Dumbest Math Theory Ever

421 thoughts on “Dumbest Math Theory Ever”

1. I didn’t get where your 381.7 W/m^2 for upwelling radiation came from, but another way to balance the energy fluxes to and from the surface is to equate 168 to the sum of sensible and latent heat plus 40 W/m^2 radiated directly to space. That leaves only 26 W/m^2 needed from geothermal for balance. I suspect that the geothermal component is even lower than that and that sensible and latent heat are underestimated.

   LikeLike

   Reply

   1. I’m not merely trying to balance numbers. I’m trying to explain why it’s ~15C at the surface with enough energy for evaporation and thermal mechanics.

      The atmosphere needs ~520 W/m^2.

      See https://phzoe.wordpress.com/2020/02/13/measuring-geothermal-a-revolutionary-hypothesis/

      LikeLike

      Reply

      1. Actually your post seems all about math and an energy budget balance:

         “315.7 + 168 – 24 – 78 = 381.7 = Upwelling Radiation”

         I wrongly attributed 26 W/m^2 as geothermal gain to the surface when the 26 W/m^2 belongs in the loss category instead. To maintain 288 K, the surface only needs to shed an additional 26 W/m^2 (plus whatever geothermal adds) while gaining 168 W/m^2 solar and losing 142 W/m^2 by sensible and latent heats, and IR directly to space. So if geothermal contributes a negligible amount, upwelling infrared needs to contribute at least the 26 W/m^2 from the surface to the air.

         168 = 24 + 78 + 40 + 26

         The atmosphere only needs 235 – 40 = 67 + 24 + 78 + 26.

         LikeLike

         Reply

      2. Zoe Phin March 4, 2020 at 7:56 pm

         The atmosphere needs ~520 W/m^2.

         Why would the atmosphere need ~520 W/m^2?
         At the top of the atmosphere Earth is losing on average ~240 W/m^2.
         To maintain the status quo only ~240 W/m^2 has to be re-supplied, what the sun is doing nicely.
         We have an ENERGY balance.
         What you (and the Greenhouse believers) are calculating is RADIATIVE balance.
         Does exist more or less on the dayside of the moon, not on Earth.
         Better calculate how much energy eg one day of sunshine delivers to the upper few meters of our oceans.
         You’ll find that a good day of sunshine delivers 20-30 MJ/m^2, enough energy to increase the temperature of the upper 5-10 m of water 1K.
         We need to think in 3 dimensions for the absorption of solar energy. The silly 2 dimensional plots are meaningless.

         LikeLike

         Reply

         1. “At the top of the atmosphere Earth is losing on average ~240 W/m^2.”

            To what is it losing 240? Matter can only heat other matter. The flyby satellite will get 240, but it’s a tiny satellite.

            I am thinking in 3D.

            The atmosphere needs ~540 to be an atmosphere, otherwise it shrinks. 240 is an AVERAGE for the whole atmosphere. The bottom needs to be warmer. It is warmer. And it got warmer from geo + solar. If we had just solar (no geo), the atmo and subsurface would dissipate the incoming 340 in 3 dimensions.

            There is a reason why the moon is below BB-balanced temperature – because the sun is dissipated into subsurface conduction. But if the moon did not have its own energy supply, the dissipation would continue to the core – decreasing surface T even more dramatically.

            LikeLike

         2. Zoe Phin March 6, 2020 at 1:38 pm

            To what is it losing 240?

            Space perhaps?
            When Earth loses ~240 W/m^2 on average, it just needs ~240 W/m^2 to MAINTAIN its temperature.
            The reason why our surface temperatures are so high is a totally different matter.

            There is a reason why the moon is below BB-balanced temperature – because the sun is dissipated into subsurface conduction.

            The moon is actually WARMER than radiative balance temperature.
            N&Z calculate ~154K, simple calc gives ~162K (albedo .11)
            So whatever you think the reason is, it’s wrong.

            LikeLike

         3. The BB temperature of the moon is 270.4K. https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

            Losing energy to space itself is strictly PHILOSOPHICAL.

            The sun’s radiation is converted into mechanical energy (kinetic energy). When the sun “goes down” for the night, the mechanics decreases. To suggest that energy is then sent to space is a double counting of energy.

            The sun’s radiative energy was used in real time for kinetic energy. That’s it. Energy was conserved.

            The corpuscular photon is a terrible and confusion-inducing paradigm.

            LikeLike

         4. Zoe Phin March 8, 2020 at 11:41 am

            The BB temperature of the moon is 270.4K.

            Utter nonsense. That’s the Effective temperature of the moon. Has nothing to do with RADIATIVE balance temperature.
            In the solar system I live in we have only one sun, so every planet/moon receives solar radiation on one half of the body at any one time. So RADIATIVE balance temperature for that half can be calculated and then averaged with the RADIATIVE balance temperature for the night side.
            For the moon this gives ~331K and 0K (~3K if you want to include Cosmic Background Radiation).
            So average BB RADIATIVE balance temperature for the moon is 167K.
            GB temperature ~164K.
            More exact calculation like N&Z gave results in a slightly lower temperature.

            LikeLike

         5. So NASA labeled it wrong? and you derived an alternative thing, and want to label it the same thing and then claim I’m wrong? Ben, if you don’t shape up, you will only get one more comment.

            LikeLike

         6. > When the sun “goes down” for the night, the mechanics decreases. To suggest that energy is then sent to space is a double counting of energy.

            How does that work? “the mechanics decreases” without energy being lost to anything (?) and that does not contradict the first law of thermodynamics? This is endlessly confusing.
            “The sun’s radiative energy was used in real time for kinetic energy. That’s it. Energy was conserved.”
            So, uh. Energy is “used” in real time for kinetic energy, therefore energy is neither added nor lost? It is “used” – what does that mean in technical terms?

            How is the law called and how does it look like (math)? Why didn’t you account for it in your surface temp simulation?

            LikeLike

         7. The sun is like gasoline in your tank. While you have gasoline, you can drive. When it runs out, your car slows down, then stops. What’s confusing about that?

            Why does anything have to be emitted? The solar energy is making the molecules dance. The dance doesn’t slow down the molecules, the lack of the sun does.

            LikeLike

         8. Radiation is turned into chemical energy by plants, in the end you get oil.
            [Now let’s say you step on the accelerator for a few seconds and then your tank is empty – let’s also do this in a vacuum to reduce complications :D; let’s also ignore and inefficiencies in the engine].
            The chemical energy in oil is turned into kinetic energy.
            The kinetic energy is turned into heat via friction between [for example] the tyres and the road.

            In short, all of the energy from the oil is turned into kinetic energy which is then entirely turned back into heat.
            Any ineffeciencies refer to energy getting converted into heat for some reason or another.

            You propose something different that also slows the car down. I have absolutely no idea what that is. Hence the question: Name of the effect and formula. What does “used” mean if it does not refer to conversion or transfer of energy? And where does the energy go? It does not disappear, does it?

            In explaining, keep in mind that Voyager is speeding along at kilometers per second without anything slowing it down or speeding it up, so that its kinetic energy is constant [if we ignore the very weak gravitational pull from the sun and planets].

            LikeLike

         9. I don’t understand why your thinking is so convoluted. Solar radiation on the ground causes thermalization: A conversion of vibrational (EM) energy into translational energy. There is no room for emission because that translational energy (what you feel as “heat”) is already due to a conversion of electromagnetic to mechanical energy – at the molecular level.

            You seem to want to undo the motion (expenditure of energy) and turn it back into EM.

            LikeLike

         10. “Convoluted” whatever.

             In your response you told me what does not happen – I am interested in what *does* happen.

             Let’s repeat this question out of the many: Why do cars slow down when the tank gets empty and Voyager does not?

             LikeLike

         11. We don’t consider large scale directed motion as the same thing as kinetic energy of RANDOMLY moving molecules. We don’t think of Voyager’s main motion as thermal.

             Why are you even focused on a literal interpretation of my analogy?

             LikeLike

2. Here we see that Earth’s geothermal energy is capable of delivering 0 °C to the surface; This is equivalent to 315.7 W/m².

   Geothermal heat is even capable of warming the entire crust to just BELOW the surface to even much higher temperatures, like 1500K or more.
   Only condition is that something else heats the surface and thus creates a “blocking” layer, that prevents geothermal heat from reaching the surface until the crust has heated up to match the new situation and the flux can reach the surface again.
   

   If we shut down the sun in this example the surface temperature would continue to drop below the JAN line until the surface temperature is such that the radiation to atmosphere/space is equal to the geothermal energy input.
   For an Earthlike Geothermal Flux this temperature would be ~50K or so.

   see eg https://en.wikipedia.org/wiki/Geothermal_gradient#Variations

   LikeLike

   Reply

   1. This is nonsense. Cold does not prevent hot from heating it. Mass from the core increseases radially, and so you would never expect all that 1500K to reach the top. You’e been on this site for a while, and have seen that CSR != CHF with logic and empirical evidence, and yet you still learned nothing, but keep repeating the same stupid talking points.

      Now apply your stupid talking points to the sun. The sun also has a tiny CHF, therefore according to you, the sun can’t heat much and its temperature “must” drop, to match its CHF.

      Unbelievable. Yes, it will drop below the January line until ~-5 C. The atmosphere shrinks, bringing matter down, and then it will rise to 0C.

      The surface warms the atmosphere, and so heat below must be depleted. The atmosphere doesn’t heat the surface. Learn heat capacity.

      LikeLike

      Reply

      1. We don’t want to mix apples with oranges here. With the suns temperature we are thinking about the photosphere. Which is the far outer atmosphere. The temperature seems to drop as you go deeper, and increase to the Corona. If I am right, and the sun goes atmosphere, ocean, rocky planet, we don’t have much of a clue how hot it is far underneath the photosphere. Not taking sides here. Just a minor note on your comparison.

         LikeLike

         Reply

         1. The conductive heat flux is still small compared to radiation.

            LikeLike

3. In support of your great post, Noether’s law and the principle of least action also call into question atmosphere warming by CO2.

   Fermat’s theorem and the principle of least action apply to the atmosphere’s response to increasing CO2 and show that a vast energy expenditure to heat atmosphere and ocean, is contrary to these laws. They oppose the notion that increasing the trace gas CO2 heats the ocean and atmosphere.

   The principle of least action states that the universe will choose the path between two states that minimises the action. This principle is a generalisation of Fermat’s theorem which requires light to take the path between two locations that minimises the travel time.

   The principle of least action can be extended to any system evolving between two states. It is the founding assumption behind Noether’ theorem that is required to explain why Einsteinian relativity does not break conservation of energy.

   Amalie Emmy Noether (she preferred the name Emmy) was a German mathematician who was born in 1882 – 13 years after my grandfather. In that time she was (sadly and inevitably) under-recognised as a female academic, but made important contributions to abstract algebra and theoretical physics that later would grow further in importance in cosmology and quantum physics.

   Noether’s theorem is fundamental. It allows calculation of the true conserved quantities for any system that is evolving according to the principle of least action. (As long as we can identify the system’s symmetries.) Noether’s theorem is used in both cosmology and quantum physics.

   The principle of least action applies to atmospheric thermodynamics. For instance, the CO2 concentration in air increases. How will the atmosphere’s state evolve as a result? Conventionally we are told that the atmosphere’s response to a small increase in this trace gas is to exert vast quantities of energy to increase the temperature of both atmosphere and ocean. This is an enormous thermodynamic response to this tiny trace gas perturbation, that transgresses the principle of least action.

   However, a response by the system rearranging its structure, changing for instance water vapour content or the emission height, or adjustment of convection or even radiative interactions, could lead the system toward a new equilibrium with much less expenditure of energy. And thus fulfil the laws of least action, Noether’s and Fermat’s theorems. Miskolczi’s hypothesis was of this nature – a rearrangement of the emission structure without temperature change.

   On the other hand, response to the tiny adjustment of CO2 amount by heating up the whole atmosphere and ocean, is the exact opposite of what one would expect in fulfilment of the principle of least action. It’s the principle of most action, and most (empty) heat and noise.

   The enormous response to CO2 is supposed to come from water vapor. Increased CO2 concentrations in the atmosphere are supposed to cause an increase in water vapor and water vapor is what is supposed to overheat the Earth.

   So far, no significant increase in water vapor is seen and no tropospheric “hotspot” has been seen to develop, which is a requirement of the CAGW (Catastrophic Anthropogenic Global Warming) speculation.

   Alarmist predictions are not coming to pass.

   “Initial” is the word that is always ‘forgotten’. CO2 has an INITIAL warming effect. A ‘missing cloud’ also has ‘an initial warming effect’, not changing everything.

   After initial warming the system adapts, probably in a way to search the way of ‘least action’. Always trying to stay close to ‘equilibrium point’ in which already many forces together created ‘equilibrium temperature’. An equilibrium temperature which will not change much by a change in just one single item. Because all other forces will react.

   In line with Fermat and Noether is Le Chatelier’s Principle in chemistry formulated in the latter half of the 19th Century. It became recognized after his death as having a much broader application and being properly a ‘law’.
   From Wiki, the broader statement of it is

   “When a settled system is disturbed, it will adjust to diminish the change that has been made to it”

   That is to say the system resists changes from applied new temperature, pressure, volume and composition. One could equally say the principle is present in Newton’s laws of motion. If I push on a stone wall it pushes back, not budging until I exceed the ‘bending’ strength of the wall. Or ‘back-EMF’ in electric motors. Similarly in economics if we increase price, demand declines and this leads to a supply surplus, resisting the price increase.

   This doesn’t seem to be distinct from Noether’s and Fermat’s theorems.

   (Contributions to this post by Tom Abbot, Wim Rost and Gary Pearse are acknowledged.)

   LikeLiked by 1 person

   Reply

4. Zoe I’m expecting you to really shake this study up and it will be a tragedy if absolutism gets in the way of that. You want to show here that this greenhouse effect is nutty and basically it doesn’t exist at all. That may be going too far. You are right to give up on photon theory. But since we can see the stars that kind of implies that there is aether between us and they, or failing that energy transmission, so that suggests for practical purposes, and probably for all purposes we can say that energy is radiated to space. You are holding your alternative theory a little too tightly. Like a $100 dollar note between the butt-cheeks. Your needle has swung against Einstein and aether denial and rightly so. But its not time to clamp down on that needle quite yet.

