Vacuum Chamber Temperature Test Debunks Climate Crisis Claims

Written by Geraint Hughes

We all know the story. Or at least, if you don’t by now you should do.  The whole world is being lied to by the BBC, CNN, EU & UN & the mainstream media about climate change.

They force feed us on a diet about the ‘climate crisis’ postulated on a runaway Radiative Greenhouse Effect. The junk science conjures up CO2 as the trigger for a catastrophic Global Mean Surface Temperature Increase –  in other words man-made Global Warming.

All reminiscent of the Nazi Propaganda machine.  Broadcasting lies on a daily basis, brainwashing enmasse. Here is another way of proving just that.

Readers may recall my previous experiments. I promised you more tests, with thermometers and other gases, involving a bigger tower  I have performed 2 sets of tests comparing the temperature differences between using my Portable Vacuum Chamber device.

Here are the results of the test in table form.

To help understand the tables RT (Room Temperature) indicates the temperature which a free standing digital temperature probe indicated the temperature was in the room.  The starting temperature was the temperature indicated on the digital probe inside the vacuum chamber as I activated the light.

This probe touches the side of the bulb with the tip resting against the inside of the chamber.   This is not perfect and in due course I will get better thermometers but this is sufficient to show that the concept of gaseous back radiant induced heating just doesn’t work.

In each instance the chamber was evacuated first, pressurised and then the light activated.

In the first test I used a Vintage Squirrel Cage bulb and pressurised the gases to 0.5 Bar.  In the second test I used a Spiral Vintage Bulb and pressurised the gases to 0.6 Bar.  The result is similar in both instances.  (My squirrel cage bulb, blew L)

You can see here that in both tests, the Bulb and container surface temperatures were cooler than in the Vacuum than with CO2 added.  In the first test after 20 minutes the Vacuum achieved a maximum temperature of 69.1 and in the second test 63.6.  Co2 on the other hand achieved a maximum temperature of 63.3 and 59.2, which was amazingly, COOLER!

Not only that, but the surface temperature rise was slower, going from 40 to 50 deg took 117 seconds in the first test and 181 seconds in the second test whereas CO2 took 177 Seconds and 247 Seconds respectively.  Wait a minute, I thought CO2 was an insulator!!!!!!  What the heck???????

I am always being told that CO2 not only increases maximum temperatures but causes faster temperature rises, due to the so called “Reduction in the rate of cooling.” Because the so-called back radiance, causes a radiant heat input.  Because CO2 supposedly acts as a layer of insulation.  IT DOES NOT.  It is a medium enabling greater heat transfer away from the warmed surface.

CO2 was never warmer, nor did it warm faster than the pure vacuum.  Its presence acted to cool the surface of the bulb.  DID YOU ACTUALLY EXPECT ANYTHING ELSE?

We are constantly being told that if we had earth in space with no atmosphere or a purely neutral atmosphere, it would experience an average GMST of 255K, but it is actually 288k, due to the presence of greenhouse gases.   Which is just a lie.

This experiment quickly shows such a theory to be completely stupid.  The addition of CO2 gas had no warming effect, only a cooling.  And when I compared Argon to CO2 I found that Argon, resulted in Warmer conditions and faster temperature rises than CO2, despite the fact that Argon is not a “Greenhouse Gas” which is actually fake and misleading terminology.

Conclusion

CO2 does not result in higher GMST  when compared to a Vacuum or Argon, yet we are being told that it does and it does so because of back radiant greenhouse effect, which is clearly complete rubbish.  Climate Crisis is a lie and this can be demonstrated in a growing number of ways.

The links to the playlist of the tests can be found here.

Test 1 Playlist

https://www.youtube.com/watch?v=KL_aKmZr4Ac&list=PLF66zq1SOYiveU3cw2KrcOPHQaX0CI6TR

Test 2 Playlist

https://www.youtube.com/watch?v=qaxfldJAXY0&list=PLF66zq1SOYitkLVH3VxVsaaT476khpcGR

The time stamps for the temperatures are as below.

1st Set of Temperature Tests (Squirrel Cage)

 

Vacuum. Room Temperature 24.9. Start Temperature on Container 30 Degree.

Vacuum – 40 Degrees – 25 Seconds.

Vacuum – 50 Degrees – 2 min 22 Seconds. = 142 Seconds. = 117 Seconds Difference.

Vacuum – 60 Degrees – 5 min 41 Seconds. = 341 Seconds. = 199 Seconds Difference.

Vacuum – 69.1 – 20 minutes.

Vacuum – 72.2 Degrees – Maximum after 40 minutes.

 

CO 2 – Room Temperature – 26.1. Start Temperature on Container Thermometer 28.1.

CO 2 – 30 Deg – 1 min 20 seconds. = 80 Seconds.

CO 2 – 40 Deg – 3 min 4 Seconds = 184 Seconds. = 104 Seconds Difference.

CO 2 – 50 Deg – 6 min 1 Seconds = 361 Seconds. = 177 Seconds Difference.

CO 2 – 60 Deg – 13 min 24 Seconds = 804 Seconds. = 443 Seconds Difference.

CO 2 – 63.3 Deg – 20 minutes.

CO 2 – 69 Deg – Maximum after 40 minutes.

 

Argon – Room Temperature – 26.2.  Start Temperature on Container Thermometer 29.0

Argon – 30 Deg – 1 min 10 Seconds. = 70 Seconds.

Argon – 40 Deg – 2 min 55 Seconds. = 175 Seconds. = 105 Seconds Difference.

Argon – 50 Deg – 5 min 30 Seconds. = 330 Seconds. = 155 Seconds Difference.

Argon – 60 Deg – 11 min 37 Seconds. = 697 Seconds. = 367 Seconds Difference.

Argon – 65.5 Deg – 20 minutes.

Argon – 70.3 Deg – Maximum after 40 minutes.

 

2nd Set of Temperature Tests – (Spiral Bulb)

 

Vacuum – 22.7 Room Temperature. 24.9 Start Temperature on Container Degree.

Vacuum – 30 Degrees – 2 min 31 = 151 Seconds

Vacuum – 40 Degrees – 4 min 44 Seconds = 284 Seconds = 133 Seconds Difference

Vacuum – 50 Degrees – 7 min 45 Seconds. = 465 Seconds = 181 Seconds Difference

Vacuum – 60 Degrees – 14 min 27 Seconds. = 867 Seconds = 402 Seconds Difference

Vacuum – 63.6 Degrees – 20 minutes

 

CO 2 – 23.0 Room Temperature – 29.0 Start Temperature on Container Thermometer.

CO 2 – 30 Deg – 2 min 34 seconds = 154 Seconds

CO 2 – 40 Deg – 4 min 58 Seconds = 298 Seconds = 144 Seconds Difference.

CO 2 – 50 Deg – 9 min 05 Seconds = 545 Seconds = 247 Seconds Difference.

CO 2 – 60 Deg –  not attained

CO 2 –  59.2 Deg – 20 minutes.

 

Argon – 23.1 Room Temperature – 31.3 Start Temperature on Container Thermometer

Argon – 40 Deg – 3 min 51 Second = 231 Seconds

Argon – 50 Deg – 7 min 08 Seconds = 428 Seconds = 197 Seconds Difference.

Argon – 60 Deg – 18 min 20 Seconds = 1100 Seconds = 672 Seconds Difference.

Argon – 60.2 Deg – 20 minutes.


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Comments (166)

  • Avatar

    geran

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    Well done, Geraint. When experiment results agree with established physics, you know you’ve done it right.

    I was temporarily confused by the two headings for the right-most columns. But after I figured it out, everything fell into place. The videos were a great help.

    Since you have correctly debunked the “CO2 can heat the planet” nonsense, expect numerous distracting criticisms from Warmists! That’s always fun to watch.

  • Avatar

    Charles Higley

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    Two reasons CO2 cools better than Argon.

    First, and the main effect in the above experiments was probably that, as a triatomic linear molecule, CO2 is has more ways to vibrate and bend than Argon, as more degrees of freedom, and thus can absorb energy in more ways.

    Second, and more of an atmospheric effect is that, if you examine the CO2 IR absorption spectrum, it has three peaks. Two are wavelengths equivalent to blackbody radiation of 400 and 800°C. Clearly, these abilities cannot be used by IR from Earth’s surface. Only solar input could activate these and such absorption would serve to dissipate solar energy in all directions.

    The 3rd peak is equivalent to –80°C. Any IR sent downward by CO2 would be reflected as that energy level is full in the surface—everything on the surface is warmer than this. The IR would be reflected and lost to space.

    Argon does have four IR peaks, equivalent to –39, –43, –60, and –95°C, also from strongest to weakest, respectively. Again, most of its downwelling IR would be reflected by the surface and lost to space. This is why they have not mentioned it as a “greenhouse gas,” besides the fact that we are not emitting it.

    Even though Ar has these absorption peaks and is over 20 times more abundant in the atmosphere than CO2, its absorption coefficient might be lower than that of CO2, while CO2 has more ways to absorb energy as heat from the atmosphere and radiate energy away at its one useful –80°C IR band.

    Just some thoughts. I have wondered why so many people do not look at these gases more closely. Many assume CO2 is a great IR absorber, while it clearly a one-trick cold pony. Last week the media was touting that SF6 (sulfur hexafluoride) was the most powerful greenhouse gas known to humanity. It turns out to have one narrow IR absorption peak equivalent to -1°C. Considering that SF6 is at 9 ppt in the atmosphere, it is irrelevant.

    It is also claimed that SF6 has an upper atmospheric lifetime of 800–3200 years. Would that be a half-life, maybe, of about 2000 years? I find that very doubtful considering the solar radiation it would encounter and the many ways that the atmosphere can clean itself, as reported a few years ago by atmospheric chemists.

    Why consider a gas’s lifetime in the upper atmosphere? It is from the upper tropical troposphere that the global warming CO2 is supposedly sending IR downward and heating the surface. Regardless of the gases there, NASA has reported that that region of the atmosphere has been gently cooling for decades.

    • Avatar

      Charles Higley

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      I must apologize regarding the Argon absorption data. The graph I accessed was not clearly labeled and involved quinilone in Argon. I will go on searching for the IR of pure Argon.

    • Avatar

      Charles Higley

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      An update
      First, no one seems to care about any IR absorption by atmospheric Argon. Even though it is the 3rd most abundant atmospheric gas, it is not mentioned in any of the IR studies (graphs) of IR absorption by the atmosphere.. Its major peaks appear to be very close to those I described above. The relevant paper was from 1952.

    • Avatar

      Gymbo

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      Talking about the peak emission wavelength temperatures is silly. What matters is the quantity of energy available for absorption in the respective bandwidths contained in any incident flux and not the so called Wien’s law calculated peak emissions.

      Certainly Earth’s surfaces are not hot enough to emit any significant radiation in the 2.7 and 4.3 micron bandwidths so CO2 will absorb nothing in these. Only the incident solar has any significant energy and this is only in the shorter wavelength of ~2.3 micron.

      But at even 303 K surface temperature there is a significant amount of energy in the flux emitted in the CO2 absorption bandwidths centred on ~15 micron.

      Talking about CO2 only absorbing due to minus 80°C temperatures is quite simply misleading and totally wrong – just look at any spectra you like for ambient temperatures – there is significant energy in the wavelengths centred on 15 micron.

      • Avatar

        geran

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        Gymbo, you may be missing the point.

        Of course 15 micron photons are emitted from Earth. But, that energy, if returned, is not able to increase surface temperatures.

  • Avatar

    John O'Sullivan

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    Thanks, Charles, good to have your insights

  • Avatar

    jerry krause

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    Hi Geraint and Charles,

    Charles wrote: “First, and the main effect in the above experiments was probably that, as a triatomic linear molecule, CO2 is has more ways to vibrate and bend than Argon, as more degrees of freedom, and thus can absorb energy in more ways.” This is basically correct but there is the factor of the rotation of the linear molecule which the spherical atom of Argon does not have.

    Gordon Barrow, a respected physical chemistry, in his introductory chemistry textbook wrote: “Inspection … shows that the analysis of translational energy provides an accurate and complete picture of the heat capacity of the noble gases, but provides a satisfactory treatment of other gases only at very low temperatures. From the experimental heat capacity results, we must deduce that the molecules of these other materials have additional ways of taking up energy that becomes important at higher temperatures. You will see that these molecules can rotate and vibrate–internal motions that are of no consequences in the simple kinetic-molecular theory of gas pressure.”

    Have a good day, Jerry

    • Avatar

      jerry krause

      |

      Hi Geraint and Charles and PSI Readers,

      I must try to explain the ‘no consequences in the simple kinetic molecular theory of gas pressure. Barrow is referring to a scientific law to which many are familiar. This law is the Ideal Gas Law (PV=nRT). Which generally describes the relationships between the pressure, volume, temperature and ‘n’ (the number of gas atoms/molecules described by the unique unit–mole). And this number quantity does not matter what the atoms/molecules are. Hence, the word ‘Ideal’. All atoms/moles of a gas are the same–ideal.

