327 thoughts on “Monckton Post: For comments on issues other than paragraph 3.”

  1. Pingback: The Blackboard » Monckton Planck Parameter: No better than pulling numbers out of a hat.

  2. Can you build a solar oven type apparatus and point it at the backradiation of 324W per m^2 and cook food with it?

    If not, why not?

  3. Bruce–
    This blog is not dedicated to teaching survivalists or campers how to building solar cookers. If you goal is to learn how to build solar cookers, you will need to find another blog or forum where that issue is considered on topic.

    I suspect you comment is an attempt at ‘arguing by asking rhetorical questions’. That’s not permitted here. If you think the answers to those questions make some point, you must do 2 things:
    1) Provide your own answer and
    2) Explain what point your question and your answer is supposed to make.

  4. 1) No
    2) If it can’t do work, it isn’t real.

    Backradiation is a fictional construct. Why is the IPCC using fictional constructs?

  5. Re: Bruce (Comment #82297)

    Can you build a solar oven type apparatus and point it at the backradiation of 324W per m^2 and cook food with it?

    If not, why not?

    Because a) it’s diffuse radiation and b) it’s already taken into account by the fact that the peanuts are around 290 K and not, say, frozen.

  6. 1) Correct.
    2) Back radiation is not a fictional construct. Any engineer who neglects back radiation in from fly ash in a furnace is going to get inaccurate answers to important design questions. Back radiation is also accounted for in home heating projects and all sorts of things.
    3) The definition of “physically real” is not “can be used to design a solar cooker”.

  7. If it can’t do work it has no impact.

    A distant star (lets say a Red Giant) may be measured at 5000K on the surface. But that 5000K is irrelevant to us on the earths surface.

    OTOH, at noon on a hot summer day you are being bombarded by 1000W/m^2 and you can “do work” with it. You can burn your skin with it. You can collect the energy with all kinds of tools (including solar cookers). It has an impact.

    The 324W/m^2 of backradiation has no impact. It does no work. You can measure the temperature. But it is irrelevant to earths radiation budget.

  8. I found the letter to be tremendous. Lucia, out of curiosity, why do you think that feedbacks in the system are positive? Is the reason because the majority of group thinkers endorse this belief? I’m just wondering because I think it is quite a reach. The first thing I think about when I picture a cloud is the fact that it blocks sunlight from hitting the earth. Another question I have is I am wondering if you would acknowledge me for claiming about a year ago that the problem with global warming is that many of the mathematical calculations are/were wrong.

    @All

    I am wondering if anyone here drives an “eco” friendly car. This would include a prius, chevy volt or a nissan leaf. Basically a car that is advertised as being good for the environment.

  9. Is “no arguing by asking rhetorical questions” a reasonable blog policy?

    “No arguing by asking rhetorical questions” is a policy that focuses commeters’ remarks and helps keep discussions on topic.

  10. Bruce–

    If it can’t do work it has no impact.

    Wrong on several.
    1) From the point of view of physics, a solar cooker does do work. (Or at least, it does nothing I can identify as “Force*distance” type work.
    2) The fact that something can’t operate as a solar cooker or any type of cooker doesn’t mean that thing can’t be taken advantage of to do work. If someone wanted to go to the trouble, they could probably gin up a system that got some type of “work” from the heat lost from our refrigerator coils, but they probably couldn’t use it to fry burgers. (It’s also probably not worth the effort- but the coils are warmer than the air in the kitchen, so they could probably do it.)

    3) The fact that something does no work doesn’t mean it has no impact. Returning to the heat coils in the back of my refrigerator– all other things being equal they warm my kitchen. That means in summer, I might want to run the air conditioner more and in winter I might want to burn less gas in my furnace. Both are impacts. Other things that do no work but have impact: Light from an LED likely can’t be used to fry hamburgers– but they can illuminate a room. That’s an impact.

    The 324W/m^2 of backradiation has no impact.

    You are merely claiming it has no impact. It exists and does have an impact relative to not being there. It is no more irrelevant to the earth’s radiation budget than back radiation from fly ash is irrelevant to the radiation budget in a furnace!

  11. Ok. Backradiation does so little work compared its claimed size that it is irrelevant.

    It is claimed that solar energy striking the earth is 168W/m^2 by Trenberth.jpg. That is 51% of claimed backradiation.

    Backradiation should do almost twice as much “work” as the solar energy that reaches the earth.

    And, my understanding is that this backradiation should exist just after sunset. Or when you are in the shade. Yet a solar cooker will not work after the sun sets or if a tree is between you and the sun even if a cooker is still pointing at the sky.

  12. I should have used 198, not 168. The reflected at surface shouldn’t be subtracted yet.

    61%, not 51%.

  13. Bruce: 2) If it can’t do work, it isn’t real.

    My body radiates heat. I very much doubt you could devise any kind of cooking apparatus to use that heat to cook anything. Therefore my body heat isn’t real. Ergo I am zombie.

    [Nyq shuffles off in search of some brains to nibble…]

  15. Backradiation should do almost twice as much “work” as the solar energy that reaches the earth.

    It’s diffuse. The sun is a point source. This makes a difference when you are trying to take advantage of the heat.

    And, my understanding is that this backradiation should exist just after sunset. Or when you are in the shade. Yet a solar cooker will not work after the sun sets or if a tree is between you and the sun even if a cooker is still pointing at the sky.

    Backradiation does exist after sunset. It’s extremely difficult to take advantage of this for a cooker because it’s diffuse rather than a point source.

  16. Diffuse? Does that matter?

    Is there a difference between 324W/m^2 from a point source versus 324W/m^2 from a non-point source when its is the effect on a surface being measured?

  17. Bruce–
    Whether a source of heat is diffuse or concentrated affects the difficulty of creating a cook stove from the heat source. Think of heat source from an electric heating element in your home stove. That generally some amount Q_dot =IR where I is current. Now, imagine instead, you took the same amount of heat created with a very, very, very long resistive wire and wond the length back and forth under a 1 km/1 km cooper plate. The heat source would be Q_dot, but if you put a cooking on that cooper plate, you would be unable to boil water. The heat would be diffused over a large area.

    You can collect and concentrate the sun energy, but it’s not really all that easy. For example: It really helps to align something to face the sun. If you were to install a panel on a north facing wall, it wouldn’t work. Similarly, if you face whatever you are using to collect the suns heat using your solar cooker, away from the sun, it won’t work. This is true even if the sun is beating down on your back causing you to get awesome blisters!

  18. Hey Monkton is quite the octopus, squirting ink while running away from his error..

    where is Josh? cartoon.

    maybe John Cook.

    the bug eyed monktopus

  19. Mosher,
    “bug eyed monktopus”
    Unkind and not needed; clearly Mr. Monckton has some health issues. No need to poke fun.

  20. lucia, the 324W/m^2 is the energy on 1 square meter of the earths surface, not the amount of energy coming from the source.

    Thats what the Trenberth diagram claims.

    Now that I think about it, a diffuse energy source would make a cooker even easier to use because you wouldn’t have to point it at the sun. As long as it was aimed at any part of the sky it would result in 324W/m^2 on the surface of the cooker.

    It would be magical. For while. And then it would get as tiresome as being in Phoenix in July. The heat would get oppressive.

    But not to worry. Backradiation is mythical.

    I kind of like this image.

    http://i53.photobucket.com/albums/g43/DerekJohn_photos/stuff/6a00d834519c3c69e200e553d605cf8834-.jpg

  21. Mockton’s correspondence with lucia sounds as though he’s writing FOR “The Team”. Dodginess, arrogance, lack of transparency, reluctance to provide citations, code, data …

  22. Lucia,

    Can you post the entire “sneeringly rude” email that Christopher Monckton referrers to?
    I want to see your ‘sneeringly rude’ mode of operation.

  23. Bruce: Now that I think about it, a diffuse energy source would make a cooker even easier to use because you wouldn’t have to point it at the sun.

    Well you can test that idea fairly easily. Do actual solar cookers work better on sunny days or cloudy days? [HINT: the answer is ‘sunny days’ and not just because sunny days are warmer].

    For a cooker you want one spot (the place you are doing the cooking) to be hotter than the surrounds. If the radiation is diffuse then it is heating everywhere more or less the same. You need to concentrate that energy in one spot so that you get higher temperatures. That is hard to do the more diffuse the energy source is.

  24. One useful thing Monckton could learn is the ability to write things opaquely isn’t proof that he’s smart (I take it as quite the opposite, the average intellect hides his mundanity via such a vessel).

    Secondly, simply because, after an explanation, you can understand what he was originally trying to drive, at after further elucidation is given, doesn’t mean it was correctly phrased at the start: in the sense of having a single, clear and unambiguous meaning.

    Personally, I could give a flip what he thinks is sneeringly rude. I believe it was Monckton himself who spoke of a critic that “he looks like an overcooked prawn”. I don’t think it’s necessary to descend to a level that is an even playing field for Monckton (grade school insults), so I suggest we not reciprocate.

  25. SteveF–
    I am 100% certain that posting would not violate copyright. However, I wrote things “inline”. So, I prefer to wait for Monckton to agree to my posting.

    FWIW: I’d like to also include color and bold etc. markup in emails. If anyone knows how to save the color, bold an so on in an email to display on a blog, I’d do that. It can probably be done with CSS, but it’s troublesome and requires me putting tags in.

  26. This is a more complete Christopher Monckton quote, given in case somebody thinks I was quoting him out of context:

    So unusual is this attempt actually to meet us in argument, and so venomously ad hominem are Abraham’s artful puerilities, delivered in a nasal and irritatingly matey tone (at least we are spared his face — he looks like an overcooked prawn), that climate-extremist bloggers everywhere have circulated them and praised them to the warming skies.

    As usual though, none of these silly bloggers make any attempt actually to verify whether what poor Abraham is saying actually has the slightest contact with reality.

    One such is George Monbiot, a scribbler for the the Guardian, the British Marxist daily propaganda sheet. What is Monbiot’s qualification to write about climate science? Well, like Abraham, he’s a “scientist.” Trouble is, he’s a fourteenth-rate zoologist, so his specialty has even less to do with climate science than that of Abraham, who nevertheless presents himself as having scientific knowledge relevant “in the area.”

    Then of course there’s the little matter of his pretense at Lordship.

    Just saying.

  27. lucia (Comment #82341),

    Just capture screen images and post them as image files (.jpg or whatever). That will preserve the color, fonts… everything.

  31. Nyq, solar cookers work better on sunny days because more energy gets through to the surface of the earth. What you seem to be missing is that the 324W/m^2 figure is the amount claimed to be reaching one square meter of the earths surface.

    If you took a 1 m^2 square box with a 1 side open to the sky, Trenberth claims 324Watts would be pouring into that opening. If you pointed the opening at the sun, you would get an extra 198W.

    But you would get that 324 Watts whether the sun was shining or not.

  32. Bruce: Nyq, solar cookers work better on sunny days because more energy gets through to the surface of the earth.

    True but even haze can reduce the effectiveness and (on the other hand) solar cookers can still work well on sunny days outside of summer or at higher lattitudes.

    A hot cloudy day is not so great for your solar cooker compared with a cooler but sunnier day (for example later in the year or further from the equator).

    Imagine a cloudy day in a hot country. Lots of radiation coming down from the sky to the ground and yet solar cookers don’t work so well. Do you really conclude that there can be no such thing as a warm cloudy day or that there is no longwave radiation when it is cloudy?

  33. Nyq, “Imagine a cloudy day in a hot country. Lots of radiation coming down from the sky to the ground…”

    From the sun. Not from backradiation. The days are long, the ground absorbs solar radiation and heats up. The nights are shorter. It cools but not by much.

  34. Just noting that at the height of the day, solar energy can be coming in at the equator at close to 1,500 W/m2 if there is no cloud or hazy conditions.

    The surface temperature, of course, does not increase by anything like the solar irradiance data says it should be. At this time of the day, thermals and evapotransporation are very strong and solid earth surface is absorbing the radiation and emitting it in the atmospheric windows at such a high rate that it does not affect the air temperature by much.

    Solar ovens are another story. You can cook a hotdog fast enough at 1,500 joules/m2-solar-oven/second.

  35. Bruce: ‘From the sun. Not from backradiation. ‘

    Yes which is why I chose that example. However your objection to the existance of back radiation (the unfeasibleness of solar cookers) works for cloudy days. Yet we know that the radiation is there for cloudy days – consequently we can’t reject its existances based on the unfeasibleness of solar cooking.

  36. I am having trouble utilzing star light in my solar cooker. Does this mean stars don’t exist either?

  37. SteveF (Comment #82330)
    September 27th, 2011 at 5:01 pm

    Mosher,
    “bug eyed monktopus”

    Unkind and not needed; clearly Mr. Monckton has some health issues. No need to poke fun.

    Well said. That was a very low blow.

  38. Wouldn’t we all agree, if you can measure radiation from something (even a diffuse source) that would be a sufficiency condition for establishing that it exists?

    … using a solar oven to heat up meat to the point it cooks is just a very crude form of radiometer if you think about it.

    They also have radiometers that measure direct radiation and ones that are designed to measure diffuse radiation.

  39. Carrick (Comment #82342)
    September 27th, 2011 at 6:23 pm

    [stuff deleted]

    Then of course there’s the little matter of his pretense at Lordship.

    Just saying.

    If you read the article, the issue is not Monckton’s peerage (aka, Lordship), it’s the right of hereditary peers (of which Monckton is not the only one contesting the matter) to sit as voting members in the House of Lords.

    The letter, sent by David Beamish, clerk of the parliaments, to Monckton last Friday and now published on the Lords’ website, states: “You are not and have never been a member of the House of Lords. Your assertion that you are a member, but without the right to sit or vote, is a contradiction in terms. No one denies that you are, by virtue of your letters patent, a peer. That is an entirely separate issue to membership of the House. This is borne out by the recent judgement in Baron Mereworth v Ministry of Justice (Crown Office).”

    Just saying.

  40. Bruce:
    Solar ovens work because they are able to focus sunlight from a point source. The backradiation cannot be focused in the same way because it is diffuse.

  41. Clearly, were any accomplished caricaturist to render the chimerical “monktopus”, and fail to incorporate, nay enlarge and highlight, those very elements of the good Lord’s visage, which we cannot fail to find most endearing–those limpid pools of blue, that languish beneath his ever arching brow–we would surely reproach his efforts, would we not?

  43. Bruce asks: “Diffuse? Does that matter?”

    Of course it matters. Put your solar cooker under a 100 W light bulb, you get heat at the focus. Put it outside on a cloudy day — where the diffuse all-sky flux is a lot more than 100 Watts — and you get nothing.

    That’s because most of that outdoor flux is coming from off-axis and therefore the cooker won’t focus it. And the tiny fraction of the all-sky flux that is on-axis is considerably less than 100 Watts.

  45. Bruce:
    “If you took a 1 m^2 square box with a 1 side open to the sky, Trenberth claims 324Watts would be pouring into that opening.”

    That’s about right. The opening would have 324 Watts coming in from the sky, and roughly 390 Watts going out from blackbody. Which means, in the absence of solar (such as at night) the box would be cooling, and in the presence of solar, in the day, the box would be warming.

    The 324 W m^-2 corresponds to a blackbody temp of about 2 C, which is the temp of the mid troposphere, from whence cometh the radiation we’re talking about.

  46. Does anyone have an algorithm or program that allows one to calculate the solar flux at any latitude at any time of the year?
    I would rather like to ‘walk’ the Pacific temperature, using small Islands and Atolls, that are at different latitudes.
    Ideally, one would pair Islands that are equidistant from the equator and examine their daily/seasonal variation in temperature with light flux.

  47. KAP, that depends on the opening angle of the box.

    324 watts is the integrated number over full sky, no?

  48. Can you build a solar oven type apparatus and point it at the backradiation of 324W per m^2 and cook food with it?
    .
    It is not so hard to imagine how to get the answer. Make the following experiment:

    – Take a cylinder (use a material with good IR absorptivity) Height 1 m , Circle surface 1 m²
    – Set it on the floor
    – Put an IR detector at the bottom of the cylinder
    – You will read much less than 324 W/m². Why? Because you will get only those IR photons which arrive approximately parallel to the axis of the cylinder.
    – Now this very small fraction of parallel photons can be useful because you can collect them in a parabolic IR reflector to concentrate them on a small surface.
    – Choose the temperature you want to attain in your cooker and its size
    – Compute the power necessary to make your cooker cook (S-B law)
    – Compute the necessary concentration factor for the parabolic IR reflector
    – Deduce the size of the parabolic IR reflector (It will be so huge that you won’t be able to get the photons parallel at this scale)
    – Conclude

    Among others, avoid confusing wave lengths and temperatures. Even if it seems incredible, microwaves, with photons much less energetic than IR can make water boil and IR has been known to melt steel.

  49. Carrick: “I am having trouble utilzing star light in my solar cooker. Does this mean stars don’t exist either?”

    What effect do you claim for star light? Do you claim 60% more energy than the sunlight that reaches the earths surface?

    198W/m^2 in from sunlight (after clouds etc) and 324W/m^2 backradiation.

    For the purposes of climate, star light does not exists.

