is sometimes like “where does the back/trapped radiation go then?” meaning that radiation must find a sort of a host to settle down. This seems to be a serious obstacle to accept the possibility that back radiation won’t warm the source.

Now, coming back the hypothetical planet-shell mutual warming process, how come nobody including the concept creators asks “where does the radiation go after leaving the shell in the direction outer space”? Because we have nothing there, so wished the creator Willis.

Well, what I would like to suggest is this. Since we have no problem with the radiation leaving the shell in the direction outer space doomed to never finding a new home, let us think that it is exactly the same way possible to have this desperate homeless back radiation in the inner space between the shell and the planet forever, still unable to warm. Please, give it a thought.

Then, once there is no problem of unclear back radiation’s destiny, we can all look at this back radiation warming thing and ask ourselves “WTF??”. Then we do not need such, you know, explanations as “yes, the temperature instantly drops when emission occurs from a blackbody with no heat capacity” (must be to the absolute zero, OMG!). Then we can recall the R.W.Wood experiment and dismiss the “greenhouse effect” as contradicting reality. Please, give it a thought.”

Joseph E Postma says:
2013/03/12 at 7:35 PM
Another comment of mine from WUWT which I want to save here:

Micro: “Or, you can simply absorb more of them, can’t you? There are more coming in every second from the original heat source. If they are delayed from exiting, are they not going to start piling up?”

Yes but photons are bosons – they don’t “pile up”, they pass right through each other. Absorption of a photon can’t induce higher temperature, i.e., material vibration frequencies, than the photon actually is. That’s not sensical or possible, by definition. Photons don’t induce or create higher temperatures than they are in the first place, because they don’t have the energy or frequency components to do so. When they come back to their own source, they only encounter matter which is vibrating at the same frequencies which produced them. These frequencies then resonate, not increase in frequency. When two photons of the same wavelength interact, they *do not* change each other’s frequency, which would correspond to a temperature increase. Likewise, if a reflected photon spectrum comes back to its source, it only encounters a material vibration frequency spectrum of the same character. These photons, being of the same vibration frequencies of the material, can not induce a higher temperature in the material because this requires a higher frequency component inputs. This is how and why radiation obeys the 2nd Law of Thermodynamics, and why radiation inside a cavity doesn’t heat itself up, but produces a blackbody spectrum, as Planck proved along ago in his discovery of quantum mechanics. It all works perfectly together to create a sane world.

Consider “Bryan’s” comment to me which I had in one of my last posts, and understand that the system of shells being theorized here then obviously violates thermodynamics & etc. I will repeat it here:
Consider the planet to have a high metal expansion coefficient, and the shell a small one. There is a small distance between them. When the planet heats up to 470, it will expand and touch the shell. Now the shell will heat up to 470. Whereas before we couldn’t, say, create steam to drive a generator turbine, now we can (using the relevant values). You get the idea I hope. The system might cool and stop touching but then it would just heat up again and touch again. With an intrinsic source that couldn’t have create steam in the first place, with this system of shells the system can now do more work than it could have in the first place, than if the input energy was just used directly. Just by using a passive shell. Obviously that has problems of the thermodynamics type.

RGB: Heat doesn’t flow from cold to hot, so, the shell never sends heat to the planet, and so it never warms up the planet. The shell just will come to the same temperature as the source, the planet. Then the shell effectively becomes the new surface of the planet, and this doesn’t require the planet to become hotter. There’s no need to invent that. The presence of a temperature differential does not mean that the hotter side of the differential gets hotter – it means that the cooler side warms up until equilibrium is established with the hotter side. This does not require the hotter side to get hotter.

Hands warmed inside gloves is a sense-perception result of reduced convective air cooling. Put your hand inside a mirror, say, a shell with a mirrored interior, and the hand will not warm up. Put it next to a mirror and the side facing the mirror will not heat up.
Temperature and radiation is all about frequency. Hotter matter vibrates at higher frequency, producing higher frequency radiation. Commensurately, higher frequency radiation falling onto matter with lower frequency vibration components will induce higher frequency vibration in the material, thus heating it up. Because higher frequencies are being introduced. You need higher frequency radiation to induce higher temperature. This is how and why radiation obeys the laws of thermodynamics, so that radiation of a particular frequency spectrum (temperature) can not induce higher frequency vibration (temperature) in matter it falls on, including its own source as a pertinent example. This is exactly the same thing as how two objects of the same temperature can not induce higher temperature in each other via conduction; neither has the higher frequency components required to do so.

Greg House says:
2013/03/12 at 8:06 PM
Gareth says, (2013/03/12 at 8:29 AM): The planet begins at 800 and the shell at zero returned to the planet and zero to space. Then the shell returns 400 to the planet and emits 400 to space.
The planet retains 200 of the 400 and reflects the other 200, along with the constant 800 to make 1000 now going to the shell. The shell then reflects 500 and emits 500 to space. The planet retains 250 and reflects 250, plus 800.(1050) …”
==========================================================

Gareth, there is no reflection in the Willis’ small universe, the precondition is that both the planet and the shell absorb and emit. It goes back and forth, but there are some inborn defects in that thing.

First, Willis wants his hypothetical cyclical process to stop at a certain point just like that, because it must reach equilibrium. However, his hypothetical process does not lead to equilibrium. This alone should be an alarm signal, but no, he has a “solution”: he declares the process done, just like that. Physics upside down.

Second, they make their radiation arithmetic, assuming that the planet always retains the same temperature, but emits more and more, which “allows” them to always add the same initial 800 (initial power) to the increasing back radiation. The bodies, however, are known to emit according to their temperatures, hence their hypothetical cyclical process would lead to an endless mutual warming, so the whole thing falls apart at this stage too. There is, however, an abstract possibility to resolve this contradiction, namely by assuming that after emitting and before absorbing the planet’s temperature always drops to the initial one and then rises again and always to a higher and higher value. I offered it, hoping it would additionally demonstrate the absurdity of the initial assumption, but guess what? Roger took it indeed as a part of their process! Unbelievable. This is their physics. (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46160)

Greg House says:
2013/03/12 at 8:13 PM
Joseph E Postma says, (2013/03/12 at 7:35 PM): “Another comment of mine from WUWT …”
=====================================================

Joe, could you give me the link, please? Thanks

Joseph E Postma says:
2013/03/12 at 8:25 PM
Here it is Greg:

http://wattsupwiththat.com/2013/03/06/notes-on-the-february-global-temperature-anomaly/#comment-1246503

Joseph E Postma says:
2013/03/12 at 9:08 PM
That two spectrums exist, one from a cooler body and one from a warmer, and that there is a temperature differential, does not mean that the hotter side of the differential warms up. That’s backwards. Skin can detect temperature changes on the order of what, 1/10th, 1/2 of a degree K? If backradiation could warm you up, you should be able to instantly feel it on your skin when you stand in front of a mirror. Backradiation is supposed to produce 50%, doublings, etc., of temperature increases. Talking 10′s of degrees K, hundreds of degrees K even. But it can’t actually even be detected at the skin sensitivity level of less than 1K. That is because the concept is physically impossible, given the explanation of frequency components and heat transfer etc.

Joseph E Postma says:
2013/03/12 at 9:20 PM
The presence of a temperature differential doesn’t mean that the hotter side of the differential gets hotter. That’s backwards. 100% backwards. It is not about directionality – radiation goes everywhere. It is about what the radiation is capable of doing, which is a function of its spectrum. It can’t do more than its spectrum says it can. It can’t create higher temperature than the temperature it is. This is how and why radiation obeys the laws of thermodynamics, particularly the 2nd, given that directionality is not known. Directionality might not be known, but it doesn’t need to be. What is known, what becomes known, are the frequency components when the photon and matter interact. If the radiation is of the same or cooler spectrum than the matter, it can’t induce any higher frequency components and therefore higher temperature in the matter.

thefordprefect says:
2013/03/12 at 10:29 PM
with ref to my post :
http://climateandstuff.blogspot.co.uk/2013/03/does-thermal-radiation-travel-from-cool.html

Joseph E Postma says: 2013/03/11 at 8:35 AM
Very simple. All he did was slow down the cooling rate because the ambient environment was changed. When you change the environment you get different thermal behaviour.

What did NOT happen was that the cooler plate caused the warmer plate to heat up some more to a higher temperature. That is what they need for the GHE.

———————————–
What is the difference between the environment of the hot plate side? The area is the same. The hot plate is the same, the hotplate temperature is the same, the ambient temperature is the same. The window is the same. The main difference is that the window is passing radiation from the warm plate in one run and in the other run it is passing radiation from a 20C ambient sink.

No true scientist says that the presence of a warm plate will heat up the hot plate.
Providing the warm plate is above the background temperature it will slow the cooling of a hot plate. If the hot plate is being heated by a fixed energy source then when a warm plate replaces a cold background the hot plate will cool to a warmer equilibrium temperature.

Gareth says:
2013/03/13 at 7:49 AM
Greg House,

Thank for the reply.

One of my mistakes (I think) is in thinking that a shell receiving 800 whatevers from the planet would emit half to space and half back to the planet.

The planet would be at 800, the emission from the planet would be 800, the shell would be at 800 and the shell would be emitting 800 at the planet and into space. The planet doesn’t heat up because the incoming 800 from the shell is equal to the outgoing 800 from the planet and the net energy emitted to space is still 800?

Alan Siddons says:
2013/03/13 at 8:19 AM
…his hypothetical process does not lead to equilibrium. This alone should be an alarm signal, but no, he has a “solution”: he declares the process done, just like that. Physics upside down.

Just wanna say that Greg House has nailed it. In a greenhouse scenario, each of the two parties — surface and shell — must continuously adjust to what the other is radiating. The initial 235 W/m² surface induces the shell to emit 235 as well, which makes the surface emit 470, which makes the shell emit 470 too — ad infinitum. This is what Gerlich &Tscheuschner’s Falsification paper was trying to get across:

The atmospheric greenhouse effect… essentially describes a fictitious mechanism, in which a planetary atmosphere acts as a heat pump driven by an environment that is radiatively interacting with but radiatively equilibrated to the atmospheric system.

Interacting with but equilibrated to. Hey wait. That’s impossible.

Joseph E Postma says:
2013/03/13 at 8:19 AM
That’s about right Gareth. Remember, no heat energy is transferred from cold to hot or between two objects of equal temperature. No net heat energy flows from the shell to the planet. The best the planet can do is heat the shell to the same temperature of the planet, and then the shell essentially becomes the new surface of the planet, emitting the 800. Between the planet and shell is a radiation field of 800, and so there’s that energy there going “back and forth” between them, but it doesn’t cause heating because it intrinsically doesn’t contain the higher energy frequency components required to do so. As the shell then loses the 800 continuously to space, this is continuously replaced by the radiation field between the planet and shell, and the energy loss in that radiation field is replaced by the planet.

Will Pratt says:
2013/03/13 at 8:39 AM
thefordprefect says:
2013/03/12 at 10:29 PM

In your experiment, you have simply increased the mass of the system, of course it will take longer to cool down.

However, adding CO2 to the atmosphere does not increase atmospheric mass. Your experiment is meaningless in the context of this discussion.

Joseph E Postma says:
2013/03/13 at 8:40 AM
My reply to Micro at WUWT (is everyone having problems posting there? or is it just me?):

Micro said: “You can protest all you want that there’s no ir reflected(in a proper ir reflector), but you’re wrong.”

My statements had nothing to do with saying that there is no IR reflection, so this objection is moot, sorry. There is IR reflection and emission from cold sources, certainly, but the nature of this radiative energy can not cause heating on its own source or a hotter source, for the reasons I’ve physically described several times. Radiation obeys the laws of thermodynamics too – not just the first one, but all of them.

No heat energy is transferred from cold to hot or between two objects of equal temperature. Remember, we can *not* say that in a temperature differential, the hotter side has to get hotter in order to warm the cool side. Rather, the cool side just warms up. No net energy flows from the shell to the planet. The best the planet can do is heat the shell to the same temperature of the planet, and then the shell essentially becomes the new surface of the planet, emitting the 235. Between the planet and shell is a radiation field of 235, and so there’s that energy there going “back and forth” between them, but it doesn’t cause heating once equilibrium is established because it intrinsically doesn’t contain the higher energy frequency components required to do so., required to induce higher frequency material vibrations and hence higher temperature. It just can’t do it. As the shell then loses the 235 continuously to outer space, this is continuously replaced by the radiation field between the planet and shell, and the energy loss in that radiation field is replaced by the planet.

We already discovered that the “shells game” idea violates thermodynamics when considering the physical effects of the interior sphere having a high metal expansion coefficient, and the shell a low one, and what would occur when they touched.

Ron C. says:
2013/03/13 at 8:49 AM
Thank you for this discussion, which has ranged from here to WUWT and TB Talkshop.

I noted and understand your point about the radiation spectrum prohbiting any increase in frequency of impacted matter of the same or higher spectrum, and therefore no increase in heat or temperature.

I am interested to know your response to something I read today on the TB site. (BTW they are searching there for the “official” physics of the GHE, and not finding anything to critique.)

“The bonds between the atoms in a molecule are somewhat elastic, like springs, and with the atoms acting like weights at opposite ends of these springs, every molecule has at least one characteristic resonant vibrational frequency. Most molecules have many different resonant frequencies, depending on vibration modes, etc.

CO2 molecules have a resonant frequency that corresponds to 15 micro meters (LWIR). When it encounters a photon of IR energy at that wavelength, it readily absorbs it and begins vibrating faster. Now, there are two ways it can lose that energy again. First, It can spontaneously re-emit IR energy at the same frequency, and lose that extra vibration. Second, (and more likely), when it bumps into another molecule, that extra vibration gives the collision a little extra “kick”, and both molecules move away a little faster. Faster molecules within a fluid equals higher heat energy. At this point, the original IR has become thermalized.”

http://tallbloke.wordpress.com/2013/03/13/wikipedia-and-ipcc-ar4-the-greenhouse-effect/#more-11617

lgl says:
2013/03/13 at 8:57 AM
All you morons, again

Try this then:
We have two equally sized radiators, and for the argument they are thin as a paper i.e only radiating from two sides. We supply enough power to keep one of them at 300K and the other at 400K when they are far apart. Then we place them very close to each other, sealing off the narrow gap between them completly so that no energy escapes from the gap. We supply the same amount of power to the radiators as before. Now, will the temperature of the radiators change? One, both, up or down?

Joseph E Postma says:
2013/03/13 at 9:01 AM
Hi Ron,

There is no question that radiative energy can heat up material that colder. That is the only thing the quote you provided describes – that radiative energy can heat up something colder. The implied, and false extension, is to then say, really it is just implied, that the CO2 or its reradiation has to heat up the original source of the radiation. This is not possible because the original source of the radiation is either warmer, or the same temperature, and so the CO2 or its radiation can’t warm up the source.

Also consider that the CO2 is already vibrationally activated in ALL modes of its possible oscillations because it is being bumped around by the other 99.96% of the atmosphere, which has itself been heated by conduction with the hot ground surface. So, the IR photons interacting with CO2 only “see” a molecule which is already oscillating. The description you quoted above is applicable only for a cold CO2 molecule in complete isolation. So, IR photons “see” a CO2 molecule which is already oscillating and vibrating, hence, they don’t induce much more of that to the molecule, if at all. What the radiation field could do is scatter of off the existing oscillation because it would resonate with it. But again, just like standing in front of a mirror doesn’t warm you, the back-scattered radiation won’t warm its own source. That radiation also leaves the atmosphere within a few milliseconds anyway after a couple of scatterings because they travel at the speed of light and the atmosphere is negligibly thin compared to that…so, even if you want to talk about “photon build up” it too is negligible, although it wouldn’t cause heating above the source temperature even if it wasn’t.

Joseph E Postma says:
2013/03/13 at 9:03 AM
lgl: The cool one will warm up. The presence of a temperature differential does not mean that the hotter side of the differential has to heat up in order to warm the cool side. The cool side just warms up.

Will Pratt says:
2013/03/13 at 9:23 AM
Ron C. says:
2013/03/13 at 8:49 AM

Ron, 15 µm has a corresponding temperature of -80º C.

http://www.calctool.org/CALC/phys/p_thermo/wien

This is “coincidently” the same temperature that CO2 sublimates via phase change, from ice to gas. Therefore it is bound to absorb strongly at these wavelengths.

The same applies to all substances. They all absorb IR strongly at their various melting/evaporating-phase change temperatures, obviously.

Alan Siddons says:
2013/03/13 at 9:31 AM
LGL, the shell doesn’t just have slightly more surface area (square meters) than the surface, it has more than TWICE because its two sides are exposed, both of which are freely radiating to their surroundings. This means that even if the shell were absorbing all of the surface’s energy it would be much cooler than the surface, radiating in the neighborhood of 117 W/m². You have to count both sides.

Slice up a cake straight out of the oven and separate the pieces. Since you’ve increased the surface area, they’ll cool down a lot faster than the whole cake would.

Joseph E Postma says:
2013/03/13 at 9:35 AM
Yes but you see Alan, now he’s going to want to add the 117 BACK to the planet to make it hotter, even though the shell is colder. lol

lgl says:
2013/03/13 at 9:52 AM
Alan
Finally you are beginning to grasp some of this.
Yes, right after you place the shell around the planet it will radiate 117 to space, but then you have a problem. The core is still generating 235, the system is only loosing 117 to space. What will that lead to? Where is the other 117 ending up?

lgl says:
2013/03/13 at 9:54 AM
Joe

And the cold one will warm to no more than 400K i presume, right?

Joseph E Postma says:
2013/03/13 at 9:57 AM
Didn’t I predict exactly what lgl would say? lol

We already went over the sequence lgl is proposing and which Willis proposed etc., right at the beginning of these comments. The sequence results in a heating progression going as (1.5)^n, which diverges to infinity, and which is ridiculous. The other 117 gets scattered right back to the shell, warming it up until it gets to equilibrium with the planet, with both at 235 and equal temperature.

The presence of a temperature differential does not mean that the hot side of the differential has to heat up in order to warm the cool side! How stupid do you have to be?

Joseph E Postma says:
2013/03/13 at 9:59 AM
“And the cold one will warm to no more than 400K i presume, right?”

This has been answered numerous times, whatever your question is referring to. The planet and shell come to the same temperature assuming they are very close together.

lgl says:
2013/03/13 at 10:22 AM
Joe
I was referring to the radiators. Will one stay at 400K and the other somewhat lower than 400K or what?
Your problem is, if we assume each side of the radiators were 1 m2, in the inital setup the effective radiating area was 2 m2 each, so in total they radiated 3827 W.
After they are placed close together the area to free space is only 1 m2 each, so 3827 W will now have to be radiated from only 2 m2. What temperature is required to achieve that? My calcs say 428K.

lgl says:
2013/03/13 at 10:29 AM
Joe
“The other 117 gets scattered right back to the shell”
Ah, more selfinvented physics.
What about when they reach the same temperature? The shell will still absorb the 235 but the planet will not, it will still ‘scatter’ back the 235 to the shell? Why this totally different behaviour when they are at the same temperature? What utter nonsense.

Joseph E Postma says:
2013/03/13 at 10:30 AM
“My calcs say 428K”

Your calcs are wrong. Cold doesn’t heat hot and and hot doesn’t have to become hotter to heat cold. QED.

Joseph E Postma says:
2013/03/13 at 10:33 AM
When they reach the same temperature, they’re in equilibrium. Temperature can not amplify itself and radiation can not induce a temperature higher than the temperature that the radiation is, because this would be a temperature amplifying itself.

lgl: Stop being an idiot.

Alan Siddons says:
2013/03/13 at 10:49 AM
“When they reach the same temperature, they’re in equilibrium.”

But that can never happen; the shell will always be colder. I would say that ‘radiative equilbrium’ is reached when the shell absorbs all the radiant energy that the surface has to offer, which amounts to 117 W/m² distributed over both sides of the shell. Given the surface area issue, the shell can’t absorb or emit more than that.

lgl says:
2013/03/13 at 10:51 AM
Joe
“Your calcs are wrong”
Show us the correct calcs then. Don’t they have to radiate 3827 watts? Or do you finally realize your physics is impossible?

Joseph E Postma says:
2013/03/13 at 10:54 AM
Alan, that is certainly more sensible than anything the “shell game” crowd has been thinking.

However, consider a thick shell. If the inside is very close, it should indeed come to nearly the same, or essentially the same if it is very close, temperature as the sphere. What occurs in general is what I descried earlier:

The best the planet can do is heat the shell to the same temperature of the planet, and then the shell essentially becomes the new surface of the planet, emitting the 235. Between the planet and shell is a radiation field of 235, and so there’s that energy there going “back and forth” between them, but it doesn’t cause heating once equilibrium is established because it intrinsically doesn’t contain the higher energy frequency components required to do so., required to induce higher frequency material vibrations and hence higher temperature. It just can’t do it. As the shell then loses the 235 continuously to outer space, this is continuously replaced by the radiation field between the planet and shell, and the energy loss in that radiation field is replaced by the planet.

Joseph E Postma says:
2013/03/13 at 10:56 AM
lgl: “Show us the correct calcs then.”

I have numerous times: (1.5)^n. It diverges to infinity and is therefore wrong, because it violates the laws of thermodynamics. Radiation interacting with itself doesn’t increase its frequency and hence its temperature stays the same; the same thing occurs for radiation interacting with matter – it can’t induce higher temperature than it is.

lgl says:
2013/03/13 at 11:10 AM
Alan
but then only 117 is emitted to space, 235 is still supplied from the core. Can’t you see if the system looses less than 235 to space it will heat up?

Joseph E Postma says:
2013/03/13 at 11:13 AM
Leif: I already explained what happens. Nothing can heat itself up with own radiation.

lgl says:
2013/03/13 at 11:14 AM
Joe
“I have numerous times”
No you have not. Tell us how much energy two 1m2 plates (two sides of course) are emiting if one is 400K and the other 300K.

Joseph E Postma says:
2013/03/13 at 11:19 AM
The question isn’t how much energy because the laws of thermodynamics are not limited to the first law. This is the mistake you can’t get passed. There’s also the 2nd and 3rd Laws. Hotter heats cooler and cooler does not heat hotter, and hotter doesn’t need to become hotter to heat cooler. The radiative reasons have been described physically several times.

As you’ve exhausted every argument you might make several times over, and the material to educate you on the matter has been provided several times over, your comments on these matters will no longer be published, to save me the time of constantly trying to instil some education in you. Best regards.

lgl says:
2013/03/13 at 11:22 AM
Good, finally you realize you were wrong.

Joseph E Postma says:
2013/03/13 at 11:23 AM
See the parting shot from an idiot everyone? That’s what an idiot does.

Simon Conway-Smith says:
2013/03/13 at 11:45 AM
No lgl, Joe just got fed up of explaining why you are wrong time and time again, with no indication from you that you were learning anything or asking rational questions based on any learning.

Alan Siddons says:
2013/03/13 at 11:52 AM
…consider a thick shell.

Right, that’s reasonable. But then you have the problem of energy distributed over the VOLUME of an actual mass. Joules per kilogram raises a substance’s temperature. With the same Joules per TWO kilograms, though, the temperature increase is less. Even if the shell’s mass is losing energy as fast as it’s gaining, then, its many molecules are clamoring for an equal share of energy, which thereby reduces the shell’s average temperature.

