Postma’s Climate Model Paper – Discussion

Here on PSI we welcome reader discussion on Joseph E Postma’s newly-published paper, ‘An Alternative Global Mean Energy Budget Model Which is Incompatible with Existing Ones.’ The paper has been accepted for review at Bulletin of the American Meteorological Society.

Any questions for Joe should be directed to him at his blog, climateofsophistry.com.

Over on Facebook, Joe is engaging in lively discussion with fellow skeptics. Here is his latest:

“A response to my paper which you guys might like to read. This isn’t from official peer-review, just another anonymous review that someone else gathered for me:-

“One of Postma’s criticisms of it is that surface radiation out and back radiation by GHGs are about twice as large as the insolar reaching the earth directly.

His hemispherical model gives 4X higher average heat flux intensities and the consequently calculated black body radiation temperatures are closer to the global actual average of about 15C.”

The hemispheric model does not merely give 4X higher heat flux…it gives the real, actual, empirical, and theoretical astrophysical, heat flux. This is a mutually exclusive difference to the fictional -40C solar input of the IPCC or K&T scheme. And yes, this is the fundamental difference to think about: is the climate created mainly by the 2X greater backradiation than solar input, or, is it created by the solar input. This is a mutually exclusive difference of absolute magnitude.

“However, the IPCC spherical heat flux numbers are only for a schematic illustration of the heat balance, and are not actually used literally in the calculations, so the difference in the input heat fluxes is irrelevant.”

As a schematic illustration of the heat balance, it represents something that doesn’t actually exist and does not actually occur. It is a schematic of a fiction. Thus, what relevance does it have to understanding reality? The fact of the matter is that these flat-Earth solar-averaged diagrams exist as schematics for understanding the supposed “heat balance” therein, and as such, they are merely schematics of fiction for understanding a fiction. Further, these diagrams (I referenced 3 in total) exist as the fundamental derivations for the concept of a climate “greenhouse effect” via back radiation. They are fundamental to deriving the concept of the climate greenhouse effect, and its functioning via back radiation 2X stronger than solar heating. But they are fictions, representing a fictionally cold sunshine diluted over the entire surface area, which is empirically impossible and certainly not consistent with astrophysical theory and how sunlight falls onto planets.

“And, although they are closer, Postma’s equivalent temperatures do not agree well with reality either.”

The difference between my diagram and the other is binary: in one, we would never experience warm sunshine let alone sunshine creating climatological/meteorological responses, whereas in the other we do get that effect from sunshine. That’s the point.

“Question: is the temperature forced on a black body at equilibrium by a given heat radiation flux the same as what a black body would be at radiating the same heat?”

My diagram and paper makes it clear: No. The real-time input is hot and non-uniform, and sunshine can directly create meteorological responses. The total energy in and out is the same, however, one cannot use the output flux as if it is equivalent to the input flux because the input flux is capable of producing physical responses that the output flux cannot. The total energies are equal, but the fluxes are not, and it is the local real-time flux which determines physical responses..

“One way heat transfer from hot to cold is a limit for conduction and convection. But radiation from the earth that excites GHG atoms can be and is re-emitted in all directions, including some back to earth at the same frequency and wave length.”

This is a common claim which is used to defend the climate greenhouse effect of “back radiation”. Let it be clear, and you read the quotation about heat flow in the paper: there is no caveat in the discussions of heat flow which state “except for radiation”. You will not find a caveat of except for radiation in any thermodynamics textbook. If you consider heat flow via conduction of molecules in a lattice, say down a metal rod, the molecules on the cooler side of the gradient also vibrate and bounce randomly in all directions thus including in the direction of the warmer side of the gradient, yet, the heat flow is only down the gradient. This is because the molecules on the cooler side of the gradient do not have the increased multiplicity they would require to send heat up the gradient. Further, conduction is mediated in any case electromagnetically by virtual photons between electrons in the lattice, and radiant heat is also mediated electromagnetically but by photons which escape the lattice on the exposed surface. It’s the same force of physics with any mode of heat transfer: electromagnetism. This is why when heat is discussed, the discussion regards heat as a universal concept where the properties of heat apply to all modes of heat transfer. There is no caveat which allows radiant heat transfer to behave oppositely of any other mode of heat transfer: it is from hot to cold only. Just as with conduction, when the photons can escape via the exposed surface instead of being trapped inside the lattice as “physical” conduction, the photons emitted from a cooler object do not have the multiplicity they would require to increase the temperature of the original source of heat.

“In any case, the back radiation is less than the outgoing so there is an increase of entropy as required by the Second Law.”

It is not only the second law, it is also the first: the first law states that either heat and/or work is required to increase temperature. The cooler atmosphere has no heat to send to the surface…let alone 2X the heat of the Sun. The Sun heats the Earth, and the heat continues to flow down the temperature gradients, creating meteorological and climatological effects along the way – this is so simple, but has been entirely destroyed as a concept by the diagrams and derivations of the climate greenhouse effect as discussed in my paper. They are just making a claim. The cooler atmosphere does not have the greater multiplicity it would require to cause an increase of temperature of the surface itself…and this idea only originates in the first place in the models where the Sun does not create the climate.

“An analogy would be dropping a rubber ball on a hard surface and for it to bounce up and back down, again and again, less high each time, until it finally came to rest. Each time on the way down it loses potential energy and gains entropy, and vice versa on the way back up, always less each time, so at the end there has been a gain of entropy, and energy has gone into the ball as heat which it then loses with more gain of entropy.”

What the greenhouse effect requires is that after the initial impulse of the ball drop, supplied by the Sun first raising the ball, then the ball bounces higher afterwards because of “back-bounce”.

Now with all that covered…this email gave me an aneurism. I do not understand what it is with these people’s inability to think, to reason, and to appreciate logical facts. I even had a PhD in physics at my university look at the paper, and then they replied with “Doesn’t the K&T diagram and yours reduce to the same thing?”

Did they not read the table? The entire point is that there are mutually exclusive concepts at play here, and thus both concepts cannot both be correct! Is the Sun so feeble that it cannot create meteorological and climatological responses directly, and so back radiation of 2X solar intensity is required to create the climate? Or does the Sunshine directly create meteorological and climatological responses due to its heat, with heat then flowing down all gradients. I do not know why this is so difficult for people to understand. One concept is based on a fictionally-cold sunshine spreading over an entire spherical surface at once, which is impossible by basic optical theory, astrophysics, thermodynamics, and geometry. The other is empirically and theoretically consistent.

Do people reason so poorly that they truly cannot cogitate, cannot comprehend in their mind’s eye and in their internal mental conceptual apparatus, the difference between cold fictional sunshine unable to create a climate vs. intense real sunshine which does? I’m flabbergasted that otherwise educated people cannot comprehend the relevance of basic things. Hence the end of the paper:

“The reader is left to consider whether or not it is relevant, or useful, or at all scientific, to value such differences.”

And so I suppose that the answer is that the reader is unable to consider and cogitate upon such differences. It simply eludes them. I assume that people are capable of thinking what I can demonstrate…but perhaps this is similar to expecting people to swim like an Olympian after the Olympian demonstrates it to them.

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Comments (225)

  • Avatar

    Barry

    |

    What a common sense approach to climate after reading this I would encourage all lay people to do the same. This is exactly the kind of papers we need to make common people understand how bizarre the ghg theory of the IPCC is. It’s nice that they call it the greenhouse effect as it makes it easy to understand if you think of ghg as a greenhouse blanket repelling some of the hot sun during daylight hours and slowing the cooling in the dark. Basically only cooling and never adding energy to the climate. This is just my own simple explanation of what I believe.

    Reply

  • Avatar

    chris

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    I don’t think that the IPCC is using flat Earth, I think that it’s worse than that. They state that the power from the sun is equal on all surfaces of the Earth at the same time. Since the Sun is a sphere then the Earth must be concave when viewed from the vantage point of the Sun. So they are using some kind of bowl Earth theory.

    Reply

  • Avatar

    Randy

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    From what I have seen they divide the flux by 2 as we’re dealing with a hemisphere that’s hit by radiation at any point in time. Then they divide by 2 again as the incidence angle goes from perpendicular to 90 degrees at the equator… so they just halve it again.

    Reply

    • Avatar

      Randy

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      Rather, 90 degrees incidence (full flux) to 180 degrees (zero) hence halved again. This would seem logical but may be a gross approximation. I’m not sure it’s a time average over 4pi r2, as mentioned here, it just works out the same.

      Reply

  • Avatar

    Randy

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    Regards the radiative gradient, I guess back radiation cannot increase MAX temperatures (add net heat to a warmer earth) but may increase AVG temps over a time window (decreases the net outbound flux)? This is not adding net energy, just delaying the cooling. Back radiation does not know about the outbound radiation, so will it not slow the rate to equilibrium, as a net balance, by ~8W/m2 or somesuch (decreasing with the cooling rate to ~zero).?

    Reply

    • Avatar

      Rick

      |

      Looking at the literature the CO2 back radiation region yields ~ 0.08W/m2 which is ~nothing? This would not slow down radiative cooling much at all.. and certainly would not add net heat energy down the gradient.

      Reply

    • Avatar

      Alan

      |

      You are spot on with this comment. If thermal energy loss is slowed then the average temperature will increase, but does not mean there is more energy input. This is why the use of average temperature and increases to it are meaningless. It is the energy input that will warm the oceans and melt ice and if the temperature stays higher for longer there is no additional energy so no effect.

      Reply

  • Avatar

    Matt Dzialak

    |

    A manuscript that has been accepted for review is not yet published literature. I certainly hope it finds its way to publication – but it still faces and uphill battle. Best of luck!

    Reply

    • Avatar

      Matt Dzialak

      |

      …an uphill battle. Not and. Damn autocorrect – it’s my enema.

      Reply

  • Avatar

    geran

    |

    For those that are not familiar with Joseph Postma, he is a great source for Skeptics who want to learn more about the actual science. He has put out several videos, now on youtube. Or, he will answer questions on his blog. He is very patient and a good teacher, if someone wants to learn.

    https://www.youtube.com/channel/UCDK7p7ivYprBgjDNvYwmk5Q/videos

    Reply

  • Avatar

    Daddy

    |

    What is the blackbody emission spectrum if CO2?
    Or do those rules not apply to matter in a gas state?

    Reply

  • Avatar

    Daddy

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    What is the blackbody emission spectrum if CO2?
    Or do those rules not apply to matter in a gas state?

    Reply

  • Avatar

    Brian James

    |

    Jan 23, 2020 Banksters Warn of Green Swan Collapse Unless They Get A Carbon Tax

    Welcome to New World Next Week — the video series from Corbett Report and Media Monarchy that covers some of the most important developments in open source intelligence news.

    https://youtu.be/V2h27ud3Oks

    Reply

  • Avatar

    Zoe Phin

    |

    I don’t see any explanation in Postma’s work why surface is ~15C.

    Is he double counting radiation in AND out?

    Stefan Boltzman Law says we can only count it on the way out.

    A hemisphere receives 480 W/m^2 over 12 hours. 240 W/m^2 is emitted during those same 12 hours, and 240 W/m^2 is retained in the surface/atmosphere matter to be retained for the 12 hours of night. At night, those 240 W/m^2 come out.

    No flat earth physics, only 12/24 hour double hemisphere physics.

    Only 240 W/m^2 is emitted during the day and during the night, hence there is not enough to create 15C … unless Postma is double counting the energy on its way in. That’s just wrong, but I don’t want to excuse him of that, and yet it’s the only way he could think he has enough energy to explain surface 15C.

    -Zoe

    Reply

    • Avatar

      geran

      |

      Zoe, your remarks indicate you don’t understand any of it.

      Joe’s model results in an average temperature of 30C, indicating there is plenty of energy to support the actual 15C.

      Reply

      • Avatar

        Zoe Phin

        |

        No, Geran, he doesn’t have 30C.

        Stefan Boltzmann Law is for EMISSION ONLY.

        If he wants 30C, he has to emit 480 W/m^2.

        If he emits 480 W/m^2 during the 12 hours that 480 W/m^2 comes in, he has 0 W/m^2 left over for the 12 hours of night.

        You’re only verifying what I’ve been saying:
        Postma believes he has 30C coming in and -18C going out. He’s probably double counting. If he wants 30C during the day, then he has 0 kelvin for the night.

        I’ve repeatedly explained this. Why is it not sinking in?

        Reply

        • Avatar

          geran

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          The 30C would be the equilibrium for a flat surface, insulated on the back, absorbing 480 W/m^2.

          As I said, you don’t understand any of this.