  50. Carrick: “using a solar oven to heat up meat to the point it cooks is just a very crude form of radiometer if you think about it”

    You can’t heat anything up with the mythical backradiation. Therefore for the purposes of climate it does not exist.

    Remember, it is 60% more energetic than solar energy that reaches the earth.

    Even if the “Backradiation Solar Cooker” is less efficient it should cook meat because there is supposedly 60% more of it.

  52. TomVonk: “Now this very small fraction of parallel photons can be useful because you can collect them in a parabolic IR reflector to concentrate them on a small surface.”

    Ok. Make the cylinder longer and put a mirror finish on the inside.

    Say 3m long. How much of that incoming IR is now available to the reflector?

  53. Bruce:

    You can’t heat anything up with the mythical backradiation

    It can be empirically measured so “mythical” is probably the wrong choice of words.

    The point with respect to starlight is your radiometer (aka solar oven) is pretty insensitive, if the threshold for detection is cooking meat.

    If you don’t understand the difference between diffuse and point-like radiation—well I wouldn’t admit to that in any forum, were I you, where you are trying to make an argument about the existence of one measurable quantity and the absence of another another equally measurable quantity.

    I think you need to reframe this either as “back radiation doesn’t matter” or “back radiation is impossible because of Bruce’s personal version of natural law.” And if you think your understanding of natural law outweighs measured observables, you should state that too.

    (All of these are to a degree “putting your foot in it”, but I think you already started that process by bringing up solar ovens.)

  54. Bruce.

    you sound like Goddard. If you are not Goddard, let me suggestthat you contact Guiness book of world records. You just out did him and are now the proud owner of the internet’s stupidest comment.

    You deny that downwelling IR exists. When you are shown measurements of it, you’re response is wonder whether solar companies are interested in harvesting this energy.

    As if, that is an argument that disproves a measurement.

    By asking what you think is a rhetorical question, you show us all that you are really not interested in observation or facts. You also, forget what Lucia says about arguing with questions.

    Well, in point of fact solar companies are trying to harvest IR.
    and yes, make a panel that would work when the sun went down.

    https://inlportal.inl.gov/portal/server.pt?open=514&objID=1269&mode=2&featurestory=DA_101047

  55. Lucia, That’s a far cry from simply asking if Kimoto was meant to serve as a ref to the RSS letter.

  56. Carrick, it is mythical if someone claims it has 60% more energy than the amount of solar energy that reaches the earth … and yet is incapable of heating anything up.

    If someone said it was 400 milliwatts/m^2 i would let it pass.

    Mosher: “When you are shown measurements of it”

    Look, they can measure the surface temperature of a distant red giant at 5000K, but it won’t effect anyone on earth will it?

    Talk about dumb statements!

    Mosher, I am giving you the opportunity to prove that this mythical backradiation can do some work … and all you claim is it can be measured.

    As for IR solar panels:

    “The sun radiates a lot of infrared energy, some of which is soaked up by the earth and later released as radiation for hours after sunset. Nanoantennas can take in energy from both sunlight and the earth’s heat, with higher efficiency than conventional solar cells.

    …

    Double-sided panels could absorb a broad spectrum of energy from the sun during the day, while the other side might be designed to take in the narrow frequency of energy produced from the earth’s radiated heat.”

    Yes, they work after sunset … but on the earth’s IR. Not from mythical backradiation.

  57. Bruce,

    You really need to learn a bit of physics my friend. Your comments reveal a lack of understanding of the basics. I am sure you believe what you say and are well intentioned, but that doesn’t make your comments technically correct.

  58. It must be Goddard, or his twin.

    Notice Bruce changes the subject from “does it exist” to can it do work?

    The change in topic of course derives from him being shown observations, but Bruce doesnt care about observations. he has an
    anti theory. Back radiation doesn’t exist.

  59. SteveF, thanks. But I understand enough to know when someone claims free energy to the tune of 60% more energy than solar is falling from the sky 24/7 to be suspicious.

    Mosher, comment #3: If it can’t do work, it isn’t real.

    http://rankexploits.com/musings/2011/monckton-post-for-comments-on-issues-other-than-paragraph-3/#comment-82299

    But thanks for the laugh about the IR solar cells needing to point down to work. I’m sure the solar panel engineers just forgot about the 324W/m^2 available 7/24 and decided to point the IR side down instead of up. You should email them. They might enjoy a chuckle at you or their forgetfulness.

  60. And no, I’m not Goddard. Or his twin. I just assume everything the IPCC said is wrong. It usually works out.

  61. And Mosher, don’t change the topic. As I said, you can measure the surface of a distant red giant at 5000K, but that is irrelevant to earths climate. That red giant can’t do work or heat up the earth.

  62. Carrick

    The point with respect to starlight is your radiometer (aka solar oven) is pretty insensitive, if the threshold for detection is cooking meat.

    I had to laugh at this. I’m imagining all the real things in the world that can’t be used to cook meat!

  63. Bruce,

    I understand enough to know when someone claims free energy to the tune of 60% more energy than solar is falling from the sky 24/7 to be suspicious.

    I am not sure what to make of this comment. You can point a radiometer at the sky and measure the net temperature (Roy Spencer blogged about this some time back)… the atmosphere is normally cooler than the surface, but by no means zero. The infrared flux from most surfaces is never close to zero because of black-body radiation (described by the Stefan-Boltzman equation)…. near zero flux would require near-zero absolute temperature.
    .
    There is always heat emitted, but the net direction of flux depends on both what is emitted and what is absorbed. So two parallel flat plates at the same temperature in a vacuum are both emitting and receiving radiation from each other continuously, in equal amount. If you drop the temperature of one of the plates, there is net radiant flux from the warmer to the cooler plate because the cooler plate emits less radiation, while the warmer place continues to emit as before. The emission from the cooler plate does not go to zero, it is just reduced (as calculated by the Stefan-Boltzman equation). If you put an even colder plate on the other side of the cooler plate, then you would see a net flux from the cool plate toward the colder plate.
    .
    As to whether or not there is “free energy” from back-radiation… well, no, the net flux continues to be from the surface to the atmosphere, because the surface is warmer (just like the two plates example above). To generate any useful energy you need to capture/use the net energy flow. Since the net flow is from the (warmer) surface to the (cooler) atmosphere, there is no possibility of “capturing” that back radiation from the atmosphere unless you have a colder ‘heat sink’ that yields a net flow from the atmosphere. So for example, if I had a bunch of liquid nitrogen, I could use the flow of radiant heat from the (warmer) atmosphere to the (colder) liquid nitrogen to generate useful energy, even though I can’t use that same radiant energy to warm a ‘solar oven’…. because the solar oven is already warmer than the atmosphere above it. Net heat flow is always from warmer to colder.

    See black-body radiation and Stefan-Boltzman at Wikipedia for much more information.

  64. I think I need a shower after reading some of the posts on here.I used to think good people outweighed the bad,now I’m not so sure after reading all the viciousness that exists in so-called climate science.

  66. MikeC–
    I knew what you meant by the word “ref”.

    So, let me try again:

    Lucia, That’s a far cry from simply asking if Kimoto was meant to serve as a ref to the RSS letter.

    Could you clarify what you are talking about? I don’t know who Monckton asked to serve as ref to the RSS letter, haven’t wondered about this at all, have written nothing about this possibility at all. What point are you trying to make?

    Agreeing that the word “ref” means “references”, I still don’t know what you are talking about. What point are you trying to make?

  70. Lucia, “I’m imagining all the real things in the world that can’t be used to cook meat!”

    Which of them supposedly put out 324W/m^2?

    SteveF, “You can point a radiometer at the sky and measure the net temperature”

    That is in dispute. As I’ve said, you can measure the temperature of a distant object, but if it won’t do any work, it has no effect on the climate.

    SteveF: “Net heat flow is always from warmer to colder.”

    So? I thought the 400W/m^2 from Mosher’s links WAS the net heat flow? Are you saying it isn’t?

  71. Stef: “Since the net flow is from the (warmer) surface to the (cooler) atmosphere, there is no possibility of “capturing” that back radiation from the atmosphere”

    So … the earth cannot be warmed by backradiation?

  72. Lucia.

    Suggest you move discussion about the misunderstanding WRT
    the emails to the other thread, where MikeC can take it up with whoever is still stuck on trivialities.

    Move this comment as well

  73. Re: STeven Mosher: Well, in point of fact solar companies are trying to harvest IR. and yes, make a panel that would work when the sun went down. https://inlportal.inl.gov/portal/server.pt?open=514&objID=1269&mode=2&featurestory=DA_101047

    Phew! At last the key existential question (if it exists then can I cook with it?) has been answered affirmatively.

    Meanwhile – Global Financial Crisis: Is it real? Sure there has been lots of talk over the past few years about the GFC but can I heat a tin of beans with it?

  74. SteveF :You can point a radiometer at the sky and measure the net temperature (Roy Spencer blogged about this some time back)”

    “Most IR thermometers have a maximum measuring distance of approximately 100 feet (30 meters), depending on atmospheric conditions.”

    “The infrared spectral range is 0.7 to 1000 μm, the range for wavelength in which infrared radiation is transmitted. For cost reasons, IR thermometers generally operate under 20 μm. Most of the IR thermometers that we carry have a spectral response of 8-20 μm.

    This range is used because it is minimally effected by CO2 and H2O in the atmosphere.”

    http://www.coleparmer.com/techinfo/techinfo.asp?htmlfile=IRTherms_faq.htm&ID=377

  75. Nyq: “Phew! At last the key existential question (if it exists then can I cook with it?) has been answered affirmatively.”

    If you read my reply you might notice that the IR is being captured from the earths surface.

    I’m sure they would point it at the sky if their actually was IR to harvest.

  76. Bruce

    So … the earth cannot be warmed by backradiation?

    No. Because in the 1-d model, the surface is warmer than the air, the the 2nd law requires is the back radiation, outlined in blue, must be less than the forward radiation, outlined in red.

  77. The world according to Mosher… take a premise of a larger point about bias which nullifies her entire argument on paragraph three and trivialize it… that’s as bad as when Mosher tried to fake those sat photos in the Peterson discussion…

  78. Bruce, move up hand up. Notice that you are displacing the whole weight of the atmosphere above you.
    Heavy isn’t it?

  79. Now that we’ve ruled out its ability to warm the surface, or cook with, or power solar panels, when would this mythical back radiation ever have an effect?

  80. Bruce–

    Lucia, “I’m imagining all the real things in the world that can’t be used to cook meat!”

    Which of them supposedly put out 324W/m^2?

    I suspect it would be as difficult to devise a cooker based on the waste heat taken from the skin surface of person running at 10 miles per hour.

    To estimate how much energy is available, I found this:
    The simple answer is 1 calorie = .001162 watt hours.
    1 nutritional calorie (1000 calories (go figure)) = 1.1620 watt hours.

    I also swagged that a person burns about 100 nutritional calories per mile. So, if they run 10 mi/hr, the burn

    10^2 cal/mi *10m/hr = 10^3 cal/hr. This is 1162 watts-hours/hour or 1162 watts. (Wow!)

    The heat must escape through their skin and is lost through convection and evaporative cooling. But it’s there.

    To estimate the runner surface area, I will assume they are cylindrically shaped: 6 ft tall with a circumference of 50″. Their surface area is roughly 2.3 m^2.

    Taking the ratio of heat lost per unit area, I get 500.W/m^2. Yet as far as I am aware, no one has designed a cooker to cook meat drawing energy from the surface of running boy’s skin. Thereby possibly proving that runner don’t exist. Or don’t generate waste heat. Or something.

  81. Bruce says:

    “I just assume everything the IPCC said is wrong. It usually works out.”

    Your physics are not so good, Bruce. But your pronounced methodology is sound, usually.

  82. for what it is worth, why not just ignore the self-styled Monckton of Brenchley? Yes, he can call himself by that style if he wishes, but it does sound pompous to an educated Brit except in very limited situations, such as replying to his Sovereign’s request for his presence at a state occason. He is not a member of the House of Lords, if that matters, because the previous government befuddled the rules. Before 1500, to be a baron was to be a big shot – the equivalent of a Buffet or Jobs. Now…who cares…so why do you guys rise to the bait? Why do you stoop to play by his rules?

  83. Lucia, a Mars ‘King size’ bar has the energy content of 2 MJ, which by some strange coincidence is the same as a stick of dynamite (2.1 MJ).
    However, do not try to blow up anything with a Mars Bar.

  84. Doc–
    Surely, if a Mars ‘King size’ bar has cannot blow anything up, then those calories can’t count. If Iiked Mars bars, I’d go gobble one up right now knowing that can have no real effect.

  85. diogenes–
    Are you suggesting we should call Monckton something other than Monckton?

    I don’t waste time worrying about Monckton’s claims about his title or status in the house of lords.. As far as I’m concerned, calling Monckton, Monckton is sort of like calling mosher mosher or Obama, Obama. I guess I’m remotely aware that Monckton might not be a last name (or maybe it is?) I don’t care either way.

  87. Bruce–
    There is no inconsistency. Those references are suggesting the average for a person going about their daily business. People generally don’t run at 10 miles/hour hour after hour after hour. Depending on activity level and size, an adult will burn roughly 1000-3000 calories a day. (Slenderizing diets advise women to eat between 1200 and 1500 calories to lose weight.) The lower level is for very small sedentary people; the high range for larger or more active people.

    When person runs at 10 miles/hour, they can burn 1000 calories in an hour. Find a treadmill with a calorie read out, run at different speeds and note the calorie read out. Also, notice that you will begin to feel warm and sweat if you run at 10 miles/hour. That happens because your body has to do something to get rid of the waste heat you generate burning all those calories.

    Vigorous exercise burns quite large number of calories; that’s why it help ward off excess weight and makes you sweat. But you still can’t cook meat on your skin.

  88. Of course my impression that Monckton frequently write things that look wrong to start out with, but take a long time to untangle is a prejudice in the non-value-judgement sort of sense. That is: based on my experience of having read quite a few of his things, selecting some to evaluate and finding they are wrong, I now tend to anticipate that Monckton’s writing will contain things that look wrong and, on evaluation, will turn out to be wrong.


    .
    I have that with most stuff on WUWT, especially of the ‘another/final nail’ variety.
    .
    It also makes one wonder what induces Watts to persevere in giving His Lordship a platform. Why would Watts do that? An enigma, I tell you.

  89. Lucia, what temperature do you think your skins gets to lucia when running compared to just everyday activity?

  90. Bruce–
    Please remember the rule about rhetorical questions. You must answer your own question and tell us what point you think your question and answer make.

  91. Ok. Skin temperature falls.

    “The reduction of skin temperature during exercise was the same throughout the year, although sweat rate was significantly higher in summer than in winter. In coloured thermographics of the skin temperature distribution during exercise of both 50 and 150 W at 10 or 20 degrees C, the skin temperature began to decline immediately at the onset of the exercise. Increased work intensities reduced skin temperature. The results suggest that fall in skin temperature during initial exercise was not due to increased evaporative cooling but to vasoconstriction, probably caused by non-thermal factors.”

    http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1478977/

    So I suspect your human body food cooker only does 50-60W/m^2.

  92. “Bruce (Comment #82445) September 28th, 2011 at 5:12 pm

    Lucia, what temperature do you think your skins gets to lucia when running compared to just everyday activity?”

    Skin temperature depends on the external air temperature, humidity and size of an individuals core body.

    The core body size can shrink to only the brain, heart, lungs, kidneys and liver. We, when cold, shout off pretty much everything else. Only a trickle of blood goes to the intestines, muscles and skin.
    When exercising were get rid of heat in all manner of ways, we ehale a lot of heat through panting, just as dog do, the amount of water loss is very high.
    We radiate heat, with blood flow directed to the surface of the skin to radiate heat. Sweating, the use of evaporative cooling is very efficient. The specific heat capacity of sweat is very high, 1 liter of water, in the form of sweat, is about each kilogram of water vaporized is about 2.25 MJ of energy.
    So for Lucia to convert her Mars Bar into something other than lipid, she needs to sweat out 1.2 liters of fluid.

    We have a bog standard hemoglobin, but Penguins have a very nice one. My hemoglobin is thermally neutral on bind and releasing oxygen. In Penguins, there is a strong heat coefficient. The binding of oxygen is highly endothermic, so the lungs are chilled by loading their hemoglobin with oxygen. When oxygen is given up at the periphery, i.e. its poor feet, it heats them up. They also have leaky mitochondria and so make more heat that way.

    As to what you actual skin temperature is, well, depends if you have acclimatized to the external temperature. Large mammals, who do not have air conditioning, change their long term physiology and have different steady state skin temperatures. In cows in the UK the skin is maintained on all but the coldest days at about 31 °C but in South Africa/Texas it climbs to about 34 °C.

  93. Bruce,

    when would this mythical back radiation ever have an effect?

    Aside from Lucia’s prohibitions on rhetorical questions, you should really, REALLY get some basics down. I tried in my earlier comments above to point you in the right direction… most everything (above absolute zero temperature) radiates energy all the time. It is a physical reality that has been completely understood for about 100 years. Once you understand that very well known and many-times confirmed principle, you will be able to see what people are talking about WRT to ‘back-radiation’.
    .
    BTW, that name (‘back-radiation’) is not well chosen; there is always net flux from warmer to colder, there is never any net backwards flow of energy from cool to warm.
    .
    When you use a word like ‘mythical’ for something that is known and understood by most everyone who has actually studied science or engineering a bit, it 1) kind of puts people off, and 2) makes people discount most everything else you say. Do yourself a favor and learn a bit of physics.

