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The runaway greenhouse effect on Venus

What the science says...

Venus very likely underwent a runaway or ‘moist’ greenhouse phase earlier in its history, and today is kept hot by a dense CO2 atmosphere.

Climate Myth...

Venus doesn't have a runaway greenhouse effect

Venus is not hot because of a runaway greenhouse.

In keeping with my recent theme of discussing planetary climate, I am revisiting a claim last year made by Steven Goddard at WUWT (here and here, and echoed by him again recently) that “the [runaway greenhouse] theory is beyond absurd,” and that it is pressure, not the greenhouse effect that keeps Venus hot.  My focus in this post is not on his alternative theory (discussed here), but to discuss Venus and the runaway greenhouse in general, as a matter of interest and as an educational opportunity.  In keeping my skepticism fair, I’d also like to address claims (sometimes thrown out by Jim Hansen in passing by) that burning all the coal, tars, and oil could conceivably initiate a runaway on Earth.

It is worth noting that the term runaway greenhouse refers to a specific process when discussed by planetary scientists, and simply having a very hot, high-CO2 atmosphere is not it.  It is best thought of as a process that may have happened in Venus’ past (or a large number of exo-planets being discovered close enough to their host star) rather than a circumstance it is currently in.

A Tutorial of Present-Day Venus

Venus’ orbit is approximately 70% closer to the sun, which means it receives about 1/0.72 ~ 2 times more solar insolation at the top of the atmosphere than Earth.  Venus also has a very high albedo which ends up over-compensating for the distance to the sun, so the absorbed solar energy by Venus is less than that for Earth.  The high albedo can be attributed to a host of gaseous sulfur species, along with what water there is, which provide fodder for several globally encircling sulfuric acid (H2SO4) cloud decks.  SO2 and H2O are the gaseous precursor of the clouds particles; the lower clouds are formed by condensation of H2SO4 vapor, with SO2 created by photochemistry in the upper clouds. Venus’ atmosphere also has a pressure of ~92 bars, nearly equivalent to what you’d feel swimming under a kilometer of ocean.  The dense atmosphere could keep the albedo well above Earth’s even without clouds due to the high Rayleigh scattering (the effect of clouds on Venus and how they could change in time is discussed in Bullock and Grinspoon, 2001). Less than 10% of the incident solar radiation reaches the surface.

Observations of the vapor content in the Venusian atmosphere show an extremely high heavy to light isotopic ratio (D/H) and is best interpreted as a preferential light hydrogen escape to space, while deuterium escapes less rapidly.  A lower limit of at least 100 times its current water content in the past can be inferred (e.g. Selsis et al. 2007 and references therein).

The greenhouse effect on Venus is primarily caused by CO2, although water vapor and SO2 are extremely important as well.  This makes Venus very opaque throughout the spectrum (figure 1a), and since most of the radiation that makes its way out to space comes from only the very topmost parts of the atmosphere, it can look as cold as Mars from IR imagery. In reality, Venus is even hotter than the dayside of Mercury, at an uncomfortable 735 K (or ~860 F). Like Earth, Venusian clouds also generate a greenhouse effect, although they are not as good infrared absorbers/emitters as water clouds.  However, the concentrated sulfuric acid droplets can scatter infrared back to the surface, generating an alternative form of the greenhouse effect that way.  In the dense Venusian CO2 atmosphere, pressure broadening from collisions and the presence of a large number of absorption features unimportant on modern Earth can come into play (figure 1b), which means quick and dirty attempts by Goddard to extrapolate the logarithmic dependence between CO2 and radiative forcing make little sense.  The typical Myhre et al (1998) equation which suggests every doubling of CO2 reduces the outgoing flux at the tropopause by ~4 W/m2, although even for CO2 concentration typical of post-snowball Earth states this can be substantially enhanced.  Figure 1b also shows that CO2 is not saturated, as some skeptics have claimed.


 Figure 1: a) Radiant spectra for the terrestrial planets.  Courtesy of David Grisp (Jet Propulsion Laboratory/CIT), from lecture "Understanding the Remote-Sensing Signatures of Life in Disk-averaged Planetary Spectra: 2" b) Absorption properties for CO2. The horizontal lines represent the absorption coefficient above which the atmosphere is strongly absorbing.  The green (orange) rectangle shows that portion of the spectrum where the atmosphere is optically thick for 300 (1200) ppm.  From Pierrehumbert (2011)

 How to get a Runaway?

To get a true runaway greenhouse, you need a conspiracy of solar radiation and the availability of some greenhouse gas in equilibrium with a surface reservoir (whose concentration increases with temperature by the Clausius-Clapeyron relation).  For Earth, or Venus in a runaway greenhouse phase, the condensable substance of interest is water— although one can generalize to other atmospheric agents as well.

The familiar water vapor feedback can be illustrated in Figure 2, whereby an increase in surface temperature increases the water vapor content, which in turn results in increased atmospheric opacity and greenhouse effect.  In a plot of outgoing radiation vs. temperature, this would result in less sensitive change in outgoing flux for a given temperature change (i.e., the outgoing radiation is more linear than one would expect from the σT4 blackbody-relation). 