      Have a good day, Jerry

  • Avatar

    WILLIAM STECKLE

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    Looking at this from a very simplistic point of view… if CO2 is an insulator why is it not used in double or triple glazed windows? Argon is and it is much more expensive. CO2 is a heat transfer medium, not an insulator.

    • Avatar

      Nick Schroeder

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      Seems to me CO2 has been used in the past for fill gas.
      I think our low E dual panes contain Krypton.
      CO2 has about half the thermal conductivity of air.
      That makes CO2 an insulator, i.e. thermal resistance, 1/U.
      Q = U A dT

  • Avatar

    Zoe Phin

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    Very nice

  • Avatar

    Zoe Phin

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    In the future people will laugh about the fact that the “greenhouse effect” turned out to be just air’s poor conduction ability.

  • Avatar

    Bob Armstrong

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    👍 I award you a Ritchie Prize http://cosy.com/Science/ComputationalEarthPhysics.html#YouTubePhysicsDemoFund
    for ” best “YouTube” quantitative experimental test of any of the non-optional classical physical computations ”

    I don’t think we are up to classical standards yet , but these are a start .

    Unfortunately the Ritchie Fund is yet unfunded , so the prize is an invitation to post your work or notices there of at the http://cosy.com/Science/ComputationalEarthPhysics.html#Disqus link .👍

  • Avatar

    Rudolf Huber

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    CO2 does not warm the atmosphere. Why is that news?

  • Avatar

    AlJones1816

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    I think this experiment rather fails to recreate the actual conditions necessary to produce a greenhouse effect. Greenhouse gases warm the surface of the planet because they set the altitude at which terrestrial radiation can escape to space. This emitting layer is cold, and because of the lapse rate the surface is warmer than the emitting layer needs to be. If you add more greenhouse gases you push the emitting layer higher, where it’s colder, so the emitting layer and consequently the atmosphere below it warm until the emitting layer is in equilibrium.

    You need to have a column of air tall enough and wide enough to have a lapse rate if you want a greenhouse effect. I support you building such a device but I think you will have a hard time with the engineering.

    • Avatar

      Zoe Phin

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      Do you understand how nonsensical that sounds? Things don’t warm up because something went to a colder location.

      • Avatar

        AlJones1816

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        I think it does not sound intuitive, but I do not think it sounds nonsensical. I will try to explain more clearly.

        If Earth had no atmosphere, radiation would move directly from the surface to space. We could calculate the planetary temperature that would be needed to balance the rate out of outgoing terrestrial radiation with the rate incoming solar radiation. It is about 255 K. The reason that temperature comes into play here is because colder things radiate less intensely than warm things (as anyone who has stood next to a campfire knows).

        But earth does have an atmosphere, and some of the components of the atmosphere absorb most of the terrestrial radiation before it goes to space, so that instead of radiating right from the surface, the energy escapes to space from the atmosphere. The layer of the atmosphere at which the energy escapes must be at the temperature that balances incoming sunlight with outgoing terrestrial radiation (255 K). A special property of our atmosphere is that it has a lapse rate – that is, the higher you go the colder it gets (as anyone who has climbed a mountain knows). Because of this lapse rate, the surface of the planet is much warmer (about 288 K) than the emitting layer of the atmosphere high above.

        This is what we call the greenhouse effect.

        If you add additional greenhouse gases to the air, the atmosphere becomes more optically opaque, and the layer at which radiation escapes must move up to a higher altitude (where the air is thin enough that a photon isn’t likely to encounter a GHG molecule and be absorbed). This higher altitude is colder (because of the lapse rate) than the 255 K needed to balance incoming sunlight. Because it is colder, it is radiating energy less intensely, and now the rate of energy coming in from the sun is greater than the rate of terrestrial energy leaving – the earth isn’t in equilibrium. The temperature must therefore warm until it reaches equilibrium (until it is 255 K). Now the old radiating layer is warmer than 255 K, and every layer of the atmosphere below it is a bit warmer than it was before.

        This is greenhouse driven global warming.

        • Avatar

          Herb Rose

          |

          Hi Al’
          If you look at a chart of the temperature according to altitude you will see the temperature decreases in the troposphere, increases the stratosphere, decreases in the mesosphere, then increases in the thermosphere. What is heating the atmosphere at the boundary of the stratosphere and mesosphere and how can the colder molecules in the lower atmosphere radiate heat through the hotter molecules in the thermosphere?
          Herb

          • Avatar

            AlJones1816

            |

            Hi Herb,

            From my understanding, the lapse rates in the stratosphere and thermosphere are inverted because these layers are absorbing incoming sunlight (the stratosphere because of the ozone layer).

            I am not sure what you mean by “how can the colder molecules in the lower atmosphere radiate heat through the hotter molecules in the thermosphere.” Radiant energy doesn’t care what the temperature is of the matter it’s heading toward. Cold things can’t radiantly heat hot things because the hot things are radiating energy more intensely than the cold thing, not because the radiation can’t move from the cold to the hot thing.

        • Avatar

          Zoe Phin

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          Al,
          As a blanket moves farther away from your body it gets colder, therefore you must warm up even more. That’s what you’re saying.

          You’re definition of “equilibrium” is actually conservation of heat flow, which is not science.

          There’s no experiment to prove what you’re saying because its just gibberish.

          • Avatar

            AlJones1816

            |

            Zoe,

            That is not what I am saying. The temperature of a blanket does not change its insulating properties. The temperature of the emitting layer of the atmosphere does change how intensely it radiates. This is a consequence of the Stefan-Boltzmann law.

          • Avatar

            Zoe Phin

            |

            “The temperature of a blanket does not change its insulating properties.”

            Sure it does. Conduction and Convection. And radiation if in a vacuum.

            Remember, you claim that removal from equilibrium (you + blanket = 98.6F), causes one to cool and one to heat up.

            “The temperature of the emitting layer of the atmosphere does change how intensely it radiates.”

            First there is no one layer of emission. There’s something like 6? distinct major layers.

            Second, you were supposed to show how emission DEFINES temperature. It doesn’t. Even the emissivity coefficient, although part of SB equation, is not defined by the equation. Emissivity is controlled by temperature (and pressure), and type of substance.

        • Avatar

          Herb Rose

          |

          Hi Al,
          Why aren’t the oxygen molecules in the mesosphere absorbing uv light and creating ozone and heat?
          I have not noticed your name before in PSI so I will repeat why the greenhouse gas theory is scientifically impossible because it violates the second law of thermodynamics.
          The definition of temperature as the mean kinetic energy of the molecules in the media being tested is wrong. 0 C water has more kinetic energy (movement) than 0 C ice where the molecules are locked in a crystal structure. 100 C steam molecules have more energy (540 calories/gram) than 100 C water molecules. This is the energy that drives a steam engine.
          A thermometer is not measuring the kinetic energy of the molecules striking it but the total kinetic energy transferred to it. Kinetic energy is 1/2 the mass of the molecules times their velocity squared. As the altitude increases both the temperature registered on a thermometer and the number of molecules (density) decrease. Is the decrease in the temperature recorded by the thermometer a result of less mass (fewer molecules) or lower kinetic energy (less velocity)?
          The way to determine the kinetic energy of the molecules is to use the universal gas law. The inverse of density (number of molecules per constant volume) gives the volume of a constant number of molecules. (The pressure referred to in the universal gas law is the pressure that confines a gas and resists its expansion.In the case of the atmosphere this pressure is provided by gravity,(not atmospheric pressure) whose force does not change significantly through the atmosphere (distance 4000 mi to 4031 mi)).When the temperature recorded by the thermometer at different altitudes is divided by the density at that altitude it shows a curve where the kinetic energy increases with altitude. It increases slowly in the troposphere where water moderates the change then increases in an exponential curve in the higher layers of the atmosphere where the kinetic energy is greater than the boiling point of water.
          The sun is the source of heat for the Earth and it is heating the atmosphere more than the surface of the Earth.
          Herb

          • Avatar

            AlJones1816

            |

            Hi Herb,

            I am not sure that I follow. The sun heats the earth the atmosphere. The atmosphere cannot heat the earth but it can make the earth warmer than it would be if the atmosphere were not there.

            In fact the atmosphere must be keeping the earth warmer than it would be in its absence because, as I said, we can calculate the equilibrium temperature we would find at the planet’s surface if it were radiating energy directly to space and we can see that it would be 255K, yet the measured surface temperature is much warmer.

          • Avatar

            Zoe Phin

            |

            Wrong, Al,
            The atmosphere is a coolant.
            Your “equilibrium” temperature dilutes the sun onto a hemisphere it doesn’t fall on, in real time!

            “The atmosphere cannot heat the earth but it can make the earth warmer”

            Wow! Warmer and heat are practically synonyms. This sentence is gibberish.

            Remove the atmosphere, it will get warmer.
            Remove the top 1km of dirt, it will get warmer.
            Remove the next 1km of dirt, it will get warmer

            Continue until you’ve reached the core.

            The atmosphere is like a heat sink on a computer.

            Blackbody radiation is for a cavity at thermal equilibrium. The Sun and Earth do not form a cavity. There is one spot between them that can be in thermal equilibrium, and NO it doesn’t control further warming, on a need-heat basis as you claim.

            The experiment shows GHGs to be coolants. You claim that their cooling power forces the surface to send it more radiation. How silly. Do you also believe a hole in a sand bag will cause stray sand to jump in the bag?

          • Avatar

            Zoe Phin

            |

            Al,
            You’ve made the classic mistake of confusing cause and effect.

            See Joule’s 2nd law.

            It’s because it’s hot that we have an atmosphere. It’s not because of the atmosphere that it’s hot.

            Aside, albedo rises for every meter the atmosphere increases. Your increase in co2 height will be offset by albedo.

            Gases have NO ability to raise temperature. NONE! Temperature makes gases.

            Never forget this.

          • Avatar

            Toto's Fan

            |

            The Empress appears to be wearing all the clothes. Literally!

          • Avatar

            HerbRose

            |

            Hi Al,
            The atmosphere is part of the Earth and the sun is indeed heating it. The surface of a satellite orbiting above the Earth facing the sun will be heated to 250 F. On the surface of the Earth that same surface will be heated to 50 F. The missing energy is contained in the atmosphere. I cannot understand how people maintain that the oxygen, nitrogen, and argon in the atmosphere cannot absorb heat when these are the gases in their ovens that transfer heat from the heating element to the food being cooked.If they do not absorb heat how do you keep the interior of a house warm in winter? The atmosphere, like in the oven, is absorbing heat from the sun during the day and radiating heat into space at night. The Earth and moon are both the same distance from the sun and receive the same energy. The Earth is habitable because the atmosphere absorbs and stores heat while on the moon the temperature goes from 250 F during the day to -250 F at night.
            Zoe’s contention that it is geothermal heat that is heating the Earth is incorrect. The Earth is in equilibrium with the energy coming from the sun and any excess heat coming from the Earth’s core is being transferred through the atmosphere where it is radiated into space just as excess heat from forest fires, which is from stored solar energy, is radiated into space. The equilibrium point between the heat being absorbed by the sun and the heat coming from the Earth’s interior is near the surface of the crust so as you dig into the Earth the temperature drops then increases. If the source of heat heating the Earth’s surface come from the core why is there permafrost?
            Herb

          • Avatar

            Zoe Phin

            |

            The statement:
            “Zoe’s contention that it is geothermal heat that is heating the Earth is incorrect.”

            is incompatible with:
            “any excess heat coming from the Earth’s”
            and
            “the heat coming from the Earth’s interior is near the surface”
            AND
            “the heat coming from the Earth’s interior is near the surface”

            “If the source of heat heating the Earth’s surface come from the core why is there permafrost?”

            Is permafrost 0 kelvin? No, it’s something like 263 kelvin.

            I’ve done the math. The sun alone cannot recreate observations of upwelling IR @ Desert Rock, NV Jan 1-19 2018:

            https://i.ibb.co/09t8gy2/out.png

          • Avatar

            Zoe Phin

            |

            Herb,
            At the peak solar of day,
            The Sun-Earth equilibrium is like 20 meters below the surface.

            At anti-peak solar of night,
            The Sun-Earth equilibrium is very high in the atmosphere, because the Earth has to heat beyond what the Sun provided.

            If the Earth had as much geothermal energy as the moon, then the Earth’s diurnal temperature would fluctuate just like the moon.

    • Avatar

      geran

      |

      Al says: ” If you add more greenhouse gases you push the emitting layer higher, where it’s colder, so the emitting layer and consequently the atmosphere below it warm until the emitting layer is in equilibrium.”

      Wrong Al, adding more gas to the atmosphere does not push the emitting layer higher. And, even it did, that would mean the surface area of the emitting layer would be greater.

      You need to learn the physics.

