Where The Temperature Rules The Total Surface Absorption

Guest Post by Willis Eschenbach

Reflecting upon my previous post, Where The Temperature Rules The Sun, I realized that while it was valid, it was just about temperature controlling downwelling solar energy via cloud variations. However, it didn’t cover total energy input to the surface. The total energy absorbed by the surface is the sum of the net solar energy (surface downwelling solar minus surface reflections) plus the downwelling longwave infrared, or DWIR. This is the total energy that is absorbed by and actually heats the surface.

According to the CERES satellite data, globally, the solar energy absorbed by the surface averages 162 W/m2. The downwelling longwave averages 345 W/m2. Conveniently, this means that on average the earth’s surface absorbs about a half a kilowatt per square meter on an ongoing basis. (And no, I have no interest in debating whether downwelling longwave radiation actually exists. It’s been measured by scientists around the world for decades, so get over it, Sky Dragons. Debate it somewhere else, please, this is not the thread for that.)

Let me note in passing that a doubling of CO2, which will increase the DWIR by something on the order of 3.7 W/m2, and which it is claimed would lead to Thermageddon, would be less than a 1% change in total downwelling radiation at the surface … which would easily be offset by a small change in total cloud cover. But I digress.

Here is the correlation between temperature and total surface absorption.

ceres total surface absorption vs temperature.png

Figure 1. Correlation of total surface absorption with temperature.

Note the similarity to the previous graph showing just the correlation between surface temperature and downwelling solar energy at the surface.

Now, to explain how this can happen I need to take another digression. I was attracted to the study of the climate, not by questions about why the temperature was changing so much, but by why it was changing so little. As a man with some experience of heat engines and governors, I found it amazing that the temperature of such a possibly unstable system could only have changed by ± 0.3°C over the entire 20th century. Why should such a world, with clouds appearing and disappearing, with huge volcanoes popping off every few decades, with winds going up and down, with storms and hurricanes appearing and vanishing, why would it be so stable in the long-term? So I started looking for some long-term kind of feedbacks that could explain it.

I was living in Fiji at the time. After literally months of fruitless searching and thinking about long-term slow feedbacks, one day I thought “Hang on. I’m looking at the wrong end of the time spectrum.” What I realized was that if there was something that kept the daily temperatures from going outside a certain range, that would, in turn, keep the weekly, monthly, annual, decadal, centennial, and millennial temperatures from going outside that same range.

And because I was living in Fiji, the answer was right above me. The daily tropical weather typically looks like this: clear at dawn, clouding up with thermally-driven cumulus clouds in the late morning, perhaps thunderstorms in the afternoon if the day is warm enough, clearing some time after dark. Lather, rinse, repeat, as they say.

I also realized that there were two variables in that scheme—the time of onset of the cumulus clouds and the thunderstorms, and the amount of each of them. I hypothesized that these factors were what controlled the tropical temperature. Since then I have amassed a lot of evidence that my hypothesis is correct, including this post and its predecessor.

There are some important things to note about this process. First, the time of the emergence of the cumulus fields and the thunderstorms is NOT dependent on total forcing. Instead, they are responding to surface temperature. When the surface is cool at dawn, clouds form later, and more sunlight comes in, warming the surface. When the surface is warmer, clouds form earlier, throttling the energy input to the system, and cooling the system back down.

As a result, the system is not affected by small changes in insolation. For example, if a volcanic eruption reduces the amount of sunshine making it through the stratosphere, the tropics cool. And when they cool, clouds form later, letting in more sunlight, and rebalancing the system.

Next, the response is based, not on average temperatures, but instantaneous temperature. As such, it is obscured by monthly or yearly temperature averages.

Finally, the response is immediate. There is no lag of days, weeks, or months. As soon as the temperature crosses some given threshold, clouds form immediately, cooling the surface. This effect is so powerful that although the morning sun is growing stronger and stronger, when the clouds kick in, the temperature can actually drop. Here’s a graph of the long-term average daily swings of a number of TAO buoys spread across the Pacific. Here are the locations of the buoys. I’ve used those on the equator because they have the most data. The TAO buoy data is available hereTAO Buoy Locations

Figure 2. Locations of the TAO buoys

These readings were taken by the automated buoys every ten minutes.

TAO daily cycles temperature

Figure 3. Daily average temperatures, equatorial TAO buoys.

In the cooler areas at the bottom of the graph, the onset of the morning cumulus field merely slows the daily warming. But in the warmer areas, when the clouds appear, the temperature actually drops. The differences can be seen clearly when they are expressed as anomalies about their individual average values, viz:

TAO daily cycles temperature anomalies

Figure 4. Daily temperature anomaly variations, equatorial TAO buoys.

Note that this “overshoot”, the ability to drive the temperature below the local initiation temperature threshold, is critical to controlling a lagged system such as the climate. It is also present in thunderstorms. They generate their own fuel once they are started, allowing them to cool the surface below the initiation temperature threshold.

Next, I divided the days into those which were warmer than usual from midnight to 5 AM, and compared them with the days which were cooler than usual during that same time span. Here’s the result:

TAO average warm and cool days warmest buoy

Figure 5. Averages of warm and cool days, one of the warmest TAO buoys

This shows the temperature control in action at one of the warmest TAO buoys. On days which start out warmer than normal, the clouds and thunderstorms form earlier and more strongly. By evening the temperatures cool towards the average value. The opposite happens when the temperature from midnight to 5 AM are cooler than usual—cumulus form later and more scattered, thunderstorms may not form at all. And as a result, the surface warms towards normal.

With that understanding, we can take another look at the graphic in Figure 1, which I reproduce here:

ceres total surface absorption vs temperature.png

Consider that this is a long-term average. This means, for example, that temperatures in the green and light yellow areas immediately outside the gray lines are not really slightly correlated with the total downwelling radiation.

Instead, it means that the number of days during which they are negatively correlated is slightly less than the number of days when they are positively correlated. However, this average conceals an important fact—the negative and positive correlations are not randomly distributed.

Instead, emergent phenomena like cumulus fields and thunderstorms occur earlier and more strongly exactly when and where the surface is hot. So those areas around the gray outlines of negative correlation are doing the same thing as the areas within the gray outlines—cooling down the hottest days and warming up the coolest days. The only difference is that the warm days are less frequent than inside the gray outlines. This puts limits on how much analysis we can do using averages, as I highlighted in “The Details Are In The Devil“.

In conclusion, let me say that the emergence of the tropical cumulus fields and associated thunderstorms are not the only temperature-linked phenomena which participate in global temperature regulation. Other phenomena include dust devils, squall lines, the Atlantic Multidecadal Oscillation, the El Nino-La Nina pump, cyclones, and the Pacific Decadal Oscillation. Likely more as well …

Me, I’m sitting on a hill in the Solomon Islands on what is scheduled to be my last day here … you’re welcome to read about it, along with the story of the Crocodile and Tufala Panadol over at my blog, Skating Under The Ice.

Best of life to all,

w.

My Strong Advice: When you comment, please QUOTE THE EXACT WORDS YOU ARE DISCUSSING so that we can all understand your thoughts and objections. Be forewarned that I’m likely to ignore your claims, hold you up to ridicule, and generally rubbish your name if you don’t have the polite kindness to quote someone’s words. I’m fed up with people saying things like “I disagree strongly with what you said”, when it is not clear who “you” is and it is totally unknown which of their statements the commenter disagrees with. If you wish to refute someone’s ideas, you need to QUOTE THEIR WORDS, and the TELL US WHAT IS WRONG WiTH THEM. Anything else is handwaving and with be referred to as such.

FURTHER READING:

Albedic Meanderings  2015-06-03

I’ve been considering the nature of the relationship between the albedo and temperature. I have hypothesized elsewhere that variations in tropical cloud albedo are one of the main mechanisms that maintain the global surface temperature within a fairly narrow range (e.g. within ± 0.3°C during the entire 20th Century). To…

An Inherently Stable System  2015-06-04

At the end of my last post, I said that the climate seems to be an inherently stable system. The graphic below shows ~2,000 climate simulations run by climateprediction.net.  Unlike the other modelers, whose failures end up on the cutting room floor, they’ve shown all of the runs ……

The Tao That Can Be Spoken … 2011-08-14

As I mentioned in an earlier post, I’ve started to look at the data from the TAO/TRITON buoy array in the Pacific Ocean. These are an array of moored buoys which collect hourly information on a variety of environmental variables. The results are quite interesting, because they relate directly to…

TAO/TRITON TAKE TWO 2011-08-25

I wrote before of my investigations into the surface air temperature records of the TAO/TRITON buoys in the Pacific Ocean. To refresh your memory, here are the locations of the TAO/TRITON buoys. Figure 1. Locations of the TAO/TRITON buoys (pink squares). Each buoy is equipped with a sensor array measuring…

Cloud Radiation Forcing in the TAO Dataset 2011-09-15

This is the third in a series ( Part 1, Part 2 ) of occasional posts regarding my somewhat peripatetic analysis of the data from the TAO moored buoys in the Western Pacific. I’m doing construction work these days, and so in between pounding nails into the frame of a building I continue to…

TAO Buoys Go Hot And Cold 2015-06-16

I got to thinking about how I could gain more understanding of the daily air temperature cycles in the tropics. I decided to look at what happens when the early morning (midnight to 5:00 AM) of a given day is cooler than usual, versus what happens when the early morning…

 

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337 thoughts on “Where The Temperature Rules The Total Surface Absorption

  1. Willis clouds are no where near the top of atmosphere. Being water droplets they will act as black body radiators just the same as oceans or ground. All of the wavelengths emitted in ghg wavelengths will be absorbed just as if emitted from the ground or water
    Thunder cloud top are still within the bulk of the atmosphere and oils be considered as part of the earth as far as if emissions are concerned. The only difference e is in SW light which will be reflected from the tops of clouds just as it is from the ground – hmm so no difference there. I do not see how clouds (low level) can affect solar gains significantly.

    • Ok clouds are not solid water so will absorb little of the same radiation. So most is reflected from presumably internal reflection in the water drops.
      So sw radiation causes only a small amount of warming of the cloud.

  2. Also I wonder if the clouds starting earlier also gives them more time to spread out due to wind ( or other factors) , thus blocking even more of the sun.

  3. Having spent much time in the tropics I am not surprised by the conclusions set out here by Willis Eschenbach. I am not a scientist but I am a trained mariner. (Have sent in more weather reports than most in the past 50 years.)
    Our ‘climate scientists’ should all be sent to live in the tropics for a while. That after all is the region receiving most solar energy. The importance of cloud cover soon becomes apparent. On a clear day you can fry an egg on the steel plates of a ship; but no chance if there is cloud cover.

    On a related subject I have read that heat generated near the centre of the earth is negligible in the matter of surface climate. I think this conclusion is incorrect. More research needs to be done on heat generation at the heart of large massively dense bodies such as our Earth. For example, I suspect gravity has more effect on temperature than is currently understood. But would welcome comments.

    • John, there appear to be three major issues around heating of the ocean from below through the seafloor:

      1. Is geothermal energy powerful enough to make a difference upon the vast ocean heat capacity?
      2. If so, Is geothermal energy variable enough to create temperature differentials?
      3. Most of the ocean floor is unexplored, so how much can we generalize from the few places we have studied?

      Some people studying this, mainly marine geologists think the effects are not trivial. An overview of research is here:
      https://rclutz.wordpress.com/2016/10/05/overview-seafloor-eruptions-and-ocean-warming/

      • It does make a difference in large ice-sheets. At the bottom, geothermal energy warms the glacier from the bottom up so that most are about -2.0C to -4.0C at bedrock while they can be -50C in the middle of the glacier.

        Ocean water is different. It becomes more dense until it gets to -1.7C, so the bottom layer of ocean/lake water is always the coldest water. If it gets warmed up it will rise up and be replaced by colder water flowing in from somewhere else. The whole planet needs to warm up so that there is no/less cold water source(s) available. Ie. the poles have to warm up to warm up the deep ocean.

      • Bill Illis December 22, 2017 at 9:08 am
        Ocean water is different. It becomes more dense until it gets to -1.7C, so the bottom layer of ocean/lake water is always the coldest water. If it gets warmed up it will rise up and be replaced by colder water flowing in from somewhere else. The whole planet needs to warm up so that there is no/less cold water source(s) available. Ie. the poles have to warm up to warm up the deep ocean.

        You seem to be confused here. Freshwater has a maximum density at 4ºC so the bottom of lakes in winter will be at that temperature whereas the surface will get colder and freeze, that’s why ice forms at the surface not the bottom. Ocean water density is more complicated and depend strongly of salinity but doesn’t have the maximum density property if salinity is constant. (Hence thermohaline circulation)

      • Phil. the pro-global warming crowd is confused about many things, especially the density of water and the thermohaline ocean circulation. Let the real oceanographers do that part because you guys always get it backwards.

      • Bill Illis December 22, 2017 at 1:33 pm
        Phil. the pro-global warming crowd is confused about many things, especially the density of water and the thermohaline ocean circulation. Let the real oceanographers do that part because you guys always get it backwards.

        Well if you’re an oceanographer, you’re one who’s got it backwards!
        Do you really think the bottom waters of a freshwater lake is always the coldest?

        Density of seawater depends on salinity and temperature:

        http://www.open.edu/openlearn/science-maths-technology/the-oceans/content-section-3.2

      • Ron, you would think that 40000+ vents would have a measurable affect, do we know if anyone is actually trying to find out how much?

      • AC, they estimate the number of seafloor vents is somewhere between 100,000 and 10,000,000. So yes, it must make a difference, but such uncertainty makes it impossible to know how much.

      • Ron Clutz December 23, 2017 at 1:15 pm

        AC, they estimate the number of seafloor vents is somewhere between 100,000 and 10,000,000. So yes, it must make a difference, but such uncertainty makes it impossible to know how much.

        While that is true, we can certainly make some order-of-magnitude estimates. I find the following:

        Conductive heat flow measured by multipenetration probe and from “Alvin” was exceptionally high (> 2 W/m2) adjacent to the thermal springs.

        And this reference shows a 10 W/m2 heat flow over an area of about 1200 m^2.

        That’s about 12E+3 watts. Assuming the biggest number of vents, 10E+6, and that they are all this large and with this large a heat flow, this gives a total heat flux of 1.2E+11 watts.

        However, the area of the ocean is about 3E+14 square metres … which gives an ocean-wide average of 0.0004 W/m^2 … I’m sure you can see the problem.

        In addition, observations show that these vents don’t change much even over decades. Makes sense, Old Faithful doesn’t change much either. As such it is unlikely that they are responsible for variations in climate.

        w.

      • Vast ranges of volcanoes hidden under the oceans are presumed by scientists to be the gentle giants of the planet, oozing lava at slow, steady rates along mid-ocean ridges. But a new study shows that they flare up on strikingly regular cycles, ranging from two weeks to 100,000 years—and, that they erupt almost exclusively during the first six months of each year. The pulses—apparently tied to short- and long-term changes in earth’s orbit, and to sea levels–may help trigger natural climate swings.

        “People have ignored seafloor volcanoes on the idea that their influence is small—but that’s because they are assumed to be in a steady state, which they’re not,” said the study’s author, marine geophysicist Maya Tolstoy of Columbia University’s Lamont-Doherty Earth Observatory . “They respond to both very large forces, and to very small ones, and that tells us that we need to look at them much more closely.” A related study by a separate team this week in the journal Science bolsters Tolstoy’s case by showing similar long-term patterns of submarine volcanism in an Antarctic region Tolstoy did not study.

        http://www.ldeo.columbia.edu/news-events/seafloor-volcano-pulses-may-alter-climate

    • “John Atkins December 22, 2017 at 3:58 am
      Our ‘climate scientists’ should all be sent to live in the tropics for a while.”

      Excellent thought!

      I think CO2 obsessed “climate team” members should also spend a few years in places; e.g. Atacama desert, Sonoran desert, High desert region of America, etc.

      After a few tropical years, followed by several years in various deserts, they should develop better understandings of what water vapor accomplishes.

  4. Willis
    ‘According to the CERES satellite data, globally, the solar energy absorbed by the surface averages 162 W/m2. The downwelling longwave averages 345 W/m2.’
    My understanding is that solar energy includes SW and LW radiation. I’m not sure I understand what you’re getting at.
    Your implication seems to be that there is some other source of LW radiation. Then you seem to be sidetracked by some Skydragon illusion.
    I don’t understand your point. Perhaps it is your phrasing that has me confused.

    • Alex December 22, 2017 at 4:06 am
      Willis
      ‘According to the CERES satellite data, globally, the solar energy absorbed by the surface averages 162 W/m2. The downwelling longwave averages 345 W/m2.’
      My understanding is that solar energy includes SW and LW radiation. I’m not sure I understand what you’re getting at.

      Solar energy contains very little radiation above a wavelength of 5 microns whereas the DWIR originates from GHGs and exceeds 7 microns.

      Your implication seems to be that there is some other source of LW radiation.

      Compared with solar yes, it comes from the emissions from GHGs.

      • Alex here is what The Sun out puts as far as we can tell.

        From Wiki.

        In terms of energy, sunlight at Earth’s surface is around 52 to 55 percent infrared (above 700 nm), 42 to 43 percent visible (400 to 700 nm), and 3 to 5 percent ultraviolet (below 400 nm).[6]

    • Its interesting that 162 W/m2 and 345 W/m2 are coming in – 507 W/m2 total – (the numbers are slightly different in different sources) but the surface acts as though it has only 390 W/m2 of energy on hand at any one time on average.

      Energy in, Energy out, Energy accumulation at any one time. Time is part of this equation.

      • Bill Illis December 22, 2017 at 9:18 am

        Its interesting that 162 W/m2 and 345 W/m2 are coming in – 507 W/m2 total – (the numbers are slightly different in different sources) but the surface acts as though it has only 390 W/m2 of energy on hand at any one time on average.

        Energy in, Energy out, Energy accumulation at any one time. Time is part of this equation.

        Thanks, Bill. You seem confused about the energy budget. The surface absorbs about half a kilowatt per square metre. It loses about 390 W/m2 via radiation, and about 110 W/m2 by a combination of evaporation and conduction/convection. There’s no “accumulation”.

        Regards,

        w.

    • Alex December 22, 2017 at 4:06 am

      Willis

      ‘According to the CERES satellite data, globally, the solar energy absorbed by the surface averages 162 W/m2. The downwelling longwave averages 345 W/m2.’

      My understanding is that solar energy includes SW and LW radiation. I’m not sure I understand what you’re getting at.

      Thanks, Alex. There are two kinds of infrared IR, sometimes referred to as “near infrared” for the one present in solar energy, and “far infrared”, also called “longwave IR” or “thermal infrared”. This is emitted by all objects except monatomic gases, with the amount radiated depending on their temperature

      Your implication seems to be that there is some other source of LW radiation. Then you seem to be sidetracked by some Skydragon illusion.

      There is an entire planet in which every solid object and most gases are sources of LW radiation. Nothing to do with sky dragons. The 345 W/m2 mentioned in my quote is the downwelling thermal (longwave) IR radiation coming down from the atmosphere.

      Regards,

      w.

      • No, it is most common to split IR in to 3 bands.
        • Near IR-A: 700 nm–1400 nm (215 THz – 430 THz) and the region closest in wavelength to the red light visible to the human eye
        • Mid IR-B: 1400 nm–3000 nm (100 THz – 215 THz)
        • Far IR-C: 3000 nm–1 mm (300 GHz – 100 THz)
        Although Wiki also shows the less common 5 bands.
        Near-infrared
        Short-wavelength infrared
        Mid-wavelength infrared
        Long-wavelength infrared
        Far infrared
        Another source
        Definition
        infrared radiation (IR)
        Posted by: Margaret Rouse
        WhatIs.com
        Contributor(s): Jessica Scarpati

        Infrared can be subdivided into multiple spectral regions, or bands, based on wavelength; however, there is no uniform definition of each band’s exact boundaries. Infrared is commonly separated into near-, mid- and far-infrared. It can also be divided into five categories: near-, short-wavelength, mid-, long-wavelength and far-infrared.

      • Also there is no “Infrared IR”, it is all Infrared Radiation and abreviated to IR.
        You know how much I just love to nit pick.

      • Willis- “The 345 W/m2 mentioned in my quote is the downwelling thermal (longwave) IR radiation coming down from the atmosphere.”

        345 W/m2 downwelling thermal IR radiation suggests an atmospheric temperature of 6.2C as measured by an IR gun looking upward. My measurements are that a low humidity clear sky is 1.1C or 320 W/m2. Increasing humidity levels raise the temperature (and increase the downwelling radiation) up to the formation of clouds when the IR measurements of the cloud bottoms are within a degree or so of the ocean surface temperature.

        Basically what I am saying is that downwelling atmospheric radiation ranges from 320 W/m2 to 467 W/m2 (which is equal to the IR from the ocean surface). The net radiation energy loss from the surface ranges from zero when completely overcast, to 147 W/m2 in clear, low humidity, sky conditions.

        My solar panels on a clear day (6 hours) easily collect 4800 W/m2 / 24 hours equals 200 W/m2 of solar insolation per hour. On a cloudy day my solar panels may not get any watts, but the ocean’s radiative net loss then is zero too.

        The energy balance all boils down to when clouds form during the day.

        Also it has been established that the ocean temperature is very stable, within a few degrees all the time, so what is the mechanism that causes clouds to form during the day? It isn’t changes in the temperature, the temperature doesn’t change.

        What it is is the Suns rays, it heats (increases the specific heat) the surface and triggers evaporation. The warmer the surface is initially, the faster the suns rays triggers evaporation. Evaporation creates moist humid air which is less dense than drier air, it is displaced by denser drier air (wind) which increases the rate of evaporation.

        The least important variable in this whole process is CHANGES in downwelling IR radiation from the atmosphere, the 147 W/m2 potential difference simply doen’t matter. It is like finding limits in calculus, when the changes get small enough they can be discarded.

      • jinghis December 22, 2017 at 7:08 pm
        Willis- “The 345 W/m2 mentioned in my quote is the downwelling thermal (longwave) IR radiation coming down from the atmosphere.”

        345 W/m2 downwelling thermal IR radiation suggests an atmospheric temperature of 6.2C as measured by an IR gun looking upward. My measurements are that a low humidity clear sky is 1.1C or 320 W/m2. Increasing humidity levels raise the temperature (and increase the downwelling radiation) up to the formation of clouds when the IR measurements of the cloud bottoms are within a degree or so of the ocean surface temperature.

        IR guns usually don’t include the CO2 region of the spectrum so the downwelling would be more than your measurement.
        http://www.fluke.com/fluke/inen/thermometers/fluke-568-2-566-2.htm?pid=56090

      • You are lumping quite a few assumptions there jinghis. Most of which do not discuss or consider Willis’s cloud/T-storm temperature control(s).

        “jinghis December 22, 2017 at 7:08 pm

        Basically what I am saying is that downwelling atmospheric radiation ranges from 320 W/m2 to 467 W/m2 (which is equal to the IR from the ocean surface). The net radiation energy loss from the surface ranges from zero when completely overcast, to 147 W/m2 in clear, low humidity, sky conditions.”

        You appear to be micro focused on a small portion of Willis’s description.

        Taking your comment on it’s face:
        If the surface’s “net radiation energy loss” is zero, with steady incoming soar radiation; that describes a closed loop scenario with constant incoming radiation; meaning steadily increasing temperatures… How does that actually work?

        “jinghis December 22, 2017 at 7:08 pm
        The energy balance all boils down to when clouds form during the day.

        Also it has been established that the ocean temperature is very stable, within a few degrees all the time, so what is the mechanism that causes clouds to form during the day? It isn’t changes in the temperature, the temperature doesn’t change.”

        I think you need to read Willis’s article and underlying theory. Perhaps again.

        You also appear to conflate sea surface temperatures with atmospheric temperatures

        Again, taking your statement at face value, you imply that sea surface temperatures are atmospheric temperatures under the assumption that sea surface temperatures are unaffected by incoming radiation.

        Leaving an impression that all tropical oceanic atmospheric areas daily temperatures are restricted to the temperature of the ocean, alone.

        “jinghis December 22, 2017 at 7:08 pm
        What it is is the Suns rays, it heats (increases the specific heat) the surface and triggers evaporation. The warmer the surface is initially, the faster the suns rays triggers evaporation. Evaporation creates moist humid air which is less dense than drier air, it is displaced by denser drier air (wind) which increases the rate of evaporation.”

        Let’s take your statements piece by piece:
        “Triggers evaporation”; Evaporation is a cooling function.
        Leaving one with the impression that cold moist air rises to displace warm moist air.
        Remember, relative humidity is affected by temperature.

        But, absolute atmosphere water content does increase; which means increased atmosphere’s water vapor that is all GHG, absorbing infrared radiation across a very large portion of the IR spectrum.

        “jinghis December 22, 2017 at 7:08 pm
        The least important variable in this whole process is CHANGES in downwelling IR radiation from the atmosphere, the 147 W/m2 potential difference simply doen’t matter. It is like finding limits in calculus, when the changes get small enough they can be discarded.”

