First Results from THE BOX: Investigating the Effects of Infrared Sky Radiation on Air Temperature

July 29th, 2010 by Roy W. Spencer, Ph. D.

(UPDATED: 3:10 p.m. July 29, 2010 with temperature difference plot)


As promised, here are the first results from my little backyard experiment to investigate the role of downwelling infrared (IR) sky radiation on air temperature. (High school students looking for a science experiment, pay attention).

It’s a heavily insulated box that — theoretically — should chill air at night to a temperature below that of the outside air. The following is a conceptual design of The Box before I built it, along with the key components:

This all came about because I got tired of being asked about the theory behind global warming, specifically, how can downwelling infrared sky radiation from greenhouse gases (mostly water vapor, to a lesser extent CO2) cause global warming of the Earth surface, when the emitting temperature of the sky is colder than the surface?

Some people are convinced that this cannot happen, since the 2nd Law of Thermodynamics says energy naturally flows from higher temperature to lower temperature. In contrast, the mainstream science community, while agreeing the NET energy flow is from warm to cold, you can still cause warming by adding more greenhouse gas to the colder atmosphere. This happens even though the IR emitting temperature of the sky “causing” that warming is 10′s of degrees colder than the surface.

[NOTE: the direct warming effect of more atmospheric CO2 is small; its the resulting indirect warming (positive feedbacks) from clouds and water vapor that has most scientists worried. But not me...I think the net feedbacks are negative.]

The Box

So, since I have two automated weather stations in my backyard, I decided to build a heavily insulated box that would contain a small amount of air, and try to reduce all the other kinds of energy exchange between that air sample and the environment to a minimum EXCEPT for the influence of the downwelling sky radiation.

The air sample and the sky would be allowed to exchange IR radiation, and the colder the infrared emitting temperature of the sky is, the colder the air in the box should become compared to the air outside of the box. More about that later.

While we might not put the debate to rest with such an experiment, we can build some intuition about the energy flows that cause day and night air temperatures to be what they are. Of course, one could simply buy a hand-held infrared radiometer and take the sky’s “temperature” directly. But since everyone (myself included) has at least some trouble conceptualizing the role of infrared radiation in weather and climate (after all, we can’t see IR radiation), I thought that letting the IR effect be measured through its influence on temperature would make a bigger impact.

So, here’s a picture of the real thing that I took this morning, after collecting data since about noon yesterday:

The wireless data processor for the cavity temperature data is the little unit on the top. It sends a new temperature measurement every 5 minutes to my desktop computer in the house.

Here’s a close-up of the cavity. There is an insulating layer of air trapped between the two thin sheets of polyethylene, which are nearly transparent to infrared energy. The temperature sensor itself can be seen below that, in the cavity, the walls of which are painted with high emissivity paint (Krylon 1502 Flat White, IR emissivity = 0.99; Note that in the infrared, black is not necessarily more emissive than white…it depends on what the paint is made of, and whether the surface is rough or smooth).

Meanwhile, my regular weather station is about 20 feet away, and it is collecting air temperature and dewpoint data on the same schedule as The Box cavity temperatures are taken:

First Data from The Box
The first 17 hours of data, from midday yesterday until 8:05 a.m. this morning, are plotted below:

When I first closed up the box with the thermometer placed in the cavity, I was surprised how hot the cavity became. The maximum temperature recorded yesterday afternoon was 158 deg. F, and that must have been the limit for the sensor, because the temperature then flatlined for about an hour.

The reason for the high temperature was some direct sunlight reflecting off of one wall of the airspace, above the cavity. Even though the cavity was painted white, it still absorbed enough energy to make the air very hot. From what I have been able to gather, it is very difficult to get the solar reflectance of white paint above about 0.9.

It is interesting to calculate what rate of energy input would be required to cause this rapid rate of warming, which was about 3 deg. F per minute. If the cavity is initially in energy equilibrium, and we start reflecting 20 Watts per sq. meter more onto the cavity walls, about 10% of that (2 Watts per sq. meter) would be heating the paint, and so the air in the cavity.

According to my calculations, that would be more than enough to explain the initial rapid rise of temperature in the cavity on its way to 158+ deg. F. My calculations are only approximate, though, since I did not take into account the heat capacity of the cavity walls (painted aluminum foil), or the increased loss of IR as the cavity warmed, or conductive losses to the styrofoam and air space above the cavity.

But what we are really interested in is what happens when the overwhelming influence of solar radiation subsides. In the above plot, look at what happens as sunset approaches. Despite diffuse solar radiation still entering The Box from the blue sky, the cavity air cools to a couple of degrees below the ambient air temperature by sunset. Then, during the night, the cavity air averages about 4 deg. F colder than the outside air. This is easier to see in the next plot of the temperature difference between the cavity and the outside air, which we see remains pretty constant during the night:

To see how even a little diffuse sunlight from the sky can cause warming of the cavity, note what happened just after sunrise this morning…even though our yard does not see direct sunlight till close to 11 a.m. (very tall trees in the way), the blue sky started warming the cavity almost immediately after sunrise.

Then, after a short while, I put a white cover from a plastic cooler over the cavity to minimize the daytime heating of the cavity. At the end of the data plot you can see this solar cover caused the cavity to cool back down to the same temperature as the ambient air.

So, we already can see the cooling effect of infrared radiation in the data…in the form of cavity temperatures colder than the air. This happens from just before sunset, until sunrise — the period when there is little or no sunlight, either direct, or diffuse from the sky. But what, exactly, is the reason for this chilling effect?

Why Was the Cavity Colder than the Outside Air Temperature?
The temperature of virtually anything is the result of a balance between (1) energy gained and (2) energy lost. As long as the energy gained exceeds that lost, the temperature will rise. This was clearly seen when I closed up The Box, and the rate of sunlight absorption in the cavity exceeded the rate of energy lost by infrared emission (and any — hopefully small — conductive losses). The temperature skyrocketed.

But once the rate of energy loss exceeds that gained, then the temperature will fall, as was seen when The Box entered the shade. Then, then rate of IR energy lost (which increases rapidly with temperature) exceeded that gained from diffuse solar radiation, and the cavity temperature fell.

So, at night when there is no solar energy available, what is to prevent the cavity from getting very cold? Outer space is supposed to emit near absolute zero, 3 K. The Box’s cavity enters the hours of darkness at something like 300 K temperature. At 300 K, and assuming an IR emissivity of 0.99, the cavity is emitting IR at a rate of just over 400 Watts per sq. meter. Assuming the box is very well insulated, and is not leaking air, what is to prevent the cavity temperature from dropping well below freezing (273 K)?

The answer is downwelling IR from the sky. During the day in the summer, the broadband infrared sky temperatures viewed from the ground generally runs about 10 – 20 deg. F cooler than the near-surface air temperatures. This source of energy must exist, because without it the temperature of a cavity in a well insulated box at night would plunge even faster than we saw it heat up when exposed to indirect sunlight. And that rapid rate of temperature rise was due to only about 2 Watts per sq. meter! Imagine what in imbalance of 400 Watts per sq. meter would do.

Instead, the sky emits at only a slightly lower temperature than the surface, so the cavity cools only a little at night: about 4 deg. F cooling out of a “potential cooling” of 15 deg. F, assuming the IR emissivity of the cavity is 1.0.

By the way, I calculate that, if the cavity emissivity was only 0.90 rather than the advertised 0.99 (we really don’t know), we could explain the entire 4 deg. F drop based upon the cavity coming to a radiating temperature equal to that of the sky.

Presumably, once drier air arrives here in Alabama in another couple months, I should see larger temperature falls in the cavity, since water vapor is the Earth’s main greenhouse gas. In the meantime, I’m open to suggestions regarding simple ways to make The Box more efficient at rejecting all sources of energy except downwelling infrared radiation from the sky.

…a radiation source which some say, does not exist. ;)

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128 Responses to “First Results from THE BOX: Investigating the Effects of Infrared Sky Radiation on Air Temperature”

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  1. MikeC says:

    OK there Mr Mad Scientist… The next step is to see if in fact it is downwelling radiation at night which is preventing the air inside the chamber from cooling further… to do that you wait til about 2 AM and block the downwelling IR and see if the air in the chamber cools further.

  2. If you block the sky, you then have replaced it with something that is warmer than the sky, and so the cavity will warm. If that object was instead a perfect reflector, then the cavity will just see itself, and stay at roughly the same temperature it was before….until it warms upr from all other sources of energy leakage. Which might be interesting to determine how fast the box loses/gains energy non-radiatively.

    • Thomas says:

      If you have easy access to some dry ice or even liquid nitrogen it might be instructive to put a plate cooled to some very low temperature above the box, far enough away that any direct heat conduction should be minimal. Then you can directly see the effect of reduced incoming radiation.

      • Good idea…I will probably do that at some point. I can’t find any IR emissivity data for dry ice, though, so I have no clue how transparent it is.