The same applies to the introductory premise, observe, for the nuclear core must of course be MUCH hotter in order to make the surface radiate 235 W/m². By extension, the surface’s total energy (surface area × W/m²) transferred to yet ANOTHER mass will further dilute the energy available for each molecule. In effect, adding a thick shell amounts to throwing dirt over the nuclear furnace in order to reduce the surface temperature. That extra dirt will make the new surface cooler than the surface was before.

Adding mass amounts to throwing dirt over a furnace to cool things down. Or so it seems to me.

Joseph E Postma says:
2013/03/13 at 11:56 AM
Yes indeed, when there is more mass this affects the differential equation for heat flow because the time constant is larger. Larger mass means it takes a longer time to heat up. If the volume was large so the outer shell surface area was much larger, then the outer shell would radiate at a cooler temperature – the same energy from the planet but over a larger surface area.

On the other hand, if the shell was very thin and the gap between planet and shell very small, and the shell material made of something with a very high thermal capacity, then again the shell would heat up slowly, but it would heat up eventually to radiate at the same energy flux density, or temperature, as the sphere, since the difference in outer area of the shell vs. the sphere is in this case very small.

Joseph E Postma says:
2013/03/13 at 12:23 PM
I just talked to an electrical engineer of 40 years at the university where I work.

I proposed this idea to him:

“Suppose you have a electrical thermal resistor in outer space, in a circuit, and the resistor gets to some temperature, say, 80C, and puts out that equivalent amount of radiative energy.

Now, put that resistor inside a shell with a mirrored interior (still in outer space so there is no air anywhere). I propose that the resistor will get hotter and hotter and hotter, and that with the power input required to get to 80C previously, now that same power can produce much higher temperature. This could, say, boil water for example, to drive a turbine generator, when it couldn’t do that before.”

His reply: “Yes I see what you’re trying to do there, but it is a violation of thermodynamics. It wouldn’t work as you propose. The reason is because the returned radiation would have “phase cancellation” with the outgoing radiation, meaning that half of the radiant energy gets destructively cancelled out from superposition. The remaining half, halves, one half from outgoing and one half from returning (and then sent outgoing again), results in unity.”

This is the same thing I’ve said elsewhere. Also note that the only way to not get phase cancellation is if one of the sources has higher frequency components of shorter wavelength, which can not combine in any phase with longer wavelengths. Only with higher frequency components can you induce higher temperature to something with lower ones. Works the same in conduction as with radiation. This is also basically all about what Claes Johnson has written about, in a lot of ways.

Will Pratt says:
2013/03/13 at 12:28 PM
Just to recap,

235 W/m2 + 235 W/m2 = (not more than) 235 W/m2

235 W/m2 * 10, 100, 1000 or 1,000,000 still = (not more than) 235 W/m2

W/m2 is a measure of energy, not heat. Energy has only a potential to heat. What determines whether that potential to produce heat is realised, is the thermodynamics of the system in question.

“CLASSICAL THERMODYNAMICS IS THE ONLY PHYSICAL THEORY OF UNIVERSAL CONTENT WHICH I AM CONVINCED WILL NEVER BE OVERTHROWN”….A Einstein.

squid2112 says:
2013/03/13 at 1:58 PM
I am glad someone included (back in comments somewhere) the topic of “No Virginia, a cold object cannot make a warmer object warmer still” … As I was going to bring that in to discussion. As I see it, Willis’ new “thought experiment” (of which I have been getting tired of lately, none of them make any sense), seems to me to be just another in a long line of attempts to get a cold object to warm a warmer object, yet again. I wish I had a quarter for every time I have seen such an attempt made, as I would no longer have to slave away at work. … just saying…

Joseph E Postma says:
2013/03/13 at 1:59 PM
Indeed.

Rosco says:
2013/03/13 at 4:02 PM
The inescapable truth is that Joe is undeniably right.

The notion of something heating itself up by “trapping” its own radiation results in a runaway mathematical series with exponential energy creation – a complete demonstration of perpetual motion which is an intellectual absurdity.

The idea that something is “heating” up whilst radiating less – exactly what IPCC physics requires – should be an intellectual obscenity to any who think about it.

The idea defies all of the basics of accepted science and has never been demonstrated.

The idea that a cold object heats hotter objects (despite the absurd justification that radiative energy from cold to hot occurs but that “net” energy goes from hot to cold ???) is intellectually absurd.

The idea defies all of the basics of accepted science and has never been demonstrated.

Actually the reverse has been demonstrated numerous times.

It is interesting that “climate scientists” make this claim about cold objects heating hot objects more yet have nothing than a similar thought experiment to justify it.

Have they never heard of Pictet’s experiment relating to the apparent radiation and reflection of cold ?

This conclusively demonstrates that the cold object did NOT cause the thermometer to heat up – the reverse was indisputably demonstrated – the thermometer dropped in temperature dramatically.

If “climate scientists” are right it should have increased.

Kristian says:
2013/03/13 at 4:02 PM
lgl says, 2013/03/13 at 11:22 AM:

Let me try to explain this to you, lgl.

We place two objects in an airless box. The one object has a temperature of 300K, the other 400K. The two freely emit according to their specific temperatures. If we were to let the thermal exchange between the two objects move all the way to equilibrium, they would both ideally end up with a temperature of 360K (not 350K). At this point, there would no longer be any heat transfer between the two objects. Up until then, though, there WOULD have been heat transfer from the hot to the cooler object. And if we plotted the RATE of this heat transfer through time, we would see that it followed a curve of exponential decay – it would start off very high but end up infinitesimally low. The hot (400K) object would cool all the way to equilibrium, while the cooler (300K) object would warm. The cooler object at no point made the hot object hotter than it originally was, even though it radiated a higher and higher flux towards it as it grew hotter itself. All it did was steadily slowing down the heat transfer rate, its own warming rate and the hot object’s cooling rate, until they were both effectively zero.

So far, so good. I hope.

Then, we change the setup a bit. This time around the two perfectly emitting objects in the airless box are different. One object is directly connected to a constant energy source supplying a power flux sufficient to heat the object’s surface to a uniform temperature of 254K. The other object enjoys no such constant supply of energy. It hypothetically holds a starting temperature of 0 K.

At the get-go, the warm object emits a flux of 235 W/m^2 towards the cold object and the cold object in return emits a flux of 0 W/m^2 towards the warm object.

What happens next? Where will the ensuing exchange process take us? What will the equilibrium state be like in this scenario?

Upon commencement, the radiative heat transfer from the warm to the cold object starts off at a prodigious rate, warming the cold object fast. But this warming once again slows down as time passes. This time the warmer object does however not cool in accordance with the warming of the colder object. The two objects do NOT meet in the middle at equilibrium. The heat transfer rate between the warm and the colder object follows the same general path as the one in the first scenario through time. But now equilibrium will not be achieved until the colder object has reached the temperature level of the warm object. At this time, heat transfer between the objects would effectively once again have come to an end.

Why, then, isn’t the warm object cooling to meet the cold object ‘half-way’ (213K) at equilibrium in this scenario?

Because it’s temperature/surface radiative flux is being maintained all along by its constant power supply. It will not drop in temperature as long as this source is active. It will not rise either as long as the source provides its constant power flux. This has to do with the internal (molecular) vibration (the level of KE) which relates to the intensity (frequency/wavelength) of the radiative flux received/emitted.

A key to understanding this outcome lies in realising that in the second scenario above (WITH power supply), as the heat transfer between the two objects WITHIN the thermodynamic system (the box) grows ever SMALLER towards equilibrium, the total system energy transmission to its SURROUNDINGS (the space around) parallelly grows ever LARGER towards that same equilibrium. In the first scenario above (WITHOUT any power supplies), this total outward system flux remains the same throughout – no extra heat generated. (In reality it would actually drop; this is after all only intended to be an idealised case cleared of any such real-world clutter. It was simply meant to prove a point.)

The heat generated and originally supplied to the warm object in other words goes into warming the COLD object (and therefore, by extension, the system as a whole) until it’s reached the constant temperature of the warm object. It does NOT (by way of ‘back radiation’ from the cold object) go into making the warm object itself warmer than it already is.

roscomac says:
2013/03/13 at 4:19 PM
At our local Mall there is an elevator with lots of mirrors.

I imagine this enclosed space with “perfect” reflection internally.

Any light in this enclosed space must, according to the arguments proposed by “climate scientists”, be continually reinforcing itself until infinite brightness.

You could turn off the light and never be in dark.

What happens to light in this scenario is an interesting thought experiment – obviously it isn’t trapped as proposed.

Joseph E Postma says:
2013/03/13 at 4:31 PM
I don’t know why they create so much bs over this when it is so obvious and so easily testable.

Martin Hodgkins says:
2013/03/13 at 4:56 PM
Joe,
That radiator analogy from lgl (comments above) convinced me you are right. The radiators can’t get any hotter with the same power input. I am pleased about that because it has been driving me mad.
Regards.

Rosco says:
2013/03/13 at 5:04 PM
I always find Willis Eschenbach a contradiction.

No matter what anyone says – he starts with a constant energy of 235 W/sq metre and proposes no other energy source exists.

His claim that reflecting this doubles the available energy ignores his original position – there is 235 W/sq metre available !

Where does the extra come from ??

Are the people who believe this so gullible that they ignore the basics of science ?

Energy can neither be created nor destroyed, merely transformed.

Can’t everyone see energy has been created out of nothing here ?

My other favourite that I see support for from even sceptics is that insolation is 342 W/sq metre.

Nobody questions the ~1368 W/sq metre solar constant but some how 3/4 disappears upon entering the atmosphere. Of course I recognize the averaging of a sphere to a disk but the only purpose this serves is to estimate how much energy needs to be emitted over a sphere to balance the solar radiation over a disk – and it is only a simplified estimate.

To use this to suggest the sun can only supply sufficient energy to heat the Earth’s surfaces to minus 18 degrees C and base computer modelling to support an unproven hypothesis is so obviously absurd it is amazing people not only accept it but will go to extraordinary lengths normally not associated with intelligent discussion or society to demonize any who find the idea flawed.

Unless “insolation” has some meaning that I cannot find it is absurd to claim that the solar radiation entering the atmosphere is 1/4 ignoring the obvious day and night – yet it stands and is defended as correct.

Radiation that is 4 times as powerful will induce temperatures that are 1.414 times higher – it is an obvious consequence of the power of 4 of the temperature in the SB equation.

Surely Joe has demolished this falsehood eloquently yet it continues.

Joseph E Postma says:
2013/03/13 at 5:08 PM
Cheers Rosco! As I pointed out, they don’t even understand elementary arithmetic.

Joseph E Postma says:
2013/03/13 at 5:09 PM
Martin, glad to hear.

roscomac says:
2013/03/13 at 5:10 PM
I’ll say it again – Willis has created energy out of nothing !

Surely that is totally obvious as well as impossible ??

Greg House says:
2013/03/13 at 6:26 PM
Rosco says, (2013/03/13 at 5:04 PM): “I always find Willis Eschenbach a contradiction.
No matter what anyone says – he starts with a constant energy of 235 W/sq metre and proposes no other energy source exists.
His claim that reflecting this doubles the available energy ignores his original position – there is 235 W/sq metre available !
Where does the extra come from ??
==================================================

As far as I understand, they mean that energy that is radiated away can be sort of caught and returned to the source (back/trapped radiation) and either slow down cooling of the source, if it is cooling, or increase the temperature of the source if the source is not cooling because of additional sort of energy, like the Willis planet. It is like you spend money, but the money spent is given to you back, so you have the same amount of money plus the things you bought. Sounds good, doesn’t it? This is a sort of fallacy that I would call “analogy fallacy”. Warmists deliver analogies, some people buy them, start then thinking in terms of those analogies and are a sort of trapped in them. They do not question the actual issue any more.

The right approach would be to ignore analogies and speculations and ask what is really physically proven, but for many it is very hard.

Kristian says:
2013/03/14 at 1:12 AM
Rosco says, 2013/03/13 at 5:04 PM:

“No matter what anyone says – he starts with a constant energy of 235 W/sq metre and proposes no other energy source exists. His claim that reflecting this doubles the available energy ignores his original position – there is 235 W/sq metre available! Where does the extra come from??”

Rosco, I think you will find from what I’ve written on this thread and on the thread at Tallbloke’s that I agree with you that Joe is right in this matter.

But here you mix up. The ‘extra energy’ comes from the constantly energy-generating core.

You seem to forget that Watt is power, not energy. It is Joule (energy) per second. 235 W/m^2 means that 235 Joule is provided to every square metre of the surface of the core planet … per second. This means that after 2 seconds, 470 Joule have been provided. And so on. Which makes you realise that the only energy generated is still coming from the nuclear power source. The point is, it needs to be continuously shed also. Well, it is. They are simply unable to see it.

The arithmetic/budgeting of the inner part of the system is not necessarily wrong. It’s the notion that it’s only the quantity, not the quality, of a radiative energy flux that matters when it comes to inducing a temperature. They also ‘forget’ that the flux generated in the nucleus at equlibrium escapes the system as a whole. It all balances. No energy is piling up anywhere.

Kristian says:
2013/03/14 at 1:15 AM
I should propbably rather say something like ‘energy delivered’, since energy can never simply be ‘generated’.

Joseph E Postma says:
2013/03/14 at 7:42 AM
Yes, “quality” is a good term to use in this case – “quality” is how the 2nd Law of Thermo comes into the picture for radiation, while Willis et al. only ever consider the 1st Law.

Alan Siddons says:
2013/03/14 at 8:06 AM
“quality” is how the 2nd Law of Thermo comes into the picture for radiation, while Willis et al. only ever consider the 1st Law.

As I like to say, the 2nd Law enforces the 1st Law. If heat (or more generally ‘thermal energy’) did NOT just raise the temperature of less energetic bodies but also made the heat source warmer, this would spark the kind of infinite heating cascade that Greg House described. Such a cycle would constitute the creation of energy out of nothing, which violates the 1st Law. So the 2nd Law is how the 1st is enforced.

Joseph E Postma says:
2013/03/14 at 8:10 AM
Two objects in physical contact, one cooler one hotter, will not heat up the hotter object. Everyone knows this. But now, just separate the two objects by a short distance, and then the warmer object will warm up even more, as it also warms up the cooler object, because there’s radiation.

If it worked like that, radiation wouldn’t obey thermodynamics, and neither does the Willis et al set up.

Think it through people…

lgl says:
2013/03/14 at 9:23 AM
[JP: Trashed because you’re just repeating the same BS idiocy over and over again.

Two objects in contact, one warmer and one cooler, will warm the cooler body and cool the warmer one. Separating them with a gap does not mean that the hotter one has to get hotter to warm the cooler one – the cooler one just warms, the warmer one isn’t heated by the cooler one or from itself.]

Rosco says:
2013/03/14 at 4:18 PM
What a lot of people ignore is that when radiating an object is losing energy and to maintain its equilibrium that energy needs to be replaced.

I’ll say it again – Willis has created energy out of nothing.

Loodt says:
2013/03/14 at 5:08 PM
Hi Joseph, I’ve been watching the comments on this website grow since you first posted this entry to your blog. I admire your tenacity to argue with patience on this very fundamental issue; a basic understanding of thermodynamics. I do want to point out a few things to you. Willis has the gift of the gab, nobody can deny his ability to spin a yarn. In the tradition of great American authors he can keep an audience spellbound with his ability to sketch the great outdoors in a masterly fashion. However, Willis has no formal eduction, and he bragged on numerous occasions about his ability to out think formally eduction persons. He get could not get his head/mind around the adiabatic process that causes air to heat up under pressure, I lost all interest to read anything technical that he wrote after reading his juvenile ramblings about that topic. This steel-ball analogy made me smile, and then I skipped to something else. I am too old to fight with fools.

Now, Anthony is driven with a desire for acceptance and formal recognition by the scientific community. His blog is clear evidence that he is doing ‘science’ as he thinks it is done at institutions of higher learning. But, like Alexander had Rasputin at his court, and the spell and hold that the latter had on the former is still being discussed and debated, Anthony is currently under the spell of Willis. And Willis, like a cocky and cheeky mongrel, likes to lash out at people far more learned and knowledgeable than himself. Did you see how Wills attacked our Piers Corbyn, an astrophysicist, with outright rude and insulting posts? Unless you have a very thick skin, and are prepared to fight dirty, stay away from Anthony’s lapdog! The man hasn’t even passed Maths, Applied Maths or Physics I.

Rosco says:
2013/03/14 at 5:36 PM
Look at Willis’s diagram. I know he didn’t claim any dimensions but it looks like the shell is twice the diameter of the steel solid.

Plug in a radius of twice into the surface area and you need a factor of four to maintain the radiation at 235 W/sq metre at the outer shell – 4 x pi x radius squared used to be the formula for a sphere – perhaps “climate scientists are proposing amending that as well.

Double radius equals four times surface area.

So to achieve the claimed 235 W/sq metre exiting at the outer shell requires NOT 470 but 1880 at the inner solid.

Outer shell 235 out + 235 backradiated = 470 x 4 times the surface area = 1880 !

What if the inner solid is a cube ?

The only way he can have the same value exiting the “shell” as supplied by the “core” is if the extra radius is zero – surely that demolishes this BS “science” ???

Joseph E Postma says:
2013/03/14 at 6:52 PM
Thanks for that Loodt – that explains the situation very well and confirms what had been my intuition as to the behaviours which have been observed. Spot on. Best regards.

Joseph E Postma says:
2013/03/14 at 6:55 PM
Rosco, MANY things demolish this BS science, and I am still trying to figure out why it doesn’t stop them! lol

Willis said the distance between the shell and sphere is small and so the error was irrelevant – however, he obviously didn’t consider that the error which is exposed when you use a large gap between the sphere and shell destroys the entire argument altogether in any case.

Rosco says:
2013/03/14 at 7:47 PM
I think they are committing deliberate fraud with sleight of hand tricks.

Their initial proposition that energy entering Earth over a disk balances energy leaving Earth over a sphere is somehow taken to mean the insolation is one quarter of the albedo adjusted solar constant ?

This figure – ~1368 / 4 = ~342 W/sq metre reduced to ~239 in to balance the ~239 out – is then used to say that there is insufficient energy to account for surface temperatures.

This is of course complete nonsense as this ignores their initial condition – the incoming radiation is over a disk not the whole sphere as the outgoing radiation is.

Surely people with PhDs can see they have completely mixed up their own initial starting point ?

The Earth only needs to radiate 239 W/sq metre average over the whole sphere to balance the average 956 W/sq metre over the disk.

If the insolation is only say 342 W/sq metre as Trenberth et al claim then the Earth would only be radiating one quarter of this or about 85.5 W /sq metre – that is what their initial claim is.

By the original geometrical construct the temperature induced over a disk has to be 1.414 times the temperature of the outgoing radiation.

239 W/sq metre is the temperature associated with approximately 255 K according to Stefan-Boltzmann.

1.414 x 255 K is approximately 360 K which is the temperature associated with 239 x 4 W/sq metre or ~956 W /sq metre.

The insolation simply has to be 956 W/sq metre over the surface area of a disk – pi x r squared – if the outgoing radiation is 239 W/sq metre over the surface area of a sphere – 4 x pi x r squared.

I simply fail to understand how this myth has become accepted and if you challenge the obvious mathematical failing you are some sort of heretic denier !!

Joe has put this succintly many times but the GHE crowd continue with their misrepresentation.

Joseph E Postma says:
2013/03/14 at 7:56 PM
Bravo Rosco, exactly

You don’t know how many climate scientists have lost their mind at me when I tell them that -18C input over a globe will NOT produce the same physical response as 1370 W/m^2 over a hemisphere – in the fake scenario sunlight can’t melt ice, in the real scenario it can. Apparently they think such a profound difference DOESN’T MATTER, and they will get downright indignant and nasty at you if you insist that sunlight only hits one side of the planet. I’ve been called so many forms of stupid for insisting that the planet is a sphere heated on one side it is mind boggling.

You have to be lying on purpose and to be purposefully covering up a huge fraud to have to yell at someone and call them names for stating that only half the Earth has sunlight, because if people this stupid are actually thinking they’re doing science, the world should have blown up by now. At the very least, keep climate scientists away from anything remotely resembling engineering.

thefordprefect says:
2013/03/15 at 9:09 PM
Will Pratt says: 2013/03/13 at 8:39 AM
In your experiment, you have simply increased the mass of the system, of course it will take longer to cool down.
———————————————————–
Second experiment
Isolated heated hot object
Distant (well 9cm) cold plate/warmplate
shows hot object temperature increase when external cold plate is replaced with idencticle warm plate.

Note that 2 external plates are used (identical pressure die cast aluminium mouldings) The mass is now constant.

The internal isolated plate is not affected by the changes in external plates other than by IR

Not the final experiment. but geting there.

thefordprefect says:
2013/03/16 at 4:59 AM
A lnk to the post mentioned
A Cool Object Reduces Energy loss from a Hot Object

http://climateandstuff.blogspot.co.uk/2013/03/a-cool-object-reduces-energy-loss-from.html

Rosco says:
2013/03/16 at 5:26 PM
Willis’s argument is total BS and it easy to prove.

Simply put another shell around his hypothetical setup and what happens ?

By his own construct his shell is now radiating to the outer shell which heats up to radiate 235 to outer space plus 235 back to his existing shell – his own statement.

The new shell is radiating equally 235 to space and 235 backradiation – this is what he said happens to his shell so it MUST be right for the new one – Willis says so !!!

So now his original shell MUST be radiating 470 to the new outer shell otherwise the maths don’t add up !

470 out = 235 to space plus 235 backradiated.

And now the core MUST be radiating at 940 because his original shell is now radiating 470 out and 470 back !!!!!

Do it again with another shell .. and another and another …..

Surely anybody can see this is impossible BS ??

If not engineers are totally dumb – they can improve the thermal efficiency of all thermal driven engines by simply building shells around the boilers and eventually do away with the need for the fuel at all – boy wouldn’t that reduce the carbon footprint ??

Rosco says:
2013/03/16 at 6:22 PM
Willis’ little model ought to be easy to demonstrate in experiment. I am certain those with the necessary vacuum chamber wouldn’t bother as they would have the intelligence to see through this little shell game.

As a final postscript Willis goes on to try to demolish reality by requiring his supposed system needs to be thermally isolated to work.

“If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.”

What utter BS. Why ?

Because we know there is a substantial temperature difference between the core and shell – it is the magical 33 degrees C of the “greenhouse effect”.

So his steel greenhouse effect – total BS anyway – cannot possibly represent Earth’s greenhouse effect because the atmosphere transfers surface heat and thermal equilibrium results – which we know is wrong.

I cannot believe how naive people like this are !

I thought temperature was a measue of kinetic energy.

If the molecules at high elevations had the same kinetic energy as the molecules at ground level the temperature of each area would still be vastly different because of the vastly reduced numbers of molecules at higher elevations.

And that is ignoring the conversion of kinetic energy to potential energy as a gas rises – which results in lower temperature by definition. This simply must be true else the molecules have gained more energy from nowhere.

These people ignore mass all the time.

Radiation is caused by the temperature of an object which has mass !

Even if a gas.

If there is radiation it simply must be in proportion to the mass – it doesn’t come from nowhere.