          Reply

          • Avatar

            Zoe Phin

            |

            For 12 hours, Geran.
            For 12 hours you can get your equilbrium of receiving and sending 480 W/m^2.

            Now you got nothing for the 12 hours of night.

            HUGE FAIL.

            As you said, you don’t understand any of this.

          • Avatar

            geran

            |

            Zoe,

            12 = 24/2

            480 = 960/2

          • Avatar

            Zoe Phin

            |

            Geran,
            960 is TSI * (1-albedo) at SOLAR ZENITH

            Postma claims a hemisphere receives 480 W/m^2. That’s what we’re discussing.

            If you want to play with 960 for an even smaller fraction of the surface, you will run into the same problem.

          • Avatar

            geran

            |

            The only problem is your inability to understand the simple model.

          • Avatar

            Zoe Phin

            |

            My inability yon understand the model?

            From Postma’s paper:

            “On the other hand, this alternative global energy budget supplies solar power over only a hemisphere which sunlight ever falls upon, giving an average forcing of 480 Wm-2 or temperature forcing of
            105 +303K (+30C, 86F), but which maximizes around the zenith at 960 Wm-2 ”

            See 480 hemisphere, not 960 hemisphere.

            But you probably already know what you’re doing!

          • Avatar

            geran

            |

            It’s just simple accounting, Zoe.

            960 – 480 = 480

          • Avatar

            Zoe Phin

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            “It’s just simple accounting, Zoe.
            960 – 480 = 480”

            Why are you subtracting a hemisphere average from an infinitesimally small zenith maximum?

            What is the significance of that?

          • Avatar

            geran

            |

            Why can’t you understand a simple model?

            (I know the answer.)

          • Avatar

            Zoe Phin

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            Geran,
            You can do all the sophistry you like. You’re not fooling the wise.

          • Avatar

            geran

            |

            Why learn when you can just make up stuff about others?

            Whatever floats your boat.

          • Avatar

            geran

            |

            How many hemispheres does a sphere have?

            480 + 480 = 960

          • Avatar

            Zoe Phin

            |

            Geran,
            One hemisphere gets 480, for 12 hours, then the other gets 480 for 12 hours. Each hemisphere has 12 hours of night. This process repeats.

            What’ your deal?
            You think the earth is flat or there’s two suns?

          • Avatar

            geran

            |

            After all my clear explanations, you still have no clue.

            Hilarious.

      • Avatar

        Zoe Phin

        |

        Geran
        What’s the average for 30C in the day, and -273C during the night?

        Is it 15C?

        LMAO

        Reply

        • Avatar

          geran

          |

          See Zoe, your lack of understanding leads to nonsense like that.

          Reply

          • Avatar

            Zoe Phin

            |

            Postma wastes his equilibrium for 12 hours of day. He needs 0C during the night to get a 15C average, but he has 0 kelvin left for the night.

            If Postma had good accounting skills like I do, he wouldn’t keep embarassing himself. I tried to help, but he banished me. I don’t want him to wear the DUNCE CAP, but he keeps insisting it’s good fashion.

          • Avatar

            geran

            |

            Zoe, where do you see 0K at night?

            Is that your imagination?

            Maybe your accounting skills are in your imagination also….

          • Avatar

            Zoe Phin

            |

            OK, Geran, tell me after receving and sending 480 W/m^2 for 12 hours of day, what quantity is left over for the night?

            Tell me, please!

          • Avatar

            geran

            |

            Zoe, you claimed you understood accounting. So if the disk receives 960, and 480 is applied to 1 hemisphere, where does the other 480 go?

            You just can’t understand a simple model.

          • Avatar

            Zoe Phin

            |

            Geran,
            The disk doesn’t receive 960. 960 is received by an Infinitesimally small ZENITH location.

            You’re now acting as if the earth receives 960 on one hemisphere. That’s what your new contrived math shoes. How dishonest of you.

          • Avatar

            geran

            |

            Wrong again, Zoe.

            The disk receives 960 W/m^2, after adjusted for albedo.

            How incompetent of you.

          • Avatar

            Zoe Phin

            |

            From Postma’s paper:

            “On the other hand, this alternative global energy budget supplies solar power over only a hemisphere which sunlight ever falls upon, giving an average forcing of 480 Wm-2 or temperature forcing of
            105 +303K (+30C, 86F), but which maximizes around the zenith at 960 Wm-2 ”

            See?
            480 hemisphere, not 960 hemisphere.

            But you probably already know what you’re doing!

          • Avatar

            Zoe Phin

            |

            “The disk receives 960 W/m^2, after adjusted for albedo.”

            Disc? Hilarious!
            It’s funny how you’ve devolved into flat earth physics.

            Yes, Geran, if the Earth was flat, then it would receive 960 W/m^2 of perpendicular solar rays, after adjusting for albedo.

            FLAT EARTHER!
            lol

          • Avatar

            geran

            |

            Now, you’re getting so daffy, you’re repeating your comments.

            Here’s my answer from above:

            It’s just simple accounting, Zoe.

            960 – 480 = 480

        • Avatar

          Zoe Phin

          |

          “Now, you’re getting so daffy, you’re repeating your comments.

          Here’s my answer from above”

          Geran, you’re really lacking in self-awareness.

          Reply

        • Avatar

          Rick

          |

          Why would temps drop to 0K? The heat capacity stored and flux rate would never get that low over 12hrs.

          Reply

          • Avatar

            Zoe Phin

            |

            Rick,
            If Postma wants his 30C then he must emit all he receives in the 12 hours he’s receiving it.

            NOTHING can be stored, if Postma wants his 30C.

            Of course in reality ~240 W/m2 is stored, and only ~240 W/m^2 is emitted.

            By SB Law this means the sun can only make the surface/(lower atmosphere) on average -18C.

            Not enough for the ~15C we observe.

          • Avatar

            Zoe Phin

            |

            Rick,
            If Postma wants his 30C then he must emit all he receives in the 12 hours he’s receiving it.

            NOTHING can be stored, if Postma wants his 30C.

            Of course in reality 240 W is stored, and only 240 W/m^2 is emitted. The stored 240 comes out at night.

            By SB Law this means the sun can only make the surface/(lower atmosphere) on average -18C.

            Not enough for the ~15C we observe.

    • Avatar

      Jerry Krause

      |

      Hi Zoe,

      Both models limit the incoming solar radiation that is incident upon and absorbed by the earth’s surface with a ‘measured’ averaged albedo and do nothing to limit the infrared radiation, being emitted from the surface (UWIR), from being totally transmitted through the same atmosphere to space. To repeat an obvious fact, the calculated average emission temperature (approximately negative 15 degrees Celsius) will be always be approximately that calculated by Svante Arrhenius in 1896 if nothing is proposed to limit the transmission of the UWIR through the atmosphere.

      There has to be some mechanism to limit the transmission of the UWIR through the atmosphere. Arrhenius stated this mechanism involved the atmospheric carbon dioxide (and now other greenhouse gases) and I state that it is the observed empirical influence of cloud which scatters the UWIR back toward the earth surface.

      But Joe, and many others, dismiss all the downwelling infrared radiation (DWIR) being measured at NOAA’a Surfrad sites because it comes from a colder atmosphere as they invoke the 2nd Law of thermodynamics. So they consider this DWIR can never do anything.

      What they overlook is that during the nighttime the surface temperature cools significantly when the atmosphere appears cloudless and cools very little during the nighttime when there is overcast cloud. If you want to say the cloud ‘traps’ the radiation, say it. But I would describe the effect of cloud as the earth surface and the bottom surface of the overcast cloud quickly comes to a state of temperature equilibrium.

      Have a good day, Jerry

      Reply

        • Avatar

          Jerry Krause

          |

          Hi Zoe,

          You asked: “Where is this emission from the top of the instrument to the atmosphere?”

          And then you answered your question: “Oh yeah, they call it DWIR.” No, it is called UWIR.

          Have a good day, Jerry

          Reply

          • Avatar

            Zoe Phin

            |

            Jerry, it’s an outrage that they call upwelling downwelling.

      • Avatar

        geran

        |

        Jerry, I’m so glad you mentioned “thermodynamics”. That’s the reason we know Arrhenius was terribly incorrect. You cannot create energy from nothing.

        The fact that the night surface cools so dramatically when the sky is clear is proof your GHE nonsense is pseudoscience. Clouds have mass. A lot of mass. Low clouds at night act like a blanket, reducing heat loss. But, clouds do NOT raise the system temperature.

        You might want to study thermodynamics, since you’ve learned to use the word in a sentence….`

        Reply

      • Avatar

        Zoe Phin

        |

        It’s the warmth that brings the clouds, not the clouds that make it warmer.

        http://www.physicalgeography.net/fundamentals/images/warmfront.GIF

        If you placed a pyrgeometer at the right side of the diagram and pointed it up, you would indeed see a reduction of outgoing radiation.

        Q = sigma*(Thot – Tcold)^4

        Warmth brings the clouds, which increases Tcold, and therefore you have less outgoing radiation.

        This has nothing to do with the clouds warming the surface. Only the thermal gradient is changed, not the surface temperature.

        The surface is just as cold, as the diagram shows.

        If clouds warmed the surface, how would this diagram even be possible?

        Do you see a lack of cloudd in the cold section?

        Reply

        • Avatar

          geran

          |

          Zoe, you continue to provide entertainment.

          That equation you tout, Q = sigma*(Thot – Tcold)^4, is pure pseudoscience. It has no application to the real world. You know so little, you don’t even get the equation correct. You make mistake after mistake, and just keep making comments.

          Hilarious.

          Reply

        • Avatar

          Norman

          |

          Zoe Phin

          While I do think you are ignorant of physics you seem several levels of intellect above geran. He can’t even process what you are saying. That is really lame. He uses 480 w/m^2 for each square meter of a spherical hemisphere and then he pretends that he is intelligent. He really is hilarious. I can get good laughs reading his clown posts.

          Reply

          • Avatar

            geran

            |

            Troll Norman, you forgot to mention the fact that you have no backgournd in physics or thermodynamics.

          • Avatar

            Norman

            |

            Geran

            Your counter post does not help your lack of knowledge of science nor your lack of logic that Zoe Phin pointed out.

            I have so much more knowledge of physics than you will ever have. I get mine from valid sources. You make yours up when you want to and call all good valid experimentally verified physics pseudoscience.

            Zoe Phin clearly pointed out you are a clown with zero knowledge of physics (except a few words you look up on the Internet to pretend to be smart like Poynting Vectors, not that you know what it means but you pretend to).

          • Avatar

            geran

            |

            Norman, your constant personal attacks are your failed efforts to make up for your lack of “knowledge of physics”. The more you insult, the more you reveal your incompetence.

            So, more please.

          • Avatar

            Norman

            |

            geran

            Your posts continue to show how ignorant you are of physics. Complete ignorant. You are worse than a person who knows nothing. At least they are humble about it and do not act like experts. All you portray is an arrogant idiot. Someone who is clueless but pretends to have knowledge that they really don’t. Zoe Phin pointed out the phony you are, now you hide behind your posts not admitting she has exposed you.

          • Avatar

            geran

            |

            Great job, Norman.

            Your constant personal attacks are your failed efforts to make up for your lack of “knowledge of physics”. The more you insult, the more you reveal your incompetence.

            So, more please.

      • Avatar

        Alan

        |

        This explanation makes no sense. If the surface is cooling then the heat has to go somewhere and it is into the atmosphere. All cooling also involves heating somewhere else, but the heated component cannot achieve a higher temperature than the heat source. The second law cannot be invoked; it applies to all heat transfer. Radiation is not thermal energy, it is electromagnetic energy and it has to be thermalised to raise the temperature when it is absorbed. This process always obeys the second law so cold to does not raise the temperature of hot.

        Reply

    • Avatar

      Rick

      |

      Not having worked through the maths on this – if the day creates sufficient for a 30C equilibrium temperature then this will be emitted according to the total heat capacity and heat capacity flux rate day and night.
      – He states the hemi-spherical in rate is 480
      – He states the averaged spherical out rate is 240
      Due to the heat lag of warming the capacitor system, the out rate will vary as a function of time far more than the fixed in rate (decreasing cooling rate). When the heat capacity is saturated will be its highest out radiation flux, this flux will decrease a with decreasing temps overnight. There will always be residual heat as each hemisphere only has 12 hrs of night. Not sure if that adds anything…

      Reply

      • Avatar

        Zoe Phin

        |

        480 in for 12 hours
        240 out for 24 hours

        “day creates sufficient for a 30C equilibrium temperature”

        If he wants 30C, he has to emit 480.
        That’s what equilibrium means.
        480 in for 12 hours – 480 out in 12 hours equals
        0 W/m^2 left over for the night.