        • Thomas says:

          The dry ice will face towards the sky while the bottom of the plate will face the box, so all you need to worry about is the material of that plate and its emissivity.

        • Anonymous says:

          Isopropal alcohol has a very low freeze point and is easy to handle. You could put isopropal alcohol in a large stainless or aluminum pot or baking pan and add dry ice. The bottom of the container would then be your low temperature radiation source. It would be easy to measure the source temperature.

    • Phil. says:

      The metallised Mylar film I mentioned should isolate the cavity from all radiation and will be of such low mass that it would be at the local atmospheric T. A double layer with an air gap should mean that the cavity doesn’t cool at all?

  3. Massim PORZIO says:

    First of all many thanks give us the opportunity to partecipate your experiment.

    I did some silly computation. Looking to the antenna which I guessed to be an UHF 6 inches length, I supposed that your chamber aperture should be 0.0232m^2 (6inches by 6inches).
    If you used polyethylene its thermal conductivity is 0.42…0.51W/(m*K). Given the thickness of 25microns and the about 400W/m^2 radiation the thermal insulation of each foil should be about 1.7…2K.

    Could this limit the nightime cooling?

    Thank you again for give us a lot of your time.

    Massimo

    • Anonymous says:

      Sorry, I did a mistake, the computed delta temperature was for both the foils not for each, as I wrote.
      (I should read my post before click on the submit button, but I always miss to do it)

  4. Loodt Pretorius says:

    From a mining ventilation perspective, you should include the following instrumentation for measurement of the air in your cavity: – A pressure sensor and a wet-bulb temperature sensor. As you are not dealing with an ideal gas but with air, that contains moisture, your dry-bulb temperature measurement on its own is meaningless.

    In addition, you failed to mention whether your box is indeed sealed. If not, during the day, the increased temperature in the box will increase the pressure of the air in the box, if not totally sealed, some of the air will escape, setting the stage for the next scenario. In the afternoon, when the outside temperature drops, the air inside the box will cool, causing the pressure to drop. With this suction acting on the air some of the moisture in the air will evaporate, and the additional cooling will be the result.

    Unless you have: pressure; and wet and dry-bulb temperature readings, talking about how the air reacts is realy very pointless.

    • yes, I also track the humidity of the air in the cavity…I use it to give some idea of whether the cavity is airtight, in which case the dewpoint would remain relatively constant. I do not think the enclosure is airtight…although I don’t want much leakage of air, for obvious reasons.

      • Anonymous says:

        And how did the humidity change throughout the day?

        If you have temperature, humidity and pressure readings you can calculate the change in entropy of the air in your box.

        Post the results of your readings: pressure; temperature; and humidity, for every 15 minutes or so, in an Excel spreadsheet, and let’s see what the guys can calculate for you!

      • Anonymous says:

        Sorry, forgot to sign in again!

        And how did the humidity change throughout the day?

        If you have the temperature, humidity and pressure readings you can calculate the change in entropy of the air in your box.

        Post the results of your readings: pressure; temperature; and humidity, for every 15 minutes or so, in an Excel spreadsheet, and let’s see what the guys can calculate for you!

  5. Brego says:

    [In the meantime, I’m open to suggestions regarding simple ways to make The Box more efficient at rejecting all sources of energy except downwelling infrared radiation from the sky.]

    All polyethylene transmits IR well. You may want to seek out black polyethylene sheeting. It is used in construction and agriculture. You could probably find it at a home improvement center or a landscaping outfit. They may even be willing to give you a small piece.

    It does get hot itself in the sun (absorbs visible), but if only used for the top barrier it might not cause any problems.

    I also agree with Loodt above insomuch as it would be nice to know the humidity in the box.

  6. Could not the lower temperature be explained more simply by the difference in albedo/emissivity between the cavity and the general surroundings? i.e. the cavity can radiate at the same rate as the surroundings at a slightly lower temperature due to the special paint. Doesn’t this explanation remove the need for back-radiation as an explanation?

    • so, if I use aluminum sheeting, which is highly reflective in the IR, then I should be able to make my own dry ice, right? ;)

    • Anonymous says:

      If my theory is correct then the temperature inside the cavity would increase if lined with foil, as due to the lower emissivity of the foil there needs to be a higher temperature at equilibrium. Easy enough to try – suggestion for another experiment. Cheers

  7. Demesure says:

    “Instead, the sky emits at only a slightly lower temperature than the surface”

    Dr Spencer,
    First, I would like to sincerely thank you for your dedication to explain the science.
    - What happens if the downwelling emissions of the sky are replaced by the emissions of a much colder plane placed slightly above the cavity (enough to minimize conduction), for example a plate cooled at 0°C with ice. Will the cavity’s temperature drop even more rapidly ?
    - Diffuse sunlight seems very efficient in raising the temperature of the cavity which is quite sensitive to a slight imbalance in radiative exchange. So do you think an optical concentrator can raise the cavity’s temperature using sky’s radiations AT NIGHT ? (imagine using round lens, Fresnel lens, parabolic mirrors, you double, trebble… the W/m2 your cavity was receiving in the above experience).

    • you cannot focus an extended source of thermal radiation.

      • Anonymous says:

        I believe that what really happen using the parabolic mirror to frozen, is just a “selection” of the “back radiation” fronts.
        I mean: when you put an insulated jar of water in the focus of the parabolic mirror all its radiation is emitted to the atmosphere, while only the “back radiation” perpendicular to the diretrix of the mirror is focused, so the incoming flux is highly reduced.
        Instead, if the jar is placed under the open sky, it emits and receives the radiation fluxes both in a wide angle of directions.

        May I be right here?

        Massimo

  8. KuhnKat says:

    Dr. Spencer,

    since there is IR coming from all over the sky, would you be averse to doing the experiment again with an opaque pipe around the window so that no direct IR can come from trees, houses, or other objects in the immediate area? I would like to guarantee only clear sky IR as opposed to blackbody from local objects. The gentlemen doing the refrigerator solar ovens specify that any IR coming from local objects prevent much cooling. They also pointed out that, obviously, low clouds would also be an issue. I am assuming last night was reasonably clear for you?

    I admit I was expecting a lower temperature.

    Thank you for your efforts to clarify this area for us.

    • the pipe would have to be highly reflective, like an aluminum sheet. The cavity already has to look through an aluminum-lined “tube” (the airspace with polyethylene film on either end) that is deeper than the cavity, so its view of the sky is already restricted somewhat..

      • Anonymous says:

        I just left a comment with a link to a gentleman in the business of solar ovens who claims he has achieved as much as 20F below ambient.

        I agree the pipe, or funnel, would need to have a reflective inner surface. The main idea is to insure there re no radiative vectors to objects in the area that enter the opening at any angle.

        Since I’m being picky, is the data recorder in the same spot as shown in the picture? Could it be moved below the top of the experiment?

        Here is another link to the same gentleman with instructions for making his funnel used for both cooling and cooking. It is interesting in that it is designed to create a line of heat, or cool, rather that a point focus. In the article he claims making ice with 47F ambient minimum night temp.

        In it he again indicates there must not be any objects in radiative line with the opening, or, in the case of his cooker, the funnel.

        http://www.solarcooking.org/plans/funnel.htm

        • OK, I added a funnel made of aluminum sheeting, and I think I can see enhanced cooling right away. We have cirrus clouds right now from surrounding thunderstorms, but it typically clears during the night. If it does, we’ll see whether we get greater cooling below ambient tonight.

  9. KuhnKat says:

    Solar oven as refrigerator.

    See this link for a soloar oven that has achieved 20F below ambient at night with the only proviso being directed to open sky with no buildings, trees, etc in the radiative path.

    http://www.josephprep.com/Provisions/Solar_Funnel/

  10. How about a few filters in front.

    Edmund Optical have UV/vis pass and IR pass filters

    Germanium IR pass filters £500! Plastic IR pass £40 to £66
    Bandpass and other high/low pass filter frequencies available

    This filters both ways so not sure how well it will work.

    It would also suggest another experiment – heating water with LW ir or visible – how well does IR heating the surface laye conduct downwards compared to visible

  11. Kevin says:

    Dr. Spencer;

    With respect, it is the speed of heat…………

    You wrote: “So, at night when there is no solar energy available, what is to prevent the cavity from getting very cold?”

    Well we need to remember that the Styrofoam which you selected to “eliminate” conduction actually has a pretty good thermal capacity (per unit volume). It also has a very low thermal conductivity (this is probably why you chose it). So it has a very low speed of heat (aka thermal diffusivity). It is “charged, filled, energized” to a temperature that is somewhere between the coldest and warmest temps of your summer day there in Alabama. This happened slowly when you took this Styrofoam out of your air-conditioned (a presumption on my part) house into your yard.

    After the sun goes down the Styrofoam is “discharging, emptying, de-energizing (probably a made up word)” back into the air in your enclosure. Since it has such a low speed of heat it can do this for a long time before its temperature starts to drop. It could probably “fill” the meager thermal capacity of your air volume with heat for many hours before its temperature drops by 1 degree.