   See over at Joannes place, where WXCycles (I call him EpiCycles) is being sarcastic towards you? I’ve got a big history with him, where I take his observations and show very clearly how they imply electrical energy moving down through the atmosphere. He won’t admit I’m right …. BUT I NEVER ONCE DOUBTED HIS OBSERVATIONS. I think he may even be a New Zealand CO2-bedwetter (in public) slumming it with the climate-rationalists. Because I’ve only seen one other fellow whom I feel right to take his observations so seriously, and so I wonder if they are two different people, or just the one talented reprobate.

   Now when he sees columns of high water vapour air but without clouds he is seeing a lot of heat retention. He just calls this “the greenhouse effect”. To my mind that column of air will be blocking more incoming joules than its retaining overnight via greenhouse alone, and so the afternoon temperature will be a somewhat equalised and if it was just greenhouse there wouldn’t be much in it. But if its truly without clouds … and we bring in the magic of latent heat …. well that column, if that heat were NOT latent would radiate this heat out and wouldn’t absorb so much during the day. Being cooler than its unrealised heat, it will absorb more. So Epi-cycles would say greenhouse, I would say a tad greenhouse but mostly this latent heat magic, and also the excellent specific heat capacity of water vapour and also of tiny floating micro-droplets …. (an issue that is under-researched.)

   But I don’t doubt the observations of Epi-cycles. How could I? They consistently prove all my prejudices correct.

   Now how about CO2? Where we don’t have good specific heat capacity like with liquid and gaseous water? Where we don’t have this latent heat capacity?

   So hard to get good data. So I go to Tony Heller for micro-data. I thought I could see some minor CO2-warming in the last five years from the satellite stuff. So I wrote a blogpost on how this could be a false positive. Then Tony showed us that even this data was rigged. But the totality of Tony Heller’s videos does yield some progress here. It doesn’t look like the extra CO2 is causing warming that we can yet detect. Or cooling that we can detect. But it does look like its taking the edge off the hottest part of the day. In the end I think this will be VERY CLEAR.

   But there has to be a reason for this and there has to be a reason for the falling intensity (not prevalence) of extreme weather events. Looks pretty solid to me. And its what you would expect because the light encounters the CO2 first before the water vapour. So there is the opportunity for the CO2 to block incoming and take the edge off the hottest of the afternoons.

   So if CO2 can block incoming. It can also block outgoing a little bit. So perhaps the extra CO2 can take the edge off the coldest mornings for the Lapp-Landers … thats a good thing. But we would have to say that this is something like what these people are calling greenhouse. As one eye blind as they are, about this “greenhouse effect”, It might be going too far to say that this is nothing-at-all. That might be taking a few good ideas and pushing them too far. Personally I wish the phrase had never been invented.

   You have done great work in the past, and you will do great work in the future. I cannot do this work. I lack your skills. I have high hopes for your efforts. You were right to recoil against all that Oligarchical bullshit science that is out there. But I feel you are getting bogged down because you have locked a few assumptions in, as the gold standard or the gospel way too early. You’ve got to ease up a bit. Backpedal a little. And I’ll tell you a symptom of this. I usually thought of Keith as a bit of a dummy. I think he got the better of you over at Joannes place. I was quite surprised. You’ve got about 50 IQ points on this fellow. I think he got the upper hand.

   Feel free to wipe this post in a few days. Get your husband to give you a good massage. Loosen up a bit. You are losing your way. Not because you are on the wrong track. But just because you are being a little absolutist about one or two things.

   LikeLike

   Reply

   1. The correlation between clouds and warmth has backwards causation. Because it’s warm, and there’s water, there are clouds. Where it’s cold, the clouds rain out or never form. It’s a little more complicated than that (dew point, etc), but you get the idea. The warmth makes the cloud and it can travel to a colder area, bringing its warmth with it. This doesn’t prove the greenhouse effect. You can take a hot pan top after you heat it and place it over an ice cube on the counter. This horizontal transfer is not the GH effect.

      Correlationists and GH effect believers are just fabricating causation where none exists.

      So I do recognize the difference between a warm cloudy night and a cool cloudless night. I just don’t think the cloud makes it warm. To oversimplify: It’s warm, therefore there’s a cloud.

      LikeLike

      Reply

      1. Hi Zoe, have you read the Roy Spencer article on GHGT? Would be good to hear you weigh in (on the Facebook thread too)… http://www.drroyspencer.com/2020/03/comments-on-dr-ollilas-claims-that-greenhouse-effect-calculations-violate-energy-conservation/?fbclid=IwAR3l1KbZmeesO5-po_OYExcb3vZwRDgND_5tjIBdRK5KShLf4RAGE6Horkw

         LikeLiked by 1 person

         Reply

         1. Roy’s rhetoric reassures his ideological mathematics.

            Energy can only flow when there’s a potential for it to flow, that is hot -> cold. But if you ignore that, anything is possible!

            “A scientific truth does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die and a new generation grows up that is familiar with it.”

            — Max Planck

            LikeLike

5. Damn that’s a frightening quote from Plank. This ‘victory by age and attrition’ is what I fear will happen with the global warming debate. In other words, the climate zealots have captured the youth and there’s nothing we can do about it.

   LikeLiked by 1 person

   Reply

6. As a retired geophysicist, I am baffled by the problem that you have brought to the fore. There would appear to be something very wrong with our ideas about the interior heat generated by the Earth.
   The interior heat is quoted as being about 47 TeraWatts or thereabouts
   The area of the earth is quoted as being 5.1007 x 10^14 square metres.
   Dividing the first by the second gives a potential flow of 0.08822 Watts per square metre at the surface.
   Applying the Stefan Boltzmann law, that equates to an average surface temperature for the Earth of 35.4 degrees Kelvin.
   Having been underground in a couple of mines, I am quite certain that the temperature was not below freezing point, probably more like 300 degrees Kelvin.
   Is my reasoning wrong or is the estimated heat flow completely wrong?

   LikeLiked by 1 person

   Reply

   1. 47 TW is the conductive heat flow, i.e. kinetic energy DIFFERENTIAL for ONE meter of height.

      I solve the problem here:
      https://phzoe.com/2020/02/20/two-theories-one-ideological-other-verified/

      (and first link within)

      LikeLike

      Reply

7. Bevan Dockery March 17, 2020 at 6:21 am

   Is my reasoning wrong or is the estimated heat flow completely wrong?

   Neither. The FLUX through 10-40 km of rock is indeed very small.
   The temperature gradient from mantle to surface is ao depending on the surface temperature.
   Without input from eg our sun the surface temperature will drop to 30-50K as you calculated.
   When the sun heats the surface, the geothermal gradient will settle at the surface near the average surface temperature.
   

   This is the reason why we have permafrost at high latitudes and a much higher temperature just below the surface in the tropics.
   On the moon we have most probably a similar flux as on the Earth. The Hermite crater where the sun never shines on its floor has a year round temperature of ~25K.
   So yes, our crust is much warmer than the flux alone could accomplish, and yes this is heat that came from within the earth. But the sun needs to provide the “blocking” layer that allows the crust the maintain its high temperature.

   LikeLike

   Reply

   1. The conductive heat flux is also tiny inside the sun! No worries about the sun though, right?

      CHF will always be small due to inverse length in the formula.

      No, it would not drop that low. It would drop to a global average of -5 to 5C.

      The diagram you showed is typical for 64N latitude. You can work out here what the result would be for that location:

      https://phzoe.com/2020/02/13/measuring-geothermal-a-revolutionary-hypothesis/

      The moon has much less energy than Earth. Not the same. Don’t cherrypick craters, the mean lunathermal is like 70-100K, while geothermal is like ~273K.

      “But the sun needs to provide the “blocking” layer that allows the crust the maintain its high temperature.”

      OK, and the sun needs geothermal to get from -42C to 15C plus evaporation and sensible heat.

      The sun can’t do it alone. The sun is not even the #1 player, as hard as it is to accept.

      LikeLike

      Reply

8. This is all really fascinating to me. One question that I have, (no need to jump on me because I’ve completely missed the science) and this is curiosity based: if we think of “space”, e.g. what surrounds the planets asteroids/debris, stars, galaxies and so on, as nothing, or something that cannot be a part of energy loss, is that the same “nothing” that the electrons of an atom are orbiting in? And if it is, doesn’t it seem logical that in order for things to form in the first place, there must be “something” to form in? What other examples can be found of the “nothing”I’ve described? It feels more plausible that the “nothing” is really “something” that we just haven’t figured out.

   And I’m sure there are lots of scientific phenomena that I would find illogical.

   LikeLiked by 1 person

   Reply

9. By your reasoning there is clearly no reason for people to wear clothes. They’re colder than our bodies, they don’t have any heat source of their own, therefore, they are not going to send any heat back to us. And yet, we all know that they do.

   LikeLike

   Reply

   1. Oh we do, do we? Please tell us which experiment(s) you refer to. Clothing KEEPS you warm, it doesn’t raise your temperature. Where did you get your “knowledge”? The climate change cult?

      LikeLike

      Reply

      1. Clothes don’t raise your temperature because your body has a control system that works to keep your temperature approximately constant. In an environment where your body is losing too much heat your control system burns more fuel to keep you warm. One feature of this is that you start to shiver. Clothes act as insulation reducing the amount of heat loss and thus allowing your body to burn less fuel in order to maintain its preferred temperature. So when someone says “clothes keep you warm” that is what they mean. It’s not that clothes are a source of energy, but rather a resistance to the flow of heat out from your body.

         LikeLike

         Reply

   2. Food for thought..
      Clothes increase thermal resistance and decrease convection cooling. If a thermal gradient exists between 2 spots with a thermal conductor between them (eg. heatsink on CPU): if you increase the temperature of the cold side, less heat will initially flow so the hot side temperature increases until the temperature gradient returns to the Kelvin/Watt.

      If the atmosphere warms the surface then warming the atmosphere will increase the surface temperature. If the surface warms the atmosphere then a warmer atmosphere reduces the heat loss of the surface until the surface increases temperature. If the sun provides the heat then it mostly heats the surface and a diminishing temperature gradient occurs with diminishing air pressure until the molecules/atoms are so sparse temperature is less meaningful.
      We’re dealing with spheres, wedge shaped columns of the spheres and very complicated mathematics. A molecule will radiate infrared (IR) in proportion to it’s temperature while absorbing IR or other heat source. All those can occur sometime during the cycle of day&night and overlap. You can’t simplify it down to a square box experiment (eg. greenhouse).

      LikeLike

      Reply

      1. “reduces the heat loss of the surface until the surface increases temperature.”

         Where did you get such a silly idea?
         Reducing heat loss does not lead to a temperature gain. Such an idea is empty rhetoric. With rhetoric, anything is possible! But in science that is not so.

         Instead of rhetoric, try to find an actual experiment that supports you.

         Here’s some more empty rhetoric:

         On a flat road, reduced breaking leads to accelarating.

         Reduced smoking leads to lung cancer.

         Any idiot can come up with a whole bunch of other nonsense with the same formula.

         Did you know that sponges make a spill wetter due to backmoisture? lol

         Please quit the nonsense rhetoric.

         LikeLike

         Reply

         1. It’s not really empty rhetoric. For a system that has both heat input from some source and heat loss to its environment a reduction in the heat loss while maintaining the heat from the source will cause an increase in temperature. This result comes from the first law of thermodynamics.

            dU = Q + W

            dU is the change in internal energy, and for simple systems where there is not a controlled phase change occurring an increase in dU is accompanied by an increase in temperature. Q is the net heat input and W is the net work done on the system. Let’s consider when W=0. Then Q = Q_source – Q_env. If you decrease Q_env then Q increases, and thus dU increases, which under most circumstances means that temperature increases. Again, there are exceptions if you want to be pedantic.

            LikeLike

10. The entire Greenhouse Gases Global Warming theory is an old Climate science

    And here is why:

    Comparison of results the planet Te calculated by the Incomplete Formula:

    Te = [ (1-a) S / 4 σ ]¹∕ ⁴

    the planet Te calculated by the Complete Formula:

    Te = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (1)

    and the planet Tsat.mean measured by satellites:

    Planet…Te.incompl…Te.compl…Tsat.mean

    Mercury……437 K…….346,11 K…….340 K
    Earth………..255 K……288,36 K…….288 K
    Moon………..271 Κ…….221,74 Κ…..220 Κ
    Mars………209,91 K…..213,42 K…..210 K

    Te.incomplete = [ (1-a) S /4σ ]¹∕ ⁴ gives very confusing results:

    Planet…Te. incomplete

    Mercury……….437 K
    Earth…………..255 K
    Moon…………..271 Κ
    Mars…………209,91 K

    When inserting the values, this calculation yields a surface temperature of about Te = 255 K. Even though this value is not quite bad for a very simple model, it still deviates significantly from the actual value of Te = 288 K.

    So the old formula yields Te = 255 K instead of the actual Te = 288 K..
    .
    But we have not finished yet:

    This formula Te.incomplete = [ (1-a) S /4σ ]¹∕ ⁴ also deviates significantly for the Mercury Te = 437 K, instead of the actual Te = 340 K

    and it deviates significantly for the Moon Te = 271 K, instead of the actual Te = 220 K.

    It is time to abandon the old effective temperature
    Te.incomplete = [ (1-a) S /4σ ]¹∕ ⁴ formula, because it gives very confusing results.

    The 288 K – 255 K = 33 oC means the by atmosphere +Δ33 oC Earth’s surface warming phenomenon, which does not exist in the real world.

    There are only traces of greenhouse gasses. The Earth’s atmosphere is very thin.
    There is not any measurable Greenhouse Gasses Warming effect on the Earth’s surface.

    The entire Greenhouse Gases Global Warming theory is based on the old effective temperature

    Te.incomplete = [ (1-a) S /4σ ]¹∕ ⁴ formula the Earth’s effective temperature calculation
    Te = 255 K which does not exist

    When applying the Effective Temperature Complete Formula

    Te = [ Φ (1-a) S (β*N*cp)¹∕ ⁴ /4σ ]¹∕ ⁴ (1)
    we have identical to the satellites measured results:

    Planet..Te.complete..Tsat.mean

    Mercury…..346,11 K……..340 K
    Earth……….288,36 K……..288 K
    Moon……….221,74 Κ……..220 Κ
    Mars………..213,42 K……..210 K

    The Earth’s Te is calculated Te = 288,36 K and it is identical to the measured by satellites the actual value of
    Te = 288 K.