At 5000 metres the temperature of the atmosphere is about – 18 degrees C. This approximates the minus 18 degrees C the so called energy balance for Earth calculation results in.

The density of air at 5000m is about 740 grams per cubic metre. Of that 740 grams 0.06 % by weight is CO2( if CO2 is a well mixed GHG) and say 5% is water vapour.

So 37 grams of water vapour and 0.44 grams of CO2 produce backradiation of 324 W/sq metre according to Trenberth et al.

I simply do not believe it – never have and never will !

A C Osborn says:
2013/03/17 at 2:52 PM
Joe any comment on this?
http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/#comment-47019
lgl says:
March 17, 2013 at 8:39 pm

Westy
My kitchen experiment was a bit more successful. 30W soldering iron heating a 0.12 m2 ‘planet’. Temp without shell, 55.7C Temp with shell, 67.4C, which means 12.4 Watts was mysteriously ‘created’ by the shell, who would have thought.
12.4 W is of course far from 30 W, but not bad considering this was not done in vacuum.

Max™ says:
2013/03/18 at 8:19 AM
In response to this comment:

“http://tallbloke.files.wordpress.com/2013/03/ipcc-v-skydragon.png

Claes Johnson/SkyDragon ‘Back Radiation’ Model
If I understand the SkyDragon/Claes Johnson radiation model correctly, there is 235Wm-2 ‘back radiation’ from the under surface of the shell (as there must be by the definition of a black body) but that radiation is not absorbed by the core. Instead, it is deflected back from the core surface. So it then gets absorbed back into the shell, as shown in my right hand diagram. The consequence of this is that the energy flow rate through shell AND core is 235Wm-2. So both shell and core are at a temperature of 254K (simple application of S-B law).

IPCC ‘Back Radiation’ Model
In contrast, the IPCC ‘back radiation’ model allows the 235Wm-2 of ‘back radiation’ from the under surface of the shell to be absorbed by the core. The consequence of this is that the energy flow rate through the core is 470Wm-2 and the energy flow rate through the shell is 235Wm-2. So the core is at 334K and the shell is at 254K.” ~David Socrates

David, I posted this earlier, your diagram does not represent the “skydragon” explanation at all.

…….. ^ …………… I’m fairly sure this is the version Postma would endorse.
…….. |
…… 235 W/m²
————————————- 254 K [235 in, 235 out]
..^..
235 (470-235) W/m²
————————————- 334 K [235 in, 235 out]
……..^
……..|
……235 W/m²

vs

…….. ^ ……………. IPCC
…….. |
…… 235 W/m²
————————————- 254 K [470 in, 470 out]
..^ …………….. 235 W/m²
470 W/m² ………v
————————————- 334 K [470 in, 470 out]
……..^
……..|
……235 W/m²

The difference is only in the amount of energy said to be absorbed by the various surfaces, if radiation from a cooler surface is subtracted from the power of the radiation leaving the warmer surface, then the skydragons are right.

If radiation from a coolder surface is added to the power of the radiation leaving the warmer surface, then the IPCC model is right.

___________

I’m not sure why your diagram has a 255 shell/255 planet and a 255 shell/334 planet being compared, so I cleaned it up to more accurately demonstrate the differences between the two explanations.

Joseph E Postma says:
2013/03/18 at 12:03 PM
Rosco: Just keep adding shells and imagine the temperature you could get?! Why don’t they extend their own arguments to explore if they’re actually meaningful or not? Because they just want their faith.

Joseph E Postma says:
2013/03/18 at 12:11 PM
@ACO: He didn’t “create” energy out of anywhere. He didn’t perform the test in vacuum. There are so many confounding errors it doesn’t mean anything. He’s just reduced convective cooling, that’s all. One thing is certain – you can not get higher temperature than the work you put in. Tell them to scale this up to a power generation unit, or to bring the plans to a power engineer or an electrical engineer for approval and design, etc. If you can get energy for free, you can use it for free. They’re just confusing themselves. If not, then why don’t they solve the world’s energy problems? Because they’re amateurs and they have no clue what they’re doing.

Joseph E Postma says:
2013/03/18 at 12:23 PM
In this diagram: http://tallbloke.files.wordpress.com/2013/03/ipcc-v-skydragon.png

there is no explanation for how radiation from a 255K source plus a 255K source can produce 334K. How does one object of temperature 255K cause another object of 255K to become 334K? It doesn’t happen. Two ice-cubes don’t heat each other up. They just “say” that 255K radiation plus 255K radiation produces 334K. That is not how reality works. The IPCC model is mentally retarded. You could melt ice to water by having two ice-walls facing each other.

I already explained how temperature and radiation works on the wavelength “quantum” level. Photons don’t interact with themselves to increase their own frequency, and you need higher frequency to generate higher temperature. 255K radiation is just that, and it doesn’t increase its own frequency – they don’t “pile up”, they go right through each other. 255K radiation from one source plus 255K radiation from another source does not produce 334K. This is the stupidest idea that has ever existed.

Kristian says:
2013/03/18 at 1:30 PM
Rosco,

I have also put forward another thought experiment at tallbloke’s, one where the shell would gain its heat not from a core planet, but from a hypothetical, evenly spread out nuclear source within the shell itself. The nuclear energy source supplies a constant wattage to the shell. If, as the Eschenbach apologetics insist, the shell emits half of this provided power outwards to space and the other half inwards upon itself, then the shell would only ever be able to rid itself of half the heat gained from the nuclear energy source.

The outcome? Unstoppable runaway heating.

To be fair, on of the proponents – David Cosserat, which BTW Joe, I feel is really misrepresenting your position on this diagramatically over at tallbloke’s – concedes that this specific shell WOULD in the end lose all its received heat through the outer surface to space (he simply arrives at that same conclusion in a more convoluted way), but for some reason fails to take the next step and see that it would ALWAYS be like this, no matter where the incoming energy is coming from. He STILL thinks the planet would necessarily heat up with the shell surrounding it.

Joseph E Postma says:
2013/03/18 at 1:50 PM
They’re just making things up at an amateur level of knowledge and understanding, and arguing towards the outcome they desire in the first place. Sophistry in other words.

Joseph E Postma says:
2013/03/18 at 2:02 PM
I have a question for everybody:

What is the temperature of an infinite amount of heat?

Will gather responses for a bit before approving them so that everyone gets a chance to answer independently.

Max™ says:
2013/03/18 at 2:06 PM
Well, my point with the diagram clarification is that an internally powered sphere+shell would never set up a black body spectrum with the same temperature between both surfaces, inverse square law makes that impossible.

Assuming the two surfaces being 334 and 254 K in both diagrams is the fair way to compare the differences.

Said difference is put simply: you, I, and modern understanding of physics all claim that radiation from a cold surface reduces the power leaving a warm surface. 235 in, 235 out.

The IPCC models and their supporters claim that radiation incident upon a warm surface adds to the internal energy no matter what temperature the source of that radiation was. 470 in, 470 out.

[JP: I think you meant 235 in, 470 out for the IPCC]

Simon Conway-Smith says:
2013/03/18 at 2:15 PM
Joe, Do I suspect the answer is similar to “How long is a ball of string?”.

Max™ says:
2013/03/18 at 5:18 PM
…….. ^ ……………What I think is correct
…….. |
…… 235 W/m²
————————————- 254 K [235 in, 235 out]
..^..
235 (470-235) W/m²
————————————- 334 K [235 in, 235 out]
……..^
……..|
……235 W/m²

235 from the internal power source to the planet surface which is equilbrated at 334 K and would emit 470 as a black body in a vacuum. The temperature difference between the shell and planet reduces the power that actually reaches the shell to 235, and it emits to space accordingly at 254 K.

235 in, 235 out.
________________

vs

…….. ^ ……………. What the IPCC thinks is correct
…….. |
…… 235 W/m²
————————————- 254 K [470 in, 470 out]
..^ …………….. 235 W/m²
470 W/m² ………v
————————————- 334 K [470 in, 470 out]
……..^
……..|
……235 W/m²

235 from the internal power source to the planet surface which is equilibrated at 334 K and emits 470 no matter what.

The temperature difference between the shell and the planet doesn’t matter, so the planet also receives 235 back from the shell, adding up to 470 in, and the 470 it emits to the shell is balanced by the 235 up and 235 down.

470 in, 470 out.

David Socrates says:
2013/03/18 at 5:51 PM
Hi Joe,

As often happens in these heated debates, I have been misrepresented by Max and Kristian with respect to my diagram at http://tallbloke.files.wordpress.com/2013/03/ipcc-v-skydragon.png .

All I was trying to do was to find a way of re-starting the somewhat meandering debate currently raging over at the Tallbloke site by asking people to concentrate on your actual proposition rather than inventing their own ‘models’ with all sorts of bizarre variations such as heating the shell, etc., etc.

So I produced two diagrams which, to the best of my understanding, represent (i) the ‘back radiation’ model that IPCC and Willis adhere to; and (ii) the Claes Johnson/SkyDragon model which you adhere to.

As I understand your proposition, it is that energy in a cavity does not enhance the temperature of that same cavity. That is a perfectly respectable scientific position to take which everyone, whether they agree with it or not, ought to be able to comprehend and debate calmly. I tried to represent your ‘standing wave’ cavity radiation diagramatically by the green u-turn energy flow loop which, in effect, does no work, because the 235Wm-2 from shell-to-core is NOT absorbed, for the reasons you have always consistently maintained (unless I have entirely misunderstood you!). So it simply does not take any effective part in the energy balance. Consequently the right hand diagram shows both shell and core at the same temperature because there is no energy enhancement, which, again, I think is your position.

My objective in producing the diagram was NOT in the first instance to argue for the left hand or right hand model, just to secure some ground rules that we all can agree on about what the two alternative propositions actually are! So my question to you now is this: do you agree that the right hand diagram (now explained to you in non-perjorative terms) is a fair representation of your position? Or, if not, in what way does it misrepresent your position (which I am certainly not trying to do) and how would you prefer to modify it so as to more fully represent it?

Then, and only then, can we move forward to a focussed and calm discussion that does not constantly divert onto irrelevancies (not, I would emphasise, ones of your making).

Best regards
David Cosserat

Rosco says:
2013/03/18 at 9:41 PM
MAX – you’ve still created energy out of nothing !

Think about adding another shell and another etc. This is no different to the original proposition as there is a surface radiating 235 out and you stick another shell over it and you have to double the core energy again to maintain the 235 out and 235 backradiated at the outer shell.

Surely anyone who thinks about this can see this creates double the energy each timeyou stick another shell over the ensemble.

30 shells results in 252 GigaWatts at the inner core – all from 235 W radioactive decay.

252 GigaWatts – i mean – c’mon !!!!

To all those who think radiation from a cold object can cause heating of a warmer object I ask that you explain Pictet’s experiment on the reflection and radiation of cold.

this experiment was conducted over 200 years ago and is reproducible.

His experiment provided the exact expected thermodynamic response that sound theory predicts.

His cold object, a container of ice water, placed at the focal point of a concave mirror caused the thermometer at the focal point of a facing concave mirroe 16 feet away to dramatically decrease its reading. Note that – 16 feet away.

That is the expected thermodynamic response – the radiation went from the hotter object to the cold one – there was no backradiation – none at all.

Why such a dramatic response I have no idea.

Rosco says:
2013/03/18 at 9:46 PM
The fourth root of infinity over sigma ? Now that would require some computing power to solve.

Or something that is completely indeterminant ?

Or ask Willis ?

Rosco says:
2013/03/18 at 10:00 PM
Perhaps I misread Max’s comment and am now totally confused.

I still think if Willis’ proposition is valid for the addition of one shell it must be valid for the addition of any number. This is surely one of the tests that must be valid for a hypothesis – reproducibility.

Logically that leads to an absurd consequence of unlimited energy from 235 W simply by “trapping” radiation.

Just because it appears it is arithmetically correct does not mean it isn’t completely absurd – which is the only possible conclusion a rational person can make.

Therefore Willis’ steel greenhouse is an absurd proposition.

I understand why his model is with the steel shell very close to the “planet”.

This is similar to the supposed GHG Earth where the additional 5 km or so of atmosphere to the radiating height where the temperature is 255 K – the 5 km is only a small fraction of the Earth’s radius – see the anology – the steel shell is really close to the original surface.

I say if the steel model is BS as logic demands it is so is the Earth’s supposed greenhouse effect.

You do not need any special physics training to see the model requires doubling the core energy for every additional shell and that is an absurd proposition.

Greg House says:
2013/03/19 at 7:13 AM
Alan Siddons says, (2013/03/14 at 8:06 AM): “As I like to say, the 2nd Law enforces the 1st Law. If heat (or more generally ‘thermal energy’) did NOT just raise the temperature of less energetic bodies but also made the heat source warmer, this would spark the kind of infinite heating cascade that Greg House described.”
====================================================

I posted this description on the Tallbloke blog a weak ago (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46123), but today, as I referred to that again, he blocked my comments and had no problem with lying straight to my face, saying that I admitted being wrong.

My comments have never been blocked anywhere, either on WUWT or on JoNova. I even challenged green radicals on the Greenfyre’s blog and no comment of mine was blocked or deleted there. I guess, some people are getting desperate.

Tallbloke has switched now from the “Willis universe” with a power engine to a model without power supply, so that the whole thing is just cooling and therefore it is not obvious that the cycle of mutual warming won’t work. Very “clever”. My demonstration is still valid, but references to it are unwelcome now there. Right, maybe people just forget it and he can fool them undisturbed.

So, meet “skeptik” Roger. I have made screenshots of every comment of mine “awaiting moderation” and his remarks he inserted: http://s8.postimage.org/93gvfkyc5/image.jpg, http://s14.postimage.org/cq7v6y9jl/image.jpg, http://s8.postimage.org/jlbip2ok5/image.jpg, http://s4.postimage.org/8mn7icnfh/image.jpg

Loodt says:
2013/03/19 at 2:05 PM
In reply to the question …What is the temperature of an infinite amount of heat?..

The answer is: as long as a piece of sting!

Temperature is a state of mass, and without specifying the mass that contains the temperature the question is basically nonsensical.

But, you knew that in any case.

Kristian says:
2013/03/19 at 2:49 PM
I tried this approach over at tallbloke’s. Maybe the simplest form of arithmetic might still work. We’ll see how it goes …

A black body receiving and absorbing a continuous incoming energy flux and accordingly reemitting (as required by definition) an equal energy flux (from whichever side of the body, you insist it’s all the same) of 470 W/m^2, would that not have an emission temperature (as per the S-B equation) of 302K?

Well, Willis’ black body shell receives, absorbs and reemits an energy flux of 470 W/m^2. But its temperature? 254K.

Remember, the absorption always occurs before the reemission. There would be no emission without the absorption first. So why doesn’t the incoming flux of 470 W/m^2 raise the temperature of the shell to 302K? It doesn’t matter if the incoming flux is coming from the outside (a sun), the inside (emitting planet) or within (‘intrastructural’ nuclear source), the flux is still absorbed completely and then reemitted to balance. Why doesn’t the S-B equation work on Willis’ shell?

The ‘IPCC model’ seems to claim the following:

# CORE absorbs 235+235= 470 W/m^2, emits 470 W/m^2, S-B temperature 334K (?!, really 302K)

# SHELL absorbs 470 W/m^2, emits 235+235= 470 W/m^2, S-B temperature 254K.

What’s wrong with this picture?

Well, if one wants to argue ‘Ah, but the NET incoming flux to the shell is 470-235= 235 W/m^2!’, even if the actually absorbed flux IS 470 W/m^2, there’s no getting around that, then one should consider this:

The NET outgoing flux from the core planet is also 470-235= 235 W/m^2. Why, then, is it OK for the planet to have a temperature of 302K, but not for the shell?

The temperature is set by the absorption of the INCOMING flux, then the emission of the OUTGOING flux is set accordingly.

Rosco says:
2013/03/19 at 5:24 PM
The discussion at tallbloke’s website is running away in a similar fashion to the runaway greenhouse effect.

I made the argument that if the original condition is valid – ignoring all errors and geometrical effects – then when the shell is added you have the same situation as the original- except for the magical doubling of energy.

In this new case you supposedly have a surface radiating 235 to space.

So adding another shell requires the original shell to now radiate 470 – which is exactly what happened in the original scenario.

This of course leads to the absurd proposition that you can create enormous energy by simply doubling it with more shells.

No-one seems to even consider that argument, instead preferring to continue with all types of variations of the same argument why it is OK to arbitrarily double the energy with the insertion of a shell.

I guess this proves that people really want to have faith in their personal truths and will defend them no matter what.

I could be completely wrong and they completely right but I thought about it long enough to come to my conclusion that the shell doubling leads to absurd consequences – it appears most commentators do not even think about the adding of new shells as per the original condition they all accept – ie double energy per shell.

Instead of even attempting to debunk this – dare I say it – obvious inconvenient truth – the discussion continues over why Willis’ proposal is theoretically OK.

I discovered that Willis originally proposed this “steel greenhouse” in 2009 !

That means it is accepted as “science” at WUWT for over 4 years without any questioning other than geometric ones – which debunk it anyway – and he has been allowed to spruik this gibberish twice yet critics of such chicanery are silenced.

I thought SKS were bad enough – it is a wake up to see even sceptical websites are so entrenched in the greenhouse effect they will not allow any voice raised against it yet allow such obvious chicanery as the steel greenhouse and its magic.

Willis always quotes that if there is a mere 170 W/sq metre insolation there is insufficient energy to prevent the oceans freezing.

I guess some people simply cannot understand they have been conned into a religious anti science belief system.

But I guess they say the same about me.

Joseph E Postma says:
2013/03/19 at 6:37 PM
Oh I see David, thanks for the explanation. Yes the diagram is quite useful in that way.

So yes, the problem is in the adding of radiative energy fluxes in a direct linear way and thus leading to the IPCC model. The IPCC model can obviously be treated as infinite plane parallel “walls” which is equivalent to one shell inside the other or a planet inside a shell. Consider if there were a radiation source heating the bottom of the lower wall with 255K radiation at the position of the lower wall, and forget about the upper wall/outer shell. By the accounting of linear adding and subtracting fluxes of the IPCC, the lower wall could never even get to the temperature at which it is being heated, because of the supposed bi-directional nature of its output.

If the lower wall received 235 W/m2 on the lower side, then supposedly it could only ever get to a temperature corresponding to 117.5 W/m2. This should intuitively indicate an error, but this invented way of adding fluxes is trusted instead. Radiative fluxes do not add with elementary-school arithmetic in their ability to perform work/induce temperature. The ability to perform work/induce vibration in matter (i.e. induce temperature) is a function of the frequency components of the radiation. It is the frequencies which induce corresponding material vibrations and hence temperature. The matter will be induced to vibrate at the frequency components of the incident radiation because this is an active forcing causing action, and the matter doesn’t care which direction the subsequent thermal energy might be emitted to. Either side of the wall will emit at the flux level corresponding to the temperature which it is at, the temperature which it has been induced. If the wall was very thin and the thermal mass was very small, both sides would be induced to be the same temperature very quickly from forcing radiation incident on one side only.

So, the equation for heat energy radiative transfer is basically q = sigma*(T_hot^4 – t_cold^4). When T_hot and T_cold come to the same temperature, there is no net energy transfer. This is what confuses the IPCC model, because they want to insist that energy is still transferring and causing heating, just as long as you balance it all out to zero in the end. This is incorrect. There is zero net energy transfer. Hence, the upper side of the wall emits at 255K, and that energy is replaced at the bottom side by the incoming radiation, and no energy is lost from the wall on the bottom side at all. Why is no energy lost from the bottom side of the wall? Because T_hot^4 – T_cool^4 = 0.

From this latter, you can write that on the “Skydragon” side, there is zero energy transfer between the shell and the core. They come to the same temperature, and no heat energy (i.e. ZERO energy) is transferred from the shell to the core. 235 goes from the core to the shell and 235 leaves the shell on the upper side. Between the shell and core is a radiation field, a blackbody spectrum, at 255K.

Joseph E Postma says:
2013/03/19 at 6:48 PM
For an object to be warmed, only another heat source of higher thermal energy (temperature) can achieve that. The same applies with any type of energy, e.g. light. If you take 2 torches of different brightness and shine the dimmer one at the brighter one, does the brighter one get any brighter? No it doesn’t. Or if you shine the a torch at a mirror which reflects the light back at itself, does it get any brighter? No! It cannot, as quantum physics rules that a lower frequency energy cannot raise the frequency of a higher frequency one, nor can the reflection of an energy source at a frequency raise its own frequency. If a cooler or same temperature body could raise another body’s temperature, we’d have an infinite energy generation machine, which is impossible.

A blanket, of course, reduces convective air cooling, and this makes your skin feel warmer. It is a very poor analogy to a radiative GHE because there is no physical correspondence in the analogy – it is just an analogy, without any actual physical connection. And the reason? As pointed out: a torch cannot make itself brighter with its own radiation. Stand in front of a mirror – do you feel any warmer? NO! This is why the shell game is wrong.

Alan Siddons says:
2013/03/19 at 6:54 PM
Allow me to repeat my first point on this blog page. If you’re holding an inert substance that gets hotter than the heat source it is touching, then you are holding onto a miracle. And if you’re holding an inert substance that RADIATES more intensely than the radiation it’s receiving you’re holding a miracle too. Indeed, a miracle is what the hocus-pocus of greenhouse physics amounts to: the target is supplied the continuous power of 235 W/m² and from that it develops a continuous power of 390 W/m². In other words it is radiating more intensely than the radiation it’s receiving.

I mean THINK of it. Think how you’d react to a salesman who claims that his Hyper-White Paint reflects more light than the light that’s falling on it. You’d run that charlatan out of town. Yet supposedly serious scientists claim that an object can absorb and EMIT more light than the light that’s falling on it. And most people take this claim for granted. The world is quite insane.

Joseph E Postma says:
2013/03/19 at 7:02 PM
And the funniest thing about it all, is that the Earth does NOT receive 235 W/m2 continuously anyway! Think of THAT! It’s all bizzaro land.

Alan Siddons says:
2013/03/19 at 7:37 PM
It’s been disgusting to witness so much “hopeful ignorance” here and on Tallbloke’s site. If physical phenomena like Eschenbach and others describe actually did exist, our ability to generate energy would be without limit. Good LORD, if a square meter on the earth’s surface can produce 390 watts from 235 with just 1% of its atmospheric gases “re-radiating” a few wavelengths on the spectrum, then imagine the power that a full-bodied continuous-spectrum Eschenbach Shell could generate. Just SHOW US, then, you silly fools. Point to a device that works as advertised. Put up or shut up.

Joseph E Postma says:
2013/03/19 at 11:52 PM
Exactly – engineer it and exploit it! We could all use such a device. I will more than happily concede everything I’ve ever said and say that I was wrong and even stupidly wrong, if they can engineer this shells game principle and create more energy than they put in. Someone already explained how to exploit it – by using a sphere and shell of metals of differing thermal expansion coefficients. If you can boil water without putting in the raw energy necessary to boil water in the first place, then you have an over-unity steam engine and you can create power with it. DO IT ALREADY! I for one would love free power.

Max™ says:
2013/03/20 at 12:43 AM
The reason I specify that the planet surface is at a higher temperature than the shell is because it is supposed to be an internally heated sphere with a shell at a non-zero distance.