Figure 2: Graph of the OLR vs. T for different values of the CO2 content and relative humidity.  For a fixed RH, the specific humidity increases with temperature. The horizontal lines are the absorbed shortwave radiation, which can be increased from 260-300 W m-2.  The water vapor feedback manifests itself as the temperature difference between b’-b and a’-a, since water vapor feedback linearizes the OLR curve.  Eventually the OLR asymptotes at the Komabayashi-Ingersoll limit.  Adopted from Pierrehumbert (2002)


One can imagine an extreme case in which the water vapor feedback becomes sufficiently effective, so that eventually the outgoing radiation is decoupled from surface temperature, and asymptotes into a horizontal line (sometimes called the “Komabayashi-Ingersoll” limit following the work of the authors in the 1960’s, although Nakajima et al (1992) expanded upon this limiting OLR in terms of tropospheric and stratospheric limitations).  In order to sustain the runaway, one requires a sufficient supply of absorbed solar radiation, as this prevents the system from reaching radiative equilibrium.  Once the absorbed radiation exceeds the limiting outgoing radiation, then a runaway greenhouse ensues and the radiation to space does not increase until the oceans are depleted, or perhaps the planet begins to get hot enough to radiate in near visible wavelengths.


Figure 3: Qualitative schematic of how the ocean reservoir is depleted in a runaway.  From Ch. 4 of R.T. Pierrehumbert’s Principles of Planetary Climate


On present-day Earth, a “cold trap” limits significant amounts of water vapor from reaching the high atmosphere, so its fate is ultimately to condense and precipitate out.  In a runaway scenario, this “cold trap” is broken and the atmosphere is moist even into the stratosphere.  This allows energetic UV radiation to break up H2O and allow for significant hydrogen loss to space, which explains the loss of water over time on Venus.  An intermediate case is the “moist greenhouse” (Kasting 1988) in which liquid water can remain on the surface, but the stratosphere is still wet so one can lose large quantities of water that way (note Venus may never actually encountered a true runaway, there is still debate over this).  Kasting (1988) explored the nature of the runaway /moist greenhouse, and later in 1993 applied this to understanding habitable zones around main-sequence stars.  He found that a planet with a vapor atmosphere can lose no more than ~310 W/m2, which corresponds to 140% of the modern solar constant (note the albedo of a dense H2O atmosphere is higher than the modern), or about 110% of the modern value for the moist greenhouse.


Earth and the Runaway: Past and Future


Because Earth is well under the absorbed solar radiation threshold for a runaway, water is in a regime where it condenses rather than accumulating indefinitely in the atmosphere.  The opposite is true for CO2, which builds up indefinitely unless checked by silicate weathering or ocean/biosphere removal processes.  In fact, a generalization to the runaway threshold thinking is when the solar radiation is so low, so that CO2 condenses out rather than building up in the atmosphere, as would be the case for very cold Mars-like planets.  Note the traditional runaway greenhouse threshold is largely independent of CO2 (figure 2 & 4; also see Kasting 1988), since the IR opacity is swamped by the water vapor effect.  This makes it very difficult to justify concerns over an anthropogenic-induced runaway.


Figure 4: The H2O–CO2 greenhouse. The plot shows the surface temperature as a function of radiated heat for different amounts of atmospheric CO2 (after Abe 1993). The albedo is the fraction of sunlight that is not absorbed (the appropriate albedo to use is the Bond albedo, which refers to all sunlight visible and invisible). Modern Earth has an albedo of 30%. Net insolations for Earth and Venus ca. 4.5 Ga (after the Sun reached the main sequence) are shown at 30% and 40% albedo. Earth entered the runaway greenhouse state only ephemerally after big impacts that generated big pulses of geothermal heat. For example, after the Moon-forming impact the atmosphere would have been in a runaway greenhouse state for ∼2 million years, during which the heat flow would have made up the difference between net insolation and the runaway greenhouse limit. A plausible trajectory takes Earth from ∼100 bars of CO2 and 40% albedo down to 0.1–1 bar and 30% albedo, at which point the oceans ice over and albedo jumps. Note that CO2 does not by itself cause a runaway. Also note that Venus would enter the runaway state when its albedo dropped below 35%.  Se e Zahnle et al 2007


This immunity to a runaway will not be the case in the long-term.  In about a billion years, the sun will brighten enough to push us into a state where hydrogen is lost much more rapidly, and a true runaway greenhouse occurs in several billion years from now, with the large caveat that clouds could increase the albedo and delay this process.

Interesting, some (e.g.. Zahnle et al 2007) have argued that Earth may have been in a transient runaway greenhouse phase within the first few million years, with geothermal heat and the heat flow from the moon-forming impact making up for the difference between the net solar insolation and the runaway greenhouse threshold, although this would last for only a brief period of time.  Because the runaway threshold also represents a maximum heat loss term, it means the planet would take many millions of years to cool off following such magma ocean & steam atmosphere events of the early Hadean, much slower than a no-atmosphere case (figure 5).