    Also almost identical are the calculated with the Complete Formula

    Mercury…….346,11 K = 340 K
    Moon……….221,74 Κ = 220 Κ
    and
    Mars………..213,42 K = 210 K

    Now everything falls in place and the entire Greenhouse Gases Global Warming theory becomes an old Climate science.

    Because Te = 255 K

    does not exist

    http://www.cristos-vournas.com.

    LikeLiked by 1 person

    Reply

11. I was blind, but now I see!
    Thanks, Zoe.

    LikeLiked by 1 person

    Reply

12. There is back radiation. You can feel it on a cloudy night. Clouds and water vapour provide the back radiation. At nightime without clouds, it can be up to 11C cooler than a night with clouds, all else being =. Deserts are colder at night without clouds than nearby places that have more water vapour. I agree with you that CO2 does not cause back radiation therefore the GHG theory is a scam.

    LikeLiked by 1 person

    Reply

    1. The averaged day AND night temperature of subtropical deserts are warmer than the tropical rainforest.

       The data does support clouds ~ warmer, but it’s a correlation fallacy with causality reversed.

       It’s because it’s warm that there are clouds. If it was cold, clouds wouldn’t form or they would rain out.

       Clouds are formed by evaporation (hot) and rain out due to condensation (cold). If clouds made it warm, one could then argue that clouds aid their own formation, which makes no sense.

       It would be odd for 0.0025 (water vapor) the mass of the atmosphere to warm anything! But the surface and other gases could easily warm water vapor.

       LikeLike

       Reply

13. Click to access PhysTodayRT2011.pdf

    Any comments to this article?

    LikeLike

    Reply

    1. The Earth has a lot mass to heat. So billions of year to heat to some foolish number.
       So one have something less mass and it would heat up faster.
       So can something with 1/billionth of mass.
       How much mass is involved per square meter in regards to earth.
       You have 10 tonnes of atmosphere per square meter.
       Earth is mostly ocean, which has 4000 meter of water per square meter, so that is 4000 tons of water which about
       4 times more earth to heat than air,. Or equal 16,000 tons of air.
       So if what had to heat was equal 1/4 of ton of air, it’s 64,040 times quicker to heat up.
       15,615 years rather than 1 billion years. to just heat air of water of earth.
       A square meter is heated by 240 watts and you increase it to 1360 watt per square meter, and so heat up about
       5 1/2 times faster:: 2839 years..
       And let’s heat up 10,000 times less than 800,000 K, or increase by only 80 K.
       So that is .2839 years or 104 days.
       Well one have blackbody surface insulated face the sun, and part heated has less thermal mass that 1/4 ton of air.
       And it will heat up rapidly to about 120 C {400 K} and it will heat to higher temperature.

       And wacky greenhouse effect theory is based upon this known fact.
       Or it’s why an ideal thermally conductive blackbody sphere at 1 AU distance from the Sun has uniform temperature of about 5 C.

       But if stop all heat from leaving earth, and there was no Sun. the geothermal energy of the radioactive Earth core will have evenually a nuclear “meltdown”- or will melt anything which one tries to insulate with. Because nuclear reaction “burns” at over million degrees- far hotter than the surface of the sun [which is 5,778 K}.
       Or any kind of fire, can only heat at the temperature it burn at.
       Or perhaps you “know” that airplane fuel can not melt steel. {but it weaken steel if gets enough hot}.
       Or jet burns at certain temperature and it’s below the temperature to melt steel.

       LikeLike

       Reply

14. Hi Zoe – why no updates to your blog ? I’m missing your articles !
    One more question : if the green plate math is incorrect, how would you write the correct energy balance for the two plates in order to determine the relevant temperatures ? I have to confess I am scratching my had since a while. Thanks. Max.

    LikeLike

    Reply

    1. Sorry, been busy. This COVID19 thing has been both a curse and a blessing for me. Children were at home while my business was booming. All work, kids, no science.

       Heating is a resonance phenomenon. Assuming the distance between plates is small, such that the view factor is 1 to 1, and there is nothing behind the 2nd plate, the two plates will come to thermal equilibrium.

       LikeLike

       Reply

       1. If heat is a resonance phenomenon, how come hot food in my microwave oven doesn’t reach ambient temperature ultrafast once oven is turned off?

          LikeLike

          Reply

          1. Time. It needs time.

             LikeLike

    2. The ISS can easily test if the greenplate effect is true …

       

       Just line up 8 panels facing perpendicular toward the sun:
       || || || ||

       Adjust for view factor. First plate should get super duper hot. Much more than the sun could make it alone … if greenplate effect was true.

       Of course, they won’t do it, because … they already know the answer, and it doesn’t match climate “scientists” claims.

       LikeLike

       Reply

15. Zoe,
    Hi my dear how are you getting along? I hope well.

    I’ve been sparring with an old friend of yours, EM F a real climate clown. I was wondering if you’d like to say hi to him again? He’s arguing with me right now on geothermal not making ocean floors warmer. So, since it’s your turf I was wondering if you’s like to set his old mind straight? Presently he denies anything I say that doesn’t go along with what he thinks is true.

    The comment is under EM F has 81 comments so far. This is his latest:

    LikeLike

    Reply

16. Thanks Zoe. What I really, really, cannot believe, is how no proper experiment is being performed anywhere in the world to definitively ascertain (debunk) this, for the benefit of both parties (“alarmists” and “deniers”). If you are right (as I believe) this is really a fraud of planetary proportions. It will throw discredit to climatologists and physicists for centuries. How is it possible that honest supporters of greenplate theory (with all its implications), trusting in their theory, are not willing to undergo any experimental verification….(ok maybe I am naive). I’d be even willing to participate to financing such experiment (but I’m a poor guy).

    LikeLiked by 1 person

    Reply

    1. Full disclosure: They think experiments have been done.

       One example is Swanson’s experiment:

       https://app.box.com/s/5wxidf87li5bo588q2xhcfxhtfy52oba

       This experiment is discussed here:

       https://www.drroyspencer.com/2019/06/a-simple-no-greenhouse-effect-model-of-day-night-temperatures-at-different-latitudes/

       I have, however, refuted it. To see the refutation, search the linked page for “pipe”, and read my comments. They cheated!

       LikeLike

       Reply

    2. Well, we know how to keep a house warm, so as to lower heating costs. And it has little to do with radiant heat loss and largely to do convectional heat loss. If want to keep a house warm or cool, large aspect is to keep the door shut {and close the windows}. And can use weather stripping {prevents convection heat loss} and double panel windows- prevents convection heat loss.
       A big aspect is having insulation in ceiling {or roof}- convectional heat loss. And better job, put insulation in the walls [and maybe in the floors}. And generally it’s a good idea to insulate the hot water plumbing {convectional heat lost}.
       Generally one should put twice as insulation in ceiling as walls, and warm air rises. And having air circulation- a ceiling fan and/or central air conditioning prevents warming building up near the ceiling and having warming air near ceiling will cause more heat loss {convectional and conductional heat loss}. And CO2 gas has nothing to do with keeping a house warm {or cool}.
       And generally Earth atmosphere insulates Earth surface- regardless of it’s trace gases. And that called greenhouse effect {though that can be called a misnomer- it has some use in terms of analogous way of explanation, but analogous ways explanation are generally inadequate}. But anyhow an actual greenhouse works because it prevents
       convectional heat loss. And if know that, the term greenhouse effect is not much of problem.

       LikeLike

       Reply

17. Great. Thanks.I’ll have a look to it. Keep this great stuff going !

    LikeLiked by 1 person

    Reply

    1. I’ll do my best.

       LikeLike

       Reply

18. Went through your refutation. Simply brilliant. Btw I love your sponge analogy.

    LikeLiked by 1 person

    Reply

    1. Thank you. Thank you!

       LikeLike

       Reply

19. I would hope that at least the heat transfer professors (not the climatologists, we know they have some difficulties…) would agree on the flaw of the “dumbest math theory ever”…so I contacted professor Howell at thermalradiation.net to ask a question in the most “neutral” way possible…but I either did not understand, or he seem to support the “back radiation” concept….
    I have put the email exchange here : https://drive.google.com/file/d/1OjpDVchnj_N29WN0C-ZSPg_nMEyVhnsJ/view?usp=sharing

    What do thermal physics professors really think ? 🙂

    LikeLiked by 1 person

    Reply

    1. He didn’t answer the question.

       It’s an unrealistic question to begin with. Keeping the shield at a fixed 20C is impossible. That would require an active system, thus destroying the point.

       It’s very important to know what objects are outside the shield, and how far away they are.

       Steady state heat flow is only established from matter to matter. The steel greenhouse assumes emission to space is a heat flow. It is not.

       Assuming nothing is outside the shell, there is ZERO heat flow.

       A better question would just be: will the inside sphere ever warm up due to the shield?

       The answer is a simple NO. An honest thermal physics professor would just say that.

       LikeLiked by 1 person

       Reply

       1. “The answer is a simple NO. An honest thermal physics professor would just say that.”

          Actually the answer is YES if the internal source puts out a constant level of energy. The equations and solution procedure you provided for this problem show exactly that. You simply applied your equations to the fixed temperature case instead of the fixed energy input case.

          LikeLike

          Reply

20. Crystal clear Zoe…thanks a lot !

    LikeLiked by 1 person

    Reply

21. Zoe, talking of experiments : what do you think of Dr Roy Spencer’s alleged “proof” of backradiation ? http://www.drroyspencer.com/2016/08/experiment-results-show-a-cool-object-can-make-a-warm-object-warmer-still/

    LikeLiked by 1 person

    Reply

    1. I think it’s fraud.
       The ice thaws at room temperature and then convects up. The hot plate wastes its energy heating formerly-ice gas. When you block this gas from arriving with a shield, the hot plate heats up. It didn’t really heat up, it was just prevented from cooling due to the mass of gas it was heating before.

       “The hot plate is kept above the ice to minimize any air convection effects on the results.”

       Completely the opposite of the truth. If the ice was above the plate, it would have eliminated convection from the post-ice gas. Sure there is still regular air being convected from hot plate, but it’s not ice gas which has greater energy due to “heat of vaporization”. So in essence, he purposely structured his experiment to yield maximum heat up – including the face of the ice being larger than hot plate. Clever guy he is.

       None of this proves heat-up due to radiative effects. Notice he doesn’t model or predict results using theoretical physics taking into consideration all types of heat transfer. It’s a shame that he stoops to this level. He knows exactly the trick he’s playing.

       Thanks, Max.

       LikeLiked by 1 person

       Reply

       1. The experiment should be repeated using cold rocks instead of ice then.

          LikeLiked by 1 person

          Reply

          1. I specifically remember from school days that the physics teacher said that on a winter night, when the window panes in your house are cold, it helps to draw the curtains. The idea being that the cold panes are then hidden from view of the IR-radiation wanting to exchange heat with them. Any comments to this?

             LikeLike

          2. Your windows are transfering heat to the air by all 3 heat transfer mechanisms. Curtains are mass. More mass cools slower. If your windows were twice as thick, it would also cool slower. Notice that your curtains do not get warmer due to backradiation from the window.

             LikeLiked by 1 person

       2. Let’s suppose the experiment is made in the vacuum so that convective effects can be neglected. While the lamp is heating the plate, the presence or absence of the shield does not affect the plate temperature increase rate. But if the lamp is switched off, then the plate starts cooling, and will do that quicker when the screen is removed from top of the sink…am I understanding correctly ?

          LikeLike

          Reply

          1. It matters what the temperature of the shield is. For if it is colder, then its presence facilitates plate cooling. In this case REMOVING it would make the plate cool slower. But if shield is same or greater temperature, then its removal would force plate to cool faster. If lamp was the only heat source, then you can expect equilibrium to form, and then everything cools to zero.

             When lamp is off, it’s easy to see that plate and shield “block” each other from cooling, i.e. plate and shield can reverse roles. Doesn’t matter though, they’re both going to zero. And the time it takes is a function of their masses and heat capacities.

             Edit: zero = external environment.

             LikeLiked by 1 person

    2. “There is no violation of the 2nd Law of Thermodynamics, which states that the net flow of heat must be from higher to lower temperature, which does not preclude cooler object from emitting IR radiation in the direction of warmer objects.”
       Let’s say somewhat true, or certainly can be true.
       “If the atmosphere didn’t exist, the Earth’s surface would lose IR radiation directly to the cold depths of outer space, which is essentially at absolute zero temperature, and emits no energy back to the Earth”

       “Cold depths of space” is fiction. The nothingness of space is more factual. Though roughly, “emits no energy back” is accurate enough.

       “; but instead the atmosphere, in effect, blocks some of that radiation, and emits some of its own IR radiation back towards the surface.”

       Well the hot thermosphere emits some radiation “back towards the surface” but it’s insignificant- though I believe it’s measurably from the Earth surface. But a question is where does atmosphere block most of that radiation and where does it radiate most of that radiation back to earth surface. I would say within the first 1000 meters of the surface. And greenhouse effect theory does not make such claims. And what else is within the atmosphere in first 1000 meter other than greenhouse gases? Also I will note that Ozone is considered an important greenhouse gas and generally regarded as being much higher than 1000 meter above the surface. It {to me} seems it doesn’t “fit” with the idea of back radiation. I would does not “fit” in similar way the whatever small effects of the hot thermosphere likewise doesn’t “fit” in the idea of back radiation.
       Another issue is what direction does most the radiation from Earth surface go? I would say the radiation radiates in hemispheric direction or most of goes sideways and little goes straight up. And there is lot’s of “things”- like a fence or tree, which are sideways. So Earth has 40% of total area in tropics, 23.5 degrees north and south latitude.
       Cut Earth in half at equator, there is 40% of area which 23.5 degree away from equator, an 30 degrees is almost half and polar region is less than 5%. Or less than 5% is roughly straight up.
       And to “inferfere” with radiation does not require the radiation to be absorbed. Or atmosphere without any gases which absorb certain certain wavelengths of IR, can still interfere with any and all radiation. Whereas in a vacuum, there is no interference by the nothing. Of course another thing, is surface would actually radiate in spherical direction- it it goes a random direction- though that is getting into weeds a bit. And continue to go into the weeds, most Earth is covered by water, and surface of water, is a crazy place- barely, explored.

       “The net effect is that the surface and lower atmosphere cannot cool as rapidly to deep space, raising its average temperature.”

       I would say all surface air and all atmosphere above it, can not cool rapidly.
       Or go back to the thermosphere, it is very hot {though it has no temperature} it’s near the “Cold depths of space” and doesn’t radiate any significant heat/energy.