  94. Bruce–
    This isn’t rhetorical: Why do you think reduced skin temperature at the onset of exercise means the body doesn’t heat at greater rate running– say a marathon– than when sitting around in a comfy chair? (Or do you think that?)

  95. SteveF:

    there is never any net backwards flow of energy from cool to warm.

    Not without work being done anyway. 😉 That’s Clausius’s Principle, though it of course deals only with “bulk materials” coupled with classical fields at high enough temperatures that e.g. quantum effects can be ignored.

    I believe the confusion by some laypeople is simply over the meaning of the word “net”, and by other’s the difference between macroscopic and microscopic physics: Namely that thermals photons can be radiated and can transfer heat between any two molecules or atoms. It doesn’t carry a card listing “allowed atoms and molecules it can couple to.” It’s only when you average over a lot of thermal photon exchanges that you get a net transfer of heat from the hotter to the cooler object.

    Neven:

    It also makes one wonder what induces Watts to persevere in giving His Lordship a platform. Why would Watts do that? An enigma, I tell you.

    What’s wrong with Watts giving him a platform? Let him talk.

  98. SteveF

    “BTW, that name (‘back-radiation’) is not well chosen; there is always net flux from warmer to colder, there is never any net backwards flow of energy from cool to warm.”

    Is it? For what ever reason, make a hollow sphere out of plutonium, internal diameter 50 cm, 7850 cm^2.
    Now the plutonium will be warm to the touch, 35 degrees, radiating at a peak of 5 micron ad a total of 400 Watts.
    Make an inner sphere out of something like polonium/bismuth. This sphere is attached by insulated wires and is only irradiated by the plutonium inner surface.
    This inner sphere has a surface area of 10 cm^2.
    As this sphere gets 400 watts, at equilibrium, it must reach a temperature of 1,360 degrees.

    So, the lower temperature outer sphere at 35 degrees, heats the inner sphere to 1,360 degrees.

    No problem with that is there SteveF?

    One could of course make the inner sphere even hotter by shrinking it. Make it a mote and the temperature gets hotter than the center of the sun.

    No problem with this model is there?

  99. “Carrick (Comment #82452) September 28th, 2011 at

    there is never any net backwards flow of energy from cool to warm.

    Not without work being done anyway.”

    i love this old chestnut.
    Carrick. HEAT is not energy nor is it a form of energy; HEAT is a description of the movement of energy between connected reservoirs.

    It is easy to make heat flow from cold to hot in any non-equilibrium system. The only thing that is important in that entropy in the whole system rises.

    I take a large lump of Nickel-Iron, nice and cold, and drop it in Earths gravity well. When it hits the ground you will find that it will be somewhat hotter than the atmosphere, even though all its collisions have been with air colder than 15 degrees.
    So the cold air heats it to melting point.
    Seen it done.
    What work has been done?
    None in the conventional sense.
    Has entropy been increased? Oh yes indeed.

  100. Carrick, my mind isn’t closed. I was at least curious as why this mythical backradiation has no effect on anything.

    Maybe Mosher or you will phone the solar company and tell them the IR side of their solar panel should be facing the sky and pick up the 324W/m^2. Make a transcript. I’m sure it will be hilarious.

  102. swine Lucia, it normally takes people days to work it out.

    Do this one. We take another Nickle-Iron rock at 4K, and drop it into the gravity well of a small black hole, just enough to put it into orbit.
    Changing the lumps linear vector into a circular orbit will of course increase its temperature. However, the black-hole has no temperature.
    So we have heating of something without the exchange of anything except gravity. Heating from an infinitely cold body to a warm one.

  104. bruce:

    Carrick, my mind isn’t closed. I was at least curious as why this mythical backradiation has no effect on anything.

    If your mind wasn’t closed you wouldn’t use words like “mythical backradiation”. Seriously other than the value in baiting people, I wonder sometimes why you post.

    As far as I can tell, you don’t really know anything (parroting what you’ve seen on other blogs doesn’t count), have much to offer of technical value, and are totally uninterested in learning anything new.

    So this puts you in the category of two-bit troll for me.

  105. Name calling instead of evidence Carrick is pathetic. But that is your style.

    If it can’t do work, but it is asserted that it supplies 60% more watts/m^2 than solar energy that reaches the earth, then it is mythical.

    The claimed energy per month is more than Phoenix gets in July per month from the sun.

    Get real.

  106. Lucia: I also swagged that a person burns about 100 nutritional calories per mile. So, if they run 10 mi/hr, the burn 10^2 cal/mi *10m/hr = 10^3 cal/hr. This is 1162 watts-hours/hour or 1162 watts.

    They measure Tour de France cyclists putting out 300+ watts of work for hours and 500+ watts for 10-20 minutes. They don’t measure waste heat but efficiencyc is said to be ~25% which would be 900 watts of waste heat for hours and 1500 watts for 10-20 minutes – which fits with your swagging.

  107. diogenes–

    Are you suggesting we should call Monckton something other than Monckton?
    I don’t waste time worrying about Monckton’s claims about his title or status in the house of lords.. As far as I’m concerned, calling Monckton, Monckton is sort of like calling mosher mosher or Obama, Obama. I guess I’m remotely aware that Monckton might not be a last name (or maybe it is?) I don’t care either way.
    unquote

    If I met the President of the US, I’d call him ‘Mr President’ as would most USians. Why? (oops, rhetorical, sorry, let me answer that) because it is polite to do so,even for one of Her Majesty’s subjects to whom the titles of a foreign power mean nothing. If I called him ‘Barack’ I would be impolite, overly-familiar, presumptuous. If I called him ‘Obama’ then I would be, in my idiolect, showing hostility unless we were fairly well acquainted (cf President Bush calling the Prime Minister ‘Blair’ as in ‘Yo, Blair!’).

    Lord Monkton is a viscount (a peer but not a member of the House of Lords) and the polite form of address is ‘Lord Monckton’. For citizens of countries where it matters to assert sturdy independence then ‘Mr Monckton’ would demonstrate the independence while still keeping the conversation on a polite level. (Incidentally, a knight is addressed as Sir [Christian Name], not Sir [Surname]. I have always thought that ‘Sir Julian Flood’ had a certain ring…. But I digress.).

    Does politeness matter, even when it’s just a foreigner who is being addressed? Yes. The politer one is in academic discourse then the more devastating the criticism. ‘Monckton is talking through his hat’ hints at dislike which might spill over into lack of objectivity in the science. ‘Thus we have demonstrated that Lord Monckton has made several errors in his calculations which results in their being meaningless in any scientific sense’ is polite, and cuts deeper than any rudely expressed disagreement. The strange people who insist on calling Prince Charles ‘Chuckles’ have failed to understand this simple point.

    Punctilious politeness is best: it makes it possible to be much ruder.

    JF

  108. Bruce,

    Here’s an interesting and simple experiment which might help you understand what’s going on with this back radiation business:

    In a shaded room, place two parabolic reflectors such that the two reflectors focus each other’s emitted radiation onto the focus of the other (I could do better with a diagram….) So if you put a bulb at one focus, the emitted parallel rays would be caught and focussed onto the second focal spot by the other reflector. Take the bulbs away, we don’t need them.

    Now put a thermometer at one focus and let everything settle down. Let’s say it reads 15 deg C, which is what we’ve got outside at the moment. Then place a block of ice at the other focus and watch the temperature as recorded by the thermometer at the first focus.

    If you don’t have the equipment, let me tell you what happens. The temperature as measured by the thermometer falls.

    Now, think about why and you’ll be a long way to understanding the concept against which you are railing.

  109. I understand that many Americans still have hangups about British nobility, just as the majority of Europeans have hangups about Americans per se (as in “they are so dumb they cannot tell the difference between Sweden and Switzerland”).
    But suffering from Garves’ Disease, which make your eyes bulge out, is not a sign of stupidity.

  110. Do this one. We take another Nickle-Iron rock at 4K, and drop it into the gravity well of a small black hole, just enough to put it into orbit.
    Changing the lumps linear vector into a circular orbit will of course increase its temperature. However, the black-hole has no temperature.
    So we have heating of something without the exchange of anything except gravity. Heating from an infinitely cold body to a warm one.

    Argh!
    Docmartyn you should read some Hawking or general relativity. A black hole small or big has not only temperature but it has entropy too. It is actually the highest entropy object in the universe.
    It radiates and evaporates.
    Besides when you “drop” something near to the horizon of a black hole, its “orbit” is far from anything circular (the space-time metrics is also far from euclidian) and unless you provide a huge amount of energy to keep it on a stable “orbit”, it will accelerate incredibly and emit gravitational waves untill it crosses the horizon. After that it will be ripped apart in a finite time by the gravity gradient of the singularity but you won’t know anything about it.
    .
    Bruce
    .
    TomVonk: “Now this very small fraction of parallel photons can be useful because you can collect them in a parabolic IR reflector to concentrate them on a small surface.”

    Ok. Make the cylinder longer and put a mirror finish on the inside.

    Say 3m long. How much of that incoming IR is now available to the reflector?

    You were on the right track here. Pity that you didn’t continue.
    With 3 m height it would be less than with 1 m. And in increasing the height it would tend to 0!
    So as you see the problem is not that there are not enough photons – there are 300 W/m² full of them.
    The problem is that there are not enough which are strictly parallel to each other.
    And as you can concentrate only those that are parallel (which would allow to achieve a power high enough to brew tea), you have a problem with making an infrared cooker.

    However it should be clear by now that the difficulty to cook meat with IR is not the lack of photons. The difficulty is that there is a negligible amount of them which are parallel to each other.

  111. Bruce, people with minimal science education really shouldn’t try and write lectures on radiative physics and post it on a blog where there are lots of technical people, then not be willing to listen to reason about it. (I did include enough information that if you were intelligent and open minded, you would be able to spot why there is such a thing as back radiation, but you prefer to sound like a broken record…boring.)

    I think your behavior fits the definition of “troll”, but perhaps you have a better choice of words. Anyway I labeled you a troll for your behavior. And it’s fundamentally no different than you describing my (accurately) labeling of you as a troll as pathetic, or claiming that SteveF appealed to authority.

    I obviously find this boring and unproductive, so I’ll let others deal with your pure silliness over “mythical” back radiation.

    Bye.

  114. Bruce:
    “198W/m^2 in from sunlight (after clouds etc) and 324W/m^2 backradiation.”

    It is not a question of diffuse or direct radiation. That’s what lenses are for.
    198 W/m² is an average. You do not use a solar oven at night. With 1000 W/m² daily maximum and a 3 m² parabolic mirror you have 3000 W available. You would need to concentrate 8 – 10 m² of ‘back-radiation’ to get 3 kW. This is where the problems start.

    You just don’t have the materials to build a lens that will concentrate radiation of 4 – 30 µm wavelength. I don’t think there is even a material that works as a lens on the whole sun spectrum of 0.3 – 4 µm. It is not a physical but a technical question why no one uses ‘back-radiation’ as an oven.

  115. lucia wrote

    quote
    Julian Flood–
    I’m not pro-actively calling Monckton Lord Monckton unless he pro-actively addresses me as Dr. Liljegren.
    unquote

    As he should. Note, however, that he hasn’t complained about people not using his title, nor, to my knowledge have you, it’s just my preference for politeness in this GW discourse. My first doubts about AGW were occasioned by the ill manners of the pro warmists — no-one, I thought, would be that aggressive if he was sure of his ground. Better to be too polite than even a little rude: it doesn’t mean you have to like or even respect the chap, it’s just a way of keeping things at an intellectual level and not an emotional one.

    Would you prefer to be addressed by your academic title? After all, you _have_ earned it.

    AMac wrote

    quote
    Concerning back radiation, Ice-at-the-parabola’s-focus is the clearest thought experiment yet. Even without a diagram.
    unquote

    The joys of a grammar school education.

    Julian Flood
    (Who is going to have trouble spelling both names — I always get them wrong.)

  116. You would need to concentrate 8 – 10 m² of ‘back-radiation’ to get 3 kW. This is where the problems start.You just don’t have the materials to build a lens that will concentrate radiation of 4 – 30 µm wavelength
    .
    Hmmm. I generally like to help people with understanding but now I begin to ask myself like Carrick if it still makes sense in the context of this specific question.
    I have tried to explain that whether parabolic mirror or a lens, it is necessary to have PARALLEL rays.
    So what counts is not how many photons fly around (e.g power in W/m²) but how many photons are PARALLEL along some direction.
    And the downwelling infrared has so to speak no parallel photons coming to a surface of 1 m² while the sun light has plenty of them and that makes the whole difference.
    Therefore the IR photons can’t be concentrated by any device and one can’t cook meat.
    This has nothing to do with technology and everything with trivial gemoetrical classical optics.
    Period.

  117. Andrewt and lucia, do you think people are made out of copper? I don’t. There is no instantaneous release of this waste heat out through the skin. The blood has to carry it to the skin etc etc.

    The body does its best to maintain its skin temperature in a very narrow range.

    100W seems reasonable for heat radiating from the skin at around 37C body temperature. I think it is reasonable to ask what temperature the skin will be to radiate the 1100W?

  118. TomVonk, the solar panel reference that Mosher provides suggests they think they can capture the IR radiating from the surface. I doubt that is a point source.

    TomVonk: “With 3 m height it would be less than with 1 m. And in increasing the height it would tend to 0!”

    Strange. The CLAIM was that there is now 324W/m^2 if IR impacting on the surface. To me it would not matter if how deep a box is as long as the opening is 1 m^2, there should be 324W going into that opening. If the sides are mirrored, the IR should bounce down to the bottom getting closer to parallel.

    Why would you claim 0 if the box is taller unless you think the sky is a point source?

    My point stands, the 324W/m^2 seems incapable of doing work. Therefore it isn’t real.

  119. Carricks standard response when he loses “Bye”

    Come on people. Its 60% more energy than the sun (averaged). More energy than Phoenix gets from the sun in July per day (which in itself is awfully suspicious).

    Surely it can do work of some kind. 8 or more hours at night with lots of energy pouring down to be used.

    I still get a laugh thinking of Carrick or Mosher calling the solar guys and telling them to point their IR solar panel at the sky at night. That would be a laugh.

  120. Bruce–
    Your novel theory about how to disprove backradiation by solar cooker has inspired me to write a post. I will be posting it later today. 🙂

  121. Julian

    Note, however, that he hasn’t complained about people not using his title, nor, to my knowledge have you, it’s just my preference for politeness in this GW discourse.

    I happen to disagree that it is necessarily more polite to constantly use titles. I agree it was the custom during the Victorian era, but it no longer is. You are correct that I have not complained that people don’t call me Dr. Liljegren, and Monckton has not complained that people don’t call him Lord Monckton. I think either of us requesting such a thing would be… hmm “in bad form”.

    What I would suggest is if your belief is politeness requires addressing people with titles, you simply adopt the practice uniformly and use it without regard to whether they make any request or not. In that light, if you pro-actively call Monckton Lord-Monckton, you should call me Dr. Lucia Liljegren (to distinguish me from Dr. Jim Liljegren, and that you refer to Gavin ad Dr. Schmidt, the Pielke’s, Dr. Pielke with the appropriate jr. and sr and so on. Of, if you call Monckton Monckton, you can go on calling me lucia.

    My standard is: if someone is saying they believe titles should be used as a matter of politeness, they should apply the standard consistently and uniformly.

  122. Lucia, how about vaccum tube solar water heaters?

    Will they work at night powered by backradiation?

    “Q: What happens on cloudy or rainy days?
    Our Solar Vacuum Tubes are specifically designed to capture UV and Infrared rays which pass through the clouds, meaning they still generate heat under full cloud cover.”

    http://www.northamericansolar.ca/faq.html

  123. Then place a block of ice at the other focus and watch the temperature as recorded by the thermometer at the first focus.

    If you don’t have the equipment, let me tell you what happens. The temperature as measured by the thermometer falls.

    .
    This seems to be related to the “solar refrigerator” effect (a solar oven turns into a refrigerator at night). I can’t say that I fully understand it, but I can try and have a go:
    .
    In the situation where there is no ice, the two lumps of air at the foci send a bit of radiation to each other through the parabolas. Everything is in equilibrium, and they stay at ambient temperature.
    .
    In the situation where there is ice in focus 1, the lump of air at focus 2 still sends a bit of radiation to focus 1 through the parabola, but now this energy is not returned (it is absorbed by the ice), and thus there is a deficit of energy – and the air at focus 2 cools.
    .
    This is true of all the lumps of air in the room (they all send a bit of radiation to the ice cube), but more so for the air at focus 2, because the parabola means that it sends proportionally more of its radiation towards the ice.
    .
    (How wrong am I?)

  124. Bruce maybe I can help explain it a little better.

    The back radiation is the ‘response’ to the earths radiation to space. It’s ‘temperature’ will always be lower than the temperature of the body it is reflecting from (the earth in this case). Think of the molecules up there as little reflectors.