      • Avatar

        Geraint Hughes

        |

        Sounds like my “TOWER” experiment needs to be posted 😀

        Yes, I have heard this drivel before, if only the column was higher, then there would be more back radiant heating, blah blah blah. Shame it doesnt happen. I will do that one this weekend me things. 🙂

    • Avatar

      Geraint Hughes

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      Alan, lay off the pills. Also, you don’t sound clever, you sound like a numpty. Your lack of knowledge and skills shows real bad.

  • Avatar

    AlJones1816

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    Zoe, “my” equilibrium temperature does account for the geometry of the earth. The equation is given as:

    T_eq = ((S_o (1-a) / 4*b ) ^(1/4)

    Where:
    * S_o is the solar constant, equal to 1362 W/m^2
    * a is the Earth’s albedo (equal to about 0.3)
    * b is the Stefan-Boltzmann constant (equal to 5.67*10^(-8) W/m^2K)

    The factor of 1/4 accounts for the fact that only one hemisphere is illuminated at once (1/2) and that the distribution of sunlight across that hemisphere depends on the angle of incidence (1/2).

    The result of this is 255 K, or -18.5 C.

    You omitted half of the sentence that I wrote. The whole thing said, “The atmosphere cannot heat the earth but it can make the earth warmer than it would be if the atmosphere were not there.” Which is simply to say that the second law of thermodynamics concerns net heat fluxes, not individual fluxes. The second law doesn’t prohibit a flow of heat from the atmosphere to earth, it merely states that the net flow must be from the warmer to the cooler. If I stand next to a burning log, my body is radiating energy toward the burning log and the burning log is radiating energy toward me. The net flow is from the log to me, but it does not mean I’m not radiating energy.

    I do not claim that GHGs have a magical cooling power. The only cooling that is occurring is because of the lapse rate. The property that GHGs have is simply that they absorb longwave terrestrial radiation, so that instead of moving directly from the surface to space the radiation must move up through the atmosphere until it reaches a layer high and thin enough to escape.

    If we wish to make an analogy, imagine a bucket of water being filled from above by a faucet, and being drained at the bottom by a small hole. The water level in the bucket will rise until the rate of flow out the bottom equals the rate of inflow at the top. The rate of outflow is controlled by the pressure and by the size of the hole. If we make the hole at the bottom of the bucket smaller, the rate of outflow will be reduced, and the water level in the bucket will rise until the pressure is high enough to make the rate of outflow again match the rate of inflow. Adding CO2 to the atmosphere is like making the hole smaller.

    • Avatar

      Geraint Hughes

      |

      Alan.

      Your bucket analogy is a total waste of time, because a more emissive atmosphere, LOSES more heat, because as it is EMISSIVE, it shoots it all out instead of holding onto it. Because the atmosphere is 3 dimensional, and shoots out that radiation in al directions, it simply isnt possible to cause the surface to be warmer, when compared to a non-emissive gas. You really need to buy my book to understand what I am talking about. Check it out. https://www.amazon.co.uk/Black-Dragon-Breaking-Frizzle-Frazzle/dp/1949267008

  • Avatar

    Zoe Phin

    |

    Al,
    “The factor of 1/4 accounts” for two hemispheres. The sun doesn’t shine on two hemispheres at once. You need to divide by 2, not 4. The surface area of a sphere is 4 times its solar intercept disc. A hemisphere is 2 times.

    “Which is simply to say that the second law of thermodynamics concerns net heat fluxes, not individual fluxes.”

    As warm heats cold, the heat flux goes to zero. You’re attempting to conserve the heat flux and you falsely call that “new equilibrium”.

    The 2nd law forbids keeping heat flux constant.

    “The second law doesn’t prohibit a flow of heat from the atmosphere to earth.”

    Yes it does. Heat flows from greater to less energy. You’re confusing heat and energy in your log example.

    “the radiation must move up through the atmosphere until it reaches a layer high and thin enough to escape.”

    Not a problem. CO2 can “teleport” (radiation) from hot surface CO2 to cold atmosphere CO2 much faster than O2 and N2 can transport their heat. Just because you don’t see the hot CO2 BELOW the cold CO2 from a satellite doesn’t mean CO2 is blocking anything. You also can’t thermally see the hot side of your air conditioner from the cold side. So what? It would go there if not for the fan.

    “If we wish to make an analogy, imagine a bucket of water …”

    Your analogy is a complete FRAUD.

    Water is an extensive property.
    1L + 1L = 2Liter

    Temperature is an intensive property.
    1C + 1C = 1 Celsius

    Water accumulates, while Temperature is AVERAGATIVE or MAXIMAL depending on heat capacity.

    Water accumulates and Heat dissipates. Heat is energy FLOW. It can not be contained and accumulated like water. Trapping heat is stopping flow – removing HEAT FLOW. Stop using fake analogies.

    “Adding CO2 to the atmosphere is like making the hole smaller.”
    Intensive properties don’t care about your hole size.

    • Avatar

      AlJones1816

      |

      Zoe,

      The first factor of 1/2 accounts for the fact that only half of the earth is illuminated at once, the second factor of 1/2 arises from the fact that sunlight is at a 90 degree angle at the equator, so 100% of light is flux hitting the ground, but near 0% at the poles (imagine pointing a flashlight straight at the ground versus laying it flat on the floor). You must integrate the dot product of the radiation field with the surface normal vector and doing so yields a factor of 1/2. That is, earth has two sides but it is also a sphere, if it were flat like a dinner plate we wouldn’t need the additional 1/2.

      You say, “as warm heats cold the flux goes to zero.” And this is exactly true for the net flux. The warm and cold will still be exchanging heat flux after they are the same temperature, the fluxes will just be equal and oppositely directed so their net is zero.

      The flow is not being contained in my analogy, the intensity of the flow is controlled by the size of the hole and the pressure exerted by the height of the water column above. The greenhouse effect does not stop energy from leaving the atmosphere, any more than a damn stops a river from flowing. The greenhouse effect dictates where in the atmosphere (what altitude) the energy is leaving from.

      • Avatar

        Zoe Phin

        |

        Al,
        The sun doesn’t shine on the dark side. If you’re claiming -18C for the day side, then you have a problem: 0 kelvin for the night side. Both for Earth and the moon, which has no atmosphere.

        “And this is exactly true for the net flux. The warm and cold will still be exchanging heat flux after they are the same temperature, the fluxes will just be equal and oppositely directed so their net is zero.”

        WRONG!

        There is NO flow without a T difference. There is NO double flow. An electron can not absorb AND emit a photon at the same Planck time.

        Think of a waterfall. If the bottom is lower than the top, the water falls. If you raise the bottom to top level (flat surface), you do not get a double opposite waterfall.

        ” dam stops a river from flowing”
        A dam constrains river flow, leading to accumulated water in a basin. Your analogy implies heat is like water, but it’s NOT! You are constraining heat flow just like a damn constrains water, but heat doesn’t work that way. Heat “water” flows right through the dam, and this never allows water (heat) to accumulate.

        • Avatar

          AlJones1816

          |

          Zoe,

          I do not think that is such a problem. The very exact reason that we do not have dramatic swings in day to night temperature (and the reason the moon does have massive swings) is because we have an atmosphere and the moon doesn’t.

          Your waterfall analogy is valid for waterfalls, but that scenario does not cover every potential for exchange. If you hand me ten pennies every minute that does not preclude me from handing you two pennies every minute. We could both hand each other ten pennies every minute. The possibilities for penny exchange are endless.

          If you are throwing pennies out the window at a rate of twenty per minute, you will go broke more slowly if I stand next to you and hand you a penny every minute than you would if you were alone.

          • Avatar

            Zoe Phin

            |

            Al,

            “If Earth had no atmosphere, radiation would move directly from the surface to space. We could calculate the planetary temperature that would be needed to balance the rate out of outgoing terrestrial radiation with the rate incoming solar radiation. It is about 255 K.”

            You’re right, the moon doesn’t have an atmosphere AND the average temperature is much colder than your math. Therefore your assumption of an Earth without an atmosphere is pure BUNK.

            The atmosphere does not heat or otherwise “make earth warmer” from -18C to 20C on the sunny side or -273C to 10C on the night side.

            Only an additional raw source of energy could do that. It’s definitely not the armosphere. Do you understand this?

            “Your waterfall analogy is valid for waterfalls, but that scenario does not cover every potential for exchange.”

            A waterfall represents a flow from PE to Kinetic Energy. It’s a much better analogy to heat, since it’s also a flow.

            Your analogy is a complete fraud because you’re trying to represent a flow with a stack. Something that happens between matter with matter itself.

            I worked on wall street for 10 years. Trust me when I tell you you would’ve made a poor analyst since you can’t tell the difference between stacks and flows.

            A flat body of water is not a double reverse waterfall.

            A flat body of water does not evaporate and condense so rapidly that it gives the appearance of stability.

            The wind is not blowing 1000 m/s East and 998 m/s West giving the illusion of 2 m/s East.

            If you claim a double exchange, you must PROVE IT by scientific observation, and not by an abstract assertion.

            “If you hand me ten pennies every minute that does not preclude me from handing you two pennies every minute. We could both hand each other ten pennies every minute. The possibilities for penny exchange are endless.”

            You can’t exchange pennies ATOMICALLY. You can’t take while you give at exactly the same time. If you give first, you “cool”, if you take first, you “heat” – which violates equilibrium.

            “If you are throwing pennies out the window at a rate of twenty per minute, you will go broke more slowly if I stand next to you and hand you a penny every minute than you would if you were alone.”

            Wrong. I’m already broke, that’s why you’re giving me pennies. Now I’m making you go broke. If I slow down my throwing, it does not make you richer.

          • Avatar

            AlJones1816

            |

            Zoe,

            The moon has a lower average temperature than the earth precisely because it has no atmosphere or oceans to retain heat at night. The averaged daytime temperature in the sunlight hemisphere will be close to the lunar equilibrium temperature (which is a bit higher than the earth’s because the moon has a lower albedo), but the nighttime side will be much, much colder than earth’s nighttime side because it cools very quickly.

            The atmosphere itself is the reason the earth doesn’t plunge into a deep freeze like the moon every night, and the greenhouse gases are what keep the surface warmer than the equilibrium temperature. The reason they do that is because the emitting layer of earth is not the surface, it’s high up in the atmosphere, and the planet has a lapse rate so that the surface is warmer than the upper atmosphere.

            The very bottom line here is that trying to test the greenhouse effect with tubes of gases will not do the trick.

          • Avatar

            jerry krause

            |

            Hi Al and /Zoe,

            First, you both need to come to grips with the fact there no such thing as an averaged temperature. There are only actual measured temperature of matter at a given place and time.

            Secondly, I have discovered that if one states “If Earth had no atmosphere, radiation would move directly from the surface to space. We could calculate the planetary temperature that would be needed to balance the rate out of outgoing terrestrial radiation with the rate incoming solar radiation. It is about 255 K.”,you have allowed someone to clearly rewrite, or ignore, known history.

            Relative to this historical problem, I have just reviewed C. Donald Ahren’s textbook Meteorology Today 9th Ed. and Ackerman & Knox’s textbook Meteorology 3rd Ed. The fact that each textbook has more than one edition is evidence that these textbooks are being commonly used in introductory meteorology courses at more than one US college or university.

            So what you both might say. The what is that I cannot find the name–Svante Arrhenius in the extensive indexes of either of these two textbooks.

            You might ask: Who is Svante Arrhenius and why should we expect to find his name in these two textbooks?

            “In 1895, Arrhenius presented a paper to the Stockholm Physical Society titled, “On the Influence of Carbonic Acid in the Air upon the Temperature of the Ground.” This article described an energy budget model that considered the radiative effects of carbon dioxide (carbonic acid) and water vapor on the surface temperature of the Earth, and variations in atmospheric carbon dioxide concentrations.” (https://earthobservatory.nasa.gov/features/Arrhenius/arrhenius_2.php)

            What one cannot read in the above link is: Arrhenius, S. 1896. On the influence of carbonic acid in the air upon the temperature
            of the ground. The London, Edinburgh and Dublin Philosophical Magazine and Journal of Science 41, 237-276.

            For Arrhenius’s Swedish paper was published in English so the knowledge of its existence has been lost because only those capable of reading Swedish every know about it.

            In these essays, as he referred to them, Arrhenius reduced the amount of solar radiation being absorbed by the earth’s surface by the average earth’s albedo which had been observed and calculated at that time. Of course, a significant portion of this albedo was due to radiation scattering by the cloud particles.

            But in calculating the effective radiation temperature of the earth’s surface to balance this reduced amount of absorbed solar radiation, did not consider that atmospheric cloud particles might scatter the longer wave IR radiation back toward the earth’s surface.

            What I just reviewed is historical fact which is clearly being ignored in these two meteorology textbook in which there is no word–Arrhenius.

            Please try to wrap your minds around this historical fact.