       LikeLike

       Reply

22. Thanks Zoe. Clearly convective effects alter completely the significance of the experiment.
    Geraint Hughes did a much better job :
    https://principia-scientific.com/bite-sized-niblets-article-number-2-greenplate-effect-fail/

    LikeLike

    Reply

23. Sorry, but I believe you skipped commenting on the essential part “hidden from view”. Is it not so that IR can only heat what is in sight? In the 80’s, the fashion was to use electric heating foil in the ceiling. This warmed up the room by IR. One negative side effect was that if you held your hands underneath the table, they would instantly feel cold. Does the mass of the table really have anything to do with it?

    LikeLike

    Reply

24. “Let’s try a theory that does not cause mental anguish and lacks empirical evidence. ”

    I think that sentence should end with “… or lack empirical evidence. “

    LikeLike

    Reply

    1. I’m not shy about English being my 3rd language. Let it be. Thanks for the correction.

       LikeLike

       Reply

25. Back on the energy budget diagram (apologies…).
    Any idea on how the proponents of this (to me) incredible diagram justify the following (in my view) nonsense :
    earth is heated with a total flux of 168 (directly from sun) + 324 (backradiated by atmosphere) = 492 W/m2
    but the only energy source (sun) delivers no more than 342 W/m2
    thus the resulting heating “efficiency” is 492/342= 1.44 > 1 in denial of the energy conservation law.
    Atmosphere is not an energy source, just an “insulator” (according to them), so how is this surplus energy created ?

    LikeLiked by 1 person

    Reply

    1. Their justification is energy balance. You’ve seen the equations, you’ve seen the rhetoric. You know it can’t be true. The math works though, but math is not physics. I created this blog precisely because the math makes absolutely no physical sense at all.

       LikeLiked by 1 person

       Reply

    2. Max, conservation of energy requires that the amount of energy entering a region equals the amount leaving*. In this diagram, the atmosphere sends an extra 324 W/m^2 to the surface, on top of the 168 W/m^2 from the sun, for a total of 492 W/m^2. However, the surface emits 390+78+24=492 W/m^2 . So energy is indeed conserved.

       * As long as energy is not actively building up in that region

       LikeLiked by 1 person

       Reply

    3. Put another way: the sun radiates some energy toward the earth’s surface, and the earth as a whole (surface + atmosphere) radiates an equal amount of energy to space. In addition, the surface and atmosphere pass some energy back and forth. This last factor results in a higher surface _temperature_, but doesn’t add any _energy_.

       LikeLike

       Reply

       1. So the cold makes the hot hotter, taking the heat …. from the hot ! what’s not to like ? Problem is, like Jarle said, reality seems not to care about such mental artifacts. Experimental evidence tells the opposite. So why climate scientists do not wake up ? or do not stop cheating ? (I honestly do not know which is the case)

          LikeLiked by 1 person

          Reply

          1. “ So the cold makes the hot hotter, taking the heat …. from the hot ! what’s not to like ?”

             Correct. What you have described also applies exactly to a blanket. It is really quite a common phenomenon, nothing strange or exciting about it.

             LikeLike

26. This ingenious way of increasing temperature has yet to be applied in a human made apparatus. Would be great if the climate scientists could explain to other scientists how this really works, but of course they can’t.

    LikeLiked by 1 person

    Reply

27. “it is really quite a common phenomenon…” so I guess that you believe in the paradigm of the “green plate”, where the 2nd added pane works as an insulator thus increasing the temperature of the 1st (irradiated) pane ? https://principia-scientific.com/greenplate-effect-it-doesnt-happen/

    LikeLike

    Reply

    1. First, I want to reiterate that there is absolutely no problem with a colder object raising the steady state temperature of a hotter object, so long as there is an even colder environment. This describes all insulation.

       The only difference for the greenhouse effect is that the heat transfer is radiative instead of conductive/convective. The “green plate” experiment you linked has many design flaws. Here is another experiment, also performed by some internet random, which claims to obtain the opposite result: https://m.box.com/shared_item/https%3A%2F%2Fapp.box.com%2Fs%2F5wxidf87li5bo588q2xhcfxhtfy52oba . Up to you which internet experiment to believe. But the theory of the green plate effect is extremely simple, well established physics.

       LikeLike

       Reply

       1. I refuted this experiment. The pipe takes up 10% view factor in the beginning. Author (Swanson) acknowledges this, and will not repeat experiment. Swanson makes no prediction what the result should be, but it doesn’t square with even his own physics. Using the pipe, I predicted his result within 1%. No GHE found. It’s on Robert Spencer’s blog. I posted link some time here. I think Max read it.

          LikeLike

          Reply

       2. Tim C – you claim Geraint’s experiment has many design flaws but you show no one.

          On the other hand, Swanson’s experiment is seriously flawed and therefore invalid as Zoe mentioned in her comment (details below) :

          The green hoax experiment is a sneaky fraud. The answer is much simpler.
          Take a look at photo 1, then photo 2. Look at the steel/lead pipe they use to lift green plate.
          It’s well within Blue plate’s view factor!
          The Blue Plate is heating the pipe! (higher heat capacity)
          Then they yank the pipe down, and the heat capacity goes down … heating the Blue plate.
          Blue emits in a radial hemisphere, and if you look at photo 1, you can easily eyeball the pipe taking up ~10% of Blue’s hemispheric view factor.
          The pipe is “holding down” Blue’s temperature.
          Then they bring green plate in … that is, they remove the pipe – and Blue heats up like 10% more.
          Experiment is a fraud.
          Further assessment :
          I took a ruler to photo 2 (top right)
          The diameter of pipe is N
          The height of pipe is 3N
          Surface area of pipe FACING plate is 3N^2*PI
          Distance from center of plate to center of pipe is ~2.8N.
          Imaginary surface area of blue plate’s emission at distance 2.8N:
          4PI*7.84N^2 = 31.36N^2*PI
          So, the pipe takes up 3/31.36 = 9.6% of BP’s radiation.
          The results of experiment showed temperature difference from lack of pipe to presence of pipe to be 116 – 106 = 10 C
          In fluxes: 1300 – 1172 = 128 W/m^2
          Flux % increase: 128 / 1172 = 10.9%
          9.6% vs. 10.9% is too close to be a coincidence. The tiny difference is also due to the fact that when the pipe is lowered it still takes up a tiny % of view factor. It’s too small for me to bother measuring it.
          The pipe is the culprit. QED.
          According to backradiation theory, blue’s final flux should have been 100% higher than green’s (400 vs 200).
          In this experiment, the result was 1300/833 (75C for GP) = 56%
          This experiment shows pipe view factor heat capacity interference and not backradiation heating. QED

          LikeLiked by 1 person

          Reply

       3. Zoe and Max,

          Your refutation is incorrect for two reasons:

          1) You claim that the presence of the pipe lowers the temperature because it is in the view factor of the blue plate — in other words, because it absorbs some radiation from the blue plate. But the green plate does the exact same thing! So if the pipe decreases the temperature, then the green plate should just decrease it further. But the opposite is observed: inserting the green plate increases the temperature.

          2) You also claim the heat capacity of the pipe is important. But heat capacity has absolutely zero influence on steady state temperatures (as long as there is no convection). To see this, just realize heat capacity is the amount of heat needed to change the temperature. But at steady state, every object receives zero total heat, deltaQ=0. So the heat capacity plays no role.

          Finally, I like your reasoning on the last point, but there’s an important mistake:

          “According to backradiation theory, blue’s final flux should have been 100% higher than green’s (400 vs 200).
          In this experiment, the result was 1300/833 (75C for GP) = 56%”

          This analysis misses the fact that both plates receive an input from the room temperature surroundings, F=sigma*(293 K)^4=418 W/m^2 . Going through the math, the fluxes for the blue and green plates Fb and Fg should obey:

          (Fb-F)=2*(Fg-F)
          Using your numbers,
          (1300-418)=2*(833-418)
          Which approximately checks out. It’s a bit strange that the factor is too high (2.13), but that could be explained if the room temperature was colder, or if there were other heat losses from the green plate.

          LikeLiked by 1 person

          Reply

          1. Tim,
             No segment of Swanson’s experiment is allowed enough time to go completely flat.

             The heat capacity of plate and pipe are different! And this matters for arbitrary time cut offs.

             If you’re going to count the environment temperature as a source, then you must also include that for the heat lamp!!! The heat lamp is only providing energy above the environment!

             So to be fair, you must remove ambient environment from everything. When you do that, you find yourself with the same problem as before.

             LikeLike

          2. I definitely agree he should have run the experiment longer to let it stabilize. But let’s be honest, you can eyeball the steady state temps from the graph, and it’s clear adding the green plate increases the steady state temperature. And steady state is independent of heat capacities.

             The expression I gave relating Fb and Fg is exact and true regardless of the heat source. It is derived from the energy balance of the green plate, which only directly receives energy from the blue plate Fb, and the environment on the other side, F.

             LikeLike

          3. Tim,
             You’re not only adding a green plate, you’re taking the pipe from 10% view to <1% view. You can't make an attribution when two things are happening at once.

             LikeLike

          4. Zoe, what we’re calling the pipe and the plate are both just pieces of metal in the vacuum chamber. Since the pipe goes from 10% to 1% of the view factor, while the plate goes from 1% to 100%, the total view factor increases. And this increases the temperature. Exactly as predicted by the theory. This is exactly the green plate effect.

             LikeLike

          5. Definitely didn’t go to 100%, but in the high 90’s %, and you showed a boost of 2.13. How can that be? We know the ambient environment was 20C initially, as shown by yellow line. You should’ve had a boost of slightly under 2.

             P.S. How many plates (layers) do you think there are in the sky?

             LikeLike

          6. You’re right that the view factor is less than 1, and now that I think about it this explains why the number is greater than 2: Fb-F=2.13(Fg-F) . Before I give the equation, it’s intuitively pretty clear: if the view factor is small, then the green plate would just remain near room temperature, and Fb would be much bigger then Fg.

             Anyway, the full energy balance for the green plate is power out=power in, or: 2Fg = V*Fb+(2-V)F . Rearranging gives Fb-F=2*(Fg-F)/V . So everything works out for a view factor of about 0.94 .

             Seems to me the bottom line is still that the green plate increased the temperature of the blue plate, as predicted by theory.

             For your question about number of layers in the sky: of course the actual atmosphere doesn’t have well defined layers, and it’s harder to analyze than the simple green plate. There’s probably some way to model it with layers, but the real way to evaluate the greenhouse effect is to dust off some calculus and solve the radiative transfer equation.

             LikeLike

          7. Do you at least appreciate the fundamental difference between the green plate effect and the greenhouse effect?

             Let’s say we receive a flux N.

             In the green plate effect; For 1 plate, this gets split into 0.5N on each side. As you keep adding plates, the hottest side gets closer and closer to emitting N.

             The upper bound in this problem is N flux. After infinite plates, there is no mystery where N came from.

             In the greenhouse effect … each layer doubles the flux. 1 layer = 2N. 2 layers = 3N. There is no upper bound.

             The atmosphere may not have well defined layers, but a 2D layer can form from looking into a 3D depth. Right?
             So using the pigeonhole principle, how many layers are there?

             Remember, climate scientists claim there’s just a handful (1, 2, or 3) of layers. But considering atmospheric mass, GHG concentration, and Avagadro’s number … how many layers are there really? And keep in mind most of the magic must be done in the lowest layer! It is the lowest layer that must deliver ~340 W/m^2 of backradiation.

             LikeLike

          8. You’re right that the green plate experiment, as performed here, can increase the flux out of the blue plate by at most a factor of 2. This is because one side of the plate is left to radiate freely, no matter how many plates are added to the other side. The fix is simple. Either place the blue plate on a very good insulator like styrofoam (in which case the outgoing flux increases proportionally with the number of plates), or put green plates on both sides (in which case the maximum flux also increases without bound). Then the temperature of the system will increase up until the next limit — either the temperature of your heat source, or due to heat losses through conduction.

             As for extending the layer model to the atmosphere, I just don’t know how to do it. It seems like a different system since it’s a continuum (not discrete layers), and the layers all have convective transport between them as well. To me the green plate is useful to prove back radiation works in a simple setting, but to analyze the atmosphere realistically you just have to upgrade to the radiative transfer equation.

             LikeLike

          9. “The fix is simple”
             How is it simple? You can’t put styrofoam or a plate on the side facing the source (heat lamp or sun).

             The GP Experiment is not in any way a proper proof of GH effect, because the whole system is bound by N flux from the source. In theory you can increase from 0.5N to N, but that is all.

             The GH effect must create over-unity flux.

             Geothermal will create 37 degrees K at the very least. And this is slightly more than the GH effect. I still think geothermal IS the real cause of so-called GH effect.

             Over-unity solved.

             BTW, do you know why bell jar temperature (yellow line) kept going up the whole experiment? It doesn’t make theoretical sense.

             Otherwise, I appreciate you explaining the experiment BETTER than its own author. The author panicked and gave up after my explanation.

             LikeLike

          10. “How is it simple? You can’t put styofoam or a plate on the side facing the source (heat lamp or sun)?”

              For example, put the green plate (plate 0) on a styrofoam backing, place n absorptive plates on the other side, and supply a flux N to plate 0 with an electric heater. If all non-radiative heat losses can be kept small, the radiative flux out of plate 0 will reach n*N+F (F is the room temperature flux). In theory, many times higher than the source flux from the heater.

              Or, if you really want the heat source to be a light instead of an electric heater, then you have to make something transparent to visible light: either the insulative backing, or the n plates. It’s hard to find materials that are perfectly transparent to visible light and have the other required properties, but that’s not a fundamental problem.

              “BTW, do you know why the bell jar temperature (yellow line) kept going up the whole experiment? It doesn’t make theoretical sense.”

              All outgoing heat from the experiment ultimately passes through the bell jar, and some of it will be absorbed, heating the bell jar. I think it just takes a long time to reach a steady state, so the temperature is still going up. This will have some effect on the temperatures of the plates. You can even think of the bell jar as second green plate, but its temperature is kept low because it’s in direct contact with the air, so it won’t be effective as the plate in vacuum.

              I think the geothermal hypothesis is a whole other discussion, for now I’d rather finish discussing the green plate physics. But you can probably guess I’m more in favor of the GHE hypothesis 🙂

              LikeLiked by 1 person

          11. Tim,
              I don’t understand what you’re saying. Plate 1 is blue plate. The radiation comes on one side of blue plate. You can’t put anything in front of blue plate, or you will block it. Putting things behind green plate won’t change anything.