        Postma doesn’t have enough.

        Reply

        • Avatar

          geran

          |

          Zoe, your inability to understand the simple model is funny enough, but not having any reading comprehension is hilarious.

          Joseph’s model clearly states: “Continuous Hemispherical System Input = 480 W/m^2”. You should look up the word “hemispherical”.

          Then, you need to learn that if the disk is receiving 960, the sphere needs to be emitting 240.

          My money says you still won’t be able to understand.

          But I enjoy the humor.

          Reply

          • Avatar

            Zoe Phin

            |

            What disk receives 960?
            The earth is not flat, and it doesn’ thave two suns shining on opposite ends.

            960 is an infinitesimally small Zenith.

            Only normal radiation counts, Geran.

            The hemisphere receives 480 over 12 hours.

            You have two basic choices now:

            You can emit all 480 back out, to get your 30C, but then you have no energy left for the night.

            Or you can emit 240 during the day 12 hrs and 240 during the night 12 hrs.

            Postma clearly says 480 NORMAL radiation hits the earth. But here you are going against him and claiming 960.

            Did you forget the geometry, or you just like pretending to win an argument?

            Postma is a dead end.

          • Avatar

            Zoe Phin

            |

            I think the problem is that you’re taking an irrelevant part of the diagram too seriously. There is no disk hovering above the earth. The 960 is ~HIGH NOON. Just because Postma had to fit text in the disk doesn’t mean it’s that size, ot had any bearing on the NON-NORMAL surface.

            We only get 480 NORMAL radiation avaraged for a hemisphere – just as Postma says below your hovering disk.

    • Avatar

      Alan

      |

      The idea that a single temperature can represent earth’s climate in any year is utter nonsense and it at the centre of the invented human cause climate change crisis. How can a single temperature represent the wide range of daily temperature and the even wider range of temperature between the poles and the tropics? It’s the equivalent to saying all men can be defined by their average height and making all clothing of one size and hoping it will somehow fit them all.

      I suggest you look up the work of Pierre-Marie Robitaille on black body radiation and Planck and Kirchhoff. You might also look up the statistical analysis of temperatures by Darko Butina.

      Reply

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      Rick

      |

      I see the dliemma but is this not just due to the assumption the earth surface is a black body zero-depth surface emitter in constant thermodynamic equilibrium (flux in = flux out to maintain (30C) Temp?). If that is his model then yes you are right.
      However it would seem he is suggesting that the in flux creates rather a steady-state T where the skyward emissivity is not 1 (gray body emitter). Temps do not have to emit all radiation to maintain those temps vs SB black body radiation. Heat can build up vs heat capacities to any depth to maintain steady T. Perhaps we are confusing equilibrium temps with steady state temps. Just a thought, though it would be good if this were explicitly stated.

      Reply

      • Avatar

        Zoe Phin

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        Rick,
        Heat capacity is not heat build up.
        Heat capacity is the amount of energy needed to raise mass by 1 degree K, so that it could then emit.

        The temp produced by the sun alone, averaged over a 24 hr period is -18C.

        The 240 stored over in the day is not acting as heat, only internal energy. At night it comes out.

        Postma never has effective 30C.

        Reply

    • Avatar

      Robert Kernodle

      |

      Zoe said:

      [[[“A hemisphere receives 480 W/m^2 over 12 hours. 240 W/m^2 is emitted during those same 12 hours, and 240 W/m^2 is retained in the surface/atmosphere matter to be retained for the 12 hours of night. At
      night, those 240 W/m^2 come out.”]]]

      “A hemisphere receives 480 W/m^2 over 12 hours. 240 W/m^2 is emitted during those same 12 hours, and 240 W/m^2 is retained in the surface/atmosphere matter to be retained for the 12 hours of night. At
      night, those 240 W/m^2 come out.”

      The quantities, 480 W/m^2 and 240 W/m^2, are NOT quantities of conserved stuff — they are RATES OF FLOW of stuff per unit area. Rates per area are NOT quantities that can be “retained”. Rates of sun intensity are NOT quantities that can “come out”, in the sense that you could remove a quart of water from a gallon of water, because solar flux is NOT a volume of something — it is, again, a RATE OF FLOW of something (energy) in a given time, for a given unit area.

      The total energy resulting from the input on the Earth sphere in 12 hours is transformed through physical processes, in such a way that the FLOW RATE of energy out of the WHOLE Earth SPHERE is 240 W/m^2. How all that energy gets transformed from the Earth hemisphere to exit at this rate over the Earth sphere is complex. You cannot reduce this complexity of energy transformations to a simple addition and subtraction of quantities that CANNOT be added or subtracted, in principle.

      You are trying to add and subtract solar fluxes to make a point that is unmakeable , because its supposed premise is NOT correct at a fundamental level, namely fluxes of solar intensity cannot be treated as additions and subtractions from one another, as I understand it.

      Reply

      • Avatar

        Zoe Phin

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        I can compute total energy equivalents from the fluxes, and it won’t change my main argument: Postma doesn’t have enough to explain surface temperatures.

        Reply

      • Avatar

        Zoe Phin

        |

        “You are trying to add and subtract solar fluxes to make a point that is unmakeable , because its supposed premise is NOT correct at a fundamental level, namely fluxes of solar intensity cannot be treated as additions and subtractions from one another, as I understand it.”

        I’m conserving solar energy. What enters is what leaves.

        Example:
        800 W over 2 hours = 40 W over 40 hours

        Have you ever charged then discharged a laptop?

        Postma is saying the battery returns more energy than it was given.

        Reply

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    Mullumhillbilly

    |

    In Postmas Fig 4, the outgoing radiation appears to be calculated as per incoming, Ie in one direction with a cosine adjustment for edge effects. Surely it should be at a normal to surface all over the sphere?
    A way to test the general budget method; does it give a correct energy budget for other planets ?

    Reply

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    geran

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    There is no “cosine adjustment for edge effects”. And the outgoing (240 W/m^2) is 1/4 of the incoming (960 W/m^2) because the sphere has 4 times more surface area than the disk.

    You might want to understand what’s happening on this planet before venturing to others.

    Reply

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    Jerry Krause

    |

    Hi Zoe,

    In 2016 I discovered PSI because Joe Postma was discussing Horace de Saussure ‘s hot box (https://principia-scientific.com/solving-global-warming-de-saussure-device-paradox/) at his site. I draw attention to this link because this essay contains an image of his hot box. Horace had seen that enclosures with glass windows, at that time (1767), got much warmer than the exterior temperature. So he reasoned that since the solar radiation passed through the glass, its energy must be somehow trapped inside the enclosure by the glass. So he designed, constructed, and tested his hot box to see how many panes of spaced glass would produce the greatest temperature. And the greatest temperature was little greater than 220oF when the box was triple glazed with 3 space panes and he pointed his hot box at the sun and even wiggled it a bit to heat its interior sides.

    I bring attention to the hot box because I consider it to be a simple instrument (radiometer) which measures the downwelling solar (DWS) incident upon it during the daytime and the DWIR during nighttime with a thermometer.

    For I have constructed two simple radiometer, very similar to de Saussure hot box, except extruded Styrofoam is used for the insulating structure and a temperature measuring device is placed beneath a sheet of aluminum foil painted black. One was triple glazed with three films of polyethylene (generally transparent to UWIR and DWIR) and another with 2 films of polyethylene with a top pane of glass to see what the influence of glass might be.

    Horace had to point his hot box at the sun to achieve its greatest temperature. I could level my boxes (latitude 45 degrees) and both would achieve, given a ‘clear sky’ within an hour of midday, a temperature of 220oF at which the radiometer had to be covered to prevent the Styrofoam from melting. It seems apparent that once Horace’s hot box had achieved its greatest temperature and had begun to cool, Horace terminated his experiment. For his experiment objective was to observe a maximum temperature.

    Now a fact is even before I had discovered Horace’s hot box, I had learned that Verner Suomi and others had designed an economical, lightweight (Styrofoam), net radiometer (two of my radiometers placed bottom to bottom) to be carried aloft by a weather sounding balloon Which measurements could only be made during the nighttime. So I tested my radiometer and found both gave similar results– interior temperatures down to 15oF below (sometimes) the atmospheric temperature just above the glazing of the radiometer. So I have to assume the surface of the painted aluminum foil was in a radiation equilibrium with the DWIR incident upon it. For it seems impossible to reason that the warmer atmosphere outside the radiometer could be cooling its interior. So it seems obvious that the glass pane was not greatly hindering the DWIR incident upon the radiometer.

    Zoe, can you question what I have observed (measured)? Do you believe I can use this observation to question your assertion that DWIR cannot be measured by an instrument?

    Have a good day, Jerry

    Reply

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      Alan

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      What happens in boxes, glass greenhouse, Tyndall’s tube and Gore’s glass jars has absolutely nothing to do with what happens in the atmosphere which has no solid surfaces to contain it.

      Reply

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      Zoe Phin

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      Jerry,
      You’re observing radiation leaving your instrument, but you believe it’s incoming.

      Reply

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        Jerry Krause

        |

        Hi Zoe,

        Previously I replied to you: You asked: “Where is this emission from the top of the instrument to the atmosphere?”

        And then you answered your question: “Oh yeah, they call it DWIR.” No, it is called UWIR. So why have you replied as if I believed otherwise.

        Read the following carefully. You seem to insist that no instrument can measuring DWIR because the instrument must be emitting UWIR. I tried to give you a simple example I have observed multiple times during the nighttime. The temperature under the aluminum foil in the insulated box being well below the temperature the ambient atmosphere. There can be only one explanation for this. It is that the emission from the painted aluminum foil is being transmitted through a pane of glass toward space.

        Now I have to ask myself and you: Why doesn’t the temperature drop further? I suppose one could suggest that the energy being conducted through the insulation from the warmer ambient atmosphere outside the box. But if the UWIR is being transmitted through the glass pane, what is preventing the DWIR from the atmosphere being transmitted through the same glass pane to balance the UWIR so that the temperature stopped decreasing.

        Now I can propose an experiment which I haven’t done but given the same general atmospheric conditions of the experiment I have described. During the night that the temperature difference between the temperature be low the foil of the radiometer and the ambient became say 15F, the ambient temperature of the atmosphere was also decreasing during that night because the ground’s surface temperature was decreasing during the night. But the ground had a much greater heat capacity because of its mass so its surface temperature could not cool near as rapidly as the painted aluminum foil inside the insulated box.

        So, my experiment would be to cover the glass pane with a sheet of Styrofoam until near sunrise and then remove it and observe how rapidly the foil cooled the inside of insulated box. Now you have to remember I haven’t done this, but it certainly is not a thought experiment that could not be simply and actually done. For it would not surprise me that temperature in the box near sunrise was still about that it was at 4pm when the ambient temperature was near the high of the day.

        Now, I am not trying to convince of anything that you could not observe if you lived in Salem OR during most summers and made the effort to construct this simple, inexpensive radiometer with which to make your experiments that I have suggested.

        Have a good day, Jerry

        Reply

        • Avatar

          Zoe Phin

          |

          Jerry, there’s two UWIR! Not one, two.
          You would know that if you read my article.

          One UWIR is from surface to instrument.

          The other UWIR is from the instrument to the atmosphere.

          There’s no serious DWIR, except perhaps inversions – but they’re no where near UWIR – otherwise there would be no inversions, obviously.

          Reply

        • Avatar

          Zoe Phin

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          Jerry,
          It’s the warmth that brings the clouds, not the clouds that make it warmer.

          http://www.physicalgeography.net/fundamentals/images/warmfront.GIF

          If you placed a pyrgeometer at the right side of the diagram and pointed it up, you would indeed see a reduction of outgoing radiation.

          Q = sigma*(Thot – Tcold)^4

          Warmth brings the clouds, which increases Tcold, and therefore you have less outgoing radiation.

          This has nothing to do with the clouds warming the surface. Only the thermal gradient is changed, not the surface temperature.

          The surface is just as cold, as the diagram shows.

          If clouds warmed the surface, how would this diagram even be possible?

          DWIR from GHGs is a false attribution. They steal the real cause:

          GEOTHERMAL!

          Reply

        • Avatar

          Jerry Krause

          |

          Hi Zoe,

          There are two radiometers. One pointed downward to measure the UWIR from the ground. The other pointing upward upward to measure the DWIR from the atmosphere and its contents. And it is true that these instruments are not simple because they have to measure the IR radiation during the daytime when there is solar radiation present. In the case of my simple radiometers near midday the maximum temperature beneath the painted foil rises to 220F and then needs to be covered to prevent the temperature increasing even more and melting the extruded Styrofoam insulation.