        Consider, your daily solar panel average:

        “jinghis December 22, 2017 at 7:08 pm

        My solar panels on a clear day (6 hours) easily collect 4800 W/m2 / 24 hours equals 200 W/m2 of solar insolation per hour. On a cloudy day my solar panels may not get any watts, but the ocean’s radiative net loss then is zero too…”

        200W/m² supplies you with energy, but 147 W/m², 73.5% of your 200W/m&#178, is inconsequential?

        A side comment: A 24 hour incoming solar panel energy average is confusing. It is a view that obscures incoming useful solar panel radiation absorption bell curve that actually occurs over a small portion of those 24 hours.
        Your solar panel production references come across as anecdotal; especially the claims regarding solar radiation on cloudy days.

        Having spent a substantial portion of my work life outdoors, all day; I can attest to getting sunburned on cloudy days.
        Yes, miles tall active thunderclouds really do minimize incoming sunlight reaching the ground; but that is one extreme, not a description for all cloudy conditions.

    • Alex, the term “solar IR” refers to the portion of the solar spectrum “below Red” or 0.7 microns and always originates from the Sun. Solar IR is often called “SWIR” for short-wave IR. Solar spectrum energy drops to near zero at 4 microns.
      The Earth also emits IR energy, but the wavelength starts around 4 microns and extends down from there, so it is called “terrestrial IR” or “LWIR” for long-wave infrared. These terms are often confused in casual discussions of Earth’s energy balance, but are completely distinct because the two curves have almost no overlap, and have opposing “directions”
      Note that the outgoing IR must always match incoming solar energy assuming net planetary albedo is constant at 0.3. “Albedo” is just means reflected. Energy absorbed is 1 minus reflected.
      Assuming the solar constant is 1360 watts per square meter, then
      Solar absorbed by Earth is 1360/4 times (1- 0.3) which equals 238 watts per square meter.
      Planetary outgoing longwave radiation (OLR) is also 238 watts per square meter on average, so Earth’s temperature can never change unless net planetary albedo changes. Net planetary albedo is VARIABLE, it is often considered a constant to simplify some calculations.

      • Excellent comment bw!

        One minor addition to your excellent graphic is the consideration that many discussions only reference water vapor’s GHG activity.

        Water’s three forms, which water vapor is one form. Liquid and solid water are both active IR absorbers, greatly increasing water’s radiative activity spectrum.

      • “so Earth’s temperature can never change unless net planetary albedo changes.”

        Er No.

        Moisture & Clouds don’t just reflect Radiation, they absorb it and prevent it getting to the Surface.
        More cloud, temp goes down, less cloud temp goes up, you only have to look at this graph to see the realtionship.

        So it is not just albedo, I also wonder how you explain Albedo’s effect on El Nino, or didn’t you notice that it also changed the Temperature?

    • I find Willis’ comments perfectly in line with my understanding of climate change. Since the early discusion by Svensmark of GCR effect on clouds,
      e.g. Space Science Reviews November 2000, Volume 94, Issue 1–2, pp 215–230
      it has been obvious that the amplification sign on solar radiance was sensitive to some other factor. Cloud initiation vs surface or lower stratosphere temperature seems to fit the bill.

  5. All those observations about the so called tropical thermostat are correct but no reason for the process is supplied.
    I recall it being general knowledge in the 1950s and no doubt it was clear to seamen for centuries previously, that in the tropics over the oceans there is a maximum temperature that can be reached and that evaporation and convection is the limiting process in action.
    So WHY does it happen ?
    It is a matter of the amount of energy required to break the bonds between molecules of liquid water so that the phase change to a gas can occur.
    Once the necessary amount of energy is provided to start the process then adding more energy goes not to raising the temperature further but instead to accelerating the rate of phase change. The same reason why water boils at 100C and adding more energy makes it boil faster without going higher than 100C. Boiling is just evaporation which occurs below the surface so that bubbles of gas can form. As we all know the 100C boiling point of water is pressure dependent and so is the evaporation point at temperatures less than 100C.
    The determining factor for the amount of energy required to start the phase change from liquid to gas is the atmospheric weight bearing down on the water surface because the tendency of that weight is to hold the water molecules together in liquid form more effectively than by relying on the forces of molecular attraction alone.
    A lower atmospheric weight sets a lower maximum temperature and a higher atmospheric weight sets a higher maximum temperature.
    As an extreme example to illustrate the point we know that if there is no atmosphere at all then no added energy is required at all since then the internal energy of the liquid is enough on its own to permit evaporation to space.
    So, if the top limit for the surface temperature of oceans is determined by the mass of the atmosphere does that not suggest something regarding the effect of the mass of an atmosphere on an entire planetary surface ?

    • One additional complicating factor (there are many) is that seawater is saltwater. This fact alters evaporation processes and kinetics. Evaporation of course leaves behind saltier water which subtly but importantly changes the thermodynamics at scales from microphysics to the thermohaline major ocean heat transport currents. But the alterations are in the kinetics (time domain). The time domain means that if evaporation rate increases (acceleration) the formation rate of higher salinity surface water also goes up, and this subtly resists more evaporation. Salt is slowing the system response down to increased heating.

      • Joel O’Bryan December 22, 2017 at 5:55 am
        One additional complicating factor (there are many) is that seawater is saltwater.

        Which as I pointed out in a comment on Willis’s earlier post high salinity water such as Red Sea and Dead Sea have higher maximum temperatures which would be expected from Willis’s hypothesis because the required evaporation rate for cloud formation would occur at a higher temperature.

      • Yes, seawater has different latent heat of evaporation compared to fresh water and the temperature response curves are extremely non-linear.
        There is also a different specific heat. Pure water specific heat is 4.18 Joules per gram per degree. Seawater is about 4.0 Joules/g.C
        That’s almost a 5 percent difference.

      • The salinity of the equatorial Pacific also generally increases east to west, just as surface temps do. Evaporation is the key.
        And like a lava lamp wax glob rising and then descending, the warmer but higher salinity water in the Western equatorial Pacific descends in the water column because it is denser despite being warmer. Thus the beginnings of an El Nino form as a warm, highsalinity blob of water descends and heads eastward riding on top of the thermo-haline boundary layer at 300-500 meters depth. Lava lamp style dynamics. Chaotic.

      • Stephen Wilde December 22, 2017 at 4:07 am
        All those observations about the so called tropical thermostat are correct but no reason for the process is supplied.
        I recall it being general knowledge in the 1950s

        Really, you were following this when you were in primary school?

    • Stephen,
      Yes it is the gas laws again, Henry’s, Charles’, Boyle’s, and, Avogadro’s Hypothesis. These applied on their own plus the knowledge that: moist air is lighter than dry air as water molecules have a lower molecular weight than Oxygen or Nitrogen; and, Infra-red cannot heat water only increase evaporation and cooling – provide all the rules needed to generate the wet and dry lapse rates — ‘green house (sic) gases’ are not needed.

      • Phil, quote an experiment that shows low levels of infrared can be used to warm a volume of water in an open environment. There isn’t one. Not one. Despite the simplicity of setting one up. Infrared is absorbed by the first water molecule it hits and if it adds sufficient energy that molecule will ‘evaporate taking the latent heat of vaporization with it cooling the volume of water.

      • What do you suppose happens if the humidity is 100%?
        Does infrared just cause evaporation and no heating then?
        No matter how much infrared?

      • menicholas December 22, 2017 at 10:45 am
        What do you suppose happens if the humidity is 100%?
        Does infrared just cause evaporation and no heating then?
        No matter how much infrared?

        Exactly, the idea that absorption of radiation by liquid water automatically increases the kinetic energy of that molecule so that it leaves the surface is false. Even if it did some of those molecules would be heading down not up!

      • Ian, do not waste your time trying to debate with those two.
        Especially do not try and talk experiments with them.
        They are true believers and ignore any contrary evidence.
        I have seen those experiments and as you say no warming.

      • At 100% humidity water molecules leave the surface and almost immediately condense but continue their convection upward. This can be seen over bodies of water in the early morning when they appear to ‘steam’. What is seen is actually small water droplets rising in convective up drafts.

      • Ian W December 22, 2017 at 6:34 pm
        At 100% humidity water molecules leave the surface and almost immediately condense but continue their convection upward. This can be seen over bodies of water in the early morning when they appear to ‘steam’. What is seen is actually small water droplets rising in convective up drafts.

        And at 100% humidity an equal number of water molecules leave the atmosphere and coalesce at the surface.

    • Stephen, what multiple places in the world are closest to the sun, so should get the most insolation, as the Radiation has less atmosphere to navigate and are also closest to “back radiation”, so they should also get the most of that as well.
      They also do not have a lot of open water to evaporate to keep them cool.
      Under those circumstances you would expect those places to have some of the highest Surface temperatures as well.

      But of course we know that Mountain tops are freezing, what could possibly so easily counteract all that warmth I wonder?

      • Conduction to cold air plus rapid upward convection would do it.
        And when the air descends again into adjoining lower levels it heats up at the dry adiabatic lapse rate so as to reduce surface cooling.
        As people are beginning to realise there is nothing in the energy budgets of Trenberth or Willis to deal with that aspect since they ignore non radiative transfers completely.
        And as ferdberple says below it is mere hand waving to suggest that there is no effect because convection is a zero sum process. The reality is that both DWIR to and from GHGs and convective overturning are net zero in themselves but both involve a delay in energy transmission through the system and so must cause surface heating.
        The thing is that if conduction and convection are doing it then the net effect on surface by DWIR must be zero and I contend that that is because convection can change to neutralise radiative imbalances as stated here:

        http://www.public.asu.edu/~hhuang38/mae578_lecture_06.pdf

        “Radiative equilibrium profile could be unstable; convection restores it to stability (or neutrality).

        Willis has observed the process of convective adjustments in action but not realised the significance. His tropical thermostat is a non radiative process but he insists on trying to describe it only in radiative terms and people here are noticing that one cannot do that.

      • Stephen, I cannot agree with “warm the surface”, only slow the rate of cooling leaving it warmer at the end of the night.
        Even air that rises and descends again is going to be colder than the surface.
        In the case of mountains, the lapse rate, dropping air temps and thinning atmosphere are quite enough to negate any extra radiation they receive. As you say radiation is only a small part of the overall energy movement equations.
        Yet the “Science” treats the Earth as one smooth solid surface.

      • AC

        I did refer to a reduction of cooling but of course that leaves the surface warmer than it otherwise would have been,
        Just like an electric fire which relies on resistance to the flow of electrons the conduction and convection between surface and atmosphere offers a resistance to the flow of radiation in and out.

      • “convection is a zero sum process” misses the point that convection transfers heat from lower to higher altitude, bypassing much of the so-called “greenhouse effect” in the process. Combine that with the phase changes, and water evaporating absorbs a lot of heat, then releases it at higher altitude when it condenses into the water droplets that form clouds, or even freezes into ice crystals that become precipitation.

    • Stephen:

      We have synergy of thought here.

      For a long time now I have considered water to be the prime planetary thermostat. This based on three properties namely: Very high latents heats. Being lighter than dry air. Having convenient ( for us) temperatures at which phase changes occur. And Additionally our gravity is just about right.

      The process is a Rankine Cycle.

      Evaporation/boiling takes place at the surface at low partial pressures — The boiler. Expansion takes place as it rises; thus doing work against gravity — The piston.
      During this process energy is dissipated, partly to Potential Energy via the work done and partly by dissipation to the surroundings.
      When all or most of the energy is used up it is exhausted to the condensation phase, which is a bit complex as it could result in either liquid water or solid ice to be formed.as further energy is dissipated.
      When sufficient liquid/ice has been formed gravity takes control and the cycle enters the feed pump/ heater phase, whereby the water descends as rain, ice, snow etc . And is compressed as it descends receiving heat from the surroundings in the process.
      At this point the water returns to the Boiler at the correct pressure for recycling.

      The overall cycle in thermodynamic terms transfers some 680 WattHrs./ Kg. Up into the atmosphere, a proportion of which reaches the cirrus clouds nudging the Troposphere radiating energy to space.

      Generally this process is independent of CO2 and reacts to any increase in external energy input by merely cycling faster rather than increasing the operating temperature.

      Hence the stable thermostat process which has served us well over millions of years.

      Where greenhouse gases are concerned these are ancillary to the above process and can easily be accommodated due to the relatively small entrapment energies involved.

      My regards
      Alasdair

      • This is exactly the mechanism that explains how water pumps heat from the ocean/lake surface waters to the upper atmosphere, bypassing the “greenhouse effect” that pretends to trap the heat near the surface.

      • Yes Alasdair, that is a reasonable analogy but not complete enough. That setup just uses the vertical plane for conversion of KE to PE which is a minor factor. In the atmosphere it is compression of gases in three dimensions that matters most since that produces vastly more conversion of KE to PE in ascent and correspondingly large amounts of heat in the subsequent descent.
        Furthermore, in adiabatic convective overturning there is no dissipation to surroundings at all because all that PE is hot heat and does not radiate so it is all preserved for repeated cycling up and down indefinitely once the atmosphere has reached hydrostatic equilibrium.

  6. For a first cut assumption at LW radiation balance during this emergent phenomenon, I would guess that the violent uplift of energy rich air to 17,000 m would enhance outbound atmospheric radiation above the uplift. Concurrently, the dried out sinking air should afford increased surface radiation opportunities around it.

    Do satellites have the spatial and temporal resolution to observe this?

  7. “For example, if a volcanic eruption reduces the amount of sunshine making it through the stratosphere, the tropics cool.”

    It may cause El Nino conditions to emerge.

    • I don’t see any interesting change at all! They are showing a graph that puts the observed temp above the models (false) …. they claim skeptical arguments are debunked in the peer reviewed publications (false) …. they claim clouds are an amplifying mechanism for CAGW (also false).

      All I see is the same propaganda I expect from the leftist rag. So …. just what is the change in tact you are referring?

  8. Hi Willis,
    Interesting reading as always. Although I generally support your thermostat theory, it seems to be active only in the tropics, so one has to wonder what the impact on the global temperature is. One interesting thing might be to use the PIRATA data (same website that you linked), which goes from -20 to +20, rather than TAO’s +/-10, to get some idea of what happens at higher latitudes. Another idea might be to guesstimate the percentual “leverage” the thermostat has, i.e. how many Joules of energy can it add or deduct to the average daily dosis received from the sun? Is it 10%, 1%, or 0.1%?
    Best regards and thanks again for your efforts.
    Frank

    • Frank, if you are going to start moving away from the tropics then the various circulations caused by the rising air in the ITCZ such as the Hadley cells and the Walker Circulation must be considered. The Corioliis effect on these circulations causes the overal atmospheric circulations and the interfaces between the air currents at different speeds causes the Rossby waves in the jet streams. Then you start seeing the entire atmosphere as a heat engine moving heat energy toward the poles where it radiates to space.

      • If this heat engine is moving “heat” (adjective) energy from the tropics toward the poles, why is it so cold at the poles ??

        Also “heat” (adjective) energy is the energy in the purely mechanical motions of real matter consisting of atoms and molecules. Only rarely can those atoms or molecules acquire escape velocity, and escape the earth into space.

        So almost NO “heat” (adjective) energy can escape to space.

        However, real physical materials that are “heated” (verb) by the “heat” (adjective) energy, which we generally refer to simply as “heat” (noun) are capable of emitting various types of “Electro-magnetic radiant” (adjective) energy which CAN escape to space.
        Such “EM radiation” (noun), which is either Long Wave (IR) or microwave, is almost invariably emitted isotropically, so usually only about half of it will escape directly to space, but conversely also only about half of any long wave EM radiation at any altitude in the atmosphere, will be directed towards the earth and become a part of the “Downwelling Long wave” (adjective) radiation that Willis, Phil , and others mention; but often don’t mention the isotropic distribution thereof.

        It has always puzzled me, that a puny 162 W/m^2 of “Total Solar Incident” (adjective) radiation energy; 98% of which lies between 0.25, and 4.0 microns wavelength reaching the earth’s surface, can “heat ” (verb) that surface causing it to emit (on average) about 390 W/m^2 of LWIR EM radiation upwards, most of which is absorbed by GHGs and clouds etc, which then give rise to 345 W/m^2 of “Downwelling Long Wave” (adjective) EM radiation; which presumably should be accompanied by a similar 345W/m^2 of “Upwelling Long Wave” (adjective) EM radiation from that same GHGs, clouds etc that must head towards outer space.
        I presume that the incoming 162W/m^2 of incident solar spectrum radiant energy (well power) is actually augmented by some additional (izzit about 83 W/m^2) of solar incoming that gets reflected, and also exits from the earth as the “albedic” (Willis adjective) component.

        But somehow that accepted 162 W/m^2 incoming solar spectrum power, gives rise to at least 345W/m^2 of exiting LWIR EM radiation.

        And more amazing is that at TOA, this all started out as 1362W/m^2 average (over a year) of TSI.

        Do these TAO buoys have a name other than TAO ??

        G

    • I have no clue if it is a parallel kind activity but I do remember many Denver summer days with clear mornings and afternoon thunderstorms. Often day after day of them.

  9. I think you can also see how this mechanism would work when we are in a glaciated state. when the northern and southern hemispheres are iced over and reflecting more light/energy and at the same time the mid latitudes are governed by this mechanism which also puts a limit on the amount of incoming energy making it to the surface, you can see how you can have a warm equatorial region on a glaciated planet.

    Conversely on a warming planet this mechanism would put an upper limit on the energy reaching any latitude and thus you would see warm northern and southern latitudes with the mid latitudes looking much as they do today.

    This would explain much of the fossil evidence that we see today

  10. And no, I have no interest in debating whether downwelling longwave radiation actually exists.
    =======!==
    Willis, a serious question. what about downwelling conduction? this is in all respects that I can see the equivalent of downwelling radiation except as to the distance over which it operates. yet nowhere is this considered in the total energy reaching the surface.

    I mean this quite seriously because I see it simply ignored as energy conducted away from the surface, stored awhile in the atmosphere the returned to the surface. thus reducing the cooling that occurred when the energy was first conducted away from the surface. if you replace conduction with radiation in the above you have the description for the greenhouse effect.

    • ferdberple December 22, 2017 at 6:02 am

      Willis, a serious question. what about downwelling conduction? this is in all respects that I can see the equivalent of downwelling radiation except as to the distance over which it operates. yet nowhere is this considered in the total energy reaching the surface.

      ferd, the amount of energy flowing from the atmosphere to the surface is minimal for a simple reason—warm air rises. This means that when heat flows from the surface to the atmosphere it is carried upwards, and is replaced by colder air … which leads to more heat going from the surface to the atmosphere. This process is one-way, by which I mean it doesn’t work in reverse.

      Thanks,

      w.

      • Cold air high up becomes warmer as it descends at the dry adiabatic lapse rate. When it reaches the surface it flows outward towards the nearest low pressure cell which contains rising warm air. During contact with the surface it becomes warmer still and reaches maximum warmth beneath the rising cell whereupon it re-joins the upward flow.
        You have previously agreed that adiabatic convection is a fully reversible process.

  11. The total energy absorbed by the surface is the sum of the net solar energy (surface downwelling solar minus surface reflections) plus the downwelling longwave infrared, or DWIR.
    ============
    no. energy is also conducted from the atmosphere back to the surface. at the molecular level not all air molecules are travelling the same speed. some will conduct energy away from the surface and some will conduct energy back to the surface. reducing the cooling that occurs from energy conducted away from the surface.

    no hand waving is accepted that this is somehow net zero. is you take the energy radiated from the surface that is absorbed by co2 the reradiated back to the surface this is also net zero.

    • If you are to include conductive and evaporative effects then you have to also include convection of water moleculles that are lighter than Nitrogen and Oxygen and will carry far more heat as latent heat up to the level where they change state to liquid water or ice and release that heat into the atmosphere.

      • Correct, but note that the portion converted to PE in adiabatic ascent does not show up as heat, cannot radiate to space and reappears as heat again during the next descent towards the surface.
        The KE / PE exchange during adiabatic ascent and descent is best seen as a discrete closed loop entirely independent of the radiative flow in and out.
        Thus once an atmosphere achieves hydrostatic equilibrium, as they all do, radiation in and out is in balance AND conduction up and down is in balance indefinitely unless the sun stops shining.

  12. The total energy absorbed by the surface is the sum of the net solar energy (surface downwelling solar minus surface reflections) plus the downwelling longwave infrared, or DWIR.
    ≠========
    air that is heated during the day and rises, descends at night and the energy is absorbed by the now cool surface. not all energy absorbed at the surface is due to radiation. I expect when looked at in detail conduction and convection play a much bigger role than has been considered in radiative theory. in effect convection and conduction mimic the GHG effect attributed to co2.

    • It seems your position is that air warms at the surface, rises, loses no energy, then falls back to the surface bringing the same net amount of energy back to the surface that it originally had when it rose in the first place.

      Is this what you are saying?

      • It goes up as moist air, the water vapour condenses to water losing the evaporation energy which radiates away and comes back down as drier air plus rain, having lost a lot of energy. Basically it works just like a steam engine, using the cold of space (3 K) as the condenser.

      • tty

        Yes, the latent heat of evaporation is radiated to space but not the PE created in lifting the air up against gravity. One can see the effect in the switch between the moist adiabatic lapse rate in uplift and the dry adiabatic lapse rate in descent. Radiation to space from the condensate throws the latent heat of evaporation out to space but the adiabatic component remains present and is recovered during the subsequent descent.

      • Yes, the latent heat of evaporation is radiated to space ,,,,

        Radiation to space from the condensate …..

        Evaporation and condensation are not radiation exchanges.

      • paulatmisterbees

        As you say, evaporation and condensation are not radiative exchanges but evaporation removes heat from the surface as latent heat which is a form of PE which is not heat and does not radiate and that PE is released as heat again at a greater height from the condensate which has radiative capability.
        The condensate radiates it out to space so that only the adiabatic portion returns to the surface and the portion that is adiabatic is set by the work needed to raise atmospheric mass against gravity to the height at which condensation occurs. That portion exists even in the absence of water vapour.

    • I’d be more apt to think that conduction occurs when wind carries heat from a warmer place to a cooler place. We see this all the time in warm fronts. This is also the mechanism of heat transport to the poles. But such, IMO would be a global cooling mechanism, as the heat is being carried to a place where it more easily radiated back to space.

      • Dr Deanster

        Only some of the energy being taken up in convection can be radiated to space. A portion of it is converted to PE which is not heat and does not radiate but which reappears at the surface as heat again when the air descends along the lapse rate slope.
        Thus the adiabatic portion is forever recycled up and down and cannot escape to space until the sun stops shining and the atmosphere falls to the ground.

        Your point about the lateral wind flows is correct. Air reaching the surface after being heated in descent at the dry adiabatic lapse rate then flows across the surface towards the nearest rising column and suppresses surface cooling all along the route.

      • However notice that the rain can’t heat up as it falls, since water is almost incompressible. The potential energy is instead partially transferred to the atmosphere by friction (this is the reason for the cold outward wind from a cloudburst, “wind shear”) and partly transferred as mechanical energy as the rain hits the ground and runs on downwards. It is this energy that wears down mountain chains to plains in a few million years.

      • of the adiabatic process is not available to heat the surface either. In the adiabatic process, the energy is used up as physical work, and it is characterized by not transferring energy to its surrounding. Thus, according to your logic …. the adiabatic system is carrying the IR trapped in molecules to the top of the atmosphere, where that radiation is released to space. In the case of water vapor, it radiates IR, and condenses releasing latent heat. In the case of CO2, it just radiates IR, but does not condense. Thus the air dries out, losing most of its heat from water vapor, and a good bit of heat from CO2 as well. As the adiabatic process descends the cooler, drier air becomes a net absorbed of IR from the surface, taking up water vapor and absorbing IR in the ranges of other respective GHGs. The physical energy of movement is not involved in the heating process at all outside of is ability to move IR and heat rich air to a place that is colder, where the IR will be released and absorbed.

        So, I’m not sure I’m following how you assume that this process would “reheat” the surface.

      • The energy of the adiabatic process is not available to heat its surroundings. Don’t know what happened to the first part of my sentence.

      • Further, if energy of the adiabatic process is transformed into heat as it descends, compresses, then that heat is lost from the adiabatic process, and reappears as IR, which is then taken up by GHG …. and radiated at the TOA.

        I’m still not following how you think this process heats the surface. Granted , sinking air, will compress and heat the surface, but now it’s getting into meteorological principles associated with high and low pressure systems.

      • ” Granted , sinking air, will compress and heat the surface”
        Precisely. Adiabatic heating. Gas laws : increase pressure of a volume of gas and the temperature will increase with no additional energy input hence adiabatic. In this case work has been done to compress the gas due to gravity.

      • In my experience, the ability of sinking air to “warm” the surface seems to be greatly influenced by the humidity. Where I live, high pressure in the summer with a constant flow of moisture heats up ….. whereas, high pressure from the arctic with no moisture is flat out cold, and doesn’t warm much at all.

  13. If the solar energy entering the earth’s atmosphere averages 162 W/m2, then the energy absorbed by the surface cannot average more than 162 W/m2. End of. Unless more energy is conjured up out of thin air against the known laws of physics.

    • Totally off topic of course, but WUWT brings up an average of four/five posts a day, and it has been a while since none of the in-house experts has written anything here about the impending ice age, and the number of years since we last saw an increase in global temperature. I feel frustrated.