    Sorry, I fail to see any proof that increasing “downwelling” radiation (due to GHGs) can cause the temperature of the surface of the Earth to slowly rise.

    Sorry if my comments about measuring the temperature of surfaces ignored your concerns about cost. I think if you look at some of the electronic suppliers on the web you can find any number of inexpensive (less than $250) sensors to measure the temp of a surface. I stand by my statement that accurately measuring the temperature of air is more difficult than accurately measuring the temp of a surface.

    Yes the synthetic diamond approach is not cheap, but over the last two decades we (the taxpayers) have spent something like 100 billion dollars to “prove” this hypothesis, pity that we could not use some of that for an empirical simulator of the atmosphere even if it was crude.

    Again, I do not disagree that gases in the atmosphere emit photons, some of which travel back to the surface and are absorbed. I disagree that this can cause the energy at the surface of the Earth to ratchet up at the end of each day. For these photons to change the outcome they must lower the “average (admittedly a dangerous word)” speed of heat through the atmosphere to an extent that compares to the cycle time (24 hours).

    I am still of the opinion that increases in “greenhouse” gases actually work to increase the speed of heat through the atmosphere since slightly more energy flows at the speed of light (very speedy) versus the speed of heat (quite a bit slower). The practical consequence of this is that the gases in the atmosphere warm slightly faster at sunrise and cool slightly faster at sunset. This effect is so small that we probably could not spend enough money to measure it. The historical temperature databases do not contain the necessary data and attempts to waterboard them until they confess to AGW has wasted 2 decades and 100 billion dollars.

    Might I suggest that you next fill the cavity between your two plastic films with carbon dioxide (check out the seltzer making machines that use the same CO2 cartridges as the BB guns that some of us had as youths) and repeat the measurements (quickly of course so Day #1 and Day #2 are similar) ?

    Cheers, Kevin.

  12. Reader says:

    Dear Dr Spencer,

    Thank you for taking this effort.

    I personally have never doubted that greenhouse gases are warming the earth’s surface because on balance it seems many people agree with this, both agw skeptics and non-skeptics alike. I always thought the difference of opinion was the sensitivity i.e. the feedback response in relation to increased warming from additional CO2.

    Consequently I’ve never questioned the science behind it. I find this very interesting however.

    Although I am now confused as to why the cavity temperature dropped below the external air temperature.

    Aren’t the greenhouse gases present day and night so any change (assuming clear skies in all cases) is directly attributable to incoming solar radiation.

    Don’t your results say more about the cooling response of earth in the absence of the sun ?

    Was the answer to this explained in your first “yes virginia” post. If so … just say yes and I will re-read it and try and understand it better.

    • Anonymous says:

      The cavity temperature dropped below air temperature because the system radiated more heat than it absorbed from the surrounding air. Once the temperature dropped enough in the cavity there was enough energy absorbed by conduction to replace the net amount radiated. Equilibrium was achieved and the cavity temperature difference stabilized. It would have stabilized at a lower temperature had back-radiation been absent.

      • Reader says:

        So it seems to me that this would suggest that the experiment provides evidence of the warming passing from cooler air to warmer air (first mentioned in his “yes virginia” post, unless there is some other explanation as to why the cavity became colder than the atmosphere.

        I would be interested seeing result graphs over a longer period though.

  13. Bill Hunter says:

    We all know heat moves from hot to cold. We also know that overcast nights don’t cool as much.

    If the cavity is at the sky temperature that is what it should be back radiation or not as if it gets colder the relatively warm sky should warm it.

    At a minimum we have radiation on demand. But this does not establish back radiation.

    The cavity cooled faster because it had less heat content with the outdoors having access to a lot more heat in storage.

    Why they stablized at a different temperature I don’t know could be relatively warm objects in line of sight of the outside sensor keeping the temperature up.

    I am not sure you can strictly test for back radiation. Radiation on demand might be like an electrical circuit with the flow of electrons increasing as the potential difference increases and maybe for all practical intents and purposes in atmospheric physics that might be the same as back radiation.

    But if there are differences between heat and energy as one poster suggested that could be very important to the greenhouse effect as greenhouse gases absorb and re-emit without converting to heat whereby blackbody’s only emit via heat, or so I have been told.

    But one interesting thing I think you could do with this set up without much fiddling is check for variation in the speed of cooling.

    An object with the same heat content conditions, humidity, ambient temperature, etc. should have different cooling rates depending upon the “net” radiative flow. That would move you one step closer. As it would establish the ability of the heated plate to get warmer in the presence of a colder plate.

    You could use an awning. If there is back radiation the temperature of the destination should affect the cooling rate in addition to changing the equilibrium temperature. You could use a glass awning to ensure you are mostly only affecting IR.

  14. Hello, Dr. Spencer,

    As an electrical engineer, and a long time out of college, I have only been able to vaguely follow the discussions here. But I do find them interesting.

    At one point you make the statement “It is interesting to calculate what rate of energy input would be required to cause this rapid rate of warming, which was about 3 deg. F per minute.” Of course, being an electrical engineer, my mind immediately thinks: why not just put a little heater (e.g., a small resistor) in the cavity and experiment to find out what power input causes the the same rate of warming? The energy output of the resistor can be measured very precisely with cheap equipment (DVM).

    But it seems like one could take this line of thinking a step further. Some of the responders bring up issues related to temperature, and the rate of change of temperature. So why not eliminate temperature from the equation altogether?

    I’m envisioning modifying your cavity experiment by placing a metal plate (aluminum or copper, maybe?) in the bottom, facing upwards. Maybe a different shape would work better, like a half-sphere or a parabolic shape. I guess you would paint this plate with your special paint.

    But here comes the (hopefully) interesting part. The plate contains a heater that uniformly heats it. The temperature of the plate is measured and precisely controlled with a feedback system. Keeping the plate at the same temperature as the surroundings (the cavity, and the external ambient) should all but eliminate losses through conduction and convection, right? And the heat capacity of the different materials should drop out of all the calculations, right?

    The heat energy emitted from the plate must be equal to electrical energy input to the plate, right? And the electrical energy input to the plate can be measured precisely.

    Just a thought . . .

  15. Hervey Bagot, Australia says:

    Dr Spencer — I once met an Israeli military officer who told me that in Roman times ice (i.e. water-ice) was made in Palestine by placing a tray of water at the bottom of a sufficiently deep, narrow, well which was otherwise dry. At night the water froze.
    Hervey Bagot
    Adelaide, Australia
    July 30, 2010

  16. benpal says:

    Thank you Dr Spencer for trying to explain climate science in simple, easy to understand terms and to provide tangible evidence through your experiment. Statistics are surely interesting, but the foundation of science has to be based on hard and reproducible facts.

  17. Gordon McKeown (UK) says:

    Dr Spencer

    Please excuse a probably niave question. Are there any long term accurate figures for night time temperatures in very dry desert regions that have a stable climate? Would they indicate the direct effect of C02 increases?

    BTW I greatly enjoyed your latest book and please keep up the good fight for sanity in the Climate debate.

  18. Mac says:

    Dr. Spencer,

    Great stuff.

    How about a radiation shield below and around the box to prevent IR from the ground and nearby objects. Well ventilated of course. Would that effect part of the correction some have worried about? Perhaps the box by itself is insulated well enough with out the bother.

    Thanks again, VERY interesting.

    PS I am an aquarian but hopefully not the type you referenced the other day.

  19. Steve Fitzpatrick says:

    Roy,

    Your kids must have had super science fair projects!

    If your sensor is a thermistor, that could be adding some heat to the inside of the box. A 2 Kohm thermistor powered at 5 volts would add about 12 milliwatts continuously.

    Might be worth a look at the sensor specs.

  20. pochas says:

    Bill Hunter says:
    July 29, 2010 at 8:46 PM

    “The cavity cooled faster because it had less heat content with the outdoors having access to a lot more heat in storage.”

    Yes, including heat from mixing the surface air with the warmer layers above.

    “I am not sure you can strictly test for back radiation.”

    Especially since the ground outside the box sees the same back radiation as the space inside the box.

    I would attribute the temperature difference to the different heat capacities as you say, but also to the fact that the outside air can mix with the warmer air above the temperature inverson. The results of this experiment will be sensitive to wind speed.