    The real question that you seem to be asking is if you can take 70Ëš heat and concentrate it to 300 degrees or so. The answer sadly is no, unless you have a very cold heat sink and use something like a Carnot engine to focus the heat.

  125. Genghis, the solar vacuum tube people think you can get to 300C from sunlight of which 51% or so is IR.

    “The temperature inside a tube can reach over 300°C while the outside remains at room temperature.”

    Mosher found places that had 400W/m^2 of IR beaming down 24 hours a day.

    Why won’t the vacuum tube warm up at night?

    I mean one of you commenters should be able to come up with an experiment proving this stuff is real instead of excuses why it won’t do anything.

  126. Haha,
    This whole starlight BBQ discussion is making me hungry. I’ll have the Tri-tip please.

    Seriously though the physics here is really pretty simple. There is a net heat gain on the surface during sunlight hours and a net heat loss to space at night. The solar cooker simply amplifies this effect. Our atmosphere is a nice warm blanket that essentially insulates us from both extremes of heat and cold via BGR scatter. A comparison between conditions on earth and the moon make this abundantly clear. In the absence of an atmosphere the solar cooker would be exponentially more effective during sunlight on the moon and it would be equally effective as a deep freezer at night.

  127. Bruce you can look at all the sites on that page and find that they all show a measured downwelling IR. Downwelling IR is real. Downwelling IR is measured and observed. Downwelling IR measured, matches Theory. When we build sensors that look up at the sky to see hot objects we have to account for this downwelling IR.

    That you don’t understand the basic physics is your problem.

    If you understood the basic physics you could move on to a smart argument. That smart argument would be about the sensitivity to doubling C02. That smart argument would be about feedbacks. Instead, you think you are convincing people by remaining unconvinced. The only thing you convince us of, is your lack of understanding, both of physics and of where the real argument is.

    So here is a question:

    A candle gives off IR.
    I can measure that IR with an IR detector.
    If I put C02 between the candle and the detector, will the detector still see the candle?

    Predict and explain your prediction.

    You wont.

  128. Bruce,

    “Genghis, the solar vacuum tube people think you can get to 300C from sunlight of which 51% or so is IR.”

    Sure and they could get hotter temperatures if they could utilize the IR portion.

    Your stumbling block seems to be our ability to capture energy and the difference between temperature and energy. Something can be very hot and have very little energy at the same time.

    This is the problem that solar collectors have. A parabolic dish can concentrate enough solar energy to melt lead but that same amount of heat can barely warm a house in the winter.

    Forget temperature and just figure out the useable energy and it gets easy.

  129. Mosher: “Downwelling IR is measured and observed. Downwelling IR measured, matches Theory”

    But Downwelling IR cannot do work it appears.

    Mosher: “When we build sensors that look up at the sky to see hot objects we have to account for this downwelling IR.”

    I don’t care that it has temperature. Distant starts have temperature. The do not affect the climate.

    Ghengis, they say the IR and UV us used by the collectors as I quoted.

    Ghengis: “Forget temperature and just figure out the useable energy and it gets easy.”

    Bingo.

    The claim is that backradiation warms the surface of the earth. Therefore there must be useable energy. Where is it?

    This is the claim: “The energy emitted back to the surface causes it to heat up more, which then results in greater emission from the surface.”

    Sounds like a perpetual motion machine … except this “energy” is magical. It can only warm the surface of the earth … it can’t warm a vacuum solar collector it seems or they would advertise it works at night (Mosher’s graph shows 400W/m^2 at night).

  130. Bruce, “The claim is that backradiation warms the surface of the earth. Therefore there must be useable energy. Where is it?”

    It is merely conservation of energy.

    Envision the surface of the moon being heated by the sun and then after the moon rotates putting up a reflector over a section of the moons surface.

    Now if you measure the temperature under the reflector and at another spot on the moons surface not under the reflector, which will will be ‘warmer’ ?

  131. The claim is that backradiation warms the surface of the earth.
    Wrong. That is not the claim.

    However, it is good to see you admit you were wrong about downwelling IR being mythical.

    So, you agree. The earth radiates heat to space in IR.
    the atmosphere and its gases, like H20 and C02, reflect/scatter some of this IR downward. That downwelling IR is measureable. The plots I showed you, show the values.

    You are making progress. Now you have to understand the rest. But its good. you are no longer rejecting observations. Im glad for that. so there is a thing we measure called downwelling IR.. and you understand that C02 is not transparent to IR.

    About that experiment.. when you put C02 between the candle and the detector.. what happens? predict and explain.

    you wont.

  132. Ghengis, I don’t doubt that the spot will be warmer for a while.

    I can put my hand over a sidewalk after the sun goes down on a sunny day and warmth will be radiating upwards from the ground. The IR solar project says they can capture that IR for a few hours after sundown. But after the sidewalk cools down there won’t be any IR to capture.

    But Mosher’s graphs show the 400W/m^2 as constant all through he night.

    Do you claim the amount of energy will be constant until then sun is back?

    If we move your reflector (and insulate it) and put it over the sidewalk. And you put your hand between the sidewalk and the reflector, yet it will be warm. But the concept of backradiation claims that even after the sun has been down all night long, if you put your hand between the reflector and the sky you should feel warmth. Nope.

  133. Mosher: “That downwelling IR is measureable. The plots I showed you, show the values.”

    But it doesn’t do anything. It won’t work with the IR solar panel or the solar vacuum tube water heater.

    It is mythical if you assign some value to it higher than the amount of solar energy Phoenix gets on an average July day.

    MosherL “The earth radiates heat to space in IR.
    the atmosphere and its gases, like H20 and C02, reflect/scatter some of this IR downward.”

    Very little. It seems imaginary. It certainly isn’t a constant 400W/m^2 as in your graphs. You should be able to run your solar vacuum water heater all night long. Can you?

    Theories are nice. But not evidence.

  134. Mosher: “If I put C02 between the candle and the detector, will the detector still see the candle?”

    How far apart? And then tell me.

  135. Bruce the graphs aren’t constant.

    The cooler the ground the less upward radiation and hence less downward reflected radiation.

  137. “Bruce

    Bingo.

    The claim is that backradiation warms the surface of the earth. Therefore there must be useable energy. Where is it?”

    Bruce,
    1) heat isn’t energy, it is a measure of the movement of energy.
    2) one can only do work from an energy gap. A pair of heat sinks at different temperatures can be used to extract work. This is why you notice the large cooling towers at power stations.
    3) one could make a collector that would work at night, collecting the back-radiation. However, to make it work your collector would have to be cooled below ambient temperature. The efficiency of a heat engine is a factor of T1/T2, in degrees K.

  138. DocMartyn, so running cold water through a solar vacuum tube collector to warm it up at night should work?

    Why doesn’t any do it? Or advertise it? Or mention it?

  139. If you all ever get Bruce to see the light on this, you will have to explain the other physics thing that has bewildered him since he first learned that his lunch-box thermos bottle would keep hot things hot, and keep cold things cold. Bruce’s question is: “How do it know?”

  143. Mosher, is there an experiment that shows how far away from the candle your IR detector has to be to not see the candle?

    Lets say it was 10 feet.

    Then the downwelling longwave claimed to be backradiation could not be more than 10 feet away right?

    If CO2 doubled … and the distance was now 9.5 feet, that would have interesting implications wouldn’t it?

  145. Don, if you stand in a desert at night as it approaches 0C, and you place you hand facing the sky … do you feel any IR?

  146. Why is it approaching 0C, there Bruce? Now go to the rain-forest. That is all the help I am going to give you.

  149. Lucia,

    http://rankexploits.com/musings/2011/monckton-post-for-comments-on-issues-other-than-paragraph-3/#comment-82422
    .
    I have two observations on this diagram.
    1. The backradiation appears to have a value which equates to 95% of the value given to solar radiation incoming to the system. Do you think this is an accurate representation?
    .
    2. The surface radiation figure of 342 W/m2 is reduced – by the atmosphere – to 235 W/m2. This therefore means that the ‘greenhouse effect’ is worth 155 W/m2. As the IPCC states that climate sensitivity equals 3.7 W/m2,

    http://www.ipcc.ch/publications_and_data/ar4/wg1/en/ch2s2-3.html

    this therefore equates to 2.4% of the ‘greenhouse effect’. Or, if you prefer, 2.4% of 33 deg C, which equals 0.8 C. Do you agree with this conclusion?

  150. Dr. Jay–
    I have sent your comment suggesting my reason for not discussing temperatures in the arctic is dishonesty to spam. I will send all future comments of that nature to spam. I told you that you are free to post data about temperatures in the arctic and explain how this data supports some theory or argument you adhere to. But I am not going to permit you to fill comments with accusations that the reason I do not let you decide which topics I will blog on is because I am dishonest.

  151. That is a very humid desert you picked there, Bruce. That should inform you of something, but it doesn’t. Carry on with your clowning.

  152. Arfur–

    Do you think this is an accurate representation?

    I can think of no reason to dispute it.

    342 W/m2

    Do you mean 390?

    This therefore means that the ‘greenhouse effect’ is worth 155 W/m2.

    I assume you are just taking (390-235)? Anyway, I don’t think I agree with your conclusion. Could you explain your reasoning in greater depths?

  154. Poor Lucia.

    Monckton is your aversion. Why doesn’t Gavin and his cohorts, who have been so wrong about so much, not get similar treatment from you?

    Couldn’t bring yourself to talk to him at dinner. Something deeper going on there.

    The track record of the English gentleman scientist is without peer. They don’t do science for money, or ideology.

    Try to get past your personal difficulty.

  155. MarkR

    Monckton is your aversion. Why doesn’t Gavin and his cohorts, who have been so wrong about so much, not get similar treatment from you?

    They do.

    Couldn’t bring yourself to talk to him at dinner.

    I don’t know why you think this. Monckton was sitting directly across from me at dinner on the evening when this photo was taken

    You can read more here: http://rankexploits.com/musings/2010/iccc-blogging-dinner-at-stefanis/
    In that post, I wrote
    “Pajamasmedia hosted the dinner. I’ve very bad with names, but among the other people I can name Roger Simon and Charlie Martin of Pajamas media, Dan Miller of Heartland, and Richard Lindzen of MIT, Christopher Monckton who sat across the table from me. ”

    I spoke to Monckton during that dinner. I also found myself sitting with him at lunch on other days and spoke to him during those lunches. Other than that, owing to “inter alia” — the fact that I live on the American continent and he does not– I have dined or lunched with the man. So, we do not generally dine or lunch together.

    If you think this issue is a “personal difficulty” in invite you to answer the techical questions above. The correct answers to those questions is entirely unaffected by my or your aversion or affection for anyone.

  156. Come on Don. Evidence. Experiments. Proof. Not just assertions. Or appeals to authority.

    “Don, if you stand in a desert at night as it approaches 0C, and you place you hand facing the sky … do you feel any IR?”

  157. Roy–
    We need to give Monckton some time.
    Nevertheless, it is clear Monckton’s most recent post (i.e. the one linked in my main blog post) did not address the technical points in my discussion of his paragraph 3 to RSS, and avoided other posts I requested he address when I posted in comments at Anthony.

  158. Lucia:

    “We need to give Monckton some time.”

    Agreed, but I watched this post for an hour before commenting. To that was I referring.

    Roy Weiler

  159. Roy–
    Well you know, differentiating or integrating functions can sometimes be a time consuming challenge. Plus, we are in different time zones. We may need to give him more time.

  161. Bruce

    “Don, if you stand in a desert at night as it approaches 0C, and you place you hand facing the sky … do you feel any IR?”

    Please follow the rules of this blog:
    a) provide your own answer to this question.
    b) tell us what point you think your question/answer make.

    Others
    Help Bruce learn how to follow the rules. When he asks a rhetorical question, do not provide your answer or your own analysis based on that answer. Instead ask him to
    a) provide his own answer to his question.
    b) tell us what point he thinks his question/answer make.

    After he does that, you can discuss his notions about what point he thinks he has made.

  162. @ Lucia “I would suggest the answers will reveal that the value of a Planck constant estimated using (18) is no more likely to be correct than the value of one pulled out of a hat.”

    Perhaps so … but talking about people pulling things out of a hat … could you elaborate on this assertion? :

    “if CO2 were to say– multiply by 20 [from its present value], the earth will almost warm enough to overcome an ice age.”

  163. Richard–

    Perhaps so … but talking about people pulling things out of a hat … could you elaborate on this assertion?

    Do you mean elaborate on

    “if CO2 were to say– multiply by 20 [from its present value], the earth will almost warm enough to overcome an ice age.”

    I will say this: That is irrelevant to the question I have asked Monckton. It is also irrelevant to the issue I discussed in my post about the errors in Monckton’s paragraph 3 . In the morning (or possibly sooner), I will move your comment to the ‘side thread’ and I am going to add an update to tell visitors that comments that do not directly address the challenge posted in the current blog post should be posted at the side thread.

  164. Lucia:
    OK, it is 1am in the UK. I am sorry if you think I spoke out of turn. I was kind of expecting a large serious of comments from your regular followers, prior to Monckton commenting or not.

    I have to tell you statements like:
    “differentiating or integrating functions can sometimes be a time consuming challenge.”
    are pretty hot. 🙂

    Roy Weiler

  165. It is not politically-correct to discuss the specifics of back-radiation. It is to be accepted as put forward by Trenberth and Kiehl with no questioning allowed or you are to be considered a non-person.

    But I, as a non-person, would like to know, what is the distance distribution of the origination point(s) for the back-radiation photons. It makes quite a difference if 50% of the photons originated within a few metres (or cms) of the instrument measuring them or if 50% come from the tropopause as Trenberth and Keihl seem to have diagrammed them.

    I would also like to know the distribution of those photons by the last atmospherical molecule which transmitted them? It makes quite a difference if more than 70% come from water vapour at the tropopause or if 50% of them are just blackbody radiation from the 98% N2 and O2 molecules immediately above the instrument.

    I would also like to know how the back-radiation intensity varies through various levels of the atmosphere, from 1 metre to 10 metres to 90 kms.

    I would also like to know what is the balance of up-radiation versus back-radiation for every 10 cms from the surface to 100 metres for example.

    I would also like to know why the temperature of the air at 2 metres only seems to notice about 0.00001% of this back-radiation per second. How does the energy actually impact the temperature level since it doesn’t seem to?

    I would also like to know how the back-radiation actually occurs since the collisional exchange rate of atmospheric molecules is 10^6 faster than than GHG relaxation rate of emission.

    I have at least 10 more questions like this and nobody seems to have to addressed them that I can find.

  166. Pingback: The Blackboard » Monckton: Answer three challenge questions in your own words.

  168. Don, Mosher, Carrick, Lucia, Ghengis, ivp0, SteveF, et alia —

    Pi is the ratio of a circle’s circumference to its diameter. I have drawn and measured many circles, and proven to my satisfaction that the value of pi is approximately 2.71828.

    I defy you to convince me otherwise.

    Without invoking some authority from beyond this thread, or engaging in knavishly complex arguments.

    Although such tactics would only prove my point.

  169. Sorry, lucia. All others have apparently given up on Bruce, so I thought I would was being kind in making another attempt. Besides, I don’t think he knows what a rhetorical question is.

    Bruce is frustrated by science. He has always been on the outside, despite a compelling desire to participate. His great ambition was to be an astronaut. Somehow, he got as far as having an interview with NASA. They asked him why he wanted to be an astronaut. He replied: “I want to be the first man to land on the moon.” They informed him that he would not be the first. He replied: “Then I will be the first man to land on the Sun.” When they pointed out that he would burn up, he replied: “I will go at night.”

  170. Lucia”
    I said serious instead of series, might as well throw all of this in junk pile.

    Rioy Weiler

  171. AMac (Comment #82543),

    I have drawn and measured many circles, and proven to my satisfaction that the value of pi is approximately 2.71828.

    LOL
    Based on this level of accuracy being acceptable to you, I do believe that you may have a future in climate science!
    .
    But seriously, sometimes you can’t save people from their own errors and misinformation. Before participating on climate blogs, I would never have guessed that was the case. “You can lead a horse to water, but you can’t make him drink.” is an important insight, whoever came up with it. Versions of the expression apparently date from at least 1175 in Old English Homilies:

    Hwa is thet mei thet hors wettrien the him self nule drinken
    [who can give water to the horse that will not drink of its own accord?]

    Human folly seems as old as humanity.

  172. Bill Illis, I think ScienceOfDoom has most of the answers, or at least has provided enough information for you to be able to work them out yourself.

    I’d like to know why you want to know the answers to all these questions, and what you think they would tell you. (That’s neither rhetorical nor a question.)

  173. SteveF,

    I don’t think you were supposed to reply to Amac. His post looks suspiciously like a rhetorical question disguised as something else. I am not going to answer him, even though I can testify that I have drawn thousands of circles by hand, and the average pi I calculated was (I am sure, not co-incidentally) 2.71828.

  174. Don Montfort,

    I don’t think you were supposed to reply to Amac. His post looks suspiciously like a rhetorical question disguised as something else.