    • It certainly can, the balance is between what enters the atmosphere and what leaves the atmosphere, what is absorbed by the surface can be more because not all the radiation that leaves the surface reaches space because we have an opaque atmosphere (particularly in the IR).

      • It certainly can’t. The surface cannot absorb more than the sun provides to the atmosphere, but it can emit it at a lower rate and so increase its temperature (for a while) until an new equilibrium is established.

      • Phillip Bratby December 22, 2017 at 10:43 am
        It certainly can’t. The surface cannot absorb more than the sun provides to the atmosphere, but it can emit it at a lower rate and so increase its temperature (for a while) until an new equilibrium is established.

        Of course it can, what do you think happens to ~300W/m^2 of measured DWIR at the surface?

      • All averages and at dynamic equilibrium:-
        The surface and near-surface are in radiative equilibrium
        Incoming solar prevents atmospheric collapse by IR loss to space. That is the work done on the system. Energy thermalized within the system can do no further work.

        All averages and at dynamic disequilibrium:-
        The system gains energy and the atmosphere expands. Work done is on the system with the energy provided directly by the sun.
        The system loses energy and the atmosphere shrinks. The work is done on space with the energy indirectly provided by the sun.

      • Phillip Bratby December 22, 2017 at 10:33 pm
        Where has that 300W/m2 come from. The sun didn’t provide it as it is only giving 162W/m2, so what is the source of all that energy?

        It’s recycled. Say for example that half of the IR leaving the surface is returned via GHGs to the surface, in order to balance the surface will be heated until 162W/m^2 leaves the atmosphere. In order to achieve that 322W/m^2 must leave the surface.

    • Phillip, you are wasting your time.
      They “believe”and will not listen, they will twist everything you say.
      Just don’t bother, I have been there and done that.

      • Yes, but it is spread over 4 times the cross-sectional area. About 25% is reflected from the upper atmosphere, clouds etc and some is reflected from the ground. The rest is the ~162W/m2 absorbed by the surface.

      • Phillip, the previso is that it is an “Average”, the actual is up to 1360 less losses and back down again for daytime and Nothing at night.
        Which the “Average” does not reflect, Nothing means Cooling, there is no continuous input as the Averages indicate.
        There is never any kind of steady state or Equilibrium as most Physics laws require, not even during the day as the Earth turns the rate of input varies from hour to hour.

      • Phillip,
        Be careful about this other misdirection from Trenberth. Technically, all 240 W/m^2 of the post albedo solar energy affects the surface temperature. Only the liquid water in clouds absorbs any appreciable amounts of solar energy. Since the water cycle closes the loop between the water in clouds and the oceans and acts between hours and weeks, relative to global yearly averages (and even monthly averages), any solar energy absorbed by clouds is equivalent to solar energy absorbed by surface of the oceans.

  14. Willis,

    “I found it amazing that the temperature of such a possibly unstable system could only have changed by ± 0.3°C over the entire 20th century. ”

    This isn’t exactly true. A smoothed 5-year average ‘anomaly’ might not vary by this much, but during each year, the planet varies over about a +/- 2C range between January and June. This is because of asymmetries where the N has more seasonal variability than the S and the planet as a whole exhibits the signature of the N hemisphere. The seasonal variability is subtracted out to arrive at the anomaly so the relatively larger variability is obscured from view. Note also that the global yearly variability is out of phase with the solar power, relative to perihelion and aphelion.

    • Thanks, co2. The part that is subtracted out, the seasonal changes, are fixed and repeat each year. As a result, removing them shows the part that is not fixed, the part that is variable. This residual is the part of interest, not the part that repeats every year like clockwork.

      Nor do we have to take a 5-year average. Peak-to-peak the maximum variation over the entire century is only ± 0.7°C. This is still only a ± 0.25% variation in temperature, indicating a very tightly regulated system.

      Best to you,

      w.

      • Willis,
        If the Earth had no atmosphere, by your definition of regulated, it would be even more tightly regulated whose temperature would be absolutely deterministic and based on the Sun, COE, the SB Law and nothing else. If anything, adding an atmosphere makes the temperature less ‘regulated’ by adding chaotic variability around the regulated mean.

        I consider what you’re calling ‘regulation’ effects to be moderation effects. These redistribute the ideal distribution of energy that would occur without an atmosphere making the tropics cooler and the Arctic regions warmer.

      • co2isnotevil December 22, 2017 at 7:36 pm

        When in fact a Planet without an Atmosphere has massive swings in Diurnal temperatures like the Moon.
        Water is our regulator or moderator as the difference between tropics and Deserts show.

      • AC,
        Yes, moderation, not regulation. Moderation maximizes the minimum and minimizes the maximum while the average remains unchanged. This differs from regulation which maintains a constant average independent of the stimulus. This distinction is crucial as regulation requires active control, while moderation is the strictly passive process of storing energy at one time and place and releasing it at another.

        Considering the atmosphere ‘active’ is the source of so much misinformation as the implicit, yet unacknowledged internal source of Joules powering the gain provides the wiggle room for an absurdly high sensitivity that literally fabricates energy out of thin air.

      • AC,
        One other thing about the Moon is that it’s much larger diurnal swing is largely the consequence of a day length that’s 28 Earth days long. If the Moon had 24 hour days, it’s diurnal variability would still be larger than Earth, but not by as much as you might think.

    • co2, saying that the GAST in 2000 was 0.3 degrees higher that GAST in 1900 is mostly a guess.
      The amount of measurement error for 1900 is unknown. Even the measurement error for 2000 has to be at least plus or minus 0.5 degrees. The null hypothesis then says that the GAST anomaly for 1900 and 2000 are not significantly different. I’d be interested to see the comparison between 2000 and 2017 in absolute degrees with realistic measurement errors.

      • Not 2000 to 2017, try 1997/1998 to 2017.
        In 1998 NOAA made the very big mistake of providing the ACTUAL Temperature that the 1997 Anomaly represented.
        It was 62.45F or 16.92C, compare that to the Hottest Year EVVAAAHH 2016 of 58.69F or 14.84C.
        Absiolutley no contest.
        Of course if you go in and check the Currently “Adjusted” temperature for 1997 you will find it is now only fiven as 57F or 14.53C.
        They blame the change on a change of Baseline, just think about that for a moment, if the baseline changes the Anomaly HAS to change as well, but the ACTUAL Temperature MUST remain the same.
        Isn’t NOAA Science a wonderful thing?
        And by the way they also said that 1998 was even WARMER than 1997.
        I can provide the links if anybody is the slightest bit interested.

  15. I don’t find it surprising that the Earth temperature as measured has changed very little during the 20th Century if you sum a very large number of positive and negative numbers which are random would you not expect to end up about zero. I find it difficult to believe that the energy flowing out from the Earth is exactly the same as the energy flowing in without a causal mechanism enabling that, the Earth gains and loses energy at many independent locations.

    • “if you sum a very large number of positive and negative numbers which are random would you not expect to end up about zero”

      No you would not, not if there is autocorrelation between the numbers, which in practice there always is in climate.

      • Yes, there are multidecadal and multicentury components to GAST.
        Any year to year temperatures are not independent samples. The null hypothesis prevails if real measurements include real estimates of uncertainty. Basically, any GAST estimate before the satellite era is not a measurement, but just a guess.

    • The positive and negative extents are far from random. The mechanism that keeps the energy flowing in and flowing out the same is Conservation of Energy. If more arrives than leaves, the planet warms and if more leaves than arrives, the planet cools. In general, they are not instantaneously the same, but are the same when averaged across a whole number of periods of the time varying stimulus. This time varying stimulus is seasonally variable solar energy per hemisphere and the period is a year.

      During the summer of a hemisphere, it will be receiving more energy than it’s emitting, while in the winter, it will be radiating more than it’s receiving. Similarly, during the day, more is received than is emitted while during night time the opposite is true.

      Instantaneously, input == output during the late morning and late afternoon and on a daily average basis, shortly after each equinox.

      Another dangerous assumption is that the two hemispheres exactly cancel. This couldn’t be any further from the truth.

  16. The idea that we can observe the total energy entering the Earth and leaving the Earth, the global temperature, is false we can try to calculate it but calculation is not observation and we would not be able to observe if the Earth is heating up or cooling down.

    • donald penman December 22, 2017 at 10:45 am

      The idea that we can observe the total energy entering the Earth and leaving the Earth, the global temperature, is false we can try to calculate it but calculation is not observation and we would not be able to observe if the Earth is heating up or cooling down.

      Donald, despite your objection the CERES satellite does exactly what you say is “false”. Its sensors observe and measure the total energy entering and leaving the earth. And while these measurements are not extremely accurate, they are very precise.

      w.

      • The CERES satellite does have observations but they are displaced temporally and spatially and there are problems joining these observations together it is not possible to observe all the Earth all the time using a single instrument.

      • donald penman December 22, 2017 at 11:25 am

        The CERES satellite does have observations but they are displaced temporally and spatially and there are problems joining these observations together it is not possible to observe all the Earth all the time using a single instrument.

        True. That’s in part why the CERES data is more precise than accurate … but I’m not clear what your point is.

        w.

    • It might be practicable to measure the amount of energy entering and leaving the Earth with a fair degree of precision. However this would not make it possible to determine “the global temperature” (whatever that is), only the energy balance of the system (of course provided we also know the geothermal heat flow with precision, which we don’t).

      • tty,
        Actually it’s practical and even relatively easy to establish the average surface temperature based on average properties. Relative to the Sun, geothermal heat flow is negligible.

        While we can’t predict the chaotic variability around the mean (PDO, AMO and other cycles) we can quite accurately calculate what the mean must be.

  17. Willis wrote: “Other phenomena include dust devils, squall lines, the Atlantic Multidecadal Oscillation, the El Nino-La Nina pump, cyclones, and the Pacific Decadal Oscillation. Likely more as well …”

    Commercial aircraft contrails.

  18. Willis, you talk as though the Down Welling IR Radiation all comes from IR radiation from the Earth being absorbed by CO2 and the 50% of it being re-radiated downward. The total radiation from the Sun consists of: IR radiation 50%; visible light 50% ( the atmosphere is transparent to all photons of light except blue light whose photons are absorbed and then re-emitted in random directions by, I believe, nitrogen, which is why the sky is blue.) The last 10% of solar radiation is UV, of which 70%, mainly the high energy A and B UV is absorbed by ozone, ( a product of life on earth,) leaving only 3% UV to reach the surface. Molecules of greenhouse gases cannot differentiate between “righteous” IR from the Sun and “evil” IR from the Earth. They will absorb the IR photons from the Sun, then re-emit them at an indeterminate time and in an indeterminate direction. T

    his means that 50% of the IR radiation from the Sun is re-radiated back out into space. Does this make CO2 a Sunshade gas as well as a Greenhouse gas?

    • J. M., I fear you are confusing near and far infrared. The sun contains “near infrared”, meaning frequencies near to visible light. This is NOT absorbed by GHGs in general.

      The other kind of infrared, “thermal” or “far” or “longwave” infrared, is what is absorbed by GHGs,

      So when you say:

      Molecules of greenhouse gases cannot differentiate between “righteous” IR from the Sun and “evil” IR from the Earth.

      that is absolutely not true. Molecules of GHGs only absorb and emit at certain frequencies in the thermal radiation band. They are transparent to other frequencies of radiation, whether those are “righteous” IR or “evil IR” …

      w.

      • Andrew, please do not confuse them with Sciency stuff.
        You have to remember that their Magic CO2 molecules take low energy, low temperature LWIR and turn half of it around with more power than all the Energy from the Sun reaching the surface put together.
        So cue any 1 or 2 of 6, Phil, tty, Trick, Paul, menicholas, Dave or even Mr Eshenbach himself to tell you the many ways that you are wrong.
        Some examples
        there are more CO2 photons
        It has been measured (like the very mixed 400 CO2 ppms have been measured and those measurements were completely discredited by the new CO2 measuring Satellite readings)
        Photons are photons
        There is no other answer, it must be GHGs and especially CO2. (even though Mr Eshenbach has admitted he thinks they play a very little role, if any at all)

        ps you forgot to add 2 other items, the Energy Chart, showing that the shorter wave Radiation has far more energy (Electron Volts) than IR and especially LWIR which is one of the weakest areas of the Radiation scale, even though Mr Eshenbach likes to call it “Thermal”, when all radiation from any object above 0K is Thermal Radiation.
        And the fact that Clouds, as well as reducing IR they also reduce those same very energetic shorter wave photons from striking the surface and more importantly entering the Oceans.

    • J. M. Davidson December 22, 2017 at 11:40 am
      Willis, you talk as though the Down Welling IR Radiation all comes from IR radiation from the Earth being absorbed by CO2 and the 50% of it being re-radiated downward. The total radiation from the Sun consists of: IR radiation 50%; visible light 50% ( the atmosphere is transparent to all photons of light except blue light whose photons are absorbed and then re-emitted in random directions by, I believe, nitrogen, which is why the sky is blue.)

      The sky is blue because of Rayleigh scattering, not absorption and re-emission, the scattering depends on 1/(wavelength)^4 so blue light is scattered about 10 times more than the red end of the visible spectrum.

      Molecules of greenhouse gases cannot differentiate between “righteous” IR from the Sun and “evil” IR from the Earth. They will absorb the IR photons from the Sun, then re-emit them at an indeterminate time and in an indeterminate direction.

      That’s exactly what they do, the near IR from the sun is only weakly absorbed by ghgs but the far IR emitted by the earth is strongly absorbed by ghgs and there’s more of it.

      • Only just seen this post.
        It appears you are not aware that the earth is more than 100 solar diameters away from the sun!
        I suggest you calculate the view factor for that. The integral under the insolation curve has to equal the integral under the earth’s emission curve otherwise we’d fry, consequently the earth’s emission in the IR far exceeds the sun’s irradiance at the same wavelengths.

  19. I love 💗 the correlation map in fig 1, between insolation and temperature. The blue parts are the most interesting – they confirm that the oceans are an excitable medium and that regions exist where, as you say, emergent reactive processes take place resulting in intermittent anti correlation between insolation and temperature – namely the ENSO.

    Even cooler is the visual confirmation that the Atlantic has its own enso. Even the Indian Ocean 🌊 has a little one.

    During a classic ENSO cycle, in the el Nino part, the ocean surface is warmer than normal due to interrupted Peruvian upwelling. But it’s also more cloudy ⛅️ than normal. Thus less insolation but higher temperature – the blue of anticorrelation in fig 1. In like manner, during the La Nina part, insolation is high due to clear cloudless skies, but equatorial Pacific sea surface temperatures are cooler due to robust reactive Bjerkes Peruvian upwelling. Thus again the anticorrelation.

    (Hint – notice that the eastern continental seaboard upwelling is responsible for all the anticorrelation in fig 1 – just saying!)

    However what is curious is that the blue anticorrelation region is limited to the central equatorial Pacific. One might have expected it to extend to the Peruvian coast to reflect the big triangular ENSO anomalies in the classic 1982/ 1997 type el Ninos. But that leads to the further thing that there are two kinds of ENSO cycles, the classic (strong Bjerknes) type and the Modoki (weak Bjerknes) type. The last true ENSO of the former, classic strong Bjerknes type was the 1997-1999 monster. However all the ENSOs from 2000 onwards have been of the weak Bjerknes or Modoki type. The restriction of the blue anticorrelation region to the equatorial central Pacific, not reaching the Peruvian coast, confirms this, since fig 1 is only from y2k onwards.

    Thus I would predict that if we had any of those fig 1 data before 2000, the blue patch on the Pacific would extend further east, connecting with the coast of Peru 🇵🇪.

    Is there any data allowing us to look pre-2000 at the fig 1 correlation map? I would love 💖 to see it.

    • ptolemy2 December 22, 2017 at 11:54 am Edit

      I love 💗 the correlation map in fig 1, between insolation and temperature.

      I’m sorry, Ptolemy, but I fear you’ve misread the graph. The correlation is between total surface absorption and temperature.

      The blue parts are the most interesting – they confirm that the oceans are an excitable medium and that regions exist where, as you say, emergent reactive processes take place resulting in intermittent anti correlation between insolation and temperature – namely the ENSO.

      Even cooler is the visual confirmation that the Atlantic has its own enso. Even the Indian Ocean 🌊 has a little one.

      Not so. Neither the Atlantic nor the Indian ocean have an ENSO phenomenon. In addition, Figure 1 doesn’t have anything to do with ENSO variations, because the ENSO in the Pacific happens right on the Equator. I had to think a bit about how to demonstrate it. Here’s a graph showing the standard deviation of the surface temperature after removal of the seasonal values:

      You can clearly see the genesis of the El Nino/La Nina variations off of the coast of Peru, as well as how far they extend out along the Equator. However, this is nothing like Figure 1.

      I note also that variations in the Gulf Stream off of the NE coast of North America are visible, along with variations in the North Pacific that may relate to the PDO.

      This is what I love about science. I come up with a new idea about how to reveal the extent of the El Nino phenomenon, and it shows plenty of other interesting stuff as well.

      w.

  20. Very interesting blog post from Willis.

    I am not sure I got Correct something in it, but just for the sake of some more clarification to me, if Willis will not mind to explain…

    Is it me or in some way it happens what I think was meant meant?!

    If I have to consider that this kind of surface-atmosphere response to radiation and RF as per this post, at some point concludes that it is and does happen in a “preemptive” mode, like in a “preemptive” response function to a forcing, like a preemptive response to the RF in this case, am I being wrong with such as understanding about your point made Willis??
    Just checking… if I have misunderstood something there.

    thanks

    cheers

  21. Research from an engineer/scientist who made a living testing gases etc for thermal etc response: https://www.omicsonline.org/open-access/the-refutation-of-the-climate-greenhouse-theory-and-a-proposal-for-ahopeful-alternative.php?aid=88698.

    https://orcid.org/0000-0002-3340-3063

    If we think of gaseous mixtures as metal alloys with vastly more degrees of freedom, we have taken a useful 1st step. In the above also are experiments showing the actuality of how noble gases handle thermodynamics.

    • From the article …

      As a spin-off of this study, the Lambert-Beer law is questioned suggesting an alternative approach. Moreover, the Stefan-Boltzmann law is relativized revealing the different characters of the two temperature terms. But in particular, the author’s recently published own work is quoted revealing novel measurement methods and yielding several crucial arguments, while finally an empiric proof is presented.

      So he’s overthrowing not just one but two well-established physical laws using “novel measurement methods” … I await his Nobel prize.

      w.

      • Willis Eschenbach
        December 22, 2017 at 2:10 pm: Having worked with Beer-lambert (light extinction) and SB I could see where he was coming from (regularities not so much laes). But circumstances usually alter cases, which the geographers and modellers who lead the warmista failed to grasp. Along with so much else.
        However, I make no claims of infallibility for the paper. Being a real scientist of the practical kind, nor does he. Just does the work and places it for perusal. Knowing some will laugh…..Oh well, that is the best medicine.

  22. Even better, what about the downwelling precipitation? It’s a real energy flux in absolute terms (liquid water enthalpy), however only the net flux (deltaH or the heat of evaporation) is of interest at the surface.

    It makes no sense to sum the total solar and the DWIR at the surface and count it as “the total energy that heats the surface”. Only the solar heats the surface, evaporation, convection, radiative heat to atmosphere (net) and the direct radiation to space cool it.

  23. ptolemy2
    December 22, 2017 at 11:54 am: Now you’re talking! This suggests changes in ocean energy net absorbance such as we have been looking for since Bob Tisdale elucudated the Enso discharge-recharge actions. Weaker Ninas may mean less input, vice-versa Nini Modoki less output, but it remains debatable, all of it…..

    The daily morning evaporation and night condensation cycles may be a huge and continuous mover of energy to where it can be radiated out, past the lower atmospheric soup.

  24. “When you comment, please QUOTE THE EXACT WORDS YOU ARE DISCUSSING so that we can all understand your thoughts and objections.”

    OK: “I’m sitting on a hill in the Solomon Islands”

    I strongly object to the fact that you’re there and I’m here freezing my butt off.

    Merry Christmas.

  25. Interesting and enjoyable discussion. Drier Texas barrier islands, which heat up fast, produce such clouds in the summer, but rarely produce rain. Maybe larger topography at work moving moisture to Louisiana where it is not needed. The waters are saltier, and have to wonder how much gets in the clouds. Harvey produced some rust problems, not altogether unusual in area even with normal winds.

    However, it’s Texas, probably adiabatic advectory. Merry Christmas!

  26. (And no, I have no interest in debating whether downwelling longwave radiation actually exists. It’s been measured by scientists around the world for decades, so get over it, Sky Dragons. Debate it somewhere else, please, this is not the thread for that.)

    345 W/m2 would make a nice input 24/7/ 365 for solar panels facing straight up. It is easy to make a panel which absorbs infra-red. Funny how no one does this…

    • Michael Moon December 22, 2017 at 3:50 pm

      (And no, I have no interest in debating whether downwelling longwave radiation actually exists. It’s been measured by scientists around the world for decades, so get over it, Sky Dragons. Debate it somewhere else, please, this is not the thread for that.)

      345 W/m2 would make a nice input 24/7/ 365 for solar panels facing straight up. It is easy to make a panel which absorbs infra-red. Funny how no one does this…

      Michael, sadly, you couldn’t be more wrong. Nobody knows how to make a “solar panel” that absorbs thermal, aka “far”, aka “longwave” infrared. This is for a simple reason. Thermal infrared does NOT have enough energy to kick any known electron out of its orbit, which is a requirement for a photo-electric device.

      Don’t quit your day job …

      w.

      PS—This is why I discourage such discussions. They tend to attract people like yourself who are absolutely positively sure enough of their position to make sarcastic remarks like “Funny how no one does this …”

      In fact what is funny is you foolishly and yet so earnestly claiming it can be done. I say again, this is not the thread for that subject.

      • Any known electron out of its orbit.

        No one has researched such a semi-conductor, because there is no energy source there. The Sun is over 5500 K. The sky is about 220-230 K. Kind of thing that makes you go, Hmmm…

      • Quote “Thermal infrared does NOT have enough energy to kick any known electron out of its orbit”.
        Or to do any other kind of “Work” at the surface of the Earth for that matter, and LWIR from a cold object does not make warmer objects warmer either.
        They can lead to a warmer object ending up warmer, but only if the 2 objects reach Eqilibrium and the Colder object has replaced an even colder object which would have meant an even lower Equilibrium.

        At no time does the Warmer object’s temperature exceed it’s starting temperature.

        During the Day/Night cycle at best it can only slow the cooling of the warmer object, God forbid that the Earth’s Surface and CO2 at -20C to -180C ever reach Equilibrium, we would not find it very comfortable.

      • Michael Moon December 22, 2017 at 4:30 pm
        Any known electron out of its orbit.

        No one has researched such a semi-conductor, because there is no energy source there.

        There’s plenty of energy there, ~300W/m^2, the problem is it’s packaged in photons which individually don’t possess sufficient energy to ‘kick out’ an electron. It was the discovery of this phenomenon which led to the development of quantum mechanics, it’s called the photoelectric effect, i.e. Einstein’s 1905 paper which won him the Nobel Prize.

    • Exactly and as you say, they would work 24hours a day, 365days a year.

      Willis says “Nobody knows how to make a “solar panel” that absorbs thermal, aka “far”, aka “longwave” infrared.” It’s because you can’t transfer energy from the cold atmosphere to the warm surface of the earth.

      • Phillip, you can transfer it OK, as I am sure the some photons reach the Earth from CO2, it just does absolutely no “work” whatsoever, especailly that of warming a warmer object.
        It is however very good at Cooling Objects as in a Solar oven tuning in to a Night time Fridge, but Mr Eshenbach doesn’t believe that it has anything to do with “Downwelling Radiation”.
        I am still not sure how a Warm object knows to send so many extra photons to a cold object when it is placed near it and make itself colder, when it carries on sending the normal number of photons to all the ambient objecst around it.
        Is cold a Hot Photon Magnet?

      • My silly experiments do one thing.
        They confirm EXACTLY what the Equations say will happen confiming the Second Law.

        Energy is transferred and absorbed, but only the cooler object warms.

      • If all energy thermalises why do none, not one single experiment demonstrate any for the hot object, not even a little bit.
        But they demonstrate cooling of the hot object and thermalising of the cold object.
        You just can’t see it can you?
        Take 2 objects at the same temperature of 62C (335K) as in my experiment, according to the first equations for output that is 10.05W/m2, a small amount, but detectable by your skin for each one.
        That energy output when placed very close together does nothing to raise either objects temperature.
        Plug the values in to the second equation and it says 0, yes zero transfer exactly as I found.
        But change the second item to ambient at 20C and the transfer is 8.8W/m2 and cooling increases dramatically for the hotter one.

        So what happened in the first case to the 10.5W/m2 going from object 1 to object 2 and the same energy going from object 2 to object 1.
        Why didn’t their temperatures change, according to you those photons must be absorbed and the temperature increased?
        You see everybody agrees and that includes Mr Eschenbach that they do not get hotter, it just slows the rate of cooling.
        So where are the 21W/m2 of energy?