              “It’s hard to find materials that are perfectly transparent to visible light and have the other required properties, but that’s not a fundamental problem.”

              Well, glass was used before Fourier and it didn’t show GHE.

              “I think it just takes a long time to reach a steady state”

              Oh, so here it will take a long time. Got it. But this isn’t true for blue plate with lead pipe. Got it.

              “You can even think of the bell jar as second green plate”

              I prefer to think of it as the first green plate, and according to theory its temperature should’ve declined after introducing a plate between it and blue plate.

              “But you can probably guess I’m more in favor of the GHE hypothesis”

              But why? You can see that 91 mW/m^2 will produce 37 Kelvin, and the GHE is within that. The GHE is not 33K. That’s an old number. It’s more like 36K. Pretty much exactly as geothermal.

              LikeLike

          12. Hi Zoe, thanks as always for the discussion.

              “I don’t understand what you’re saying. Plate 1 is blue plate. The radiation comes on one side of blue plate. You can’t put anything in front of blue plate, or you will block it. Putting things behind green plate won’t change anything.”

              Let me clarify. Place the blue plate directly on top of an insulator. Above the plate put one or more green plates, let’s say n of them. Since there’s no direct line of sight to the blue plate, the options to heat the blue plate are 1) wire an electric heater onto the blue plate, 2) make the green plate(s) transparent and shine a light through, or 3) make the insulator transparent and shine a light through.

              Call the amount of heat delivered to the blue plate N. At steady state, the blue plate will reach a temperature where it emits a flux of n*N+F . That is, if heat leakage through other means (like conduction) is negligible. I can derive this if you’d like.

              Regarding the time to reach steady state: again, I agree with you the experiment should have been run a little longer. But looking at the graph it looks like everything, including the bell jar, is pretty close to reaching steady state at each step.

              “But why? You can see that 91 mW/m^2 will produce 37 Kelvin, and the GHE is within that. The GHE is not 33K. That’s an old number. It’s more like 36K. Pretty much exactly as geothermal.”

              For a surface that is roughly a black body, an input flux of 400 W/m^2 will result in a steady state temperature of (400W/m^2/stefan boltzmann constant)^.25 = 289.81 K. Adding an additional 91mW/m^2 will take the temperature to 289.83 K .

              LikeLiked by 1 person

          13. Tim,

              “options to heat the blue plate”
              1) You’re only allowed to use IR, such as a lamp or the sun
              2 & 3) This was tried as early as late 18th century. No GHE. Fourier pointed this out regarding de Saussure’s (spelling?) experiments.

              Why can’t anyone do this experiment TODAY?
              Why won’t IPCC or NASA settle this?
              Why won’t GHE enthusiasts?

              “Adding an additional 91mW/m^2”

              Very clever, but that is fraud. It’s the Sun that is added to Earth. Who said you can add the tiny flux 2nd? Nay, first you find the Temperature produced by Earth, THEN you add the sun’s flux.

              Keep in mind this 91 mW/m^2 = 37K is for top-of-the-atmosphere! Not the bottom of it!!!

              37K is the minimum produced by Earth alone. It is there. You can’t get rid of it. You don’t get to pretend Earth only adds 0.02K to the sun. That is fraud.

              Go 3km below the surface where it’s above 100C. Zero sun, and still 91 mW/m^2 of geothermal flux. How does 91mW/m^2 support that temperature? Can’t use SB Law, but you can use what I found here at my blog. You’re welcome 🙂

              “But looking at the graph it looks like everything, including the bell jar, is pretty close to reaching steady state at each step.”

              And yet at every step it went up. It really should’ve went down after green plate. Remember, by green plate theory, adding more plates makes the first plate hotter, while it makes the last plates cooler. However, if the blue plate didn’t reach equilbrium to begin with, that wouldn’t be the case. In other words, the blue plate had more to go without green plate!

              LikeLike

       4. Max, you asked about flaws in Geraint’s experiment. The most important is he makes no attempt to analyze parasitic heat losses through conduction, either through the air or support structures. Neither does the experiment I linked, to be fair.

          A more concrete flaw is this: the two discs are placed rather far apart, I’d estimate by about R/2, where R is the radius of the disc. That means the green plate only absorbs about 33% of the radiation from the blue plate, and the blue plate only absorbs about 33% of the radiation from the blue plate. So at most 1/9 of the blue plate’s outgoing radiation can be returned!

          LikeLike

          Reply

28. Small correction:

    That means the green plate only absorbs about 33% of the radiation from the blue plate, and the blue plate only absorbs about 33% of the radiation from the GREEN plate.

    LikeLike

    Reply

    1. “This means” you are fudging parameters to suit the observation.

       You’re supposed to start with parameters to predict, not post-dict the result.

       LikeLike

       Reply

29. Tim C, I tend to believe Geraint’s experiment gives the proper answers, as opposed to Swanson’s. Reason is because I don’t trust anyone, therefore I tried to do an experiment myself. I’ve tried to reproduce the “infamous” two panes thought experiment of Dr Roy Spencer from a few years ago, but considering the effect of convection (in addition to radiation) to see what the expected temperature jump should be, and what it actually is. You can imagine what results I got. But since I did things in a hurry I may have done something wrong. Here it is : setup, calculations, results. I took A LOT of measurements with no discernible “backradiation heating”, I am afraid.
    Work permitting, I plan on refining things a bit.

    https://drive.google.com/file/d/1RzWwLYskAzAXYpTtOkgtDgQ5QtRF4w0Y/view?usp=sharing

    LikeLike

    Reply

    1. Max, kudos for performing the experiment and actually analyzing all heat transfer mechanisms. Unfortunately I don’t have time to go through exactly what’s going on in the spreadsheet, or to learn the physics involved in estimating convective heat transfer. But I would imagine that doing this experiment in open air, the influence of conduction and convection through the air would totally dominate over radiative effects and make it hard to draw conclusions.

       One indication that this is the case is the temperature of the plate. If your plate is fairly perpendicular to the sun, it should receive about 1000 W/m^2. If it only lost heat through radiation, the temperature would then stabilize to at least 90 Celsius, significantly higher if you account for infrared radiation from the environment and non-unity emissivity of the plate. Since it only reaches 50 C, this indicates other heat loss mechanisms are dominant.

       LikeLike

       Reply

30. Tim,

    for air convection I used a textbook empirical correlation that provides the convection coefficient for vertical plates in still air as a function of the DT (wall-air) that provides 4.2 – 4.3 W/m2/°C for our case. From school I recall a suggested value around 5 W/m2/°C for quick calculations, so we are in the ballpark.

    Measurements I’ve shown were taken in the morning of June 26 around 8.30-90 am (daylight saving time) here in Lignano Sabbiadoro Italy. At that early hour you don’t get 1000 W/m2 I think. I’ve tried to calculate what the radiation intensity could be, and I have found this resource (downloadable spreadsheet) : https://www.nrel.gov/grid/solar-resource/clear-sky.html, that provides 550/650 W/m2 so again that seem to be in the ballpark.

    My results in a nutshell are :
    expected backradiation heating(*) in the vacuum (theoretical) of the front pane (front pane directly irradiated, back pane “insulating”) : +3.7°C
    expected backradiation heating(*) in calm air of the front pane of the couple (front pane directly irradiated, back pane “insulating”): +2.0°C
    actual average measured backradiation heating in calm air : 0.0°C +/-0.5°C
    The +/-0.5°C was in my view determined by the small breeze puffs that sometimes occurred, although I’ve always tried to wait until there was zero wind.

    (*) according to “consensus climate science” calculations

    I plan on doing more tests (work permitting) but so far the results are consistently averaging out to a flat zero backradiation heating. The good thing is that this experiment is so cheap and easy to do, so I – and anyone else – can do and repeat it forever. On the other hand the calculation model enables you to calculate exactly what you should expect to see, without needing any vacuum.

    I’ll be glad if anyone can help me out to correct any errors I may have made in the calculation model…but apart from background assumptions maybe there are not so many.

    LikeLiked by 1 person

    Reply

    1. Wow, Max, just looked at your experiment.

       First time I see your scientific mind at work, and it’s excellent!

       Not saying it’s fully correct, but also don’t see any serious flaws. You seemed to have taken all the details into account.

       Just like the videos of people on infrared camera in my article here … you found essentially no difference in T. What a pleasant non-surprise!

       LikeLiked by 1 person

       Reply

       1. 🙂 thx Zoe – you make me blush. “Not saying it’s fully correct” …. of course I know, that’s why I posted it here….hoping to get some hints from my physics Teacher 🙂

          LikeLike

          Reply

          1. Thank you, but I’m just an economist, financial analyst, businesswoman, certified quant, data programmer. I only took 7 physics/astronomy courses.

             And I’m biased: I like the result. A few things I don’t know:

             1) How free convection works in a small gap (that’s my current lack of knowledge). I think you’re right for the open sides.
             2) Wind speed, if any (safe to assume zero, maybe. It’s definitely equal for both)
             3) Did you determine insolation from temperature?
             4) Actual emissivity of plates (saw 0.92 from a google search, not 0.95)
             5) Humidity (although it should affect both equally)

             I also like to keep evidence as simple as possible:

             https://youtu.be/Sth61H7FZSQ

             LikeLike

    2. Max, I have to say I’m impressed by the analysis you’ve done. You have models for all of the relevant heat transfer mechanisms, and you’ve solved the models mathematically, which is more than the other experiments linked here have done.

       Still, how can we account for the difference in result between your experiment and the one I linked? Namely, that they measured a temperature change consistent with the theory, while your measurement disagreed with the theoretical prediction. It seems to me the obvious answer is that modeling convection is hard, and introduces a big question mark into your model. Their experiment is much less influenced by complicated factors like convection, and so seems to be a clearer test of whether back radiation can increase temperature.

       LikeLike

       Reply

       1. Tim,
          Did you not notice that incoming radiation didn’t get split in two? The sun maxed it out at ~49C for one plate. You already agreed that N is the maximum possible for this type of experiment, and here it is right before your eyes.

          LikeLike

          Reply

31. Zoe, you said “you are just an economist…etc etc” – no worries ! nobody’s perfect 🙂

    Jokes apart, obviously you seem to know your stuff pretty well (better than many Physics Phd’s…my idea based on my MS in mechanical engineering – although that happened so long ago…).

    Back to your questions :

    1) that was my doubt as well, and that’s why I decided not to reduce the double plates distance too much (4 cm seems a good compromise to keep the view factor sufficiently high)

    2) wind speed : I’d need to improve the experiment by using a wind meter otherwise that parameter is very difficult to control. Wind speed / pattern is a big factor : in certain conditions it affects temperatures by 2-3 degrees ! If a small puff comes “from the back”, it cools down mainly the single pane, because of the presence of the back pane acting as screen…This can create significant errors. Also, using some wind screens around may help..but complicates things.

    What do you think of my measurement of the “sky temperature” (around 0-5°C) with the infrared thermometer ? Does it make sense to use this temperature as (part of) the background ? I was wondering if the temperature of the microwave background makes more sense.

    3) yes, correct. What is important in my view is to use a ballpark value for the insolation (not necessarily very accurate) since then I obviously apply the same insolation to both panes to estimate the backradiation heating that I should measure if theory was true

    4) noted. I’ve played a bit by changing various “uncertain” inputs (emissivity, convection coefficient, background temperatures) +/-20% to see the effects on the expected backradiation heating, but it is quite small +/- 0.4°C in comparison with the 2.0°C increase expected.

    5) noted. I decided not to worry about it, because it’s not affecting the convection coefficient correlation (although it probably should). However as you say it should be affecting both panes in the same way.

    I also love simple evidence…! like that of the video you linked. I just wanted to get a sense of the effects of the various factors, in order to see to what extent a very simple amateur experiment can provide an answer.
    What I would like to do the most, is to debunk – beyond any doubt – that Dr Roy’s tought experiment, because I am fed up with those stupid tought experiments 🙂 Galileo should have taught something to modern scientists.

    LikeLiked by 1 person

    Reply

    1. The sky temperature is correct for radiative-only transfer from objects at the surface to the atmosphere. Your IR thermometer measures an averaged imaginary layer where radiation can not penetrate further. In other words, a sky wall, even though it’s just molecules at various distances above, a fog.

       No, CMB has no role here. It’s smaller and beyond reach.

       LikeLike

       Reply

    2. What I would like to do the most, is to debunk – beyond any doubt – that Dr Roy’s tought experiment

       What I ask GHE believers is how backradiation can warm our ~3-4 km deep oceans (> 70% of Earth’s surface).
       Although considered cold, the temperature of the deep oceans is always >270K, which is at least 15K above the infamous 255K. So if you believe that the atmosphere “further heats” the surface (eg Lacis ea 2010) you also believe that our cold, thin atmosphere heats our 3-4 km deep oceans. Entirely new physics required 😉

       If you accept reality, that the deep oceans are “hot” for the same reason continental crust is hot (geothermal), the picture becomes clear instantly. Sun only has to increase the temperature of the mixed surface layer a little, the atmosphere reduces the energy loss to space, and yes backradiation does play its role in this reduction.
       The solar heated surface (land and ocean) provides an insulation layer that keeps geothermal heat in, unless the surface temperature drops like at night or in winter or at high latitudes for the oceans.

       LikeLiked by 1 person

       Reply

32. Tim,

    I think you are right on convection, if this parameter is not managed well. However with all the caveats about it, in hundreds of measurements taken in different times and hours, I have never noticed a STABLE trend of backradiation heating, that should be well visible (2 degrees or so). This makes me think that Zoe Phin (but also Joe Postma and other guys) are right, and most climate scientists are wrong. Like Zoe said : why are NASA/IPCC not making a PROPER experiment to stop this discussion forever ? It’s just because it’s “established physics” as you/they claim ? I doubt it.