    I think Lucia sometimes cuts otherwise well behaved commenters some slack about answering rhetorical questions…. asking them? Not so much. 😉
    Rhetorical questions are too often just a way to make other people do a lot of work for nothing. Answering your own rhetorical questions is honest, since you have to lay out your own thinking (or the complete lack thereof), before asking other people to do a lot of work.

  175. Bruce.

    The candle is close enough to the IR detector to be seen by the detector. That is, the detector detects IR from the candle.
    Now, interpose, C02 between the candle and the detector and
    predict. Will the detector still be able to see the candle? explain your answer and why you think what you think.

    If you do not understand or have any idea how the presence of C02 will change or not change the transmission of IR, then you can say that.

    Then. can you imagine any device which is built on the physical principle this experiment would demonstrate?

  176. lucia,

    I have heard that Lord Monckton usually orders the overcooked prawns. Do you recall what he had for dinner? Lighten up lady. You are OK.

  178. Don Monfort–
    He had shrimp. I noticed partly because I am allergic to shrimp (and lobster, crab, prawns and all crustaceans).

    I’m going to move things now. (The reason is merely to keep this thread for technical issues.) 🙂

  179. SteveF–
    Ordinarly, I cut people slack for answering. But in this case, Bruce seems to be asking lots and lots. So I would like people to try to help me house break him.

  180. Correct Don. You don’t feel the backradiation because it is not and never will be 60% more Watts/m^2 than solar energy. It is not and cannot be 324W/m^2 because you cannot think up any experiment to prove it exists.

    And Don, you are as poor a psychoanalyst as you are at coming up with an experiment to prove backradiation exists in the values claimed by Trenberth. But I understand you are frustrated by people not accepting pronouncements by “authority figures” at face value.

  181. Mosher, “Will the detector still be able to see the candle? explain your answer and why you think what you think.”

    Mosher, I cannot answer your question unless you explain all the parameters. I see that I have frustrated too many of you by asking for a simple experiment to prove backradiation exists … so you change topic.

  183. Bruce:

    Mosher, is there an experiment that shows how far away from the candle your IR detector has to be to not see the candle?
    Lets say it was 10 feet.
    Then the downwelling longwave claimed to be backradiation could not be more than 10 feet away right?
    If CO2 doubled … and the distance was now 9.5 feet, that would have interesting implications wouldn’t it?

    #############################

    we are not talking about downwelling IR yet bruce. we will get to that. First, we need to see how much you know about the basic physics of IR and transfer through a gas.

    Here is what you have agreed to.
    1. Downwelling IR exists.

    Now, I am asking you to answer a simple question. Suppose a candle is close enough ( say 3 feet ) to an IR detector that the detector sees the candle. Now change the gas between the candle and detector. put C02 in the way and predict what happens.
    explain your prediction with the physics you understand.

    When you have shown us that you grasp the basic physics of the transfer of radiation through various gases, then we can talk about
    why this is important. But first things first. predict and explain your prediction

  184. Carrick (Comment #82547)
    September 29th, 2011 at 7:33 pm
    Bill Illis, I think ScienceOfDoom has most of the answers,
    ———–

    I have not seen one answer to the questions I asked at ScienceofDoom.

  185. easy bruce.

    The detector is 3 feet away
    the detector can see the candle.

    Then interpose C02 between the detector and the candle
    predict what will happen and explain your prediction.

    SteveF you may not answer.

    This bruce is not a rhetorical question. Answering this question will tell us if you understand basic physics of radiation transfer.

    Downwelling will come. but if you fail algebra you dont get to go to calculus.. and diffy q is miles away

  186. Bruce,
    My request:

    Please follow the rules of this blog:
    a) provide your own answer to this question.
    b) tell us what point you think your question/answer make.

    is a simple one.

    Steve F has correctly identified the reason for the rule

    Rhetorical questions are too often just a way to make other people do a lot of work for nothing. Answering your own rhetorical questions is honest, since you have to lay out your own thinking (or the complete lack thereof), before asking other people to do a lot of work.

    Questions are allowed, and people who aren’t constantly asking rhetorical questions are cut quite a bit of slack. But you started out this thread by trying to argue by rhetorical questions, and but seem to have reverted to ‘trying to make points by asking questions’ in:
    Bruce (Comment #82378) , Bruce (Comment #82381) , Bruce (Comment #82382) , Bruce (Comment #82391) , Bruce (Comment #82416) , Bruce (Comment #82417) , Bruce (Comment #82425) , Bruce (Comment #82445) , Bruce (Comment #82478) , Bruce (Comment #82483) , Bruce (Comment #82487) , Bruce (Comment #82492)

    Owing to the frequency with which you seem to try to ‘make a point by asking a question’, I am asking others to help me train you to follow the rules about rhetorical questions.

    Some people call the method of trying to make points by asking questions Socratic. Socrates was put to death.

  187. SteveF,

    Methinks that some people take this climate science/stuff, and themselves, too seriously. With the exception of military science and the quasi-sciences of finance and economics, I am pretty sure that I have as little formal science training , as our friend Bruce. But I am capable of learning. I find learning experiences to be most valuable, when they allow for the free exchange of ideas and information. I often read lucia’s blog, because I believe that she is very intelligent and liberal, in the sense of allowing unfettered conversation. I have never before seen this foolishness about rhetorical questions being verbotten. My bad.

    What is this:

    “About that experiment.. when you put C02 between the candle and the detector.. what happens? predict and explain.”

    Ain’t that a rhetorical question, that has been repeatedly asked? Where is the admonishment?

    I admire and respect lucia (and mosher), but has the often less-than-correct, but always interesting and entertaining Lord Monckton got them rattled? Why not pick on the Nobel Pize winner, Al Gore, instead? Sorry, rhetorical. I withdraw the question.

    You spelled my name wrong. However, I am thinking about changing it back to the former spelling. When my forefathers came to this continent, much more than four score and seven years ago, they dropped the “t”, for some reason. Maybe they thought it sounded too “froggy”. Actually, it was De Montfort. Descended from Simon De Montfort, a lieutenant of William the Conquerer, whose namesake became the Earl of Leicester. I am going to start claiming that I outrank His Highness Monckton of Brenchly/Benchly/ whatever.

    I always enjoy your level-headed posts.

  188. Bruce, I think you and other people are making too much out of this. It’s quite simple. Everything with a temperature above absolute zero radiates energy. You, the ground, etc. If you are in the desert and attempt to measure the temperature of the sky by infrared thermography, you get a reading of about -18 deg. C IIRC. That’s pretty cold but not -273, so the sky IS radiating heat down at you. You won’t be able to feel it by holding your hands out as you will be losing net heat, but you’d be losing heat a lot faster if the sky wasn’t radiating anything (it would be the equivalent of a -273 deg. C surface.) If you were an alien with a body temperature of -30 and you landed your spaceship in the desert and hopped outside without protection, then you might indeed feel the heat radiating down from that warm -18 degree surface up there. It might be somewhat masked though by the searing zero-degree air surrounding you…

  189. Bruce.

    I did better than give you an experiment showing that back radiation exists. I pointed you the website where the data from all the sensors is. Downwelling IR is measured day in and day out.

    I trust we are not back to you denying observation data. I though we could progress to your prediction for the experiment.

    But, if you would like to go back to basic math and start over we can start back over with the observations of downwelling IR. Note I am not asking you to answer questions about whether it can do work. First we need to see that you get basic physics. However, if you want to go back to the observations and claim that those reading are a conspiracy or something. Oh, those stations are the same ones that measure the sunlight you like to talk about..

    So. 3 feet candle, c02… predict and explain.

  190. Bruce,

    “Correct Don. You don’t feel the backradiation because it is not and never will be 60% more Watts/m^2 than solar energy”

    I don’t feel the back radiation, because it is not the only variable that determines surface temp. The net IR at night is considerably negative. That is why it cools off at night. Duh! Why should I feel any discernible warmth from IR radiation at night, particularly in the desert, where generally ‘green house’ gases (primarily water vapor) are scarce?

    Think about it Bruce. Think about the rain-forest. Hint: It ain’t the trees.

  191. Don–
    It’s my rule. Everyone asks rhetorical questions sometimes. But there are some people who start trying to make all their points by asking rhetorical questions. The ‘points’ are generally obscure, and it shifts the burden onto other people.

    I only remind people when someone is pretty clearly arguing by rhetorical question. When used as a practice it is very disrespectful of other people’s time. Bruce is the guilty one on this thread.

    On Mosher’s question– I’m not sure it’s rhetorical. It might be. The difficulty is that when someone like Bruce starts arguing by posing lots of rhethorical question, other people start responding that way, and conversations quickly become very stupid.

    I know Mosher will stop after reading the other message.

    Why not pick on the Nobel Pize winner, Al Gore, instead?

    I discussed Gore’s lame climate 24 hour thingie.:)

  192. Bruce. Note that the brighter bulbs in the class have eagerly raised their hands and know the answer without any more detail. That should tell you that you dint know what is relevant.

    For extra credit there will be a question about how these physical principles are used daily to save lives.. But first, Bruce, show us what you have learned ( ya, stop googling frantically and answer)

  193. Mosher, how much CO2? The current amount or 100%? Thats a parameter request, not a rhetorical question.

    NASA thinks backradiation heats up the surface of the earth.

    “However, a significant fraction of the longwave radiation emitted by the surface is absorbed by trace gases in the air. This heats the air and causes it to radiate energy both out to space and back toward the Earth’s surface. The energy emitted back to the surface causes it to heat up more, which then results in greater emission from the surface. ”

    http://earthobservatory.nasa.gov/Features/Clouds/

    If it can heat up the earth, it must be able to heat up my hand or warm water in a solar vacuum water heat. Yet it cannot.

  194. Bill:

    I have not seen one answer to the questions I asked at ScienceofDoom.

    What I was trying to say is ScienceOfDoom provides a framework that you can use to answer your specific questions. Obviously he didn’t’ (and isn’t) going to entertain your exact questions unless he sees a particularly relevant reason for this.

    Like anybody else, I would need to understand why these particular questions are so burning for you, and how knowing them would change your view or understanding in any way, before I made much personal effort to answer them for you. (After all I do research for a living… asking us to do your research for you for free, is like asking a house painter to paint your house for free because he’s gotten good at it, and it shouldn’t be a big deal for him to paint your house for you).

    And like anybody, I would wonder why you aren’t striving to answer them yourself.

  195. Mosher: “I pointed you the website where the data from all the sensors is. Downwelling IR is measured day in and day out.”

    Its probably just the outgoing IR bouncing back from dust and outgoing IR from higher points around the sensor.

  196. lucia,

    OK, I get it. The actual rule is: annoying, obtuse, knuckleheads are not allowed to use rhetorical questions. It’s your blog:) I like you.

  197. 3 feet. Candle. IR Detector. 100% CO2.

    The candle goes out. No oxygen.

    But seriously. It would depend on the wavelength of infrared that the detector is tuned to, and the wavelengths of infrared coming from the candle. And how much water vapor is in the air. And the temperature of the air.

    I’d like to see an experiment. Experiments trump theory if they are well designed. The world is full of elegant theories that fail spectacularly when put to the test.

  198. Bruce,

    Compare the charts below:

    http://www.srrb.noaa.gov/cgi-bin/surf_check?site=desr&mos=July&day=03&year=2011&p5=dpir&p6=upir&p9=rnir&ptype=gif

    http://www.srrb.noaa.gov/cgi-bin/surf_check?site=desr&mos=January&day=03&year=2011&p5=dpir&p6=upir&p9=rnir&ptype=gif

    Look up the diurnal temp differences for those dates. Look at the differences in humidity for the respective dates. Look at “net infrared”. Doesn’t it look like back radiation from water vapor influences the surface temperature, making it warmer than it would be in the absence of water vapor? I will help you Bruce; yes, it obviously does. That is why they call water vapor a greenhouse gas. CO2 is another greenhouse gas. It’s got something to do with the physics, Bruce. I don’t get the whole thing either, but they got us on that one, Bruce. You are fighting the wrong battle, and getting your behind kicked up between your ears. Your type give skeptics a bad name. Stop it.

  199. Bruce. Assume The candle is not in the C02.
    Picture a detector at one end of a tube.
    The detector is pressed up against the end of the tube.
    the tube is sealed. The candle sits at the other end.
    You test the set up before putting the C02 in the tube.
    the detector sees the candle. So it sees the wavelength
    that the candle emits.
    Now you fill that tube with C02. the candle stays burning
    predict what will happen as you fill the tube with C02
    and explain in terms of physics why you believe what you
    believe. D = detector. — a sealed tube. Candle outside the
    tube. It stays lit. The detector is tuned to see the IR from the candle. what happens and why

    D—————–!

  200. Don.

    Yes. There are good skeptical arguments INSIDE the science of
    the climate. By inside the science I mean the science that comes after radiation physics. The physics of how visible light moves through various materials is well understood. The physics of how other parts of the EM spectrum move through various materials is also well understood. Its the science that helps us . radars, and heat seeking missiles, and satillities, and microwave ovens, and cell phones and wifi. Any time you build a device that transmits radiation through the atmosphere you have to take into account the modelcules in front of the energy you are sending. What i am trying to do is invite bruce to the real debate.

  201. Bruce.

    you will get to see the experiment. But first
    1. you denied the existence of real observations. That gives us pause. It gives us pause because we hear people asking to see evidence all
    the time. I showed you and you still persisted.

    no cookie

  202. Don, the claim is that this backradiation can warm up the earth, but will not affect things like vacuum tube solar water heaters. Again, the graphs you show me CLAIM 350W/m^2 at night. Yet you can’t do anything with it. It is measured but has no effect.

    There is the famous diagram that claims 60% more backradiation energy that energy from the sun. Yet we know the Sun’s energy can do work while the backradiation does none.

  203. One other thing. Backradiation is not measured. It is calculated.

    “In order to calculate the incoming LW irradiance at the detector, the temperature of the pyrgeometer body must be known. This is measured using a thermistor (type YSI44031), which is located beside the cold junctions of the thermopile. The downward (or upward) longwave radiation is then calculated using the following formula :-
    LW = Uemf/S + ( 5.67*10-8 * Tb4 )
    where Uemf is the output voltage from the thermopile, S is the calibration constant of the instrument, and Tb is the pyrgeometer body temperature, measured by the thermistor, in degrees kelvin. Note that for an upward facing pyrgeometer, the thermopile output voltage will in most instances be negative. This is beacause the upwelling irradiance from the pyrgeometer is likely to be greater then the incoming irradiance from the sky.”

    http://badc.nerc.ac.uk/data/cardington/instr_v7/pyrgeometer.html

  204. Bruce,

    OK, you honesty don’t get this stuff. Bruce, at night your vacuum tube solar heater is losing more IR than it is receiving from back radiation. It ain’t gonna get hotter. Look at “net infrared” on those charts. It should be clear that the water in the tubes gets a lot colder overnight, in the absence of back radiation from a significant amount of water vapor/greenhouse gas. Compare the diurnal surface temperatures of July 3, with January, 3. It’s the water vapor. It’s a greenhouse gas, bruce. The big one.

  205. steve mosher,

    I think Bruce is about to get it. It will be interesting to see if he apologizes, or just slinks away.

  206. Bruce (Comment #82587) September 30th, 2011 at 12:34 am One other thing. Backradiation is not measured. It is calculated.

    Good point – and you know what? Mass – that is another thing that is effectively calculated rather than directly measured. I’ve got a mercury thermometere. You know what IT measures? Temperature? No the length of a column of mercury. Lots more examples – we can have hours of fun.

  207. Bruce
    .
    Strange. The CLAIM was that there is now 324W/m^2 if IR impacting on the surface. To me it would not matter if how deep a box is as long as the opening is 1 m^2, there should be 324W going into that opening. If the sides are mirrored, the IR should bounce down to the bottom getting closer to parallel.

    Why would you claim 0 if the box is taller unless you think the sky is a point source?
    .
    I am very patient in answering questions. What I find EXTREMELY annoying is when people don’t read the answer and reask the same question. This is exactly what you are doing in this quote – you are being annoying on purpose (can’t see exactly why).

    Did you notice that I wrote that the cylinder had a GOOD IR ABSORPTIVITY ? If you did then why are you writing about reflexion and “bouncing down”? And if you didn’t then why are you asking questions when you are not interested in answers.
    Why is it necessary that the cylinder has a good IR absorptivity? Because I am interested by how many IR rays are parallel. And why am I interested by how many IR rays are parallel? Because all (I stress all) concentrating devices like mirrors and lenses can only concentrate parallel rays. So if there are few parallel rays, there is little IR available to concentration. And then I have not enough photons concentrated on a small surface to cook meat.
    QED.
    .
    Illis
    .
    what is the distance distribution of the origination point(s) for the back-radiation photons.
    It depends on the frequency (big distance for little absorbed frequencies and short one for strongly absorbed) and the distribution is exponential. If you take the well known 15µ and measure on ground, all these photons originate from the few first dozens of m of the atmosphere and much (exponentially) more in the first meter than in the last meter.
    This distance increases with decreasing density so it is bigger as you go up. It is infinite in vacuum.
    .
    I would also like to know the distribution of those photons by the last atmospherical molecule which transmitted them? It makes quite a difference if more than 70% come from water vapour at the tropopause or if 50% of them are just blackbody radiation from the 98% N2 and O2 molecules immediately above the instrument.
    .
    This depends again on frequency and the question is very similar to the previous one. Let’s take again the 15µ. Here it’s CO2. For other frequencies it will be a mix of H20 and CO2 and again for others it will be H2O alone. N2 and O2 emit little and very low frequency so can be neglected (at least for the conditions in the Earth’s atmosphere).
    For strongly absorbed frequencies it is not important whether it is CO2, H2O, CH4 or whatever – all photons coming to ground will originate in the first few meters of the atmosphere. So if you measure on ground, the IR will come from the first meters and the majority will come from H2O because there is much of it.
    .