            Have a good day, Jerry

    • Avatar

      Geraint Hughes

      |

      Well Done Zoe, keep it up.

      Anyone got any ideas about how to perform a spherical experiment which actually shows this? I got a few ideas, but I’m not too sure. Do a drawing and send it me on you-tube or facebook and I will see if I conjure something up.

  • Avatar

    Zoe Phin

    |

    Al,

    “The averaged daytime temperature in the sunlight hemisphere will be close to the lunar equilibrium temperature”

    You’re looking at data only for the equator. The moon is much colder than your stupid blackbody calculations, even using Earth’s higher albedo, thus proving they are worthless.

    “the planet has a lapse rate so that the surface is warmer than the upper atmosphere.”

    The lapse rate is -g/Cp, bottom to top, hot to cold. You can’t reverse the lapse rate and prove the reverse, any more than can you boil water and then claim the steam made it boil.

    “The atmosphere itself is the reason the earth doesn’t plunge into a deep freeze like the moon every night”

    LOL! No, the Earth has ~6 times the geothermal energy of the moon. The atmosphere is not a raw source of energy, nor does it take solar energy and somehow multiply it with your junk science.

    “The very bottom line here is that trying to test the greenhouse effect with tubes of gases will not do the trick.”

    Right! No scientific experiment could possibly disprove your junk science. How convenient.
    Your cult believes GHGs are twice as powerful as the sun. We can put 100% CO2 in a bottle, compress it to 1000 atmospheres (i.e. give you all the advantages you need), radiate it with IR and NOWHERE will radiation exceed what was fed in, i.e no increase in temperature via CO2 absorption “power”. But such a fact will never disprove your crackpot belief system because you like the conclusions of crackpot belief system.

  • Avatar

    AlJones1816

    |

    Zoe,

    the sunlight side of the moon will be the part of the body that is in equilibrium with incoming sunlight. You and I have already discussed that there must be factors in the equilibrium temperature calculation to account for this. I am not only considering the equator – and in fact where the sun is directly overhead will be much, much hotter than the moon’s equilibrium temperature, the poles will be much colder. Averaged across the surface the daytime temperature will be close to the equilibrium temperature. The average lunar temperature includes both day and nighttime temperature and so it is much, much colder than the equilibrium temperature.

    You say: “LOL! No, the Earth has ~6 times the geothermal energy of the moon. The atmosphere is not a raw source of energy, nor does it take solar energy and somehow multiply it with your junk science.”

    The earth’s interior heat flux is undoubtedly larger than the moon’s, but it is nowhere near large enough to produce the temperature of the lower atmosphere. Nowhere do I claim that the atmosphere is a raw source of energy.

    You say: “We can put 100% CO2 in a bottle, compress it to 1000 atmospheres (i.e. give you all the advantages you need), radiate it with IR and NOWHERE will radiation exceed what was fed in, i.e no increase in temperature via CO2 absorption “power”.”

    This is because the mechanism you describe is not the mechanism of the greenhouse effect. In my original post I described the mechanism. Greenhouse gases do not produce energy that was not already present in the system, they just determine where the energy can exit the system.

    • Avatar

      Zoe Phin

      |

      “The earth’s interior heat flux is undoubtedly larger than the moon’s, but it is nowhere near large enough to produce the temperature of the lower atmosphere.”
      Sure it is.

      Do you know how geothermal heat flux is
      determined? Comparing heat flux to radiation is fraud.

      “Greenhouse gases do not produce energy that was not already present in the system, they just determine where the energy can exit the system.”

      According to you, the sun provides only up to -18C. If you want it warmer, you need to add energy. But now you deny GHGs do that, and so you’re left at -18C. Do you have a brain?

      • Avatar

        AlJones1916

        |

        Zoe,

        As I’ve been saying, GHGs don’t by themselves raise the temperature. GHGs cause radiation to be emitted from high in the atmosphere. Adding GHGs doesn’t generate any energy that isn’t already in the system, it just slows the rate of energy loss to space until the new radiating layer warms up to 255 K again.

        • Avatar

          Zoe Phin

          |

          Al,
          Please put down the crack pipe.

          “GHGs cause radiation to be emitted from high in the atmosphere.”

          Emitting radiation causes cooling.

          The Earth receives ~240 W/m^2. That’s the energy in the system (excluding geothermal, which climate scientists deny).

          The surface is at 15°C, which is 390 W/m^2.

          You can’t slow the loss of 240 and get 390. You are claiming an additional 150 W/m^2, while at the same time denying it.

          You’re sick. Get help.

          • Avatar

            AlJones1816

            |

            Zoe,

            Your math seems to be missing some pieces. The earth receives ~340 W/m^2 at the top of the atmosphere. About 100 W/m^2 of this is reflected directly off of the atmosphere or earth’s surface, and does not contribute to heating in any way. 240 W/m^2 is absorbed by the surface or ground, and 240 W/m^2 is radiated back into space either from the top of the atmosphere or directly from the planet’s surface. The net energy budget is 0. If we add greenhouse gases, the energy budget becomes slightly positive and the planet warms.

            It is important to understand that irradiance, measured in W/m^2, is a flux. Adding greenhouse gases doesn’t lower the amount of energy leaving the earth – all of the energy eventually leaves, it reduces the flux (rate of flow) until thermal equilibrium with incoming sunlight is regained by warming the atmosphere.

          • Avatar

            Zoe Phin

            |

            Retard,
            240 W/m^2 is quite shy of 390 W/m^2, don’t you think?

            One needs to add 150 Joules EVERY second, to get to 390. Where does that come from? What atomic/chemical process creates that?

            “until thermal equilibrium with incoming sunlight is regained by warming the atmosphere.”

            A co2-in-bottle experiment should be able to prove that. It should make the IR source warmer than it would be without the CO2.

            What you’re saying is that the hot surface can’t heat the cold atmosphere because all that heat is returning back to the surface.

            Cold radiatively forces hot to get even hotter.

            Nonsense.

          • Avatar

            AlJones1816

            |

            Zoe,

            There is no difference to make up. The earth receives ~340 W/m^2 at the top of the atmosphere, the earth loses ~340 W/m^2 at the top of the atmosphere. It is in radiative equilibrium unless something perturbs the flux in or out. Increasing GHG concentration perturbs the flux by raising the emitting layer higher in to the atmosphere where it is colder. Colder things radiate less intensely than hot things, and so the flux is reduced. This creates an imbalance between incoming and outgoing energy, and so the temperature of the atmosphere increases until balance is restored.

            Here is a diagram of the earth’s energy budget from NASA, let me know if some part of it still does not make sense to you:

            https://upload.wikimedia.org/wikipedia/commons/b/bb/The-NASA-Earth%27s-Energy-Budget-Poster-Radiant-Energy-System-satellite-infrared-radiation-fluxes.jpg

          • Avatar

            Zoe Phin

            |

            Where’s the scientific experiment proving your balance-seeking thermal enhancing process?

            “Colder things radiate less intensely than hot things, and so the flux is reduced. This creates an imbalance between incoming and outgoing energy, and so the temperature of the atmosphere increases”

            Uhuh. Here you are saying that GHGs decrease tempetature at the top, and increase temperature at the bottom.
            This is a decrease in entropy, which is a violation of the 2nd law.

            You are too stupid to understand physics. All you can do is repeat a nonsense narrative and believe it’s true.

            Aside,
            Your energy budget has a problem. The 398.2 is dependant on the 340.3, and vice-versa. Where is the origin? It can’t be the sun, because the sun only provides 239.9 (after albedo, obviously) !!!

          • Avatar

            Zoe Phin

            |

            Al,
            In normal science:
            Hotter things emit more radiation.
            Colder things emit less radiation.

            But in climate “science”,
            emitting less radiation (to space), is proof that something on the inside is getting hotter.

            How dumb.

          • Avatar

            AlJones1816

            |

            Zoe,

            The concept of planetary radiative equilibrium is not controversial, you can read more about it on Wikipedia:

            https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature

            I think you are misunderstanding me. GHGs affect temperature only because they afect the altitude at which radiation is emitted to space. The layer of the planet where radiation is emitted to space must be in radiative equilibrium with incoming sunlight, which is true for earth. On earth, because we have an atmosphere, the emitting layer is not the surface, but is the upper atmosphere.

            Adding more GHGs does not decrease the temperature of the atmosphere, it raises the emitting layer up higher. Because of the atmospheric lapse rate, this higher layer happens to be colder. If the lapse rate were reversed (temps went up with altitude), adding GHGs would actually cause the atmosphere to cool! In fact, in the stratosphere, where the lapse rate is inverted, this is actually what we observe:

            http://images.remss.com/data/msu/graphics/TLS_v40/plots/RSS_TS_channel_TLS_Global_Land_And_Sea_v04_0.png

            And I will again repeat that your arithmetic is incorrect regarding the energy budget; please refer to the diagram from NASA I provided. 340 W/m^2 arrive at the top of the atmosphere, 340W/m^2 leave at the top of the atmosphere. I do not know where your 390 W/m^2 figure comes from.

          • Avatar

            Zoe Phin

            |

            Al,

            “The concept of planetary radiative equilibrium is not controversial”

            The mistake has been around so long, it stopped being controvertial. Retards just repeat nonsense until they become desensitized.

            “I think you are misunderstanding me.
            GHGs affect temperature only because they afect the altitude at which radiation is emitted to space.”

            There is no single altitude of emission.

            https://www.pnas.org/content/pnas/115/41/10293/F1.medium.gif

            And emission doesn’t follow SB law.

            How many times do you need to be told that before it sinks into your head?

            Earth’s emission to space (240) is a mass-weighted average for the whole height of the atmosphere.

            There is no single height of emission. Tu comprende?

            “And I will again repeat that your arithmetic is incorrect regarding the energy budget”

            I don’t count the albedo. Why do you care about it?

            “I do not know where your 390 W/m^2 figure comes from.”

            It comes from ten years ago 398.2 being 390. Now it’s 398.2. It’s the surface emission to the atmosphere.

            NASA’s energy budget has a circular dependency problem. It’s worthless.

            The atmosphere has something like 503 W/m^2 going in and out, while the sun only provides 239.9 (again, albedo removed).

          • Avatar

            AlJones1816

            |

            Zoe,

            You are correct that there is not a single emission altitude, that is a simplification on my part. One must use a program like MODTRAN to work through propagation of radiation through the atmosphere. The idea is conceptually identical. Imagine instead of a single emitting layer that there is a single “average” emitting layer. Adding GHGs to the atmosphere raises this average emitting layer to a higher altitude.

            I have read the article you cited carefully, thank you for linking it. The author is quite mistaken. In all energy budget diagrams, such as the one I showed you from NASA, the flux leaving the TOA is the same 340 W/m^2 that is received from the sun. Do you sums yourself using the diagram, they will add up. Here is a very nice and simple explanation of how the sums work out:

            http://rabett.blogspot.com/2017/10/an-evergreen-of-denial-is-that-colder.html

            There is no violating physics going on here.

          • Avatar

            Zoe Phin

            |

            Al,

            Why do you care about so much about albedo? It’s not used energy. It goes in, and gets reflected out.

            “There is no violating physics going on here.”

            Plenty of violations of physics. Betty Pound in the comments section points out many.

            The most glaring and obvious one is the blue plate emits a total of 533, when it only receives 400. Where did it get the extra 133? From the green plate? Where did the green plate get it? Again you have an internal energy creation loop.

            While you may think you’re conserving energy flux with the outside source, you’re violating entropy (2nd law) internally.

            You can’t sacrifice one for the other.

            You have ZERO experiments to back up your junk science.

            You learned nothing.

            ““The Discovery of Global Warming”

            Wow, that explains it. You’re an imbecile and a sucker for self-serving narrative.

          • Avatar

            Zoe Phin

            |

            Al,
            The other problem with Rabett is a violation of the law of conduction. A single blue plate that receives 400 W/m^2 will, over time, radiate 400 W/m^2 from all sides.

            You can’t add fluxes from two sides of the plate to begin with.

            Also,
            The higher and lower frequency photons of 400 W/m^2 vs. 200 W/m^2 do not just magically disappear. This plate will have an absorbtivity of 0 in some frequency in some ranges, while its emissivity will exceed 1 in other ranges (those common to a doubled 200 W/m^2).

            With two plates, you have another problem. Electrons can not absorb and emit photons at the same time. The blue plate on the green plate side is sending 266 W/m^2 and receiving 133 W/m^2 at the same time. What does that mean? The electron jumps between 266 and 133, producing an average of 200.

            This is junk science, but to a person lacking critical thinking skills this looks legit. LOL

            Math is not physics. While physics uses math, math is not bound by physics.

            Rabett sets up a purely mathematic trick devoid of observational physics.

          • Avatar

            AlJones1816

            |

            Zoe,

            I do not know where you are getting 533 W/m^2 from the diagrams. There are 400 W/m^2 going into the system and 400 W/m^2 going out of the system – energy is conserved.