    Regarding Swanson’s experiment, with all the due respect I simply do not trust it, for the reasons said by Zoe.

    ciao
    Max

    LikeLike

    Reply

    1. Max,
       Are you familiar with the infamous experiment shown on TV by Al Gore and Bill Nye? Total fake. Different experiment though. But the point is, if there was any experiment that could be conducted in support of the warming hypothesis, it would be totally unnecessary for these guys to fake it. Keep up the good work! If time allows, maybe you could also replicate Roy Spencer’s experiment with IR-lamp and ice-cubes, where you replace the ice-cubes with cold pebles or something like that? (Something that does not evaporate)

       LikeLiked by 1 person

       Reply

       1. Thanks Jarle. I agree. I also recall that other experiment of Dr Spencer. I also thought of reproducing something similar. The truth is…I don’t have much time. Still too far from retiring 🙂

          LikeLike

          Reply

          1. Max,
             I think you may be accused of doing your experiment “wrong”. You should measure the T of backsides of panel 1 (closest to sun) in both setup A & B. Please find out those measurements when you get a chance. There should be a warm-up, but it’s not the GH effect.

             LikeLike

    2. Max and Jarle,
       I would like you to take a look at this document about the new James Webb Satellite:
       https://safir.jpl.nasa.gov/BeyondSpitzerConf/proceedings/sessionB/Parrish.pdf

       Look at PDF Page 26. This is a thermal map of the satellite. May need to zoom in.

       Do you guys see what I see?

       LikeLike

       Reply

       1. …those panels are just the same color as the thermal shield i.e. NOT warmer, even if there is sun insolation + shield back-radiation ? Not so sure…there is another image at page 20 that looks different…

          LikeLike

          Reply

          1. Oh i missed page 20. But It’s not called a result. I think it’s just a graphic space filler for the text.

             Yeah everything gets ~just as hot, weather its the central part or the shielded periphery.

             The “plates” don’t make it warmer than the non-plate central part.

             Also the obvious, as I’ve said to Tim: Green-Plate Enthusiasts claim that incoming radiation (N) gets split in two: 0.5N from each side. This is plainly wrong. N is reached on the source side. Your experiment shows this. This PDF shows it. Swanson’s experiment doesn’t provide info to show it. Geraint Hughes’ experiments shows this.

             We have at least definitively proved that the green plate experiment is a dumb and worthless analogy for the greenhouse effect. The GH effect claims over-unity, i.e >N. But you already knew this.

             LikeLiked by 1 person

          2. “Also the obvious, as I’ve said to Tim: Green-Plate Enthusiasts claim that incoming radiation (N) gets split in two: 0.5N from each side. This is plainly wrong. N is reached on the source side.”

             Zoe, I would say that you still have to include flux from the environment. A plate receiving N (~500 to 1000 W/m^2) from the sun also receives radiative flux F on each side from the surroundings (F~400 W/m^2 from each side -> 2F~800 W/m^2 total). So the total outgoing flux from each side should be N/2+F , neglecting conduction+convection. With N=700 W/m^2 and F=400, this corresponds to a temperature of 66 C.

             (I made a mistake in an earlier comment somewhere and forgot to divide N by 2)

             LikeLike

          3. Very funny, Tim. The environment is a sink, not a source.
             See recent comments about James Webb Satellite here.

             Also, NASA will tell you that an Earth-orbiting satellite will achieve a max average temp of ~121C (depending on proximity to sun) somewhere on its surface. If what you say is true (it’s not), then we should add the Earth’s ~240 W/m^2 onto the ~1360 W/m^2 from the sun, and get 1600 W/m^2 -> 133.6C.

             And that would be wrong.
             https://science.nasa.gov/science-news/science-at-nasa/2001/ast21mar_1/

             https://letstalkscience.ca/educational-resources/backgrounders/temperature-on-earth-and-on-space-station

             You don’t add cold to hot!

             LikeLiked by 1 person

          4. “Also, NASA will tell you that an Earth-orbiting satellite will achieve a max average temp of ~121C (depending on proximity to sun) somewhere on its surface. If what you say is true (it’s not), then we should add the Earth’s ~240 W/m^2 onto the ~1360 W/m^2 from the sun, and get 1600 W/m^2 -> 133.6C. And that would be wrong.”

             121 C is the maximum temperature of a patch of the ISS surface directly facing the sun, which receives 1360 W/m^2. A patch on the other side of the ISS wouldn’t see the sun at all, and other parts of the ISS would see an oblique angle, so they receive less solar flux.

             The same goes for the flux the ISS receives from the earth.

             Now, here’s the important part: it’s geometrically impossible for a given patch of the ISS surface to directly face the sun and earth at the same time. To receive the full flux from the earth, the patch would have to be facing straight down, so it wouldn’t be facing the sun.

             Therefore there’s always a tradeoff between flux from the sun and flux from the earth. The maximum possible flux is from fully facing the sun. In this case there will be very little flux from the earth — a small amount, depending on geometry, but the author of that web page probably didn’t want to bother calculating what it would be.

             LikeLike

          5. Incorrect. When the satellite is generally behind Earth (but not obscured by Earth), it will receive fully from the sun, and from the Earth. If you have any doubt, just look at a lamp, and hold your phone a little off to the side so as not to obscure the lamp. You will see both lights. QED.

             But it’s funny that you would cherrypick a situation where the satellite would be directly in between sun & earth, thus causing both sources to shine on two different sides.

             LikeLiked by 1 person

          6. Zoe, I encourage you to think about it a little more. Keep in mind that even if a surface has line of sight to the earth or sun, the flux it receives is reduced by a factor of cos(theta), where theta is the angle between the surface normal and the line of sight.

             It might be helpful to look at this diagram https://imgur.com/DZ3kX0v . Try to draw a line segment (which must be short and very close to the earth) which is directly facing both the earth and the sun. You’ll find it can’t be done.

             LikeLike

          7. “reduced by a factor of cos(theta)”
             Very true. But you forget that a satellite is ~100 miles up and the Earth’s radius is ~3400 miles. The incline angle is a tiny reduction from direct earth shine, when directly facing sun.

             Most importantly you forgot that Earth doesn’t just shine radially out from its center.

             Here is what things look like:

             S <—————————– Sun
             EEEEEEEEEEEEEEEE

             The right-most E can shine on the satellite with an incline angle so small it may as well be direct.

             Think about it some more.

             LikeLiked by 1 person

          8. Thinking about it more, ultimately it comes down to maximizing the view factor of the earth, while maintaining a perpendicular line of sight to the sun. The situation you’ve drawn indeed has the maximum possible view factor under this condition. But it’s still considerably less than 1/2.

             I’m pretty sure the author of that webpage didn’t think about any of this, and just made an estimate based on the solar flux.

             LikeLike

          9. Yeah, people at NASA are lazy or stupid, and you know better. So find me a source that says otherwise … good luck.

             “But it’s still considerably less than 1/2”
             100mi height with a 4000mi radius objects … your geometry skills are terrible, sorry to say. Try again.

             You know what else is not added to satellite? Solar shortwave reflected from Earth. I forgot to mention that. That would bump T even higher, by your theory, not reality.

             LikeLiked by 1 person

          10. “ You know what else is not added to satellite? Solar shortwave reflected from Earth. I forgot to mention that. That would bump T even higher, by your theory, not reality.”

              What? Of course reflected sunlight would also increase temperature! That is the basis of solar cookers, which are routinely used to reach hundreds of degrees Celsius.

              LikeLike

          11. Solar cookers focus light from a large area to a small area, not unlike a magnifying lens. Completely wrong concept. But please tell me what unobserved max temperature your satellite is at now? Please inform NASA 🙂

              LikeLiked by 1 person

          12. You requested a source with information about heating of a satellite by planetary infrared. Try the appropriately named book Spacecraft Thermal Control https://books.google.com/books?id=_dpkAgAAQBAJ . Which includes this rather self explanatory figure: https://i.stack.imgur.com/I08Nr.png . I will try to find the accompanying text later

              LikeLike

          13. Tim, all that is absolutely correct for finding the overall surface temperature of the spacecraft, but the topic at hand is the MAXIMUM temperature. You seemed to have missed this crucial distinction. I was always talking about:

              The AVERAGE MAX is still ~1361 W/m^2 => ~121C

              The MAX MAX (perihelion) is ~1407 W/m^2 => ~124C

              LikeLiked by 1 person

          14. Zoe, you can build a solar cooker using a cardboard box and aluminum foil on the walls. You’ll do better with fancy focusing optics, but all you need is reflected sunlight.

              Another NASA source citing 125 Celsius in low earth orbit: https://nepp.nasa.gov/DocUploads/C5E0869C-0469-4D11-9FAA8012C8F52351/Environmental%20Testing%20Survey.doc

              LikeLike

          15. Oh you got 125C you say?

              Congratulations. You found the perihelion temperature prediction back in 2004 when scientists thought insolation was 1370 W/m^2 (not ~1361 as we know today)

              1370 * (1 + 2*0.017) = 1417 W/m^2 => 124.45C

              And this doc is for the moon, which can be a little closer to the sun, round up and you got 125C. No room for any extra Moon radiation.

              “you can build a solar cooker using a cardboard box and aluminum foil on the walls”

              Great. And if the spacecraft has a crevice/dip/concavity in it, it can also go above normal insolation max. Still no Earth radiation to boost beyond that.

              LikeLiked by 1 person

          16. I figured you would wave it away, even though you demanded to see a temperature a few degrees higher than 121 C earlier.

              In reality there are many effects that can push the maximum temperature higher or lower, so this is a terrible way to try to prove or disprove any one effect. You will always be able to attribute the maximum temperature to something else. For example: finite reflectivity of the surface will always lower the max temperature below the calculation. Meanwhile, emissivity <1 will increase the max temperature.

              LikeLike

          17. Tim, the central point was …

              “In reality there are many effects that can push the maximum temperature higher”

              AND NONE OF THEM INCLUDE ADDING EARTH’S RADIATION AS YOU CLAIMED. BECAUSE IT DOESN’T HAPPEN.

              So why try to debunk my central point, the ONLY reason I brought it up to begin with, when you know you can’t?

              Direct sun (or whatever maximal source) will set the max, regardless of whatever properties of the surface you can imagine.

              There is one exception: If the matter is internally heated, by conduction or convection.

              I’m sleepy. G’night, Tim.

              LikeLiked by 1 person

       2. Layer of shield facing the sun is supposed to be warmer than the part protruding from the shield, yet the model (supposedly verified by experiment) shows the opposite?

          LikeLiked by 2 people

          Reply

          1. Yup. You got it. Although there might be a difference of angle to the sun. I’ll even claim that its equal, not greater. Still, I think this is good evidence.

             LikeLiked by 1 person

33. Just wanted to add one element on convection : I mentioned that, in the experiments, when there was a puff of air “from the back”, the front-double panel could show a higher temperature even by two degrees (for limited periods). So there it is … the backradiation effect, one may say ! Well… I don’t think so. Air is moving at the back of the single pane, but not (or much less) at the back of the front-double pane.
    I have tried to simulate this fact by doubling the film coefficient on one side of the single pane (from 4.3 to 8.6 W/m2/°C) that may well be the case with 1-2 m/s of air speed.
    Result : single pane will cool down by approx 3°C (keeping insolation constant), so that’s the “apparent” backradiation that one may think could exist ! But – for the reasons said – I think it doesnt.

    LikeLike

    Reply

34. Zoe, regarding the backside temperature reading : for the single pane, no problem to measure it, temperature drops min 0.2 °C max 0.6°C less than front, for the double panes there is not enough room to measure accurately…

    LikeLike

    Reply

    1. Can you measure via IR thermometer INTO 4cm gap at an angle?

       LikeLike

       Reply

       1. …I get questionable results as it’s difficult to control the measured spot …I’ll try again anyhow

          LikeLike

          Reply

35. Tim : “t seems to me the obvious answer is that modeling convection is hard, and introduces a big question mark into your model. Their experiment is much less influenced by complicated factors like convection, and so seems to be a clearer test of whether back radiation can increase”

    but modeling difficulties have nothing to do with the measured results, that show consistently absence of heating due to backradiation. There is no temperature increase whatsoever. My model only confirms that – if the theory was true – you should well see the heating. The heating is not seen, regardless of model complexity or uncertainty.

    LikeLiked by 1 person

    Reply

    1. By the way, have you measured the sky temperature at different angles? I would guess you would get a much higher temperature when pointing at the horizon, than when pointing straight up, as there is more atmosphere in that direction. I would also be curious how increasing the sky temperature in your model (say to 10 or 15 C) effects the prediction.

       I appreciate you taking the time to do this experiment, and discussing it with me. It’s fascinating, even if I’m not yet convinced that it provides a good measurement of back radiation.

       LikeLike

       Reply

       1. Tim – today is a cloudy day here…no chance to do anything. I also appreciate talking with competent and honest people that challenge my “credence”…well I do think it’s more than a credence 🙂

          LikeLike

          Reply

          1. …but here is the simulation of the effect of the sky at 15°C (maybe with clouds ?, but I assure you there was always a clear, blue sky when I did the tests) : 1.3°C backradiation heating.

             https://drive.google.com/file/d/1VCSKBMXiXixJ1LuzLpqsjhLnKAJBJhaG/view?usp=sharing

             LikeLike

36. Just for the record, I took a further series of measures today (maybe 50 shots or so) afternoon from 3 to 4 pm, temperature went up close to 60°C sometimes (when zero wind and 30°C air temp), I still have clean/sort them out but again zero backradiation heating…despite switching over several times the positions of the black panes (just to avoid any possible problem with different emissivity…etc). I’ll post the number when I’ll have time.

    LikeLiked by 1 person

    Reply

37. One last comment : main reason why I made the model, is to be reasonably sure that we are not in a situation where the backradiation heating is too low to be measured…like in the case of the lightbulb experiment of Joe Postma (“light bulbs disprove the GHE” :https://climateofsophistry.com/2013/10/26/greenhouse-fraud-20/). In this specific case due to view factors you would not be able to appreciate any effect (in my view)…

    LikeLike

    Reply

38. “but modeling difficulties have nothing to do with the measured results”

    This is true, but modeling difficulties complicate the _interpretation_ of the measurement. The Swanson experiment shows a strong positive result, in that adding a green plate increases the temperature substantially. This can be ascribed to radiative effects fairly certainly, since conduction and convection are heavily suppressed. If the vacuum is very good, we can say convection has zero effect, without any need to model anything.

    Your experiment measured a null result — a second plate did not increase the temperature of the first. But we know that convection and conduction are substantial and affect the temperature. Without modeling these effects, it’s difficult to say whether back radiation has zero effect, or whether its effect is suppressed by other heat transfer mechanisms.

    Already your modeling suggests that the theoretical temperature enhancement is fairly small (2 degrees), much smaller than would be expected in a vacuum. If there is any uncertainty in the model, it becomes even more difficult to interpret.