    I would also like to know how the back-radiation intensity varies through various levels of the atmosphere, from 1 metre to 10 metres to 90 kms.
    .
    Roughly proportional to density. So maximum at 1 m and near 0 at 90 km. This is typically what any good radiation transfer calculator computes. It can almost be done by hand if you take only a few very broad frequency bands.
    .

    I would also like to know what is the balance of up-radiation versus back-radiation for every 10 cms from the surface to 100 metres for example.
    .
    If you take slices of 10 cm then both are almost equal.
    Any IR that is absorbed in this slice will be reemitted. The reemission being isotropic, there is as much reemitted “up” than reemitted “down”.
    However as the density decreases when you go up, for every slice there will be slightly less photons incoming from above than photons incoming from below. There will be a net flux from down to up.
    Here again different photons will have very different stories.
    Strongly absorbed frequency photons will stop and go in every slice while little absorbed frequency photons will fly through many slices before being absorbed and reemitted.
    .
    I would also like to know why the temperature of the air at 2 metres only seems to notice about 0.00001% of this back-radiation per second. How does the energy actually impact the temperature level since it doesn’t seem to?
    .
    I don’t understand well the question.
    For all practical purposes ALL IR absorbed in your first 2 m is reemitted (see answer above).
    It is actually the very purpose of the LTE (local thermodynamic equilibrium) to stabilize the gaz at a temperature where absorption=emission. If you are out of equilibrium and have absorption > emission then the gaz will heat up and if you have absorption < emission then the gaz will cool down.
    When you are at LTE, it doesn't matter how much IR you send in the gaz – it will have ALWAYS absorption=emission so it won’t “notice” the IR at all. There will be neither heating nor cooling – not even 0,00001%, it is exactly zero.
    .
    I would also like to know how the back-radiation actually occurs since the collisional exchange rate of atmospheric molecules is 10^6 faster than than GHG relaxation rate of emission.
    .
    Congratulations 🙂 Excellent and largely misunderstood point.
    The answer is that there is simply a huge amount of molecules, it’s a matter of statistics.
    To understand everything about this point it is just enough to realize that collisions do BOTH heating and cooling.
    So, like in the above answer, the LTE will be realized at such a temperature where the collisions cool and heat CO2 at an equal rate. Once that happens, the conservation of energy makes simply necessary that absorption=emission regardless how fast or slow an individual molecule collides with another.
    If you compute the fraction of CO2 molecules that will have no collision in 0,1 second, you will get an extremely low %tage. But when you multiply it by the huge number of molecules there is, you will see that there is exactly enough molecules for the emission observed in LTE.
    .
    And before you ask it, yes when the density becomes very low, the above is no more true and we have a non LTE condition. Things become then extremely complicated and you must use QM to answer questions. Classical and easy equilibrium statistical thermodynamics work no more there.

  208. Lucia #82527,

    Yes, I did mean 390 W/M2, not 342, thanks for spotting that; it was late for me but that’s not an excuse for poor proof-reading.

    My reasoning goes:
    According to the diagram…
    390 W/M2 leaves the Earth’s surface.
    235 W/M2 leaves the atmosphere to space.
    Therefore the ‘Greenhouse Effect’ equates to 155 W/M2.
    The IPCC states that a doubling of CO2 equates tot 3.7 W/M2.
    3.7 is 2.4% of 155, therefore climate sensitivity is 2.4% of the GE which, in terms of temperature, is accepted as being 33 deg C.
    Therefore climate sensitivity, according to the diagram and IPCC, is 0.8 C (2.4% of 33).

    Sorry for the delay (Roy), I was asleep.

    As for the first point, the diagram shows back-radiation as being a value equalling 95% of solar radiation. That seems highto me.

    Also, the diagram depicts back-radiation as being almost twice the amount that is provided by solar radiation at the surface. That also seems high to me.

  209. TomVonk (Comment #82592)
    September 30th, 2011 at 3:14 am
    ——————

    Thanks. I think that answers my questions pretty well.

  210. Arfur, “As for the first point, the diagram shows back-radiation as being a value equalling 95% of solar radiation. That seems highto me.

    Also, the diagram depicts back-radiation as being almost twice the amount that is provided by solar radiation at the surface. That also seems high to me.”

    If you consider that absorbed solar and latent heat from the surface account for ~70 % of the energy in the atmosphere, it make more sense.

    TomVonk,

    In LTE, increased water vapor would tend to smear the CO2 spectrum with the collisional heat transfer. So local irrigation or wet land restoration should provide more local cooling at the cost of warmer low temperatures. Why is water’s stabilization of temperature so controversial?

  211. Dallas,

    Thanks for your input. It appears to me that the combination of absorbed solar and latent heat from the surface adds up to 28% of the total energy in the atmosphere, not 70%. (78+67=145, which is 28% of the total energy input of 519 W/M2, made up from 350+78+67+24) Either way, according to the diagram the amount of back-radiation absorbed by the planet is almost twice as much as that absorbed by the sun.
    Would you care to comment on my question regarding climate sensitivity? I’d appreciate your input on that also.
    Regards,

  213. Arfur, depends on how you look at it. The 390 includes back radiation re-radiation. If you use nets like most of the normal world it is about 70%. If the simple cartoon was broken down by source, it would be easy to see that the down welling is roughly equal to direct radiation at the surface, that 60 to 70 percent of the downwelling is basically buffered solar warming with about 30% due to the Greenhouse effect as related to OLR.

  214. Don: “It should be clear that the water in the tubes gets a lot colder overnight, in the absence of back radiation”

    But the graphs you and Mosher post show backradiation doesn’t change all night long.

  215. TomVonk, Mosher was kind enough to provide a reference to a solar panel maker that will harvesting terrestrial infrared. I’m assuming that isn’t a point source either. Pointing their panels at the sky should work if the backradiation is actually there.

  216. Bruce, I suggested that you compare the net infrared on the two example charts and also look at the differences in humidity. If you don’t want to do that, you fail. You are not accepting the help that is being graciously given to you.

  217. I’ve posted this before (and it was pointed out to me that it is missing moist enthalpy energy levels and perhaps thermals energy transfer and I guess energy absorbed and released by the land surface) but the real 2 metre air temperature change (per second) is much more stable than the measured radiation flows per second over a 24 hour period.

    The SurfRad station at Table Mountain CO on Nov 17, 2009.

    http://img140.imageshack.us/img140/4109/tablemountainall.png

    http://img12.imageshack.us/img12/3225/tablemountainnets.png

  218. “The Down-welling Long-wave Radiation (DLR) flux (W.m-2) is defined as the thermal irradiance reaching the surface in the thermal infrared spectrum (4 – 100 µm). It is determined by the radiation that originates from a shallow layer close to the surface, about one third being emitted by the lowest 10 meters and 80% by the 500-meter layer”

    http://postel.mediasfrance.org/en/BIOGEOPHYSICAL-PRODUCTS/Downwelling-Longwave-Radiation/

    Ahhh. Just the temperature of the air immediately above you.

  219. Brucey,

    “thermal irradiance reaching the surface”

    Look at the charts Brucey. What happens if you turn off the downwelling thermal irradiance that reaches the surface of your little water heater thingy? I will help you, Brucey; it will be IR thermal irradiance all outgoing and no incoming, and it will get damn cold.

  220. Don, vacuum insulated tube. They work when its cold out for solar energy. They don’t work with backradiation.

  221. Another thought …

    The human body radiates about 50W/m^2 and I’ll admit the palm of the hand probably radiates more until it gets cold and vasoconstriction sets in.

    Thats way less than 400W/m^2 claimed for backradiation on some days. On a cold winters night you should be able to put your hand towards the sky and feel this supposed infrared the same way you can feel the sun’s similar amount on a sunny winter day.

  222. So Brucey, does the water in the vacuum insulated tubes on your rooftop solar heater stay warm overnight? I will help you, again; no it doesn’t. And if you could somehow turn off the downwelling backradiation /”thermal irradiance reaching the surface”, it will get a lot colder. There is a good reason why they call it “thermal” irradiance, instead of “no-work” irradiance.

    I hear that the IPCC’s Black Propaganda Department is paying professional actors good money to portray stubbornly stupid deniers, on various climate discussion boards. Do you know anything about that, Brucey?

  223. lucia,

    I made the following post, which has not appeared. Am I in moderation?

    So Brucey, does the water in the vacuum insulated tubes on your rooftop solar heater stay warm overnight? I will help you, again; no it doesn’t. And if you could somehow turn off the downwelling backradiation /”thermal irradiance reaching the surface”, it will get a lot colder. There is a good reason why they call it “thermal” irradiance, instead of “no-work” irradiance.

    I hear that the IPCC’s Black Propaganda Department is paying professional actors good money to portray stubbornly stupid deniers, on various climate discussion boards. Do you know anything about that, Brucey?

  224. Don Monfort–
    No. You are not moderated.

    I suspect the problem is traffic is very heavy right now. Some comments may be getting lost when the server fails.

    I increased memory resources at Dreamhost to reduce the problem (and will reduce them again later to avoid having to pay $100 a month every day of the year! Dreamhost lets you adjust instantly. So, I can adjust for a day.)

  225. Don–
    I looked in the spam trap and I found 3 comments sent there– two were yours one was Bruce’s. I’ve released them. Let me know if more comments get lost.

  228. Bruce: What is your prediction on Mosher’s candle – IR detector setup?

    That’s the place where the conversation has repeatedly stalled. There seems no point in attempting complex ideas in quantitative terms instead of tackling simple, well-understood systems, qualitatively.

  229. Don (Donny would be childish): “does the water in the vacuum insulated tubes on your rooftop solar heater stay warm overnight?”

    Supposedly.

    “Not affected by temperature and thus can operate in the coldest climates”

    “Double walled evacuated tubes prevent heat loss”

  230. AMac, As I said. It depends. The demonstration I’ve seen show the candle faintly even after the tube is filled with CO2.

    But I would be more interested in the difference between 0% CO2 and a very small amount … not 100%.

  232. OK Brucey, I get it now. You got those new miracle one-way double-walled vacuum tubes that allow thermal radiation in, to heat up the water, but prevent thermal radiation from getting out. How do it know, Brucey? Don’t take everything you see in an advertisement literally.

  233. MikeC
    How is discussing a line of argumentation to the point of inescapable conclusion and ultimate resolution equivalent to “obsessing” over “a guy”? “Obsessing over a guy” is what my teenaged daughter does.

  236. … and your teenage daughter’s obsessing over a guy will not lead to a prescription for ACE inhibitors

  237. bender:

    How is discussing a line of argumentation to the point of inescapable conclusion and ultimate resolution equivalent to “obsessing” over “a guy”? “

    I’m thinking it’s because of the inconvenience of the truth in the matter (on this topic, CM lacks the expertise for his opinions to matter).

    Lucia gets it from both sides for attacking their icons.

  239. MikeC
    “your teenage daughter’s obsessing over a guy will not lead to a prescription for ACE inhibitors”

    Stop the science, folks. MikeC has a policy for us to implement.

  241. Sorry Carrick, the man is no icon of mine… but what is pretty clear is that one in the middle can create a niche for him/herself and be just as full of it as the extremes

  243. gosh… I thought that was a vague glancing blow at everybody… but my thoughts on the issue was addressed in the last thread… about how it cannot be summed up with a tidy equasion…

  244. Bruce.

    You were suppose to predict and explain. If you think that 0% and a small amount is interesting why?

    How does C02 interact with IR. What is this science called and how is it used to protect our lives. How is that physics used to detect even very small increases in C02? everyday. perhaps even where you are right now

    You have to demonstrate your understand of the basic physics, rather than simple google things that are not on topic.

    The reason for this is we cannot even begin to talk about what you claim. here is what you claim

    A Downwelling IR does not exists ( well we busted that one)
    B. It does no work

    B is a claim of positive knowledge. There are three positions you can take on downwelling IR ( well after you are busted for denying its very existence )

    1. It exists and it causes some changes to the energy balance
    2. It exists and I dont know enough about physics to really say anything.
    3. It exists ( after embarrasingly be shown this ) but it does no work.

    #1 is the position of AGW
    #2 is just an honest admission of no knowledge
    #3.. your position, is a claim of positive knowledge. you know it can do no work.

    Consequently, We are left with examining your actual knowledge of how things work. This will be tested by asking you questions about the physics we are discussing. When you dodge and weave you leave the possibility of being in camp #2 behind. You also paint yourself into a corner when it comes to explaining the physics you know that backs up your claim that it can do no work. If you dont even understand the basic physics of how radatiation propagates, the next steps will really be beyond you. But.. you will have lost the opportunity to scurry back to position number 2

  245. Mosher, #4. It does no work, therefore it is irrelevant. You claim it warms up the earth, but not anything else.

    Don, the thermal radiation you claim exists, but does no work .

    “This is because the upwelling irradiance from the pyrgeometer is likely to be greater then the incoming irradiance from the sky”

  246. Dallas, #82602

    Dallas, I’m just looking at it at face value – as would any member of the public. There is no mention of back radiation re-radiation. The numbers are presented as a ‘fait accompli’. I’m not sure how to take your ‘like most of the normal world’ comment but, on face value, the K-T diagram implies that back-radiation plays a larger part of the surface warming than solar radiation. I’m not saying you are wrong with your explanation, just that the diagram does not explain your explanation.
    It seems Lucia is busy at the moment, do you have an answer for the climate sensitivity question?

  247. steve mosher,

    We look like a couple of dummies here. We have been talking to a freaking rock. And the fact that the rock talks back does not excuse our foolishness. I’m out.

  248. Arfur, That drawing by K&T sure looks that way don’t it? K&T were doing a radiation balance study. Mainly concerned with the TOA and the surface, not what was happening in the middle. So it is pretty confusing as far as what is heating what. The NASA radiation budget is the one that should be used for just general information. Too many people read too much into the K&T drawing.

    By normal world I mean heat flows from warm to cold, proportionally with the difference in temperature. If the sky has a temperature, it will reduce the flow of heat from the surface proportional to temperature. It is not that simple, but pretty close. The sky gets its “temperature” from solar either directly or from surface temperature rising through convection(involving latent and sensible heat), conduction (sensible heat only though it causes convection) or radiating to space. The percentage the sky gets from surface radiation is the 390W/m^2 minus 324W/m^2 minus 40W/m^2 that passes through unabsorbed, equals 24 W/m^2 at equilibrium. That is how it is shown on the NASA cartoon, with some minor changes. K&T uses “thermals” instead of conduction for some odd reason.

    So do I have an answer for the sensitivity problem? I have several LOL, just which one is right, dunno.

  250. MikeC:

    Sorry Carrick, the man is no icon of mine… but what is pretty clear is that one in the middle can create a niche for him/herself and be just as full of it as the extremes

    I don’t know if this helps or hurts, but I lump you in with me: curmudgeonry old men. I definitely don’t lump you in with the idol worshipers.

  251. MikeC says:
    “I thought that was a vague glancing blow at everybody”
    .
    But you previously referred to “one in the middle”. And when you said “one” I assumed you meant “one”. So I wanted to know what “one” you were referring to. Now you say it is every one.
    .
    Never mind. Really.

  252. Arfur, That drawing by K&T sure looks that way don’t it? K&T were doing a radiation balance study. Mainly concerned with the TOA and the surface, not what was happening in the middle. So it is pretty confusing as far as what is heating what. The NASA radiation budget is the one that should be used for just general information. Too many people read too much into the K&T drawing.

  253. For more information,Arfur,

    By normal world I mean heat flows from warm to cold, proportionally with the difference in temperature. If the sky has a temperature, it will reduce the flow of heat from the surface proportional to temperature. It is not that simple, but pretty close. The sky gets its “temperature” from solar either directly or from surface temperature rising through convection(involving latent and sensible heat), conduction (sensible heat only though it causes convection) or radiating to space. The percentage the sky gets from surface radiation is the 390W/m^2 minus 324W/m^2 minus 40W/m^2 that passes through unabsorbed, equals 24 W/m^2 at equilibrium. That is how it is shown on the NASA cartoon, with some minor changes. K&T uses “thermals” instead of conduction for some odd reason.

    So do I have an answer for the sensitivity problem? I have several LOL, just which one is right, dunno.

  254. K&T uses “thermals” instead of conduction for some odd reason.

    If you ever piloted a glider, you’d know why. Thermal exist. The are most evident on sunny days when the sun heats the earth but the air is cooler aloft. Due to hydrodynamic instability, the warm air ‘wants’ to rise. However, it can’t all rise uniformly in a plane, because that would just leave vacuum at the surface. So, instead, warm air will rise in sort of ‘plumes’, with cooler air falling around it. The plumes are the thermals and glider pilots with their very low weight gliders look for them because the warm air is rising fast enough that they can glide up and gain altitude.