I posted a simple example for the case of a sphere with a 100 m radius and the addition of shells 1 m above the surface or the last shell.

________________________

4*pi*100^2 =
125,663.7*235 = 29,530,969.5/128,189.5 = 230.3

4*pi*101^2 =
128,189.5*230.3 = 29,522,041.8/130,740.5 = 225.8

4*pi*102^2
=130,740.5*225.8 = 29,521,204.9/133,316.6 = 221.4

4*pi*103^2
=133,316.6*221.4 = 29,516,295.2/135,917.8 = 217.1

_______________________

Whether you have four shells or one shell, if you have a shell located 4 m above an internally heated sphere with a 100 m radius which emits 235 W/m^2, then that shell will only receive 217~ W/m^2 and accordingly must emit that much or less.

Joseph E Postma says:
2013/03/20 at 4:37 AM
This is a very good point for Max to make because it indicates that the outer shell always has to be cooler than the inner sphere, as a necessary condition for equilibrium. Of course, as the difference in radii diminishes, the difference in equilibrium flux of the outer shell relative to the interior sphere diminishes.

David Socrates says:
2013/03/20 at 5:41 AM
Joe,
Thanks very much for your replies of 2013/03/19 at 6:37 PM and 2013/03/19 at 6:48 PM.

The first thing to explain is that I am a hardline skeptic on climate change. My motive here, as elsewhere, is to find a definitive way of proving to warmists and to ‘luke warmists’ alike that they are wrong: that CO2 concentration in the atmosphere has NO EFFECT AT ALL on mean surface temperature. Not just that “sensitivity to an increase in CO2 concentration is low” – but that it is ZERO! I have come to this conclusion long ago, not by diving deeply into quantum mechanics or radiative transfer theory but because there is simply no sign of the earth warming dangerously, despite all the uninformed hype on the subject. We all form our initial prejudices in that kind of way, long before we dive deeply into a subject. But the key to success in science is always to remain open to being wrong. So I endorse very much the spirit of your comments at 2013/03/19 at 11:52 PM: I will more than happily concede everything I’ve ever said and say that I was wrong and even stupidly wrong, if they can engineer this shells game principle and create more energy than they put in. Me too! And I would do likewise if the warmists or luke warmists were ever to be able to prove their case for global warming due to increases in atmospheric CO2!

I think that the reason there is so much heated debate over things like Willis Eschenbach’s ‘steel greenhouse’ core-warming model is because most people, even experienced scientists and engineers like me, are unskilled in the specific physics needed to analyse the issues at hand. This often leads to people on all sides in the debate making ridiculous and/or naive objections to things other people have said with the best of intetions and in all innocence. For example, nobody at all (except perhaps for a few way-out eccentrics!) would claim that ANY Willis-style shell model could really cause unlimited energy to be produced from a fixed input energy flow. So to criticise any such a model on such grounds is fine as a way of pulling people up short and making them think harder – just so long as it is intended as robust repartee and not meant to humiliate.

So could I suggest that we make a fresh start, remembering that we skeptics really are all on the same side – not opposing warriors? In that spirit, I would first ask you to clear up the following…

My diagrams on the TB site show what I intended to be a pictorial description of the difference between the IPCC ‘Back Radiation’ model on the one hand, and the Claes Johnson/SkyDragon model on the other. You in your response have kindly confirmed that these diagrams are quite useful (thanks!) [NOTE: in the original IPCC diagram there was an error: the left hand core temperature should have been 302K not 334K, now corrected.]

Yet several commenters over on the TB site have said (unhelpfully – i.e. without any elucidation!) that the right hand diagram “does not represent correctly” the Claes Johnson/SkyDragon position.

So is there anything about my right hand diagram that you disagree with, or feel could be improved? Also is there anything about the text in my accompanying blog comment that you feel is wrong or that you could improve on? Particularly the following paragraph:

Claes Johnson/SkyDragon ‘Back Radiation’ Model
If I understand the SkyDragon/Claes Johnson radiation model correctly, there is 235Wm-2 ‘back radiation’ from the under surface of the shell (as there must be by the definition of a black body) but that radiation is not absorbed by the core. Instead, it is deflected back from the core surface. So it then gets absorbed back into the shell, as shown in my right hand diagram. The consequence of this is that the energy flow rate through shell AND core is 235Wm-2. So both shell and core are at a temperature of 254K (simple application of S-B law).

If so, I am more than happy to try to agree a better diagram and an improved form of words with you.

Alan Siddons says:
2013/03/20 at 7:46 AM
“the planet surface is at a higher temperature than the shell…”

Which kind of ties in to what I was trying to get across before. Consider, for example, that the nuclear core in Eschenbach’s scenario might well be radiating 50 million W/m² by itself. If this core happened to be the surface in question, with no other mass on top of it, 50 million watts/m² would be what it emits. Dump enough rocks and dirt on it, though, and you could reduce the surface’s emission rate to almost zero — which in fact is pretty close to what the Earth’s internal heat does emit. A mass whose surface has been reduced to 235 W/m² was just convenient for Eschenbach’s argument. But introducing more mass in the guise of a dense shell will reduce the temperature even more, thus also reducing the emission. More molecules are now competing for the same amount of heat, after all.

In my view, the 2nd Law only specifies that heat always seeks to minimize a temperature difference. Given the impediments of the inverse square law, or the amount of mass to travel through and the particular conductivity of that mass, however, the 2nd Law does NOT connote that a heat source and its target will eventually reach the same temperature. No, between Eschenbach’s hot nuclear core and an upper surface, a permanent temperature GRADIENT will exist instead — and when such a gradient is CONSTANT one can regard the two locations as being in thermal equilibrium because no further transfer is possible. Same deal with a suspended shell. In short, “thermal equilibrium” doesn’t necessarily imply the attainment of equal temperatures.

Joseph E Postma says:
2013/03/20 at 8:09 AM
I have a few comments to moderate right now and you’ve probably noticed there’s been a significant time delay in approving comments. I am currently on travel in India again this week and the time-zone difference for me is 12 hours, so I spend a lot of the first week fighting abject delirium and struggling to form coherent thoughts. Somehow I still manage to hit the ground running and write c++ GUI code, handle spacecraft equipment, perform functional test procedures of hours and hours of extremely boring documentation, analyze data and write summary reports of the findings, etc. etc. Suffice it to say, non-essential work gets sidelined and my free-time is spent either working more, or just trying to stay awake so that I can sleep at the appropriate time.
So, just in case you were wondering. Will get to moderating comments my tomorrow morning…10 hours from now. Time to go see what the kanteena (sp?) made for dinner tonight and then pass out in a spice-augmented sweaty delirious haze

Greg House says:
2013/03/20 at 10:01 AM
Another warmist article on WUWT by Ira Glickstein: http://wattsupwiththat.com/2013/03/20/how-well-did-hansen-1988-do/.

Quote: “…it would be reasonable to assume that Climate Sensitivity is closer to 1 ⁰C than 4 ⁰C.” And later Ira Glickstein, PhD says, (March 20, 2013 at 8:07 am): “I am pretty sure Climate Sensitivity is somewhere between 0.25 ⁰C and 1 ⁰C. If Atmospheric CO2 had zero effect, the Earth Surface temperatures would be over 30 ⁰C cooler, because the Atmospheric “Greenhouse Effect” is real.”

David Socrates says:
2013/03/20 at 10:25 AM
Joe,

I have now had time to re-read this post + comments from the beginning and, if you permit, have a few comments of my own to make which I will post as I complete them:

Andrew says, 2013/03/09 at 11:59 PM: Please do not resort to name calling as the layman out there will turn off and not bother to read your comments. The bigger cause here is to ensure that people understand that the fear of CAGW and GHE are greatly exaggerated. You must keep as many people onside as possible. I understand that this may be frustrating to you but keep trying to explore methods to widen this message to other scientific and media outlets.

Whilst sympathising with the frustrations vented here, I feel Andrew is absolutely right. The way forward is a clearer explanation of your position.

Joseph E Postma says, 2013/03/10 at 8:47 AM:
The error has nothing to do with the 0.3% problem Willis commented on.

Agreed!

The error is in the rejection of quantum mechanics and the known behaviour of radiation trapped inside a cavity, and also basic arithmetic. This is a process of exploring methods to widen the message: if nicely telling people that 1+1 does not equal 3 doesn’t work, then maybe yelling at them will?

With respect I genuinely think that yelling at people is a counterproductive approach, however cathartic for you guys! The right way forward is to try harder to convince people like me who have a very strong scientific background but lack the quantum-level knowledge. If you succeed, I (and I am sure others such as TB) will be your strongest advocates.

Remember, as a hardline skeptic I would be delighted to be convinced. But it has to be done by intellectual persuasion, not by shouting and ranting and calling people names.

Simon Conway-Smith says:
2013/03/20 at 12:05 PM
Joe, When in Bangkok last week, I avoided all those little street food stalls, and kept to the main shopping centre eateries. I do like Indian though, but have never been there to taste the real stuff. Don’t forget through, have your ice cubes ready in case it hasn’t been heated well enough

Simon Conway-Smith says:
2013/03/20 at 1:06 PM
Joe, Just been reading the SkepticalScience ‘repost’ of your debunking of the GHE hypothesis, particularly the argument you make that we have to treat the Earth as a sphere with only half being irradiated by the sun at any one time. They seem to think this is not necessary and undergrad level stuff (funny how it appears in the IPCC reports as ‘gold standard’ then), but what made me really narked was the statement…

The atmosphere and/or ocean help smooth the diurnal temperature difference very well.

I defy anyone to stand outside for 24 hours and say that the temperature was the same, or changed little, during all that time! Simply unbelievable!

For them to say you “did not like a simple model of Earth’s radiative balance where we approximate the Earth as a sphere with uniform solar absorption” is to totally fail to understand your point, and their error. It’s not just that you didn’t like it, but that it’s fundamentally wrong.

Joseph E Postma says:
2013/03/20 at 6:14 PM
David Socrates @ 2013/03/20 at 5:41 AM

Indeed, there’s been no warming above the 1930′s values even though CO2 has increased, what 50% or more since then. Hello.

OK so for the Skydragon diagram. We need to look at the REAL, actual equations for heat flow:

q = k*(T_hot – T_cool) {conduction}
q = s*(T_hot^4 – T_cool^4) {radiation}

These are directly physically analogous to other equations in physics:
Force:
a = 1/m * (F2- F1)
Electricity
I = 1/R * (V2 – V1)

When two opposing forces are equal in magnitude, it does not mean that the forces do not exist, but it does mean that they do not do any work. Same thing with two opposing voltages of equal magnitude – it does not mean the voltages don’t exist, but it does mean that they can’t induce any current.

And so with temperature, when you have two equal temperatures “facing each other”, it does not mean that they don’t exist or that the thermal radiation from them doesn’t exist, but it does mean that they don’t do any work or induce any temperature, or cause any heat energy flow.

This is where the IPCC supporters think that Claes and the Slayers say that there is no such thing as backradiation or that photons must “have to know” to not be emitted in a specific direction. This is very uninformed and dis-knowledgable about physics, to say such a thing. The backradiation DOES exist, just like one force opposing the other force exists, but it doesn’t do any work. The heat energy flow equation, just like the force equation, equates to ZERO.

What the IPCC supporters do is insist that the backradiation must perform work or heat flow, which is equivalent to saying that one of the forces must do work even through it doesn’t actually induce any acceleration/displacement. Really, they’ve just thrown the whole fundamental physics of energy, work, and the basic set of physical relations out the window.

They WANT backradiation to “do work”, which is impossible by the definition of heat flow, because they can’t figure their way out of treating sunshine as freezing cold over the entire planet. I provided the way out (and other Slayers did too) by pointing out the fact that sunshine is NOT cold and is NOT distributed over the entire planet. (More on this in a coming moderation reply to another comment up the line here). But essentially, they want to keep the assumption of cold sunshine because this is the only vector by which CO2 can be vilified and climate alarm be created, via the manufactured GHE.

So, on the Skydragon diagram, how much energy is sent from the shell back to the planet? Zero. Why? Because q = s*(T_hot^4 – T_cool^4), and T_hot and T_cool are either equal so that q = 0, or, the shell is the cooler meaning that the q only goes from the planet to the shell. Does the shell emit to towards the planet? Sure. Does it induce any work or heat flow? No! Push your hands together, hard. Do the opposing forces exist? Yes. Does either perform any work? No. Why? Because the displacement is zero.

The supposed “knowledge” required by the photons is created right at the inner surface of emission: when photons are emitted at the inner side of the shell, they are immediately confronted with incoming photons from the planet of equal frequency. What happens then? They cancel each other out, in terms of net energy flow. (The photons pass through each other in fact, but the net energy they carry is balanced to zero.)

Now, the shell will have some mass and will contain some internal energy, but, even if it was nearly mass-less, as soon as energy is emitted from the outer side of the shell and truly lost from the shell, there will be a slight deficit in the thermal energy balance between the shell and inner sphere, and so the energy gets replaced into the shell, and this happens continuously – in thermal equilibrium, the shell will lose heat energy on the outside just as fast as it is replaced on the inside. Of course as Max points out, because the shell is larger than the sphere, the actual Wattage and temperature will be less from the shell; but, the total energy will be the same.

So, what you could draw is 235W/m2 emission from the inside of the shell, but then cross it out or something because that emission, when facing incoming energy of the same magnitude and frequencies, can’t actually induce any work or heat flow to occur. Why? Because of the basic equations of physics as shown above. The outer side of the shell CAN freely lose bonafide heat energy out to infinity, and as this energy is lost from the shell, it is continuously resupplied by the planet.

Joseph E Postma says:
2013/03/20 at 6:16 PM
Alan @ 2013/03/20 at 7:46 AM

I would just make the clarification that increased mass in the form of increased density, while the surface areas stay the same, would just slow down the speed at which thermal equilibrium is acheived. If the surface area increases, then indeed the flux and temperature must decrease.

Joseph E Postma says:
2013/03/20 at 6:18 PM
Greg @ 2013/03/20 at 10:01 AM

They just make stuff up as they go along as it suits them. Of course they say that NOW – their models have all failed!

Joseph E Postma says:
2013/03/20 at 6:27 PM
Simon @ 2013/03/20 at 1:06 PM

Yes, they completely missed the point. They are directly saying that it doesn’t matter how we treat the sunshine and Earth at all. It is NOT as simple as making averages, as they like to claim and say that I don’t understand what an average is. THEY do not understand what an average is, the morons. First, 24hr sunshine at -18C is not what exists. It simply isn’t what is real, so, why would it be considered a valid starting point? Second, you can’t average-down the power input in a non-linear response system! -18C for 24 hours can’t melt ice; actual, real sunshine CAN melt ice. Isn’t that an important difference? Especially considering latent heat storage and the temperature plateau created from it? Of course it is important. In fact in my last paper I actually modeled REAL daily sunshine in real time as it actually exists, shining on an ice-ball Earth. -18C for 24 hours of course would never melt the ice ball Earth. The reality-based model melted the ice and maintained an average water temperature of +14C! And the effect of latent heat was to help hold the temperature higher than what it would have been without latent heat.

Joseph E Postma says:
2013/03/20 at 6:40 PM
Here is a plot Simon made for Max:

I ran a spreadsheet starting with the core at 100m radius emitting 235W/m2, all the way up to a 400m radius shell in 1m increments and plotted the radius vs. W/m2. It produces the classic inverse log (square? -JP) graph, which is what you would expect…

Joseph E Postma says:
2013/03/20 at 10:48 PM
Following up briefly on Me at 2013/03/20 at 6:14 PM:

If you push against a wall, energy is spent, but no work, another quantification of energy, is performed. This is the nature of work and energy which seems paradoxical to the IPCC supporters and so it is rejected – they consider that any energy which is spent must result in something, i.e. work, being done. But this is not true, and it is not what physics says. You can spend lots of energy pushing against a wall, but that energy does not perform any work (another form of energy)…it doesn’t cause anything to happen.

So, likewise, backradiation can spend lots of energy shining back to its source, but, it doesn’t perform any work. The equation for thermal physics says so, just like the one for mechanical physics. In radiative thermal terms, this lack of work means that no temperature is induced to raise. The physical reason at the quantum level is due to the frequency components analysis I’ve discussed previously. This is how and why radiation obeys the 2nd Law of thermo.

Greg House says:
2013/03/21 at 6:28 AM
Joseph E Postma says, (2013/03/20 at 10:48 PM): Following up briefly on Me at 2013/03/20 at 6:14 PM:
If you push against a wall, energy is spent, but no work, another quantification of energy, is performed. …
==========================================================

Hi Joe,

I like this explanation very much. It is clear and understandable for everyone.

David Socrates says:
2013/03/21 at 6:43 AM
Joe,

Thanks very much for your carefully considered and well articulated replies of 2013/03/20 at 6:14 PM and 2013/03/20 at 10:48 PM. A number of points:

Radiation need not do work
Your point about radiation not necessarily “doing work” is spot on. I recently published an article on the TB site that insisted that ~333Wm-2 of radiation had to circulate round and round between the base of the atmosphere and the surface (simply as a consequence of their both being at around 288K) but that it did no work. This produced howls of rage from some skeptics who in my opinion have a quite unjustified phobia about back radiation and just deny that it exists. This despite the fact that my diagrams clearly showed the downward flow being more than balanced by the upward flow from the earth’s surface to the atmosphere, thus avoiding any possibility of a misunderstanding about a “cooler body being able to warm a warmer body”.

Or so I thought!

This controversy got so heated that it actually spawned another article on pyrgeometers (installed at ~190 meteorological centres around the world specifically to measure IR back radiation) where some commentators even went to the extent of arguing that the devices didn’t work and/or that pyrgeometer manufacturers were apparently in some kind of an international conspiracy with warmists. How about that for cognitive dissonance?

So, for me it’s official – back radiation exists
I have encountered exactly the same problem more recently in responding to others on the TB SkyDragon blog, some of whom still deny the existence of back radiation. That is one of the reasons for contacting you now. Whether or not skeptics accept your arguments, your position on the existence of back radiation puts the lie to the idea that “only warmists” support back radiation. I tried and tried in my article to explain that such people had to get over their phobia, otherwise they could not reasonably expect to be able to engage properly with sophisticated warmists without being ridiculed.

Layer models rule OK
I am also relieved to see you agreeing that it is legitimate when examining these energy flow issues at a simple level to use a layered ‘model’ that assumes that all radiating surfaces are of the same area, even though in a real set of concentric surfaces, this of course is not strictly the case. Why people think that, say, a 0.3% variation in surface area makes any difference to the fundamental principles under discussion is beyond me. It smacks of sophistry, trying deparately to find a flaw in a basic argument that they don’t like by putting up a nit-picking and diversionary objection.

However…
…even if we now set completely aside those skeptics who still deny back radiation (phew!), there is obviously still a long way to go to resolve the real issue between ‘slayers’ and ‘luke warmists’ of whom there are many distinguished examples (Lindzen, Spencer, Singer).

The prize for you and your supporters is to win the luke warmists over. I believe that will then naturally result in the general public also being won over in increasing numbers. Otherwise you will have an ever steeper uphill task.

All that of course assumes that you are right! As I said before, as a hard line skeptic I very much hope you are. For example, it would cut out all the tedious ‘climate sensitivity’ discussions about how much warming results from a doubling of atmospheric CO2.

The Diagram
I think there is still a long road ahead to convince enough people you are right…or wrong. As a very modest next step forward, can I suggest that we at least agree a simple diagram that represents the Slayer case. In your responses you have not explicitly rejected my previous diagram but you did say “…what you could draw is 235W/m2 emission from the inside of the shell, but then cross it out or something because that emission, when facing incoming energy of the same magnitude and frequencies, can’t actually induce any work or heat flow to occur.” That was essentially what I intended the u-turn arrow in my diagram to imply but I wonder whether the picture should look a bit more symmetrical, such as this one?

Let me know what you think.

The Physics
You say: The physical reason at the quantum level is due to the frequency components analysis I’ve discussed previously. This is how and why radiation obeys the 2nd Law of thermo.. Please direct me to this. I will study it (yet again!) and report back.

A C Osborn says:
2013/03/21 at 7:58 AM
Joseph E Postma says:2013/03/20 at 10:48 PM Following up briefly on Me at 2013/03/20 at 6:14 PM:
and
Greg House says: 2013/03/21 at 6:28 AM

Joe, what you have written does not appear to be quite true. Isn’t some Work done, the atmosphere around you would be warmed up by the excess heat that your body would generate expending all that energy.

Alan Siddons says:
2013/03/21 at 8:16 AM
“…increased mass in the form of increased density, while the surface areas stay the same, would just SLOW DOWN the speed at which thermal equilibrium is acheived. If the surface area increases, then indeed the flux and temperature must decrease.”

Increased density delays thermal equilibrium? To the contrary, a dense material often HASTENS heat transfer because vibration (heat) spreads across its closely-packed atoms so fast — much like kinetic energy passing through a Newton’s Cradle:

No material is absolutely conductive, however, so any material is somewhat insulative too. Surround a conductive metal rod with insulation and touch one end to the flame of a Bunsen burner. Heat will travel to the other end but it won’t get as hot as the flame. With a longer rod the other end will be cooler still, even though the surface area (the rod’s cross section) hasn’t changed. A point to stress is that the far end of a long rod won’t EVENTUALLY get as hot as a short rod. No, it will NEVER get as hot. Otherwise, insulation (i.e., something less than 100% conductive) would do no earthly good, for the same amount of heat would spill out of a thick insulator as out of thin. This kind of temperature reduction is more a matter of the distance heat has to cover rather than of surface area.

Just as density can assist transfer, though, sparseness is able to thwart it. Indeed, the most dramatically “interrupted” gradient occurs within a material that’s nearly as light as air: Aerogel. You can comfortably press your finger on top of this tenuous stuff while on the bottom, a mere fraction of an inch away, a propane torch is blazing right at it.
http://inhabitat.com/exciting-advances-in-insulation-with-aerogel/aerogel-insulation-2/

Again, the surface area difference with or without aerogel is negligible. It’s simply that this uniquely disorderly material won’t let heat cross the distance, so there’s an enormous temperature difference between one side and the other — the same as you’d see with a more conductive substance over a greater distance.

As I say, then, there’s the temperature-reducing impact of simply adding mass. A candle can bring a teaspoon full of water to a boil but never a whole pot. And there’s the diminishing return that conductive limits bring about. Neither of these instances involves a mere delay. It’s not a matter of taking more time for a previous condition to obtain but of resetting the temperature where equilibrium occurs.

No big deal, though. I’m just trying to emphasize that these imaginary full-transfer scenarios are inherently unrealistic. But let me add that heating-by-reduced-emission scenarios are also in error. Even NASA admits that the Earth is constantly radiating away all of the radiant energy it receives from the sun. There’s no sign of atmospheric insulation. I hope Tallbloke is listening.

squid2112 says:
2013/03/21 at 1:40 PM
Joe, you are making things so incredibly perfectly clear to me it is amazing. This is awesome!