          You have to do experiments to gain experience about the instruments used.

          As I write this I see I need to ask JD a question about his experiences with his $60 radiometer with which he measures a temperature of negative 65 C of the atmosphere.

          Have a good day, Jerry

          Reply

          • Avatar

            Zoe Phin

            |

            Jerry did you read my article “Why up not down” or not?

            “One pointed downward to measure the UWIR from the ground. The other pointing upward upward to measure the DWIR from the atmosphere and its contents.”

            The UWIR to the instrument heats the instrument. The instrument emits 2nd UWIR to the atmosphere. You for some reason call this 2nd UWIR : DWIR.

            Why?

            Boltzmann and Planck had ONE standing wave per frequency between two opposing walls. He did not have two way photon streams in “equilibrium”. If such a thing existed, then SB law and Planck Law would have an additional 2x factor. They don’t.

            CLAUSIUS WAS WRONG!

            Your equipment is heating the air. Your equipment is radiating to the atmosphere. That is what you are measuring. Your electronic equipment can only measure a differential. YOU NEED A FLUX. A flux only exists due to a disequilbrium. If you see a flux, there is NO equilibrium between 2nd UWIR and DWIR! All you see is 2nd UWIR.

            I hope that helps.

          • Avatar

            geran

            |

            Zoe, you don’t know squat about radiative physics. You just keep making comments about things you don’t understand, and continue to avoid learning.

            You keep throwing out incorrect and inaccurate garbage, like the equation you provided, above: Q = sigma*(Thot – Tcold)^4

            How many things are wrong with your equation?

  • Avatar

    Rick

    |

    “the cross-section of intercept of solar power by Earth’s disk is averaged over the sphere of the Earth, giving the division by four as the ratio of a disk area to that of a sphere’s area with the same radius, such as to average the disk-intercept over the entire sphere”
    I don’t think they do this – it’s /2 for a hemisphere and /2 for the average incidence for a hemisphere, for an instantaneous point in time flux-rate, not a time-average on a sphere. I’m not sure where this comes from? Mathematically it’s the same so the question is – is it a valid approximation?

    Reply

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    Alan

    |

    I hope that Joseph is successful in getting this paper published. It occurred to me recently that the term “backradiation” has been created entirely to define the supposed greenhouse effect. It does not exist in anywhere else, and strangely, front and side radiation do not seem to exist.

    I disagree with Joseph’s view that energy in equals energy out; conservation of energy applies which allow for storage or release of energy. If energy in equals energy out applies then there would be no change in the total thermal energy of the earth’s system and hence there would be no long term change in the climate, just local variations in weather.

    Joseph is correct to question the reasoning ability of the climate scientists, but this also applies to the general population who soak up all the nonsense of human caused climate change. It is as though education is completely wasted on them. The problem is as basic as not understanding the difference between temperature and thermal energy. Everybody thinks they understand this because weather is an obsession of humanity and because the temperature is important to us when we leave home. They don’t understand that thermal energy is only applied once to a system and just because temperature does not fall immediately does not mean there is additional warming. A boiled kettle retains its heat and will heat the air around it but no there is no additional energy input driving this. It is not heating, it is actually cooling of the kettle. Similarly, the sun heats the surface and when it cools it heats the atmosphere. The atmosphere does not heat the surface, and even if there are local atmospheric events which could result in the atmosphere being warmer than the surface, the difference in mass and specific heat of air compared to the land masses and water show that heating from the atmosphere will be very minor. Most people think that insulation increases temperature, but it does not, it slows down heat loss, which is not the same at all. Most people do not understand the very basics of physics.

    Reply

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      Herb Rose

      |

      Hi Alan,
      You are incorrect when saying that the atmosphere doesn’t heat the Earth’s surface. People believe that since the oxygen and nitrogen in the atmosphere do not absorb the visible spectrum of light that they are not getting energy from the sun. This is nonsense every object (even gas molecules) absorb radiated energy. In the case of nitrogen and oxygen molecules they absorb uv radiated by the sun and these wave lengths contain more energy than the visible spectrum. The ionosphere and ozone layer are a result of nitrogen and oxygen molecules absorb uv energy (940 kjoules/mole N2, 490 kjoules/mole O2) and converting it into kinetic energy of the atoms and molecules.The reason the atmosphere is less dense at higher altitudes is because of the greater kinetic energy of the atoms/molecules.
      The visible and infrared light from the sun comes from the sun’s surface. The uv light is produced in solar flares. The grand solar minimum will reveal if our climate is a result of the visible spectrum (varies .1% in sunspot cycles) or by the uv heating the atmosphere.
      Herb

      Reply

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      Finn McCool

      |

      Alan,
      When someone says, ‘You see…I have a skill set. I can see things that no one else can, I can see the inside of how things work that no one else can.’, I tend to suspect that they are getting their ideas mixed up with their capabilities.
      If Postma was made out of chocolate, he would have eaten himself by now.

      Reply

      • Avatar

        Zoe Phin

        |

        Hey!
        Yeah I cringed when I read that.
        Too much hubris. He’s flying too close to the sun, if you get the reference. That’s why he thinks the sun is enough.

        Reply

        • Avatar

          geran

          |

          Zoe, did you also cringe when you tossed out that inaccurate equation:

          Q = sigma*(Thot – Tcold)^4

          You can’t even find all the mistakes in that equation.

          Hilarious.

          Reply

  • Avatar

    Jerry Krause

    |

    Hi Zoe,

    You asked: “Where is this emission from the top of the instrument to the atmosphere?”

    And then you answered your question: “Oh yeah, they call it DWIR.” No, it is called UWIR.

    Have a good day, Jerry

    Reply

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    Carl

    |

    Looking at the Earth from space as an independent, self-contained heavenly body its “surface” is the atmosphere. The amount of IR radiation that this “surface” of the Earth is emitting out into space, as measured by satellites, averages about 200 W/m^2. When you combine this with the about 40W/m^2 from the ground and sea surface through the “atmospheric window” you get the magical 240 W/m^2 number that the “greenhouse effect” hypothesis is based on.

    Using the SB formula this produces a black-body temperature of -18C, which Astro-Physicists called the “Earth’s Effective Radiation Temperature.”

    Some 75% of the “surface” of the Earth–its atmosphere–is contained within the Troposphere and miraculously the average temperature of Tropospheric air top to bottom globally is about -20 C (from the International Standard Atmosphere,) which you must admit is pretty close to the calculated “Effective Radiation Temperature” of the Earth.

    Based on these points of scientific data one should conclude that “greenhouse gases” in the atmosphere are not causing the atmosphere to retain any excess thermal energy as the “greenhouse effect” hypothesis asserts. Why? Because the average temperature of the “surface” of the Earth is pretty close to the “Effective Radiation Temperature” of the Earth.

    So, where does the “greenhouse effect” hypothesis go wrong? For some unexplained reason, rather than juxtaposing the out-going LWIR radiation of the entire planet with the average temperature of the entire atmosphere (its “surface”), it only juxtaposes it with the hottest layer of the atmosphere, i.e., surface level air and thus declares that the “surface” of the Earth is “33 C warmer than it should be.”

    Were the “greenhouse effect” hypothesis to compare the “Earth’s Effective Radiation Temperature” to the coldest layer of the atmosphere–the Tropopause (which is about -60C)–we would be debating “why is the “surface” of the Earth 42C colder than it should be!

    In other words, when calculating the “Earth’s average ‘surface’ temperature” why not average the temperature of the entire Troposphere top to bottom globally rather than just its hottest layer? Were you to do this you would quickly realize that “greenhouse gases” in the Troposphere are not causing the Troposphere to retain any excess thermal energy what so ever.

    Reply

  • Avatar

    Jerry Krause

    |

    Hi Carl,

    After reading your entire comment, I went back to the first sentence. “Looking at the Earth from space as an independent, self-contained heavenly body its “surface” is the atmosphere.” When I first read this sentence I could not understand it. And I still do not understand it. So I ask: What is your definition of ‘surface’?

    Have a good day, Jerry

    Reply

  • Avatar

    Rick

    |

    An odd statement keeps appearing in some of the work here that references the 2nd law of thermodynamics, namely energy (photonic or otherwise) cannot travel UP a temperature (energy) gradient. This is incorrect, of course. The photons from any source are not psychic. They are not communicating. They just emit per their temperature.
    The 2nd Law in reality means no NET energy can flow from a cold body to a hot body. Net energy is flowing from the earth to the atmosphere – but this doesn’t mean no energy can flow from the colder atmosphere to the warmer ground.
    As such the colder body merely slows down the rate of cooling and the ultimate stead-state/equilibrium temperature. Such is implicit in the equations, which are net (aggregate) calculations of flow rate ensembles – not the explicit directions of every individual particle. We can thus increase AVG temps of a warmer body for a cooling period – but this is not adding net energy up a gradient, similar in effect to insulation.

    Reply

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      Zoe Phin

      |

      Rick,
      “The photons from any source are not psychic. They are not communicating.”

      Actually they are “psychic”. In Boltzmann’s and Planck’s photon gas box, only certain wavelengths can form. And those possible wavelengths are based on the dimensions of the box. So separation distance is known, and therefore some photons are allowed, and others not.

      You’re reciting Clausius’ 1855 statement. He was shown to be wrong regarding radiation.

      Reply

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        Rick

        |

        Zoe, what on earth are you talking about..

        Reply

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          Zoe Phin

          |

          Oops, here’s the correct link:
          http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

          Only some wavelengths will form and that is depended on the dimensions of the cavity. Those formed wavelengths are called photons.

          It’s as if one side of the box knows how far away another side is. And only emits photons of those wavelengths.

          If you don’t think that qualifies as “psychic”, then what does?

          Reply

          • Avatar

            Zoe Explainer

            |

            Idk how a blackbody spectrum makes psychic photons any more than your understanding is just misguided. Blackbodies are an entirely statistical process. It is not that one photon tells the next “I am wavelength x, therefore you should be wavelength y”

            It is rather that each photon will have a probability of being conceived with a certain probability distribution, so the resulting spectrum is determined by the temperature or geometry or whatever is spawning the photons.

            Look at the Casimir effect (SUPER INTERESTING) it is a measurable attraction between two plates brought closely together because the lack of certain wavelengths of virtual photons being impossible between the two plates. They are literally attracted by the zero-point energy of the electric field, very cool:

            https://en.m.wikipedia.org/wiki/Casimir_effect

            A side note, there’s two types of people in this world. One is people that do not understand something and are driven to the ends of the Earth (ironic statement) to try an understand it. These are the real scientists who make careers out of it. The other type of people are those who don’t understand it, try to come up with their own misguided/misinformed explanations and parade those as fact on fake news forums like these. It’s been a pleasure diving into these forums to see so many people who believe they are correct but will sadly never be as long as they continue down their current paths.

        • Avatar

          Zoe Phin

          |

          “I was about science. Have you heard about it?”

          Sorry If I sounded condescending

          Reply

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    Jerry Krause

    |

    Hi Rick,

    You concluded: “We can thus increase AVG temps of a warmer body for a cooling period – but this is not adding net energy up a gradient, similar in effect to insulation.”

    First I assume AVG in this case is average. Next, what I write is for you to confirm that I have correctly understood (in my words) what you wrote.

    I imagine there are two houses in a climate where there are no clouds. One house is very well insulated and the other has no insulation.

    First, after sunrise the radiation of the sun begins to heat the natural environment and the atmospheric temperature increases to its maximum value at 4pm and then slowly begins to cool because the soil has absorbed most of the solar radiation that has been absorbed by the surface and atmosphere (for the density of the soil is much greater than the density of the atmosphere in contact with the atmosphere at its base. Hence it principally the ground surface which warms the atmosphere and not the direct solar radiation. And for the atmosphere to warmed by the surface requires that the surface be warmer than the atmosphere above it. But surface warmed by the solar also begins to conduct energy down into the soil which has cooled during the nighttime. This conduction of energy down into the soil stores this energy as the temperature of the soil increases during the day. But a observed fact is the soil’s surface begins to cool shortly after midday when the surface is much greater than that of the atmosphere 1.5 meter above the surface. Assumed calm atmosphere. So until maybe 4pm surface cools to the atmosphere’s maximum temperature, after which the ground surface begins to be ‘heated’ by the warmer atmosphere over it and warmer soil beneath it. And because of the density difference between the atmosphere and ground, the ground is the major supplier energy to the surface and as the atmosphere is cooled by surface a layer of colder air is formed at the surface which insulates the atmosphere above it from the cooling effect of the cooling surface which is being cooled by the energy being emitted from its surface. Hence the surface is the principal cooling agent of the atmosphere during the nighttime.