      • It appears I posted in the wrong place.
        So here goes again.
        If all energy thermalises why do none, not one single experiment demonstrate any for the hot object, not even a little bit.
        But they demonstrate cooling of the hot object and thermalising of the cold object.
        You just can’t see it can you?
        Take 2 objects at the same temperature of 62C (335K) as in my experiment, according to the first equations for output that is 10.05W/m2, a small amount, but detectable by your skin for each one.
        That energy output when placed very close together does nothing to raise either objects temperature.
        Plug the values in to the second equation and it says 0, yes zero transfer exactly as I found.
        But change the second item to ambient at 20C and the transfer is 8.8W/m2 and cooling increases dramatically for the hotter one.

        So what happened in the first case to the 10.5W/m2 going from object 1 to object 2 and the same energy going from object 2 to object 1.
        Why didn’t their temperatures change, according to you those photons must be absorbed and the temperature increased?
        You see everybody agrees and that includes Mr Eschenbach that they do not get hotter, it just slows the rate of cooling.
        So where are the 21W/m2 of energy?

    • Michael, as Mr Eshenbach says, no one knows how to make an Infra Red Panel.
      But isn’t it surprising that all those Watts/M2, even more than the Sun’s also can’t warm water in Solar Heating Panels to keep it hot all night.
      In fact a Night Radiator Panel is used to Cool Water.

      • For those of you who think that cold things can’t warm hotter things, i would ask that you lay off the snark long enough to really think long and hard about what your position is and how shallow your statements appear.

        You are flat out wrong! You make intemperate comments flowing from word salads of incomplete, seriously over simplified and imprecise language.

        I am in a warm house right now, being heated by the cold air outside. Impossible? Well millions of warm homes are heated by winter air and they heat summer air from their cold air conditioning too. It’s called a heat pump.

        The magic is very simple. The cold thing and the hot thing are in a system that is NOT ISOLATED. External work is being provided which pumps cold joules into my warm house.

        The earth-atmosphere is a system that is NOT ISOLATED. Work is being applied to it constantly. In such a system cold can provide energy to warm all day long.

        If you can’t get past that, then you have no business diving into quantum mechanics, or snark for that matter.

      • Switch off the Electricity completely and see how much warmer your house gets due to Cooler things, work is being done you do not get something for nothing.

        You seem to have a major problem accepting Reality, in fact after our last discussions I am not sure why I bother except you are getting at people for having opposing views just like last time

        Why the hell don’t you actually try the Experiments for yourself instead of using words like snark.
        It will cost you about £35 for a decent Temperature Measuring Device with Probes and many hours of devising objects, waiting and measuring.
        Come back with your results as I have done on 3 seperate Threads now.
        This is what I have tested
        2 cooling objects of the same Temeperature.
        2 cooling objects one hotter then the other where both are above ambient Temperature ie one replacing ambient.
        2 cooling objects one hot (above ambient) one cold (below ambient).
        1 heated object at Equlibrium with ambient & 1 cooling above ambient ie one replacing ambient.
        1 heated object at Equlibrium with ambient & 1 cooling below ambient.
        both items heated and above ambient, but one cooler than the other ie one replacing ambient.

        None of them made the hotter object any hotter by even 0.1C.

        The only way that any of these can warm the hotter object is if they heat the air between them thus changing the ambient experienced by the hotter object which is not by Direct Radiation.
        I have tested it and does not happen if the air between them is stirred by a low level fan.

        So the only thing that happens is the hotter object can cool more slowly and END UP at a higher Equilibrium Temeperature.

        If you should find something that works (Please do not go down the light bulb in a box bullshit) let us all know so that we can confirm it and then we can accept what you are saying.

        Otherwise stop with the superioriority attitude and stupid analogies.

      • A.C. you say….

        “Switch off the Electricity completely and see how much warmer your house gets due to Cooler things, work is being done you do not get something for nothing”

        But that is exactly the point you are always missing. The electricity is the work that pumps the cool things energy into the warm thing. It is not a stupid analogy.

        It is exactly what is happening on earth. The sun is providing the work to allow the cool thing to heat the warm thing.

        Doing a radiation experiment between two beer cans in air with a fan blowing is not an experiment. It’s a waste of beer.

        Not trying to convince you, by the way. Just hoping that your ramblings don’t drag in any innocents who come here to learn.

      • I have just re-read your last statements.
        “Work is being applied to it constantly. In such a system cold can provide energy to warm all day long.”
        So you are admitting Back Radiation doesn’t do it.
        It requires Work to make it happen.
        Thank you.

      • A.C. says…

        “”I have just re-read your last statements.
        “Work is being applied to it constantly. In such a system cold can provide energy to warm all day long.”
        So you are admitting Back Radiation doesn’t do it.
        It requires Work to make it happen.
        Thank you.””

        Probably a dozen people have said this to you over the last week, in 2 dozen different ways. I am not ‘admitting’ anything. I am REPEATING something which you choose to ignore.

        WORK IS REQUIRED TO MOVE ENERGY FROM COLD RESERVOIR TO WARM RESERVOIR.

        Nobody in this blog has EVER claimed otherwise, yet you persist in setting up experiments to prove your strawman about warming up your beer on a counter with ice cubes and a fan with foil wrapped round things.

        THE SUN PROVIDES THE WORK

      • OK, got it now, it is absolutely NOTHING to do with CO2 Back Radiation, it is all to do with the Sun acting as your heat pump does.
        Great, all we need now is the actual mechanism that does it.
        Perhaps you can also provide that.

        PS your CAGW mates are not going to be very pleased with you after that statement.

      • A.C says…

        “”OK, got it now, it is absolutely NOTHING to do with CO2 Back Radiation, it is all to do with the Sun acting as your heat pump does.
        Great, all we need now is the actual mechanism that does it.
        Perhaps you can also provide that.

        PS your CAGW mates are not going to be very pleased with you after that “”

        It has everything to do with back radiation. That’s the cold thing. Earth surface is the warm thing. Insolation is the work thing. You want the math? Willis has provided it in wonderful examples at least twice.

        You are wrong about my AGW proclivities. I am strongly in the camp that believes CAGW is bad science. I don’t believe you fight it with more bad science, however. That is what many people who deny CAGW are pushing and what you have bought in to.

        Redirect your passion for experimentation to first understanding the 100 years of thermodynamic practice you are trying to wash away.

        Who knows, you could be on to something but it is evident in your writings that you don’t yet understand what you are trying to disprove.

        Believe it or not we’re in the same school and if you come up with an experiment with significant results you will find great support here no matter what it proves. But it must be based on sound principles for both the science in play and your lab techniques.

        You just can not do valid radiation experiments in atmosphere with a fan on your subject and tin foil tents. You can not get provable results that way.

      • It has everything to do with back radiation. That’s the cold thing. Earth surface is the warm thing. Insolation is the work thing. You want the math? Willis has provided it in wonderful examples at least twice.”

        You obviously missed the fact that I used the Equations.
        They will tell you how much energy is leaving each object.
        The second one will tell you how much energy is transferred from the Earth’s Surface to the CO2, but when you use it to find out what the energy transfer is in the other direction it give a massive Negative value.
        Because that is what happens. Although the first part of the equations tell you how much energy is leaving the Cold Object it causes no Heat Gain, as in measured by it’s Temperature in the Hotter Object, it only cools.

        PS there is no insolation work thing at night, someone switched it off, just like I suggested you do with your house work thing.

      • I did miss your equations. Could you point them out to me in this post?

        As for night time, well i don’t think the sun stops at night. Copernicus pretty much put that one to rest.

      • That is because they are not on this post, they are on the previous post on the same subject.
        They are not my Equations they are the ones posted by Ed Bo and Mr Eshenbach and working examples are shown on the Link provided by both
        Strangely none of the examples show how the Cooler object makes the hotter object hotter, they only show the Heat Transfer between the Hot to the Cold. The energy from the Cold to the Hot object is glossed over by saying “The Energy from the cold object is absorbed by the hot object”, absolutely no mention of it being Thermalised and raising the temperature of the hot object.
        I used the nice Calculator provided by Mr Eshenbach to double check my results and they do so.
        Watts/M2 output by Surface at 288K = 390
        Watts/M2 output by CO2 at 223K = 140
        Transfer of Energy from Surface at 288K to CO2 at 223K =+250W/m2.
        Transfer of Energy from CO2 at 223K to Surface at 288K = -271W/m2.
        You can check the values for yourself here
        http://www.efunda.com/formulae/heat_transfer/radiation/calc_2bodies_enclosure.cfm

        And just for clarity plugging in the Steel Greenhouse when the Shell reaches 120K
        Sphere to Shell = +56688242 W/m2
        Shell to Sphere = -113376485.4 W/m2

        As to your vapid remark about the Sun, tell that to the Night Side of the Surface that it shouldn’t be cooling because the Sun is shining on the other side of the world.
        Or to the Dark Side of the Moon at around 90K, that it should warm up because the sun is shining on the bright side.

      • My silly experiments do one thing.
        They confirm EXACTLY what the Equations say will happen confiming the Second Law.

        Energy is transferred and absorbed, but only the cooler object warms.

      • As the Energy out is directly controlled by the Temperature it means that no rise in heat means no increase in output either, as is proposed by Phil up thread when he quotes the Steel Greenhouse fantasy physics prcatically doubling the output of the surface to CO2.

      • Your calculator link describes a system of two isolated bodies inside a shell. That is not the system in the steel greenhouse. It is not the system of earth-atmosphere. If does not have a warm body energy source.

        B.T.W. all of the figures being used in these discussions are average flux values that are valid over reasonable integration intervals, i.e. long enough that daily, seasonal, or even annual variations are averaged out. No fair making the sun set.

      • Also, saying that ‘being absorbed’ somehow glosses over ‘thermalization’ is a slayer trick. They claim cold IR gets inside a body, wanders around and leaves without doing anything.

        Absorbed energy is in the energy budget of the absorber. It can’t get out by magic.

        The entire thermalization strawman is a diversion required by slayers to explain where BR goes.

      • Your reading comprehension has failed you again.
        It is not my Calculator, it is Mr Eshenbach’s.
        He provided it and I used it, it is not my fault if it is inappropriate.
        So take it up with him.
        But you should notice that the Equation it is based on is the Exact one he showed along with the link on his post.

        As to using “Averages”, they can be used to hide a multitude of sins, and additional information.
        Like ignoring night time cooling.

      • If all energy thermalises why do none, not one single experiment demonstrate any for the hot object, not even a little bit.
        But they demonstrate cooling of the hot object and thermalising of the cold object.
        You just can’t see it can you?
        Take 2 objects at the same temperature of 62C (335K) as in my experiment, according to the first equations for output that is 10.05W/m2, a small amount, but detectable by your skin for each one.
        That energy output when placed very close together does nothing to raise either objects temperature.
        Plug the values in to the second equation and it says 0, yes zero transfer exactly as I found.
        But change the second item to ambient at 20C and the transfer is 8.8W/m2 and cooling increases dramatically for the hotter one.

        So what happened in the first case to the 10.5W/m2 going from object 1 to object 2 and the same energy going from object 2 to object 1.
        Why didn’t their temperatures change, according to you those photons must be absorbed and the temperature increased?
        You see everybody agrees and that includes Mr Eschenbach that they do not get hotter, it just slows the rate of cooling.
        So where are the 21W/m2 of energy?

      • GRR, it is a real pain inding the right place to post today. Last try.
        If all energy thermalises why do none, not one single experiment demonstrate any for the hot object, not even a little bit.
        But they demonstrate cooling of the hot object and thermalising of the cold object.
        You just can’t see it can you?
        Take 2 objects at the same temperature of 62C (335K) as in my experiment, according to the first equations for output that is 10.05W/m2, a small amount, but detectable by your skin for each one.
        That energy output when placed very close together does nothing to raise either objects temperature.
        Plug the values in to the second equation and it says 0, yes zero transfer exactly as I found.
        But change the second item to ambient at 20C and the transfer is 8.8W/m2 and cooling increases dramatically for the hotter one.

        So what happened in the first case to the 10.5W/m2 going from object 1 to object 2 and the same energy going from object 2 to object 1.
        Why didn’t their temperatures change, according to you those photons must be absorbed and the temperature increased?
        You see everybody agrees and that includes Mr Eschenbach that they do not get hotter, it just slows the rate of cooling.
        So where are the 21W/m2 of energy?

      • Has it occurred to you that your experimental technique could be a problem, or that you are applying your results incorrectly?

        For example I remember one where you added a fan. I remember that none were done in a vacuum. I don’t remember if you ever provided an energy source to warm object. All of these things are problematic. If you are doing these in your kitcben, have you taken the radiation of: the room, your body, your lighting, your electronics?

        As to results you have stated emphatically that the only effect you have seen is that a cold body slows the cooling of a warmer one. Yikes! That is exactly the point we have been trying to get you to see.

        What you seem to be missing is that you are proving the 2nd law and we are talking about a NON ISOLATED SYSTEM that does not and can not violate the second law because it is not applicable to a system that is undergoing work from an external source.

      • I can see that you just cannot face facts.
        The excuses you are coming up with as to why the Experiments may be wrong are laughable.
        The most laughable of all is that I did not do them in a vacuum.
        Which part of the Earth’s Surface and the TOA is in a Vacuum exactly.
        The only criteria is that it all takes place in the AIR, just like all the Downwelling Radiation and the Surface.
        The only part that doesn’t is the Solar input at the TOA, which has absolutely nothing to do with Heat Transfer between 2 objects within the Atmosphere.
        Sorry, I will no longer waste time discussing the subject with you.

      • The point of the vacuum is to ensure you are setting up your subjects to be isolated from all the influences that are superfluous to the subject under investigation.

        If you don’t do that then you must account for all those extraneous heat transfers in the setup. When you don’t isolate from, or account for those things you haven’t created a repeatable experiment to make your point.

        Sorry. Thought you might have known that.

        For example, the fan blowing on your beer cans might just create more conductive flow than radiation flow. The radiation from your own irradiance might overwhelm the radiation transfer you are investigating. The distances involved in your set up will influence conduction flows. The nature of surfaces will be influential.

      • I have never in my life seen anyone clutch at as many straws as you.
        Your latest explanation of what I could be doing wrong is another ridiculous one.
        The Hotter Objects were at EQUILIBRIUM with the fan running, get it, any Blowing away had already been accounted for.
        Why don’t you just give up you haven’t got a clue what you are talking about and are making a foot of yourself.

        Do you actually think that anything happening in the Atmosphere does not have to accommodate many more variables than I am introducing and yet you still try to trash simple tests that should show warming, that have absolutely no trouble demonstrating the expected cooling.

        Just go away with your stupid straws, you are really starting to annoy me now the novelty has worn off.

      • For system with insolation i, surface radiation s, output radiation o, and back radiation b, with an atmosphere that exhibits a returned surface radiation percentage A. Assume b and i are absorbed at the surface and reemitted with no loss.

        This is the math…

        Surface radiates

        s = i+b

        Back radiation is

        b = sA
        b= (i+b)A
        b-bA=iA
        b=iA/(1-A)

        Surface radiates

        s = i+b
        s = i + iA/(1-A)

        Output radiation is

        o = s(1-A)
        o = [ i + iA/(1-A) ] (1-A)
        o = i – iA + iA
        o = i

        For A=1, total reflection, surface explodes.
        For A=0, total transparency, s=i=o and b=0
        For all 0<Ai>o and b not=0

        Try it with i=240 and A=0.3846. You’ll get b= 150, s=390, o= 240

        You have been experimenting with i=0 and everything bathed in unknown radiation from your lab so while theoretically at equilibrium i=0, b=0, s=0, o=0, a not very meaningful or satisfying result, you will never get there because your whole experiment will flat line at room temperature.

        The really important thing here is that all of the above is based solely on first principles, conservation of energy. You don’t have to invoke quantum mechanics or IR or photon interactions. Nothing. It works with demonstrable contemporary measurements.

        Anyone who declares an alternative to this math and accompanying assumptions must show that their theory/assumptions does not provoke violations of this first principle analysis!

        You are free to proclaim that BR does nothing or even that it doesn’t exist. Just remember that whatever you come up with must obey this simple set of equations while acknowledging the contemporary measurements.

        You are free to proclaim that your experiments disprove all this too. I expectantly await your new theory with the math that shows us how it all works.

      • Paul Bahlin, has anyone ever told you that you haven’t the first idea what you’re wittering about?

        I haven’t seen such a heap of old donkey droppings as your posts for a very long time!

        Admit it, you’re just taking the p!ss, aren’t you?

      • catweazle666 December 24, 2017 at 2:18 pm

        Paul Bahlin, has anyone ever told you that you haven’t the first idea what you’re wittering about?

        I haven’t seen such a heap of old donkey droppings as your posts for a very long time!

        Admit it, you’re just taking the p!ss, aren’t you?

        TRANSLATION: In other words, the Weasel can’t find one single thing in Paul’s post that he can quote and show is wrong … so instead he turns nasty and starts throwing mud.

        Folks, if you wish to refute someone’s claim, saying it is “donkey droppings” just makes people point at you and laugh. The only way to refute what Paul says is to QUOTE WHAT YOU THINK IS WRONG and then TELL US WHAT IS WRONG WITH IT.

        You seem to actually believe that throwing excrement at Paul affects his reputation and not yours … bad news …

        w.

      • ”TRANSLATION: In other words, the Weasel can’t find one single thing in Paul’s post that he can quote and show is wrong … so instead he turns nasty and starts throwing mud.”

        So let’s see shall we, Eschenbach?

        ”External work is being provided which pumps cold joules into my warm house….
        In such a system cold can provide energy to warm all day long….”

        And that utter BS is just from one post.

        Heh, ”…cold Joules”…”cold can provide energy to warm”… Yeah, right!

        Prat.

      • Tens of million heat pumps are heating homes today, pumping energy from cold outside air into warm and cozy houses at half the energy required for conventional furnaces. Hundreds of millions of refrigeration units and air conditioners do the same thing all over the world.

        They all pump from cold to hot all day long.

        Someone who doesn’t know that is astoundingly ignorant. Or, maybe just a troll. Calling me stupid for pointing it out? Well that’s just priceless.

      • “Tens of million heat pumps are heating homes today”

        And even more cooling electronic systems and microprocessors, of course.

        Joules are joules, they are no more hot or cold than they are red or circular.

        And heat pumps pump from hot to hotter, I have dealt with them in applications where the inputs were at some hundreds of of degrees centigrade and output radiators were red-hot, and I was dealing with solid state Peltier effect pumps in microprocessor applications decades ago.

        So go teach your grandmother to suck eggs.

      • You failed in setting the parameters right at the beginning.
        You used the word ASSUME.
        There is no assume, b may be absorbed by the surface, but as I keep on keeping on showing, it does nothing to the surface temperature, which is the ONLY thing that controls the output.
        There is no
        i + b
        Or are you suggesting that all the b photons go in and back out so fast it is impossible to measure a change.
        In which case it still hasn’t done anything to the surface temperature.
        Because if there was the temperature would go up and I have shown that that does not happen, the Equations say it does not happen, I have told you before the maths may be perfectly Ok, but it is useless if incorrectly applied.
        Just like your stupid cash flow analogy, it does not consider debt and it does not consider unacceptable money or IOUs which all exist in the real world.

      • I can assume anything i want. That’s how science works. You assume a set of constraints, build the math to describe it, then test it in the real world.

        My test is that this set of assumptions and equations perfectly conforms to what we observe with contemporary measurement.

        You are doing the same thing. You are assuming back radiation does nothing. That’s fine. The next step is to build a mathematical representation to describe your theory using your assumptions. Then test it.

        You’re testing it without a mathematical representation and that’s fine too. Many a discovery has been made by raw experimentation and even sometimes by accident. But, when you reach a point where you are convinced that you are onto something, and you appear to be there, then you must describe what you have found with more than handwaving or digs at the opposing theory.

        I am waiting, at this point, for your equations. It aint hard. I’ve shown you mine.

      • Mr Bahlin, you have spent too much time around Mr Eshenbach, your Ego, Hubris, Superiority Complex, Condescension are almost as bad as his.
        Unlike you I do not need to make up fantasy Science to apply fantasy maths to.
        I just used the good Old Fashioned Equations supplied by Actual Scientists that Ed Bo and Mr Eshenbach referred me to.
        They show Energy Output, Heat Transfer and that Energy Output does NOT = Heat Transfer unless it is from a Hot Body to a Cold Body.

        You seriously need to get over yourself.

  27. Back when I was a soaring pilot the moment when cumulus clouds began to form in the morning was a very big deal. It meant the difference between a jolly afternoon zipping along the cumulus cloud streets and a short 20 minutes and back on the ground.

  28. Willis, We should over decades be able to measure the clouds coming up earlier, if co2 is a forcing ?

    Do the satellites and data exist to test the hypothesis ?

    • Well, Stevek, you should, over decades be able to measure the clouds coming up earlier if the global temperature was going up over that time period. Not sure what that would tell you about CO2 though (unless you are convinced that CO2 and only CO2 drives temperature in which case, why bother, you already have all the answers)

      • SmartRock, I suppose clouds coming up earlier would point to something is causing them to come up earlier. It could be higher temps. The higher temps could be caused by c02 or some other factor.

    • stevek December 22, 2017 at 4:59 pm

      Willis, We should over decades be able to measure the clouds coming up earlier, if co2 is a forcing ?

      Do the satellites and data exist to test the hypothesis ?

      Good question. Not as far as I know. Some problems. First, the change in CO2 during the satellite era equals a forcing change of a watt/m2 or so, and in the tropics total downwelling radiation is 600 W/m2.

      Next, the only dataset that comes even close is the TAO buoy dataset, and the funding and maintenance have been drying up for those buoys.

      Finally, we don’t know how much of a change in forcing there might be from other sources (volcanoes, aerosols, dust storms, etc.)

      w.

  29. Very interesting, thought provoking post, Willis. Creative use of public domain data. Thank you.

    Quite a few one-dimensional comments though (not unusual and not unexpected in an open blog)

  30. Just can’t believe in a hypothesis where the only source of cooling for earth is through IR radiation to space and adding more of an IR radiant is going to heat things. How about the low ability of nitrogen and oxygen to radiate because of their dipole nature allows them to hold on to this energy longer and they are in fact the greenhouse gasses.

  31. Very Interesting, makes perfect sense. Kind of reminds me of a similar phenomena of the dew point. As the radiational cooling causes surface temperature to fall below the dew point and condensation forms, which releases heat and raises the surface temperature slightly above the dew point. The transitional states of H2O are no doubt a magical regulatory process that make the climate of this blue marble we live on so stable.

  32. I’ve made some observations with regards to fog over a couple of rivers that I have pondered for years and wonder if those observations could be pertinent to satellite readings.

    Case one is a dam release that occurs for power generation in hot summer afternoons where general air temp is in the 90’s F and measured water temp is very low 40’s F. there will be a layer of cold fog about 5 ft thick over the river for about 2 – 4 miles. You can reach your hand up and feel the 90 F air while the fog bank may be around 50 +/-. Visibility in the fog is about 15 – 20′ max and you need to periodically stand up to view the shoreline to see where you are in the river. I assume this is caused by the cold water and humid air above it causing the moisture in the air to condense. It’s quite a treat the first few times you float this stretch of moving water.

    The second case is the opposite. On a different river on some nights when the river temp is warm (75 – 80 F ?) and air is still after dusk you can see the plumes/streams of moisture rising off the water as the fog begins to form (much lighter than where the fog formed from air air above it as noted earlier). Its really interesting to visually see the streamers of moisture rising in the light’s beam. There’s a lot of it. The river is a drop-pool character so the water is continually mixed every 1/4 mile or so. But what atmospheric factor causes the streamers/plumes one night and not the next (a couple of days later) when it seems conditions are similar?

    The thought in my mind on a much larger scale would be is in calm air ocean environment and evaporation of the magnitude as I witnessed in the river and the heat content held by the layer of air near the surface (~ 1 – 15′ or so). Would this layer of high humidity moisture laden air affect the satellite readings. If so, what would be the effect of barometric pressure on a different day with all other things being equal? I guess in a nutshell, is there any chance that a layer of saturated air covering the surface of a body of water would affect a satellite reading for temp or emissions or absorption?

  33. QUOTE THEIR WORDS, and the TELL US WHAT IS WRONG WiTH THEM
    the should be then

    The letter “i” in the all-caps “with” is in lower case.

    You mean like that?

    ;-)

  34. Since CERES is built on top of ERBE, this view of the energy budget is pertinent:

    Credit: Image courtesy NASA’s ERBE (Earth Radiation Budget Experiment) program.

    • Thinking about the head post and in context of the ERBE graphic Ron shared it would seem the process described by Willis is likely part of the “carried to clouds and atmosphere” by latent heat [at least partially]. Are there data to quantify how much thermal energy is lofted over the CO2 emission layer? It would seem as though a process the cooled the surface in such a way as demonstrated above were to also help cool the planet we would be able to observe [if we had minute by minute data rather than daily averages]. Or, as Wllis put it, “However, this average conceals an important fact—the negative and positive correlations are not randomly distributed.”

      • Randy, the ERBE diagram shows the atmosphere absorbs 15% of emitted surface energy,which is added to 16% of incoming solar energy absorbed by the atmosphere. Both are then part of the OLR.

    • Ron Clutz December 23, 2017 at 6:51 am

      Since CERES is built on top of ERBE, this view of the energy budget is pertinent:

      It is pertinent, but it is not complete. Why? Because it is only showing the NET energy flows, not all of the energy flows. Here’s my diagram which shows all the energy flows.