    You might wonder how the pilots see these things: They look for the very small cloud just starting to form at the top of the thermal. That’s the water in the moist air condensing when it reaches a point where the temperature as dropped enough for the water to condense out.

    This isn’t the only mechanisms for heat transfer from the lower atmosphere to the upper atmosphere, but my understanding is pretty big. K&T probably used the image of thermals because they knew many looking at their cartoon-figure would recognize thermals.

  255. Lucia, I do know what a thermal is, but it is hard to find thermals that don’t entrain some moisture. So are his thermals only ones that do not include a water phase change? Separating heat transfer effects makes more sense to me rather than using a kinda ubiquitous term, thermodynamically speaking.

  256. Dallas– Now that I don’t know. Obviously, my description of thermals includes water molecules being displaced upward.

    I try not to read too much into cartoons. They are meant to illustrate certain things, but not others. But really, there isn’t a lot of pure conduction in the atmosphere. The air is generally moving around.

  258. Dallas & Lucia,

    Dallas, thanks for the further info. If you think that some information has been omitted from the K&T ‘cartoon’ then I can accept that as an explanation why the numbers don’t make sense of themselves.
    Dallas & Lucia,
    Thermals – they don’t have to contain any moisture. Dry thermals can and do exist, it is just that they are easier to see when the air is moist for the reasons Lucia gave. Towering clouds, eg cumulus, give good indication but lenticular clouds actually pinpoint the peak of standing (or mountain) waves, which can stretch for hundreds of miles and can be particularly useful for a glider pilot.
    The question of conduction in the atmosphere is interesting. I would surmise that conduction may be a larger cause of heating than many think. As only 0.04% of atmospheric molecules can absorb and emit radiation in the relevant wavelengths, it may be that much of the warming could be due to low energy collisions. Just a thought.
    Lucia, I gave you my reasoning (for my first question) in #82593. Would you care to give your opinion please?

  259. Arfur–
    Without sitting down and doing numbers, my first reaction is that you seem to assume temperature is linear with total forcing. That’s not right for total forcing, and it’s not right for the increment where we are.

  260. Don Mon(t)fort,

    Sorry about misspelling your name.

    BTW, I have nothing against the French, no matter how they choose to spell their names. Great cooking and good wines come from France. A little edgy about Americans though. 😉

  261. Lucia,

    That’s not right for total forcing, and it’s not right for the increment where we are.

    For sure that is right for total forcing, but I am not so sure that a linear approximation is not pretty close to right for modest size increments (eg, 1% change in total forcing). Maybe I did not understand what you wrote.

  262. Arfur, I don’t think K&T’s numbers are all that wrong, just not shown in a manner you can make much use of. As Lucia said the temperature is not linear with total forcing. To get an estimate from the cartoon you could take the 24 W/m^2(radiative absorption) / [(%78 (latent)+ %24(thermal)+%24(radiative)+%67 (solar)] giving you 12.4% of the 33C temperature of the atmosphere is due to radiative warming of the atmosphere or 4.1 degrees if none of the values changed with a change in radiative forcing or are higher due to radiative forcing. Then 4.1*(24+3.7)/24 equals 4.73 degrees. The difference would be 0.6 degrees which would be the minimum climate sensitivity to a 3.7W/m^2 increase in forcing if everything else remained the same. Of course an increase in radiative forcing would change the other values, but it does say, yep, changing the radiative forcing will change the temperature. Change all the % to partials, add lapse rate and albedo, next thing you know you have a basic equation for climate sensitivity. Unless I screwed up of course.

  263. SteveF– Well… I’m reading what Arthur said quickly, on a Friday night. There are no equations, and I’m not entirely sure I know precisely what he is trying to explain. This is not his fault. If someone wants an quick answer to a question posed in comments on a Friday afternoon, well… not a great time!

    I have not looked at the number to figure out the impact of non-linearity. All I am saying is: I haven’t looked at that. The issue “strikes” me. I think Arfur should look into that issue.

    I don’t know what the answer will be when he looks at it.

    Out of curiosity, Arfur, do you mean climate sensitivity with no feedbacks? Or with feedbacks? Because without doing numbers, or checking carefully, it reads as if no CO2 feedbacks sensitivity is 0.8C with your argument. That’s just a little lower than the roughly 1C people generally bandy about. SteveF is likely right the non-linearity wouldn’t make a huge difference– but 20% is not “huge”. So, it may be what you are doing is a decent first order estimate for “no feedbacks”.

    But… Friday afternoon– I’m not doing numbers!

  264. no matter how they choose to spell their names

    We who speak and spell English can’t really criticize other languages for any spelling oddities.

    Through, bough, rough, though, cough, thorough, ….

  265. Arfur, Sorry I did read the cartoon wrong, the radiative impact of surface temperature on the atmosphere is (390-324-40) =26. So it would be 26/(78+24+26+67)=0.133 0.133*33=4.4 would be the radiative impact, so increase the radiative forcing by 3.7W/m^2 (26+3.7)/(78+24+29.7+67)=0.149 0.1495*33=4.92 4.92-4.4=0.52 C which would be the minimum sensitivity to an extra 3.7 W/m^2 forcing. Still it is just a cartoon and K&T used a model to determine that there was an energy imbalance.

  266. SteveF,

    For sure that is right for total forcing, but I am not so sure that a linear approximation is not pretty close to right for modest size increments (eg, 1% change in total forcing). Maybe I did not understand what you wrote.

    I think Lucia is right and more into chillin’ that calculating. The 3.7 W/m^2 is just one percent of the gross forcing but 13% of the radiative forcing of the atmosphere due to surface temperature. It depend on how you badly you want to determine the world’s fate based on a cartoon 🙂

  267. lucia (Comment #82538)

    Richard–
    Perhaps so … but talking about people pulling things out of a hat … could you elaborate on this assertion?
    Do you mean elaborate on
    “if CO2 were to say– multiply by 20 [from its present value], the earth will almost warm enough to overcome an ice age.”

    Yes

  268. Pingback: Collide-a-scape » Blog Archive » Collide-a-scape >> The Monktopus vs the Lucianator

  269. Lucia, SteveF and Dallas…

    I agree that any non-linearity may make a difference but I also agree with SteveF that the difference will be tiny at the small percentage of 2.4%.
    I don’t think we need to over-complicate this. All I was pointing out is that the K&T diagram (cartoon ,if you prefer) indicates the GE is 155 W/M2. That is a radiation figure. The temperate figure for the GE is 33 C. As 3.7 is 2.4% of 155, the figure of 0.8 C would seem correct but that is, as you say, a non-feedback measurement. I suppose the next question would be ‘by how much would feedbacks change that figure?’. But that is possibly a rhetorical question and I know you don’t like those! Another point is that even though it is a no-feedback answer, it assumes the 3.7 figure is correct. I will look into non-linearity.
    My thanks to you, SteveF and Dallas for the discussion.

    ps, Dallas, I agree that K&T used a model:) Can you explain what you mean by 3.7 equalling 13% due to surface temperature?

  270. Arfur,
    “ps, Dallas, I agree that K&T used a model:) Can you explain what you mean by 3.7 equalling 13% due to surface temperature?”

    The change in radiative absorption impacts the atmosphere first which impact the surface, so the change in absorbed OLR, the 26 W/M^2, is the value of concern, the surface follows (3.7/26).

  271. What’s missing from the KandT diagram is time and space and the quantum level that it operates at. It is far too simplistic to represent the real physical world.

    Each photon in the back-radiation (or more accurately, the energy represented by the photon) will have spent time in more than 1 million individual molecules before it can be measured as back-radiation. And then it might spend time in another 1 million individual molecules before it is re-measured for the second time as back-radiation (and maybe another 3 or 4 times) before it random-walks its way to the upper atmosphere where it starts to have a 50% chance of escaping directly to space.

    It takes days for that photon to finally escape from the Earth. On Venus it might take 100 days. For energy generated inside the Sun, it takes an average 200,000 years.

    We have to move this down to photons and molecules and collision rates to understand it better. Climate science seems to believe the frequency is the most important thing. But the energy represented by the back-radition photon will have changed frequencies thousands of times before being measured as a CO2 15 um frequency back-radiation photon.

    It is a million times more complicated than the KandT diagram.

  272. Sorry to stray off topic, but I was flabbergasted by something I just read:

    http://online.wsj.com/article/SB10001424052970204226204576602524023932438.html

    The most flabbergasting part; our energy policy is based on a fantasy:

    When it was Mr. Hamm’s turn to talk briefly with President Obama, “I told him of the revolution in the oil and gas industry and how we have the capacity to produce enough oil to enable America to replace OPEC. I wanted to make sure he knew about this.”

    The president’s reaction? “He turned to me and said, ‘Oil and gas will be important for the next few years. But we need to go on to green and alternative energy. [Energy] Secretary [Steven] Chu has assured me that within five years, we can have a battery developed that will make a car with the equivalent of 130 miles per gallon.'” Mr. Hamm holds his head in his hands and says, “Even if you believed that, why would you want to stop oil and gas development? It was pretty disappointing.”

  273. StevenF’

    Actually, the De Montforts were not French, but Norman/Norsemen/Vikings. Their food and wine was not so good, but they could defend themselves.

  274. Dallas, this is getting very interesting, if a little confusing!
    .
    Given the caveat that it is indeed a cartoon and we are assuming the 324 W/M2 figure for back-radiaiotn is correct, then am I right in thinking you mean that a 13% (actually 3.7/26 is 14%) increase in radiative forcing will lead to a climate sensitivity of 1.6% of the GE? (.52/33=0.016) for a doubling of CO2? I’m not sure why, but the 13 (14)% seems to be a red herring but I’m not clever enough to spot why! If, however, you are correct, then doubling the CO2 will have negligible effect on global temperature, especially given the logarithmic effect of CO2 absorption.

    What am i missing? 🙂

    Oh, ps, I still think there is something wrong with the idea (cartoon) that the atmosphere provides almost twice as much radiation to the planet’s surface as the sun does…

  275. Arfur, Oh, ps, I still think there is something wrong with the idea (cartoon) that the atmosphere provides almost twice as much radiation to the planet’s surface as the sun does…

    Then you are thinking correctly, it don’t. Globally, the whole thing doesn’t make much sense because the tropics are optically opaque. There is zero direct radiative impact, it is all sensible and latent heat with solar input. Actually, the sun inputs 67 W/m^2 on average directly to the atmosphere and 78 W/m^2 via latent and a large percentage of the 24 W/m^2 of sensible (conductive or thermals depending on who drew the cartoon). That provides 70 odd percent of the energy in the atmosphere. So back radiation due to surface temperature radiation (cooling at night) is only 30 percent of the energy if that. The only thing added CO2 is going to change is that 30 percent with a little impact on the others.

    The big 390 up and the 300 whatever it is arrows just tend to confuse. One might say, exaggerate the situation.

  276. BTW, Arfur, this smart old dead guy Angstrom discover that the atmosphere provides, “nearly” as much energy as the sun. Mainly buffered solar since the Greenhouse effect is a two way street.

  277. Bruce (Comment #82696)
    September 30th, 2011 at 1:50 pm
    Mosher, #4. It does no work, therefore it is irrelevant. You claim it warms up the earth, but not anything else.
    Don, the thermal radiation you claim exists, but does no work .

    #####

    Bruce> please describe the experiment you did to confirm that it does no work, or more generally that it does not change the energy balance of the planet.

  278. well Don

    The demonstration is here

    http://www.youtube.com/watch?v=kGaV3PiobYk

    Its just a neat little demonstration for people who dont get transmission windows and how C02 is opaque to IR.

    For folks who wonder how this gets used..

    http://www1.eere.energy.gov/femp/pdfs/fta_co2.pdf

    basically, IR does not transmit freely through C02. That fact is used to create C02 sensors. The sensors are able to detect small amounts merely by looking at changes in transmission.

    “The technology most often used in CO2
    sensors is non-dispersive infrared spec­
    troscopy. It is based on the principle
    that every gas absorbs light at specific
    wavelengths. Carbon dioxide sensors
    calculate CO2
    concentrations by
    measuring the absorption of infrared
    light (at a wavelength of 4.26 microns)
    by CO2
    molecules. The sensor appara­
    tus incorporates a source of infrared
    radiation, a detector, and electronics to
    detect the absorption. Air from the area
    being monitored diffuses into a chamber
    that has a light source at one end and a
    light detector at the other. Selective
    optical fibers mounted over the light
    detector permit only light at the 4.26­
    micron wavelength absorbed by CO2
    to pass through to the detector. As
    CO2
    levels rise, more infrared light is
    absorbed and less light is detectable”

    So, To recap.

    1. Downwelling IR exists. ( we showed that )
    2. We have a theory that explains why we get downwelling IR
    namely, radiative physics explains how IR propagates ( reflects/scatters/is aborbed) through
    various gases.
    3. That physics is known and we use it every day to build devices that work. ( like C02 sensors in high rise buildings)

    If you meet a skeptic, like bruce, who denies these things. Then my suggestion is to try to educate them. If they refuse to learn these basic facts, then you cant even begin to discuss HOW having a more opaque atmosphere, slows the cooling of the planet.

  279. steve mosher,

    The candle thing is simple and interesting, but if our rock were still here, I am sure he would say that it confirms his foolishness. He would say there is no work being done there.

  280. Work is done when a rock or any matter is heated as the molecules constituting that matter vibrate faster.

  281. Don.

    The point of the demonstration was this. Bruce has no idea how the physics of radiative transfer work. he denied that downwelling IR existed. Then when showed that it does, he argued, without evidence, that it could do no work.

    If he cannot understand how c02 and other gases obstruct radiation leaving the earth, then ther is no hope of trying to get him to understand effective radiating altitude. And no hope of explaining that the “work” it does is not “heating” the surface.
    People speak imprecisely when they say that downwelling IR “heats” the surface. Downwelling IR is an effect. It is the effect of an opaque atmosphere. As the atmopshere becomes more opaque to IR, the height at which radiation finally escapes t(o space is raised. As that height is raised, earth is effectively radiating from a colder point, that in turn results in a warmer surface. The surface cools less rapidly ( effectively “warms”) when the atmosphere becomes more opaque.

    Metaphor time. You have a thermos. With a shiny container. The vaccum between the container walls ensures that the heat is not lossed by convection. So loss comes from radiation. The shiny surface reflects radiation, so the coffee cools less rapidly. The reflected radation doesnt “warm” the coffee. The coffee cools less rapidly. end of metaphor

  282. Mosher: “please describe the experiment you did to confirm that it does no work, or more generally that it does not change the energy balance of the planet.”

    The whole conversation was an attempt to find out some proof it does do work. Remember, it is your side that claims that it warms the planet. If there was experimental evidence it does do work, I’d love to read about.

    However, having waited in vain for such evidence, I noted that no one tries to use the energy claimed in any device.

  283. Everyone please check your definitions of heat and work before continuing in this discussion.

  284. Mosher, the candle is still somewhat visible on the monitor. Just not brightly visible. Heat appears to still make it through the CO2.

    “People speak imprecisely when they say that downwelling IR “heats” the surface.”

    Correct. It doesn’t.

    “The height at which radiation finally escapes to space is raised.”

    How far for each ppm extra of CO2? 1cm? 1m?

    From the quote I have above where 1/3rd of the “back radiation” comes from 10m above the sensor, I’m not sure what is actually being measured … since your graphs show only miniscule changes over a day no matter what happens.

  285. steve mosher,

    Maybe I am another rock, but I don’t entirely buy your explanation. You seem to be saying that downwelling IR has no local immediate effect, at the surface. When I was in the soldiering business, I spent a great deal of time sleeping outdoors, on the ground. Cloud cover=warmer, clear sky=cooler. That is a local effect.

    That is what I was trying to get our pet rock to understand, by pointing to something he could actually feel. See the discussion above, about the SURFRAD charts (comment 82582). At the same desert locale there was a measurable and significant difference in the diurnal temp variation, depending on how much humidity was locally present. Dryer air=less downwelling IR=relatively colder night. Humid air=more downwelling IR=relatively warmer night.

    I would say that the downwelling IR is heating the surface, at the same time the surface is losing heat by emitting IR. The surface is not getting net warmer, because it is emitting more than it is receiving. Same goes for the reflected radiation on the coffee, in the thermos. More radiation is escaping, than is being reflected back. Imagine that you could turn off the emission of IR from the surface, while it still received the downwelling IR. Wouldn’t it get warmer? Or what if the surface were at absolute zero and it was exposed to the downwelling IR?

    Maybe I am nitpicking. I will have another scotch and call it a night.

  286. Yes Don, there is that local effect. But when we talk about global warming in the long term we are talking about the effect that occurs due to the increasing opacity of the atmosphere

  287. Bruce. You have failed miserably at the simplest of tasks. from denying the existence of a measured entity to failing to understand the basics of RTE. Sorry. no cookie for you

  288. And Bruce. Check my comments all the way at the beginning where I explained that this was not about ‘warming’ the surface.
    That is your misunderstanding. Sorry but you dont know addition or subtraction. There is no point explaining division to you.