  21. Massimo PORZIO says:

    Dear Dr. Spencer,
    I’m back again with the polyethylene (PE) thermal conductivity issue, because maybe I’m not been clear in my former message. That is, I know that you made the pre-chamber to stop the air convection and conduction, but maybe you missed one aspect of that design which is the IR backward emission of the upper PE film due to its own temperature. Despite PE has a discrete transmittance at the SW IR bands, being a solid it emits IR radiation depending on its own surface temperature.
    I did a check about using my own thermopile based IR thermometer.
    I bought a thin PE bag sized 210 by 297 mm (the one used to protect the A4 sized documents) and I checked its IR transmittance using my left hand this way: I sighted the thermometer to my naked hand and I measured 35.5°C then I put my hand into the bag and I measured 35.2°C, good enough I thought.
    Then placed the thermometer so it was sighting the wall at 28°C and I disposed the air gun of my lab surface mount device reworking station so that it emitted a flow at 110°C with the blow perpendicular to the thermometer and I verified that (as imagined) the thermometer didn’t detect anything.
    I turned of the air gun, and I put the bag perpendicular to the thermometer too, so it was parallel to the wall, and the thermometer reported a little less than 28°C, 27.2°C (I guess because of the two layers of the bag of course).
    At last I filled the bag with the air flow from the air gun. The bag warmed in few seconds and the thermometer immediately measured it to be 35.7°C, so I stopped the air flux. And the thermometer still measured the temperature of the PE, until the PE returned at the ambient temperature.
    I wondered of that because the thermometer has a Fresnel lens in front of the thermopile which is made of PE. So I investigated how the thermometer compensates for that lens, and I discovered that there is a thermometer inside the sensor which correct the additional IR flux due to the ambient temperature.
    I believe that the upper PE film of your box could be the culprit of most of the reduced nighttime cooling, because it adds an IR flux due to the internal PE surface temperature, which is the external ambient temperature at the external surface minus the delta temperature consequence of its thermal conductivity.

    What do you think about?

    Have a nice day.

    Massimo

  22. Andrew S says:

    Dr. Spencer.

    Everything in this post is my (subjective) opinion and can very well be wrong. Keeping this in mind, I’ll start with needed terminological clarification. I think people here have different definitions of what “back radiation” is. For the purpose of this post I define “back radiation” as ability of colder object to increase temperature of warmer object by way of thermal radiation.

    There are couple of (highly subjective) observations. First, you say “…a radiation source which some say, does not exist”. I believe this to be a straw man. I followed comments to your latest posts relatively close. I do not remember anyone claiming cold objects (the sky included) refuse to radiate towards warmer objects. Even if there are such people, proving them wrong does not advance your original “Yes, Virginia” statement in any way.

    Second, again, IMHO, typical argument in support of compatibility of Second Law and “back radiation” claims that opponents of theory confuse and mix what “temperature” and “heat” are. Argument then proceeds in great detail using terms “heat”, “flux”, “Poynting vector”, “emissivity” etc. I think, it is this argument that conflates heat and temperature. The argument cannot have bearings on Second Law because Second Law is expressed in terms of temperature and mention neither flux nor emissivity.

    Thermal radiation have two essential properties: flux and temperature. Flux defines how fast radiation can warm “target” object. (Two heaters can warm an object faster than one of them.) Temperature of radiation, on the other hand, puts an upper limit on result temperature of the subject of heating. (No amount of heaters can make the object warmer then heaters themselves.)

    This second property is often lost in the discussion. I think, it is incorrect to say “324 W/m2 is 324 W/m2, no matter where it comes from”? It does matter, because the temperature of radiation cannot be ignored.

    I put aluminium foil screen as close to my face as I can without actually touching it. “Back radiation” tells me I should feel heat coming at my face from my mirror image and I don’t.

    IR thermometers unable to take direct temperature measurement of objects colder then thermometer’s sensors.

    I’ve yet to see single experiment that demonstrates existence of “back radiation” as in “Yes, Virginia”.

    • Anonymous says:

      Maybe you can feel the *lack* of back radiation by facing a mirror freshly removed from the freezer so that it is not at the same temperature as the room. The back radiation phenomenon in atmospheric physics occurs because the near-surface air and the sky beyond the near-surface air are at disparate temperatures. According to Newton’s Law of Cooling, the heat flow is driven by the temperature DIFFERENCE.

      • Andrew S says:

        Huh? What is the relation between temperature of mirror and temperature of reflected object (my face)? Sorry, I did not get your point. The mirrors are good at reflecting radiation. They are terrible radiators by themselves.

    • Anonymous says:

      ***Thermal radiation have two essential properties: flux and temperature. Flux defines how fast radiation can warm “target” object. (Two heaters can warm an object faster than one of them.) Temperature of radiation, on the other hand, puts an upper limit on result temperature of the subject of heating. (No amount of heaters can make the object warmer then heaters themselves.)***

      This is the paragraph that is incorrect. Radiation does not have temperature. The amount radiated depends on the temperature of the heater, but the radiation itself does not have temperature. A heater can make the object warmer than the heater, if the object has no way of losing heat.

      • Andrew S says:

        “A heater can make the object warmer than the heater, if the object has no way of losing heat.”

        Would that mean that heat will flow from colder object to warmer one? Any reference of this happening?

        Of course, term “temperature” I use loosely in relation to electromagnetic radiation. Call it wavelength or frequency. In that sense I call solar radiation “hot” and atmospheric GHG radiation as “cold”

        • Anonymous says:

          Once again, the same problem.
          Infrared radiation refers to a specific band of frequency and wavelength. In that sense IR radiation from the sun or from green house gas are the same. They have no temperature, they just contain energy. REPEAT: there is no distinction between IR radiation from the sun and those from the green house gases (one is not “hot” and the other “cold”). The only difference is quantity.

          Radiation contains packets of energy. An object with high temperature emits many packets of energy. An object with low temperature emits few packets. The packets, when it hits an object, transfer that energy to the object regardless of its temperature. Thus, a low temperature object can emit IR radiation to warm up a higher temperature object (if the higher temperature object has no way of losing the extra energy).

          When you say heat flowing, you are lapsing back into a liquid metaphor which does not work well when it comes to radiative warming. It is much helpful (more precise) to think of energy transfer.

  23. Tom Davidson says:

    Dr. Spencer: You can make a very effective insulation with multiple alternating layers of aluminum foil and tissue paper. Such is used in large Dewar flasks used for transporting bulk liquid helium. The foil reflects radiation and the tissue prevents conduction and convection. If the air space between the layers of foil is evacuated, so much the better. The conductive leads for your thermal sensors may provide a ‘heat leak.’ If you can measure temperature remotely to the required precision, I recommend that. A control consisting of an identical box with the polyethylene sheets raised 2-3 cm above the cavity (to allow exchange of air) would also be helpful. Make sure the boxes have environments that as as similar as possible.

    • Anonymous says:

      Roy, listen to Tom. He’s right, I’ve used it before. Ten layers of foil, shiny side out, seperated by crumpled craft paper (grocery bag paper) in my case. You want the paper to touch the foil in the least number of points as possible and randomly spaced through the layers to minimize any conduction. Great insulator! Stop 99% of any heat before it ever gets to the styrofoam. That should do it.

      My test was of 1 sq.ft places on electric stove on medium, hand softly on top, temp 700F+. Abandoned test after few minutes with hand still fine but low sheet of paper smoking and about to catch on fire! Great stuff. R factor must be really high for it’s merely 1/8″ thick.

      • I suspect if you replace the paper with styrofoam, it would perform about as well. I’ve now got a few layers of foil separating the outside of the box (which is foil covered) from the inside cavity.

        • Anonymous says:

          Styrofoam? Well maybe, that would be true especially if the surface of the styrofoam was rough on both sides. It’s the micro scale foil to air interface that make it work so very well. Many layers of air gaps in the paper for minimum conduction and convection and many powers of emissivity to minimize the radiation (like ~0.10^5) for five layers of aluminum foil. Just don’t know if foil works at all if in contact with a flat surface.

          Sounds like you have plenty on the outside of the box and conduction through the polyethylene is going to be your limiting factor by conduction anyway. :)

          Real neat experiment! We’ve needed that, if not just for the fun. Thanks Roy.

          - wayne

  24. Willywolfe says:

    I live at an altitude of 7,400 feet in NM where it is very dry most of the year. From my solar engineering class many years ago I recall that the usual admonition about freezing is that a collector can freeze at around 40 F nighttime temperature. However, up here on dry cool nights I have observed after washing my car (it is a white car) in the evening that water on the car will freeze at nighttime air temperatures well into the 40s. I am guessing the downwelling radiation at this altitude with dry air is much lower. Perhaps if you repeat the experiment at high altitude it would result in lower temperatures in the box.

  25. Aero says:

    If you need absolutely best insulation, Aerogel is way to go. Yes, it is “quite” expensive.

    http://en.wikipedia.org/wiki/Aerogel

  26. RH says:

    Try turning the box upside down at sunset tonight.

  27. mkelly says:

    Well done from a fellow SHS’er. I really enjoyed this. It is a pleasure to relearn this information.

    I have never doubted that back radiation exists. It can be measured so it must exist.

    My contention is that CO2 absorbs/emits at 15 micro so it cannot heat the earth. I believe your experiment shows that. The cavity temperature is lower that the “surface” temperaure at night. Presumably there was CO2 above your box radiating down. But as CO2 is at 200 K (-99 F) it cannot heat the cavity.

    If the cavity had gone to a higher temperature than the “suface” temperature at night then we have a story.