      It nets out to the same flows as the NASA diagram, but it includes all of the radiation flows, not just the net flow.

      w.

      • The net matters because heat flow is defined as the net difference. Saying back radiation is absorbed by the surface is misleading since the heat flow is the other way.

      • Ron Clutz December 23, 2017 at 2:47 pm

        The net matters because heat flow is defined as the net difference. Saying back radiation is absorbed by the surface is misleading since the heat flow is the other way.

        Misleading? Ron, how do you think they got the net values? You say correctly that heat flow is the “net difference” … but the net difference of what? They didn’t start with the net flows.. They started with the energy flows that you call “misleading”, clearly a wrong move on your planet.

        Next, if someone doesn’t know the difference between ENERGY flows and HEAT flows, they shouldn’t be in the discussion …

        Finally, the two energy flows in opposite directions are what is actually physically happening. On what planet is describing what is actually physically happening “misleading”?

        Regards,

        w.

      • Willis:

        Thank you for another stimulating entry.

        I agree with the basic thesis and would like to add some numbers. In your energy flow diagram, I would call attention to the orange “latent heat” arrow. I submit that this is where your magic occurs.

        From planetary average rainfall, and the fact that 29 w/m^2 produces 1 mm per day, we know that roughly 80 w/m^2 goes to evaporation.

        In the moist enthalpy calculation, the ratio of L, the heat of evaporation of water, to RT, the specific heat due to molecular motion, is a staggering 18 to 1. This means that evaporating water consumes large quantities of heat and produces no temperature change, though density does change.

        A very interesting paper by Kumar and Arakeri
        http://www.sciencedirect.com/science/article/pii/S2210983815001273
        shows that natural evaporation can have a rate constant of 33 mm per day (equivalent to 970 watt/sq^m) per 0.1 atm partial pressure difference. For the differences between temperature and dew point typical of Honolulu I find roughly 6 mm per day of evaporation, equivalent to loss of ~180 watts/cm^2.

        The impact on vapor density change (which drives convection) from evaporation as compared to heating with an amount Q is significantly less by roughly a factor of Q/ 9R vs. Q/ 5/2 R.

        I cannot quantify the factor (heat of evaporation)/heating which would tell us when cloud formation sets in, but the paper by Bond and Struchtrup on evaporation seems to offer come possibilities. https://www.ncbi.nlm.nih.gov/pubmed/15697379

        The phenomenon of diurnal cycling you describe for tropical cloud formation is well described by Eastman and Warren. https://atmos.washington.edu/~rmeast/Eastman&Warren_2014.pdf

        Thanks again and Merry Christmas to all.

    • Ron Clutz:

      Thank you for posting ERBE’s reasonably realistic view of energy fluxes in the climate system.

      What is critically important in the energy budget is the NET heat transport by all mechanisms. The customary “climate science” representation of radiative transfer as a pair of upward and downward fluxes over-idealizes what is actually isotropic radiation by each parcel of matter. They are wholly unconstrained by the energy budget, since ANY pair of opposed fluxes with the SAME net value will do.

      In reality, the magnitude of those inseparably-paired components is not set by the FLUX of energy through the system, but by the system’s internal STORAGE of thermal energy, as evidenced by the temperature. Orthodox presentations, such as Willis’, obfuscate that physically crucial distinction and create the popular illusion that LW radiation, rather than moist convection, is the principal means of heat transfer from surface to the atmosphere.

      • 1sky1 December 23, 2017 at 4:57 pm

        Ron Clutz:

        Thank you for posting ERBE’s reasonably realistic view of energy fluxes in the climate system.

        What is critically important in the energy budget is the NET heat transport by all mechanisms. The customary “climate science” representation of radiative transfer as a pair of upward and downward fluxes over-idealizes what is actually isotropic radiation by each parcel of matter.

        I say again, as I said to Ron: the “representation of radiative transfer as a pair of upward and downward fluxes” is simply a description of the physical reality. It is not “idealized”, it is how the physics of it actually works. Yes, it is isotropic … but the horizontal part of it is basically net-zero and is left out of such overall budgets.

        In fact, what is “idealized” is the flow of heat. We cannot measure that flow, it is an idealized concept.

        They are wholly unconstrained by the energy budget, since ANY pair of opposed fluxes with the SAME net value will do.

        Oh, please, that’s just silly. We can MEASURE the fluxes, so no, we don’t just get to pick numbers with the “same net value”. Yes, the net upwelling from surface to atmosphere is about 70W/m2 … but that does NOT allow us to claim that the surface is radiating 890 W/m2 upwards and the atmosphere is radiating 820 W/mw downwards. That’s ludicrous.

        1sky1, how do you think NASA got the net numbers in the budget Ron put up there? Do you honestly think they measured the net flux? Get real. The very name “NET” should give you a clue … NET means that it is THE UPWARD FLUX MINUS THE DOWNWARD FLUX … are they “over-idealizing” the situation too?

        You desperately need to read a thermo text before you embarrass yourself further …

        w.

      • 1sky1, much as I respect many posts from Willis, he has a blind spot here, and attacks rather than reconsidering. Radiation is energy all right, but not automatically heat. I get exposed to kwatts of radio waves all the time, but it doesn’t raise my temperature since my body is transparent to that radiation. Radiation is only heat if the exposed material absorbs it. Solar radiation is high energy, aborbed by both land and sea and warmed by it. The bit of far longwave radiation from the cool atmosphere is not comparable. The Trenberth diagram misleads people to think the atmosphere has twice the power to heat earth’s surface.

      • Ron Clutz December 23, 2017 at 7:04 pm

        1sky1, much as I respect many posts from Willis, he has a blind spot here, and attacks rather than reconsidering. Radiation is energy all right, but not automatically heat. I get exposed to kwatts of radio waves all the time, but it doesn’t raise my temperature since my body is transparent to that radiation.

        Wow. Just … wow.

        To start with, the human body can and does absorb radio waves. Here’s a typical quote, this one from the WHO:

        Radars usually operate at radio frequencies (RF) between 300 MHz and 15 GHz. They generate EMFs that are called RF fields. RF fields within this part of the electromagnetic spectrum are known to interact differently with human body.

        RF fields below 10 GHz (to 1 MHz) penetrate exposed tissues and produce heating due to energy absorption.

        Your foolishness doesn’t end there, however. You make another equally ludicrous claim:

        Radiation is only heat if the exposed material absorbs it. Solar radiation is high energy, absorbed by both land and sea and warmed by it. The bit of far longwave radiation from the cool atmosphere is not comparable.

        So … your claim is that LW radiation is NOT absorbed by the land or the sea? Say what?

        Kirchoff’s Law says that absorptivity is equal to emissivity. My bible for many things climatish, including the emissivity (which is equal to the absorptivity) of common substances, is Geiger’s The Climate Near The Ground, first published sometime around the fifties when people still measured things instead of modeling them. He gives the following figures for IR emissivity at 9 to 12 microns:

        Water, 0.96
        Fresh snow, 0.99
        Dry sand, 0.95
        Wet sand, 0.96
        Forest, deciduous, 0.95
        Forest, conifer, 0.97
        Leaves Corn, Beans, 0.94

        In other words, Ron, you don’t have a clue what you are talking about, and on top of that, you are so taken with your own incorrect opinion that you don’t even know that you are talking nonsense. Instead, you seem to believe you’re an expert on the subject.

        You’re welcome to continue to parade your ignorance, but my advice would be to ponder the First Rule of Holes … it’s not doing your reputation any good.

        w.

      • …Willis, he has a blind spot here, and attacks rather than reconsidering. Radiation is energy all right, but not automatically heat.

        Apart from the issue of thermalization of radiation, the blind spots here are numerous. Foremost is the failure to recognize the clear difference between energy and its rate of flow in time, i.e., power. The former can be stored, like money in the bank, or put into flux, like income or expenditure. Sustained power, on the other hand, has to be continually produced. Without additional sources of power, which are absent in the atmosphere, it cannot be multiplied.

        To make this crucial point clear to laymen, note that a constant amount of money in the bank–the monetary equivalent of climate equilibrium–is maintained by ANY balance of household income and expenditure. The sustained spending power of a household depends solely on its income, the equivalent of thermalized insolation. No matter how often spouses may exchange checks, the equivalent of radiative exchange between surface and atmosphere, their total income and bank deposits don’t change. These fundamental properties of energy budgets patently contradict the mistaken notion that backradiation is an additional “forcing,” like a “second sun,” peddled by wannabe “climate scientists.”

        In arguing that we don’t just get to pick numbers with the “same net value,” Willis totally misses that crucial mathematical point being made about energy-conserving budgets. And he is plainly unaware that all that is measured by pyrgeometers is “the resistance/voltage changes in a material that is sensitive to the net energy transfer by radiation that occurs between itself and its surroundings.” As is the case with much of CERES data, the directional radiative fluxes are not measured, as he claims, but inferred from a set of assumptions. By posting mocking dismissals of scientific basics he merely reveals his own lack of schooling.

      • ” By posting mocking dismissals of scientific basics he merely reveals his own lack of schooling.”

        On the other hand, insisting that waste energy that has been rejected can never be used again might reveal a retinal infirmity of your own.

      • 1sky1, thanks for drawing some implications often overlooked but obvious when you point them out. Indeed, solar radiation is not only much higher quantum energy, but is also a sustained addition to earth’s energy reserves. Atmospheric IR fluxes are a dissipative process, unsustainable in itself, and thus not a “second sun.” And I take the point of your financial analogy: For the earth as for an household, it is a good thing when the rate of income exceeds the rate of spending. Having energy in reserve comes in handy not only for rainy days, but for cold, snow stormy Christmas days like this one.

      • Ron Clutz December 25, 2017 at 4:39 am

        For the record, my example was radio waves, not radar waves which are actually in the microwave part of the spectrum.

        Thanks, Ron. For the record, “RADAR” is an acronym for “RAdio Detection And Ranging” …

        w.

  35. Let me note in passing that a doubling of CO2, which will increase the DWIR by something on the order of 3.7 W/m2, and which it is claimed would lead to Thermageddon, would be less than a 1% change in total downwelling radiation at the surface … which would easily be offset by a small change in total cloud cover.

    See, there’s the problem with climate modeling in a nutshell. The attribute with the largest partial derivative drives any changes in the output of a function, but also magnifies measurement errors for that attribute. There’s simply no way to accurately measure what actually drives the temperature (which is the clouds), and the attributes you can measure accurately (C02) are largely irrelevant.

    In math form:

    if T = f(C02) + g(CloudCover) + h(SomethingElse), and the partial derivative of dT/dg(CloudCover) is greater 10 times that of dT/dh(SomethingElse) and dT/df(C02), then you can generally simplify the equation to T = g(CloudCover) when you are looking to see what the change in temperature is going to be due to some natural phenomena.

    This is metrology (sic) 101.

    Good luck actually measuring cloud cover to 0.1% accuracy.

    Peter

  36. “However, it didn’t cover total energy input to the surface. The total energy absorbed by the surface is the sum of the net solar energy (surface downwelling solar minus surface reflections) plus the downwelling longwave infrared, or DWIR. This is the total energy that is absorbed by and actually heats the surface.”

    You did not mention the meridonial heat transport. In your new analysis the meridional heat transport in the atmosphere is taken into account by the downwelling longwave IR. But the heat transport in the ocean is missing.

    • Thanks, P. I am not analyzing where the heat goes after it is absorbed. Part is radiated, part is conducted, part goes into latent heat, part is moved horizontally … I’ve not discussed any of those.

      I’m looking at the relationship between temperature and what is absorbed.

      Regards,

      w.

      • “Thanks, P. I am not analyzing where the heat goes after it is absorbed. Part is radiated, part is conducted, part goes into latent heat, part is moved horizontally … I’ve not discussed any of those.”

        Willis, my point is that you cannot neglect it. The energy balance at the surface is

        ASRS+LWSin-LWSout-Thermals= 0

        with absorbed solar radiation at the surface ASRS, long wave radiation in and out LWSin and LWSout and Thermals= sensible heat+latent heat. The thermals are equal to the radiation imbalance ASRS+LWSIn-LWSOut and are approximately 0.7*ASRS. The thermals correlate better with ASRS than with LWSIn. Therefore I approximate Thermals= const*ASRS= 0.7*ASRS. Then LWSOut is then given by

        LWSout= ASRS*(1-const)+LWSIn

        For const=0, I find for CERES 4.0 2001-2016 a correlation factor for avg.globe corr= 0.59. It increases to 0.75 for const = 0.7. This is a little improvement of your analysis. Happy new year!

  37. Atmospheric heat dynamics are sufficiently complex to allow any politically mandated belief system about it to be forcefully imposed by a sufficiently well-paid mafia.

  38. Really Willis?

    “In other words, Ron, you don’t have a clue what you are talking about, and on top of that, you are so taken with your own incorrect opinion that you don’t even know that you are talking nonsense. Instead, you seem to believe you’re an expert on the subject.

    You’re welcome to continue to parade your ignorance, but my advice would be to ponder the First Rule of Holes … it’s not doing your reputation any good.”

    Maybe you should take a break. Enjoy the holiday season.

    • Extreme Hiatus December 23, 2017 at 7:49 pm

      Really Willis?

      “In other words, Ron, you don’t have a clue what you are talking about, and on top of that, you are so taken with your own incorrect opinion that you don’t even know that you are talking nonsense. Instead, you seem to believe you’re an expert on the subject.

      You’re welcome to continue to parade your ignorance, but my advice would be to ponder the First Rule of Holes … it’s not doing your reputation any good.”

      Maybe you should take a break. Enjoy the holiday season.

      Extreme, it appears you have some objection to what I wrote. However, it is totally unclear what that objection might be. Yes, I get tired of people vociferously insisting that all the physics and thermo textbooks ever written are wrong … so might you if you had to deal with that kind of nonsense day after day.

      Ron is claiming that LW IR is not absorbed by land or water. That’s so bad it’s not even wrong, as the most minimum inquiry would establish … but nooo, he sticks to his guns and doubles down on dumb.

      I’m not willing to pretend that that is anything but parading his ignorance. I’m not willing to act like his opinion is anything but terminally misguided. Playing along with that nonsense just makes it multiply. I was serious when I told him he’s not doing his reputation any good, and in doing so, I’m trying to improve his reputation

      I’m trying to save a man from his own foolishness, and somehow I’m the bad guy?!? I was just repeating the old maxim:

      It is better to remain silent at the risk of being thought a fool, than to talk and remove all doubt of it.

      Finally, I’m enjoying the holiday season greatly … what gives you the impression I’m not?

      Best Christmas wishes to you and yours,

      w.

      • ==> Willis Eschenbach (December 23, 2017 at 7:26 pm)

        Willis,

        Firstly, I wish you and yours the very best for the Season and a very Happy New Year! ;-)

        Willis, I want to question this (Your prioritisation of Kirchhoff’s law), as I’ve tried to get a good answer from other sources (I’m sure there would be a simple and sensible answer, that you might offer. One that I’ve failed to find! ;-)

        When researching this several years ago, I came across a PHD physicist who retracted a paper because of this exact issue. He had conflated emission and absorption by prioritising Kirchhoff* over Planck**! He had made the mistake of assuming Kirchhoff’s law to be true for all frequencies and materials but this is not the case.

        I mention this only because it appears to be an often confused issue, even for the highly trained. I’m not even remotely in the category of these professionals or yourself and that is the point, are we – are you – sure that what you are saying is clearly understood, even by those whom we would expect to know these things?

        In my “unwashed” layman’s terms, snow is a white body for sunlight (SW) but a black body for heat (LW). And even in the “IR” spectrum as I understand it, there is some ill-defined overlap between SW and LW.

        Willis, in response to Ron Clutz (December 23, 2017 at 7:04 pm) you say:

        “Kirchoff’s Law says that absorptivity is equal to emissivity…”

        And yet you quote figures for IR emissivity and list fresh snow as 0.99. Yet Ron Clutz had spoken about sunlight*** and therefore, strictly speaking your figure differs from the actual value, which has a maximum of 0.30 absorptivity for sunlight (SW) – that he was presumably speaking of.

        The observation of a material-specific (Snow in this case) difference with regard to absorption and emission is not inconsistent with a good emitter simultaneously being a good absorber. The differences is that absorption and emission are dependent on the wavelength range.

        In the context provided above, what is it that you are specifically disagreeing with in Ron Klutz’s statement?

        cheers,

        Scott

        *Kirchhoff’s law – “Emission and Absorption are equal”
        **Planck’s law – “Emission is dependent upon the wavelength and the absolute temperature”
        ***”Radiation is only heat if the exposed material absorbs it. Solar radiation is high energy, absorbed by both land and sea and warmed by it. The bit of far longwave radiation from the cool atmosphere is not comparable.” – Ron Klutz

      • ==> Willis Eschenbach (December 23, 2017 at 7:26 pm)

        Willis,

        Firstly, I wish you and yours the very best for the Season and a very Happy New Year! ;-)

        Thanks, much appreciated.

        Willis, I want to question this (Your prioritisation of Kirchhoff’s law), as I’ve tried to get a good answer from other sources (I’m sure there would be a simple and sensible answer, that you might offer. One that I’ve failed to find! ;-)

        When researching this several years ago, I came across a PHD physicist who retracted a paper because of this exact issue. He had conflated emission and absorption by prioritising Kirchhoff* over Planck**! He had made the mistake of assuming Kirchhoff’s law to be true for all frequencies and materials but this is not the case.

        Sorry for the confusion, I didn’t state the full case. Kirchoffs Law says that emissivity = absorptivity at a given frequency. If it doesn’t absorb at that frequency, it doesn’t emit at that frequency, and vice versa.

        w.

  39. Willis, you wrote:

    The total energy absorbed by the surface is the sum of the net solar energy (surface downwelling solar minus surface reflections) plus the downwelling longwave infrared, or DWIR. This is the total energy that is absorbed by and actually heats the surface.

    This particular subject is apparently one around which you’ve got a real problem wrapping your head. The energy that “actually heats the surface” is the SOLAR heat flux. The SUN heats the surface, Willis. Not the atmosphere. The atmosphere is COOLER than the surface, so does not heat it. The Sun is WARMER than the surface, so does heat it. The DWLWIR is really just a mathematically derived “hemiflux”, its upward counterpart being the surface UWLWIR “hemiflux” (simply directly calculated from the average surface temperature). These two conceptual “hemifluxes” together make up the full RADIATIVE FLUX from the surface to the atmosphere and space. IOW, the DWLWIR is distinctly part of the surface radiative heat LOSS.

    Heat IN: Q_sw = 165 W/m^2
    Heat OUT: Q_lw + Q_cond + Q_evap = 53 + 24 + 88 = 165 W/m^2

    (…) a doubling of CO2, which will increase the DWIR by something on the order of 3.7 W/m2 (…)

    No, it is ASSUMED to do so. THEORETICALLY. All Else Being Equal.

    • Kristian December 24, 2017 at 5:44 am
      The DWLWIR is really just a mathematically derived “hemiflux”,

      Except when it’s measured.

      Here’s the measured downwards IR flux (in this case measured near the north pole).

      • Oh great, glorious and knowledgeable Kristian, please tell us what physical process the instrument undergoes in order for it to “calculate” it’s reading?

      • Rob Bradley, a description from Wiki.

        “A pyrgeometer is a device that measures near-surface infra-red radiation spectrum in the wavelength spectrum approximately from 4.5 µm to 100 µm.

        It measures the resistance/voltage changes in a material that is sensitive to the net energy transfer by radiation that occurs between itself and its surroundings (which can be either in or out). By also measuring its own temperature and making some assumptions about the nature of its surroundings it can infer a temperature of the local atmosphere with which it is exchanging radiation.

        These instruments generally have no spectral (frequency/wavelength) measurement capabilities – they use a single (non-frequency resolved) resistance/voltage measurement. They are constructed to be sensitive to the infra-red radiation spectrum that extends approximately from 4.5 µm to 100 µm, thus excluding the main shortwave (solar) spectrum.

        Since the mean free path of IR radiation in the atmosphere is ~25 meters, this device typically measures IR flux in the nearest 25 meter layer.”

        Note the last sentence, does it sound like IR Photons can get from the area of the Atmosphere where they are free of Water Molecules down to th Surface?

        Look up “the mean free path”.

      • Thanks AC, you have addressed my concern with Kristian saying: ” it isn’t measured, as in ‘physically detected’.”
        ….
        Something has to physically impinge on the instrument.

      • Kristian December 24, 2017 at 6:06 am
        Except it isn’t measured, as in ‘physically detected’. It’s calculated. By the instrument.

        It’s called ‘calibration’, I suppose you think that the temperature measured by Pt resistance thermometers or by Thermocouples is ‘calculated’ too.

      • A C Osborn December 24, 2017 at 8:18 am
        Since the mean free path of IR radiation in the atmosphere is ~25 meters, this device typically measures IR flux in the nearest 25 meter layer.”

        Note the last sentence, does it sound like IR Photons can get from the area of the Atmosphere where they are free of Water Molecules down to th Surface?

        Why exactly do you think this matters, we want to know the downwelling IR at the surface?

      • Rob Bradley December 24, 2017 at 8:45 am
        Thanks AC, you have addressed my concern with Kristian saying: ” it isn’t measured, as in ‘physically detected’.”
        ….
        Something has to physically impinge on the instrument.

        Yes it’s called Infrared light.

      • Rob Bradley says, December 24, 2017 at 8:45 am:

        Thanks AC, you have addressed my concern with Kristian saying: ” it isn’t measured, as in ‘physically detected’.”
        ….
        Something has to physically impinge on the instrument.

        Yes, the instrument physically detect the instantaneous EXCHANGE of radiative energy at the sensor surface. This is NOT the IR from the cool atmosphere to the warm sensor, much less the IR from the cool sensor to the warm SURFACE of the Earth. The instantaneous exchange produces a voltage which is translated into a flux value. This SHOULD be common knowledge.

      • Phil. says, December 24, 2017 at 8:48 am:

        It’s called ‘calibration’, I suppose you think that the temperature measured by Pt resistance thermometers or by Thermocouples is ‘calculated’ too.

        Time for a simple 101 lesson in how DWLWIR is “measured” in the everyday world for little Phil.

        https://en.wikipedia.org/wiki/Pyrgeometer#Measurement_of_long_wave_downward_radiation
        Measurement of long wave downward radiation

        The atmosphere and the pyrgeometer (in effect its sensor surface) exchange long wave IR radiation. This results in a net radiation balance according to:

        E_net = E_in – E_out

        Where:
        E_net – net radiation at sensor surface [W/m^2]
        E_in – Long-wave radiation received from the atmosphere [W/m^2]
        E_out – Long-wave radiation emitted by the sensor surface [W/m^2]

        The pyrgeometer’s thermopile detects the net radiation balance between the incoming and outgoing long wave radiation flux and converts it to a voltage according to the equation below.

        E_net = U_emf / S

        Where:
        E_net – net radiation at sensor surface [W/m^2]
        U_emf – thermopile output voltage [V]
        S – sensitivity/calibration factor of instrument [V/W/m^2]

        The value for S is determined during calibration of the instrument. The calibration is performed at the production factory with a reference instrument traceable to a regional calibration center.

        To derive the absolute downward long wave flux, the temperature of the pyrgeometer has to be taken into account. It is measured using a temperature sensor inside the instrument, near the cold junctions of the thermopile. The pyrgeometer is considered to approximate a black body. Due to this it emits long wave radiation according to:

        E_out = σT^4

        Where:
        E_out – Long-wave radiation emitted by the earth surface [W/m^2]
        σ – Stefan-Boltzmann constant [W/(m^2 K^4)]
        T – Absolute temperature of pyrgeometer detector [kelvins]

        From the calculations above the incoming long wave radiation can be derived. This is usually done by rearranging the equations above to yield the so-called pyrgeometer equation by Albrecht and Cox.

        E_in = U_emf / S + σT^4

        Where all the variables have the same meaning as before.

        As a result, the detected voltage and instrument temperature yield the total global long wave downward radiation.

        (My boldface.)

        Yes it’s called Infrared light.

        No. They’re called IR photons.

      • First: “it isn’t measured, as in ‘physically detected’.” ” Your statement is flat out wrong. The instrument measuring the DWLWIR is PHYSICALLY DETECTING THE IR PHOTONS STRIKING THE SENSOR.
        .
        .
        .
        .
        Next: “This is NOT the IR from the cool atmosphere to the warm sensor,’” ……Actually it is IR from the atmosphere. The IR photons are emitted by the atmosphere, and strike the sensor. The photons don’t know what “temperature” is……..

        PS, you can call it “infrared light” or you can call it IR photons, there is no difference between the two.

      • Rob Bradley said, December 24, 2017 at 3:54 pm:

        Your statement is flat out wrong. The instrument measuring the DWLWIR is PHYSICALLY DETECTING THE IR PHOTONS STRIKING THE SENSOR.

        Nope. It detects a radiative heat flux. The sensor in question is a so-called “thermal detector”. If the sensor is cooler than the air layers above, then the heat flux is INCOMING, which will induce a positive voltage signal, and if the sensor is warmer than the air layers above, the heat flux is OUTGOING, inducing a negative voltage signal. This is basic stuff that you can find out by simply reading what the manufacturers of these particular radiometric instruments are writing.