            A Watt is 1 Joule per second, so 400 W/m^2 means that the plate receives 400 Joules of energy every second per square meter on one of its sides. Both of the plate’s sides are radiating, so the surface area radiating is twice the surface area receiving. The flux on a single side must therefore be reduced by half. If you do as Petty Pound suggests in the comments and assume both sides of the plate are radiating 400 W/m^2, you’ve got more energy going out than is coming in. Either the plates have to cool until the net energy out equals the net energy in or you have violated conservation of energy.

            I also want to say that your combative tone is not necessary. We are all just trying to learn and help each other understand.

          • Avatar

            Zoe Phin

            |

            Al,
            “I do not know where you are getting 533 W/m^2 from the diagrams. ”

            His solution was blue plate emits 266 from both sides and green plate emits 133 from both sides. Since left blue and right green is 266+133=400, he thinks this is valid. But blue plate is emitting 266 from both sides and this is invalid.

            “Both of the plate’s sides are radiating, so the surface area radiating is twice the surface area receiving.”

            Heating is a resonant phenomena. All electrons in all molecules will, over time, resonate with the input flux.

            https://thumbs.gfycat.com/ShockedReliableIndigobunting-size_restricted.gif

            The input area is 5, the output area is 46. Yet all resonate with the input.

            Did you see Betty’s radiator video?

            How about observations?

            “The temperature on the side of the International Space Station (ISS) facing the Sun can reach 121°C, while the side facing away from the Sun can go as low as -157°C.”

            121C = 1955 W/m^2
            -157C = 15 W/m^2

            Total = 1980 W/m^2

            The sun’s flux near the Earth is 1361 W/m^2

            According to you, ISS temperatures are impossible!

            Obviously ISS is more complicated than a plate, and it does spend some time in earth’s shade – so we don’t expect it to emit 1361×2 W/m^2. But the simple fact that it emits more flux than is input, proves that Eli Rabett is a con man.

            “We are all just trying to learn”
            You are definely not. You’re protesting learning and repeating junk science.

          • Avatar

            AlJones1816

            |

            Zoe,

            I believe that this is the crux of the misunderstanding:

            “Heating is a resonant phenomena. All electrons in all molecules will, over time, resonate with the input flux.

            https://thumbs.gfycat.com/ShockedReliableIndigobunting-size_restricted.gif

            The input area is 5, the output area is 46. Yet all resonate with the input.’

            A molecule is not an area, you are conflating irradiance (W/m^2) with power (W). The individual molecules will resonate with the input, but the irradiance must be spread out over the radiating area.

            The ISS is a three dimensional object that is being heated sporadically by the sun, rather than receiving continuous flux as do the plates or as does the earth on its daytime side. The ISS is similar to the moon.

          • Avatar

            Zoe Phin

            |

            Al,

            “A molecule is not an area, you are conflating irradiance (W/m^2) with power (W). The individual molecules will resonate with the input, but the irradiance must be spread out over the radiating area.”

            LOL. You think the radiating area is not composed of molecules?

            You don’t believe radiation is formed from an electron dropping to a lower energy level?

            You just said they are all resonating!

            Honestly, I don’t know how someone can look at that gif i provided and conclude what you just said.

            According to you, the “balls” should all be a pale bluish-red, or purple, but they are not. The molecules on the far left side have as much internal microscopic energy as those on side facing the heat source.

            The ISS space station has 6 faces, and we can say that NORMAL solar radiation powers only one face. If all faces had equal area, the maximum output should be 1361/6 = 227 W/m^2, ice cold.

            And yet we know some place has 1955 W/m^2!

            I honestly don’t know how you can be so stupid in the face of observational evidence!

            A first year physics textbook will show a steady-state conduction through a metal bar. Let’s pretend the average tempetature is that of the farther end, and the bar is a perfect cube. Well, according to you, we would have to convert the temperature to W/m^2, divide by 6, and convert back to temperature.

            And yet no student is even taught to consider the 6 faces of the metal bar and its output vs input surface area. Why do you think that is?

        • Avatar

          jerry krause

          |

          Hi Al,

          Evidently you do not believe historical facts. I addressed a comment to you and Zoe. And there has been no; I did not know that.

          Have a good day

          • Avatar

            AlJones1816

            |

            Hi Jerry,

            apologies, I only saw your comment just now. I agree that Arrhenius should be discussed more, since his contribution to our understanding was pretty revolutionary. If you have no read it, a really really fantastic history of Arrhenius’ discovery can be found in the eBook “The Discovery of Global Warming” by science historian Spencer Weart:

            https://history.aip.org/climate/index.htm

        • Avatar

          jerry krause

          |

          Hi Zoe,

          Evidently you do not believe historical facts. I addressed a comment to you and Al. And there has been no; I did not know that.

          Have a good day, Jerry

        • Avatar

          Geraint Hughes

          |

          The higher up in the atmosphere the gas, the more energy goes out to space, the less that returns to the surface, making it all the more impossible for the surface to be warmer, because when convection brings the air back down to the surface, it will have lost much more heat compared to a non-emissive gas, resulting in cooler conditions. Also, you are ignoring the fact that the earth atmosphere has a huge daily bulge, the bulge is so big the atmosphere when viewed from afar and above in the UV spectrum (so you can see it) is noticeably a great distance from the surface, like several hundred miles. https://www.facebook.com/photo.php?fbid=10212500279817133&set=a.1558034325907&type=3&theater The atmosphere is not some perfect sphere, when you see it like this, you can see how much of the atmospheric emissions, TOTALLY miss the earth. This alone makes Back Irradiance a total non-starter. Hey non-probs pal.

  • Avatar

    Herb Rose

    |

    Both Al and Zoe are correct when they say the other is wrong. The greenhouse gas theory is complete nonsense and the contention that geothermal energy is heating the Earth’s surface is equally ridiculous. The. grand solar minimum will show both theories are wrong.

    • Avatar

      Zoe Phin

      |

      Herb,

      “and the contention that geothermal energy is heating the Earth’s surface is equally ridiculous.”

      Not at all. In fact, it’s a measured truth.

  • Avatar

    Pierre D. Bernier

    |

    Geez guys, cut it out…
    When you use the right science you get the right results. From the Ideal Gas Law, PV = nRT you get T = PV/nR = PM/dR where T=temperature, P=pressure, M=molar mass, d=density and R=gas constant(8.3145).

    From NASA itself, Venus near surface data, T=740…
    P=9200, M=43.45, d=65, T=739.7 ~ 740. No runaway GHGE there. All explained.
    From NASA itself, Earth near surface data, T=288…
    P=101.3, M=28.97, d=1.225, T=288.1 ~ 288. No GHGE there. All explained.

    There is not, never has, never will, anywhere, be any GHGE. NEVER EVER !!! Not even on planet Beta in solar system Zeta !!! NEVER EVER !!! It is not only the Ideal Gas Law, it is the Universal Gas Law !!! It’s all about adiabatic auto-compression due to gravity.

    From Robert Ian Holmes
    https://pdfs.semanticscholar.org/dd17/e71f5e2c8ca3de2c4cc110ac49380b118612.pdf?_ga=2.202317654.619044460.1568144103-1210803403.1563538073

    Heck. It even works for Antarctica, T=224…
    P=68.13, M=28.97, d=1.06, T=223.9 ~ 224. No GHGE there either. All explained.

    Furthermore…

    WHAT IS WRONG WITH THE CO2 HYPOTHESIS?

    The greenhouse effect theory claims that the various infrared-active gases dramatically alter the atmospheric temperature profile which will vary with height, temperature, latitude and atmospheric composition.

    To summarise, it predicts that more CO2 leads to slower cooling rates in the troposphere and faster cooling rates in the stratosphere (Figures 8).

    According to the current climate models, temperatures stop decreasing with height in the tropopause and start increasing with height in the stratosphere and the reason is because of ozone heating (Figure 9).

    https://globalwarmingsolved.com/2013/11/summary-the-physics-of-the-earths-atmosphere-papers-1-3/

    The Connelly’s D vs P fits (Figure 10) did not require any consideration of the concentration of carbon dioxide, ozone or any of the other infrared-active gases. This directly contradicts the greenhouse effect theory.

    Furthermore, the CO2 hypothesis cannot explain the troposphere/tropopause phase change in molar density behaviour. Also, if the ozone in the tropopause heats up the atmosphere, how come it is more easily compressible (Figure 10) ?

    If greenhouse gases are supposed to keep the atmosphere warm, why is it that in the mid troposphere high concentrations of water vapour cools that atmosphere faster at night then with less water vapour in it ?
    All of this shows that the temperatures at each height are completely independent of the infrared-active gas concentrations, which directly contradicts the predictions of the greenhouse effect theory. So, if there is no greenhouse effect, this also disproves the man-made global warming theory.

    • Avatar

      jerry krause

      |

      Hi Pierre,

      You clearly do not believe in giving numbers a unit. “R=gas constant(8.3145).” In my Handbook of Chemistry and Physics a table of several values of R for various units is given.

      I do find 8.3144 but after this decimal number is times 10^7 (8.3144 X 10^7). So, I question what system of units you are using since you give no units for P, V, n. Only for temperatures do you use the unit K.

      Please clarify the units for all the variables which you used to calculate the temperatures .

      Have a good day, Jerry

      • Avatar

        Pierre D. Bernier

        |

        Are you sure you’re not a priest ? Read Holmes document you’ll get the units.

    • Avatar

      Zoe Phin

      |

      Pierre,
      Have you looked up Joule’s 2nd law yet?

      Without a solar or geothermal source a planet will not be hot enough to have gases to compress in the first place.

      By solar alone, Venus’ atmosphere would be contracted by ~10 times, leaving it at 9 bars of pressure.

      The ideal gas law doesn’t work in reverse.
      The ideal gas law correctly stated, should be:

      nRT -> PV

      Look at a simple phase diagram chart. Move up the axis in each direction. You will see that pressure is a solidifying force, and temperature is a gasifying force.

      Pressure alone can’t explain temperature!

      Why do you think Pressure is on the y-axis? By convention, we plot the depedant variable on the y-axis.

      Think about it!

      Holmes’ math works really well, but it doesn’t explain the cause. The cause can only be solar and/or geothermal.

      • Avatar

        Pierre D. Bernier

        |

        You tried for 6 months to pass your geothermal thing on Joe Postma’s site. Holmes prooves you wrong and Joe is on the same page. Cut it out !

        • Avatar

          Zoe Phin

          |

          Pierre,
          Holmes does not prove me wrong. Holmes is modeling an EFFECT which is absolutely correct. But he doesn’t properly explain the cause.

          What can I say? Joseph is incorrect.

          I’ve measured Total Solar Energy received at 6 SURFRAD sites over a span of 1 year – 2018.

          Then I compared it to Total Upwelling IR Energy emitted by surface (and received by sensor).

          There is a stupendous mismatch. Something needs to fill that gap.

          Alarmists believe it’s “Downwelling” IR. But we both know that doesn’t exist. What does exist is Upwelling-from-Pyrgeometer-IR-that-came-from-the-bottom-of-Pyrgeometer that is falsely called “Downwelling”.

          This leaves the question open. Where does the extra energy come from for Upwelling IR? Hmm…

          In Joseph’s paper, he thinks that solar is enough because he matched one single solar maxium with upwelling IR. One single point! All the other time, the upwelling exceeds the solar by a large margin.

          Be wise, Pierre.

        • Avatar

          Zoe Phin

          |

          Pierre
          Since Postma censors my comments, I can’t reply to your last comment on climateofsophistry. I’m posting it here instead:

          Pierre,
          “No need for Geothermal”
          What are you doing?

          Please think before you act.

          “Downwelling infrared”
          LOL.

          That’s exactly the problem. Solar Input is less than both “Downwelling” IR and Upwelling IR.

          A pyrgeometer measures Net IR. There is no “Downwelling” IR – it’s Upwelling-from-pyrgeometer-IR.

          You need to explain Upwelling IR solely from Solar Input.

          – Zoe

        • Avatar

          Zoe Phin

          |

          And my censored comment to Robert:

          Robert,
          Seems self-explanatory to me. I use two data columns available in the SURFRAD data: Solar Downwelling and Upwelling IR. Both in W/m^2.

          The data is for every minute. In one year there is 1440 * 365 data points.

          I add up all of the data points and then multiply by 60 (seconds) to get Joules per m^2 – Total Energy per m^2.

          If the sun was enough, conservation of energy would have resulted in Downwelling Solar = Upwelling IR.

          The sun is not enough. I need to add GeoWm2 * 60 * 1440 * 365 to Total Solar Downwelling in order for it to match Total Upwelling IR.

          I could have also subtracted Solar Upwelling from Solar Downwelling, but that would only bolster my point.

          • Avatar

            jerry krause

            |

            Hi Zoe,

            I do not deny there must be a geo-thermal contribution to to upwelling IR but I do not know it magnitude relative to that of solar radiation

            However, I know there is a third radiation column In the SURFRAD data. The downwelling IR. Many ignore this as it seems you have.