However, I do have just one question for you. I am imagining presenting to my father your backradiation/work scenario, which I totally get, especially within the framework that the backradiation could not make a warmer object warmer still. However, when I envision presenting this argument, I anticipate the rebuttal to be “but it can make cooler objects warmer”, which would imply that there is a GHE. Not all things on the ground are necessarily going to be warmer than the air above all the time. Wouldn’t down welling long wave radiation make something cooler than itself, warmer? And if so, how does this fit within the GHE argument?

Max™ says:
2013/03/21 at 6:40 PM
Similarly, if back radiation was going to do anything, it would be reducing the work available to be done by radiation leaving the warmer surface.

If you push on a wall while someone else pushes back the other way, does that somehow make you push the wall harder?

ThePhysicsGuy says:
2013/03/22 at 3:13 AM
Joseph,

I participated in the WUWT discussion and had several exchanges with tallbloke. One of tb’s claims was, and I quote:
Radiative energy doesn’t care where it is coming from or going go, or what the temperature is on either end. If I light a candle on the earth during the day, the sun ends up warmer than it would be if I didn’t light the candle. Of course the reverse is true as well, the candle ends up warmer than if there were no sun. Since NET heat flow is from the sun to the candle, no thermodynamic laws are broken … but that doesn’t mean that the light from the candle is not absorbed by the sun. It is definitely absorbed, and the sun ends up warmer because of that radiation. Physics. Don’t leave home without it.

I am a professional civil engineer by trade, so my physics expertise is more towards Newtonian physics, but our core physics requirements at university included a semester of thermodynamics. What tb and lgl are violating is the basic Clausian statement of the Second Law of Thermodynamics…pure and simple. If a cooler body can warm up a warmer body, then that “warmed up” warmer body will in turn warm up the cooler body (the cycle continuing without end) violating both the First and Second laws since one would be creating energy out of nothing. And if a cooler body could indeed warm up a warmer body, then by the same logic, bodies of equal temperature would warm each other up because they are after all, receiving energy from each other. If this were indeed true, it would seem there would have been precise lab experiments performed by physicists demonstrating this amazing phenomenon.

Joseph, could you comment on a post by Dr. Roger Pielke Sr. regarding the Greenhouse effect and the 2nd Law? His short explanation on why there is no violation is as follows:

With regards to the violation of the second law, what actually happens when absorbing gases are added to the atmosphere is that the cooling is slowed down. Equilibrium with the incoming absorbed sunlight is maintained by the emission of infrared radiation to space. When absorbing gases are added to the atmosphere, more of emitted radiation from the ground is absorbed by the atmosphere. This results in increased downward radiation toward the surface, so that the rate of escape of IR radiation to space is decreased, i.e., the rate of infrared cooling is decreased. This results in warming of the lower atmosphere and thus the second law is not violated. Thus, the warming is a result of decreased cooling rates…….The bottom line here is that when you add IR absorbing gases to the atmosphere, you slow down the loss of energy from the ground and the ground must warm up.

I don’t quite know what to make of it since atmospheric physics are beyond my expertise.

Joe. I understand your frustration and passion for this subject, but I suggest taking the high road. Just address the errors in your methodical rational way and leave it at that……for what its worth.

John in France says:
2013/03/22 at 3:20 AM
If you push against a wall, energy is spent, but no work, another quantification of energy, is performed.
“I like this explanation very much. It is clear and understandable for everyone.”

I agree with Greg House (2013/03/21 at 6:28 AM) on this and as one with little science background of any account can confirm that it is a precise summing up of the situation such as this that we all need.
So why the difficulty in getting the message through to people who are evidently not the morons you are calling them? Don’t forget we are speaking not only of people like Anthony and Willis, but Jo Nova, David Evans, Richard Lindzen, Chris Monckton, to name but very few. These people have been active in debunking the global warming / anti CO2 brigade for how many years now? Like it or not, we’re on the same side.
The nagging question is why do Watts and his ilk so aggressively defend such an untenable theory, displaying the same attitudes and indulging in the same exclusion tactics as the pseudo scientists they decry?
I suspect that the key is to be found in Monckton’s approach. Most of his arguments are based on the figures actually given by the IPCC and other alarmist groups; he then proceeds to take them apart and show that even by these low standards they do not in any way stack up and therefore CO2 induced global warming is no cause for concern.
I think that adherence to the GHE at least gives them some common ground to engage with alarmists (and to show outsiders they are not cranks) and our rejection of that common ground in their eyes takes it away from under their feet.

ThePhysicsGuy says:
2013/03/22 at 3:41 AM
Joseph,

Oops! The first quote I mentioned above regarding the candle was a Willis Eschenbach quote, (not tallbloke), made on February 6, 2013 at 2:33 pm, on the WUWT, “The R.W.Wood Experiment” post. Maybe you can make an edit for me?

Kristian says:
2013/03/22 at 6:37 AM
It’s sad seeing self-appointed ‘skeptics’ over at Tallbloke’s succumb to the exact same tactics as the alarmists employ when their dogma is challenged. They remain completely indifferent to countering arguments, just restating their axioms as truth over and over again, not addressing what the opposition is saying at all, only building strawman arguments to tear down, willfully ‘misunderstanding’ time and time again the points made by people who disagree with them. It’s frustrating. But I guess that’s their aim.

I totally sympathise with how you feel, Joe.

Max™ says:
2013/03/22 at 9:29 PM
Crosspost, as I think this is a rather insurmountable disproof of the whole “the inner surface will emit 470 W/m^2 after adding a shell” nonsense.

“Isn’t that a bit like saying you can add half a cup of water at 5 deg C to another half a cup of water at slightly less than 5 deg C and getting a full cup of 10 deg C?” ~Arfur

Indeed.

It is also like saying if you take a sphere with the previously mentioned 1 million square meter surface area (8920~ m radius) and it has an internal power supply which raises the temperature until the surface emits 235 W/m^2 (total 235 megawatts) then you could do the following:

1. Add a shell 1 meter above the surface (so the radius of the shell is 8921~ m) and the shell will then emit 235 W/m^2 (total more than 235 megawatts) and the original surface will warm until it emits 470 W/m^2 (total 470 megawatts) with no other changes made to the system.

2. Fill the gap between the shell and the surface with the same material as the shell and surface are made of, maintain the 235 W/m^2 emission from the shell surface and the original surface–now at a depth of 1 meter–will be at or near 302 K.

_______________

If this is the case, it seems to me that if we start with the original sphere+shell, assuming the sphere surface is emitting 470 W/m^2, we should be able to remove 1 meter of material inside the sphere and ask what the temperature would be afterwards.

If it is true that adding a 1 meter gap + shell above the surface raised the surface temperature until it doubled the emissions, then adding another gap within, such that the surface is a new shell, would also double emissions, right?

So the new surface would emit 470 and receive 470, totalling to 940 W/m^2.

Do that again, now with 1 “outer” and 2 “inner” gaps we have a total of 1880 W/m^2 leaving the inner surface, right?

But above I suggested that an internal power source, spherical, and 4000 m across could supply 1138 W/m^2 which would give 235 W/m^2 through a naive approximation of conduction following an inverse-square law.

Now we find that simply removing layers of material beneath the original surface+shell would produce more energy?

Why isn’t this being engineered and utilized?

Max™ says:
2013/03/22 at 9:34 PM
Note: that should be “235 gigawatts”, I mentally corrected it to a more sensible sounding value, but 8920~ m radius gives 1 billion m surface area.

Joseph E Postma says:
2013/03/23 at 6:31 AM
OK I’m crazy busy here in India testing spacecraft, writing code, banging my head against the laptop, etc etc. I’m just going to approve all pending comments. Either a comment will really somehow have to grab my attention for me to respond, or email me directly or keep trying to get my attention on anything you want my attention on Cheers.

Joseph E Postma says:
2013/03/23 at 6:36 AM
@Squid 2013/03/21 at 1:40 PM

There is nothing wrong with warmer things heating cooler things, but the atmosphere is generally cooler than the Earth and a warm atmosphere heating a cooler ground is not what the GHE is about in any case. Tell you Dad that q = k*(T_hot – T_cool), when T_hot = T_cool, means that q = 0, and never that T_hot increases itself in order to heat T_cool.

Simon Conway-Smith says:
2013/03/23 at 6:46 AM
TPG: Just ask someone; when they buy something for 3$ and hand over a $5 note, they get $2 in change. How much do they now have? By the cooler heating warmer ruse, they would now have $7!! Absurd isn’t it? When heat energy is emitted, it is not replicated, but LOST! Any back radiation, even if possible/real, is a fraction of that emitted. I offer the very simple maths calculation that back-radiationists get so very wrong, which is if 1 unit of heat energy is radiated from the surface and if 0.x is radiated back, then the net heat energy remaining at the surface is -1 + 0.x, i.e. cooler, and NOT +1 + 0.x (warmer). To clsaim +1 +0.x is to magically create energy out of nothing. You do wonder how so many ‘scientists’ can make such a fundamental maths error.

Simon Conway-Smith says:
2013/03/23 at 6:51 AM
JiF: The problem with the Moncktons of this world, however much we admire their resolve and ability to take apart and demolish the alarmists case (and long may they do so, and this is no way a criticism of their ability to do so), is that by accepting that there is some level of back-radiation/GHE, it provides a legitimacy to the alarmists case. It’s this false legitimacy, or pseudo-science, that needs to be demolished, as with that done, there is no case to be made – at all.

Joseph E Postma says:
2013/03/23 at 6:58 AM
@ThePhysicsGuy 2013/03/22 at 3:13 AM

The quote:

“Radiative energy doesn’t care where it is coming from or going go, or what the temperature is on either end. If I light a candle on the earth during the day, the sun ends up warmer than it would be if I didn’t light the candle. Of course the reverse is true as well, the candle ends up warmer than if there were no sun. Since NET heat flow is from the sun to the candle, no thermodynamic laws are broken … but that doesn’t mean that the light from the candle is not absorbed by the sun. It is definitely absorbed, and the sun ends up warmer because of that radiation. Physics. Don’t leave home without it.”

assumes the intended result to be true in the first place, and so is logically, mathematically, and physically meaningless. One can not just violate the laws of thermo and hide it, and then just forget you did the deed…the deed was done! lol A torch shone another torch does not make either or both torches shine brighter! A candle does not burn brighter in the day time vs. the night time (due to solar energy “adding” to it).

The first sentence is correct. Everything else is wrong. Look at this equation: a = 1/m * (F2-F1). F2 and F1 exist, and are real, and “press” in the direction they’re intended, but if they’re equal and opposite, they DO NOTHING. Isn’t it amazing? A real application of force, and a real expenditure of energy maintaining that force, can result in NOTHING happening.

Now lets look at a similar equation: q = k*(T_hot – T_cool). If T_hot = T_cool, NOTHING happens, i.e. q = 0. Is it not amazing? Real temperature, with the real energy it represents, causes NOTHING to happen if they contact each other. NOTHING can happen even though energy is present and is interacting with another entity.

In these equation, the only action that WILL and CAN occur, is the action from the stronger to the weaker, and, never does the stronger need to become stronger in order to affect the weaker – this idea is nowhere in physics…NOWHERE. The presence of a force differential does not mean that the stronger force will naturally become stronger as it affects the weaker force – it just affects the weaker force.

Now another one: q = A*s*(T_hot^4 – T_cool^4). This works the exact same way. T_Hot does not need to become hotter to warm the cooler thing. If the temperatures are equal, the radiation DOES NOTHING. It is just as simple/amazing as the other examples: the presence of energy, this time in the form of radiation, can cause NOTHING to happen because there is no force/temperature/energy differential. By ignoring this, you can create arguments to violate thermo with radiation even though the physical analogues are strictly never considered possible of doing so.

You said: “If a cooler body can warm up a warmer body, then that “warmed up” warmer body will in turn warm up the cooler body (the cycle continuing without end) violating both the First and Second laws since one would be creating energy out of nothing. And if a cooler body could indeed warm up a warmer body, then by the same logic, bodies of equal temperature would warm each other up because they are after all, receiving energy from each other. If this were indeed true, it would seem there would have been precise lab experiments performed by physicists demonstrating this amazing phenomenon.”

Thank you. Exactly. Thank you for being rational.

The quote from Pielke is identical to the “shells game” which has been discussed at length here. It is just as wrong and for the exact same reasons. He also simply does not comprehend the difference between low-power for 24hrs vs. high power as a cosine function into a non-linear system etc etc.

Joseph E Postma says:
2013/03/23 at 7:00 AM
@Max 2013/03/22 at 9:29 PM

“Why isn’t this being engineered and utilized?”

Indeed. Make it work people…demonstrate it!

Kristian says:
2013/03/23 at 7:30 AM
John in France says, 2013/03/22 at 3:20 AM:

“So why the difficulty in getting the message through to people who are evidently not the morons [Postma is] calling them? Don’t forget we are speaking not only of people like Anthony and Willis, but Jo Nova, David Evans, Richard Lindzen, Chris Monckton, to name but very few. These people have been active in debunking the global warming / anti CO2 brigade for how many years now? Like it or not, we’re on the same side.

The nagging question is why do Watts and his ilk so aggressively defend such an untenable theory, displaying the same attitudes and indulging in the same exclusion tactics as the pseudo scientists they decry?

I suspect that the key is to be found in Monckton’s approach. Most of his arguments are based on the figures actually given by the IPCC and other alarmist groups; he then proceeds to take them apart and show that even by these low standards they do not in any way stack up and therefore CO2 induced global warming is no cause for concern.

I think that adherence to the GHE at least gives them some common ground to engage with alarmists (and to show outsiders they are not cranks) and our rejection of that common ground in their eyes takes it away from under their feet.”

Very well put! I think you’ve hit the nail on its head here. Tallbloke is NOT a stupid man. I am quite certain he knows his ‘official’ position on this particular subject is totally and utterly mistaken. The logical/physical flaws in his argument are simply too ridiculously elementary. But, simple and harsh truth of the matter, he feels he has to ‘adhere’ to at least the theory behind the radiative GHE not to get marginalised, losing all his ‘credibility’ and ‘impact’ as a profiled skeptic blogger.

Alan Siddons says:
2013/03/23 at 7:38 AM
Arfur’s cup of water example is apt and also hints at the arithmetic of radiant energy transfer. If 200 W/m² is radiating onto a blackbody that’s independently radiating 100, that body will proceed to radiate 200 W/m², of course, not 300. Transfer is accomplished by DIFFERENCE, not by addition. Similarly, then, if 100 W/m² radiates onto a blackbody that’s independently radiating 200, nothing happens. Otherwise you’d have an example of lesser energizing greater and a magical mutual heating cycle would commence: energy ex nihilo.

Greg House says:
2013/03/23 at 8:15 AM
Kristian says, (2013/03/23 at 7:30 AM): “Tallbloke is NOT a stupid man. I am quite certain he knows his ‘official’ position on this particular subject is totally and utterly mistaken. The logical/physical flaws in his argument are simply too ridiculously elementary. But, simple and harsh truth of the matter, he feels he has to ‘adhere’ to at least the theory behind the radiative GHE not to get marginalised, losing all his ‘credibility’ and ‘impact’ as a profiled skeptic blogger.”
===============================================================

I do not know, what Tallbloke knows, but generally, if someone knows that “greenhouse effect” is BS and nevertheless promotes this BS, why not assume that the image of a “skeptic” is just designed to fool people more effectively?

Their message is like “look, we are skeptics, and if we skeptics say that the GHE is real and CO2 does produce warming, then you can be sure that it is correct”.

sunsettommy says:
2013/03/23 at 8:34 AM
Settled science: New paper questions usefulness of core IPCC radiative forcing concept
http://hockeyschtick.blogspot.com/2013/01/settled-science-new-paper-questions.html

sunsettommy says:
2013/03/23 at 9:21 AM
Man-made global warming theory is falsified by satellite observations
http://hockeyschtick.blogspot.com/2013/03/man-made-global-warming-theory-is.html

Joseph E Postma says:
2013/03/23 at 9:22 AM
Nice…yes that one has been discussed here too I think.

A C Osborn says:
2013/03/23 at 1:19 PM
Greg House says:
2013/03/23 at 8:15 AM Do you think Tallbloke could be playing devil’s advocate?
As at times he seems to jump the fence and say something against the Back Radiation and then jump back again and says something supporting the current argument.
Have you seen the latest TFP experiment, to my thinking it disproves back radiation and to their minds it proves it?

Truthseeker says:
2013/03/23 at 5:25 PM
Joe, you say …

“OK I’m crazy busy here in India testing spacecraft …”

So you ARE a rocket scientist!

Joseph E Postma says:
2013/03/23 at 8:35 PM
Hi Truthseeker,

I don’t actually do rocket propulsion design. Rocket propulsion is an old problem and has largely been figured out…it’s not really that complex. I work with the things that go on top of the rocket

hehe

Simon Conway-Smith says:
2013/03/24 at 3:51 AM
What, like how the aerodynamics are affected when Michael Mann is tied to the nose-cone, or the heat transfer equations when he falls off back to earth from the troposphere (now *that’s* back-radiation)?

Greg House says:
2013/03/24 at 8:19 AM
A C Osborn says, 2013/03/23 at 1:19 PM: “Do you think Tallbloke could be playing devil’s advocate?”
============================================================

Well, I suggest you read my previous comment here (2013/03/19 at 7:13 AM) and the screenshots I posted there, in case you missed it.

Generally, I would divide warmists (meaning this term technically) into victims and perpetrators. Even intelligent people sometimes lack critical thinking ability and fall into demagogic traps, particularly when appeals to authority are involved in the argumentation, so, it does not seem right to me to classify them automatically as liars or morons. Often I am not sure who I am talking to, but Tallbloke does not seem to be a victim to me.

Kristian says:
2013/03/24 at 12:22 PM
Greg,

I’ve now read your original post on Tallbloke’s, the one you linked to, and it’s very clear you made a simple mathematical mistake in you step 3). Tallbloke is actually quite right in pointing it out to you. His stepwise back-and-forth sequence of energy/heat exchange between planet and shell until the former has asymptoted to a steady-state temperature of ~302K and the latter to a temperature of ~254K is in fact internally consistent. Its only problem is that it doesn’t describe a real, physical process. It is merely a case of mathematical trickery.

The main absurdity in his approach lies in the fact that he simply postulates that the shell, upon reception and absorption of the original 235 W/m^2 from the planet, does NOT acquire the corresponding emission temperature of 254K, but rather one of 213K. Why? Who knows why? But one explanation would be that he thinks that half the received energy (235 Joules per second per square metre) is somehow absorbed at the inner surface and the other half at the outer surface of the shell (probably some unspecified conductive property (whatever conductive thermal resistance might be doing in a model trying to emulate ‘the radiative GHE’ in the first place) lets through half and only half the amount from one side to the other). Either way, effectively what he does is splitting the received and absorbed power density flux (235 W/m^2) in two. 235 W/m^2 comes in. 117.5 W/m^2 goes out. His reasoning? The shell has two surfaces. Even though one is not facing a cold reservoir at all, only a hot reservoir. Can you believe that?! That’s the entire argument. That’s what we’re up against.

To be clear: Only reflection serves as radiative insulation. Conductive insulation serves as … conductive insulation. Conduction holds no part in ‘the radiative GHE’. It takes part in the conductive/convective ‘GHE’, or rather ‘the real atmospheric effect’, where the atmospheric mass weighs down upon the surface, working to suppress convective cooling fuelled by the incoming radiation from the Sun. Did Willis Eschenbach ever state that the shell to some extent reflected incoming radiation, or that the shell exercised some sort of conductive resistance of any significance to the heat flow through the shell to space? If he did, he would not have modelled ‘the radiative GHE’ as is. If he did NOT say that, but rather the opposite, then he WOULD have modelled the concept behind ‘the radiative GHE’. But then there would be no restriction to the energy flow from core planet to space. And hence, his model would’ve proved that the radiative GHE does not work. It is zero.

Greg House says:
2013/03/24 at 8:12 PM
Kristian says, (2013/03/24 at 12:22 PM): “Greg,
I’ve now read your original post on Tallbloke’s, the one you linked to, and it’s very clear you made a simple mathematical mistake in you step 3).”
=============================================================

The math in my step 3 was 600+1200=1800, no mistake.

If you mean physics, however, then you should have read my next comment where I addressed his point: http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46135. You were active on that thread, I wonder how you could have missed it.

I also suggest you start thinking about why he allows you to disagree with him, but blocks my comments. Critical thinking on, please. And about why he lied straight to my face saying that I admitted being wrong.

OK, I’d better tell you just now why your comments are welcome there, since I do not really expect you to get it: because your comments are… you know… they will do no harm to warmists.

Joseph E Postma says:
2013/03/24 at 8:47 PM
I see both Greg and Kristian making valid points.

The cycling of energy back and forth to asymptote at the desired value is baseless and senseless. Early on here we discussed the math of that sequence and proved it to be physically impossible and logically meaningless. Of course this doesn’t stop anyone from believing in it because the goal of pretending a GHE is the intention – not actually doing a critical analysis of anything.

Kristian says:
2013/03/24 at 11:30 PM
Greg House says, 2013/03/24 at 8:12 PM:

“The math in my step 3 was 600+1200=1800, no mistake.

If you mean physics, however, then you should have read my next comment where I addressed his point: http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46135. You were active on that thread, I wonder how you could have missed it.”

Greg, I’m not going to get into an argument about this one with you. We both agree his exercise is absurd. Your 600+1200=1800 should simply be 600+800=1400. That’s it. You can’t add 600 to the 400 you added in the first round. The 400 has now become 600. You have to remove the 400 and replace it with the 600. Then you get 600+800=1400 in the next step. Silly, I know, but mathematically (or like you say, ‘physically’, correct).

If you look at it this way: The planet initially sends out 800 W/m^2. This represents radiative cooling. It is replaced by another 800 W/m^2 from the nucleus. Then it receives 400 back from the shell, absorbs it and reemits the 800 PLUS the 400. Equals 1200. Now the 1200 W/m^2 is the radiative cooling flux, but upon leaving the surface it is only replaced by the next ‘pulse’ of 800 W/m^2 from the nucleus. You might say the 400 coming in is still there. But in that case, there will only be an increase in the incoming flux from the second to the third step of 600-400=200 W/m^2. Again, 200+1200=1400. 1400 is the number. Not 1800.

Greg House says:
2013/03/25 at 5:41 AM
Kristian says (2013/03/24 at 11:30 PM): “Greg, I’m not going to get into an argument about this one with you. […] Again, 200+1200=1400. 1400 is the number. Not 1800.”
=================================================================

Well, as I already told you, I answered that (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46135), but I do not see any sign that you read it. Your radiation arithmetic ignores physics. According to physics, bodies radiate according to their temperature. To justify your and Tallbloke’s approach, it would require a constant increase in temperature of the planet, then decrease, then increase again, then decrease again and so on, always a drop to the initial temperature. To radiate every time (the initial 800) + (an amount equal to back radiation from the shell) requires the temperature to drop to the initial level every time, there is no way around it. I would say, if it works this way, it is certainly a new word in physics.

And look at that, Tallbloke did indeed say this “new word”! Here: “yes, the temperature instantly drops when emission occurs from a blackbody with no heat capacity.”! (http://tallbloke.wordpress.com/2013/03/10/entering-the-skydragons-lair/comment-page-1/#comment-46160)

So, as you can see, your hypothetical cyclical process of warming by back radiation inevitably leads to an absurdity, one way or another. This proves that your hypothesis is wrong.