  289. Yes Steven, I get that stuff about the increasing opacity of the atmosphere. That’s why I didn’t question that part of your explanation.

  292. TomVonk (Comment #82592)
    September 30th, 2011 at 3:14 am

    Why is it necessary that the cylinder has a good IR absorptivity? Because I am interested by how many IR rays are parallel. And why am I interested by how many IR rays are parallel? Because all (I stress all) concentrating devices like mirrors and lenses can only concentrate parallel rays.

    Best not to make such absolute statements, ellipsoidal mirrors do not concentrate parallel rays!

  293. steven mosher (Comment #82810)
    The next step then is best explained by Ray P’s paper.
    http://geosci.uchicago.edu/~rt…..RT2011.pdf
    ————————————————
    Great paper – he is excellent at explaining complex concepts. And I like the way he emphasized the ca. 200-year historical development of infrared radiative transfer theory and its application to other planetary atmospheres. Thanks for the link.

  294. On average, each square metre of the surface,

    – receives 6.1×10^20 solar photons per day; and,

    – emits 58.9×10^20 IR photons per day (yes, there is more of them).

    It is going to take time for those all those IR photons to make their way up to 5 kms where it is -18C and is the effective Earth TOA longwave emission temperature. At this 5 km level, each square metre is only:

    – emitting 36.4×10^20 IR photons per day.

    I think of the backradiation more like an accumulation since it takes time for those photons to make their way through all those molecules, (whether they are highly intercepting/specific frequency GHGs or molecular collisions or land surface rock/soil/vegetation accumulation or whatever).

    The N2 and O2 molecules are at the same average temperature as the CO2 or H20 molecules since collisions happen at a much, much faster rate than the relaxation/emission timeline. Near the surface, each CO2 molecule collides with another molecule 7 billion times per second (the vast majority of those will be with N2 or O2 molecules with CO2 giving up some of the energy that it just absorbed in the 15 um frequency to the other molecules).

    CO2 absorbs a 15 um photon, immediately jumps to an energy state of 350K and promptly collides with a 287K N2 molecule, loses its energy to the N2 molecule and goes back to ground state – N2 is now at 288K.

    Eventually, the N2 molecules at 257K at 5 kms high presumably has to collide with a CO2/H20 molecule and give back the extra energy so that photon emission can occur at the 5 km high level at 255K by the CO2/H20 molecule.

    So, a one-time time delay that continues on forever give or take changing circumstances. There is more accumulation next to the surface and less as one moves up the atmosphere. It is not back-radiation but a general reduction in the energy/radiation as one moves up the atmospheric column and all the molecules are involved in this.

  295. Mosher, you pointed me to an experiment where you claimed 100% CO2 blocks all IR. It didn’t. It did seem to block the IR fromt he hottest part of the candle (1100C to 1400C) but not the cooler parts.

    I doubt the earth is on fire.

    And then you bail.

    But I ams still curious about your claim:

    Mosher: “The height at which radiation finally escapes to space is raised.”
    Me: “How far for each ppm extra of CO2? 1cm? 1m?”

    Don: “At the same desert locale there was a measurable and significant difference in the diurnal temp variation”

    Almost none in the winter. And July 3 was one of the few days where a daytime minor spike was really noticeable..

  296. Steven, I get it. Trust me.

    You don’t know how to train knuckleheaded recruits. You are trying to get Bruce to think, but he wants to feel. If he cannot put his hand out and feel the warming effect of downwelling IR, at the surface where he lives, he ain’t gonna care too much about where IR exits the atmosphere, way up there where he doesn’t live.

    I have been attempting get him past thinking that IR don’t matter. If I had him in my charge I would make him walk guard duty every night, until he experienced an epiphany. It wouldn’t take too many nights for him to come to me and say, “Sir, I see the light! I mean I feel the warmth from that IR stuff. Halalooya!”

  297. Bruce,

    Maybe I am wrong in trying to make the point, but the point was that on Jan 3, the night time temperature did not decline much, due to back radiation from the high humidity/cloud cover. On July 3, it was dry and it got a lot colder, relative to the daytime temp. You could put your hand out on those days and feel the relative temperature differences, from day to night. Whatever.

    Have you been to scienceofdoom. He explains basic concepts pretty well. Read this stuff and related sections, and I think you may benefit:

    http://scienceofdoom.com/2010/07/31/the-amazing-case-of-back-radiation-part-three/

    here are semantics which can confuse those less familiar with thermal radiation.

    If we consider the specific terminology of heat we can all agree and say that heat flows from the warmer to the colder. In the case of radiation, this means that more is emitted by the hotter surface (and absorbed by the colder surface) than the reverse.

    However, what many people have come to believe is that the colder surface can have no effect at all on the hotter surface. This is clearly wrong. And just to try and avoid upsetting the purists but without making the terminology too obscure I will say that the radiation from the colder surface can have an effect on the warmer surface and can change the temperature of the warmer surface.

    Here is an example from a standard thermodynamics textbook.

    http://scienceofdoom.com/2011/01/30/understanding-atmospheric-radiation-and-the-%E2%80%9Cgreenhouse%E2%80%9D-effect-%E2%80%93-part-three/

    As emission of radiation is reduced due to increases in absorbing gases, with all other things being equal, the planet must warm up. It must warm up until the emission of radiation again balances the absorbed radiation (note 1).

    Another way to consider the effect is to think about where the radiation to space comes from in the atmosphere. As the opacity of the atmosphere increases the radiation to space must be from a higher altitude. See also The Earth’s Energy Budget – Part Three.

    Higher altitudes are colder and so the radiation to space is a lower value. Less radiation from the climate means the climate warms.

    As the climate warms, if the lapse rate (note 3) stays the same, eventually the radiation to space – from this higher altitude – will match the absorbed solar radiation. This is how increases in radiatively active gases (aka “greenhouse” gases) affect the surface temperature (see note 4).

  298. Don: “If he cannot put his hand out and feel the warming effect of downwelling IR”

    Or make it do work …

    And this “downwelling” is a misnomer right? It is supposedly in all directions. Up down left right etc etc.

    How much of the claimed measured IR is actually going down? TomVonk claimed it wasn’t actually going straight down.

    “As the opacity of the atmosphere increases the radiation to space must be from a higher altitude.”

    How much higher for each ppm of CO2? 1 meter. 10 meters.

    And wouldn’t a higher altitude imply a larger surface area making it more efficient?

    I’d like to get back to the failed 100% CO2 experiment that let IR through. I wonder if the CO2 got warmer? And what convection would do.

  300. For arguments sake, lets say at 280ppm CO2 absorbs all of its IR within 100 meters of the surface radiating IR. And 380ppm absorbs all of its IR within 99.9 meters. I don’t that would increase the height at which heat was radiated into space. It would lower it.

  301. The downwelling IR is the part that goes down, Bruce. Since it is radiated in all directions, one of those directions is “down”, right?

    The charts in comment #82582 show you how much downward radiation was measured near the surface, Bruce. Do you think they just sit there every day and make that stuff up? If you can’t believe that they are measuring downwelling IR, then you are wasting your time thinking about this any further. You already have your answer.

  302. Charts, science of doom and gloom… KISS
    Bruce the Moon has no atmosphere. It is freezing cold at night. Try and figure out why we are not freezing cold at night. Obviously some of that daytime heat must be getting trapped by the atmosphere.

  303. Don, they are actually using an equation based on black body emissivity and the IR is usually less than the IR flowing out of the instrument. And we know the instrument doesn’t warm up the earth.

    “In order to calculate the incoming LW irradiance at the detector, the temperature of the pyrgeometer body must be known. This is measured using a thermistor (type YSI44031), which is located beside the cold junctions of the thermopile. The downward (or upward) longwave radiation is then calculated using the following formula :-
    LW = Uemf/S + ( 5.67*10-8 * Tb4 )
    where Uemf is the output voltage from the thermopile, S is the calibration constant of the instrument, and Tb is the pyrgeometer body temperature, measured by the thermistor, in degrees kelvin. Note that for an upward facing pyrgeometer, the thermopile output voltage will in most instances be negative. This is beacause the upwelling irradiance from the pyrgeometer is likely to be greater then the incoming irradiance from the sky.”

    Richard: “Try and figure out why we are not freezing cold at night.”

    Deserts can drop 50C from day to night because of the absence of water vapor. Jungles rarely drop much overnight. CO2 is usually considered the same.

    Figuring out the effect of CO2 divorced from water vapor is the tricky part.

    Remember in the really big experiment temperatures have not risen for 13 years while CO2 has risen.

  304. Dallas (#82785-6)

    [“The big 390 up and the 300 whatever it is arrows just tend to confuse. One might say, exaggerate the situation.”]
    .
    I agree. T’is way more complicated than that!
    .
    Thank you (and Lucia and SteveF) for an interesting, informative and courteous discussion.

    Regards,

    Arfur

  305. whoa Richard,

    Settle down boy. Bruce has never been to the moon. He won’t believe what he experiences here on Earth. Why should he be impressed by what allegedly happens on the moon? He will just tell you that it is freezing cold at night a lot of places on Earth too.

  306. OK, Bruce:

    “Figuring out the effect of CO2 divorced from water vapor is the tricky part.”

    So are you admitting that water vapor is a greenhouse gas?

  307. Humid air holds a lot more heat.

    Don, do you agree that CO2 seems particularly weak (non-existant?) as a GHG in the absence of water vapor?

  308. It is effect is smaller, it is not non-existant, water vapor acts as a positive feedback, their blocking effect on the spectrum is at different frequencies.

  309. Bruce,

    Now that you realize water vapor and C02 are greenhouse gases, I suggest you ask Mosher that question. Biased Bugs is winging it, and I don’t want to do that. Mosher will give you an honest answer that is not IPCC propaganda. You can also do some more reading on scienceofdoom regarding climate sensitivity and feedbacks, and also check out two, who in my opinion are credible skeptic scientists, Richard Lindzen, and Roy Spencer. I will just say that I ain’t worried about getting roasted by C02. Keep at it Bruce.

  311. Don, then we could discuss albedo.

    “An increase of 1% in albedo results in a decrease
    of about 0.6K in Te.”

    http://sait.oat.ts.astro.it/MSAIS/2/PDF/168.pdf

    I would then suggest that a 1% decrease = a .6K rise.

    “We correlate an overlapping period of earthshine measurements of Earth’s reflectance (from 1999 through mid-2001) with satellite observations of global cloud properties to construct from the latter a proxy measure of Earth’s global shortwave reflectance. This proxy shows a steady decrease in Earth’s reflectance from 1984 to 2000, with a strong climatologically significant drop after 1995. ”

    http://www.realclimate.org/palle_et_al06.jpg

    http://www.sciencemag.org/content/304/5675/1299.abstract

  312. Bruce.

    I said nothing about 100% of C02 blocking IR. I asked you to admit the existence of downwelling IR and do the following:

    PREDICT and EXPLAIN what would happen if you interpose C02 between a IR source and an IR detector.

    The fact that you were unwilling or unable to do so tells me that you are either ignorant or willfully obtuse and incapable or learning.

    You first claim that Downwelling IR does not exist. you then deny the observations when shown, you then claim that you know it can do no work.

    You could claim that you dont know if it can do any work. But you didnt. You claimed it can do no work.

    To test your understanding of the core physics I posed a simple demonstration. You could not predict the outcome. you dont understand the physics, you merely google. So, excuse me but I will answer Don’s questions because he is open to discussion.

  313. Owen.

    You are welcomed. That paper makes it all pretty simple. he even addressed the “saturation” argument. What people dont realize is that the GHGs do not have to block all the IR to raise the temperature at the surface. Also, the effect C02 has changes as you go higher in the atmosphere.

  314. Explain how C02 blocks incoming energy, Bruce. Does it have anything to do with that radiative transfer stuff, that you have been denying, up to now? You refuse to engage in honest discussion, Bruce. Everybody can see that.

  315. Bruce (Comment #82845)
    October 2nd, 2011 at 4:16 pm
    Don, I think we first need to address the amount of incoming energy blocked by CO2. More CO2 may cool the earth.

    Why, there is no ‘blocking’ of IR by CO2?

  317. Mosher: “I said nothing about 100% of C02 blocking IR. ”

    Yet you posted a link to a video claiming CO2 blocks all IR.

    Don: “Explain how C02 blocks incoming energy, Bruce. Does it have anything to do with that radiative transfer stuff, that you have been denying, up to now? You refuse to engage in honest discussion, Bruce. Everybody can see that.”

    Don, just because I attack your theory with your own theory doesn’t mean I buy everything your theory claims.

    I find it fascinating the way mosher seems to think the only way earths energy budget changes is because of CO2.

    But for those CO2 fans …

    “A conventional estimate is that soil and vegetation take in roughly 120 billion tonnes, or gigatonnes, of carbon each year through the natural process of photosynthesis.

    The new study, published in the science journal Nature, says the uptake could be 25-45 percent higher, to 150-175 gigatonnes per year”

    http://www.physorg.com/news/2011-09-global-carbon-benchmark.html

  318. Yet you posted a link to a video claiming CO2 blocks all IR.

    no. I asked you to predict the outcome. you could not. The co2 does not block all the IR. It doesnt block all the IR in the atmosphere either. try again.

    How do C02 sensors that use IR detectors work to detect minute increases in C02?

  319. I find it fascinating the way mosher seems to think the only way earths energy budget changes is because of CO2.

    Wrong:

    Earths energy budget can change in many ways. C02 is only ONE component. The other components include
    1. changes in TSI
    2. changes in aerosols
    3. changes in albedo
    4. changes in ALL the GHGs

    C02 by the last recknoning is perhaps 50% of the sum total of changes in forcings.

    So, wrong again. bruce. Wrong about the existence of downwelling IR. ignorant about the physics of RTE.
    wrong about all the causes of global warming.

    Keep going, your up for a Goddard

  320. yes bruce closing the C02 budget is one of the uncertainty’s of climate science. However,

    “The new study, published in the science journal Nature, says the uptake could be 25-45 percent higher, to 150-175 gigatonnes per year”

    That tell you one thing. do you know what it tells you about crabon sinks?

    Do you know that it has nothing to do with the physics of RTE?

    And how do those C02 sensors I pointed to you work? That is, how do they use measurements of IR to deduce C02 concentrations?

    give it your best monktopus effort.

  321. Don.

    You will love the stuff Bruce linked to to prove his point

    http://apollo.lsc.vsc.edu/classes/met130/notes/chapter2/sw_atm2.html

    “Therefore, the atmosphere is fairly transparent to incoming solar radiation.”

    hehe so much for more C02 keeping us cool.. but wait..
    There is more.. here is the source bruce cited

    Google is not his friend

    http://apollo.lsc.vsc.edu/classes/met130/notes/chapter2/lw_atm2.html

    Opps theres that pesky back radiation

    And your question to Bruce is covered here

    http://apollo.lsc.vsc.edu/classes/met130/notes/chapter2/spectrum_interactive.html

  323. Thanks for the laugh mosher.

    I’ll give you a hint … “fairly transparent”. is on the shortwave radiation page.

    I didn’t make up the claim that solar energy is blocked by GHG’s, but since you believe in it, try for a second to believe in all of it, not just the parts that adhere to your failed theories.

    37% of solar energy is near infrared. 11% is far infrared.

    http://apollo.lsc.vsc.edu/classes/met130/notes/chapter2/plank_sun_closer_look.html

    Either CO2 blocks infrared or it doesn’t.

    http://eesc.columbia.edu/courses/ees/slides/climate/absorption.gif

    Your obsession with the failed CO2 theory clouds your judgement.

  324. Mosher: “The height at which radiation finally escapes to space is raised.”

    You never did answer my question.

    How far for each ppm extra of CO2? 1cm? 1m?

  325. Mosher:

    “1. changes in TSI
    2. changes in aerosols
    3. changes in albedo
    4. changes in ALL the GHGs

    C02 by the last recknoning is perhaps 50% of the sum total of changes in forcings.”

    Maybe you could quantify that.

    Albedo changes seem to be huge. 2.5% drop in the late 1990s.

    You remember the late 1990s?

    “”Earth’s climate is driven by the net sunlight that it absorbs,” says Philip R. Goode, leader of the New Jersey Institute of Technology team, Director of the Big Bear Solar Observatory, and a Distinguished Professor of Physics at NJIT. “We have found surprisingly large—up to 20 percent—seasonal variations in Earth’s reflectance. Further, we have found a hint of a 2.5-percent decrease in Earth’s albedo over the past five years.””

    http://media.caltech.edu/press_releases/12131

    “In the May 6, 2005, issue of the journal Science, the CERES Science Team reported Earth’s shortwave albedo has been steadily declining since the Terra CERES instrument began making the measurement in February 2000. Over the 4-year span (2000 through 2004), the CERES instrument measured an albedo decrease of 0.0027, which equals 0.9 watt of energy per square meter retained in the Earth system. ”

    http://earthobservatory.nasa.gov/IOTD/view.php?id=5484

  326. If a .27% drop in albedo over 4 years = .9W/m2
    Then a 2.5% + .27% drop should equal 9.233W/m2 over 9 years.

    Not trivial. Is it 3x or 4x or 5x CO2’s current effect claimed?

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