        Next: “This is NOT the IR from the cool atmosphere to the warm sensor,’” ……Actually it is IR from the atmosphere.

        Nope. A thermal detector warmer than the air layers above does NOT detect IR from the atmosphere. It detects the radiative flux moving OUT.

        PS, you can call it “infrared light” or you can call it IR photons, there is no difference between the two.

        Oh, yes there is. “Radiation” can be described MACROscopically, or it can be described MICROscopically. A radiative flux is MACROscopic “radiation”, while photons are MICROscopic “radiation”.

        Who said the sensor is “warm?”

        If we’re talking about a pyrgeometer, the sensor (a thermal detector) is “warm”, but not necessarily warmER than the air layers above. A “quantum detector” is a different creature altogether, and is normally cryogenically cooled.

      • Rob Bradley said, December 25, 2017 at 6:02 am:

        You assumed the sensor was thermal, and forgot the fact that there are other ways (photonic) to detect IR (…)

        *Sigh* No, Rob. What did I just write in my previous comment? “If we’re talking about a pyrgeometer, the sensor (a thermal detector) is “warm”, but not necessarily warmER than the air layers above. A “quantum detector” is a different creature altogether, and is normally cryogenically cooled.”

        PS, “radiation” can be described either macro or microscopically, so your distinction is moot.

        No, it’s all-important.

        Rob Bradley said, December 25, 2017 at 6:15 am:

        a flux of photons (microscopic) is “radiation” …

        No. A “photon” can be described as “radiation”, but only in a strictly MICROscopic sense. A “radiative flux”, on the other hand, is “radiation” in a strictly MACROscopic sense. It is the average movement (spatial direction and frequency) of ALL individual photons within some specified part of the universe. As a consequence, a radiative flux is – by definition – UNIdirectional. It moves spontaneously from hot to cold :)

      • Tsk, tsk, tsk Kristian, you moved the goal posts by bringing up the “pyrgeometer.” You first mentioned “instrument” with no qualification. When I questioned your post about “calculation” you then qualified the instrument to be a pyrgeometer. So you immediately discarded a photonic sensor which I then brought up. You see a photonic sensor in fact does physically detect incoming IR.

        Then you go on to say ” A “quantum detector” is a different creature altogether, and is normally cryogenically cooled.”

        Night vision goggles are not cryogenically cooled.

        Now, for the last take down, you post ” “radiation” in a strictly MACROscopic sense.”

        That is not true, for example cosmic rays are “radiation” and by definition are not sourced locally, are from an external source, and are not “averaged movement because they are individualized. A perfect example of this is the OMG particle: https://en.wikipedia.org/wiki/OMG_particle

        You state that a radiative flux is: ” by definition – Unidirectional. It moves spontaneously from hot to cold :)”

        ..

        Your problem is that the definition of radiative flux is ” is the amount of power radiated through a given area, in the form of photons or other elementary particles, typically measured in W/m2″ ( https://en.wikipedia.org/wiki/Radiative_flux ) There is no mention of “direction” and there is also no mention of “net” in the defintion. Your definition does not encompass the fact that objects that a cold object can radiate towards a warm object. If you think that a cold object cannot radiate towards a warm object, you are welcome to do this experiment. Take an LED flashlight, and put it in your freezer for several hours. After it becomes cold, remove it from the freezer, turn it on, and see if the light it emits will illuminate the floor at your feet. It will, and proves a “cold” object and “radiate” to a warm object, as the photons it emits will bounce off the “warm” floor and impinge upon your retina, activating the neural connections to your cortex.

      • PS Kristian, you don’t even need to cool down the LED flashlight. Try this. At night in your place of residence, turn on a light. Look at a place on the floor where the light illuminates it. Unless the floor where you live is above 98-99 degrees F, the photons you are observing have been emitted by an object cooler than 98-99F, absorbed by your retina which is notably warmer than your floor.

        See? Your eyeballs are observing radiative flux moving spontaneously from cold to hot.

      • Kristian December 24, 2017 at 3:25 pm
        Phil. says, December 24, 2017 at 8:48 am:

        “It’s called ‘calibration’, I suppose you think that the temperature measured by Pt resistance thermometers or by Thermocouples is ‘calculated’ too.”

        Time for a simple 101 lesson in how DWLWIR is “measured” in the everyday world for little Phil.

        You can cut the gratuitous insults (a sure sign you’re losing the argument).

        The analysis you quoted fro Wikipedia is correct and indicates that the unknown quantity is the LW radiation received from the atmosphere, all other parameters being eliminated by calibration.
        Just like measuring the temperature of a flame using a thermocouple, correct for (or design to be insignificant) the conduction and convection from the junction, use a radiation shield to minimize the radiation loss (the biggest error term) or correct for it. If you didn’t do those corrections if you measured a flame temperature at ~1100K your measurement would be about 100K too low.


        “Yes it’s called Infrared light.”

        No. They’re called IR photons.

        Your source refers to it as ‘IR radiation’, I don’t see the need to invoke ‘photon’ since we’re not analyzing the process on the basis of individual QM interactions.

      • Rob Bradley said, December 25, 2017 at 4:57 pm:

        Tsk, tsk, tsk Kristian, you moved the goal posts by bringing up the “pyrgeometer.” You first mentioned “instrument” with no qualification. When I questioned your post about “calculation” you then qualified the instrument to be a pyrgeometer. So you immediately discarded a photonic sensor (…)

        Hahaha! Sorry to remind you, Rob, but the world doesn’t revolve around you. You’re a latecomer to this party. It started long before you arrived. So you see, you don’t own it. And you don’t get to dictate what it’s about and what not.

        It all started out with pyrgeometers and what they allegedly measure. Pyrgeometers are, after all, the specific radiometric instruments being used daily all around the globe to “measure” the atmospheric “back radiation” to the surface. And the DWLWIR values provided by these instruments are distinctly COMPUTED internally, based on the physical signals they actually do detect: 1) The radiative heat flux at the sensor surface, and 2) the sensor temperature. This is all common knowledge, Rob. Undisputed and totally uncontroversial. So I’m afraid you’re just being a little child here …

        You see a photonic sensor in fact does physically detect incoming IR.

        Er, a quantum (‘photonic’) sensor physically detects photons, yes. I’ve never said otherwise. It seems you’re fighting a straw man, Rob ;)

        Then you go on to say ” A “quantum detector” is a different creature altogether, and is normally cryogenically cooled.”

        Night vision goggles are not cryogenically cooled.

        OMG! How uninformed can a person get …!? Seriously? Night vision goggles!!?

        Now, for the last take down, you post ” “radiation” in a strictly MACROscopic sense.”

        That is not true, for example cosmic rays are “radiation” and by definition are not sourced locally, are from an external source, and are not “averaged movement because they are individualized. A perfect example of this is the OMG particle (…)

        What does this have to do with anything!? If you’re detecting a photon (MICRO), you’re not detecting a radiative flux (MACRO).

        You state that a radiative flux is: ” by definition – Unidirectional. It moves spontaneously from hot to cold :)”

        Your problem is that the definition of radiative flux is ” is the amount of power radiated through a given area, in the form of photons or other elementary particles, typically measured in W/m2″ (…)

        There is no mention of “direction” and there is also no mention of “net” in the defintion.

        The “direction” gives itself. It’s where the AVERAGE (“net”) movement of radiative energy through the given area goes. You need to read up on this topic, Rob. You come off as a guy who’s just frantically looking up whatever’s on the table just in order to have something to say about it. Without any previous contextual understanding.

        Your definition does not encompass the fact (…) that a cold object can radiate towards a warm object. If you think that a cold object cannot radiate towards a warm object (…)

        I don’t. So leave it be. You need to distinguish between what happens at the MICRO level (quantum effects) and what’s happening at the MACRO level (thermodynamic effects). These two descriptional levels simply relate to different aspects of reality.

        Take an LED flashlight, and put it in your freezer for several hours. After it becomes cold, remove it from the freezer, turn it on, and see if the light it emits will illuminate the floor at your feet. It will, and proves a “cold” object [can] “radiate” to a warm object, as the photons it emits will bounce off the “warm” floor and impinge upon your retina, activating the neural connections to your cortex.

        Hahahaha!! You’re completely ignorant about this subject, aren’t you? LOL!

        Rob Bradley said, December 25, 2017 at 5:07 pm:

        PS Kristian, you don’t even need to cool down the LED flashlight. Try this. At night in your place of residence, turn on a light. Look at a place on the floor where the light illuminates it. Unless the floor where you live is above 98-99 degrees F, the photons you are observing have been emitted by an object cooler than 98-99F, absorbed by your retina which is notably warmer than your floor.

        Hahahahaha!

        See? Your eyeballs are observing radiative flux moving spontaneously from cold to hot.

        Er, no.

        I think we’re done here. You’re not an interesting person to discuss this particular topic with, since you obviously have no real understanding of what we’re in fact discussing … Bye.

      • Phil. said, December 26, 2017 at 8:39 am:

        You can cut the gratuitous insults (a sure sign you’re losing the argument).

        You don’t have an argument, Phil. You look at that plot of yours and seriously believe that is shows a measured DWLWIR value, when it is CLEARLY just a mathematically DERIVED one. This factual circumstance isn’t really a point up for debate. That’s just the way it is. Everything beyond that will be nothing but a bunch of waving of hands.

        The analysis you quoted fro Wikipedia is correct (…)

        So why are you still debating it!? It CLEARLY states that the DWLWIR isn’t something that the instrument actually DETECTS physically in any way. It COMPUTES it, based on 1) a two-way transfer assumption, 2) a BB sensor assumption, 3) the voltage signal given by the direct detection of a radiative heat flux at the sensor surface, and 4) the detected sensor temperature. There’s no direct measurement of DWLWIR here, Phil. Sorry I have to be the one breaking it to you. Look up “thermal detector”.

        “Yes it’s called Infrared light.”

        No. They’re called IR photons.

        Your source refers to it as ‘IR radiation’ (…)

        As I already pointed out in a previous comment: “Radiation” can be described MICROscopically (photons) and MACROscopically (radiative fluxes). It isn’t more complicated than that …

        (…) I don’t see the need to invoke ‘photon’ since we’re not analyzing the process on the basis of individual QM interactions.

        Well, if we’re analysing the process on a MACROscopic level (thermodynamic/thermal effects), there IS no spontaneous radiative flux from cold to hot. The radiative transfer is from hot to cold only. MACROscopically. Thermodynamically. Whenever you’re talking about “IR radiation” from cold “impinging” on hot, then you’re – by definition – talking about photons. Then you’re distinctly operating in the MICROscopic (quantum) realm. And there are no thermal (thermodynamic) effects in the quantum realm …

      • Kristian December 27, 2017 at 7:13 am
        Phil. said, December 26, 2017 at 8:39 am:

        You can cut the gratuitous insults (a sure sign you’re losing the argument).

        You don’t have an argument, Phil. You look at that plot of yours and seriously believe that is shows a measured DWLWIR value, when it is CLEARLY just a mathematically DERIVED one. This factual circumstance isn’t really a point up for debate. That’s just the way it is. Everything beyond that will be nothing but a bunch of waving of hands.

        The analysis you quoted fro Wikipedia is correct (…)

        So why are you still debating it!? It CLEARLY states that the DWLWIR isn’t something that the instrument actually DETECTS physically in any way. It COMPUTES it, based on 1) a two-way transfer assumption, 2) a BB sensor assumption, 3) the voltage signal given by the direct detection of a radiative heat flux at the sensor surface, and 4) the detected sensor temperature. There’s no direct measurement of DWLWIR here, Phil. Sorry I have to be the one breaking it to you. Look up “thermal detector”.

        Thanks for your apology for your insult. (sarcasm mode off)
        The response of the pyrgeometer obeys the following relationship:

        E_in = U_emf / S + σT^4

        Where:
        E_in – Long-wave radiation received from the atmosphere [W/m^2]
        U_emf – thermopile output voltage [V]
        S – sensitivity/calibration factor of instrument [V/W/m^2]
        σ – Stefan-Boltzmann constant [W/(m^2 K^4)]
        T – Absolute temperature of pyrgeometer detector [Kelvins]

        So U_emf is a measured quantity, S is a measured calibration factor, and T is a measured temperature.
        Which leaves E_in, if as you appear to believe that E_in isn’t Long-wave radiation from the atmosphere what do you propose is the alternate source of that energy flux?
        This is based on well established radiation heat transfer mechanisms which are successfully used everyday by engineers.
        We also know by independent methods that there is downwelling IR radiation, for example spectroscopic methods, thermal imaging etc.

        Well, if we’re analysing the process on a MACROscopic level (thermodynamic/thermal effects), there IS no spontaneous radiative flux from cold to hot. The radiative transfer is from hot to cold only.

        Not true, net flux is from hot to cold however there is radiative flux from cold to hot.
        For example measure flux from an object at 0ºC using a cryogenically cooled detector, get about 300W/m^2.
        Now place a target at 20ºC where the detector was, it’s still receiving 300W/m^2 of incoming radiative flux.

        Just as indicated by the equation from your source:
        E_net = E_in – E_out
        We can measure each of those quantities independently for such an instrument.

      • Phil. said, December 27, 2017 at 8:27 pm:

        Thanks for your apology for your insult. (sarcasm mode off)

        What insult? Drop the holier-than-thou attitude. You obviously do need an introductory course on this subject. I simply call it like I see it. Your confusion (ignorance?) still shines through in what you write.

        So U_emf is a measured quantity (…)

        Not really. U_emf is simply the particular voltage output signal resulting FROM 1) the prior calibration of the instrument, and 2) the direct input of the radiative heat flux at the sensor surface. The calibration is between the radiative heat flux at the sensor surface and the resulting voltage output signal. IOW, the actually detected physical phenomenon here is the radiative heat flux. Notice what the article says:

        “The pyrgeometer’s thermopile DETECTS the NET RADIATION BALANCE* between the incoming and outgoing long wave radiation flux and converts it to a voltage according to the equation below.

        *[that’s the “radiative heat flux”, Phil.]

        E_net = U_emf / S

        (…)

        The value for S is determined during calibration of the instrument.”

        Which leaves E_in, if as you appear to believe that E_in isn’t Long-wave radiation from the atmosphere what do you propose is the alternate source of that energy flux?

        There IS no such flux, Phil.!! That’s the whole point! It is PURELY an ASSUMED entity with a COMPUTED value! You still don’t get it!?

        The instrument detects the radiative heat flux at the sensor surface and the sensor temperature only. That’s it! Everything else is just based on the ASSUMPTION, and on the MATHEMATICAL FORMALISM, that the actually detected radiative flux is somehow made up of TWO internal, opposing macroscopic fluxes. And so, even if these hypothetical “hemifluxes” did in fact exist as discrete, independent macroscopic entities, you couldn’t possibly see or feel any of them separately, distinct from its counterpart. Within one and the same radiative heat transfer. You can only ever ASSUME their presence, and from this, CALCULATE their magnitude. How hard is this to comprehend!?

        This is based on well established radiation heat transfer mechanisms which are successfully used everyday by engineers.

        No. It is based on a well-established mathematical formalism that is successfully used every day by scientists and engineers alike – the two-way model of radiative transfer, the natural extension of the caloric “Theory of Exchanges”, proposed by Pierre Prevost back in 1791:
        http://www.oxfordreference.com/view/10.1093/oi/authority.20110803100344457

        However, this is NOT how radiation and radiative transfer REALLY behave macroscopically. Radiation and radiative transfer are most realistically described macroscopically today as statistical (probabilistic) phenomena:
        “The radiation field consists of a large number of particles or quanta distributed in space and time moving in various directions with differing energies. A description of such a system is necessarily statistical in nature and rests on the introduction of a six-dimensional phase space, the direct product of configuration space and momentum space constructed from the position and momentum co-ordinates.”
        https://www.science.mcmaster.ca/radgrad/images/775CourseResources/775_ch3.pdf

        Radiative flux:
        “The monochromatic flux F_v is the integral of the product I_v cos(θ) over all solid angles. This is the net flow of energy perpendicular to some surface dA whose normal makes an angle θ with respect to the observer. Note that flux is a vector. It depends on θ. Intensity is a scalar. We often separate the flux into two parts, an ingoing and an outgoing part. F = F^+ – F^-. (…)”
        http://www.astro.sunysb.edu/fwalter/AST341/idef.html

        Notice how the bidirectional approach is strictly a mathematical formalism, optionally employed ‘after the fact’, as a simplifying method of calculation only. The two “hemifluxes” are NOT themselves in any way a necessary (naturally emergent) “statistical” result …! They only “appear” after you have first made the conscious choice to divide the field into two directional hemispheres. You simply have two DIRECTIONS in which to look. And so you naturally incorporate this perspective into your MATHEMATICAL analysis of the transfer. But there is only ONE actual macroscopic movement of radiative energy through the radiation field – the radiative (heat) flux.

        We also know by independent methods that there is downwelling IR radiation, for example spectroscopic methods, thermal imaging etc.

        Nope. Still confused, I see. If you cool your detector to below the temperature of the air layers above it, what will happen? You set up a natural HEAT TRANSFER from those warmer air layers to your cooler detector. And THIS is what you’ll detect. Still. The colder you manage to make your detector, the “cleaner” the heat transfer will be. Ideally, in order for you to measure (detect) a macroscopic radiative flux (W/m^2) from the atmosphere to your instrument (that is, NOT to the surface!!) that matches the magnitude of the ones computed by the thermal detectors found in pyrgeometers, your detector’s temperature needs to be as close to absolute zero as possible. This is why total flux values given by radiometric instruments with quantum detectors are normally considerably lower than the equivalent values provided by pyrgeometers (with thermal detectors). The radiative heat flux they experience simply isn’t “clean” enough.

        This isn’t very complicated, Phil. That’s why I’m so surprised by your continued confusion on this topic.

        (…) net flux is from hot to cold however there is radiative flux from cold to hot.

        Again, no. There is no radiative FLUX, no MACROscopic movement of radiative energy, from cold to hot.

        You NEED to be able to distinguish between the MICRO and the MACRO aspects of reality in order to get this. Photons and radiative fluxes are NOT equal (or equivalent) physical phenomena. They’re NOT the same thing!

        For example measure flux from an object at 0ºC using a cryogenically cooled detector, get about 300W/m^2. Now place a target at 20ºC where the detector was, it’s still receiving 300W/m^2 of incoming radiative flux.

        Hehe, no, that’s exactly it. You won’t be able to measure the same flux if your detector gets warmer, Phil.

        E_net = E_in – E_out
        We can measure each of those quantities independently for such an instrument.

        Nope. You will HAVE to turn them both into radiative HEAT fluxes before you can ‘see’ them as separate macroscopic entities. Which you can only do if you cool your detector down towards absolute zero. But then you will STILL only observe a net movement of radiative energy. And that net movement STILL isn’t moving from the cool atmosphere down to the warm surface. That is something that simply isn’t happening. The (warmer) surface only LOSES radiative energy to the (cooler) atmosphere above. Your detected flux distinctly moves from the cool atmosphere to your much, much colder detector. IOW, you’re no longer looking at the same thing, the situation we started out discussing – the radiative transfer between the atmosphere and the surface.

        In conclusion, get the distinction between the MICRO and the MACRO descriptions of radiation and radiative transfer sorted out in your head, and you will hopefully start seeing what I’m talking about.

      • Phil, do you believe as Mr Eschenbach does that two identical cooling objects at the same temperature do not make each other warmer?

      • Kristian December 28, 2017 at 4:46 am
        Phil. said, December 27, 2017 at 8:27 pm:

        “Thanks for your apology for your insult. (sarcasm mode off)”

        What insult? Drop the holier-than-thou attitude. You obviously do need an introductory course on this subject. I simply call it like I see it. Your confusion (ignorance?) still shines through in what you write.

        “So U_emf is a measured quantity (…)”

        Not really. U_emf is simply the particular voltage output signal resulting FROM 1) the prior calibration of the instrument, and 2) the direct input of the radiative heat flux at the sensor surface. The calibration is between the radiative heat flux at the sensor surface and the resulting voltage output signal. IOW, the actually detected physical phenomenon here is the radiative heat flux. Notice what the article says:

        “The pyrgeometer’s thermopile DETECTS the NET RADIATION BALANCE* between the incoming and outgoing long wave radiation flux and converts it to a voltage according to the equation below.

        *[that’s the “radiative heat flux”, Phil.]

        As the article you quoted says it’s the net result of two fluxes: “the NET RADIATION BALANCE* between the incoming and outgoing long wave radiation flux” (emphasis mine)

      • Kristian December 28, 2017 at 4:46 am
        Nope. You will HAVE to turn them both into radiative HEAT fluxes before you can ‘see’ them as separate macroscopic entities. Which you can only do if you cool your detector down towards absolute zero. But then you will STILL only observe a net movement of radiative energy. And that net movement STILL isn’t moving from the cool atmosphere down to the warm surface. That is something that simply isn’t happening. The (warmer) surface only LOSES radiative energy to the (cooler) atmosphere above. Your detected flux distinctly moves from the cool atmosphere to your much, much colder detector. IOW, you’re no longer looking at the same thing, the situation we started out discussing – the radiative transfer between the atmosphere and the surface.

        OK little Kristian, you seem to be laboring under the delusion that an object only emits energy by radiation if it ‘sees’ a target that is colder than it is. In reality the object is always emitting the same quantity of radiation regardless of its surroundings, what counts is how much radiation it receives in turn from its surroundings.

        There IS no such flux, Phil.!! That’s the whole point! It is PURELY an ASSUMED entity with a COMPUTED value! You still don’t get it!?

        No there are in fact fluxes emanating from both the object and the surroundings depending on the temperature of each. I’m afraid it’s you who doesn’t get it.

      • Kristian, if you had to identical hot steel plates, say 1000 K each about 10 cm apart, would you stick your hand between them? Net flux is zero. Shouldn’t be uncomfortable, right?

      • So maybe you don’t want to stick your hand in there. How about we just suspend a small plate between those two white hot ones. Tell me what happens to that plate in your net 0 flow.

        BTW just to make it simple, put the whole mess in a vacuum.

      • Phil. said, December 29, 2017 at 11:51 am:

        As the article you quoted says it’s the net result of two fluxes: “the NET RADIATION BALANCE* between the incoming and outgoing long wave radiation flux” (emphasis mine)

        *Sigh*

        No. That’s the two-way ASSUMPTION, Phil. The preconceived idea that the observed one-way transfer is in fact just the difference between TWO opposite transfers, that there are really two fluxes making up the one. Even though you only ever observe the one.

        You will note how they are careful to point out that what the instrument actually DETECTS is very much a UNIdirectional transfer of energy, the radiative (heat) flux. But they still CALL this unidirectional transfer of energy “the net radiation balance between the incoming and outgoing long wave radiation flux”. Because they adhere to the two-way model of radiative transfer, and because they naturally employ the mathematical formalism of two conceptual “hemifluxes” (which are really just mathematically derived radiative expressions of each body’s temperature).

        IOW, you focus on what people SAY about physical observations, specifically on people’s theoretical interpretations. While I concentrate on the actual physical OBSERVATIONS.

        This whole discussion started out with me telling Willis that the radiometric instruments allegedly “measuring” the atmospheric “back radiation” to the surface, the pyrgeometers, are NOT actually ‘measuring’ (as in ‘physically detect’) this quantity at all. They CALCULATE it based on two theoretical assumptions plus the detection of two OTHER physical quantities – 1) the radiative heat flux at the sensor surface, and 2) the sensor temperature. You then came on trying to defend Willis’ silly ‘misunderstanding’ by posting a chart obviously showing plotted pyrgeometer data (you gave no source, though). So, either you’re being deceitful, or you’re ignorant about how these instruments actually produce their DWLWIR data …

        And now this discussion has been going on for a while, and you STILL have your heels digged in. Is it because you just don’t want to admit the reality of Willis’ (and your) initial misconception, or is it because you still don’t understand what this discussion is even about …!?

        Phil. said, December 29, 2017 at 12:08 pm:

        OK little Kristian, you seem to be laboring under the delusion that an object only emits energy by radiation if it ‘sees’ a target that is colder than it is.

        No, Phil. You’re not this stupid. So why are you deliberately ‘misreading’ (or not reading at all) what I’m writing? Why are you not addressing what I actually say, but rather your own, privately constructed straw man? Have you finally realised you’re losing the argument …?

        Micro (quantum) vs. macro (thermo), Phil. Photons vs. radiative fluxes. That’s what you need to sort out …

        In reality the object is always emitting the same quantity of radiation regardless of its surroundings, what counts is how much radiation it receives in turn from its surroundings.

        Yes, that’s the mathematical formalism, Phil. But it is not how MACROscopic reality actually works. It is not a realistic description of macroscopic reality. It is strictly a conceptual one. It is purely a simplifying model of reality. Based on the geometrically conceived concept of “view factors”, plus the quantum concept of photons. However, photons are not MACROscopic phenomena. They are distinctly MICROscopic phenomena. And so they only exist OUTSIDE the “thermodynamic limit”.

        In macroscopic reality, IN the “thermodynamic limit”, and thus governed by the Laws of Thermodynamics, there is just the UNIdirectional transfer of radiative energy between two radiating objects – the radiative (heat) flux. This is ALL we will ever observe macroscopically. No matter how much you twist and turn, you cannot escape this simple fact of life.