    Yes I understood you were showing that back radiation existed since the cavity did not go to 32 F. But there is a difference between not going as low and an actual increase in temperature.

  28. Anonymous says:

    again the same question?… how do you take care of the temperature effects due to blocking of turbulent heat transfer?..

    and any comments on this?
    http://mindeavour.blogspot.com/2010/07/comments-on-backradiation.html

    • The point is to block all sources of energy, except radiative. So, we purposely try to minimize turbulent and conductive transfers of heat.

      Not sure what the link you provided has to do with the current discussion.

  29. Schiller Thurkettle says:

    There is remarkable agreement between the results of this experiment, and the experiences of those using ‘solar ovens’ which, properly oriented, act as night-time ‘radiative refrigerators’.

    Links:

    http://solarcooking.org/radiant-fridge.htm and
    http://www.provident-living-today.com/Alternative-Refrigeration.html

    BTW many thanks for investigating the physics involved in these phenomena. I have long been exasperated by claims that ‘we can’t explain it, therefore humans are causing it’.

  30. johnd says:

    How much difference would height above the ground make? Data from Geiger indicates most difference in air temperature during daylight hours but still some during night time.
    Should any difference be detectable by your test between readings taken at ground level and say from the top of a 2 story structure?

    • during calm wind conditions, there can be several degrees C difference between air temperature a few inches above the ground, and at standard observing height (5 or 6 feet), with the air next to the ground being colder. In my case, if I could get the box higher, it would have a more unobstructed view of the cold sky, and should go to a somewhat lower temperature.

  31. Derek says:

    Errr, I assume the air temperature sensor is getting some regolith (and vegetation) released heat, particularly at night (relative cooling by day I would also assume), and the cavity sensor is getting none, or very little (day and night).
    Would this be bound to produce a higher air temp than cavity temp at night, and slightly lower (air temp) during (at least the early part of) the day. ?

    If you have allowed for this, and I had missed it on my quick skim through so far, then please accept my apologies.

    yours,
    Derek.

  32. dp says:

    Drop your box to the bottom of a dry well such that the principle radiation comes from a narrow portion of the sky. Non-dry wells are known to freeze over even when the surface temperature is quite a bit warmer.

    • Mikael Lönnroth says:

      Here’s another nice page that shows the 24h measurements from both a pyranometer (radiation from the sun, cloud effects nicely visible) and a pyrgeometer (downwelling LW radiation).

      http://www.chilbolton.rl.ac.uk/weather/radiation_sensors.htm
      http://www.chilbolton.rl.ac.uk/weather/radiation_sensors_yesterday.htm

      Assuming the pyrgeometer readings are correct, Chilbolton experiences more or less constant downwelling LW radiation about 300-400 W/m2 throughout the 24h cycle.

      It would be interesting to know more about how these pyrgeometers are calibrated and what the science/methods are behind determining the initial calibration reference?
      Maybe this could provide some further evidence to the existance of downwelling radiation?

      • yes, all of these sky-looking radiometer measurements are evidence of “back radiation” (or, more properly, just downwelling sky radiation). In fact, since the instrumentation uses a thermopile, it is actually based upon the temperature difference across a detector, so when it measures a warmer sky versus a colder sky, it is measuring the temperature increase of an internal plate on one side of the thermopile from COLD sky radiation…direct evidence of what some claim does not happen.

  33. Terry says:

    Roy I haven’t thought this thru completely yet, but it seems wrong to me. The outside air and the air in the box should always be in radiative equilibrium. Since the outside air is still warm due to conduction and convective processes from soils, buildings etc then heat should be transferred by IR to the isolated air in the box and keep it at the same temp as outside. If not then there must some absorption going on in either the plastic film or the air space between them, that prevents IR going in but is transparent enuf at other wavelengths to cool. Doesn’t seem right to me.

    • IR *IS* being transferred from the air into the box, at least along the line of sight of the cavity. But very little of that originates from air near to the box…it is an accumulation of very weak IR emission through the depth of the atmosphere.

  34. Terry says:

    Re above, LPDE does have a strong absorption band at 1500 cm-1. Im thinking that this could be the reason.

  35. Joletaxi says:

    Just a question

    What,if any, shoud be the difference off t°, if the experiment would be ,in place off the box, too put the sensor in a very dry desert(T° could be very chilly there at night)?

    Anyway, I apologize to have say that D. Spencer had fouled me with the first proposal,I shoud havre say that he had give me a good lesson.
    best regards

  36. RH says:

    OK. I did a scaled down version of your experiment last night, but turned the box upside down. My box was about 689 cubic inches in size, my styrofoam was 1 inch thick. I used a laboratory thermometer through the box and styrofoam at the top of the box with the scale outside the box so it could be read each hour. I rounded the temperatures to the nearest half degree. My air temperature was taken from a Davis Vantage Pro2 about 6 feet above the box. I ran the exercise from 5pm to 11pm, but I think you can get the point.

    Results:
    Time Box Temp Air Temp
    5pm 32C 23.5C
    6pm 30C 23C
    7pm 28C 22.5C
    8pm 20C 20C
    9om 19C 19C
    10pm 19C 17.5C
    11pm 16.5C 14C

    I think the temperature dropped in your box because you had 4 inches of styrofoam between the temperature sensor and the ground, and the ground is the heat source at night. The air outside your box was measuring heat traveling from the ground to outer space.

    I would suggest that someone else do this experiment with more than one box at the same time to control the experiment. My hypothesis is that the heat is NOT coming from greenhouse gases in the atmosphere, but from the ground. I’m not buying into downwelling infrared.

    • Of course there is IR radiation flowing up from the ground. No one claimed there wasn’t. The current version of the box is heavily insulated and IR-reflective (foil-covered), specifically to exclude other sources of energy as much as possible.

  37. JAE says:

    As I’ve said many times, I think there is way too much focus on radiation and not enough on simple heat storage effects/thermalization and convection. Because of the predominant roles of thermalization and convection, it is really unlikely that small increases in radiation due to more GHGs have any measureable effect on temperature near Earth’s surface. That is probably one reason it not hotter in Atlanta than in Phoenix in summer, despite there being more than three times as much GHGs in Atlanta (for anyone here that still believes in the “cold desert night myth,” it is also hotter at night in Phoenix, BTW). In fact it is probably colder in Atlanta simply because so much heat gets tied up in the latent heat of evaporation there.

    Now, perhaps ironically, it may be a whole different story at TOA, where increases in GHGs will lead to more radiation and thus more/faster COOLING!

  38. RH says:

    Amen JAE.

  39. harrywr2 says:

    JAE,

    Mean temperatures in Georgia are higher then mean temperatures in Arizona.
    http://www.esrl.noaa.gov/psd/data/usclimate/tmp.state.19712000.climo

  40. Why not make a box similar to the one you are using but without the window with two thin sheets of polyethylene. In other words; a box with 6 similar sides. Then no radiation can enter and no radiation can exit.
    Put that box alongside the original one. Then at least you will be able to say with conviction that your theory:
    “If you block the sky, you then have replaced it with something that is warmer than the sky, and so the cavity will warm. If that object was instead a perfect reflector, then the cavity will just see itself, and stay at roughly the same temperature it was before….until it warms up from all other sources of energy leakage. Which might be interesting to determine how fast the box loses/gains energy non-radiatively.”
    is correct

  41. Pehr Bjornbom says:

    Dear Dr. Spencer,

    If you would heat the cavity electrically you could regulate its temperature to be kept equal to the air temperature. The cavity in such case could neither loose nor gain heat to or from the air through convection or conduction. Then you would not even need the polyethylene film cover because ventilation of the cavity would not change its temperature or add or remove energy from it.

    The cavity would loose energy by radiation and gain energy by electrical power, P (W), and by the back radiation I_b W/m2. Hence you could calculate the back radiation according to

    I_b = 0,99*sigma*T^4 – P/A ( W/m2)

    where T (K) is the air temperature and A (m2) is the area of the cavity opening.

  42. Steve Fitzpatrick says:

    Pehr Bjornbom,

    Great idea! But you would need a pair of very closely matched temperatures sensors if there was no polyethylene film to avoid ventilation of the cavity. Even a tiny fraction of a degree difference in the sensor calibrations could lead to big heat errors if a significant volume of air inside the cavity were being swapped with outside air (eg. on a breezy night). You could buy relatively inexpensive thermistors (a couple of dollars… or Euros.. each) and find two with nearly identical resistance versus temperature curves.

    Or just keep a single thin thin film of polyethylene in place.

  43. Gord says:

    Dr. Spencer,

    Congratulations Dr. Spencer, you have dis-proved Back Radiation heating of the Earth’s surface and the fantasy “Greenhouse Effect”.

    The Cavity you constructed is just another form of Solar Oven and you achieved the increased cooling at night that the Physics Dept. at Brigham Young University demontrated in their experiments. The similarity is stunning.