P.S. In case you do not understand why a hypothesis is wrong if it is proven to lead to an absurdity, you might need to read about “reductio ad absurdum”.

Joseph E Postma says:
2013/03/25 at 6:12 AM
Greg I think that Kristian was simply explaining the math of their process, while maintaining in the end that it was baseless – not agreeing with it.

Greg House says:
2013/03/25 at 6:36 AM
Joseph E Postma says (2013/03/25 at 6:12 AM): “Greg I think that Kristian was simply explaining the math of their process, while maintaining in the end that it was baseless – not agreeing with it.”
==========================================================

I have a quite different impression. Maybe he does not agree with the notion of warming by back radiation for some other reasons, I do not know, but here he clearly said, as Tallbloke did as well, that you should always add a new portion of back radiation to the initial 800, thus ignoring the rise in temperature of the planet (ironically, the rise in temperature is exactly what they are trying to “prove” with that planet/shell thing, LOL). So, I countered with “temperature jumping back and forth” argument, referring to my previous comment. He ignored it, no reaction. Now I have repeated this argument on this thread. Let us see, what he has to say about it. My guess is, he would ignore it and just stick to his arithmetic or maybe start counting protons, like apples and oranges.

Joseph E Postma says:
2013/03/25 at 6:55 AM
I think he was just trying to help demonstrate their math.
He said: “Its only problem is that it doesn’t describe a real, physical process. It is merely a case of mathematical trickery. … Can you believe that?! That’s the entire argument. That’s what we’re up against. … Did Willis Eschenbach ever state that the shell to some extent reflected incoming radiation, or that the shell exercised some sort of conductive resistance of any significance to the heat flow through the shell to space? If he did, he would not have modelled ‘the radiative GHE’ as is. If he did NOT say that, but rather the opposite, then he WOULD have modelled the concept behind ‘the radiative GHE’. But then there would be no restriction to the energy flow from core planet to space. And hence, his model would’ve proved that the radiative GHE does not work. It is zero. …”

Etc.

Joseph E Postma says:
2013/03/25 at 7:18 AM
So, the question of “what is the temperature of an infinite amount of heat”, is of course a trick question. The temperature of the heat is whatever you would like to define it as, and the concept of an infinite heat source/sink is a very common and useful one in thermodynamics. The point is that in thermodynamics, when modelling systems with infinite amounts of available heat energy, not even this much energy can be used to heat itself up to higher temperature! Is that not an amazing restriction of what energy and temperature can do? The GHE and the shells game throws these restrictions of thermodynamics out the window, and openly, proudly so…without ever actually demonstrating it to be true. Adherents just say things…and then assume it to be true.

What do two objects in contact with each other do? Let’s use the shells game. The outer shell is actually in contact with the inner sphere. At first the inner sphere was “free”, then we came by with this cold shell and stuck it over the sphere. What happens to the sphere? What happens to the shell? Does the shell cause the sphere to heat up because it “blocks cooling”? No, two objects in contact, one hotter and one cooler, simply causes the cooler object to heat up. That’s all that would happen to the shell – it would heat up to the temperature of the surface of the sphere (assuming the thickness of the shell was negligible). Back-conduction occurs entirely analogously and truthfully as back-radiation, but, back-conduction doesn’t cause cold to heat hot! You can’t just say that back-radiation exists but not back-conduction…although they will. You could recreate the entire GHE argument with the shells game using back-conduction. But we all know intuitively that nature does not behave this way. We also know that two torches shone on each other don’t make each other shine brighter – a fundamental premise of the GHE and the shells game!! – but, in this case rational and every-day knowledge is discarded for the religious precepts that the GHE and CO2 vilification offers instead.

It doesn’t need to be repeated – the shells game GHE is absurd, mentally retarded, grotesquely sophistical, and rationally intellectually offensive. It is amazing and disgusting that humans can be this stupid. I am always reminded of Hanlon: “Never attribute to malice that which can be adequately explained by stupidity, but don’t rule out malice.” But hey…it’s not like we don’t have thousands of years of historical precedents of the stupidity of man. Just look at religion… It’s amazing we survive at all…really, don’t you think?

Greg House says:
2013/03/25 at 7:29 AM
Joe, the whole model drops dead immediately if you ask, why they assume that the back radiation is absorbed and re-radiated and then absorbed on the other side and so on changing temperature. There is no physical proof behind this assumption. You can as well assume any absurd thing and draw a “model” on paper. You can assume 2×2=5 and convey this message all your life long. People still apply for a patent for perpetuum mobile.

As Willis came up with that thing on WUWT, I restricted my comment to pointing out that it is nothing else than a fictional story not worth commenting on for lack of physical basis. But, since some people stick to it, I thought “why not”. Maybe reductio ad absurdum could help. So, here we are. Of course, they do not like it and will refer to other absurdities to support what is falling apart, like Tallbloke did.

Joseph E Postma says:
2013/03/25 at 7:39 AM
SHINE A GOD-DAMNED TORCH AT A MIRROR, RIGHT UP NEXT TO IT, AND THE TORCH DOES NOT SHINE BRIGHTER!

Joseph E Postma says:
2013/03/25 at 7:40 AM
You can’t even detect a fraction of a degree change in skin temperature when standing in front of a mirror! Let alone a doubling of temperature…!

Greg House says:
2013/03/25 at 7:46 AM
Joseph E Postma says (2013/03/25 at 7:39 AM): “SHINE A GOD-DAMNED TORCH AT A MIRROR, RIGHT UP NEXT TO IT, AND THE TORCH DOES NOT SHINE BRIGHTER!”
==========================================================

What about 2 mirrors facing each other and a torch in the middle? (lol)

David Socrates says:
2013/03/25 at 7:51 AM
Hi Joe,

I am still waiting to hear whether you are happy that the improvement to my original diagram now adequately represents your position.

Cheers.

David

Max™ says:
2013/03/25 at 8:23 AM
Yeah, I was trying to bring the discussion to the back-conduction comparison and was told by tb to “stop throwing curve-balls three at a time”, and that I should stick to one point… I’m ok with that, not so much about letting someone else choose which point I get to argue though.

Been taking screenshots of my unedited posts when I get a chance, or resurrecting them with lazarus, but anyway, enough about that.

The shell can not cool through radiation emitted towards the planet, so the shell can not warm the planet up, simple enough, right?

Joseph E Postma says:
2013/03/25 at 9:08 AM
Check your email David…I can’t access that link from the server I’m going through (it blocks almost everything). Send me the image via response to email and I will post it and comment.

squid2112 says:
2013/03/25 at 10:10 AM
What about 2 mirrors facing each other and a torch in the middle? (lol)

Under the GHE scenario/principals, would this not create a runaway optical feedback? Please don’t try this at home or it could lead to a rift in the space-time continuum and destroy the universe!

Joseph E Postma says:
2013/03/25 at 10:43 AM
Oh (David) all I needed to do was use your link and insert it as an image link…will repeat here:

So yes that is probably all that needs to be shown. Just think for example if the shell and core were touching – you would still have exactly what you’ve shown on the RHS. Separating the shell and core with a gap doesn’t suddenly totally transform thermal physics and thermodynamic behavior and laws. I mean could you imagine if, when two objects of the same temperature are touching, nothing happens, but then if you separate them by a small distance, one of the sides starts spontaneously heating itself? Wouldn’t that be great, a wonderful discovery, a simple physical fact all young boys (and girls) leaned on their own at some point fiddling with stuff? Radiation is simply conduction at a distance, mediated in the exact same manner – just with real photons as opposed to virtual photons.

If you could somehow incorporate the actual physically factual heat flow equation of q = A*s*(T_2^4 – T_1^4), so that q > 0 when T2 > T1, and q = 0 when T2 = T1, then this might help explain what the “loop” means. Because I can just imagine people getting upset over the loop, and so this is why it would be good to refer to actual physics and math.

Joseph E Postma says:
2013/03/25 at 10:48 AM
Because there is NO actual loop – there is in fact q = 0. If they were touching you wouldn’t actually need to show a loop, because q = 0 if they’re touching and the same temperature. Likewise the case for radiation: q = 0, and so there is no actual loop occurring. In the touching “conductive” case, yes energy is going every which way by contact, but there is no real “loop”, there is just q = 0.

So to repeat, the loop isn’t real. Or, it is as real as it would be if the shell and core were touching. So you see the potential problem there? People will look at the diagram and take it seriously as gospel truth and then freak out about it, when in fact the “loop” was never meant to indicate anything actually occurring at all, because q = 0.

Kristian says:
2013/03/25 at 11:35 AM
Joe,

I think the loop is not what people (the Escenbach followers) would primarily object to in the above diagram. I’m quite certain they would rather object to the red arrow beside the loop, going from the core to the shell. That one is in fact what this argument is all about. They claim that the red arrow would be zero (or not be there at all) in the no-temperature-gradient scenario. In their minds it could only be 235 W/m^2 if there were a temperature difference between the core and the shell of (in this particular case) 48K (302-254K). That is their central claim.

I’ve tried to explain it to them using a fairly simplistic approach. To no avail. They seem (quite) unwilling to embrace any such concept: Apparent heat transfer when q is supposed to be 0.

Could you perhaps make an attempt?

ThePhysicsGuy says:
2013/03/25 at 1:18 PM
Hi Joseph,

I’ve seen claims that the operation of a “microbolometer” (special thermal imaging camera) proves that cool objects can indeed warm up warmer objects. See article:

http://www.climateandstuff.blogspot.co.uk/2012/05/cool-body-can-transfer-measurable-heat.html

So from what I understand, the claim is that the microbolometer sensor plate is “warmed up” by objects as low as -40 degrees C (the lowest temperature detection limit of the device), and the sensor plate is at ambient temperature (i.e., it is un-cooled). I am completely skeptical, but don’t know enough about the device (or enough about thermodynamics) to see how the wool is being pulled over my eyes. Are you familiar with this claim and/or the device, and could you provide comment?

Thanks.

Max™ says:
2013/03/25 at 4:15 PM
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Max™ says:
2013/03/25 at 4:16 PM
“Vacuums are good insulators. That’s why Mr Thermos got a patent.” ~tb

Joseph E Postma says:
2013/03/25 at 7:15 PM
@Kristian @ 2013/03/25 at 11:35 AM

So, they reject the red-arrow because if the temperatures are equal then there is no heat transfer (which would be correct), but then they say that self-radiation or radiation from a cold source can heat up something warmer (which is mindbogglingly incorrect). There is no consistency anywhere in their thoughts. At one point they go nuts strictly following a definition of heat transfer, and in the next they completely reverse the directionality of heat flow. If they obey q = 0 when there is no gradient, why do they disobey the directionality of q when there is a gradient, such that the hotter side of the gradient heats up!!!?

Look at the top of the shell – it is emitting 235. Therefore, it can not be strictly in thermal equilibrium with the core. QED. It is as simple as that. Their starting definition by which they can ignore the red 235 is incorrect, and obviously so, but they will want to hold to that strict definition so that they can create an alternative instead. As Max has been pointing out, the shell has a larger area than the core – even if it is quite thin. No matter what, the shell has larger area than the core, hence, there HAS to be a small (or large) difference in temperature between them, a small gradient. And even despite that, if the shell were a wall and the core a wall, the shell IS losing 235 on the outside, and hence this will create a differential between the core and shell, even if surface area wasn’t a factor. And this all happens continuously of course – exactly as 235 is being emitted from the shell, 235 is being filled back in to the shell. What if the shell and core (either as spherical or plane walls) were touching? It would still apply exactly as described here.

Joseph E Postma says:
2013/03/25 at 7:21 PM
@ThePhysicsGuy @2013/03/25 at 1:18 PM

When these devices detect “cool”, it is a LACK of signal that they detect, a “negative voltage”. The devices are calibrated in this way. The cooler the object, the more negative the “q” is between the camera and source. The hotter the object, the more positive “q” is between the source and camera (positive q when the source is hotter than the camera, and vice-versa). Their interpretation is a lie.

Does standing next to a wall of ice heat you up? Does standing next to a mirror heat you up? Nope.

Kristian says:
2013/03/26 at 2:50 AM
Joe, you said: “(…) if the shell were a wall and the core a wall, the shell IS losing 235 on the outside, and hence this will create a differential between the core and shell, even if surface area wasn’t a factor. And this all happens continuously of course – exactly as 235 is being emitted from the shell, 235 is being filled back in to the shell. What if the shell and core (either as spherical or plane walls) were touching? It would still apply exactly as described here.”

You’re right of course. And this is what I have been trying to tell them. But they simply seem oblivious to the green arrow going out to space from the top plane (outer surface of the shell) in D. Socrates’ diagram. They seem to think that this should be zero in the event of planet and shell attaining equal temperature, because the apparent heat transfer between them at that point would have reached 0. And so the planet’s surface in their mind would no longer be able to shed the constantly provided heat input from the planet’s nucleus. In other words, they seem to think that all that matters is the rate of heat transfer between planet and shell, that the rising temperature of the shell does not matter at all, that as the heat transfer from planet to shell grows smaller, the temperature of the shell also grows higher, closer to that of the planet (that’s why the heat transfer is slowing down). And by doing so, they willfully (?) neglect the exceedingly basic fact that the shell having a temperature would have to radiate accordingly to space. And how would this constant cooling flux be replenished if not from the core planet?

thefordprefect says:
2013/03/26 at 6:53 AM
Joseph E Postma says: 2013/03/25 at 7:21 PM
When these devices detect “cool”, it is a LACK of signal that they detect, a “negative voltage”.
———–
Look at the drawing for a bolometer. There is an IR absober. It has IR only focussed on it (the camera lens is made from germanium – which is opaque to visible light – I have added the transmission property to the post indexed above). The IR absober temperature is measured and the this value is output to the video processor.
Can you please explain what you mean by “detect cool” I keep stressing there are no cool rays there is only Thermal radiation (between 2u and 15um) that can get through the lens. This thermal energy adds to the energy from the ambient conditions and changes the temperature of the IR absorber – a -273C temperature adds no energy, a -20C adds more and the absorber warms above the temperature that would have occurred at -273C. a source of 1500C (within the range of the cameras calibrated sensitivity) adds even more energy so the absorber is warmer still.
Whatever the temperature of the object that is focussed on the absorber (above -273C) energy is added to it and its temperature rises (it is not that -273C cools the object because that would require cooling rays to be focussed)

——-
Does standing next to a wall of ice heat you up?

No but it heats you up more than standing next to a wall of frozen oxygen,
———–
Does standing next to a mirror heat you up? Nope

You need to tell the manufactures of survival bags made of aluminized plastic, and the manufacturers of thermos flasks, this rather outstanding fact! IR normally lost to ambient will be reflected back and will reduce the loss of IR from your body

.

Joseph E Postma says:
2013/03/26 at 7:25 AM
Don’t be a complete sophist moron Prefect – back-radiation doesn’t heat you up and thermos flasks etc. don’t generate increasing temperatures. Yes, there is some process going on with IR detectors, just as there is with any kind of radiation detector. But just because you can convert radiation into a signal doesn’t mean it heats up hotter things! Your own radiation from the mirror doesn’t heat you up, for example. You can detect “cold” with a thermometer or even your hand – this does NOT mean that the cool thing is heating your warmer hand or warmer thermometer, lol.

What I should have said is that I don’t know specifically how IR detectors work, but what I do know is that they do NOT work by cold heating hot – hahaha what a stupid idea.

Why doesn’t standing NEXT to an aluminium blanket warm you up? Crickets. Why does such a blanket only warm you up if you’re wrapped in it and the breeze is sealed off from getting inside? Because it is just trapping warm air, like a normal botanist greenhouse, and this has nothing to do with the atmospheric GHE. The behaviour of these survival blankets is almost perfect proof that the atmospheric GHE doesn’t exist and that backradiation doesn’t heat itself up. Get this crap figured out before you start spewing such vomitous mental garbage around here. You really don’t know how a thermos works? It is because convective and conductive loss to the outside flask is reduced to a minimum, occurring only at the inner-cylinder contact-point with the outer cylinder at the lip, because between the inner flask and the outer there is a vacuum, which can’t conduct or convect heat away from the surface of the inner flask.

Never has a thermos heated itself up with its own contents so it makes ZERO sense why a thermos has EVER been presented as evidence supporting the GHE – a thermos has NEVER, ever, anywhere, at anytime, in the history of thermoses, from here to the moon, heated itself up. A thermos thus proves that the atmospheric GHE doesn’t exist because a thermos has NEVER EVER done what is claimed it is analogized to do in the atmosphere! But hey why should we expect GHE faithers to actually have a rational comparison of something in reality with which to actually compare the GHE? That is how stupid such people are – that they will use an example of something WHICH DOES NOT DO SELF-HEATING to then pretend that the non-existence of self-heating proves self-heating. Mentally freaking retarded…

Simon Conway-Smith says:
2013/03/26 at 7:52 AM
I saw ‘thefordprefect’s comment and immediately thought “oh no, he’s placed himself right into Joe’s 12-bore missile target area”

As explained to a professor friend of mine (who believes Hansen is an upright scientist, and Houghton is a saint), that the ‘blanket effect’ is a HUMAN understanding of “warming” and not the PHYSICS (i.e. scientific) one. The human understanding needs qualification and explanation – you do *feel* warmer, but that is RELATIVE to the temperature you would otherwise have been without it, as the blanket (thermos) reduces the RATE of heat loss by reducing conduction and convection.

It really is very simple!

Joseph E Postma says:
2013/03/26 at 7:58 AM
“right into Joe’s 12-bore missile target area” – LOL, nice, love it.

Greg House says:
2013/03/26 at 8:01 AM
thefordprefect says (2013/03/26 at 6:53 AM): “Does standing next to a wall of ice heat you up?No but it heats you up more than standing next to a wall of frozen oxygen,”
==========================================================

“A wall of ice dies not heat you up, but it heats you up more than…” is a contradiction.

Greg House says:
2013/03/26 at 8:13 AM
Joseph E Postma says (2013/03/26 at 7:25 AM): “Yes, there is some process going on with IR detectors, just as there is with any kind of radiation detector. But just because you can convert radiation into a signal doesn’t mean it heats up hotter things!
===============================================================

Joe, I think this IR-Thermometer example should be dealt with properly, because it might sound convincing to unprepared people, including politicians and journalists. Talking to this audience we can not really argue by quantum mechanics or even by a mirror example, it would be not enough, we need to address this issue directly.

Fred Kilger says:
2013/03/26 at 8:21 AM
To maintain the radiation of 235 W/m^2 to space, these models assume (without any reason) a constant radiation source (fed from an unlimited energy source) inside of the planet.
This constant flux source forces (in a stationary state ) constant radiation i_PH (Planet -> Shell) = i_HS (Shell -> Space), or (SB:) T_P^4 -T_H^4 = T_H^4 –T_S^4 -> T_P^4 = 2* T_H^4 –T_S^4.
The temperature of the planet is always greater than the temperature of the shell. Putting more shells around the planet increases his temperature to arbitrary levels.

Only the (unrealistically constructed) constant radiation source (inside of the planet) determines in these models the temperature of planet and shells.

A constant radiation source can be applied to suns but never to real planets.
A real planet can only radiate energy out of the energy storage of ground, sea and atmosphere. This system is never in a stationary state, all temperatures, radiation fluxes are functions of time. Radiation of energy out of a storage element always lowers the temperature of this element. Constant radiation sources do not exist.

squid2112 says:
2013/03/26 at 8:36 AM
@thefordprefect says:

You need to tell the manufactures of survival bags made of aluminized plastic, and the manufacturers of thermos flasks, this rather outstanding fact! IR normally lost to ambient will be reflected back and will reduce the loss of IR from your body

Ah, but it did not heat anything, as that would require additional energy to enter the system, and since you cannot simply “create” energy from nothing, the reflective properties of the aluminium plastic did NOT in fact heat anything!

squid2112 says:
2013/03/26 at 8:42 AM
Further on the aluminum plastic blanket. If what TheFordPrefect is saying were true, then I could simply wrap my house in aluminum plastic during the winter and turn my heater off, but being careful not to burn my house down because of the continuous heating that the blanket would provide. Seems like another rift in the space-time continuum coming…..

Joseph E Postma says:
2013/03/26 at 8:51 AM
…rift in the space-time continuum… lol love it.

thefordprefect says:
2013/03/26 at 12:43 PM
A thermos flask contents remains warm because of the vacuum stopping conduction, the flask stopping convection to ambient, and the reflective inner container reflecting the radiation from the glass (the reflective material is 1 glass wall away from the liquid, which will therefore be cooler than the liquid) back to the liquid.

Assuming the “space blanket” is not in contact with the skin it too reflects IR back to the body It is a surface coating so will be reflecting at the same temperature as the body. neither of these 2 examples are similar to the iron greenhouse!

The microbolometer is extremely simple in operation. It is warmed by ambient or a temperature controlled heater. It looses heat to ambient. It gains heat from whatever is focussed onto the IR receiver plate via the germanium lens+ power loss in the readout circuit+power from any heater+ heat from other sources in the camera etc.
It periodically calibrates itself against a standard plate at fixed temperature. Thus the internal sources of heat are known. The only variable is the IR (2um to 13um) coming through the lens

All this IR can do is to add to the energy hitting each cell, it cannot subtract. It may be coming from a 1500degC source or from a -200degC

Perhams section 2.3 of this doc will help?
http://download.ebooks6.com/Uncooled-Carbon-Microbolometer-Imager-download-w43166.pdf

thefordprefect says:
2013/03/26 at 12:47 PM
Apologies strike out
“from the glass (the reflective material is 1 glass wall away from the liquid, which will therefore be cooler than the liquid) back to the liquid. ”
The radiation reflected will be from the liquid and will be at the same “temperature”

Rosco says:
2013/03/26 at 6:11 PM
I just love the way people make all sorts of stupid claims about “trapping” radiation – just making stuff up and ignoring centuries of proven and demonstrable science for – well – stupidity.

In an atmosphere energy transfer by radiation is trivial compared to diffusion/conduction to the air and convection – surely the well documented rate of cooling shown for the Moon’s surfaces demonstrate this indisputable fact ?

These so called “radiation” blankets act primarily by preventing contact with a convecting atmospher – trapping already warmed air.

I have yet to see a rock or any other inaminate object heat itself up by trapping radiation – or even by excellent thermal insulation – without an energy source all that happens is that objects tend to reach an equilibrium temperature/energy level with the surrounding environment.

You could hardly claim any animal fur could possibly be considere any sort of radiation “trap” yet the same appears to be especially effective at reducing convection of the warmed air layer trapped within the fur.

Or, perhaps it is the animal’s lack of understanding about radiation which keeps them feeling warm when in fact they are radiating all their warmth away and dying right before our eyes ?

Rosco says:
2013/03/26 at 6:59 PM
Really – who cares about the infinite number of different thought bubbles that are being submitted as “proof” that something that is absurd is real.

Accept Willis arguments are real. Add another shell and do the math.

A “new” shell around his original is exactly what he prescribes – a shell capable of radiating 235 to space and 235 backradiation.

Do the arithmetic as he prescribes and then add more and more until you arrive at infinite temperature of the original planet.

This obviously is absurd so the original proposition is absurd.

QED.

Joseph E Postma says:
2013/03/26 at 8:16 PM
Prefect said: “neither of these 2 examples are similar to the iron greenhouse!”