    I have not yet considered the two houses and I will come back to do this as I have something else I need to do now.

    Have a good day, Jerry

    Reply

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    Rick

    |

    Perhaps the above is because people confuse kinetics (the actual micro processes) with the bulk statistical outcomes of thermodynamics.

    Reply

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    Jerry Krause

    |

    Hi Rick,

    Continuing to the houses. From experience I know that the inside temperature of both houses will eventually become warmer than the maximum atmosphere’s temperature. This because the roof quickly warms to it maximum temperature shortly after midday as did the ground’s surface temperature. But the inside temperature of the insulated will warm to and beyond the atmosphere’s temperature more slowly but even 10 inches of fiberglass insulation cannot prevent it from as warm, if not warmer, than the maximum atmosphere’s temperature. It just that the uninsulated house will warm more rapidly and beyond the maximum atmosphere’s temperature. Hence, I conclude the maximum inside temperature of the uninsulated house becomes greater than that of the insulated house.

    But, without a doubt, the uninsulated house will cool faster and to a lower temperature than the insulated house will. Hence, the maximum interior temperatures of both houses were the same while the minimum temperature of the insulated house was greater than that of the uninsulated house. So the average temperature of the insulate house was greater than that of the uninsulated house because the insulated house cooled slower during the nighttime because of insulation.

    And it is a common observation that during an overcast night the nighttime temperature cools very slowly so that the average temperature of that day is greater than one with a cloudless nighttime. But the importance of this one overcast night is that at sunrise the next morning the temperature begins at a greater temperature than it had the morning before. So, if the atmosphere clears shortly after sunrise this next morning, the afternoon maximum temperature is likely (must be) to be greater than that of the previous day.

    The lesson I hope one sees if that clouds warm, increase average temperature, by slowing cooling during the nighttime. Which I considered was what I was to take away from you concluding statement.

    Have a good day, Jerry

    Reply

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    Rick

    |

    Hi Jerry
    I was just saying that if you have a wall of say 100C facing a wall of say 10C then to ask the wall at 10C to “not radiate” because it’s facing a hotter wall would be absurd. A body with a heat capacity at temp T will radiate at its temperature regardless – it knows nothing of any other body. If warm bodies did not absorb radiation from cooler sources or surroundings then you wouldn’t need to know the cooler temperature to calculate the net flux rate – it would be constant, which it is not.
    Now people take a statistical equation to assume that kinetically no energy can be emitted by the cooler side to the warmer side, which is not true. Photons do not phone ahead. Hence do not confuse kinetics with thermodynamics. It is true the cooler wall cannot increase the overall temperature of the warmer wall – but they will nevertheless exchange energy and establish a (diminishing) net flux gradient until they equilibrate to an ensemble temperature according to their individual starting temperatures.

    Reply

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      geran

      |

      “If warm bodies did not absorb radiation from cooler sources or surroundings then you wouldn’t need to know the cooler temperature to calculate the net flux rate – it would be constant, which it is not.”

      Rick, warm bodies do no absorb emitted flux from colder sources. And a “net flux” is only a mathematical construct. It has no meaning in reality, since fluxes can not be treated with simple arithmetic.

      Reply

      • Avatar

        Rick

        |

        Hi Geran
        Given a warmer and a cooler body – energy is constantly being transmitted in both directions according to their temps. You saying the warm body must “reject” the cooler body’s radiation? I am not sure about this – that would mean that there is a digital switchover for backbody thermal equilibria – e.g. if we are at 30C and the sun is dimmed by 50% at mid-day due to cloud cover – the earth would cool at the same rate as if it wasn’t there?
        Thermodynamics does not apply to individual molecules – it applies to the net flow of energy in the entire system.
        But, I enjoy debate and happily stand to be corrected.

        Reply

        • Avatar

          Zoe Phin

          |

          Rick,

          “Given a warmer and a cooler body – energy is constantly being transmitted in both directions according to their temps.”

          EVIDENCE?
          Boltzmann and Planck had one standing wave per wavelength between two opposing walls.

          ONE!

          What experimentalist discovered something different?

          Reply

          • Avatar

            Rick

            |

            Zoe
            But it’s right there in the derivation of the equations for radiative heat transfer… the net radiative heat transfer from one surface T1 to another T2 is the radiation leaving the first surface minus that arriving from the second surface. Else why would the rate depend on T2? The absorptivity goes down with increasing T, emissivity goes up.

          • Avatar

            Zoe Phin

            |

            The difference doesn’t mean the lesser emits to the greater.

            If you raise the water level at the bottom of a waterfall, there is less height to fall.

            This doesn’t mean the water fall raises its height so that water can fall the same height as below. Only charlatans use Clausius net heat flow statement in 1855 – he was obviously referring to conduction.

            Kirchoff first made cavity experiments in 1859.

            How would Clausius have any idea about cavity radiation 4 years before any experiments?

          • Avatar

            Zoe Phin

            |

            same height as before

          • Avatar

            geran

            |

            Rick, which “radiative heat transfer” equation are you referring to?

            The bogus one: Q/A = sigma*[(Thot)^4 – (Tcold)^4]

            Or, Zoe’s even funnier one: Q = sigma*(Thot – Tcold)^4

            IOW, how do you prefer your pseudoscience, normal or extreme?

          • Avatar

            Zoe Phin

            |

            The first equation is correct. I did make a mistake in writing it.

          • Avatar

            geran

            |

            Sorry Zoe, but your mistake was trying to fake a knowledge of physics.

            But pompous mistakes make for great comedy.

            More please.

        • Avatar

          geran

          |

          Yes Rick, energy would be transferring between the two bodies, but NOT “heat”.

          Photons can move in opposing directions, but not “heat”. “Heat” is the transfer of energy from hot to cold.

          Your example of the clouds/Sun does not fit because Sun is hotter than Earth.

          Reply

          • Avatar

            Rick

            |

            Geran
            Yes – this is what I am saying: heat (i.e. the NET flow of energy as a result of temperature differences) cannot go from a cooler to hotter object. But energy can still go from cold to hot, as it must do. Hence the colder object can only leave the hot object warmer than it would be if the cold object weren’t there. Thus it raises its average temperature over time. A cold atmosphere can also leave the earth warmer than it would be without the atmosphere because it is blocking something even colder from view – in terms of direct energy gradient.

          • Avatar

            Zoe Phin

            |

            “But energy can still go from cold to hot, as it must do.”

            No, there is no flow of energy from cold to hot.

            Energy only flows “downhill”.

            A thermister won’t detect a double flux between two objects at thermal equilibrium. Hence it’s not there.

            A thermister can’t tell you if it’s between two bodied at 10C, or 100C, or 1000000C.

            The MEASURED flux is ZERO in all cases.

            That’s why IR thermometers also have a pressure-based thermometer (non-radiation) to tell air temperature.

            Only using this non-radiative means allows you to guess what the radiation is.

            If Clausius’ fools were correct, a thermister would detect their double energy flows. It does not. PERIOD.

          • Avatar

            geran

            |

            Rick, photons going from cold to hot do not raise “its average temperature”.

            That erroneous thinking would mean you could bake a turkey with ice cubes.

            Are you one of the clowns that believe you can bake a turkey with ice cubes?

          • Avatar

            geran

            |

            Zoe, maybe you meant “thermistor”.

            Of course, with your lack of knowledge, who knows?

          • Avatar

            Rick

            |

            Geran
            Now you are contradicting yourself. Stop repeating your bum chum’s words and think for yourself.

          • Avatar

            geran

            |

            Where did I contradict myself, Rick?

            PS Running out after false accusations is NOT debate. But maybe such standards are only for adults….

          • Avatar

            Norman

            |

            Zoe Phin

            I wonder if you will ever answer the question as to what is your source that you use to declare energy can only flow one way? I have NOT read that in any valid science textbook on radiant heat transfer. You are just making things up.

            You are confused with forces acting on particles. Yes in a gravity field a mass will only flow one way and in an electric field a charged particle will only flow in one way. Photons have no charge they have self contained fields and are not dependent upon some external field to determine the direction of their flow.

            I ask but am not sure you will ever provide any valid source for your fantasy physics.

            I think this is the blog where people just make up whatever idea they think is valid. Then you have the blog clown, Geran, that calls all valid science pseudoscience. He can’t understand the radiant heat transfer equation is used all the time in engineering applications. He is gone from the real world of science and dwells in his fantasy world where he is a super genius and all others are not worthy to exist in his space. He is a severe case of Dunnng-Kruger effect. He makes up ideas and think they are real physics. When intelligent posters point out he is wrong he diverts into nonsense comments. Real similar to his identical twin JDHuffman

            Keep making up your own ideas. I guess it makes you feel valuable or something. Don’t be surprised that only on this bottom dwelling blog of conspiracy crackpots and delusional posters that no one in the science world will ever consider any of your insights as valuable.

          • Avatar

            Zoe Phin

            |

            Norman,

            “I have NOT read that in any valid science textbook on radiant heat transfer.”

            What evidence do these textbooks provide?

            What experiment shows two way flow?

            Assertions in textbooks are not good enough.

            Boltzmann and Planck showed the opposite.

            If there was two flowx Planck’s equation would have a 2x factor.

            The density of a photon gas in a box would be double!

            A thermistor doesn’t detect two way flow.

            That’s two lines of evidence.

            P.S. I can’t believe you bothered to “debunk” an ANALOGY. How pathetic.

          • Avatar

            Norman

            |

            Zoe Phin

            As expected you were totally unable to find even one source of evidence that your idea is correct. You twist some ideas to try and convince yourself your thought errors are correct. Although I do think you are much more intelligent than Geran I do not think you also suffer from Dunning-Kruger
            The textbooks are the result of many experiments, some are explained in textbooks, most will be found in journals on the specific topic. You are highly unscientific. You have no evidence to support you claims. You use distorted and twisted logic to convince yourself you know what you are talking about. The truth is you know nothing of physics and make up most of your material and are so impressed with you imagination you will then no longer use rational thought to look for flaws in your ideas but peddle them as if they were actually factual (which they are not). As long as your giant ego exists the truth will forever evade you.

            The early scientists did accept IR was a two-way transfer (using logical inference on what IR is). Your ideas are make-believe. Fantasies of your imagination.

          • Avatar

            Zoe Phin

            |

            Norman,
            Surely you can find one experiment that MEASURED two-way normal photon flow between two objects with pure temperatures.

            MEASURED, not inferred.

            If it is inferred, it doesn’t count.

            The instrument must have a change of state.

          • Avatar

            Norman

            |

            Zoe Phin

            Note to you. I asked you for your source of information that there is not a two-way flow of radiant energy. So far all your evidence is that all textbooks are wrong. So you would have me accept that all the textbooks are wrong but somehow you are the only correct one. Why should I consider this a valid proposal?

            I can give you a real simple test to prove radiant energy flows two ways. Sit in a totally dark room with a light source (off for the moment) behind you. You have a painting on the wall in front of you that you can’t see at the time. It is dark. Switch the light on. The energy flows out from the light toward the wall with the painting. Now some of this light hits the painting and is reflected back to your eyes (the detectors). The energy is moving against the gradient toward your eyes and also back to the source of the energy.
            I have also done my own research with FLIR instrument. This one does not just detect temperatures of objects, it forms images of them based upon the IR reaching the array inside the instrument. There is a hot pipe (around 300 F) that is heating a hand rail (around 100 F…ambient is around 80 F). The FLIR forms an image (as I stated) so you can “see” the hot pipe’s image in the camera. Then you rotate it 180 degrees to the hand rail and it “sees” this as well. The IR can only be coming from the cooler rail as it has this image in the camera. The IR is moving toward the hotter pipe against the gradient. You can peddle your false physics all day. It is still wrong. The only hope is you are intelligent enough to realize you are not correct and work to correct your misunderstandings. If you are too arrogant to do this you will not be a useful source of information on the current climate debate. Better to have informed intelligent skeptics than those that reject established science and make up their own.

          • Avatar

            Zoe Phin

            |

            Norman is insanely stupid.

            “The energy is moving against the gradient toward your eyes and also back to the source of the energy.”

            1) The source doesn’t get brighter

            2) My eyes were never emitting light, so there is no movement “against the gradient”.

            3) Visible light can be added to Infrared. Those fluxes can be added. That’s all your showing in stupid example.

            4) You had to show that the colder painting increases my eyes’ temperature.

            Your FLIR example is incomprehensible. I can’t make sense of what you’re even describing. Perhaps you have photos or video?