    Solar Cookers and Other Cooking Alternatives
    http://solarcooking.org/research/McGuire-Jones.mht
    ——
    The so called “Greenhouse Effect” is credited with increasing the Earth’s average temperature from -18 deg C to +15 deg C because of the colder atmosphere Back Radiation heating of the Earth.

    Using the Stefan-Boltzmann Law, -18 deg C (255K) will produce 239.75 w/m^2 and +15 deg C (288K) will produce 390.10 w/m^2.

    That means there is 390.10 – 239.75 = 150.35 w/m^2 CONSTANTLY heating the Earth’s surface, Day and Night, to maintain the +15 deg C average temperature.

    (Trenberth says it is more than double this amount and uses 324 w/m^2 for the Back Radiation heating of the Earth.)

    With a drop in temperature (compared to night air temperature) inside the Cavity of between about -3 deg F and -5.3 deg F, your experiment proves that this is pure fiction.

    By the way, this is how Infrared Thermometers work.

    The Thermistor in these units drop in resistance when pointed at the cold atmosphere.

    And, all direct measurements of Back Radiation uses instruments that have cryogenically cooled IR detectors, far below the -20 deg C average atmospheric temperature, to make direct measurements possible.

    Just like the 2nd Law of Thermodynamics states:

    “Second Law of Thermodynamics: It is NOT POSSIBLE for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.”

    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html#c3

  44. Jean-Christophe LOISON says:

    « Despite PE has a discrete transmittance at the SW IR bands, being a solid it emits IR radiation depending on its own surface temperature.”
    I’m not sure that the PE emits a lot in the IR as emissivity at a given wavelength is function of the absorbance. By the way, how can we estimate what is the maximum atmosphere height which influence the cavity with downwelling sky radiation.

    Jean-Christophe LOISON
    France

    • Massimo PORZIO says:

      Hi Jean-Christophe,
      <>
      I’m not a scientist and my thermodynamic background goes back to school when I get my degree in electronic engineering. So maybe I’ve been wrong there, but the experiment I did with the PE bag seems to evidence that the bag with the hot air inside emits additional IR in the long wave band.
      Today I did a new experiment, I placed into the A4 sized bag a K thermocouple and I put the air gun 45° on one external side of the bag, and the IR thermometer perpendicular on the other side. I heated up the bag until the thermocouple based thermometer shown about 39°C, then I removed the hot air source and I read the IR thermometer. In fact, the IR thermometer always shown a temperature below 33°C, while the thermocouple one was showing 36°C. But that could be due to the thermocouple thermal capacitance and the PE thermal conductivity. Anyways, on the other side of the bag (where the IR thermometer was aimed) there was a wide white cardboard which the IR thermometer reported to be at 27.5°C. To check the two thermometers I measured in the middle of the cardboard with the K thermocouple and it reported just 24.7°C, so the IR thermometer shows about 2.8°C above the K. Since the K thermometer was calibrated last year, and compared with one of the same type I have it is just +/- 0.1°C. I guess that my IR themometer is a little high, but the measurements were done to see if the heated PE increases the relative temperature not to get an absolute value.
      That’s what I see and I think is easy to replicate by your own.

      Have a nice day.

      Massimo

  45. Fulco Bohle says:

    Roy,

    Wat happens if you fill the cavity with 100% CO2 or 100% N2.
    I think the more CO2 the steeper the cooling and warming curves and higher peeks. N2 is the battery and CO2 the collector/cooler.

    Fulco

  46. KuhnKat says:

    Dr. Spencer,

    “From what I have been able to gather, it is very difficult to get the solar reflectance of white paint above about 0.9.”

    I read recently that “white” is only relative to visible wavelengths. “White” is infrared’s “black”. There is a lot of energy in the near infrared from the sun. This may be the issue.

    • Anonymous says:

      I agree, and I add that there is “white” and “white”.
      Our eyes are not so skilled detecting colours.
      One meaningful example is in front of anybody who reads this post. The white background of this post has not a continuum spectrum from 380 to 780nm but just three peaks of emission in the middle of the red, green and blue ranges.
      The reflectance of a white paint depends on its pigments absorbance spectra.

      Massimo

  47. KuhnKat says:

    Here is an article from the Paint and Coating industry talking about absorption and pigments:

    http://www.pcimag.com/Articles/Cover_Story/645bb044256a7010VgnVCM100000f932a8c0____

  48. Bill Illis says:

    There is a surface radiation (SURFRAD) measuring station at Oxford Mississipi (Goodwin Creek) which is not far from UAH and is at the same latitude etc.

    It does measurements of all kinds of upwelling/downwelling radiation and provides the data in very easy to use/access/plot fashion (delay is only 48 hours or so).

    Here is July 29th 2010, net solar, upwelling IR, and downwelling IR charted for Goodwin Creek if you want to double-check against your data.

    http://www.srrb.noaa.gov/cgi-bin/surf_plot?ptype=gif&site=3&date=29-jul-2010&list=5+6+13

    Lots of different plots available at:

    http://www.srrb.noaa.gov/surfrad/pick.html

    My comment is the upwelling IR is always higher than the downwelling IR and the average of the two (which would be something like the black-body radiation of the atmosphere at 2 metres) closely equates to the air temperature when plugged into the Stefan Boltzmann equation. As one goes up higher in the atmosphere, there is less and less net energy in a given volume of air (until the tropopause anyway). The question becomes is the upwelling and downwelling IR coming from the molecules within several metres of the instruments or is it coming from much farther distances. The perspective of what we are talking about changes considerably based on the distance the IR photons are travelling.

    • Anonymous says:

      “The perspective of what we are talking about changes considerably based on the distance the IR photons are travelling.”

      Interesting question. I also want a part of that answer!! The bottom of the atmosphere is the most dense with, generally, the most CO2 and WV. It is also usually the hottest. Why are we seeing an alledged average temp of radiation of somewhere kilometers above the surface when we ought to be reading the molecules emissions immediately above us?

      Are we actually seeing the molecular bond LW and conflating it with blackbody radiation, which is unequivocally connected with the temperature of the emitting body?? I never understood whether there was a separation between these two sources in the papers I have read and the commentary on them.

      • the total downwelling IR is contributed to by some frequency intervals where the emitting temperature is very high because of strong absorption/emission, to frequencies where the atmosphere is almost transparent and you are seeing close to the “cold” of outer space.

        • Anonymous says:

          “the total downwelling IR is contributed to by some frequency intervals where the emitting temperature is very high because of strong absorption/emission, to frequencies where the atmosphere is almost transparent and you are seeing close to the “cold” of outer space.”

          This paragraph seems to allude to both very high temperaturs and very low temperaturs. Are you just saying that the range of temperatures represented by the total downwelling IR is quite broad??

          What is your position on whether the molecular bond radiation of GHG’s is temperature based as black body radiation?

  49. (Sorry for reposting, but it’s been “waiting in moderation” for a couple of days and I thought maybe the moderator missed it)

    Here’s another nice page that shows the 24h measurements from both a pyranometer (radiation from the sun, cloud effects nicely visible) and a pyrgeometer (downwelling LW radiation).

    http://www.chilbolton.rl.ac.uk/weather/radiation_sensors.htm
    http://www.chilbolton.rl.ac.uk/weather/radiation_sensors_yesterday.htm

    Assuming the pyrgeometer readings are correct, Chilbolton experiences more or less constant downwelling LW radiation about 300-400 W/m2 throughout the 24h cycle.

    It would be interesting to know more about how these pyrgeometers are calibrated and what the science/methods are behind determining the initial calibration reference?
    Maybe this could provide some further evidence to the existance of downwelling radiation?

    • Anonymous says:

      Both days are obviously partly cloudy. Note that when the sun “appears” around noon, it is radiating at about 1200 Wm-2. If you add the 400 Wm-2 backradiation, you have a total radiation of 1600 Wm-2. That’s equivalent to a blackbody at 137 C (278 F). I think something’s wrong….

  50. Bryan says:

    Cold surfaces can radiate to a warmer surface.
    Warm surfaces will radiate more to the colder surface.
    Cold surfaces cannot radiate HEAT to a colder surface.
    Why do I say that?
    Because HEAT is a process by which energy is transferred from a hotter substance to a colder substance.
    Heat has the thermodynamic property of being able to do WORK.
    If the energy transfer is incapable of doing work it is not heat.
    Some gases like H2O vapour and CO2 have a radiative effect.
    Water vapour is by far the most important of the radiative gases.
    But the important additional physical effects H2O bring are evaporation and condensation in phase change.

    On a cloudy night temperatures do not fall as low as on a clear night.
    Why is this?
    Some say this effect is mainly due to backradiation.
    But consider this;
    the clouds contain condensing water vapour.
    The release of 2,260,000J per kilogram of H2O vapour keeps the temperature of the clouds at a higher value.
    Now the convection currents of air (the main method of heat transfer in the atmosphere ) will be much weaker if there is little temperature difference between the Earth surface and the clouds.
    The insulating properties of the night sky is improved by the presence of the clouds.
    Now add in if the radiative contribution to and from the clouds you have the picture we can all recognise.
    It is important to separate the important from the less important factors affecting heat transfer in the atmosphere.