        There aren’t two separate macroscopic radiative fluxes (W/m^2) inside the ONE macroscopic transfer of radiative energy between the surface and the atmosphere, Phil. There are PHOTONS flying in ALL spatial directions, but the macroscopic transfer of energy between the two regions in question is only the net of ALL these photons, their individual paths and frequencies. You don’t statistically average all photons within the integrated radiation field and get TWO net movements. This only happens once you MENTALLY/CONCEPTUALLY/MATHEMATICALLY decide to divide the radiation field into two hypothetical hemispheres, pointing in two opposite general spatial directions.

        There are two spatial DIRECTIONS in which to look, Phil. Because there are two radiating regions ‘facing’ each other. But if you find yourself immersed in the radiation field ‘between’ the two, it doesn’t matter in which direction you look: up, down, back, forth, or to the sides – you will ALWAYS experience/detect photons coming in and going out. Why? Because you’re in the middle of a thermal photon cloud!! Where photons fly around everywhere! At all times. That’s what they do!

        Look, even a SINGLE object in space will constantly absorb its own emitted photons, both internally and at its surface. It’s a prerequisite for it to reach thermal equilibrium and for it to be able to emit a thermal flux of radiative energy to its surroundings according to its temperature (blackbody radiation). That doesn’t mean, Phil., that it thereby produces TWO opposing internal/external radiative FLUXES! Photons and fluxes are NOT the same thing! They are NOT directly relatable! They exist at fundamentally different descriptional levels of reality! It seems you labour under the delusion that they are somehow one and the same thing …! That if you have the one, then you automatically have the other. You don’t. If you have the direction and intensity (frequency) of ONE individual photon, or even a thousand, YOU DO NOT THEREBY KNOW IN WHAT DIRECTION THE RADIATIVE FLUX, OF WHICH THE PHOTONS ARE BUT A MICROSCOPIC PART, MOVES, Phil.!

        No there are in fact fluxes emanating from both the object and the surroundings depending on the temperature of each. I’m afraid it’s you who doesn’t get it.

        No. You’re confused. Photon vs. flux! Micro vs. macro. Read up! Get the distinction!

      • While you think about how that little plate reacts to 50,000 W/m² coming at it from both sides, answer this paradox for me…..

        The white hot plates have no net radiation flux so by your postulate those inside surfaces are not radiating, yet they are white hot which means they are giving off visible light.That,s still radiation right? Explain to me how that works. Shouldn’t they go dark in your theory?

      • Paul Bahlin said, December 29, 2017 at 1:50 pm:

        (…) if you had to identical hot steel plates, say 1000 K each about 10 cm apart, would you stick your hand between them? Net flux is zero. Shouldn’t be uncomfortable, right?

        That depends. Is your hand also at 1000 K? You see, there is no thermal flux of energy between two objects at the same temperature, but there IS one between an object at 1000 K and a human hand at, say, 300 K held 4-5 cm away from it.

      • So make it a small plate between two very large white hot plates. Put it in and describe what happens.

      • Pounding the table is a sure sign you can’t answer the question.

        It’s pretty simple really. Two white hot plates facing each other in a vacuum. According to Kristian there is only net radiation flow in the world so in that postulate you have no flow between the two plates, ergo no radiation, yet they are both emitting white light. Why?

        Answer it without childish name calling if you can.

      • Crickets!

        While you are avoiding answering my last question, here is another one to throw against the albedo = f(T) theory…..

        Two equally energetic bodies placed in a vacuum, with only net energy flow =0 between them, violates 2nd law of thermodynamics because it is decreasing entropy in an isolated system by creating asymetrical radiation fields.

        Of course my statement eminates from a feeble mind so you can ignore it if you like.

      • A.C. rants….

        “Sometimes I wonder if you are actually a human being or a BOT.
        Both plates are emitting the equivelent 51,000W/M2, so if there is a “Flux” where the hell is it going and why aren’t they both getting much hotter?”

        Read the thread for God’s sake. Kristian says there is ONLY NET FLUX, in this case zero. I claim net flux is zero too. But, and this critical to get into your head before you start regurgitating idiotic statements, My net is the result of two equal and opposite energy flows. He denies such a flow regime is possible.

        I say if there is no flow in his regime explain why they emit white light.

      • Sometimes I wonder if you are actually a human being or a BOT.
        Both plates are emitting the equivelent 51,000W/M2, so if there is a “Flux” where the hell is it going and why aren’t they both getting much hotter?
        Because 2 objects at the same Temperature DO NOT make each other hotter.
        So all this crap about Photons are Photons and when absorbed MUST raise the temperature of the absorbing body is just that, CRAP.
        It does not happen!
        Just as photons from a Cold Body DO NOT make hot bodies hotter.
        I am talking from experience and you are talking from bullshit Maths.

      • Paul Bahlin said, December 30, 2017 at 12:10 pm:

        The white hot plates have no net radiation flux so by your postulate those inside surfaces are not radiating (…)

        No, that’s by YOUR postulate. You’ve created a straw man, Paul.

        (…) yet they are white hot which means they are giving off visible light.That,s still radiation right? Explain to me how that works. Shouldn’t they go dark in your theory?

        They look white hot to YOU. Are your eyes at ‘white hot’ temps? Do you even understand how heat transfers work …!?

        Paul Bahlin said, December 30, 2017 at 9:07 am:

        So make it a small plate between two very large white hot plates. Put it in and describe what happens.

        What difference does it make? My hand, a small plate. Same thing. There will be spontaneous heat transfers from hot things to cold things. How hard is this?

        Paul Bahlin said, December 30, 2017 at 4:03 pm:

        According to Kristian there is only net radiation flow in the world (…)

        No, in the MACROSCOPIC world.

        (…) so in that postulate you have no flow between the two plates (…)

        That’s right. There is no flow – no flux, no net movement – of radiative energy.

        (…) ergo no radiation (…)

        No. Saying there’s no net movement of radiative energy is NOT equal to saying there is “no radiation”. There is still radiation, as in “photons flying around everywhere”. If you’ve got completely calm air, does that mean the air molecules aren’t still flying around at ferocious speeds in all directions? You need to distinguish between the MICRO and the MACRO aspects of reality, Paul.

      • Kristian says…

        “No. Saying there’s no net movement of radiative energy is NOT equal to saying there is “no radiation”. There is still radiation, as in “photons flying around everywhere”. If you’ve got completely calm air, does that mean the air molecules aren’t still flying around at ferocious speeds in all directions? You need to distinguish between the MICRO and the MACRO aspects of reality, Paul.”

        You can’t have it both ways. Your macro micro concept is a diversion. You have stated repeatedly that there is only one flow, cold to hot and that it is not a vector sum of opposing flows, just a unitary vector.

        Extend that thinking to net=0 and there is no flow from either plate. You can’t suddenly turn it on. Take two white hot plates at nearly equal temps. You say there is one vector, hot to cold. Yet it would have to be very small, nearly zero. Still, the hotter plate is white hot. Explain how you get a white hot surface emitting, say 50 W/m² in your photon fog.

        Now as the plates reach equilibrium that hot to cold vector gets smaller and smaller, approaching 0, right? Then at equilibrium it becomes a 50,000 W/m² monster facing down an equal one from the other side, where an instant ago there was nothing.

        BTW, don’t try to answer by telling me I am stupid. Makes good snark but it mostly shows up as pounding the table for lack of a cogent answer.

      • Kristian December 30, 2017 at 8:52 am
        Phil. said, December 29, 2017 at 11:51 am:

        “As the article you quoted says it’s the net result of two fluxes: “the NET RADIATION BALANCE* between the incoming and outgoing long wave radiation flux” (emphasis mine)”

        *Sigh*

        No. That’s the two-way ASSUMPTION, Phil. The preconceived idea that the observed one-way transfer is in fact just the difference between TWO opposite transfers, that there are really two fluxes making up the one. Even though you only ever observe the one.

        It’s not an assumption, there are two fluxes in opposite directions which can be physically separated.

        This whole discussion started out with me telling Willis that the radiometric instruments allegedly “measuring” the atmospheric “back radiation” to the surface, the pyrgeometers, are NOT actually ‘measuring’ (as in ‘physically detect’) this quantity at all. They CALCULATE it based on two theoretical assumptions plus the detection of two OTHER physical quantities – 1) the radiative heat flux at the sensor surface, and 2) the sensor temperature. You then came on trying to defend Willis’ silly ‘misunderstanding’ by posting a chart obviously showing plotted pyrgeometer data (you gave no source, though).

        I didn’t think it was necessary since if you click on the image it shows the source on your computer.

        There aren’t two separate macroscopic radiative fluxes (W/m^2) inside the ONE macroscopic transfer of radiative energy between the surface and the atmosphere, Phil. There are PHOTONS flying in ALL spatial directions, but the macroscopic transfer of energy between the two regions in question is only the net of ALL these photons, their individual paths and frequencies. You don’t statistically average all photons within the integrated radiation field and get TWO net movements. This only happens once you MENTALLY/CONCEPTUALLY/MATHEMATICALLY decide to divide the radiation field into two hypothetical hemispheres, pointing in two opposite general spatial directions.

        So take an incandescent light bulb and place a 45º red dichroic mirror in its light path you’ll see red light reflected to the side and the rest of the spectrum carries on undeviated. Place another bulb pointing in the opposite direction and you’ll see another red beam reflected to the other side. Clearly there are two physically separable beams of light traveling in both directions.

      • Mr Bahlin, you are a hypocrite for calling out other people qyestioning YOUR eintelligence and credibility.
        You have done so with practically evry one of your I am so superior condescending remarks.
        I should know as I have been on the end of a lot of them.

        You seem to be avoiding Kristian’s last remark “No. Saying there’s no net movement of radiative energy is NOT equal to saying there is “no radiation”. There is still radiation, as in “photons flying around everywhere””
        I see that you are now admitting that there is NO HEATING between objects of equal temperature and yet when I conducted that very experiment you called it names, told me I was wasting my time and suggested the only thing to do was drink the beer.
        What has changed your mind?

      • A C Osborn December 31, 2017 at 2:42 am
        Sometimes I wonder if you are actually a human being or a BOT.
        Both plates are emitting the equivelent 51,000W/M2, so if there is a “Flux” where the hell is it going and why aren’t they both getting much hotter?
        Because 2 objects at the same Temperature DO NOT make each other hotter.
        So all this crap about Photons are Photons and when absorbed MUST raise the temperature of the absorbing body is just that, CRAP.
        It does not happen!
        Just as photons from a Cold Body DO NOT make hot bodies hotter.
        I am talking from experience and you are talking from bullshit Maths.

        Let’s clearly define the situation:
        A steel plate heated to white heat (~1500K) in a flame, background at room temperature (~300K)
        Add a similar plate nearby and parallel to the original in the flame, when they equilibrate they will both be at a higher temperature than the original one was.

      • Kristian December 31, 2017 at 3:40 am
        No. Saying there’s no net movement of radiative energy is NOT equal to saying there is “no radiation”. There is still radiation, as in “photons flying around everywhere”. If you’ve got completely calm air, does that mean the air molecules aren’t still flying around at ferocious speeds in all directions? You need to distinguish between the MICRO and the MACRO aspects of reality, Paul.

        Except unlike molecules, photons are not continually hitting each other and bouncing off in different directions. For the most part they continue traveling in the direction they were originally traveling in, regardless of any others heading in the opposite direction.

      • A C Osborn December 31, 2017 at 10:03 am
        You really haven’t got a clue.

        You are the clueless one, the concept is extremely simple.
        The single plate achieves a temperature below the temperature of the flame because of losses to its surroundings. The temperature reached is a result of the balance between incoming heat via conduction/convection and loss of heat to its surroundings. When a parallel plate is placed near it the area for heat transfer by conduction/convection doubles, however the area available to the surroundings does not double because of the overlap between the plates and the interchange of radiation between them. Consequently the temperature of the two plates is higher (and nearer to the flame temperature) than the single plate. QED

      • Paul Bahlin said, December 31, 2017 at 4:14 am:

        You can’t have it both ways. Your macro micro concept is a diversion. You have stated repeatedly that there is only one flow, cold to hot and that it is not a vector sum of opposing flows, just a unitary vector.

        Yes. MACROscopically.

        Extend that thinking to net=0 and there is no flow from either plate.

        Correct.

        You can’t suddenly turn it on.

        Yup. As soon as there’s a temp difference, you “turn it on”. It’s called “heat transfer”, Paul. Maybe you should consider reading a book.

        Take two white hot plates at nearly equal temps. You say there is one vector, hot to cold. Yet it would have to be very small, nearly zero.

        Correct. It all depends on the temperature difference.

        Still, the hotter plate is white hot.

        No. It’s just “hot”. That is, it has a high temperature, a high content of internal energy. It just LOOKS “white hot” to YOU.

        Explain how you get a white hot surface emitting, say 50 W/m² in your photon fog.

        MICRO vs. MACRO, Paul. Photon vs. radiative flux.

        Now as the plates reach equilibrium that hot to cold vector gets smaller and smaller, approaching 0, right?

        Yes.

        Then at equilibrium it becomes a 50,000 W/m² monster facing down an equal one from the other side, where an instant ago there was nothing.

        Huh? No. Equal temps, no net movement of thermal radiative energy, no “flow”.

        Again, I have to ask you, Paul:
        Do you understand the distinction between bulk air and air molecules? Between the MOVEMENT of the bulk air and the MOVEMENT of its air molecules? Do you see how we’re in fact addressing two fundamentally different aspects (descriptive levels) of reality here? MACRO vs. MICRO.

      • So white hot objects only look like they are hot to my eyes. HaHaHaHaHaHaHaHaHaHaHaHa

        You really believe this crap?

      • If you put two white hot plates facing each other in a vacuum There are only three ways to get no radiant exchange in the gap.

        1. neither radiates anything to the gap. Which results in objects with one surface at something like 1000K and the other at 0K. Laughable.

        2. They radiate opposing equal energy flows to the gap. which results in a system that is completely in balance.

        3. They radiate equally but the energy being poured into the gap (around 100kW/m²) just makes photons wizz around (your words) in a gas like air particles and neither interior surface is affected in any way by the wizzing. The white hot surfaces don’t really radiate they just look white to my eyes and it’s all due to the micro-macro thingy which allows you to pick scale values as needed to explain away the fact that you can not provide an accounting for your fantasy at either scale.

        I will leave it more astute readers than either of us to decide who needs to read a book. Me, I’m just laughing.

      • Phil. said, December 31, 2017 at 5:54 am:

        It’s not an assumption (…)

        Yes, it is.

        (…) there are two fluxes in opposite directions which can be physically separated.

        No, there aren’t.

        How long are we going to let this charade go on, Phil. You have nothing at this point, just arguments by assertion.

        I didn’t think it was necessary since if you click on the image it shows the source on your computer.

        I’m talking about the instrument being used. Not readily gathered from the link alone.

        So take an incandescent light bulb and place a 45º red dichroic mirror in its light path you’ll see red light reflected to the side and the rest of the spectrum carries on undeviated. Place another bulb pointing in the opposite direction and you’ll see another red beam reflected to the other side. Clearly there are two physically separable beams of light traveling in both directions.

        Phil., it seems you are still under the delusion that what I’m saying is that objects somehow aren’t RADIATING once they’re faced with other objects.

        Once more,

        Macroscopic reality doesn’t look like this:

        It looks like this:

        That final one is what statistical mechanics teaches us – ALL the microscopic movements/actions average out to ONE bulk movement/property. Not two. One.

        You get two only when YOU – in your mind – decide to distinguish between two general DIRECTIONS in space …

        A “radiative flux” is ALL (!!!) photons within the thermal radiation field at once. Their AVERAGE intensity and their AVERAGE direction. Not the average of just the ones that YOU have chosen to include!

        MICRO vs. MACRO, Phil.

      • Kristian, there is no need to make the distinction between “macro” and “micro” viewpoints. All radiative transfer phenomena can be described from the “micro” viewpoint.

      • Phil. said, December 31, 2017 at 10:08 am:

        “If you’ve got completely calm air, does that mean the air molecules aren’t still flying around at ferocious speeds in all directions? You need to distinguish between the MICRO and the MACRO aspects of reality, Paul.”

        Except unlike molecules, photons are not continually hitting each other and bouncing off in different directions. For the most part they continue traveling in the direction they were originally traveling in, regardless of any others heading in the opposite direction.

        *Facepalm*

        Always this “counterargument”. And just as irrelevant as always.

        I’m not describing what I’m describing to Paul to try and convince him that molecules really behave like photons. I’m doing it to make him understand how the MICRO and the MACRO realms are NOT directly relatable aspects of reality. In order to get this message across, it matters not one bit whether the microscopic constituents of the bulk entity we’re discussing crash into each other or not.

        My point is only to show how dividing a thermal radiation field and its NET movement of radiative energy into TWO opposite “fluxes” is just as arbitrary, just as much a mental (mathematical) construct, and nothing else, as deciding to define a volume of air as the net of TWO opposing “winds”, simply because you COULD say that, at any one time, about half the air molecules making up the air is moving in ONE general direction (within one “hemisphere”), and the other half in the exact opposite general direction (the “other” hemisphere). Yes, you could say that, in the next instant, the molecules moving within the one hemisphere will no longer be the same as the ones doing it in the previous instant, because they’ve all changed their direction from collisions in the meantime. But the total number would still be the same. It, after all, AVERAGES OUT. And so the “wind arrow” would basically be unchanged. (Photons inside a thermal photon gas/cloud are also not conserved over time. And they DON’T move in just TWO spatial directions.)

        Here’s the equation for convective heat transfer (look familiar?): Q = hA (T_h – T_c)

        MICRO vs. MACRO, Phil.

        A cooler atmosphere is not REALLY sending a radiative FLUX (W/m^2) down to a warmer surface, Phil. This is totally misunderstood. Only mathematically so. Not physically. It WOULD if the surface were cooler than it. And the average magnitude of this flux WOULD’VE been ~345 W/m^2 if – and only if – the surface were at absolute zero. However, it isn’t … It’s a POTENTIAL flux.

      • Kristian said:

        My point is only to show how dividing a thermal radiation field and its NET movement of radiative energy into TWO opposite “fluxes” is just as arbitrary, just as much a mental (mathematical) construct, and nothing else, as deciding to define a volume of air as the net of TWO opposing “winds”, simply because you COULD say that, at any one time, about half the air molecules making up the air is moving in ONE general direction (within one “hemisphere”), and the other half in the exact opposite general direction (the “other” hemisphere).

        Arbitrary? Not in the least. There are actually two different flows of energy.

        We can demonstrate this in a variety of ways. For starters, you can use night vision glasses to look at the two objects. The fact that you can see them both indicates that both are individually radiating energy.

        But no, this isn’t like the wind at all. With the wind, you CAN’T have two opposing flows because the two flows physically interfere by bouncing off each other. It’s like having two opposing flows of automobiles—it can’t happen, they’d crash into each other and the flow in either direction is stopped.

        But this is NOT true for photons. The two opposing flows of photons pass right through each other and don’t interfere.

        Finally, the equation for radiative heat transfer is simple. It is:

        Q = ( σ ε T_h ^ 4 ) – ( σ ε T_c ^ 4 )

        where sigma is the Stefan-Boltzmann constant and epsilon is the emissivity.

        Those two terms on the right are the two opposing actual physical flows that Phil and I are trying to get you to notice. One is the flow from the hot object, the other is the flow from the cold object. Both of those are very real flows—we can and do measure them individually.

        The NET flow, of course, is the larger minus the smaller, and always flows from hotter to colder. However, the fact that we can subtract one from the other doesn’t make them imaginary or not physical. Those flows are given by the Stefan-Boltzmann equation, and they are both assuredly real.

        Seriously, amigo, you desperately need to read a thermo book. You are embarrassing yourself here, people are gonna start pointing and laughing.

        w.

      • Paul Bahlin said, January 2, 2018 at 9:12 am:

        So white hot objects only look like they are hot to my eyes.

        No. Read what I write. Hot objects are objectively hot. What your eyes see, however, is specific to you eyes. Do you seriously believe otherwise …?

      • Kristian, so, according to you, if I used a sensor that is cooled to near absolute zero, I could measure the radiative flux (W/m^2) emitted downward from the “cold” atmosphere. Now, if I then raise the temperature of this sensor to above the temperature of the “cold” atmosphere, this radiative flux vanishes.

        How do the molecules of the “cold” atmosphere realize the sensor is no longer cold and stop emitting? You got some pretty clever little photons coming from the upper atmosphere Kristian.

      • Paul Bahlin said, January 2, 2018 at 9:37 am:

        If you put two white hot plates facing each other in a vacuum There are only three ways to get no radiant exchange in the gap.

        1. neither radiates anything to the gap. Which results in objects with one surface at something like 1000K and the other at 0K. Laughable.

        2. They radiate opposing equal energy flows to the gap. which results in a system that is completely in balance.

        3. They radiate equally but the energy being poured into the gap (around 100kW/m²) just makes photons wizz around (your words) in a gas like air particles and neither interior surface is affected in any way by the wizzing. The white hot surfaces don’t really radiate they just look white to my eyes and it’s all due to the micro-macro thingy which allows you to pick scale values as needed to explain away the fact that you can not provide an accounting for your fantasy at either scale.

        *Sigh*
        Listen to me, dimwit: MICRO vs. MACRO. Quantum vs. thermo. Photon vs. radiative flux. Learn the distinction!

        “Micro-macro thingy”! Hahahahaha!!

      • Seems to me you have a theory that you can feed enormous energy into even a tiny gap and it doesn’t do anything at all. You are using your scale fantasy to concoct a quantum effect that has no rational result in the physical world.

        Explain where all that energy goes, with something more substantial than handwaving. Every time yiu call me a name, you prove that you can’t.

      • Ralph Dave Westfall said, January 2, 2018 at 9:37 am:

        Kristian, there is no need to make the distinction between “macro” and “micro” viewpoints. All radiative transfer phenomena can be described from the “micro” viewpoint.

        Heard of “statistical mechanics”, Ralph?

      • Ralph Dave Westfall said, January 2, 2018 at 10:08 am:

        Kristian, so, according to you, if I used a sensor that is cooled to near absolute zero, I could measure the radiative flux (W/m^2) emitted downward from the “cold” atmosphere.

        No, you would measure A radiative flux from the cool atmosphere to your much, much colder detector.

        Now, if I then raise the temperature of this sensor to above the temperature of the “cold” atmosphere, this radiative flux vanishes.

        Yup. Or, rather, it now moves the other way. It’s called (radiative) heat transfer.

        How do the molecules of the “cold” atmosphere realize the sensor is no longer cold and stop emitting?

        They haven’t stopped emitting. MICRO vs. MACRO. Quantum vs. thermo. Photon vs. radiative flux.

        I’m only talking about the radiative flux (W/m^2), Ralph. Not the photons. Just like, when I’m talking about wind, I’m talking about BULK air masses, not individual air molecules.

      • Using probability and statistics to describe the behavior of quantum objects is perfectly acceptable Kristian. Using these tools does not change the viewpoint from “micro” to “macro.” The average kinetic energy of a set of helium atoms is perfectly acceptable from the “micro” viewpoint.

      • Kristian January 2, 2018 at 9:08 am
        Phil. said, December 31, 2017 at 5:54 am:

        “It’s not an assumption (…)”

        Yes, it is.

        “(…) there are two fluxes in opposite directions which can be physically separated.”

        No, there aren’t.

        There certainly are, there are two physically separable sources and the flux from each of them can be independently measured.

        How long are we going to let this charade go on, Phil. You have nothing at this point, just arguments by assertion.

        I didn’t think it was necessary since if you click on the image it shows the source on your computer.

        I’m talking about the instrument being used. Not readily gathered from the link alone.

        It was included at the site not really that hard to find:
        “PIR The PIR is an Eppley radiometer that measures incoming radiation with wavelengths between 3.5 and 50 µm. This instrument is also called a pyrgeometer. Units are Watts/m**2”

        So take an incandescent light bulb and place a 45º red dichroic mirror in its light path you’ll see red light reflected to the side and the rest of the spectrum carries on undeviated. Place another bulb pointing in the opposite direction and you’ll see another red beam reflected to the other side. Clearly there are two physically separable beams of light traveling in both directions.

        Phil., it seems you are still under the delusion that what I’m saying is that objects somehow aren’t RADIATING once they’re faced with other objects.

        You’re constantly saying that there’s only one macroscopic flux which is not separable, if that were so then this experiment would not be possible. If they are independently radiating dependent on their temperature then there is no problem using a pyrgeometer to measure the downward flux

        You get two only when YOU – in your mind – decide to distinguish between two general DIRECTIONS in space …

        A “radiative flux” is ALL (!!!) photons within the thermal radiation field at once. Their AVERAGE intensity and their AVERAGE direction. Not the average of just the ones that YOU have chosen to include!

        There are two separate sources, that gives us two separate radiation fluxes, which can be used to determine the net flux, or you can measure the net flux and one of the two and so determine the other.

        MICRO vs. MACRO, Phil.

        Nonsense, Kristian

  40. Kristian – At some stage, the penny should drop. Just not holding my breath. Never mind, it is Christmas Day here in New Zealand. God Bless and keep all of us as we seek the truth…..