            I gave you my sources:
            Boltzmann, Planck, Thermistors.

            I linked Rick to photon gas in a box physics. Have you read it?

          • Avatar

            Norman

            |

            Zoe Phin

            Again you still have failed to show a valid source of information that proves your one-way conjecture for energy flow.

            Your points lack reasoning or logic. It is like you just threw them up from the bile of your unscientific mind.

            I realize it is not possible to reason with you but it does not hurt to continue trying.

            YOUR POINTS: “1) The source doesn’t get brighter (ME: now what evidence do you have to prove this point? None, you just made that up).

            2) My eyes were never emitting light, so there is no movement “against the gradient”. (ME: this is really bad logic, it does not matter AT ALL that your eyes are not emitting, they are detecting, goodness you are not logical at all! The light from the painting is moving against the gradient from the actual light source to reach your eyes!!! Think would you, your horrible logic can be painful!!!)

            3) Visible light can be added to Infrared. Those fluxes can be added. That’s all your showing in stupid example. (ME: No that is not what my example is showing at all!!)

            4) You had to show that the colder painting increases my eyes’ temperature.(ME: no one would not have to show that at all to prove radiant energy is moving against the gradient of the light source, only that your eyes can detect the energy moving in the direction toward the light source. Are you really that stupid or are you trying to be funny?)

          • Avatar

            Norman

            |

            Zoe Phin

            YOU: “Your FLIR example is incomprehensible. I can’t make sense of what you’re even describing. Perhaps you have photos or video?”

            It is really a simple thing. Do it yourself. Have a hot plate with another plate a little away from it (to create a space you can put your IR camera between). The hot plate will increase the temperature of the other plate. Now put your camera between them facing the hot plate. The IR emitted by the hot plate is detected by the camera and gives an image and temperature. Rotate the camera to the colder plate and the IR camera will pick up the energy from this plate as well moving against the gradient toward the hotter plate. I do not know why such a simple reality boggles your mind.

            Now I would ask you to give any evidence that all the textbooks on the topic of radiant heat transfer are wrong and your fantasy made up physics is the correct one. The textbook physics is used on all actual heat transfer engineering and design. It seems to work quite well as they use this information to build and design heat transfer equipment. So you have the entire wold of heat transfer engineering telling you that your ideas are wrong, but you think posting your fantasy physics on this blog makes them valid, correct or right. I will stick to the valid science. You can waste your time trying to convince people you are a genius. The rest of us know the sad reality.

          • Avatar

            Zoe Phin

            |

            Norman,
            The back of the camera is preventing normal radiation from going from hot to cold plate. There is no normal radiation going “against the gradient”.

            This is not an example of backradiation heating.

            Secondly, you’re not providing any evidence, you’re just making an assertion. Link to some real text, photos, or video.

          • Avatar

            Zoe Phin

            |

            Norman is an imbecile.

            “The hot plate will increase the temperature of the other plate.”

            Yes! If you just left the camera facing the cold it would get warmer!

            But you decided to rotate it twice, and thought there’s some extra magic effect when you turned back to cold.

            Cold would’ve gotten hot all on its own had you left the camera alone the whole time

            Gosh, you’re so pathetic.

          • Avatar

            Rick

            |

            Geran: “Photons can move in opposing directions”
            Thankyou – that is what I am saying. I clearly described heat flow as the NET flow of energy. I said that a colder object cannot add heat (as defined) to a hotter object. I said that a colder object can raise a hotter objects AVERAGE temperature because it affects the NET heat flow out of that object, but you need to understand what an average is – it slows the rate of cooling (as has been shown in the latest article on PSI). It does not add heat. It’s all very simple. Nobody is talking about cooking turkeys with ice cubes so I think you have issues with terminology.

          • Avatar

            geran

            |

            Yes Rick, photons can move in opposite directions. But not “heat”. My issue with terminology is that I prefer the accepted definitions.

            “I clearly described heat flow as the NET flow of energy.”

            Rick, that description is wrong. “Heat” is NOT the net flow of energy. “Heat” is the transfer of energy from a hot object to a cold object. There are many ways to say it, but “heat” always has TWO components — 1) Transfer of energy, and 2) From hot to cold.
            If you don’t have both components, you don’t have “heat”.

            “I said that a colder object cannot add heat (as defined) to a hotter object.”

            There is nothing that “adds heat”. You can not “add heat”. “Heat” is a thermodynamic process. It is not something you can hold in your hand. “Heat” is the transfer of energy from hot to cold.

            “I said that a colder object can raise a hotter objects AVERAGE temperature because it affects the NET heat flow out of that object…”

            And that is wrong. A colder object can NOT raise the temperature of a hotter object. That would be a violation of the Second Law of Thermodynamics.

            “…but you need to understand what an average is – it slows the rate of cooling”

            No, an “average” does not slow the rate of cooling. You may be thinking of insulation?

            “It does not add heat.”

            Correct, you can not “add heat”.

            “Nobody is talking about cooking turkeys with ice cubes…”

            Wrong Rick, I was talking about that very thing here:

            https://principia-scientific.com/postmas-climate-model-paper-discussion/#comment-31193

          • Avatar

            Zoe Phin

            |

            Rick,
            Temperature doesn’t depend on anything “down the line”, so to speak. Joule, Maxwell, and Boltzmann proved that. Sorry you are stuck in early 1800s physics.

          • Avatar

            James McGinn

            |

            Geran:
            “Heat” is NOT the net flow of energy. “Heat” is the transfer of energy from a hot object to a cold object. There are many ways to say it, but “heat” always has TWO components — 1) Transfer of energy, and 2) From hot to cold.
            If you don’t have both components, you don’t have “heat”.

            James:
            So, Geran, if heat always has the two components you mentioned above what do you call–if anything–the flow of energy is from cold to hot?

            IOW, what do you call it when you have 1) Transfer of energy, and 2) From cold to hot?

            James McGinn / Genius

          • Avatar

            geran

            |

            James, you copied/pasted correctly, but you forgot to read and understand.

          • Avatar

            geran

            |

            James, taunting me just makes you look like an incompetent windbag. Is that your choice in life?

            “IOW, what do you call it when you have 1) Transfer of energy, and 2) From cold to hot?”

            Let’s use two walls in a hallway for an example. One wall is 20º warmer than the other wall. Both will be emitting energy, but photons going from the hotter wall and absorbed by the colder wall will be called “heat”. The photons going from the colder wall to the hotter wall will be called “photons”.

          • Avatar

            Zoe Phin

            |

            According to Boltzmann and Planck, the “fathers” of radiation, there will not be a two way flow of photons, regardless of Einstein (a guy who never conducted an experiment in his whole life) says.

            What will form between hot and cold are ONE standing wave per frequency. They will add up to the difference between temperature of hot and cold.

          • Avatar

            geran

            |

            That’s generally correct, Zoe. Photons from hot to cold will be absorbed (called “heat”). Photons from cold to hot will be reflected by hot and form standing waves. Standing waves transfer no energy.

            That’s one reason the “blue/green plates” nonsense is pure pseudoscience. The colder plate cannot raise the temperature of the hotter plate.

          • Avatar

            Zoe Phin

            |

            Geran,
            The allowable wavelengths of “photons” are determined by separation distance. Hot knows the distance to cold. Hot knows what to send cold. And it only sends some “photons”, and not others.

            This is what we learned from Boltzmann and Planck.

            Yet some people believe Hot and Cold aimlessly both send whatever they like.

            Maxwell revolutionized EM theory by debunking the corpuscular theory of light.

            Einstein brought it back, though therr was no need. Now many are confused.

          • Avatar

            James McGinn

            |

            Geran:
            Photons from hot to cold will be absorbed (called “heat”). Photons from cold to hot will be reflected by hot and form standing waves. Standing waves transfer no energy.

            James:
            Seriously,?

            Standibg waves?

            Who else subscribes to this characterization? Or, are you just speaking for yourself?

            Standing waves.

            Seriously?

            https://anchor.fm/james-mcginn

          • Avatar

            geran

            |

            “Who else subscribes to this characterization?”

            James, everyone understands the concept, except the uneducated and incompetent.

            Seriously.

          • Avatar

            James McGinn

            |

            “Who else subscribes to this characterization?”

            James, everyone

            Nobody but you has ever mentioned standing waves. Obviously it is just a figment of your imagination.

            Name one other person who supports your standing wave hypothesis?

            Geran:
            understands the concept, except the uneducated and incompetent.

            James:
            Name one other person who believes in your magical standing waves. Just one!

            Put up or shut up

            James McGinn / Genius

          • Avatar

            geran

            |

            James, if you weren’t so uneducated and incompetent I wouldn’t have to help you so much.

            See if you can find an adult to explain this to you:

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

          • Avatar

            James McGinn

            |

            Rick:
            “But energy can still go from cold to hot, as it must do.”

            Zoe:
            No, there is no flow of energy from cold to hot. Energy only flows “downhill”.

            James:
            Zoe, energy flows in all directions constantly. So, it does flow uphill. The net flow is from hot to cold.

            Again, radiation travels in all directions. The net flow of radiation over time is from hot to cold.

            It really is this simple.

            This obsession you all have with “one way” flow makes you all look ridiculous.

            James McGinn / Genius

          • Avatar

            James McGinn

            |

            Geran:
            James, if you weren’t so uneducated and incompetent I wouldn’t have to help you so much.

            Geran:
            Address the issue, you evasive POS. You stated that standing waves prevents the transfer of energy from cold to hot.

            Nobody else is going to support you on this. You are on your own.

            None of you slayers actually understands any of this. You are all engineers who have mistaken your equations for reality.

            Don’t tell me other people believe it unless you are able to provide names and references?

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

            I know what a standing wave is, you lying POS. Now show us that standing waves block the transfer of energy. You can’t.

            Geran, you are just a confused engineer.

            James McGinn / Genius

          • Avatar

            Zoe Phin

            |

            James,
            Why do you disagree with Boltzmann and Planck?

            Their two formulas explain emission. This hasn’t changed in over 100 years.

            Neither Boltzmann nor Planck recognize two-way photon flow between two opposing walls in their radiation oven.

            They only recognize ONE standing wave per wavelength.

            If there really were two-way photon flow, then the photon gas density in the oven would be double what they derived. It is not.

            I will take Boltzmann and Planck over people who never understood real science, and simply extended Clausius’ 1855 statement to radiation.

          • Avatar

            James McGinn

            |

            Zoe:
            Why do you disagree with Boltzmann and Planck?

            James:
            I don’t disagree with these gentlemen. You are confused as to what they are saying.

            Zoe:
            Their two formulas explain emission. This hasn’t changed in over 100 years. Neither Boltzmann nor Planck recognize two-way photon flow between two opposing walls in their radiation oven.

            James:
            You’ve been confused by semantics. Neither Boltzmann nor Planck did not fully realize that energy moves in all directions at all time. ENERGY MOVES IN ALL DIRECTIONS AT ALL TIMES. The net flow of energy is from hotter to colder.

            That’s it. It’s that simple. All of this crap about one way, two way, standing waves–it’s all crap.

            Zoe:
            They only recognize ONE standing wave per wavelength.

            James:
            Standing waves DO NOT BLOCK THE FLOW OF ENERGY.

            The Public Believes Plainly Dumb Things About the Atmosphere
            https://anchor.fm/james-mcginn/episodes/The-Public-Believes-Plainly-Dumb-Things-About-the-Atmosphere-eacv0k

            James McGinn / Genius

          • Avatar

            geran

            |

            James, did you know that accusing someone of lying when they’re not lying is immature?

            You have now added immaturity to your uneducated and incompetent resume.

            You stated: “Now show us that standing waves block the transfer of energy. You can’t.”

            Obviously you were unable to find an adult to help you read and understand the link I provided. Because, the last line of the opening section stated: For waves of equal amplitude traveling in opposing directions, there is on average no net propagation of energy.

          • Avatar

            Zoe Phin

            |

            James,

            “Neither Boltzmann nor Planck did not fully realize that energy moves in all directions at all time. ENERGY MOVES IN ALL DIRECTIONS AT ALL TIMES. The net flow of energy is from hotter to colder.

            That’s it. It’s that simple.”

            Facepalm!

            Please tell us which experiments proved two-way normal photon flow between two opposite objects.

            Tell us who originated the theory you aspouse.

            If Boltzmann and Planck were wrong then their respective formulas would both have an additional 2x factor.

            Yes, Boltzmann and Planck did NOT recognize B.S, and that’s why they got correct results and considered the fathers of radiation theory.

          • Avatar

            James McGinn

            |

            Geran:
            James, if you weren’t so uneducated and incompetent I wouldn’t have to help you so much.