    Finally, I would argue that a null result is harder to interpret than a positive one. If your experiment sees a change, something real is clearly happening. If you don’t see a change, it’s possible that there is still an effect, but it’s smaller than your noise and uncertainties.

    LikeLike

    Reply

    1. Tim, I have done many simulations (sensitivity analyses) of the effects of any errors in the model input parameters. I think we both agree that convection is the biggest one. So have a look at this “really-worst-case-scenario” where I’ve assumed to be off by a factor of 3 in the convection coefficient. Having worked in the heat exchangers industry, I tell you that there is a very hard chance that such well tested industry-practice correlations are so far off. The spreadsheet also tells you that this is an impossible scenario, because I think you can’t have 1060 W/m2 insulation at 7:30 in the morning (std time), so the model is taking away by convection 400 W/m2 more than it should be and this of course attenuates the backradiation heating. Nevertheless, even with this, I haven’t been able to reduce the expected backradiation heating below 1°C. Having done hundreds of measurements under the most different conditions, positions, switching panes over, etc, I can tell you that this 1°C is A LOT for this IR thermometer that is very sensitive to the minimal change, and I see no reason why I should not be able to detect any heating trend whatsoever.

       https://drive.google.com/file/d/1Jd1NIwuNuM2XJfyb6tw0Ar3KGmAoJk1d/view?usp=sharing

       LikeLiked by 1 person

       Reply

    2. Tim I had a look to Swanson’s experiment setup…maybe one day I will try to reproduce it. Do you have any idea on how he was able to achieve the vacuum inside. I have noticed there is no forcing/locking system for the jar-gasket sealing surface (Swanson didnt provide any details at all). Same for all the previous experiments. A proper flange would need either bolts (not possible here) or some wedge system to force the contact and ensure a strong initial preload/sealing. Nothing is shown nor mentioned, although he mentions leakage problems for the thermocouples but not for that large contact surface ? This is a relatively large flange, and I am surprised it can work and achieve even a minimal negative pressure just with the jar laying on top (self weight).

       LikeLike

       Reply

       1. Max, my understanding is that a bell jar maintains a tight seal using the force of the vacuum pressure, which pulls the jar against the lid. The experiment reaches 50 microtorr, which is in the medium vacuum range. I’ve never worked with medium vacuum myself (I’ve used ultrahigh vacuum and vacuum cleaners, nothing in between), but apparently it’s not so hard to achieve.

          LikeLike

          Reply

       2. I’m going to buy for my 8yo son …
          https://www.amazon.com/dp/B07YYMFQQQ/

          LikeLiked by 1 person

          Reply

          1. Ha, there are actually a lot of cool experiments a kid could do with that if it works all right

             LikeLiked by 1 person

          2. Thanks, Tim. I’m also looking for a greenhouse effect kit. Can you find it?

             LikeLiked by 1 person

39. Yes that makes sense…there seem to be a few kits around…not as cheap as my setup, I spent 9.43 euros total (IR thermometer excluded). Stuff for kids apart :-), a good jar+plate+vacuum pump system may come close to 1000 usd.

    LikeLike

    Reply

40. Zoe, here it is. 74.99 USD.
    Established physics.
    https://www.wardsci.com/store/product/8890195/the-greenhouse-effect-exploring-the-theory-kit

    LikeLike

    Reply

    1. This is probably the other greenhouse effect. You know, the one concerning greenhouses 😉 climate scientists themselves stress that without a lapse-rate, there is no atmospheric warming effect due to IR-absorbing gases.

       LikeLiked by 2 people

       Reply

       1. Exactly right, Jarle. I need the kit for the over-unity radiation effect.

          I’m sure Max was joking 😄

          LikeLiked by 1 person

          Reply

       2. 👍 too bad — I need that big expensive vacuum jar then 😉

          LikeLike

          Reply

41. Zoe, I owe you the info on the backside temperature of the front pane setup B (double panes arrangement). Not sure how precise the measurement is, however this morning I got front/back 41.7/40.4, 40.6/40.2, 46.2/45.8 °C. Too windy this morning, with fluctuating temperatures and sometimes TA >> TB (3-4 °C) and vice versa, erratic results but with again no trend.

    I’m thinking of making some tests inside to ensure there is no moving air. Maybe in the next days.

    LikeLike

    Reply

42. …I’ve tried to register in the wardsci.com website because that kit must be fun….so I wanted to ask more info, maybe to get the instructions with the description of the experiment. But I am from Italy and not allowed to register. Too bad…

    LikeLike

    Reply

43. Today was probably the best day so far for my tests (I know, I know, there is no vacuum…so take it for what it’s worth…) because it was very calm, almost no wind, and with a nice blue clear sky with no clouds. Just a few slight puffs sometimes. So I took the largest number of shots ever (41). Then I did two simulations, one with the “zenith” sky temperature (1°C), the other with the sky temp lowest at the horizon as suggested by Tim (12°C). In the former case, calculated backradiation warm-up is 2.2°C, in the latter 1.8°C, well substantial in both cases.
    Temperature readings were sometimes slightly higher for arrangement A (single pane), sometimes slightly higher for arrangement B (double panes). I was surprised (well, maybe not) to see that again the average difference TB-TA (actual backradiation warm-up) was…0.0°C.
    You can also see the graph that tells the story well : no trend whatsoever.

    Here is the link to the pdf :
    https://drive.google.com/file/d/1DPBqahnVgGubtC-mnvBYsKhO7Knte3Dl/view?usp=sharing

    Here is the link to the source excel file – in case you would like to use it, see all the raw data, or to find any errors (always possible, obviously). Be advised that you need to run the solver to find the power balance for both setup A and B :
    https://drive.google.com/file/d/1uhQwrHv3oX73tOdZ2jjkcM0rxnIAAJe2/view?usp=sharing

    LikeLiked by 1 person

    Reply

    1. Excellent. I think the front pane facing sun is now settled science.

       LikeLike

       Reply

44. …OMG ! A myth collapses… I found the canonical depiction of the greenhouse effect even in a prestigious (?) textbook like D.Schroeder’s Introduction to Thermal Physics…

    Here is the link to page 306 Chapter 7 :
    https://drive.google.com/file/d/15th5Jx70PtKYVEzJ92bfUihd_2buZZLp/view?usp=sharing

    How can poor climatologists be ever right with such masters 🙂

    LikeLike

    Reply

    1. Math is math?

       http://rabett.blogspot.com/2017/10/an-evergreen-of-denial-is-that-colder.html?m=1

       LikeLike

       Reply

45. …yes, and it’s not physics 🙂

    LikeLike

    Reply

46. Max, I’m not too proud to admit I can’t explain why your experiment shows no temperature increase from the second plate. Still, there are about a hundred ideas I would like to try and holes I would try to poke in the experiment before I really took seriously the idea that the greenhouse effect is in doubt.

    The main reason for this is: we would have to totally rewrite the laws of thermodynamics to not have a greenhouse effect*. You’ve done the calculation yourself in your model: the green plate effect is a simple consequence of the energy conservation, and the Stefan Boltzmann law. We would need a new theory of thermodynamics which gives up energy conservation to explain the lack of a green plate effect. It seems far more likely that there is some effect in your experiment from convection, or scattered sunlight, or some other variable, which obscures the green plate effect.

    The second reason I am very hesitant to doubt the greenhouse effect is: the earth itself is a convincing demonstration of the effect. Radiation spectra taken from the ground and from space are consistent with the theory of the greenhouse effect, see e.g. http://climatemodels.uchicago.edu/modtran/modtran_iris.jpg . What’s more, one can prove with very simple physics that the average temperature of the earth’s surface could not exceed ~255 K without a greenhouse effect, see https://wattsupwiththat.com/2021/06/04/mathematical-proof-of-the-greenhouse-effect/ .

    All this said, it’s a well done experiment. But there is room for confusion in the modeling. And given the enormous strength of evidence for the greenhouse effect, both theoretical and observed, I think there is a lot of work to be done removing confounding factors, before this experiment can seriously challenge the status quo. I would certainly to be interested to discuss more and see where things lead. And I really appreciate the time you are putting into this.

    * of course, Zoe and a few others will disagree here. But basically every physics professor would agree, including the authors of every thermodynamics textbook of note, and even including the most skeptical of catastrophic global warming (e.g. William Happer, Freeman Dyson)

    LikeLike

    Reply

    1. Tim, why is it over 100C 3km below the surface?

       LikeLiked by 1 person

       Reply

       1. “ Tim, why is it over 100C 3km below the surface?”

          A combination of energy from radioactive decay in the earth, and energy left from the earth’s formation.

          If you were touching a 3 km thick rock and someone put boiling water (100 C) on the other side, would you notice?

          LikeLike

          Reply

          1. Why not extend the greenhouse theory further? Add two-way subsurface conduction, and you can pretend greenhouse effect also does what you claim acreation/nuclear energy does. The math works out the same!!!

             The sun delivers 0 W/m^2 to ~10 meters below the ground, and ~60 meters below ocean. Why is it on average 15+ C at these levels?

             Did you know that Earth’s surface is several kilometers below the top of the atmosphere?

             “If you were touching a 3 km thick rock and someone put boiling water (100 C) on the other side, would you notice?”

             Of course you would. The gradient is 25-30C per km. So you would get 10 to 25C. That’s quite significant. And if the geothermal flux was 0 W/m^2, then you would get 100C!!!

             LikeLiked by 1 person

          2. Zoe, the outside of my oven is just a little warm right now when the inside is 200 C, and that’s through only a few centimeters of insulation. I guarantee you would not observe a temperature change on one side of a 3 _kilometer_ rock when heating the other side by 100 C. Which is exactly why geothermal heat has a tiny contribution to surface temperatures.

             LikeLike

          3. Tim, but it’s ~72C 2km below. That’s from the 100C 1km away, is it not? (plus a very tiny nuclear in between)

             Your oven is mostly empty, how is that like bulky matter?

             Even so, you think your oven floating in space would only be 2.7K?

             LikeLike

          4. Zoe, that’s what insulators do. The temperatures of the two ends don’t much affect each other, and the insulator takes on a temperature gradient between the two ends. If you put the end of a good insulator in boiling water, that end will be at 100 C, your end will be at room temperature, and there will be a gradient between. If you heat that end to 200 C instead, your end will still be room temperature, and there will be a steeper gradient. Either way, the other end doesn’t change the temperature of your end.

             So in answer to the oven-in-space scenario: the oven will still get hot, because it has a heat source and is well insulated from the cold environment.

             LikeLike

          5. Tim, the surface is not an “end”. The “end” would be the top of the atmosphere. We’re ~15km below TOA, and air without any GHGs is a good insulator. Think!

             This insulation is not the greenhouse effect, any more than is 3km of dirt sitting above 100C.

             “the other end doesn’t change the temperature of your end”

             Not true. Don’t you believe the conduction will equal the radiation?

             “well insulated from the cold environment.”
             If your oven is set to 200C, I assure you its exterior will not be 2.7K (the temperature set by radiation from all distant objects in space).

             LikeLiked by 1 person

          6. “ If your oven is set to 200C, I assure you its exterior will not be 2.7K (the temperature set by radiation from all distant objects in space).”

             Oh, you meant the exterior. True, for a real oven the outside will be hotter than the temperature of the environment (in space 2.7K). This is because the insulation isn’t perfect. With more ideal insulation around the oven (for example, three kilometers of rock), the outside temperature could very nearly approach the temperature of space.

             “ Don’t you believe the conduction will equal the radiation?”

             Yes. At steady state, heat gains balance heat losses. Conduction through a very good insulator is negligible, so all other heat fluxes must balance.

             If you have an insulator sitting around, it will settle to room temperature, at which point heat gains = heat losses. If you stick one end in boiling water, the other end will stay at room temperature, as there is virtually no additional heat delivered to that end.

             This is a pretty easy experiment. Stick one end a long wooden rod or rock in a pot of boiling water. The other end will not increase appreciably in temperature.

             LikeLike

          7. “We’re ~15km below TOA, and air without any GHGs is a good insulator. Think!”

             Well, stagnant air is a good insulator. However, air without any GHGs is not any insulation at all for the radiation being emitted by the surface of the Earth. Without the GHE the average temperature of the surface can be no higher than the effective radiating temperature (accounting appropriately for surface emissivity) of the planet.

             So, if we could somehow keep the albedo fixed while “turning off” the GHE then the average temperature of the surface could be no higher than ~255K (or whatever number is appropriate accounting for surface emissivity).

             In fact, the convection of the atmosphere actually acts as a short circuit for the transfer of radiative energy from GHGs to space. If we left the GHE on and could somehow turn convection off then the average temperature of the surface would be even higher than it is now. So the convective action of the atmosphere actually acts to cool the surface.

             LikeLike

    2. Do the radiation spectra really show us much beyond this? https://youtu.be/fpx7hsoYEt4

       LikeLike

       Reply

       1. Tim, the most elegant, and – let me add – obvious (once you know it) explanation of earth surface temperature is that provided by Zoe in her blog.

          Have a look to this to-scale diagram (more or less) of earth and atmosphere temperatures (link below).
          I think that any kid – without knowing thermodynamics – would agree that earth inside frigging hot temps play a role in determining earth surface temperature…and not those poor, light, and cool gases ….
          How on earth is it possible that almost no one realizes it.

          https://drive.google.com/file/d/16ssnjHOfQNHxmzysYpy2B78iuw8q-emH/view?usp=sharing

          ok – granted – my above “explanation” is not very scientific … but Zoe is here for that 🙂

          As for my experiment : what was a bit disturbing was the effect of air outside, that in my view determines that temperature jumps (due to any minimal puff). This made me nervous. But I’m going to repeat the test indoors, I just need some powerful lamps instead of the sun. Convection will be much more stable, and can be much better represented by the correlation I use, and the environment at fixed temperature will be more controllable than sky/ground uncertainties.

          You seem concerned but there is no need to re-write any laws of physics : it’s just that they are wrongly applied. My revised spreadsheet (not yet shared) uses the same laws applied differently, to get the correct (=measured) temperatures.

          LikeLiked by 1 person

          Reply

          1. That sounds like a good next step for the experiment. Putting it inside also removes uncertainty about the sky temperature, and scattered sunlight.

             Re geothermal, see my response above to Zoe. The inside of the earth is several thousand C, but that’s separated from us by 100 km of rock. By 3 km down, it’s only 100 C. Can you feel 100 C applied to the other end of a 3 km rock you’re touching? Of course not, you can barely feel 100 C through a wooden spoon left in boiling water.