No shit. Which is why they shouldn’t be used as an analogizing example of any GHE! But, they’ll continue to do so anyway. Just amazing.

An IR receiver does not function by cold heating hot. Interpreting it that way is pure sophist BS.

Joseph E Postma says:
2013/03/26 at 8:31 PM
That’s WHY these people and their ideas exist – because they’re so absurd! The more absurd they get, paradoxically the more people believe them, because people are so retarded. Hitler knew this all too well:

“Make the lie big, make it simple, keep saying it, and eventually they will believe it.”

“What good fortune for governments that the people do not think.”

“The receptivity of the masses is very limited, their intelligence is small, but their power of forgetting is enormous. In consequence of these facts, all effective propaganda must be limited to a very few points and must harp on these in slogans until the last member of the public understands what you want him to understand by your slogan.”

“If you tell a big enough lie and tell it frequently enough, it will be believed.”

“The great masses of the people will more easily fall victims to a big lie than to a small one.”

“Through clever and constant application of propaganda, people can be made to see paradise as hell, and also the other way round, to consider the most wretched sort of life as paradise.”

“But the most brilliant propagandist technique will yield no success unless one fundamental principle is borne in mind constantly and with unflagging attention. It must confine itself to a few points and repeat them over and over. Here, as so often in this world, persistence is the first and most important requirement for success.”

And one from Einstein:

“Two things are infinite: the universe and human stupidity; and I’m not sure about the universe.”

In other words, it is just as I have been saying. When confronted with the truth, GHE faithers just get even more stupid as a strategy to perpetuate the lie! Just keep on repeating it, and repeat the idea in stupider and stupider ways that make less and less sense. Such as a thermos being an example of the GHE when a thermos has NEVER warmed itself up! Do you see how stupid and retarded that is? To offer as proof something which does not even do what is claimed? I mean you got to give it to human stupidity at some level – it is a winning strategy. It’s what the controllers of society count on…and they’ve been quite successful. It also makes them pathetic of course, to find self-value and self-identification in controlling extremely stupid people. How sad and pathetic. Yah…a thermos is proof of the GHE – how RETARDED do you have to be!?

Max™ says:
2013/03/27 at 2:19 AM
^ ___________ ^ ___________ ^
1 ___________ 100 _______ 234.99
Shell ________ Shell _______ Shell
1 ___________ 100 _______ 234.99
v ____________ v __________ v
| ____________ | ___________|
^ ____________ ^ __________ ^
235 _________ 235 _______ 235
Core_a _____ Core_b ____Core_c

Core_a loses 234 to shell, 1 to space, total out = 235
Core_b loses 135 to shell, 100 to space, total out = 235
Core_c loses 0.01 to shell, 234.99 to space, total out = 235

In the last step, the shell is unable to cool through radiation emitted back towards the core, so it can only cool by emitting to space.

The core is unable to cool beyond the 0.01 difference due to the inverse-square law, so the temperatures will be very similar.

David Socrates says:
2013/03/27 at 2:19 AM
Joe,

Re. your comments of 2013/03/26 at 8:31 PM above, I think you are fighting a straw man.

As we witness everywhere, blogs are full of awkward people who seem to take a delight in talking rubbish. Fighting them is a waste of time and achieves nothing. The people that climate skeptics have to persuade are the uncommitted citizens who have been told by their governments that there is a very serious problem that needs addressing. That is why I am interested in finding simple ways to explain complex issues. I don’t subscribe to Hitler’s or Einstein’s view that the general public is stupid. Ignorant of science maybe. But not ignorant of everday life.The best way forward is by reasoned argument with reasonable people. Surely our job, here and elsewhere, is to debate the best way to explain to ‘the ordinary voter’ exactly how they have been conned over 3 decades by an unholy alliance of special interests.

Looking at the physics is essential, but it is only part of a much bigger project, is it not?

Joseph E Postma says:
2013/03/27 at 3:00 AM
Great points David. The problem is how to utilize the techniques of propaganda for spreading truth…because you can’t just tell the truth, you have to make it hip, trendy, appeal to emotion, etc. Make it NOT look like the rational truth in other words. Make it an idea that’s just fun to have, and that makes you feel like you’re with everyone else.

thefordprefect says:
2013/03/27 at 7:20 AM
Mr Joseph E Postma
If you would care to explain how a uncooled thermal imaging camera works when photographing cold objects. This would add to the understanding about how cold objects add/do not add to the energy of hot objects.

Can you please provide an explanation.

Here’s anothe bolometer explanation (note that this one is cooled to minimise NOISE)
search: Detection of Light Lecture 10 bolometers
or search this one:
R. Westervelt Imaging Infrared Detectors II
No where in documentation I have read does it mention that you subtract heat from the bolometer when the source is below the temperature of the bolometer – energy is always added to the bolometer.

from the second doc.
“Bolometers operate by sensing the temperature rise associated with the
absorption of radiation. Bolometers are one of the oldest types of radiation
detector, and they have many advantages. Bolometers respond to absorbed
energy, and can be made sensitive to a very wide range of wavelengths. Because
they sense heat rather than photocarriers, bolometers are insensitive
to the photo carrier population and dynamics.”

Alan Siddons says:
2013/03/27 at 7:32 AM
Re: the Thermos Argument. If one asserts that the GHE works by trapping heat, then one is directly implying that the Earth radiates less energy than it gains — much like a thermos, which “bottles up” the heat it’s given. But that’s the very point, for satellites observe that the Earth radiates to space all the energy it gets from the sun. 235 W/m² or so continuously go in, 235 W/m² or so continuously go out. Obviously this is not how a thermos performs. This doesn’t bother Believers, however. Although they acknowledge that an Earth with a greenhouse effect sheds the same amount of thermal energy as an Earth without one, they argue nevertheless that the Earth is trapping heat. The Greenhouse Effect is thus like the biblical Bush That Burns Yet Is Not Consumed. Sort of a miracle.

Joseph E Postma says:
2013/03/27 at 8:02 AM
Well the first thing for you to learn is that cold things don’t heat up hot things. Try it yourself. Go around your house, the town, the country side, other countries, etc., and try to find something cold that heats up a hotter thing. Please report back on your findings. Actual findings – not just interpretations or assurances from other sources.

Whatever the process is, you will always have the limitations of the 1st Law of Thermodynamics, and then the 2nd and 3rd etc. Cold will never heat up hot. The reading suggestions easily fall under the physically-real framework provided earlier: what is the direction of ‘q’? What you need to do is develop an explanation of the bolometer which does not make the statements that cold heats up hot or that it is a demonstration of the GHE or that it violates the laws of thermodynamics – meaning precisely that cold does not heat up hot. One thing will be clear – a bolometer is not an example of cold heating up hot, at best, it will be an example of measuring the directionality of ‘q’. Does a bolometer violate the laws of thermodynamics? No of course not, nothing does. Therefore, precisely, they are not an example of cold heating up hot. And again, what the heck does a boloemter have to do with something HEATING ITSELF UP WITH IT OWN ENERGY!? Again, this is just another retarded analogy to something which doesn’t even do what the GHE is said to do in the first place(!) – a bolometer doesn’t heat itself up with its own reflected energy, so therefore this has nothing to do with the GHE and is in fact another disproof of the GHE…

You know one time I was told that the “bubbles from cavitation of a propeller on a nuclear submarine” were evidence of the GHE. You just have to say something retarded and then you have evidence for the GHE.

Bryan says:
2013/03/27 at 9:32 AM
thefordprefect

Dont get mixed up between a thermopile and a bolometer.
They work on completely different physics principles.

A thermopile has a voltage induced if the two dissimilar materials are at different temperatures.

A bolometer senses the change in resistance of the sensor due to temperature change.
The sensor forms one part of an initially balanced Wheatstone bridge
No current will flow if the external object is at the same temperature as the sensor – the null point.
If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the Wheatstone bridge is out of balance and a current flows through the sensor.
If the object is at a higher temperature; the resistance increases as the temperature rises the Wheatstone bridge is again out of balance and current flows but in the opposite direction.
If calibrated against a known temperature the current can then represent temperature.
This change in resistance is linear near the null point but if moved too far from the null point the change in resistance is non linear and will give rise to errors.

Greg House says:
2013/03/27 at 9:33 AM
@thefordprefect

I find the points about IR-Thermometers interesting, but at the same time we can take it closer to the GHE question. What I would like to know, is this: how much warmer gets the sensor when the IR-Thermometer is pointed to the sky, let’s say, at night? Because, you know, I would disregard the whole thermodynamics immediately, if the GHE were alleged to be like 0.000001C. Let us just clarify that before we get deeper into the issue about how it is possible that colder bodies do not warm warmer bodies but IR-sensor still detects IR from colder bodies. Let us address the core issue: how much. I am looking forward to scientifically proven numbers.

Joseph E Postma says:
2013/03/27 at 9:46 AM
Well I wouldn’t encourage the creation of more sophistry myself Greg, but I’ll let you deal with it I’ll quote Bryan’s response again because it has all the info necessary, and it basically means the thing measures the direction of ‘q’, as I said…

A bolometer senses the change in resistance of the sensor due to temperature change.
The sensor forms one part of an initially balanced Wheatstone bridge
No current will flow if the external object is at the same temperature as the sensor – the null point.
If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the Wheatstone bridge is out of balance and a current flows through the sensor.
If the object is at a higher temperature; the resistance increases as the temperature rises the Wheatstone bridge is again out of balance and current flows but in the opposite direction.
If calibrated against a known temperature the current can then represent temperature.
This change in resistance is linear near the null point but if moved too far from the null point the change in resistance is non linear and will give rise to errors.

Greg House says:
2013/03/27 at 9:57 AM
Joseph E Postma says (2013/03/27 at 9:46 AM): “Well I wouldn’t encourage the creation of more sophistry myself Greg, but I’ll let you deal with it ”
===========================================================

Thanks Joe . Because, if you remember the R.W.Wood experiment, he did not engage in an endless theoretical exchange, but instead demonstrated that the whole thing is at best negligible. It was enough for the warmists back then.

Last time I asked an allegedly knowledgeable warmist about those numbers, he suddenly did not want to discuss it any longer. Let us see, what thefordprefect can fetch.

Alan Siddons says:
2013/03/27 at 10:04 AM
Re: Bolometers and the like. I thought it was pretty simple. Warmer to cooler on the instruments induces a current so the needle registers a temperature. Cooler to warmer reverses the instrument’s current, which makes the needle register another temperature. The point is, you’re just seeing an electrical response, not a transfer of more thermal energy to less thermal energy.

Loodt says:
2013/03/27 at 10:22 AM
Hi Joseph,

To get back to this IR meter being pointed at various objects in the house, outside, and the fridge door, I think we have to go where it all started.

http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

I saw this post by Dr Roy Spencer in 2010. He placed a metal object next to his electric heater and read an increase in the temperature of heating element. There, fixed in his mind, is proof of back radiation. An Eureka moment, famous for life and for future generations, and he blogged about this new ‘Scientific”proof!

Our good Dr is not an engineer and has never dealt with ventilation and heating problems, and evidently doesn’t know how an electric heater works.

The hot electric metal element, heated to about 1200 to 1500 degrees Celsius is covered by a ceramic outer casing. When I grew up our heating elements had the electric wires on the outside, coiled along a ceramic core. Obviously it was too dangerous, must have started some fires, and the metallic heating element is nowadays hidden inside the ceramic cover than glows when the heater is on. If you placed a metal tube around the ceramic core, it would heat up the inner core, the metal heating element would melt, and the heater would stop working.

This our Dr did not understand, reduce the ability of the heating element to dissipate heat and it would heat up until the metal melts.

Ever since, we have proof of immaculate conception, and back radiation, and the disciples can out and spread the word.

I give you further links about the musing of this esteemed Dr about this very topic:

http://www.drroyspencer.com/2010/07/first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature/

http://www.drroyspencer.com/2010/08/help-back-radiation-has-invaded-my-backyard/

http://www.drroyspencer.com/2012/02/more-musings-from-the-greenhouse/

http://www.drroyspencer.com/2012/02/yes-virginia-the-vacuum-of-space-does-have-a-temperature/

So, my advice to you is to forget about the non-entity that is Wills, and start with Dr Roy Spencer. And being an American he can be neither a Sir, Duke, Earl, or Lord.

Joseph E Postma says:
2013/03/27 at 10:58 AM
A person that thinks that empty space has a temperature is a complete unscientific idiot. It is amazing such people can be associated with science and call themselves scientists. What did he say, that the temperature of the vacuum of space is the temperature of the CMB? What if there was no CMB…then what?

Alan Siddons says:
2013/03/27 at 11:17 AM
Correction to the above: “…you’re just seeing an electrical response, not a transfer of less thermal energy to more thermal energy.” Sorry.

Greg House says:
2013/03/27 at 11:25 AM
Loodt says (2013/03/27 at 10:22 AM): “I give you further links about the musing of this esteemed Dr about this very topic: …”
=============================================================

The links are dead (I wonder why…), but there is another way:

1. http://87.248.112.8/search/srpcache?ei=UTF-8&p=first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature%2F&rd=r1&fr=yfp-t-708&u=http://cc.bingj.com/cache.aspx?q=first-results-from-the-box-investigating-the-effects-of-infrared-sky-radiation-on-air-temperature%2f&d=4677512377075159&mkt=de-DE&setlang=de-DE&w=acdjW_nG5xhPFjdP_OFleEWvs72sWiHG&icp=1&.intl=de&sig=tESnetizxiz8Uuc4W4o7ow–

2. http://87.248.112.8/search/srpcache?ei=UTF-8&p=%2Fhelp-back-radiation-has-invaded-my-backyard%2F&rd=r1&fr=yfp-t-708&u=http://cc.bingj.com/cache.aspx?q=%2fhelp-back-radiation-has-invaded-my-backyard%2f&d=4686866851040923&mkt=de-DE&setlang=de-DE&w=Iru0kvDN3ZZTV3QO2E5VpLGtuRjL2mUC&icp=1&.intl=de&sig=be4.drVh5ok57sgc9MZhIA–

3. http://87.248.112.8/search/srpcache?ei=UTF-8&p=http%3A%2F%2Fwww.drroyspencer.com%2F2012%2F02%2Fmore-musings-from-the-greenhouse%2F&rd=r1&fr=yfp-t-708&u=http://cc.bingj.com/cache.aspx?q=http%3a%2f%2fwww.drroyspencer.com%2f2012%2f02%2fmore-musings-from-the-greenhouse%2f&d=4774690300368764&mkt=de-DE&setlang=de-DE&w=JoqIiozDz8F7R2dHAmlc86ta4TUSQL4m&icp=1&.intl=de&sig=Iz65v9nYul9avSaB1RTrVg–

4. http://87.248.112.8/search/srpcache?ei=UTF-8&p=Yes%2C+Virginia%2C+the+%E2%80%9CVacuum%E2%80%9D+of+Space+Does+have+a+%E2%80%9CTemperature%E2%80%9D&fr=yfp-t-708&u=http://cc.bingj.com/cache.aspx?q=Yes%2c+Virginia%2c+the+%e2%80%9cVacuum%e2%80%9d+of+Space+Does+have+a+%e2%80%9cTemperature%e2%80%9d&d=4706520609263305&mkt=de-DE&setlang=de-DE&w=0CvwrGzEBk9cR3xPdGJsw091uB0mlVxX&icp=1&.intl=de&sig=vBtIqlrReVTLTppFx6_TbA–

I recommend saving them, just in case.

thefordprefect says:
2013/03/27 at 1:00 PM
A bolometer is a devive for converting radiation into heat and then measuring the effect of the heat for example by a resistance change. The bolometer does not care what wavelength it is (providing it is within its 2 to 13um range) – it can only convert the radiation to heat.
We have a warm bolometer array (warmed from its use of power to digitise the resistance of each pixel) This will obviously be emitting as black body radiation from each pixel (it is designed that way) in all directions. Some of the pixels radiation will pass through the lens and be focussed on an object
I think most will agree that black body objects above absolute zero will radiate energy. So whatever the temperature of the object that the radiation from the pixel hits will radiate back to that pixel. No radiation of this type can cause cooling. There is zero radiation at -273°C there is some radiation at -272°C. There is no such animal as -274°C – there is no negative radiation!
Mr Postma the brings into play a numerical value q which depends on the temperature between the two objects. The question is where does this differencing occur.
The pixel will warm the remote object but by a negligible amount (compare the pixel size to the object). There canl be no differencing in the space between the pixel and object. It can only occur at the bolometer.
Assume the object is 0.3 metres away from the camera then energy leaving the bolometer pixel will not know what temperature the object is until 2ns later (2ns=2*10^-9s the time it takes to travel 2×0.3m) and similarly for the radiation leaving the object. Obviously each source of radiation does not know where it will land so the energy in transit can only depend only on the temperature of the pixel or the object.
The bolometer is at a stable temperature so:
energy lost [conduction, convection and radiation]=energy gained from the electronics and ambient)
BUT this is not true! The effect of the stuff in front of lens has not been considered.
Put a lens cap on – there is thermal radiation from the cap equivalent to its temperature.
Point it at ice at 0°C – there is radiation being emitted at +273Kelvin.
Point the lens at absolute zero – that would work, no radiation now!
Absolute zero is not an option but the camera does self calibrate by inserting a warmed plate in front of the bolometer array but again this is radiating this time at a known temperature
Consider the case when the bolometer is colder than the object then obviously the energy from the object gets focussed on the pixel and it warms – q is obviously positive
Consider the case where the pixel is hotter than the object BUT the object is above -273°C and therefore emitting radiation. The radiation still gets focussed on the pixel. This is of course positive radiation there is no negative radiation. The bolometer does not care what wavelength of radiation is hitting it.
Lets assume that because the pixel is hotter than the objet the object’s radiation gets rejected. But then all objects below the temperature of the pixel will get rejected and so there would be no image. This is not what happens.
Lets assume that the cooler radiation cancels an equivalent amount of energy in the pixel (q is negative). But how does this happen there is no difference in the effect of long or short wave radiation to a bolometer it is all just energy. There is no physical mechanism for cancellation. This is not what happens

David Socrates says:
2013/03/27 at 1:55 PM
Alan Siddons says, 2013/03/27 at 7:32 AM: Although they acknowledge that an Earth with a greenhouse effect sheds the same amount of thermal energy as an Earth without one, they argue nevertheless that the Earth is trapping heat. The Greenhouse Effect is thus like the biblical Bush That Burns Yet Is Not Consumed.

Although I agree with the sentiment, you have to be careful that you are not misinterpreted. The constant flow of energy through the earth-atmosphere system from the Sun and out again to be lost to space is quite capable over time of heating the surface and atmosphere to any arbrarily higher temperature level than the surface would have if the earth had no atmosphere. It all depends on the ‘resistance to flow’ offered by the paths through which the atmosphere’s Kinetic Energy flows. In practice, it is the mean value of the atmosphere’s ‘conductivity’ that gives us the Atmospheric Thermal Enhancement and, thereby, our elevated mean surface temperature of ~288K.

So two points:

1. Although a thermos contains a heat source, the heat diminishes until the temperature of the contents reaches ambient. I do not think that the more sophisticated warmist scientists use the ‘thermos’ argument nowadays so criticising them using the ‘thermos’ misconception may be a bit counter-productive.

2. Turning to the steady-state energy flow scenario, where the earth is in energy flow balance and exhibits an enhanced mean surface temperature, it is indeed the atmosphere’s ‘resistance’ to heat flow that keeps its fund of kinetic energy at an enhanced level and, therefore, at an elevated temperature. So the earth-atmosphere system is indeed ‘trapping heat’ by virtue of being a giant ‘resistive conductor’ of energy.

I hope you are not suggesting that, in this latter scenario, the earth’s atmosphere is not ‘trapping heat’ because it obviously is (take the atmosphere away and the mean surface temperature drops to a similar level to that of the Moon). The real issue is what is the physical mechanism?

Not GHGs because radiation is simply a transport mechanism and not a STORE of energy. It is a fallacy to assume it can ‘bootstrap’ itself into a storage role. Instead the correct answer is good old thermodynamics – specifically the resistance to energy flow up through the atmospheric column and out to space caused by the limited speed of conduction/convection of heated air and latent heat carried upwards in the water cycle.

Max™ says:
2013/03/27 at 6:22 PM
I was gonna say, it seems like someone has confused cooled and uncooled bolometers.

Joseph E Postma says:
2013/03/27 at 6:53 PM
thefordprefect – read Bryan’s answer.

Greg House says:
2013/03/27 at 7:16 PM
@thefordprefect

You have forgotten an important scientific issue, let me repeat it. What I would like to know, is this: how much warmer gets the sensor when the IR-Thermometer is pointed to the sky, let’s say, at night? Because, you know, I would disregard the whole thermodynamics immediately, if the GHE were alleged to be like 0.000001C. Let us just clarify that before we get deeper into the issue about how it is possible that colder bodies do not warm warmer bodies but IR-sensor still detects IR from colder bodies. Let us address the core issue: how much. I am looking forward to scientifically proven numbers.

Max™ says:
2013/03/27 at 7:27 PM
The shell can cool radiatively, right?

It can not cool by radiating towards a warmer surface, it is receiving more radiation than it loses in that direction.

It doesn’t “know” that the surface in that direction is warmer, but across distances this short it takes a couple of nanoseconds per meter, so any photons leaving are more than replaced immediately.

As such, no, the surface can not be treated as existing in isolation and radiating as though there are no other sources of radiation nearby, that is the first mistake Willis made, and it is commonly repeated.

___________

In fact this is what will happen when you add the shell:

Initially the shell will be cold, it will absorb close to 234.99 W/m^2, it will emit a small fraction of that amount inwards and outwards.

Then it will warm until it is emitting say, 100 W/m^2, then it will absorb 134.99~ W/m^2 from the planet, and emit 100 W/m^2 outwards.

Finally when it and the planet are in radiative equilibrium it will be emitting 234.99~ W/m^2, it will be absorbing a net 0.01 W/m^2 from the planet, and it will emit 234.99~ W/m^2 outwards.

Thus the emissions outwards will balance perfectly with those from the planet with no temperature increases from unphysical greenhouse effects.

Greg House says:
2013/03/27 at 7:30 PM
Greg House says (2013/03/27 at 11:25 AM )”The links are dead (I wonder why…), but there is another way…”
=======================================================

The links Loodt provided (2013/03/27 at 10:22 AM) are alive now.

thefordprefect says:
2013/03/27 at 7:31 PM
Joseph E Postma says: 2013/03/27 at 6:53 PM
thefordprefect – read Bryan’s answer.

——————
I assume you mean the one where he describes a whetstone bridge for measuring the resistance change caused by temperature difference?