            Why have you ignored it after NOAA made its measurement part of the SURFRAD project.

            Have a good day, Jerry

          • Avatar

            geran

            |

            Jerry, if you’re trying to make sense of the bogus “energy balance”, I can understand your frustration.

            Fluxes can not be added/averaged. They are not scalar quantities. That makes the “340 W/m^2 downwelling” meaningless pseudoscience.

          • Avatar

            jerry krause

            |

            Hi Geran,

            I commented to Zoe because she (I assume) referred to measured data which I have not read you doing. And I admitted to anyone that read my comment that we (I) cannot know what portion of the measured downwelling IR radiation is contributing to the radiation balance at the earth’s surface when the measured air temperature (or the measured upwelling IR is not changing from one minute to the next or from one hour to the next. And I can see how the measured downwelling IR does suddenly increase to the value of the upwelling IR. Which I must conclude that this sudden increase is the result of scattering by cloud.

            If this sudden increase is not caused by scattering by cloud, you must explain by what if is caused.

            For I consider it would be stupid to argue that this sudden increase is the result of any instrument malfunction (regardless of what is actually being measured).

            Readers can read and see the nonsense of what you write.

            Have a good day, Jerry

          • Avatar

            geran

            |

            Jerry, unfortunately you were unable to understand my point. Adding to your frustration, you were also unable to make a point.

            Have a great day.

          • Avatar

            Zoe Phin

            |

            Jerry,
            A pyrgeometer measures Net IR.

            There is practically no Downwelling IR (inversions only),

            The “Downwelling” IR is actually Upwelling-from-pyrgeometer-IR.

            Again, one needs to explain why Upwelling IR is so much greater than Solar. Only geothermal can rationally plug in the gap.

      • Avatar

        Pierre D. Bernier

        |

        Joule’s second law states that the internal energy/total energy of a gas depends on its temperature and not on its pressure and volume.

        Have you looked at Boyle’s, Charles’s, Gay-Lussac and Avogadro,s ?

        Pump a bike’s tyre. P goes up, n goes up, V is constant and T goes up. Nothing to do with Joiules law.

        • Avatar

          Zoe Phin

          |

          Pierre,
          V is not constant if the tire was flat to start with.

          The T in your tire will match T outside.

          Only when air is pushed through the tiny nozzle will there be a temporary heating effect. It will leave as fast as it came.

          You’ve already assumed it’s hot enough outside for gases to exist.

          If it’s not hot enough to have gases, there is no gases for gravity to compress. What is so hard to understand?

          • Avatar

            Pierre D. Bernier

            |

            Now you are really wiggling ! Start with a tyre that is not completely deflated. V stays constant. Connollys have D = aP +c. Since D = P/RT then T = P / aRP + cR. Since a tends to 0 (0.0004) and c tends towards 4.0 then T = P / cR. If P goes up, T goes up. End of story.

          • Avatar

            Zoe Phin

            |

            Pierre,
            The atmosphere is not enclosed and being pressed.

            The ideal gas law uses CONSTANT pressure. Did you notice a delta P term in formula?

          • Avatar

            Zoe Phin

            |

            Pierre,
            In the bicycle tire pump example, you are taking the abundant air from the outside and putting it into a confined space.

            In the atmosphere, you don’t have extra outside air !!!

            Now modify your tire example to incorporate such a reality. You’re only allowed to pump in and out the air already in the tire.

            Compression, Decompression.
            Heating, and Cooling.

            The maximum T is set by available air.

            What creates air?

            Temperature and availability of gaseous molecules to outgas from the surface.

            Because it’s so hot, we so much air.

            On Venus, it’s hotter, and so there’s more air for gravity to compress & decompress.

            Get it?

          • Avatar

            Pierre D. Bernier

            |

            Zoe

            You are desperate and willing to say just about anything to get out of it. You are either dishonest or have no FKNN idea of what you are talking about. NOW YOU ‘RE OUT !!!

          • Avatar

            Zoe Phin

            |

            Wisdom is the antidote to frustration, Pierre.
            I’m not being dishonest, I’m being 100% sincere.

            You’re the one that thinks there’s a delta-P term in the ideal gas law. You’re the one that thinks the tire analogy means something.

            I agree that if we add air to the atmosphere it will get hotter. But where would the air come from and by what means?

            It would come from a raise in temperature leading to outgassing from ocean/ground. This raise in temperature will be perfectly reflected by changes in the pressure. You will then think the pressure is what caused the T change.

          • Avatar

            Herb Rose

            |

            Zoe, You still cannot comprehend that the pressure that confines the atmosphere and resists its expansion is gravity not atmospheric pressure. Atmospheric pressure is the weight of the gas molecules in the atmosphere. The weight of something is its mass times gravity. Until you understand that gravity attracts all objects equally and they fall at the same rate (excluding wind resistance) you will never understand physics. Look up the acceleration due to gravity in Wikipedia.

          • Avatar

            Zoe Phin

            |

            Uh.
            Herb, I’m surprised John let’s you publish articles at PSI.

            P = F/A

            Area = Volume / Height

            P=Fh/V
            P=mgh/V
            P=(m/V)gh
            P=dgh

            Where d=Density

            My husband paid $700 3 years ago so we could take a hot-air balloon ride for 2.

            Given that ideal gas law has constant P for the surface air, how were we able to rise?

            Easy, P=dgh

            We lowered our density by heating the air. Since P is constant, h must increase. And that is indeed what happened. We defeated gravity by reducing our density and thereby raising our height.

            P.S. You’re an idiot.

        • Avatar

          Herb Rose

          |

          Hi Pierre,
          You shall learn that Zoe adjusts physics to conform to her theories.
          She believes that gravity preferentially pulls on dense objects. When you pump up a tire the gas is expanding (going from higher pressure to lower pressure) into the tire therefore the gas is getting hotter as all expanding gases do. You must adjust the laws of physics (PV not equal to nRt) because Zoe has had them amended to comply with her theories.
          Herb

          • Avatar

            Zoe Phin

            |

            Herb,
            “She believes that gravity preferentially pulls on dense objects.”

            It does. That’s why hot air or helium or hydrogen balloons rise.

            The ideal gas law is:
            PV=nRT

            Did you notice that the pressure term is P, and not delta P?

            That’s right the pressure is constant. So why would you think your tire analogy is applicable?

            As Postma would say, that is sophistry.

          • Avatar

            Herb Rose

            |

            Hi Zoe,
            What the universal gas law states is that if you add/subtract more gas molecules or increase/decrease the temperature of the gas molecules either the pressure of the gas, the volume of the gas, or both will increase/decrease. It is not talking about the temperature of the entire gas container.
            In a scuba tank when you compress the gas into the tank the tank is submerged in water to cool it. If this is not done you get a partial fill as the pressure drops when the gas cools. The tank does not get hotter because the kinetic energy of the gas molecules increases but because there are more gas molecules (greater mass) striking the tank, transferring more kinetic energy to it. Think of the air molecules as identical hammers pounding on the walls of the container. The greater the number the more heat is transferred even when they all have the same velocity. The temperature in a valley is greater than the temperature on top of a mountain because the air is denser at the lower altitude and there are more molecules transferring kinetic energy to your skin or a thermometer. The thermometer does not measure the mean kinetic energy of the molecules but the total heat being transferred to the thermometer.
            Herb

          • Avatar

            Zoe Phin

            |

            Uh.H erb, I’m surprised John let’s you publish articles at PSI.

            P = F/A

            Area = Volume / Height

            P=Fh/V
            P=mgh/V
            P=(m/V)gh
            P=dgh

            Where d=Density

            My husband paid $700 3 years ago so we could take a hot-air balloon ride for 2.

            Given that ideal gas law has constant P for the surface air, how were we able to rise?

            Easy, P=dgh

            We lowered our density by heating the air. Since P is constant, h must increase. And that is indeed what happened. We defeated gravity by reducing our density and thereby raising our height.

            P.S. You’re an idiot.

          • Avatar

            Herb Rose

            |

            Zoe,
            In your equation m is the mass of the Earth, not your balloon, h is the height from the center of the Earth, not your altitude. Since neither of these change your reasoning is bullshit as usual.

          • Avatar

            Zoe Phin

            |

            Herb,
            m is the mass of the earth?
            You’re a nutcase.

            “h is the height from the center of the Earth”

            Well, then by that stupid reasoning, we need to use “g” for the center of earth as well, no? That would be ZERO.

            According to you the pressure is ZERO.

            I hope John recognizes that you’re mentally ill and bans you for good.

  • Avatar

    Pierre D. Bernier

    |

    Can anyone explain this to me ?

    The Inconvenient Spotlight

    A spotlight is made of a filament enclosed in a sealed vacuum globe, a back polished parabolic aluminium reflector and a front transparent glass lens. When the power is set on, the filament glows to about 3000K and emits light. We can see the light because it passes through the front lens. We can also feel some Infrared light passing through the lens because the temperature of the lens reaches about 200 C which is 473K. According to the Stefan-Botlzman law, 473K is the equivalent of emitting 2838 W/m2. This 2838 W/m2 is emitted externally and internally from the lens where it will be reflected by the back reflector. Now the lens receives both the filament energy and the internally reflected energy for a total of 5676 W/m2. If the Greenhouse Effect is true then according to the Stefan-Boltzman law this is equivalent to 562K or 289 C. Now, that lens has to emit more energy to cool off, 5676 W/m2 outward and inward which will also be reflected by the back reflector. So, now the lens receives 8514 W/m2 which is equivalent to 622K or 349 C. This continuous heating cycle must result in continuous increases in temperature causing increases in radiative emissions and it has no end. If the Greenhouse Effect is true then no spotlight should ever be operational. They will all blowup instantaneously. Since they do work then the Greenhouse Effect is not real.

    • Avatar

      AlJones1816

      |

      Pierre,

      The lens would have a total flux of 2838. Assuming flow is only moving backward and forward, and not radially along the disc (and it certainly would be), this means the “back” radiation is only half of the lens’ total surface area, or 1419 W/m^2. The lens is not a blackbody, and so it will only absorb some tiny fractional part of the 1419 W/m^2 reflected back at it (assuming 100% of the light hitting the reflector bounces back to begin with). Your repeated doublings should be halvings. First you’d have 1419 extra, then you’d have 1419+702, then 1419+702+351… etc etc. (in reality much, much less because the lens is not a blackbody). The lens will be hotter than it would be without the reflector, but it will not keep warming infinitely.

      • Avatar

        Zoe Phin

        |

        Al,
        If the reflector sends 1419 back, then emitter should get it and raise flux by 1419. The next time emitter sends to reflector, it will be 1419 higher!

        That’s the part you don’t want to understand.

        • Avatar

          Pierre D. Bernier

          |

          Thx Zoe,

          AlJones1816… I’m not saying it, the clymists are. They draw a line in the sky and half the light goes up, half comes down. Look at their diagram before commenting stupid details.

          • Avatar

            AlJones1816

            |

            Pierre,

            when you say “half the light goes up, half comes down,” you are exactly right, but that is not what you said in your previous post about the spotlight. In that post you said that 100% of the light goes forward and 100% goes backward.

      • Avatar

        geran

        |

        Sorry Al, but that’s incorrect. If a two-sided surface is emitting 2838 Watts/m^2 due to its temperature, then it is emitting 2838 Watts/m^2 from BOTH sides. It’s entire surface is emitting the same, but the total flux emitted remains 2838 Watts/m^2.

        Radiative physics can be confusing to the uninformed. That’s why so many folks are easily fooled by the GHE nonsense.

        • Avatar

          AlJones1816

          |

          Geran,

          I agree that radiative physics can be confusing. If a two sided object is emitting 2838 W/m^2 from both sides, it is emitting 5676 W/m^2. If a two sided object receives only 2838 W/m^2 on one side, it cannot emit 5676 W/m^2, it can on re-emit 2838 W/m^2 as 1419 W/m^2 from each side. Imagine the plate is 1 m^2 per side. The incoming light is 2838 W over a single square meter (hitting one side), but the outgoing light is 2838 W over 2 square meters (coming out of both sides), so 1419 W per square meter.

          • Avatar

            geran

            |

            Al states: “”I agree that radiative physics can be confusing. If a two sided object is emitting 2838 W/m^2 from both sides, it is emitting 5676 W/m^2. “

            Al, you are still confused. Radiative fluxes do NOT add. If the 2-sided object is emitting 2838 W/m^2 from both sides, it is only emitting 2838 Watts/m^2. You are confusing “flux” with “energy”.

            If each side of the object is 1 sq. meter, then it is emitting 2838 Watts from each side, or 5676 Watts total. But, it is NOT emitting 5676 Watts/m^2.

            Energy “adds”, flux does not “add”.