  51. JAE says:

    Hmmm. A couple of commenters have posted links that show the downwelling radiation in the UK and in Mississippi. Both show a rather constant radiation of about 350-400 Wm-2 throughout the day. Now, the amount of water vapor in the air probably varies considerably during the day. So, if the amount of this downwelling radiation is a function of the amount of GHGs in the air, why is it so constant?

    • the day-to-night change in total integrated water vapor is not that great, plus you are looking up at the bottom where most of the water vapor exists. If you look down from the top of the atmosphere, there is much greater sensitivity of upwelling IR on small changes in water vapor content.

  52. wayne says:

    Roy you said:

    “Presumably, once drier air arrives here in Alabama in another couple months, I should see larger temperature falls in the cavity, since water vapor is the Earth’s main greenhouse gas.”

    Of coarse water vapor is the main greenhouse gas, always somewhere between 1-4% of the atmospher, but at what ratio is the capacity of water vapor over carbon dioxide to radiate and heat? Is it ~2% which equals ~400/20000(2% wv) or is that skewed due to each being on its own logarithmic curve and the intersection would need to be determined? If on logarithmic curves, is it feasible to know the capacity ratio between the two at a point of concentrations? Do you know where such charts are located on the web?

    That is, how much of the down radiation you are measuring above from non-co2 gases? I’ve never heard that stated to get a good feeling for what we are talking about in co2′s case.

    • I don’t know the answer for downwelling IR at the ground. For upwelling IR at the top of atmosphere, the answer you get depends on your assumption. Water vapor is a bigger effect, but by just how much depends on your assumptions and what variables you use. After all, the “greenhouse effect” describes a process more than it does a measurement you can make that has a specific value with physical units.

      • Anonymous says:

        Gotcha. I agree with the aspects of greenhouse gases at that level you stated, no skeptism there. Maybe further research will tell us in the future how they relate numerically. I know, probably line-by-line with a dash of broadening tossed in. Stopping here, my head hurts from trying to see too deep in the radiative murk today. :)
        -wayne

  53. Loodt Pretorius says:

    Hi Roy,

    Using the graph you supplied, I used a an on-line psychrometric calculator to establish the relative humidity of the ambient air at tea time, 4 o’clock: -
    Dry temperature 91 F = 32.77 C
    Wet bulb temperature 76.5 F = 24.72 C
    This gives a relative Humidity of 52.39% at an alitude of 200m

    (Seeing that you did not supply the elevation of your observations I assumed relative low and flat country of say 200 metres AMSL – Above Mean Sea Level)

    The humidity ratio is 0.0168 kg/kg.

    You did not supply the dimensions of the cavity and for purposes of illustration I used 200x300x500mm which will give a volume of 30 litres or 0.03 cubic metres.

    The specific volume of air at the above conditions is 0.911m3/kg which results in the 30 litres containing 0.033kg of air when you filled the cavity. The air in the cavity has a mass of 32.94g and the humidity ratio is 0.0168 kg/kg. So the air in the cavity contained 0.55 grams of water when you filled it, or shall we say isolated it from the rest of the air when you enclosed it in the cavity.

    When the air in the cavity heated up the moisture also heated up. You did not share the physical observations of the covering sheet with us. Did it bulge or balloon out when the temperature reached 158 F (70 C)?

    If we can get the dimensions of the cavity, the alitude of the observation station and the moisture content of the air in the cavity, we will be able to start making some sense of the results.

    Last word, using a Psychrometric chart or calculator the condtions for ice forming in a desert can be looked up. You need the dew point to be below 0 C. It is actually quite fun to look-up. A low humidity i.e. dry desert air, is required.

    • I’ve already done some of the calculations you talk about. But the enclosure is not air-tight, and I do not know at what rate air slowly ventilates through. It would be difficult to do an accurate energy and moisture budget without a lot of additional information.

  54. Bryan says:

    Roy

    What about the role of the main air gases N2 and O2?
    They would seem to be the perfect insulators.
    Since they do not radiate in the IR they would continually interchange their KE for PE as they move up and down in the Gravitational Field.
    If we add in the convection process which could be turbulent we might even have net energy returning to the Earth surface on occation!

    • If there were no greenhouse gases in the atmosphere at all, the atmosphere would gain energy from the solar heated ground by conduction only. Eventually, the atmosphere would become isothermal with height, so there would be no convection. It’s interesting that such a small proportion of greenhouse gases can have such a large effect. But the COOLING effect that the evaporation of water from the surface, and the clouds it creates, and the convective heat transport, all keep the GHE effect on surface temperature much lower than it would otherwise be.

      • Anonymous says:

        Er, Roy, aren’t you forgetting convection???

      • Anonymous says:

        Roy, are you seriously proposing that an atmosphere comprised mainly of hydrogen and oxygen (as you say, without greenhouse gases) would not have ANY convection?
        Surely, you don’t mean this?

  55. Ronald says:

    There is a IR atmospheric window inbetween 8 and 13 micometers of a 70% transmitance. A black body emits in this window 62.5 W/m2 at 300K. A 70% leaves only an emission of 44 W/m2 (figures have bin obtained though grafic calculations, so maybe not very exact).

    If amosphere is not transparent to most of the IR, why this discussion on downwelling IR?

    A loss of 459 W/m2 would be like having 10 or 20 air conditioners in your roof space. Nice dream for a hot summer night.

  56. Brego says:

    Dr. Spencer, I was thinking about your experiment this morning and I think you now need to perform a second step.

    Your first results demonstrate that The Box is adequately insulated so that the air in the cavity can cool below, and not be warmed by the surrounding ambient. But that doesn’t demonstrate the existence of downwelling IR. For that you need the next step.

    Do you have access to a walk-in cooler or freezer? I think you now need to start the experiment with the air in the cavity in The Box much colder (~20F)than outdoor evening ambient. Then place the box outdoors after dark (no diffuse solar)and then record (hopefully) the air in the cavity warming through the night. With the ambient being unable to warm the cavity the only other explanation would be downwelling IR.

    That should put an end to all but the most stubborn criticism. (maybe? :) )

  57. Bryan says:

    Brego

    Your proposed experiment would show that there is a possibility of clouds say at -10c increasing the temperature of the air in the Box at -20c.
    Only those who think a that the clouds do not radiate at all would be confounded.
    There is a much larger group group who think that the clouds at -10c cannot raise the temperature of the surface (at say 0c) despite radiating onto it.

  58. Loodt Pretorius says:

    Hi Roy,

    On 2 August I posted some comments, the gist is that the air you isolated contained moisture (was humid), a fact clearly illustrated by all those silicon gel sachets that the Japanese, Taiwanese, Koreans, Chinese, Malayan, etc. include in their packaging when shipping electronic goods.

    We also established that the cavity is not airtight and that there in no pressure build-up due to the increased temperature of the air in the cavity as the excess air mixture can vent to the outside and that therefor the ambient air pressure prevails in the cavity. The fact that the cavity temperature stayed for more than an hour at the peak temperature of 158 F (70 C) gives us comfort that there was indeed enough time for the pressures to equalize.

    Another assumption we can make is that the air that escaped from the cavity contained moisture and that during the temperature rise in the cavity the Humidity Ratio of the air inside the cavity stayed the same. Using the psychrometric calculator we can establish that under these conditions (same Humidity Ratio) that the wet bulb temperature of the air in the cavity was 32.33 C, the relative humidity percentage decreased to 8.35% but tellingly the specific volume m3/kg of air changed to 1.02.

    Basically, some of the air in the cavity vented, leaving slightly less water behind. The grammes of moisture for the hot air inside the cavity is 0.49, from a starting point of 0.55.

    Now when the air cools down, a negative pressure will be generated in the cavity, how the air in the cavity reacts under that ‘suction’ is a topic for another day.

    Ps.The engineer’s toolbox is a handy site that explains heating and ventilation issues rather nicely, and also links to online psychrometric calculators.

    • we meteorologists cut our teeth on moisture variables, measuring moisture, etc. I prefer dewpoint temperature since it is an absolute measure of the moisture content of air, independent of any temperature variations.

  59. JAE says:

    Bryan says up there:

    “What about the role of the main air gases N2 and O2?
    They would seem to be the perfect insulators.
    Since they do not radiate in the IR they would continually interchange their KE for PE as they move up and down in the Gravitational Field.
    If we add in the convection process which could be turbulent we might even have net energy returning to the Earth surface on occation!’

    And Roy says:

    “If there were no greenhouse gases in the atmosphere at all, the atmosphere would gain energy from the solar heated ground by conduction only. Eventually, the atmosphere would become isothermal with height, so there would be no convection. It’s interesting that such a small proportion of greenhouse gases can have such a large effect. But the COOLING effect that the evaporation of water from the surface, and the clouds it creates, and the convective heat transport, all keep the GHE effect on surface temperature much lower than it would otherwise be.”