             Put another way: the heat inside the earth has no good way of getting out. In fact, that’s why it stays so hot. There is some energy input from radioactive decay, and no good way to shed heat to the surface/space, resulting in high temperatures. (In fact this is very similar to the greenhouse effect: it’s just insulation)

             “ You seem concerned but there is no need to re-write any laws of physics : it’s just that they are wrongly applied.”

             Part of the laws of physics is applying them correctly. The known laws of thermodynamics, correctly applied, predict the greenhouse effect. Curious to see your calculation though.

             LikeLike

          2. Insulation is only a temporary barrier, if you left your oven on long enough the exterior of the oven the kitchen the ceiling cavity roofing and the space above would all distribute the heat from inside the oven. A long rod of iron in a blacksmiths furnace glows cherry red at one end but cold enough to hold at the other end but will get so hot over time that you cannot hold it at all. Wrap the iron bar in insulation and this process is accelerated, the greater the thermal resistance of the insulation the faster the process. Plunge a dense rod of rock kmls into towards the earths centre, insulated by dense solid matter with a molten heat source and wait a few million years and I think there would a heat transfer end to end.

             LikeLiked by 1 person

          3. ross, in your example the other end of a blacksmiths iron will get hot, but never as hot as the end in the fire. Make the iron longer, or use a better insulting material, and the end will barely heat up at all.

             LikeLike

    3. The main reason for this is: we would have to totally rewrite the laws of thermodynamics to not have a greenhouse effect

       It’s the other way around. To make the GHE work you must be able to explain how the backradiation of our cold, thin atmosphere has heated the deep oceans to at least 270K.
       Requires totally new physics 😉

       see https://phzoe.com/2020/03/04/dumbest-math-theory-ever/#comment-4275
       and previous comments I made in this thread.

       LikeLiked by 1 person

       Reply

       1. A wooden object such as a spoon , being of low density is a poor conductor of heat, a metal spoon or granite utensil will heat more rapidly. If you insulate the handle of a metal spoon with say rubber the heating of the metal handle within that insulation becomes even more efficient.

          LikeLike

          Reply

          1. @ ross July 15, 2021 at 9:33 PM

             Agree, but I don’t see how this is relevant to our oceans.

             Oceans are heated by 2 sources, solar and geothermal.
             On a good day in the tropics solar delivers some 25-30 MJ/m^2, which is enough energy to warm the upper 5-10 m 1K. At night this energy is lost again to atmosphere/space.
             Due to the seasons solar warming penetrates max. ~200-300 m into the oceans before the cooling at the surface prevents further conduction downward.
             At the ocean floor the crust delivers ~100 mW/m^2 to the water directly in contact with the floor.
             As soon as bottom warmed water leaves the floor this warming stops.
             100 mW/m^2 is enough energy to warm the average ocean column 1K every ~5000 year.
             It should be obvious that rising bottom heated water can’t reach the surface, wherever the sun maintains a warm mixed surface layer.
             This is almost every where:
             

             So the temperature of the deep oceans is the result of a balance between geothermal heating and cooling at (very high) latitudes, mostly AntArctic Bottom Water sinking to the ocean floor.
             Solar just increases the temp of the mixed surface layer a little.
             So for >70% of Earth the surface temp is the sum of the deep ocean temp plus what the sun can add to that.
             Since the deep oceans are solely heated by geothermal the temp of Earth’s surface is mostly caused by geothermal, the sun just adding the last 10-20 K.

             For the last ~80 MY the deep oceans have been cooling down, eventually resulting in the current ice age with its glacials and inter glacials.

             LikeLiked by 1 person

47. Thanks Tim and Ben for your informative comments, as Im not a scientist I am well out of my depth here and keep my comments brief, my line of thinking is that geothermal and undersea volcanic activity could be an under estimated heat contribution to deep ocean temp. A small additional heat contribution could further discredit the accuracy of climate models. It seems to be a very under researched area of the climate debate. Thanks again to all commenters.

    LikeLike

    Reply

    1. my line of thinking is that geothermal and undersea volcanic activity could be an under estimated heat contribution to deep ocean temp

       Actually it is the ONLY heat source for the deep oceans 😉
       Once you realize this the whole climate system makes sense again.
       The incredibly stable climate on Earth is due to the enormous heat capacity of our oceans. It takes a very long time to change the temp of the deep oceans even 1K.

       Be glad you’re not a climate scientist otherwise you would believe that cold air can heat 4 km deep oceans from above 😉

       See https://tallbloke.wordpress.com/2014/03/03/ben-wouters-influence-of-geothermal-heat-on-past-and-present-climate/
       for some more info.

       LikeLiked by 1 person

       Reply

48. –Rocky
    August 14, 2022 at 5:20 PM

    “We’re ~15km below TOA, and air without any GHGs is a good insulator. Think!”

    Well, stagnant air is a good insulator. However, air without any GHGs is not any insulation at all for the radiation being emitted by the surface of the Earth. —

    Earth’s ocean is also a good insulator.
    The average temperature of the ocean determines global climate.
    Our ocean average temperature is about 3.5 C
    And our average ocean surface temperature is about 17 C
    {and average land is 10 C -giving global average surface temperature of about 15 C],
    Most sunlight reaching Earth’s surface is also going thru the transparent ocean surface.
    70% of earth surface is ocean, and about 80% of surface of tropical zone is ocean.
    Most of sunlight reaching the surface, reaches 40% of the surface within the tropical zone.
    And tropical ocean is the world’s heat engine, the tropical ocean heat entire world.
    The heat from the tropical ocean warms Europe by about 10 K. Europe average temperature
    is about 9 C, and without being warmed by tropical water, Europe’s average temperature
    would be below 0 C.
    Any atmosphere would be controlled would be controlled by 70% of it’s surface. Or ocean surface
    temperature controls global air temperature.
    The 17 C average temperature, make the global average temperature higher and smaller 30% of
    land surface is making global average temperature cooler.
    Most of the Sun’s shortwave radiation passing thru the atmosphere and passes thru the ocean and
    the sun’s energy is “trapped” or absorbed by the ocean.
    The ocean can trap heat from thousands of years. The atmosphere can trap heat for days,
    More than 90% of all global warming is warming the 3.5 C.
    One could think of the ocean “robbing” the heat. But you also say without the ocean more
    than 90% of all global warming would not happen.
    If understand it, you will understand ii is the latter.
    And that average temperature of our ocean is global temperature.
    The coldest of 3.5 C makes global climate be an ice house global climate.
    And ice house global climate is another word for an Ice Age.
    Our Ice Age is called the Late Cenozoic Ice Age.
    And the warmest time in interglacial periods has had ocean average temperature of
    4 C or warmer.
    The amount of heat to make the ocean .5 C warmer is enormous.
    And effect of having ocean be .5 C is warmer enormous as would having the ocean
    being .5 C cooler.
    The recent warming of Earth has been about .05 C increase in the ocean average temperature.
    The so called “end of world” would be an ocean with average temperature as warm all other warmest
    periods of all other interglacial periods- an ocean with average temperature of 4 C or warmer.

    LikeLike

    Reply

    1. Not sure what all of that has to do with what I am attempting to discuss. Again, without a GHE the average temperature of the SURFACE, including any contributions from the oceans, cannot be higher than the effective radiating temperature with appropriate consideration of the surface emissivity. I’m not talking about the depths of the oceans or the heights of the tropopause, I’m talking about the surface.

       LikeLike

       Reply

       1. “Not sure what all of that has to do with what I am attempting to discuss. ”

          You said:
          “However, air without any GHGs is not any insulation at all for the radiation being emitted by the surface of the Earth.”

          The ocean is way better insulation than all GHGs in the atmosphere or could *ever be”.
          Better insulation than even the massive Venus atmosphere.
          GHGs in Earth’s atmosphere are small bit players.

          LikeLike

          Reply

          1. You said:
             “However, air without any GHGs is not any insulation at all for the radiation being emitted by the surface of the Earth.”

             Indeed, I did say that. It is true.

             “The ocean is way better insulation than all GHGs in the atmosphere or could *ever be”.”

             The ocean is no insulation at all for the radiation being emitted by the surface to space given that the ocean does not reside between the surface and space.

             I think you are confused about insulation versus a heat sink.

             LikeLike

          2. “I think you are confused about insulation versus a heat sink.”
             The Ocean is transparent surface and can be called am enormous heat sink, trapping heat, retaining heat. And Venus atmosphere as transparent heat sink is small compared to Earth’s ocean.
             But topic is greenhouse, which transparent to Shortwave and less transparent to Longwave IR.
             1 mm above ocean transparent surface, longwave is completely blocked, whereas with atmosphere about 40 watt per square meter on average radiates directly into space.
             What could call a rocky surface as heat sink, but it’s not transparent to sunlight like a greenhouse.
             So, a rocky surface itself does not absorb much energy from the Sun and rocky surface of earth does warm to rest of planet. The ocean warms Europe, Earth’s rocky surface does not.

             LikeLike

49. @Max Polo – I think it would be a good idea to try to simplify your system. First, convection complicates things quite a bit since the heat transfer due to convection can overwhelm that due to radiation. Given your purpose is to prove that there is no backradiation effect it is important to remove other confounding factors. Another issue is the energy source. It would be much more controlled to heat the plate with a controlled power source, say by resistive heating with a current passing through. Finally, simplifying the geometry would also be useful. The easiest geometry to compare to theory would be spherical because the calculation of the view factors is the simplest. However, this is not an easy geometry to experiment on. Hence, the next best thing would be a cylindrical geometry. Pass the current through a round bar, measure temperature, then bring a cylindrical tube into place, and measure the temperature to steady state again. Again, it would be best if all of this were done in a vacuum and the longer the bar and tube the easier it will be to compare to theory.

    LikeLike

    Reply

    1. Thanks, Rocky, wise suggestions. The thing is, it was way easier, cheap, and convenient, to do what I did. Regarding the heat source, I also tried to use a powerful electric lamp (1kW halogen) instead of the sun, but I couldn’t be sure that irradiation distribution would be uniform so I gave up with this method. The sun is much more reliable and realistic in my view. I have recently refined the calculation model and made a further series of tests. The convection effects can be well controlled and duly accounted for. Even assuming large errors in the convection coefficient and/or other input parameters, the calculation shows that I should measure from 2°C to 2.5°C back-radiation heating with the chosen geometry. I never ever came close to measuring that much effect, with a thermometer that is sensitive to 0.1°C, and that feels even the slightest puff of wind in terms of temperature changes. Back-radiation heating doesn’t show up. This, despite some solar irradiation “leaking” to the rear pane (approx top 25% of rear pane surface, due to sun altitude, is short-wave irradiated). I have now accounted for this additional energy in the model, reaching a very good agreement between the calculated and the measured rear pane temperature. So I have more and more evidence that all the parameters and their effects are under control, and still, zero back-radiation heating is measured. I have a much more detailed report ready maybe in a couple of weeks, work permitting. On the other hand, the revised model that correctly predicts reality is quite simple and works very well…I’ll show it in the revised report.

       LikeLike

       Reply

       1. “The thing is, it was way easier, cheap, and convenient, to do what I did.”

          Agreed, but it is also far more difficult to interpret the results because you are unable to isolate the different effects. For example, let’s say you had an electric circuit with two resistors in parallel, red and blue. You are trying to measure the resistance of the red resistor, which has a large resistance in comparison to the the blue one. The most accurate way to measure the red resistance is to take the blue resistor out of the circuit entirely. Furthermore, if the blue resistance is very small, then it may be impossible to measure the red resistance if your instruments are not sensitive enough. Radiative and convective heat transfer sort of act like the red and blue resistors in your experiment.

          “I also tried to use a powerful electric lamp (1kW halogen) instead of the sun, but I couldn’t be sure that irradiation distribution would be uniform so I gave up with this method.”

          That’s really no better than using the sun if you are interested in doing a controlled experiment.

          “The sun is much more reliable and realistic in my view.”

          Perhaps more reliable than a heat lamp, but not more reliable than a controlled, directed, and measured heat source. With the heat lamp, you don’t know how much of its energy is actually being absorbed by the panel. The same goes with the sun.

          “The convection effects can be well controlled and duly accounted for.”

          I’m skeptical of that claim. Duly accounting for convection would require highly detailed numerical calculations.

          “Even assuming large errors in the convection coefficient and/or other input parameters, the calculation shows that I should measure from 2°C to 2.5°C back-radiation heating with the chosen geometry.”

          Can you please show me the details of that calculation? I’d like to see if I agree with your methods.

          “I never ever came close to measuring that much effect.”

          That’s not surprising to me with the geometry that you have. It is dominated by convective heat transfer. Thus I highly doubt that your calculations are correct.

          “Back-radiation heating doesn’t show up.”

          I suggest you try a vacuum. I’m willing to bet a significant amount of money that it will show up.

          “This, despite some solar irradiation “leaking” to the rear pane”

          You also have radiation to the rear pane from the surroundings. Did you account for that? Do you know what the temperature of the surroundings was in your experiment?

          “I have now accounted for this additional energy in the model, reaching a very good agreement between the calculated and the measured rear pane temperature.”

          Again, I’d love to see your calculations. I’m highly skeptical that you have accounted for the radiative and convective heat transfer properly. It’s clear that you are not testing the simple plate configuration that has been analyzed by others.

          LikeLike

          Reply

50. “We see that the screen is “trapping” a lot of human radiation from reaching the IR camera, and we expect an extra 296.8 W/m² greenhouse effect, bringing us up to 55°C. Merely placing a screen in front of us should make us feel as if we’re stepping inside a sauna.”

    I don’t understand your reasoning here. Prior to putting the screen in place you were receiving radiation from the IR camera and whatever else the screen is blocking. You were also emitting radiation to those things. With the screen in place, you are now receiving radiation from the screen. If the screen is emitting more radiation than what it was blocking then you would feel warmer, otherwise you would feel cooler. If we approximate the room as being at ~25C and the screen as a blackbody radiator, then in order to be receiving an extra 296.8 W/m², the screen would have to be at ~65C. A screen that hot might make you feel like you are in a sauna.

    LikeLike

    Reply

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out /  Change )

You are commenting using your Twitter account. ( Log Out /  Change )

You are commenting using your Facebook account. ( Log Out /  Change )

Cancel

Connecting to %s

Notify me of new comments via email.

Notify me of new posts via email.

Δ