This is jus a means of measuring the resistance change caused by thermalisation of IR. (This is certainly not the only means and definitely not the method used in an imaging array)

What I need explained to me is what happens if the pixel is hotter than the object focussed on it. Basically what causes its TEMPERATURE to change. not how that change is measured.
Cheers

Joseph E Postma says:
2013/03/27 at 7:39 PM
q ~ (T2^4 – T1^4)

The direction and magnitude of q is what changes the resistance and the current. “Thermalisation” does not mean that cold is heating up hot. That’s not what thermalisation means. The sensor is only heated proper when the balance of q is positive coming into it, else q is negative and that is what happens when the pixel is hotter. Alan and Bryan probably know more about this than me, so hopefully they can comment. But what is plainly true is that nothing violates the laws of thermodynamics, and hence, precisely, a bolometer can not be an example of something cold heating up hot. Nor do they heat themselves up with their own radiation so their operation isn’t even a relevant concern in regards to the GHE in any case.

thefordprefect says:
2013/03/27 at 7:59 PM
Joseph E Postma says: 2013/03/27 at 7:39 PM
q ~ (T2^4 – T1^4)

sensor is only heated proper when the balance of q is positive coming into it, else q is negative and that is what happens when the pixel is hotter.
———————————–
Apologies I still do not understand!
are you suggesting that if the object is cooler then the pixel is cooled? q is negative?
If so please tell me how a pixel is cooled below its ambient without assuming negative radiation

Joseph E Postma says:
2013/03/27 at 8:05 PM
q can be negative but the sensor, I believe, is attached to a large heat sink, so, it doesn’t cool even though q is negative. As more and more radiation comes in, from warmer and warmer sources, q becomes less and less negative until it is positive (when the source is hotter than the sensor). The heat sink, being very large, still would’t change. Hoping for Alan or Bryan to comment in more detail.

thefordprefect says:
2013/03/27 at 8:40 PM
Joseph E Postma says: 2013/03/27 at 8:05 PM
q can be negative but the sensor, I believe, is attached to a large heat sink, so, it doesn’t cool even though q is negative.
———————
tfp.
The resistive sensor is conductively isolated as from the heat sink. For the bolometer resistance to change the sense element must change in temperature. In some bolometers there is even a mirror under the sense element to reflect IR back to the sensor.
————————
jp. As more and more radiation comes in, from warmer and warmer sources, q becomes less and less negative until it is positive (when the source is hotter than the sensor). The heat sink, being very large, still would’t change.
——————.
tfp.
you use the term “radiation comes in” this as I stated in my post 2013/03/27 at 1:00 PM can only have a positive thermal effect. If you open a fridge door it feels cold because of a LACK of heat radiation (o.k. and because of cold air escaping) not because of cold radiation!
At 0K there is no radiation and no heating. at 1K there is radiation and therefore heating.

The bolometer measures radiation at all wavelengths (filtered usually into 2um to 14um bandwidth).
It converts this radiation to heat The wavelength is not material to this type of bolometer.
All changes in temperatures that the bolometer reaches are effectively referenced to zero incoming radiation i.e. the source is at 0K.
All other temperatures of source add proportionally more radiation to the pixel.
The pixel temperature will therefore always be above that with no incoming radiation (from 0K source)
The temperature of the pixel will always depend on balance of heat loss and heat input.
Heat input from the source beyond the lens is the only variable (internal heating is calibrated out).
The focused source radiation will control the pixel temperature – higher temperature of source = higher temperature of pixel.

Joseph E Postma says:
2013/03/27 at 8:52 PM
prefect – all you have to do is use the laws of thermodynamics. Cold doesn’t heat up hot. The type of “pixels” I use in my field do not detect radiation by increasing in temperature – they convert photons to electrons. Bolometers might do something different than that and I’m not an expert on them. But the only thing you do need to know, is that cold doesn’t heat up hot. Not only that, a bolometer does not heat up via its own radiation, and so however they function has nothing to do with the GHE in any case. Therefore this is a useless discussion. We’re discussing how many teeth are in a horses mouth when the claim is that tigers can eat airplanes. Bolometers don’t heat themselves up with their own radiation, therefore they offer no support for the GHE. If a bolometer is pointed at a mirror, it does not spontaneously begin reporting higher and higher temperature.

A bolometer senses the change in resistance of the sensor due to temperature change.
The sensor forms one part of an initially balanced Wheatstone bridge
No current will flow if the external object is at the same temperature as the sensor – the null point.
If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the Wheatstone bridge is out of balance and a current flows through the sensor.
If the object is at a higher temperature; the resistance increases as the temperature rises the Wheatstone bridge is again out of balance and current flows but in the opposite direction.
If calibrated against a known temperature the current can then represent temperature.
This change in resistance is linear near the null point but if moved too far from the null point the change in resistance is non linear and will give rise to errors.

thefordprefect says:
2013/03/27 at 9:30 PM
Joseph E Postma quotes: 2013/03/27 at 8:52 PM
If the object is at a lower temperature; the sensor resistance loses heat and resistance goes down and the …
————————————
But the question is – Why does the temperature go down?
Bolometers are in thermal balance with all energy inputs so power into the array is part of this balance as is the focused radiation from the source

I think any scientist would agree that cold cannot heat up hot, However in the world of radiation balance the starting temperature should be the ambient temperature with added radiation at zero (-273°C) – “zero temp” – any change from this source temperature adds more energy and the temperature rises from its “zero temp”.

So if there is no ghg then the earth sees TSI plus near zero radiation from space this will give be its “zero temp”.
Add ghg and the radiation from above is perhaps from a temperature of -50°C (223K) This adds to the energy heating the earth (at perhaps -18C). The temperature will now be heated above its “zero temp”.
This may seem like cold heating hot but really its cold heating “zero temp”.
But what is important is that the new radiative balance allows the earth to warm to a new higher temperature above “zero temp”

There is no breaking of any laws.

Joseph E Postma says:
2013/03/27 at 9:46 PM
“Add ghg and the radiation from above is perhaps from a temperature of -50°C (223K) This adds to the energy heating the earth (at perhaps -18C). The temperature will now be heated above its “zero temp”.”

No, that is not correct at all. And it has nothing to do with bolometers either because bolometers do not heat themselves up or report higher temperatures from their own radiation coming back to them. Being surrounded by ice or mirrors with your own radiation coming back does not heat you up or add to your heating. This is of course what bolometers prove as well.

Max™ says:
2013/03/27 at 10:09 PM
Just to help Joe out a little: essentially tfp is claiming that the Hubble and Webb would work flawlessly in infrared wavelengths even if they weren’t cooled, because the cool sources must add energy to the ccd pixels and whatnot.

thefordprefect says:
2013/03/27 at 10:50 PM
[trashed]

tfp, bolometers don’t violate the laws of thermo. Cold doesn’t heat hot. CCD’s absorb photons and create current just fine without being caused to heat up etc.

thefordprefect says:
2013/03/27 at 11:03 PM
Max, although this is for astro purposes this is helpful on bolometer detectors and why they are sometimes cooled (noise reduction)
http://home.strw.leidenuniv.nl/~kenworthy/teaching/dol2011/10_DOL_Bolometers.pdf
http://home.strw.leidenuniv.nl/~kenworthy/teaching/dol2011/11_DOL_Bolometers_part_2.pdf

Joseph E Postma says:
2013/03/27 at 11:29 PM
Reducing noise isn’t about cold heating hot.

Max™ says:
2013/03/28 at 12:58 AM
Joe also went over the actual mechanism: there is a sink tied to the device to stabilize it, the detector pixels are isolated, and the default with no radiation incoming is treated as 0 K, any level of radiation coming in will reduce that emitted by the device, this can be used to work out the temperature of a distant source.

thefordprefect says:
Your comment is awaiting moderation.

2013/03/28 at 6:15 AM
thefordprefect says:
2013/03/27 at 10:50 PM
[trashed]
————–
well so much for open debate. Just because you disagree with my sciencience you trashit. This is exactly what you claim about warmist science!
Very disappointed!
If you are accusing someone of fraudulent science you should at least let them dig there own hole.

thefordprefect says:
Your comment is awaiting moderation.

2013/03/28 at 6:19 AM
Max™ says:
2013/03/28 at 12:58 AM
Joe also went over the actual mechanism: there is a sink tied to the device to stabilize it, the detector pixels are isolated, and the default with no radiation incoming is treated as 0 K, any level of radiation coming in will reduce that emitted by the device, this can be used to work out the temperature of a distant source
—————
What is the mechanism for reducing the radiation emitted by the device. There is no cancellation the bolometer does not respond to phase of radiation.
Please explain to me how an object above absolute zero but below bolometer temperature reduces the outgoing radiation.

But please keep it simple.

Tall bloke too?

http://tallbloke.wordpress.com/2013/02/04/david-cosserat-atmospheric-thermal-enhancement-part-i-the-great-debate-begins/comment-page-1/#comment-43406

thefordprefect says:Your comment is awaiting moderation.

Bryan says: February 10, 2013 at 12:38 pm

…This is because DLWIR cannot be measured directly. Instruments that claim to measure DLWIR are measuring an INFERRED value. The pyrgeometer is a typical device

From the nrel site linked in my post

PRECISION INFRARED RADIOMETER Model PIR The Precision Infrared Radiometer, Pyrgeometer, is intended for unidirectional operation in the measurement, separately, of incoming or outgoing terrestrial radiation as distinct from net long-wave flux. The PIR comprises a circular multi-junction wire-wound Eppley thermopile which has the ability to withstand severe mechanical vibration and shock. Its receiver is coated with Parson’s black lacquer (non-wavelength selective absorption). Temperature compensation of detector response is incorporated. Radiation emitted by the detector in its corresponding orientation is automatically compensated, eliminating that portion of the signal. A battery voltage, precisely controlled by a thermistor which senses detector temperature continuously, is introduced into the principle electrical circuit.

Isolation of long-wave radiation from solar short-wave radiation in daytime is accomplished by using a silicone dome. The inner surface of this hemisphere has a vacuum-deposited interference filter with a transmission range of approximately 3.5 to 50 µm.

SPECIFICATIONS

Sensitivity: approx. 4 µV/Wm-2. Impedance: approx. 700 Ohms. Temperature Dependence: ±1% over ambient temperature range -20 to +40°C. Linearity: ±1% from 0 to 700 Wm-2. Response time: 2 seconds (1/e signal). Cosine: better than 5%. Mechanical Vibration: tested up to 20 g’s without damage. Calibration: blackbody reference. Size: 5.75 inch diameter, 3.5 inches high. Weight: 7 pounds. Orientation: Performance is not affected by orientation or tilt. ————————-

This looks to me as if it is measuring the heating effect (thermopile) of radiation hitting the dome of the sensor (transmission 3.5 to 50um. The thermopile of course generates a voltage dependant on the temperature difference between one side and the other The non-dome side is not exposed to external radiation so no effect there. However, the nondome side temperature must be measured and compensated. The instrument also compensates for its own generated IR. No assumption of BB radiation is assumed. It is the ACTUAL heating effect of IR radiation of narrow or wide bandwith hitting the sensor that is the cause.

This all sounds as accurate as a liquid in glasss thermometer to me.

If the radiative “temperature” is less than the receiver temperature then the thermopile still measures – see my series of posts about thermal imaging – the camera microbolometers sitting at 20+C shows temperatures down to -40C

The R. W. Wood Experiment posted 9/2/2013 on wuwt

george e. smith says: February 9, 2013 at 12:30 pm …Perhaps a single atom or molecule in free flight in space, does not emit thermal radiation; but that is not an exception, because such a non interractng isolated atom or molecule by definition is at zero Kelvins, so is not expected to radiate. And very sparse collections of molecules that collide less frequently, and at lower velocities, such as gases, would naturally be expected to emit (or absorb) less of the thermal radiation, than denser more frequently colliding collections of “particles”.

Measurements using a thermal imaging camera show invisible water vapour at 100C play this over paper and the paper radiates at near 100C Water vapour is not black body (radiates at discrete wavelengths some outside the sensitivity range of the camera)  paper is black body and hence shows as BB heat. water vapour and thermal imaging http://climateandstuff.blogspot.co.uk/2012/12/water-vapour-and-thermal-imaging.html .

post wuwt 31/1/13

<i>Konrad says: January 30, 2013 at 1:57 pm   thefordprefect,  I note in your IR imaging that while water vapour does not read at 100C is is still slightly visible, reading 28C.</i> ——- Looks as if blogger has lost the videos one showed Steam at 28C rising from the boiling water – the still is from the same run ——– <i>Experiment 1. (low cost)  Build two insulated containers with potassium chloride salt lenses for lids. Under a clear dry night sky (desert conditions would be best) fill both containers with dry 30C gas. CO2 in one container 1, N2 in container 2. Which container cools Faster? Is the answer – A. Both containers cool at the same rate.  B. Container 1 cools faster because of the greater IR emission from CO2.  C. Container 2 cools faster because of the greater thermal conductivity of N2</i> ——– This is not a good experiment as you describe it: the insulated walls/salt winow will be heated by the gas to gas temperature. These will then radiate in all directions, with BB radiationprofile, eventuall passing the window to space. same for both boxes. The CO2 will additionally radiate at specific wavelengths but the large proportion of escaping radiation will be from the warm insulation/salt. the CO2 will additionally intercept some of the wavelengths from the walls of the box/window preventing this escaping The hot Co2 will radiate in all directions. this wold reduce the radiation escaping at the absorption frequencies.

I think i would suggest that the CO2 gas will cool slower.    ——— <i> Experiment 2. (high cost)  Attach two gas cylinders with regulators one CO2, one N2, to two 10m long lengths of 5mm PVC tubing. Coil most of the tubing through an insulated container full of hot water. Attach the two open ends of the PVC tubes to two retort stands in front of a cool wall. Set gas flow from both tubes to 1 L/s. Observe the gas flowing out of both tubes with a high quality IR camera capable of seeing beyond 15um. Are the results – A. Both tubes are visible as warm, but both gas plumes are undetectable.  B. Both tubes are visible as warm and the CO2 gas plume is also visible</i> ————— both tubes will be visible at same temperature. N2 will be invisible. CO2 will show up as warm plume but on a camera adjusted for BB radiation the temp will be much less than the actual temp.. ————-   <i>Or perhaps this –  “What would happen to convective circulation in the lower atmosphere if the atmosphere contained no radiative gases?”</i>

Convection does not depend on ghgs. The ground would radiate as a BB and the lw radiation would all escape without absorption with no ghgs. The ground/sea transfers heat (conducts) from/to the non ghg atmosphere and eventually equilibrium will be reached when radiation from the ground = radiation from the sun. the ground temp will be the same as the lowest layer of atmos temp and presumably the amos will cool at adiabatic lapse rate from this temp

http://wattsupwiththat.com/2013/01/28/matt-ridley-a-lukewarmers-ten-tests/#comments

NOAA experiment at WUWT Comment

Hopefully the thermometer enclosures are proof against radiative effects.

The solar radiation is capable of heating a suitably insulated surface to 1000s deg C. A warm building/hot trash burner certainly is less of a radiation hazard for the internal thermometers.

The enclosures are designed to reduce radiative effects and to measure air temperature.

wuwt post

Why do people assume that because a green house gas is increasng then so should temperature follow this increase exactly?

The GHE is small but insidious.

Look on it as a drip of water entering a jacuzzi with no safety overflow – if the jacuzzi is switched off then a slow rise in water height can be measured. Switch on the turbulance. Does this stop the drip?
The water level will now move up and down at random but – if you measure the height you may decide that the water level is flat or even falling – but is the total volume really static?
Can you leave it dripping an go away for a year? Or will it eventually overflow id the jacuzzi is off orON?

I have made a totally nonscientific, non-predictive simulation of hadcrut3v temperatures using sine waves. This shows that a 60year cycle is on its way down and this is more than capable of holding the temperature increase due to GHGs – BUT the GHG effect is still there and on the next upswing the temperature starts to rise wit a vegance.
the plot is here:

but remeber this is simply showing that an underlying trend can be masked by a sinusoid with no trend – it is not meant to predict temperatures!!
All the figures for this are availble as a spreadsheet.

Just because temperatures and GHG levels do not follow each other does not mean that the undelying temperature trend does not match the insiduous drip, drip, drip of a trend caused increasing GHGs.

This is not rocket science. It is just logical!

CA posts

Posted Sep 21, 2012 at 12:11 PM | Permalink | Reply
Your comment is awaiting moderation.
joannenova
Posted Sep 21, 2012 at 10:40 AM | Permalink
would you care to ask Watts why he lost his cool over his pet surface stations when his data was used before he published?
thefordprefect
Posted Sep 21, 2012 at 12:14 PM | Permalink | Reply
Your comment is awaiting moderation.
Steve: around 20% identified themselves as “skeptic”, but some of these responses were fraudulent. The actual number of respondents appears to be much less than that. My guess is that over half of the “skeptic” responses were fake.

How do you KNOW
are you sure some of those calling themselves non-sceptic were fraudulent. (Fraudulent is not the right word much too emotive – but that’s what you are trying to do -whip up a bit of name calling)

Tisdale on sea temperatures

posted this at WUWT
I doubt it will be published despite the claim that few posts are censored

All oscillations are just that – one moment you are warming the next cooling. The overall effect is neutral.

The only additional input to sea temps according to you is? Why do they keep warming.
Solar is just about constant – something must be adding energy,

If you have air at 35C above water at 30C the only way the water can cool is by evaporation and also the air will warm the surface layer which will mix.
If you have air at 25C above water at 30C evap will still occur but now the water will lose energy to the Air.

Thus the energy content of the water will be affected by the air temp. The air temp IS affected by GHGs.

Lewandowsky Censors Discussion of Fake Data

http://climateaudit.org/2012/09/10/lewandowsky-censors-discussion-of-fake-data/

Amazing stuff from the great McIntyre (with arch-accolytes Watts and Tallbloke) complaining about moderation on Lewandowsky’s blog.

None of these have any right to complain about censorship. So many people banned from their blogs, so many posts removed.
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Lets look at WUWT for hints of conspiracy

iron sun 57 hits
New mission to study crossed magnetic streams and magnetic portals …
wattsupwiththat.com/2009/09/01/new-mission-to-study-crossed…The phenomenon of paramagnetism is not evidence of an ‘iron sun’. Again, you are out on the uttermost fringe of the worst pseudo-science dressed up with words and …
New book: Slaying the Sky Dragon | Watts Up With That?
wattsupwiththat.com/2010/11/29/new-book-slaying-the-sky-dragonI’m obviously not the only one who wishes you didn’t go into ad hominem attacks. REPLY: No, I think the Iron Sun theory is in fact “nutty”, which is my honest …
Volcanoes and Water | Watts Up With That?
wattsupwiththat.com/2010/04/18/volcanoes-and-waterWell….i’ve read that link and it’s like an ‘Iron Sun’ ala Oliver Manuel site I think. I just don’t know enough about it but like Oliver and Tallbloke but don
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watermelon

Poo Gloo Gaia Pans | Watts Up With That?
wattsupwiththat.com/2011/01/09/poo-gloo-gaia-pans[now does anyone here see how the sneaky, via the back door, watermelon infiltration world works – Joseph the serial polluter] teacher : now children… don’t use the …

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marxist

China Warns EU’s Green Fanatics Of Global Trade War | Watts Up …
wattsupwiththat.com/2011/12/22/china-warns-eus-green-fanatics-of…I’m also sure that China is intelligent enough to already know that “global warming” is only a another marxist lie! After all, they have been marxist long enough to …
Bad News (Polar) Bears? | Watts Up With That?
wattsupwiththat.com/2010/10/20/bad-news-polar%c2%a0bearsI am calling them Marxist because they are Marxist, that is the founders are real honest to goodness Marxists. They literally want the destruction of property rights and …
Linking health, wealth, and well being with the use of energy …
wattsupwiththat.com/2009/10/12/linking-health-wealth-and-well…There is no question in my mind that the global warming scam is nothing more than an attempt by the Marxist of the world to redistribute the wealth of those nations who …
First They Came For The Scientists…. | Watts Up With That?
wattsupwiththat.com/2010/07/07/first-they-came-for-the-scientistsThus I was a non-marxist using marxist tactics to achieve anti-marxist ends. Some might say it was unethical of me to do that, but in reality, since I, as a libertarian
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New world order

Lord Monckton barred from Copenhagen conference – pushed to the …
wattsupwiththat.com/2009/12/17/lord-monckton-barred-from…Today, as I lay in the snow with a cut knee, a bruised back, a banged head, a ruined suit, and a written-off coat, I wondered whether the brutality of the New World Order …
Stephan Lewandowsky’s slow motion Psychological Science train …
wattsupwiththat.com/2012/09/05/stephan-lewandowskys-slow-motion…05/09/2012 · I’m a bit of a latecomer to this affair, as Lucia and Jo Nova took an early lead on pointing out the many problems with the survey methodology (or lack …
Climate Craziness of the Week: Soylent Green Earth Sim | Watts Up …
wattsupwiththat.com/2010/11/06/climate-craziness-of-the-week…They were both lefties trying to bring in a new world order for the good of all. Fortunately they were stopped. Hey PandR, let us not forget which side we’re supposedly …
Madrid 1995: Was this the Tipping Point in the Corruption of …
wattsupwiththat.com/2012/09/06/madrid-1995-was-this-the-tipping…06/09/2012 · Such an illusion it is only possible if the majority of the earth´s population become “Gammas”, servants or slaves of the new world order.
Take Examiner.com’s First Annual Survey on Global Warming …
wattsupwiththat.com/2009/10/31/examiner-coms-first-annual-survey…Mark.R (01:11:48) : My question was do you think AGW is a smoke screen for the new world order?. ——————-Ditto. Halleluya. Nail. Head.
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Common purpose

The Telegraph “gets it” about Climategate investigations and …
wattsupwiththat.com/2011/04/24/the-telegraph-gets-it-about…The entire corrupt Common Purpose controlled edifice is crumbling away and about time too. The BBC is nothing but a propaganda organ for whichever British government is in …
Phil Jones does an about face on “statistically significant …
wattsupwiththat.com/2011/06/11/phil-jones-does-an-about-face-on…How Jones has the gall to call himself a “Professor” is beyond me. He is a Common Purpose stooge taking orders from Brandon Gough, the UEA chancellor.
Socialists of the World Unite for Youth Climate Conference | Watts …
wattsupwiththat.com/2011/04/13/socialists-of-the-world-unite-for…In the UK, this agenda is being implemented by Common Purpose. The chancellor of the UEA is a senior Common Purpose figure. Join the dots and try to understand the politics …
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World Government

Monckton’s Mexican Missive #3 | Watts Up With That?
wattsupwiththat.com/2010/12/09/moncktons-mexican-missive-3The world-government Secretariat: In all but name, the UN Convention’s Secretariat will become a world government directly controlling hundreds of global …
Obama Poised to Cede US Sovereignty in Copenhagen, Claims British …
wattsupwiththat.com/2009/10/16/obama-poised-to-cede-us-sovereignty…And what it says is this, that a world government is going to be created. The word “government” actually appears as the first of three purposes of the new entity.
.The Copenhagen Climate Change Treaty Draft – wealth transfer …
wattsupwiththat.com/2009/10/03/the-copenhagen-treaty-draft-wealth…And to think I have always consistently stated that AGW was the means by which the Fabian Socialists will implement a world government. Looks like my assessment was spot …
Durban: what the media are not telling you | Watts Up With That?
wattsupwiththat.com/2011/12/09/durban-whatReporting to the world government: From 2013/14, the world government will oblige Western nations to prepare reports and submit them to it every two years.
Global Warming and “The Early Spring” | Watts Up With That?
wattsupwiththat.com/2009/04/05/global-warming-and-the-early-springThis is all driven at a high level, which is why all the media are playing the same tune. The New World Order needs a One World Government..