    Now, this is exactly what I have been trying to communicate for years! Both Roy and Bryan are caught in a possibly hopeless brain-freeze stigma that will not allow them to look at all the options/variables/scenarios (the “other point of view, as it were).

    It is very well-established, and I don’t think even debated, that the “greenhouse gases” help thermalize the “non-greenhouse gases,” thereby increasing the temperature of the WHOLE DAMN ATMOSPHERE! (You just can’t have a “hot” CO2 molecule beside a “cold” N2″ molecule for more than microseconds). Roy is correct, IMHO, that a VERY SMALL AMOUNT OF GHGs can have a very large effect (thermalization, which is orders of magnitude faster than radiative effects–and cooling at TOA). This why the desert air is even hotter than the Gulf air on a midsummer day (and night).

    SO, the thing that needs proving is whether the addition of more GHGs has any effect, whatsoever, on the temperature of the air. There is absolutly NO empirical evidence to support such a concept, of which I am aware. AND THAT IS WHY ROY IS DOING THIS EXPERIMENT! LOL.

    And Bryan has another very important point: the N2 and O2 slow the amount of heat loss to space.

  60. editor says:

    I would think temperature changes would affect air density and composition in confined spaces, perhaps humidity changes or even barometric changes would want to be measured, and may take measurements at the atomic level to make any sense of the data (electron microscopes do not fit in your car).

    Since it is impractical to make measurements of moving atoms at the atomic level, you would have to have faith in your ability to predict patterns.

    From the minds eye, you could think that atoms bouncing off each other in the heat, would eventually stop bouncing against each other, and bounce out of the confined space, creating an event similar to a vacuum, perhaps the larger air mass wants to keep warm and “senses” or pulls the energy out of the confined space.

    The measurements are also of not much use unless different containers using different insulating materials are used, to rule out the special effects of some metals and synthetic that might have photochemical reactions due to sunlight or oxidation effects that are not visible to the naked eye, or must be measured over time.

  61. John Millett says:

    “the box is meant to minimize the rate of heat flow from *everything* except IR radiative loss by the cavity, and IR radiative gain from the sky.”

    IR loss from the surface and “gain” from the sky also occur outside the box. Convection occurs outside the box but not inside the box. Surface emissivity is higher in the box than outside.

    Aren’t these differences a sufficient explanation of the temperature difference inside and outside the box?

  62. Roy, you asked for suggestions to isolate the inner sensor from other outside influences. I suggest you use aerogel sheet for the box (better insulation than styrofoam) and space blanket for the double layer screen (reduce solar shortwave influence).

  63. Loodt Pretorius says:

    Roy,

    Let’s look at the construction of your box, a cooler box with a window.

    100mm styrofoam, painted white and then black, inside.

    Do you seriously think that a mere thin layer of paint will stop the exchange of heat? If so, you might have hit the jackpot! Coleman, the maker of cooler boxes, maybe still the biggest in the world unless the Chinese have overtaken them, would be prepared to pay a fortune for that magic potion. Just think of the commercial spin offs, a cooler box that you can fill with ice in spring, keep outside and still use the ice in Autumn to mix the last of your summer cocktails!

    Unfortunately, somehow I think that Coleman, cooler box designers, would have tested a black lining for their boxes, and if successfull, the camping stores would have stocked the ‘Black Knight’ line by now.

    A cooler box filled with water will have ice in it if left outside north of Yellow Knife in winter, and a cooler box filled with ice will have water in it if left outside in the Kalahari, or any other desert of your choice, in summer.

    The thermal conductivity is expressed in W/(m*degree C).

    A good conductor of heat is copper with a thermal conductivity of 401, Aluminium is 250 and Iron comes in at 80. A heat insulator, like syrofoam has thermal conductivity of 0.033. Now look at the formula again, the thermal conductivity is expressed in a unit length, a thin layer of paint will hardly have the characteristics to stop the transfer of heat! Beefing up the thickness of the styrofoam makes more sense!

    Again, most of this information can be obtained from the heating and ventilation section of the engineer’s toolbox.

  64. Fred Staples says:

    There are some interesting links on this thread, Roy. The link to the Earth System Research Laboratory plots up-welling and down-welling radiation in (more or less) real time for a number of sites.

    The development of the solar oven/refrigerator developed at Brigham University has been mentioned approvingly by your Secretary of State. It can maintain a temperature differential of 11degreesC and has produced ice.

    What puzzles me is that not many (as far as I can see) doubt the existence of IR back-radiation, so who are “they” that you are trying to convince. It is easy to prove that back-radiation exists by comparing outgoing surface radiation from the real earth and the bare rock earth, where the radiative energy output must equal the solar input.

    The real earth surface radiates at about 288K, and the bare-rock at about 255K, so the difference is 63% of the solar input. Only the back-radiation from the atmosphere can balance this increase to maintain the surface temperature. The back-radiation is, after all, the negative term in the Stefan-Bolzmann heat transfer equation used by heating engineers (whose equations have to work in the real world).

    The contentious point is not the existence of back-radiation, but what it is capable of doing. It cannot directly heat the surface of the earth, because it originates from a colder region. It can reduce the outgoing radiant energy from the surface, which will allow the sun to warm the earth. An increases in atmospheric absorption might produce an AGW effect on temperature, but this might or might not be detectable. Any increase in atmospheric temperature might increase radiation to space, counteracting the warming.

    Woods designed his twin greenhouse experiment to test this theory. Could you repeat this, Roy, with better instrumentation and variable IR absorption?

    • Anonymous says:

      You are forgetting conduction to deeper layers of the earth. This will give a similar result to so-called greenhouse gasses. In fact, it probably IS the effect blamed on GHG’s.

      You are also ignoring the heat capacity of the atmosphere itself. Then there is the ocean heat capacity.

      Trying to compare a bare earth to a full system and then claiming only the GHG’s in the full system make a difference is obtuse.

  65. Reader says:

    Roy,

    any chance we can get a writeup of the status of posts, discussions & experiment results from the initial “yes virginia” post until now ?

    I’ve become lost somewhere.

    In yes virginia I thought you were talking about temperature moving from cold to hot surfaces.

    People seemed to provide plausible reasons why your example was flawed. You then proposed an experiment which I assumed was designed to prove this.

    Now following (or trying to) follow that discussion I am confused.

    Do we regard the air as a surface. Were you trying to show that a colder air pocket (surface) would transfer heat to a warmer air pocket (surface) ?

    Is the air regarded as a surface ?

    If downwelling radiation exists, why is the box colder than the air.

    How can you claim evidence for downwelling radiation when the reason for the box cooling is ???? Is the air really not a surface and transparent and the surface is really the stratospheric layer which is colder and hence that is where the boxes heat is going ?

    Does the air actually hold heat or does the heat just pass through it ? Does it mean temperature readings of air are temperature readings of the heat that is currently passing through it (i.e either radiating down or being reflected up). How is the heat transported ? Is it attached to molecules that are rising or descending ? What molecules ?

    Is co2 present at the surface layer or does it rise rapidly and only contained in the upper atmosphere (well obviously not) so why isn’t the box temperature rising if it has a blanket of CO2 over it and the blankets temperature is warmer that the cavity ?

    Is there any merit in what you are doing here or are you in some tangential way trying to appease the AGW overlords by trying to appear totally objective to get science funding ?

    Where are the points of agreement here. Where is the actual proven science in all this discussion and which parts of it are theoretical or debated ?

    Sorry, but I find this has completely confused me. And in trying to follow it I’ve just become more confused.

  66. JAE says:

    Virginia still needs some data to make up her mind, I surmise. :)

    I still don’t think that the “atmospheric greenhouse effect” can be demonstrated empirically, although I support a continued effort to do so (and, importantly, until this happens, the “atmospheric greenhouse effect” is still an hypotheisis and not even a scientific theory, according to “consensus” scientific principles).

    I still think the “atmospheric greenhouse effect” can be refuted empirically simply by comparing Atlanta to Phoenix. If the whole world consisted of the climate of Phoenix, then maybe the excess radiation from CO2 would make some difference. But most of the world is humid, and it is clear that there is a water-regulating system that controls temperatures on this planet. That is not to refute the concept of back-radiation; indeed it is simply a property of matter (but it is an effect, not any cause). Mother Nature (i.e., God) has checks and balances that far surpass human understanding. The planet and your silly ass could not possibly be here, otherwise.

    Perhaps Miskolski and Eschenbach are the closest humans to explain the physics of the real world. If so, we humans don’t have much effect.

    • Jon H says:

      “empirically simply by comparing Atlanta to Phoenix.”

      Prevailing winds in Phoenix tend to be from the East. So winds will be coming in over sun-warmed desert. Prevailing winds in Atlanta are from the northwest.

      Not really that hard to figure out.

  67. Gail F says:

    Looks like you are collecting “orgone energy,” ha ha