          • Avatar

            AlJones1816

            |

            Geran,

            Thanks for correcting me. That sentence is in error. The remainder of my post I believe is correct. If you input 100 W over one side of a disc with a surface area of 1 square meter and the disc radiates out of each side, 2 square meters of disc will radiate 100 W of energy equally, or 50 W/m^2.

          • Avatar

            geran

            |

            And thanks for admitting your error.

            The only way to learn is to admit your errors. Typically folks that have swallowed the GHE koolaid cannot admit there mistakes.

      • Avatar

        Geraint Hughes

        |

        Hey Al.

        I know an experiment that proves you totally wrong. When you get a halogen light bulb and stick it in the middle of carambola, (i.e a 3d star shaped mirror arrangement so that all light is refelcted right back to the filament emitting the light) the temperature of the filament does not rise. Read this. https://principia-scientific.org/light-recycling-disproves-greenhouse-gas-theory/ You need to know, CO2 does not induce higher surface temperatures and that greenhouse theory is a total lie.

        • Avatar

          AlJones1816

          |

          Geraint,

          The “light recycling” process you describe is indeed raising the temperature of the filament. That is entirely how it works. You try to describe it as “increas[ing] the uniformity of temperature of the filament.” But that is another way of saying it is raising the temperature of the filament.

  • Avatar

    geran

    |

    AlJones1816, the “blue/green plate” nonsense died about a year ago. So, it’s fun to see it trying to come back to life.

    For the ideal situation–perfect conductors, vacuum, black bodies, no losses, etc.–the correct solution would have the two plates at the same temperature:

    https://postimg.cc/image/jcotys8e3/

    Intutitively, it is easy to understand if the plates start in full contact. If they move a slight millimeter apart, there would simply remain the same temperature.

    A more rigid explanation involves both radiative physics and thermodynamics, which are typically not understood.

    • Avatar

      Geraint Hughes

      |

      You sound like you read my book 😀 I did an example using cubes, then splitting the cubes. Why do people think that seperating an object, will suddenly make the first back surface of the first object hotter because of some back irradiance, its bonkers? It will only be cooler, because SOME of that heat will be lost out to the sides, albiet at 1mm it will be a small loss.

      • Avatar

        geran

        |

        I haven’t read your book, but I’ve read enough of your posts and comments to know you “get it”.

        Hope you sell millions of copies.

    • Avatar

      AlaJones1816

      |

      Geran,

      the plates in contact will be exchanging heat through conduction, while the plates apart will exchange heat only through radiation (assuming they’re in a vacuum). Their temperatures will indeed be different pulled apart than they were together. One plate will now have a flow of energy from one side of the first plate, while the first plate will have a flux directly from the heat source.

        • Avatar

          AlJones1816

          |

          Geran,

          Do you see in your example that the green plate at equilibrium has a flux of 600 W/m^2? The equilibrium temperature is not 244 K, it is higher.

          • Avatar

            geran

            |

            What I see is your continued inability to understand the physics.

            You demonstrated above you did not understand flux. Let’s call that “Error 1”.

            Now, you can not understand the simple diagram. The green plate doesn’t have a “flux of 600 W/m^2”, as you stated. The green plate absorbs a net 200 W/m^2 and emits a net 200 W/m^2, just as the diagram indicates.

            Let’s call this your “Error 2”.

          • Avatar

            AlJones1816

            |

            Geran,

            Mistake #2 from my would apparently be color blindness. I meant the blue plate. The blue plate is radiating 600 W/m^2, the green plate 400.

            I believe that your 400 for the green plate is in error, though, since it receives 200 W/m^2 from the blue. The green arrow from blue to green (?) should be 50 with a corresponding left facing green 50 arrow on the left side of the blue plate.

          • Avatar

            geran

            |

            Al states: “The blue plate is radiating 600 W/m^2, the green plate 400.”

            Sorry Al, but you’re wrong again. Both blue and green plates are emitting 200 W/m^2, just as the simple diagram indicates.

            That’s really two more errors, but let’s just call it “Error 3”.

            The rest of your comment is so nonsensical I can only consider it “Error 4”.

            I suggest you earnestly study the simple diagram, and ask responsible questions if necessary, before making more erroneous statements.

          • Avatar

            Zoe Phin

            |

            Al,
            Seriously, why would any self-respecting person believe a theory that originated on a blog by a cowardly scientist that can’t even use his own name?

            His crackpot theory is not in a textbook, not in a peer-reviewed paper, and has no experimental support.

            Why would you believe this moron?

          • Avatar

            AlJones1816

            |

            Geran,

            The net flux to “space” in the diagram is 400 W/m^2, but the plates are exchanging energy before it is emitted to space. You can’t simply cancel out the flux between the two plates. You have the blue plate receiving 400 from the heat source, and you have it emitting a blue 200 arrow from either side. This is right. Next you have the green plate receiving 200 from the blue arrow, but you have it emitting 200 from green arrows on either side. This is wrong. Those green arrows should be 100 each. You then have to balance this out by sticking in another 200 green arrow being emitted from the blue plate. This is also wrong. You’ve made energy appear out of nowhere.

            Because I have too much time on my hands, I’ve made you an illustrative gif. It starts at time t=0, when the heat source to the left is “switched on” and ends at t=n-1, shortly before the plates reach equilibrium. The energies all balance out as you will see, but the flux is affected by the presence of the second plate:

            https://gfycat.com/villainousunconsciousblackbird

          • Avatar

            geran

            |

            Al’s Error #5: ”Next you have the green plate receiving 200 from the blue arrow, but you have it emitting 200 from green arrows on either side. This is wrong. Those green arrows should be 100 each.”

            Wrong again, Al. A surface emits based on its temperature. The temperature is 244K. The emission is 200 W/m^2, as indicated in the simple diagram.

            Al’s Error #6: ”You then have to balance this out by sticking in another 200 green arrow being emitted from the blue plate. This is also wrong. You’ve made energy appear out of nowhere.”

            Wrong again, Al. The green arrow coming from the blue plate is the reflected flux. The net flow from blue to green is the blue arrow, 200 W/m^2.

            For someone that claims to be color blind, you somehow can tell the difference between blue and green. But you are still confused by the physics.

            And your GIF violates the laws of physics by requiring the blue plate to absorb the flux from the downstream green plate. In simple terms, a canoe does not float upstream. Let’s call your GIF “Error #7”.

          • Avatar

            AlJones1816

            |

            Geran,

            The temperature of the blue plate would not be 244 K, someone has simply drawn the number beneath it. The blue plate would indeed radiate based upon its temperature.

            Just to clarify, because it is not obvious at all, the back side of the blue plate in your diagram is a mirror? It is not a blackbody? And the front is a blackbody? I’m not familiar with the obviously storied history behind your diagram’s construction.

            How is the blue plate absorbing photons from the green plate a violation of any physical law?

          • Avatar

            geran

            |

            Al’s Error #8: ”The temperature of the blue plate would not be 244 K…”

            The blue plate has to emit 200 W/m^2 from both sides. The only temperature that will do that is 244 K.

            ”Just to clarify, because it is not obvious at all, the back side of the blue plate in your diagram is a mirror? It is not a blackbody? And the front is a blackbody?”

            The blue plate is a black body. A black body is an imaginary concept. By definition, a black body absorbs all incoming flux. BUT, an imaginary concept does NOT get to violate the laws of thermodynamics. If the incoming flux has incompatible wavelengths, the incoming flux will be reflected.

            ”How is the blue plate absorbing photons from the green plate a violation of any physical law?”

            Both plates are identical black bodies. If the green plate is downstream in the energy flow, its emitted photons will be incompatible with the upstream surface. “Cold” can not warm “hot”.

          • Avatar

            Zoe Phin

            |

            Al,
            You’re a big joke.
            In your gif you show 100,50,25,12.5,6.25,….

            That’s all a subset of the SAME 200 W/m^2 you sent out to green plate.

            Since they are all SUBSETS, you can’t just add them to 200+50+6.25 !!!

            Do you understand that?

            You’ve progressively cut smaller slices of the same cake and now you believe you have more cake!

            LMAO

            Now do you see that the blue plate emits 266 on both sides?

            That’s 533.33 from 400 !!!

            You violated the 2nd law.

            Have you learned anything?

  • Avatar

    AlJones1816

    |

    Geran, the blue plate is emitting 200 W/m^2 from both sides because you’ve imposed a nonphysical condition about the behavior of a black body. A black body is specifically defined as an “idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.” For some reason your black body blue plate is acting like a perfect mirror on one side and reflecting 100% of the radiation it receives in a specific range of wavelengths.

    How can photons be “incompatible?” This is not a physically realistic assumption you are making.

    • Avatar

      geran

      |

      Errror #9: “the blue plate is emitting 200 W/m^2 from both sides because you’ve imposed a nonphysical condition about the behavior of a black body.”

      No Al, the blue plate is emitting 200 W/m^2 because it is at a temperature of 244 K. This is just very basic physics. If you can’t do the calculation, there are online calculators that can do it for you.

      Error #10: How can photons be “incompatible?” This is not a physically realistic assumption you are making.

      It’s not an “assumption”, Al. It is established physics. If photons are not compatible with the absorber, they will not be absorbed. That’s why you can see a tree. The tree is reflecting visible wavelengths that are not compatible.

      You now have 10 errors, but have only acknowledged one. It’s time for you to admit your don’t know crap about the subject, and demonstrate a willingness to learn. Otherwise, I’m just wasting my time.

      • Avatar

        AlJones1816

        |

        Geran,

        The plate’s temperature is set by the irradiance needed to balance incoming energy. The blue plate doesn’t need to balance 200 W/m^2 from the sun, it needs to balance 400 W/m^2 from the sun.

        “It’s not an “assumption”, Al. It is established physics. If photons are not compatible with the absorber, they will not be absorbed. That’s why you can see a tree. The tree is reflecting visible wavelengths that are not compatible.”

        Can you describe for me it is then that I can see trees when it’s cold outside? Surely the cold tree’s photons are incompatible with my hot eyes.

        • Avatar

          AlJones1816

          |

          Oops, just made mistake #3 – I see the reflected light from the tree, not emitted. Add one to the counter.

        • Avatar

          geran

          |

          Obviously you overlooked my last paragraph:

          You now have 10 errors, but have only acknowledged one. It’s time for you to admit you don’t know crap about the subject, and demonstrate a willingness to learn. Otherwise, I’m just wasting my time.

          • Avatar

            AlJones1816

            |

            Geran, I’m happy to acknowledge errors when they occur, which is why I count 3 in this thread. The remaining 7 you allege are not errors on my part. To reiterate:

            The plate’s temperature is set by the irradiance needed to balance incoming energy. The blue plate doesn’t need to balance 200 W/m^2 from the sun, it needs to balance 400 W/m^2 from the sun.

            At equilibrium Ein = Eout, therefore:

            Ein = 5.6710^(-8) W/(m^2K^4)T^4 with T in units of K

            T^4 = Ein/(5.67*10^(-8)

            T = (Ein/(5.67*10^(-8)))^(1/4)

            For Ein = 400 W/m^2:

            T = (400/(5.67*10^(-8)))^(1/4) = 290 K

            Let me know if there’s an arithmetic error in there but otherwise I think we’ve concluded this branch of the discussion.

          • Avatar

            geran

            |

            Al, your incompetence matches your inability to learn.

            You don’t even understand how the simple S/B equation applies. If the plate is receiving 400 Watts, it must emit 200 Watts from each side, at equilibrium. Since its sides are 1 square meter, it is emitting 200 W/m^2.

            That corresponds to a temperature of 244K.

            Learn some physics, or remain a clown, your choice.

          • Avatar

            AlJones1816

            |

            Geran,

            The mistakes do keep racking up, number 4 now. You are correct, the equilibrium temperature of the single plate would indeed be 244 K. Thanks again for continued patience and for pointing out errors while I catch up.

            Do the math using the gif I made, with 265.625 W/m^2 per side of the blue plate at equilibrium and you will get an equilibrium temperature of ~262 K. The presence of the green plate makes the equilibrium temp of the blue plate higher.

            Slowly but surely we are converging on the answers.

          • Avatar

            geran

            |

            Wrong again, Al. That’s now Error #12.

            You keep making mistakes that reveal your ignorance of the relevant physics. But instead of realizing that your beliefs are wrong, you just keep plunging ahead.

            Your devotion to the incorrect plates solution reveals your faith in a false religion. And your continued efforts to pervert and corrupt reality just makes you another climate clown.

            You need to take responsibility for your 12 errors and demonstrate an interest in learning. Otherwise, I’m just wasting my time.

          • Avatar

            AlJones1816

            |

            Geran,

            I believe I’ve happily admitted to each of my errors I’ve identified. The errors have not been errors in the overall premise of my argument, nor its conclusions. and so the overall point I’ve made still stands. I can be wrong about some thing and right about others.