            Geran:
            Address the issue, you evasive POS. You stated that standing waves prevents the transfer of energy from cold to hot. Nobody else is going to support you on this. Plainly, you are talking out of your ass.

            None of you slayers actually understands any of this. You are all engineers who have mistaken your equations for reality.

            Don’t tell me other people believe it unless you are able to provide names and references?

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

            I know what a standing wave is, you lying POS. Now show us that standing waves block the transfer of energy. You can’t.

            Geran, you are just a confused engineer.

            James McGinn / Genius
            The Public Believes Plainly Dumb Things About the Atmosphere
            https://anchor.fm/james-mcginn/episodes/Much-of-Science-Involves-Models-That-Have-Been-Dumbed-Down-to-Pander-to-the-Public-e9c1vd

          • Avatar

            James McGinn

            |

            James:
            Neither Boltzmann nor Planck did not fully realize that energy moves in all directions at all time. ENERGY MOVES IN ALL DIRECTIONS AT ALL TIMES. The net flow of energy is from hotter to colder. That’s it. It’s that simple.

            Zoe:
            Please tell us which experiments proved two-way normal photon flow between two opposite objects.

            James:
            ENERGY MOVES IN ALL DIRECTIONS AT ALL TIMES. No experiments proved two way. Moreover, no experiments proved ‘one way.’

            Zoe:
            Tell us who originated the theory you aspouse.

            James:
            Boltzmann and Planck.

            Zoe:
            If Boltzmann and Planck were wrong then their respective formulas would both have an additional 2x factor.

            James:
            Boltzmann and Planck are correct. You are confused.

            Zoe:
            Yes, Boltzmann and Planck did NOT recognize B.S, and that’s why they got correct results and considered the fathers of radiation theory.

            James:
            Boltzmann and Planck knew full well that energy moves in all directions at all times.

            James McGinn / Genius
            The Public Believes Plainly Dumb Things About the Atmosphere
            https://anchor.fm/james-mcginn/episodes/Much-of-Science-Involves-Models-That-Have-Been-Dumbed-Down-to-Pander-to-the-Public-e9c1vd

          • Avatar

            geran

            |

            James, you’re just copying/pasting the same comments again. All that indicates is your inability to learn. Hince you remain uneducated and incompetent.

            But, you’re hilarious.

            More please.

          • Avatar

            Zoe Phin

            |

            James,
            Hilarious.
            The guy who didn’t read Boltzmann and Planck’s derivation is telling the girl who did read what it’s all about.

            You shove stupidity in the face of those who did their homework.

            You’re an arrogant idiot.

          • Avatar

            geran

            |

            James, when I answer your questions, you still don’t learn. Which proves me correct — you’re uneducated and incompetent.

            I enjoy being proven correct.

            More please.

          • Avatar

            Zoe Phin

            |

            James,
            You remind of that kid in 8th grade who summarized the movie version of a book. After finishing, the teacher gave him an F, because he obviously didn’t read the book.

            That’s you.

          • Avatar

            geran

            |

            James, my exact wording is in my comments above. As well as the quote I supplied. You can’t learn, thus proving me right again, that you are uneducated and incompetent.

            Prove me right some more, please.

    • Avatar

      James McGinn

      |

      Rick:
      people take a statistical equation to assume that kinetically no energy can be emitted by the cooler side to the warmer side, which is not true. Photons do not phone ahead. Hence do not confuse kinetics with thermodynamics. It is true the cooler wall cannot increase the overall temperature of the warmer wall – but they will nevertheless exchange energy and establish a (diminishing) net flux gradient until they equilibrate to an ensemble temperature according to their individual starting temperatures.

      James:
      Rick, well stated. This has been explained to over and over again, to no avail.

      Some people are determined to remain ignorant and there is nothing you can do to educate them.

      Simple Experiment Proves Moist Air is Heavier (not lighter) Than Dry Air
      https://anchor.fm/james-mcginn/episodes/Simple-Experiment-Proves-Moist-Air-is-Heavier-not-lighter-Than-Dry-Air-eabt10

      James McGinn / Genius

      Reply

      • Avatar

        Rick

        |

        Yes, I’m outa here… these guys are a joke.

        Reply

        • Avatar

          Zoe Phin

          |

          When climate junk scientists couldn’t convince real scientists, they too said we’re outa here.

          Then they went to stupid politicians and textbook publishers, and now you’re brainwashed, and don’t know what real science says.

          Mission accomplished.

          The sad part of reality is that sometimes persistence can triumph over truth.

          Reply

    • Avatar

      Jerry Krause

      |

      Hi Rick,

      From the beginning I understood what you were writing, but I did have to work at a little. So merely tried to say the same thing and different way so maybe more people might understand the principles to which you refer.

      Somewhere I just quoted the explanation (the theory) for what you and I observe to be occurring. It is pure magic to those who disregard the quantum mechanical physicists because they (these physicists) admit they do not understand their theories, but the fact is that these strange theories do explain many things about small bodies matter that the classical physics of large bodies of matter could not explain.

      Because you have not mentioned what I quoted I suspect (but do not know) you are not familiar with this theory of Einstein which Richard Feynman taught his students at Caltech.

      “Thus Einstein assumed that there are three kinds of processes: an absorption proportional to the intensity of light, an emission proportional to the intensity of light, called induced emission or sometimes stimulated emission, and a spontaneous emission independent of light.” (pp42-9, The Feynman Lectures on Physicsptio

      One absorption, two emissions. Thus, I understand, in the case of your two walls, the emission of the 10C absorbed by the wall of 100C is by induced emitted back toward the 10C wall plus the radiation that 100C wall is also spontaneously emitting toward to 10C wall according to its temperature according to S-B radiation law. Which due to this spontaneous emission the 100C wall is cooled. Now the 10C wall which was formally cooled by its spontaneous emission toward the 100C wall is warmed by the absorption of the two emissions from 100C wall. And back and forth these three processes occur until the temperatures of the two walls become equal. But the these three process do not stop even though the changes of temperatures do.

      Have a good day, Jerry

      Reply

      • Avatar

        geran

        |

        Jerry, you’ve yanked that quote out of context, meaning you have no idea what you’re talking about. Notice the context (bold my emphasis): “Now Einstein proposed that when such an atom has light of the right frequency shining on it…”

        Feynman was talking about higher energy photons, not the low-energy photons associated with thermal energy.

        You could admit that you don’t know what the heck you’re talking about, but instead, you entertain us with your hilarious pseudoscience.

        More please.

        Reply

      • Avatar

        Jerry Krause

        |

        Hi Rick,

        I see that I did not emphasize at all that the absorption radiation was (is) proportional to the intensity of the incident radiation.

        Which creates another question: What happens to the portion of the incident radiation which is not absorbed? Somewhere in Feynman’s lectures he taught that opaque matter, which does not transmit the radiation, reflects most all of the incident radiation except for the tiny portion which is absorbed at the surface and in a very, very shallow surface layer. It is only because of the achievements of these modern quantum mechanical physicists and the facts of what is observed that I accept such statements. But the knowledge of these men was proven by this small group of physicists who designed three bombs, which were constructed according to these designs, and which worked the first three times they were tested.

        Have a good day, Jerry

        Have a good day, Jerry

        Reply

      • Avatar

        Jerry Krause

        |

        Hi Geran,

        Where in your quote: “Now Einstein proposed that when such an atom has light of the right frequency shining on it…” is there any definition of what the right frequency? I considered it was common knowledge this was the condition for any atom, molecule, to absorb any radiation.

        Have a good day, Jerry

        Reply

        • Avatar

          geran

          |

          Jerry, the source is the same as yours. Are you disputing Feynman’s words?

          Don’t start off in another direction. You took something out of context. Either you didn’t understand it, or you were trying to be deceptive, or both.

          Address the issue that you got caught, or do you want to be a clown, running from your own comment?

          Reply

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    geran

    |

    Jerry, local clouds are weather, not climate. Over the entire planet, the net effect of clouds is to reduce temperatures. During the day, clouds reduce incoming solar more than they restrict cooling at night.

    Reply

    • Avatar

      Jerry Krause

      |

      Hi Geran,

      Climate is merely the average of weather. So if you wish to understand climate you must first understand weather. Don’t in your reasoning overlook this fact. And don’t overlook the fact that climates of different locations are greatly different.

      Have a good day, Jerry

      Reply

      • Avatar

        geran

        |

        And low clouds, at night, at one location, cannot be extrapolated into anything meaningful about Earth’s climate.

        Don’t overlook that fact.

        Reply

        • Avatar

          Jerry Krause

          |

          Hi Geran,

          Maybe you need to this study the impact clouds have before you make such a statement. R.C. Sutcliffe, a meteorologist, wrote (Weather & Climate, 1966): “Taking an overall average, about 50 per cent of the earth’s surface is covered with cloud at any time.”

          Have a good day, Jerry

          Reply

          • Avatar

            geran

            |

            Much better, Jerry. “50%” has significance over just one location.

            And that significance would, of course, be that an increase to 60% would mean a decrease in insolation and corresponding reduction in surface temperatures.

            Very good.

  • Avatar

    Rick

    |

    Hi Jerry
    Yes your uninsulated house will have much greater temp variation (max and min) vs the insulated. The same is true with soil depth and the heat capacity there depends if the soil is damp or not.. the temp variation is greatest at the surface and much more lag > 1,5m down.
    The warmest days will be those with no clouds and clouds that move in at night that clear by morning.

    Reply

  • Avatar

    T. C. Clark

    |

    Place a 1000 watt light bulb and a 1 watt light bulb on opposite walls in an enclosed room. Will the 1 watt bulb add any illumination to the opposite wall? Yes, but your eye probably cannot discern it…need special equipment to measure the small contribution of the 1 watt bulb. Can you cook a turkey with an ice cube? O course not – but take a turkey frozen to -20C and surround it with ice at 0 C and you can begin to thaw the turkey…..using ice. Heat experiments don’t have to be about higher temps….temp exists down to absolute zero….which has never been attained….and probably cannot be attained.

    Reply

    • Avatar

      geran

      |

      T.C., that quaint analogy is like poison to GHE believers. They try to claim that low energy photons can warm Earth’s surface. They claim that photons, even though they are low energy, would be “adding energy”. They don’t have enough technical background to recognize the flaws in their “logic”.

      Most people realize that you can’t bake a turkey with ice, even though the ice is emitting photons. In fact, ice emits “hotter” photons than the CO2 15μ photon, which makes the analogy even funnier. So the analogy is just one more easy way to debunk the GHE nonsense.

      Reply

  • Avatar

    tom0mason

    |

    Joseph Postma has a new video at his site describing the peer review and how his paper was rejected.
    My comment at his site https://climateofsophistry.com/2020/01/26/ams-official-sun-does-not-create-earths-weather/#comment-61095 is

    Joseph, you have shown that AMS is beyond logic and science.
    The paradigm that AMS subscribes to, namely that atmospheric recycling heat energy to provide the climate, is just illogical madness!
    The 3 steps to madness —
    1. The idea of reducing the solar input to a theoretical average spread evenly over a flat surface (representing the TOTAL Earth’s surface area).
    2. Heat (as radiant energy) can go from a cooler body to a warm one.
    3. Observational evidence is of no consequence as the math fits.

    This is a sad moment but it shows how much the AMS is in the thrall of the current madness and hysteria.

    Reply

    • Avatar

      Zoe Phin

      |

      Postma still doesn’t understand the difference between 12 and 24 hours.

      Reply

      • Avatar

        josh

        |

        zoe could you make a model that includes all the variables you talk about and see if you can get it admitted?

        Reply

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    James McGinn

    |

    The reality is that meteorologists don’t give a shit about the facts of the atmosphere. Thermodynamics is just one of a number of approaches that they absolutetely refuse to consider. You guys are just scratching the surface.

    Meteorology’s theoretical thinking is just marketing. It has evolved over the years to coincide with what the public generally assumes. Consequently they are not going to do anything that would risk revealing to the public that it actually is nonsensical.

    Meteorology is not an actual science and it has actually never been anything but a long, continuing conversation about a subset of physical phenomena.

    Meteorology doesn’t debate. They don’t do experiments. And they are extremely careful and even skilled at using imprecise terminology so that debate and experiments never threaten their percieved credibility.

    The fact that you guys have isolated yourselves by conceptually focussing on the one area of science you understand comprehensively–thermodynamics–makes it easy for them to dismiss you.

    There is a larger puzzle of truth that needs to be put together one piece at a time. You guys have discovered one of these pieces. Congratulations. Now you all need to stop waving your hands in the air declaring that you have the whole puzzle. You don’t. You have one piece!