  41. “According to the CERES satellite data, globally, the solar energy absorbed by the surface averages 162 W/m2.”

    How much of that 162 Watts is conducted and convected to the air in contact with the ground? Half?

    • Crispin in Waterloo December 24, 2017 at 12:00 pm Edit

      “According to the CERES satellite data, globally, the solar energy absorbed by the surface averages 162 W/m2.”

      How much of that 162 Watts is conducted and convected to the air in contact with the ground? Half?

      Trenberth puts the number at about 20 W/m2 or so. From their paper Earth’s global energy budget
      Kevin E. Trenberth, John T. Fasullo and Jeffrey Kiehl, 2008
      , (SH = sensible heat):

      The SH is available from the reanalyses for all years, and in W m-2 ranges from 15.7 to 18.9
      globally, 26.3 to 27.5 over land and 11.8 to 16.0 over the ocean. The value in KT97 was
      computed as a residual and was unrealistically high at 24 W m-2. Here we adopt values of 17, 27
      and 12 W m-2 for the globe, land and ocean, and even with uncertainties of 10%, the errors are
      only order 2 W m-2.

      w.

  42. A. C., you keep asking for experiments showing back radiation. You’ve been pointed to this one more than once, but no reply … so I’ll point to it again:

    Slaying the Slayers

    How about you repeat that very experiment and tell us what you find out before you start going on about some vague description of something you did?

    w.

    • I am a sceptic, and thus I am sceptical of the position adopted by the group to whom you refer. But you might like to ponder on the point that I raise below regarding the extremely small volume of water that DWLWIR is absorbed in. The contrast with solar irradiance could not be more stark.

      See richard verney December 25, 2017 at 1:28 am

    • Well I can see your absolute need to protect Mr Eshenbach’s reputation, but if some words offended they could have been snipped. So I can see exactly what you are doing
      I will just write and post it again and it will still destroy his reputation.
      So thanks mod and happy Christmas to you too.

  43. Actually on second thoughts the boiler in the warehouse will get warmer as the water in the rads get’s hotter, but that doesn’t have anything to do with back radiation either.

  44. One other experiment.
    The glass mirror obviously did not work.
    How about a sheet of Ali foil, shiny side towards the heated object.
    Did the object get warmer, yes.
    Was it direct back radiation, NO.
    Proof.
    Remove the foil, add a slow speed fan to circulate the air.
    Allow the object to reach it’s new EQUILIBRIUM and then add back the foil.
    How dear, no heating whatsoever.
    No magic back radiation, no reflected radiation.
    Zilch, nada, nothing.
    Exactly the same as the box, it heats the cooler air which changes the ambient conditions around the object.
    Job done another myth exploded.
    How many more can you take one asks?

    • A C Osborn December 24, 2017 at 2:34 pm
      One other experiment.

      Unfortunately you’re not very good at experiments, and your explanations of your procedures are terrible, you’d fail any freshman lab class.

      The glass mirror obviously did not work.
      How about a sheet of Ali foil, shiny side towards the heated object.
      Did the object get warmer, yes.
      Was it direct back radiation, NO.

      Despite the fact that Al foil reflects ~95% of the IR incident on it.
      You also claim to have measured the temperature of the foil, what method did you use and how did you do that without any radiation influencing your measurement? You also claimed to have measured the air temperature between the foil and the heated object, again what method and how did you avoid the effect of the radiation?

      Proof.
      Remove the foil, add a slow speed fan to circulate the air.
      Allow the object to reach it’s new EQUILIBRIUM and then add back the foil.
      How dear, no heating whatsoever.
      No magic back radiation, no reflected radiation.
      Zilch, nada, nothing.
      Exactly the same as the box, it heats the cooler air which changes the ambient conditions around the object.
      Job done another myth exploded.

      You added cooling by forced convection, it’s a totally different experiment.
      A more reasonable approach would have been to replace the foil with a non-reflective material, say black cartridge paper. Which if you’re right would increase the temperature of the object, whereas if it’s the result of reflection the temperature would be lower than in the presence of Al foil. In that case only one parameter is changed and all others kept the same a good experimental design philosophy.

      It’s typical of your approach to experiments, when one is suggested you change it out of all recognition.
      For example I suggested an experiment which has been carried out for at least 70 years: measure the temperature of a flame with a thermocouple, then surround it with an open ended quartz tube to act as a radiation shield. Usually results in 100+ºC increase in temperature. You proposed measuring the temperature of a 34ºC object and putting a glass jar over it, you then said that there wasn’t a 100ºC change!
      What a surprise!

  45. Willis,

    I am going to try to explain this to you one more time.

    The Sun is Hot. The Sky is Not Hot.

    “Climate Scientists” claim Back Radiation.

    When a pyrgeometer is pointed at the sky, it measures a Temperature. Not, importantly, a Flux, but simply a Temperature.

    Flux means Heat Transfer, as in, something cold gets hotter, and something warm gets colder.

    The Sun heats the Earth, and also the Earth’s atmosphere.

    The Sky cannot heat itself, nor the surface of the Earth. All the energy comes from the Sun.

    CO2 absorbs IR from the surface. It absorbs all it can at about 3 meters from the surface, all of which energy is Thermalized immediately, which means the CO2 molecule gains velocity and bounces off neighboring molecules, transferring energy to them in Brownian Motion.

    A far different thing happens at the TOA. CO2 at the TOA is opaque to IR, absorbs it, and re-radiates it down, sideways, and up. The down part gets thermalized at some lower altitude. Sideways and up can escape to space.

    The big question is, at just what altitude does the atmosphere cease to be opaque to CO2? Increasing CO2 raises that altitude slightly, which means the Earth’s atmosphere is freely radiating to space at a slightly higher altitude, which means a slightly lower temperature, which means Earth’s atmosphere contains some more energy, which means temperatures at the surface might be slightly higher due to the lapse rate.

    Hope you followed that, and tell all your friends, as, if this is beyond comprehension, you just should not talk about things you do not understand.

    Trying to help.

    Michael

    • Michael Moon December 24, 2017 at 8:25 pm

      CO2 absorbs IR from the surface. It absorbs all it can at about 3 meters from the surface, all of which energy is Thermalized immediately, which means the CO2 molecule gains velocity and bounces off neighboring molecules, transferring energy to them in Brownian Motion.

      All the energy is not thermalized immediately some of it is emitted. Also the CO2 molecule does not gain velocity, it vibrates at a different frequency, collision with an air molecule causes an increase in the KE of the air molecule and a decrease in the vibration of the CO2.

    • Michael Moon said “When a pyrgeometer is pointed at the sky, it measures a Temperature. Not, importantly, a Flux, but simply a Temperature.”

      I disagree. The only time you can define and measure a temperature is when a system is at equilibrium. At equilibrium the radiation field and the kinetic fields are consistent and not changing with time, which almost never happens in the atmosphere.

      In using a pyrogeometer, radiation from someplace is incident on a black surface in which is (typically) embedded a thermocouple. You actually measure voltage arising from heating or cooling of the thermocouple referenced to (or in) the body of the device. Based on comparison with black body sources you interpret the voltage as being consistent with radiation from a body of a certain temperature. If the radiation is not thermal, say from a laser, all you know for sure is the heat intensity, in say, watts/m^2, but not the temperature of the source. A thermocouple based device is not particularly sensitive but it has a broad spectral response so can be used for both thermal and athermal sources.

      Been there, done that.

      Merry Christmas to all.

  46. According to the CERES satellite data, globally, the solar energy absorbed by the surface averages 162 W/m2. The downwelling longwave averages 345 W/m2. Conveniently, this means that on average the earth’s surface absorbs about a half a kilowatt per square meter on an ongoing basis. (And no, I have no interest in debating whether downwelling longwave radiation actually exists. It’s been measured by scientists around the world for decades,

    I have never seen anyone argue that DWLWIR does not exist. The issue is what does it do?

    These figures should raise questions because of the absorption of of EMR in water. The depth to which EMR is absorbed is wavelength dependent.

    Because of the wavelength of Solar Irradiance, virtually no solar insolation is absorbed in the top millimeters of the oceans. In practice, for the main part it is absorbed in a volume of water occupying about 10 metres in depth, say between 1 metre to 11 metres. This means that the energy of some 162W/m2 of incoming solar is absorbed in a volume of 10 cubic metres (ie., 1m x 1m x 10m).

    By way of contrast, due to the wavelength of DWLWIR, some 90% of it is fully absorbed in about 10 microns of vertical penetrative depth. Because DWLWIR is omnidirectional, such that some is intercepting with the oceans at 10 degrees, some at 20 degrees, some at 30 degrees, this means that almost all DWLWIR is fully absorbed in no more than 5 microns. This means that the energy of some 345W/m2 of incoming DWLWIR is absorbed in a volume of 0.000005 cubic metres (ie., 1m x 1m x 5/1000,000m).

    So if K&T energy budget cartoon is correct, one has twice as much energy absorbed in 2 millionths of the volume. If that does not raise eyebrows, it is difficult to know what will. This begs the question, what exactly is all this DWLWIR doing in such a small volume of water?

    Fortunately, for life on this planet, solar irradiance, because it is absorbed over such a large volume, slowly hears the oceans. If solar irradiance was absorbed in the same way as DWLWIR, then the oceans would have boiled off, from the top down, long ago.

    • Richard, as you say “I have never seen anyone argue that DWLWIR does not exist. The issue is what does it do?”

      In that context, where Heat Transfer does not occur from cold to hot, how much of the Atmosphere is actually warmer than the Surface Water?
      Add to that the Heat transfer is Higher to a colder body than a warmer body, so more heat is transferred up than down
      The other key part of the whole concept, especially regarding CO2 is the one I mentioned yesterday “the mean free path of IR radiation in the atmosphere is ~25 meters”.
      This suggests the the IR from the CO2 only area (no H2O) of the Atmosphere has absolutely no chance of ever getting back to the surface, as every 25 or so metres (probably longer in less dense parts of the atmosphere) the Photon will have lost it’s energy to another molecule and it’s “downwardness” halved at each collision.
      So it can only warm the Atmosphere immediately below it, that is assuming that you believe a cold area can warm a warmer area. Which I do not.

    • Richard poses this fair and thought provoking question…

      “This begs the question, what exactly is all this DWLWIR doing in such a small volume of water?”

      Here is my take. It got me to thinking about the same dilemma on land with insolation. It’s the same question right. What is all that SW doing in such a small volume, because it doesn’t penetrate any more than LW in water, maybe less. Why doesn’t the dirt explode?

      I model a solid or liquid as two parts; the very thin surface, and all the stuff underneath it. Now it’s only the surface that can radiate; the other stuff has to conduct to it.

      The surface has to be thought of as a speed of light responder to its incident energy and it radiates according to only its temperature. It receives energy from two sides; underneath slowly, on top speed of light.

      With this model in mind, think about incoming radiation to a surface as a modulator that controls the rate at which the ponderous conductivity underneath gets to transfer body energy to its skin.

      The skin-body differential temperature controls the conduction magnitude and direction to the body. If the skin temp exceeds the body temp it immediately radiates enough to maintain balance (of itself).

      That’s the logic i would use to answer your question. The surface and body are dancing in a delicate balance to the music of the radiation. The skin only gets hotter if the conduction and radiation outflows can’t keep up.

      • Thanks your response.I agree that

        It receives energy from two sides; underneath slowly, on top speed of light.

        Whether there is simply an IR exchange, without absorption, I do not know, and I need to think more about your response.

      • i think there has to be absorption. Anything less is a reflection and then you haven’t affected skin temperature or energy budget.

        When tou think about it, make a cartoon with skin seperate from a body and connect these with a q bar vector.to represent conduction.

        I make the top of the body and bottom of skin have same temp. Then it is easier to see how a 10 micron skin can instantaneosly respond to incoming radiation cranking up T of skin on top while controlling ΔT on the uppermost part of the body which is then the driving force for the magnitude of q bar.

        With a model like that it is easier to wrap your head around a mechanism that can control outbound q bar without adding heat to anything more than the skin temp.

        You can even think of that skin with some dx thickness and let dx approach 0. That makes it even easier to grasp.

    • richard verney December 25, 2017 at 1:28 am

      I have never seen anyone argue that DWLWIR does not exist. The issue is what does it do?

      These figures should raise questions because of the absorption of of EMR in water. The depth to which EMR is absorbed is wavelength dependent.

      Because of the wavelength of Solar Irradiance, virtually no solar insolation is absorbed in the top millimeters of the oceans. In practice, for the main part it is absorbed in a volume of water occupying about 10 metres in depth, say between 1 metre to 11 metres. This means that the energy of some 162W/m2 of incoming solar is absorbed in a volume of 10 cubic metres (ie., 1m x 1m x 10m).

      By way of contrast, due to the wavelength of DWLWIR, some 90% of it is fully absorbed in about 10 microns of vertical penetrative depth. Because DWLWIR is omnidirectional, such that some is intercepting with the oceans at 10 degrees, some at 20 degrees, some at 30 degrees, this means that almost all DWLWIR is fully absorbed in no more than 5 microns. This means that the energy of some 345W/m2 of incoming DWLWIR is absorbed in a volume of 0.000005 cubic metres (ie., 1m x 1m x 5/1000,000m).

      So if K&T energy budget cartoon is correct, one has twice as much energy absorbed in 2 millionths of the volume. If that does not raise eyebrows, it is difficult to know what will. This begs the question, what exactly is all this DWLWIR doing in such a small volume of water?

      Richard, first, why should that raise eyebrows? The same is true whatever the DWLWIR hits, a rock, a tree, a person, the ocean. It’s absorbed at the skin … so what?

      Second, if it is NOT absorbed in the water as you seem to be arguing, we’re left with two questions:

      1. Why isn’t the ocean frozen? It’s only getting about 165 W/m2 from the sun or so, and it’s radiating at about 390 W/m2 plus losing another 110 W/m2 to evaporation and sensible heat. Obviously, if that were the whole story it would have frozen long, long ago … so what is the source of the ~340 W/m2 energy that’s needed to balance the equation?

      2. If the energy of the ~340 W/m2 of DWIR is NOT going into the ocean, then what’s happening to it? We know that it’s not evaporating the surface water because even if it were providing 100% of the energy going into the evaporation, that only accounts for ~80 W/m2 of the absorbed energy … what’s happening to the rest if it isn’t warming the ocean? It can’t just vanish, and we know it isn’t reflected. Where in your hypothesis is it going?

      Regards,

      w.

      • Willis Eschenbach said, December 25, 2017 at 11:03 am:

        1. Why isn’t the ocean frozen? It’s only getting about 165 W/m2 from the sun or so, and it’s radiating at about 390 W/m2 plus losing another 110 W/m2 to evaporation and sensible heat.

        Nope. It gains 165 W/m^2 worth of HEAT from the Sun. This is the heat input to the surface, and is what needs to be balanced by parallel heat losses from the surface. It loses ~53 W/m^2 to the atmosphere and space via radiation, 24 W/m^2 to the atmosphere via conduction, and 88 W/m^2 to the atmosphere via evaporation. That’s [53+24+88=] 165 W/m^2 worth of OUTGOING HEAT from the surface. And we have balance!

        Temperature -> thermodynamics -> heat fluxes. Conceptual (mathematically derived) “hemifluxes” and hypothetical radiant emittances won’t do, I’m sorry. Because they’re not themselves thermodynamic quantities …

        No, the atmosphere INSULATES the surface, Willis. It doesn’t HEAT it som more … It insulates it by being warmer than its outside surroundings. Like all insulation does. It has a thermal mass, it CAN be warmed, via heat transfer from the body being insulated. The atmosphere can, space can’t.

      • Thanks your response. I do not claim to know the answers, but the point I raise is interesting. I understand the point you make, but you should also be asking yourself, why have the oceans not boiled off, from the top down, long ago?

        One cannot compare a substance like water, which evaporates, with a substance like rock, but I will pick up on the point you make about skin.

        Materially, we know what some 162W/m2 of incoming solar does when it is absorbed by the oceans in a volume of around 10 cubic metres, ie., it heats the oceans up to around 20 degrees. In the equatorial/tropical region where there is approximately double the incoming solar, it heats the oceans up to around 34 degC (eg Red Sea), although more typically only to about 30 to 31 degC, before evaporation, winds and oceanic currents cut in to restrict the warming reaching above 30/31 degC.

        In one of your posts on ARGO, you identified the role of evaporation in capping temperatures in tropical oceans, and that is with circa 162W/m2 of incoming solar being absorbed in a volume of 10 cubic metres, so what the heck does some 345W/m2 of incoming DWLWIR being absorbed in a volume of just 0.000005 cubic metres do? Where, within the body of water, does the evaporation come from? Consider the orders of magnitude we are dealing with.

        As regards your point regarding skin. I am presently in Norway, it is around minus 12 degC. There is only about 5 hours of daylight. When the sun is out, the skin on your face and hands feels warm, notwithstanding the very low solar insolation given the winter geometry. There cannot be much energy given the very high Northern Latitude. In a couple of weeks time, I will be in Southern Spain. On a cloudy day, it will probably be about plus 14 deg C, and there will be more DWLWIR hitting my body than there was solar insolation in Norway, and yet my skin will feel cold, not warm as it did in Norway.

      • An interesting thing about human temperature sensing is that our skin doesn’t ‘feel’ temperature. It senses direction of energy flow. In the winter sun of Norway you were gaining energy. You felt warm. In Cloudy Spain you were losing energy and felt cool.

        Ever notice how walking into the same temperature house in winter and summer feels, respectively, warmer and colder. It’s your energy direction skin sensors at work.

      • Another good example of your skin sensors is when you walk down an aisle of freezer cases in a grocery store. You feel really cold because you are radiating like crazy to a very cold set of ‘walls’. Walk down an aisle of canned goods and you feel nothing in essentially the same air.

        We feel comfy at 21C because we are in balance there.

        Funny but true story… I got heat stroke symptoms one morning in Florida last summer before sun up. It was 72⁰F and dew point was 72⁰F. Couldn’t shed energy at all in dead calm air. Temperature is sometimes not a very useful comfort reading.

    • richard verney December 25, 2017 at 1:28 am
      By way of contrast, due to the wavelength of DWLWIR, some 90% of it is fully absorbed in about 10 microns of vertical penetrative depth. Because DWLWIR is omnidirectional, such that some is intercepting with the oceans at 10 degrees, some at 20 degrees, some at 30 degrees, this means that almost all DWLWIR is fully absorbed in no more than 5 microns. This means that the energy of some 345W/m2 of incoming DWLWIR is absorbed in a volume of 0.000005 cubic metres (ie., 1m x 1m x 5/1000,000m).

      So if K&T energy budget cartoon is correct, one has twice as much energy absorbed in 2 millionths of the volume. If that does not raise eyebrows, it is difficult to know what will.

      But what you have neglected to consider is that it is those few microns of water which emit the thermal IR.
      So if the water surface is at 25ºC it will be emitting ~440W/m^2

    • A C Osborn December 25, 2017 at 3:03 am

      The other key part of the whole concept, especially regarding CO2 is the one I mentioned yesterday “the mean free path of IR radiation in the atmosphere is ~25 meters”.
      This suggests the the IR from the CO2 only area (no H2O) of the Atmosphere has absolutely no chance of ever getting back to the surface, as every 25 or so metres (probably longer in less dense parts of the atmosphere) the Photon will have lost it’s energy to another molecule and it’s “downwardness” halved at each collision.

      What does the ‘CO2 only area (no H2O) of the Atmosphere’ have to do with anything, we’re talking about the total DWIR at the surface, it doesn’t matter how high in the atmosphere it originates from, the key point is that IR from the surface is absorbed by GHGs in the atmosphere and some of it is re-emitted and is absorbed by the surface.

  47. I think you are on to something very important. Your figure 3 is very nice. I would suggest that you divide your data into more groups than two for your figure 5. Perhaps you will be able to find out how much stronger the cooling/warming process gets when the days are really hot/cold compared to only somewhat hot/cold.
    I do not really understand why you limit the areas where the cooling process is at play to green and yellow of figure 1.
    Clouds are formed at the Swedish west coast as well during the day, at least in our too short summer. We can possibly expect that the radiation reflected from the clouds out into space is a little higher warmer days here as well.
    Best regards
    Johan

  48. I always thought, well not always but for some time now, that convection is rules the day moving heat from the surface upward. Radiation plays second fiddle. No?

  49. Willis Eschenbach

    I found it amazing that the temperature of such a possibly unstable system could only have changed by ± 0.3°C over the entire 20th century.

    First of all best wishes.
    Imo you’re viewing only part of the system. The whole system is running at maximum power already, no chance of overheating. Just realize that since the last major re-heat some 85 million year ago the deep oceans have been cooling down some 15-20K in spite of slightly increasing output from the sun.
    So no, the system is not possibly unstable.

    Willis Eschenbach December 25, 2017 at 11:03 am

    1. Why isn’t the ocean frozen? It’s only getting about 165 W/m2 from the sun or so, and it’s radiating at about 390 W/m2 plus losing another 110 W/m2 to evaporation and sensible heat.

    The oceans aren’t losing that energy, they are transferring most of it to the atmosphere. Only radiation through the atmospheric window is lost directly to space. Only thing that counts is the amount of energy that is eventually lost to space, mostly from the atmosphere. Is supposed to be ~equal to the solar input.

    Take a step back and consider the entire System Earth.
    Hot interior insulated by the crust which is HOT.
    Deep oceans sitting on this hot crust, having a shallow solar heated surface layer that almost completely blocks energy loss to the atmosphere.
    Finally the atmosphere that insulates the (oceans) surface and just reduces energy loss to space.

    • Ben Wouters December 26, 2017 at 5:17 am Edit

      Willis Eschenbach

      I found it amazing that the temperature of such a possibly unstable system could only have changed by ± 0.3°C over the entire 20th century.

      First of all best wishes.
      Imo you’re viewing only part of the system. The whole system is running at maximum power already, no chance of overheating. Just realize that since the last major re-heat some 85 million year ago the deep oceans have been cooling down some 15-20K in spite of slightly increasing output from the sun.
      So no, the system is not possibly unstable.

      Agreed. The Constructal Law requires that the system evolve to max power.

      Willis Eschenbach December 25, 2017 at 11:03 am

      1. Why isn’t the ocean frozen? It’s only getting about 165 W/m2 from the sun or so, and it’s radiating at about 390 W/m2 plus losing another 110 W/m2 to evaporation and sensible heat.

      The oceans aren’t losing that energy, they are transferring most of it to the atmosphere. Only radiation through the atmospheric window is lost directly to space. Only thing that counts is the amount of energy that is eventually lost to space, mostly from the atmosphere. Is supposed to be ~equal to the solar input.

      Yes, the oceans are indeed losing that energy. It is true that they get back something on the order of half of it, but it is assuredly lost.

      My question was directed to those who, unlike you, claim that there is ZERO energy returned to the ocean by the atmosphere. They make all kinds of claims about why this is … but my question remains:

      IF, as many people claim, there is no energy going from the atmosphere to the ocean, why is it not frozen solid?

      Best to you,

      w.

      • If one places a film which blocks LWIR, say just 50 cm above a shallow tray of water,, how long does it take for the surface skin to freeze?

      • Willis Eschenbach December 26, 2017 at 12:49 pm

        Yes, the oceans are indeed losing that energy. It is true that they get back something on the order of half of it, but it is assuredly lost.

        I intended lost to mean “lost to space”.
        The energy the oceans lose at the surface isn’t lost for System Earth, only the radiation through the atmospheric window is. Together with the energy that the sun delivers directly to the atmosphere (~20%) this energy is used to keep the atmosphere of the surface: hydrostatic equlibrium (HE).
        This atmosphere in HE loses energy to space together with what the surface radiates through the atmospheric window. This should more or less balance with incoming solar less reflected radiation.

        To understand why the oceans are so hot, we should realize that the temperature of the DEEP oceans (lets say below the permanent thermocline) is completely caused by geothermal energy. The sun only increases the temperature of a shallow surface layer, thus creating an impenetrable barrier for all geothermal energy entering through the ocean floor, except at (very) high latitudes. This makes their temperature a balance between geothermal heating and cooling at high latitudes. The last 84 million years this “balance” resulted in a cooling rate of roughly 1K/5 million year.
        Given this cooling, Earth must be at an energy balance that results in the maximum temperature possible.

      • richard verney December 26, 2017 at 6:15 pm
        If one places a film which blocks LWIR, say just 50 cm above a shallow tray of water,, how long does it take for the surface skin to freeze?

        I assume by ‘blocked’ that you mean that the film does not emit IR either, in which case it would need to be at something like liquid nitrogen temperature. So I would say that it wouldn’t take very long to freeze.

      • Willis Eschenbach December 26, 2017 at 12:49 pm

        IF, as many people claim, there is no energy going from the atmosphere to the ocean, why is it not frozen solid?

        It did answer your question just below your post. Awaiting a reaction.
        Given your ideas about Cb’s driving the Hadley circulation, are you aware of the Hydrostatic Equlibrium (HE) the atmosphere is in, and how this mechanism is driving a lot of the weather systems on planet Earth?
        HE makes it virtually impossible that a trace gas like CO2 can have a noticeable influence on our climate.

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