            I think that the discussion is more productive if we treat each other with mutual respect, instead of you taking a patronizing attitude toward me. If that doesn’t seem like anything you’re interested in I’m happy to conclude here.

          • Avatar

            geran

            |

            Of course you want to “believe” that nonsense, Al. But your beliefs do not align with reality.

            I called them “errors”, but I was being nice. In reality, you got caught trying to deceive. You were trying to pervert and twist physics to match your beliefs. You claim that you are willing to learn, but you provide no evidence of your claims.

            Admit, and learn from, your 12 errors, or waste someone else’s time.

          • Avatar

            AlJones1816

            |

            Geran,

            thanks for your time so far, I don’t think the discussion has anywhere productive left to go. If you ever want to discuss the science some more I would be happy to do that. I’m not interested in responding to personal attacks.

          • Avatar

            geran

            |

            Try to spin your way out of it, Al.

            But your record here speaks for itself.

          • Avatar

            Zoe Phin

            |

            Al, you pathetic dumbf*ck, not only does the whole system need to follow the 2nd law, but so does every SUBprocress.

            The blue plate can not emit 533 Joules every second from the 400 Joules it receives in the same second.

  • Avatar

    AlJones1816

    |

    Zoe,

    the blue plate receives 400 W/m^2 and the blue + green plate emit 400 W/m^2 – energy is conserved. Energy can bounce around within the 2 plate system without violating the 2nd law. The blue plate has a flux higher than 400 W/m^2 because some of the energy it “sends off to space” is being returned back to it from the green plate.

    • Avatar

      Zoe Phin

      |

      Al,
      “Energy can bounce around within the 2 plate system without violating the 2nd law.”

      Aww, how cute. You just fabricated your own physics.

      “The blue plate has a flux higher than 400 W/m^2 because some of the energy it “sends off to space” is being returned back to it from the green plate.”

      Returning back energy does not add energy. Everything you return back is just a subset of the original. You can’t ADD a subset onto the original. But that is exactly what you do: 200+50+6.25+…

      Energy doesn’t accumulate by bouncing around between things in a zig zag fashion.

      • Avatar

        AlJones1816

        |

        Zoe,

        Returning back energy adds energy because the sun is a constant heat source. When the body emits a joule of energy it does not contain N-1 joules of energy because the sun immediately replaced the lost joule, it contains N joules. If the green plate sends a joule back to the blue plate it now has N+1 joules.

        • Avatar

          Zoe Phin

          |

          No, Al,
          The problem is about what happens to every 400J that is sent. You can not bungle it up with the next 400 J. You are fouling up every single 400 J.

          Since your timer keeps adding a second, nothing changes. The whole bungling needs to be divided by # of seconds.

          You can not escape the fact that you’re adding a subset of energy to the the original energy, PER second.

          Again, I ask you: why would any self-respecting person believe a theory that originated on a blog by a cowardly scientist that can’t even use his own name?

          His crackpot theory is not in a textbook, not in a peer-reviewed paper, and has no experimental support.

          Why would you believe this moron?

          • Avatar

            AlJones1816

            |

            Zoe,

            I think the confusion is coming from the idea of adding “a subset of energy to the original energy.” This is not what is happening. You are adding a subset of the original energy to new energy. The blue plate is receiving a constant flow of energy from the sun. It is never receiving less than 400 W/m^2 from the sun. It never has less than 400W/m^2. If it receives some radiation from the green plate, it is receiving more than 400 W/m^2 total. We aren’t talking about discrete steps but about constant flows.

            “Again, I ask you: why would any self-respecting person believe a theory that originated on a blog by a cowardly scientist that can’t even use his own name?”

            I don’t think this is Eli Rabbet’s theory, I think this is basic physics and math. Eli Rabbet just provided a nice illustration for us to look at.

          • Avatar

            Zoe Phin

            |

            Al,
            “This is not what is happening. You are adding a subset of the original energy to new energy.”

            Your gif doesn’t show “new” energy.

            Now you’re mixing 400J from the first second with 400J of the next second. But you’re going to do the same thing with the next 400J, you did with the first.

            You can do this a million times, but don’t forget to divide by a million seconds – and you’re still gonna end up with an average of what you did with the first second – ADDING a subset of energy back onto the original.

            “I don’t think”
            First honest thing you’ve said.

            “I think this is basic physics and math.”

            Oh really? Basic physics is found in textbooks. What textbook did Eli use?

            Eli apparently doesn’t give credit from where he copied his idea from, nor does he use his actual name: Joshua Halpern.

            Hmm, a fraudster who doesn’t use his own name nor give credit. What an effing joke.

            I pity those who fall for his fraud.

          • Avatar

            Zoe Phin

            |

            Al,
            “You are adding a subset of the original energy to new energy.”

            You are incredibly stupid.

            For 1 second you add:
            200+50+6.25

            For 10 seconds you will add:
            2000+500+62.5

            For 100 seconds you will add:
            20000+5000+625

            For n seconds you will add
            200n+50n+6.25n

            You only have 400n/2 to work with for n seconds, and you always end up sending more than you have to green plate.

            Why? Because you’re adding a subset of the original energy back onto the original energy.

            You simply don’t want to acknowledge reality, because you love the conclusions of your pseudo-science.

          • Avatar

            AlJones1816

            |

            “Your gif doesn’t show “new” energy.”

            It does show this. A Watt is a joule per second – it represents a flux. If the pate is 1 square meter, every second the sun puts in a whole new 400 joules of energy that weren’t there before. The plate radiates out 400 joules of energy. There is never a time after we turn the sun on when the plate contains fewer than 400 joules of energy. If we add the second plate and it starts returning some of the radiation, there will never be a time when the plate has less than 400 + green plate’s joules.

            Take your garden hose and turn it on, let a nice stream develop across your yard. Now dig a hole underneath the stream. After the hole fills with water and the stream starts running again does the hole shed water and refill again every second, such that the water level is constantly fluctuation? Or does it maintain a constant volume of water within itself? This is the concept of equilibrium.

            The questions that you have for Eli I am afraid you will have to take to the man (rabbit?) himself.

          • Avatar

            Zoe Phin

            |

            Al,
            “It does show this”
            Don’t lie.

            “whole new 400 joules of energy that weren’t there before. The plate radiates out 400 joules of energy”

            What are you talking about?
            The sun is far away, the plates and space are right next to each other. It’s faster to drain your plate system then to receive another batch from the sun, but we can pretend that the plates are drained JUST before another batch arrives.

            You are gaining 400J every second, and losing 400J WITHIN the SAME second, just before a new batch arrives.

            “There is never a time after we turn the sun on when the plate contains fewer than 400 joules of energy.”
            Then you’re not emitting 400J to space.

            “Take your garden hose and turn it on, let a nice stream develop across your yard. Now dig a hole underneath the stream. After the hole fills with water and the stream starts running again does the hole shed water and refill again every second, such that the water level is constantly fluctuation? Or does it maintain a constant volume of water within itself? This is the concept of equilibrium.”

            What are you smoking?

            Imagine the hole is 400mL.

            The first problem is that your original problem stipulates that the hole empties every second.

            The second problem is that you’re arguing that the back-runoff causes the hole to become 533mL.

            You’re so desperate to defend your religion that you feel no shame saying whatever gibberish you need.

            Don’t you understand what you look like?

          • Avatar

            Zoe Phin

            |

            Al,

            “The questions that you have for Eli I am afraid you will have to take to the man (rabbit?) himself.”

            Wait. What? YOU claimed it’s basic physics. That must mean YOU saw it in a textbook.

            If you are just repeating Eli’s words that it’s basic physics, then you’re an idiot repeating a liar who is now covering for a liar.

  • Avatar

    AlJones1816

    |

    Zoe,

    The sun doesn’t fire out discrete packets of energy like a t-shirt cannon every second, it’s a continuous stream of photons. The sun doesn’t wait until the blue plate empties itself before supplying more energy, the blue plate has to keep shedding energy or else it’s not in equilibrium. The presence of the green plate means that to stay in equilibrium the blue plate has to shed the sunlight and the light from the green plate, so its equilibrium temp is higher.

    • Avatar

      Zoe Phin

      |

      I replied in the wrong thread. Read my comment below.

      If your system is accumulating energy, then it’s not sheding 400J per second.

      At time 0, system receives 400Joules.
      At time 0.999999s, the system has already sent out 400Joules.

      If the system gets rid of its old energy any time after 1 second, then it’s not emitting 400 Joules per second.

      You can slice and dice this all you want, and it still scales to prove you wrong.

      Read more below.

  • Avatar

    Zoe Phin

    |

    Al,
    “The sun doesn’t fire out discrete packets of energy like a t-shirt cannon every second, it’s a continuous stream of photons.”

    A continuous stream of photons that adds up to 400J per second.

    “The sun doesn’t wait until the blue plate empties itself before supplying more energy”

    The blue plate doesn’t wait for the sun to supply it more energy. It’s required to stream 400J per second.

    “and the light from the green plate”
    The green plate doesn’t emit nee raw energy, but a part of the 400J yhe blue plate already had.

    Gosh you’re an imbecile. If you want, you can run the analysis for 1/400th of a second with 1 Joule, and you still make the same mistake.

    The system sheds energy as fast as it receives it. There is no accumulation of anything, you filthy lying idiot.

    • Avatar

      AlJones1816

      |

      Zoe,

      Please be polite. I have not insulted you or been rude to you throughout our discussion. Please show me the same courtesy.

      The system is indeed shedding energy as fast as it receives it, and it is receiving energy from the sun and energy from the green plate, so it has more energy to shed than it would if it were receiving energy from the sun alone and so has a higher equilibrium temperature.

      • Avatar

        geran

        |

        Al, you are STILL spreading falsehoods. You haven’t a clue about the relevant physics. You just try to distort things to fit your false beliefs.

        The energy back from the green plate is already “in the system”. It is not able to raise the temperature of the system beyond the 244 K.

        No wonder people get frustrated with your refusal to learn and your avoidance of reality.

        • Avatar

          AlJones1816

          |

          geran,

          The green plate is elevating the equilibrium temperature of the blue plate. I showed this conclusively in the gif I provided earlier. I don’t think it is physically meaningful to say that one object cannot affect another’s temperature merely because they are in the same system.

          • Avatar

            geran

            |

            Wrong again, Al.

            You did not show it “conclusively”. Your arithmetic was correct, but your physics was incorrect. You don’t understand the physics, and you refuse to learn. When you violate the laws of physics, you MUST backup and correct your thinking.

            The correct solution does not violate any laws of physics:

            https://postimg.cc/image/jcotys8e3/

          • Avatar

            AlJones1816

            |

            Geran,

            the “correct” solution you provide does not violate any laws of physics (from my understanding of the diagram), it just imposes the condition that the back side of the blue plate be a perfect mirror. My solution does not violate any laws of physics, it just models the back side of the blue plate as a black body, just as it models the front side as a black body.

            There is no law of physics I am aware of that says, “an object cannot absorb any wavelength of light that is emanated from a colder object.” If the colder object is radiating light in the wavelengths that the hotter object absorbs in, then saying that nothing happens to the hotter object is to say that energy is absorbed and then just vanishes, and this would violate the second law. You get around this by imposing the “perfect mirror” constraint, but this is a non-physical condition (there is no perfect mirror in nature that I’m aware of) that you are imposing simply to avoid an otherwise inescapable conclusion.

          • Avatar

            AlJones1816

            |

            I suppose a good question to answer would be, “can a thermal imaging camera detect radiation from an object cooler than the camera’s sensor?” (i.e. could I point a thermal camera out the window of my heated home and capture an image of the outside in the winter?)

          • Avatar

            geran

            |

            No Al, that is a irrelevant question. It indicates, once again, your ignorance of the issues. Until you learn the basics, you are not ready to understand thermal imaging cameras.

            You’re still confused by the simple diagram. You can’t learn. You’re “stuck on stupid”.

          • Avatar

            AlJones1816

            |

            Geran,

            you once again decided to forgo civility and mutual respect and make personal attacks in a condescending tone. You should try to be a better person than that. I think that this concludes our discussion.

          • Avatar

            geran

            |

            Trying to play the abused one, again?

            You try to pervert and corrupt physics, get caught at it, then try to feel sorry for yourself.

            You’re not fooling anyone but yourself, clown.

      • Avatar

        Zoe Phin

        |

        Al, you repeat the same false narrative regardless of how many times you are shown to be wrong. You don’t deserve courtesy. You deserve contempt. You are a low life.

        • Avatar

          AlJones1816

          |

          Zoe,

          if you are unable to be civil I think we can conclude this discussion. If ever you change your mind and want to talk about the science in a mutually respectful manner I’ll be happy to do that.

          • Avatar

            Zoe Phin

            |

            Al, you ignore criticism and simply regurgitate the same story in the vain hope that asserting last makes you look right.

            That is the definition of being uncivil. You are now trying to play the victim.

            There’s nothing to discuss with you. The green plate thought experiment is a well known internet hoax.

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