    Meteorology has failed to understand storms
    https://anchor.fm/james-mcginn/episodes/Meteorology-has-failed-to-understand-storms-e91i9b

    James McGinn / Genius

    Reply

    • Avatar

      Jerry Krause

      |

      Hi James,

      You wrote: “Meteorology doesn’t debate. They don’t do experiments. And they are extremely careful and even skilled at using imprecise terminology so that debate and experiments never threaten their percieved credibility.”

      You know I have more than once referred PSI readers and you to the data of at least 5 USA government projects which have been measuring fundamental meteorological factors and radiation data for more than a decade and most for two or three decades. Yet, I cannot remember you commenting about any of this data.

      So speak for yourself James.

      Have a good day, Jerry

      Reply

      • Avatar

        Jim Stupid

        |

        Don’t even try Jerry, he’s a lost cause

        Reply

    • Avatar

      Coward Jim

      |

      The “revolutionary” “scientist” Jim McGinn is back!

      For the record, he shied away from an email thread we had after I flooded him with literature from the past 100 years about exactly what he’s been foolhardily investigating with ill-informed thought experiments and silly conclusions. Maybe if Jim knew how to research properly he would make it to this century in scientific knowledge.

      Reply

  • Avatar

    John12755

    |

    New Reader:
    Zoe-99.9
    Geran, Rick, and the rest of the opposition-0.1
    Your winner is…

    Reply

    • Avatar

      Zoe Phin

      |

      Talking someone’s identity is a crime. Shame on you.

      Reply

      • Avatar

        Lol Rekt

        |

        Yeah maybe if I applied for a loan or something.

        Anyways where are you, Morrisville or Richardson

        Reply

      • Avatar

        Ironic Much

        |

        Also BTW super ironic how you mention that you don’t use python because it’s too abstract for you but you claim to have debunked climate change? Silly Zoe

        Reply

        • Avatar

          Zoe Phin

          |

          I have over 10 years experience working with command line CSV file processing.

          I spent 2 years researching climate.

          My longest python code is 6 lines long, and I copied it from a manual, so I could do one thing – extract data and manipulate it the only way I know how.

          If I put my mind to it, I’m sure I could figure out python as well. I’ve had no need.

          Who are you?

          Reply

        • Avatar

          Finn McCool

          |

          On the contrary, Ironic Mulch. The shell scripts Zoe writes are masterpieces in using awk and other bash commands to do her data analysis.
          One thing a programmer never does is criticise the language of choice to perform analysis. Her code is on her website. If you don’t understand it or disagree with it, I’m sure she can help you out.

          Reply

          • Avatar

            Finn McDouche

            |

            Nah just a crack at her intelligence. She of all people should know that there are more complicated things than linear regression, especially if you trim your data to the last 200 years. It’s effectively similar to using stock data from the past 10 minutes when you compare it to geological times

  • Avatar

    Rosco

    |

    Wow !!

    And I got banned for supposed insulting comments but anything I’ve ever said pales into insignificance compared to the arrogant ego trips I’m witnessing here !

    A sphere with Earth’s radius irradiated by the solar radiation is continually irradiated by ~1.75 x 10^17 watts – continuously.

    Half of the Earth’s surface area lies between 30°N and 30°S with a minimum radiative flux of ~830 W/m2 – continuous influx.

    Three Quarters of the Earth’s surface area lies between 41.41°N and 41.41°S with a minimum radiative flux of ~719 W/m2 – continuous influx.

    86% of Earth’s surface lies between 60°N and 60°S with a minimum radiative flux of ~478.8 W/m2 – continuous influx.

    Note I haven’t factored in dawn to dusk values but the ratios are the same.

    All my values were calculated by a spreadsheet using a series of spherical cap heights to give 6371 equal surface area segments of 40030.17 square metres over the hemisphere and the appropriate cos value of the incidence normal at each calculated cap angle. With the variation of 1 kilometre cap height the variation of the cos of the angle of incidence over each segment area is very small 1/6371 = ~1.6 x 10^-4.

    But of course it is simply wrong to dismiss the fact that geothermal energy exists and it should be accounted for !

    Reply

    • Avatar

      Zoe Phin

      |

      But by continuous, you really mean 12 hours.

      The dark side of all those latitudes get ZERO!

      Reply

      • Avatar

        Rosco

        |

        No I mean continuous.

        From dawn the insolation increases to a maximum when the sun is at the zenith and then it decreases. Meanwhile east of what erver point you choose the same is happening.

        Atmospheric heating lags several hours behind and the irradiated surface begins to cool slowly. There is significant retained heat in the surfaces especially the oceans.

        Yes of course the unlit side of a sphere gets zero – anyone should comprehend that.

        There is ample stored energy in the heated land and ocean surfaces to maintain temperatures over a 12 hour night.

        The minimum temperature where I live is regularly 23 – 25 °C in summer – the air rarely cools to less than that whilst our average maximum is ~29 – 33 °C. I live 1.5 kilometres from the Pacific on the east coast and from a minimum of 23°C at 5:00 am just before the sun rises it regularly reaches 29 °C by ~8:00 am.

        Surely Robert Wood had it right when he ascribed the retention of temperatures in the atmosphere to the low radiating power of a gas..

        Reply

        • Avatar

          Zoe Phin

          |

          Rosco
          “There is ample stored energy in the heated land and ocean surfaces to maintain temperatures over a 12 hour night.”

          Yes, because 240 W/m^2 is stored.
          Yes, because geothermal is active too.

          If Postma wants 30C, then he has to emit 480 W/m^2 as he’s receiving 480 W/m^2 – all in 12 hours.

          Reality is correct, but Postma is not.

          Why is that so hard for Postma fans to understand?

          “where I live …”
          We’re analyzing Postma’s model, not making real life observations.

          Reply

    • Avatar

      Zoe Phin

      |

      Rosco,
      Could you ask Postma if he really thinks he’s got 30C on one side and then -18C on the other?

      Because he’s clearly copying energy. If he wants 30C on one side, he’ll actually have 0K left over for night time, for the reasons I already gave.

      Reply

      • Avatar

        Rosco

        |

        Nobody think we have 30C on one side and then -18C on the other !

        Seriously that is one of the stupidest comments I have ever seen.

        We all should know the minus 18°C exists at ~5000 m – it actually does exist there whether or not the hemisphere is in day or night.

        Personally I don’t get why you destroy your perfectly reasonable arguments with aggressive nonsense like this.

        Averages do not explain anything much at all. Real time flows indicate reality. With no GHGs at all the Moon’s surfaces heat to ~120°C and have sufficient heat capacity to reduce the cooling rate by radiation alone to less than 0.6°C per Earth hour.

        If he wants 30C on one side, he’ll actually have 0K left over for night time, for the reasons I already gave. – CRAP !

        Reply

        • Avatar

          ZoeBrain =Brick

          |

          Don’t bother mate people like this are dumb as bricks. Just as complacent too

          Reply

        • Avatar

          Zoe Phin

          |

          The warming rate is also 0.6C per Earth hour.

          What you call crap is your abandonment of reason.

          Reply

          • Avatar

            Zoe Tard

            |

            I think you just fail to see past the simplest models and get caught up when a naive assumption (like a model that assumes ideal x, y, z) is actually in slight contradiction with what really happens (which is the point of a goddamn model)

            You really wanna see how the earth heats up? How bout you run a simulation of every molecule on the surface and in the atmosphere and the probability that the molecule absorbs a photon from the near-blackbody sun, and then simulate how that system evolves over time? Does that seem feasible?

        • Avatar

          Zoe Phin

          |

          Ask Postma if He thinks there’s a TRANSITION from a max of 30C to -18C.

          Does he not understand that to have 30C you have to emit 480 W/m^2?

          E = esT^4

          Reply

          • Avatar

            geran

            |

            Zoe, you still can’t understand Postma’s graphic. And you’re confused about Geraint’s post. You can’t seem to get anything right. Yet you just keep commenting, wthout having a clue.

            Hilarious.

            More please.

          • Avatar

            Zoe Phin

            |

            The graphic is a physical impossibility.

            He has two hemispheres emitting 240, yet believes one hemisphere can be 30C.

            SB Law is only for emission, and not for absorption. If he wants 30C, he needs to emit 480, which leaves nothing for the night. There’s no 240 physically available for the night, yet he has it there anyway (bottom/middle right).

            He has no right to put 30C anywhere in his diagram.

          • Avatar

            geran

            |

            Still not understanding are you, Zoe?

            You can’t figure out simple things, yet you believe you know it all.

            Hilarious.

            More please.

          • Avatar

            Jerry Krause

            |

            Hi Zoe and PSI Readers,

            Galileo is said to have stated (as translated by someone: “I have never met a man so ignorant that I couldn’t learn something form him.”

            Zoe wrote: “SB Law is only for emission, and not for absorption. If he wants 30C, he needs to emit 480, which leaves nothing for the night.”

            Zoe recognizes Joe’s model’s problem. When one averages the temperature of a day, there is no longer a ‘natural’ 24hr temperature oscillation.

            Have a good day, Jerry

          • Avatar

            geran

            |

            Sigmund Einstein Jones IV once said: “If you can’t understand the issue, at least come up with something irrelevant”.

            Postma’s model is NOT indicating the planet is emitting at 30C. The 30C is indicating the equilibrium if there were a one-sided plate absorbing/emitting 480 W/m^2.

            It’s hilarious that you two have to make up false things about Joseph’s work, yet you ignore all the pseudoscience of the GHE model.

          • Avatar

            Zoe Phin

            |

            OK, Geran, then what average temperature is achieved on Postma’s hemisphere in daylight?

          • Avatar

            geran

            |

            Zoe, the model does NOT address daylight temperature. You still can’t understand the simple model. Earth’s average surface temperature, 288 K, is the same in the model. The simple model just indicates there is no need for artificial “warming”. The “33 K” is nonsense.

          • Avatar

            Zoe Phin

            |

            “Zoe, the model does NOT address daylight temperature.”

            Or night temperature.
            Glad you admit there’s nothing useful in his model.

            “Earth’s average surface temperature, 288 K, is the same in the model.”

            No, 15C is not in the model. How could it be? You’ve already said there’s no day time temperature, so how could you know what overall avg his model yields?

          • Avatar

            geran

            |

            See, you can’t understand the simple model.

            Hilarious.

  • Avatar

    CD Marshall

    |

    Relevant or not Phillip Mulholland did work on the CERES image and came up with an average 27.9C for the Hadley Cell at the Equator where most of the DWR falls, covering 50% of the planet and supporting the 30C surface temperature at the Equator.

    The Hadley cells occupy 50% of the surface area of the globe, and in total intercept 60.9% of the sun’s insolation.
    The Ferrel cells occupy 41.75% of the surface area of the globe, and in total intercept 36.29% of the sun’s insolation.
    The Polar cells occupy 8.25% of the surface area of the globe, and in total intercept only 2.81% of the sun’s insolation.

    https://wattsupwiththat.com/2019/05/19/calibrating-the-ceres-image-of-the-earths-radiant-emission-to-space/

    I really think they were on to something that could propagate real time DWR even more accurately if they could calculate cell/surface and temperature average and I think it could be an evolution step in Joseph’s model, in my humble and simple opinion.

    Reply

    • Avatar

      Zoe Phin

      |

      The keyword is “equator”

      Reply

      • Avatar

        Zoe Omg

        |

        Wait Zoe are you a flat earther?

        Reply

        • Avatar

          Zoe Phin

          |

          No, I’m simplifying to a double hemisphere 12/24 hr dynamic.

          Postma is a half-earther.

          Reply

      • Avatar

        CD Marshall

        |

        Keyword “real time solar irradiance”

        Zoe, I don’t recall Joseph saying his model was perfect just better than the current ones. The model shows real time input which would naturally be mainly at the Equator (30C).

        No model would be better than monitoring real time solar input on the Earth 24/7 but we are not there yet.

        Reply

        • Avatar

          geran

          |

          CD, Zoe can’t find anything wrong with Postma’s model, so she has to make things up. She’s still miffed that Postma banned her because she was acting like such a child. Over on another thread, she is competing to be the “Most Incompetent”. On this thread, she is competing to be the “Most Immature”.

          She could likely win both competitions….

          Reply

        • Avatar

          Zoe Phin

          |

          Chris,
          The hemispheric input ranges from 960 to 0 W/m^2, with 480 W/m^2 being the average. He’s claiming 30C for the whole hemisphere. To achieve that, 480 W/m^2 must be emitted. This leaves 0 W/m^2 for 12 hours of night.

          Postma is a dead end.

          Reply

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