Figure 5: Radiative cooling rates from a steam atmosphere over a magma ocean. The radiated heat is equal to the sum of absorbed sunlight (net insolation) and geothermal heat flow. The plot shows the surface temperature as a function of radiated heat for different amounts of atmospheric H2O (adapted from Abe et al. 2000). The radiated heat is the sum of absorbed sunlight (net insolation) and geothermal heat flow. The different curves are labeled by the amount of H2O in the atmosphere (in bars). The runaway greenhouse threshold is indicated. This is the maximum rate that a steam atmosphere can radiate if condensed water is present. If at least 30 bars of water are present (a tenth of an ocean), the runaway greenhouse threshold applies even over a magma ocean. Note that the radiative cooling rate is always much smaller than the σT4 of a planet without an atmosphere


Venus likely underwent a runaway or “moist greenhouse” phase associated with rapid water loss and very high temperatures.  Once water is gone, silicate weathering reactions that draw down CO2 from the atmosphere are insignificant, and CO2 can then build up to very high values.  Today, a dense CO2 atmosphere keeps Venus extremely hot.

Last updated on 11 April 2011 by Chris Colose.

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Comments 51 to 100 out of 266:

  1. Rosco - how do you find out who is right? You measure it. See which number matches reality. Do you accept that this is the way science questions are settled?
  2. Tom Curtis @ 48 says "You have not raised any interesting questions about the moon. You have merely cited a vaguely remembered maximum temperature. Apparently you base all your reasoning on the assumption that the maximum temperature is the only relevant temperature, but I am disinclined to follow you in that absurdity." But this is precisely what the Stefan-Boltzman formula relates - the temperature for a given radiative flux - well actually it is the reverse if we're being strictly correct - it gives the radiative flux for a given temperature.
  3. scaddenp @51 Certainly - you cannot argue with accurately measured observations. If the energy input is 342 W/sq m which results in ~5.5 C check out the temperature on the moon which has no atmosphere. NASA provides all the facts - here is a link - "During the day the temperature on the Moon can reach 253 Fahrenheit (123 Celsius), while at night it can drop to -387 Fahrenheit (-233 Celsius). The Earth, which has an atmosphere, has a much more comfortable range of temperatures." As I said - We have an anomaly that I find very interesting.
  4. Well sorry about the link in the previous post. It worked in June. - works. Google it and you'll find it is a little bit more than ~5.5 C. Draw your own conclusions. As I said - We have an anomaly that I find very interesting.
  5. Rosco, You're comparing the average temperature of earth that has a day/night cycle of 1 day, to the max and min temperature (not even average max or min temp) of the moon where the day/night cycle is 29 days. And it is a surprise that they differ?
  6. You miss the point - once equilibrium is reached the only way to increase the temperature is to increase the energy input - simply pumping in the same amount simply maintains the temperature. So if the moon is irradiated by 342 W/sq m as is claimed for Earth its temperature is ~5.5 C - Simple indisputable physics.
  7. 56, Rosco, No, it's temperature is 5.5C or less due to the moon's albedo. Of course, the moon has no atmosphere, so there is no opportunity for a greenhouse effect. The mean daytime temperature is 380K. The mean nighttime temperature is 120K. The overall mean temperature is 250K. This corresponds to an energy input of 221 W/m2, which is not surprising. It suggests an albedo of 221/342 or 0.65, which certainly fits with how bright the moon appears at night. Simple, properly applied physics. What does this have to do with Venus, by the way?
  8. All of this has everything to do with Venus. How does the mean daytime temperature get to 380 K on the moon with no atmosphere but the earth, irradiated by the same irradiation, is minus 18 ? What does the mean temperature on the moon have to do with anything ? It is either illuminated (`1368 W/sq m) and hot or dark (~0 W/sq m) and cold.
  9. Sorry should have said 255 K or minus 18 C.
  10. Properly applied physics says that the temperature is proportional to the radiative flux - Stefan-Boltzman. It is a simple equation with only 2 variables ! 342 W/sq m = ~278.68 K or ~5.5 C Maximum. 1368 W/sq m = ~394.12 K or ~ 121 C Maximum.
  11. 50, Rosco, Your statement is nonsensical. The time over which the energy is distributed is irrelevant. Your inability to understand the factor of four is astounding. If you wish, lets use a day's worth of energy. As you stated, a Watt is measured in Joules per second, or an amount of energy delivered per second. As already stated, the total energy received by the earth in one second is 1.748310 x 1017 W. There are 86,400 seconds in an earth day, so if we multiply the two we get the total amount of energy (in Joules) received by the earth in 24 hours. 86,400 seconds times 1.748310 x 1017 W... or rather, 86,400 seconds times 1.748310 x 1017 J/sec gives 1.51053984 x 1022 Joules. If we divide that by the total area of the earth (again, from before, 5.1120196 x 1014 m2) we will get the average energy per square meter delivered to the earth in one earth day. So 1.51053984 x 1022 Joules / 5.1120196 x 1014 m2 gives us 2.9548788 x 107 Joules/m2. That is the amount of energy in Joules per square meter delivered to the earth in one earth day. But we'd like to get that number in J/sec/m2 (which is W/m2), so we'll divide by the number of seconds in one earth day, or 86,400 seconds. 2.9548788 x 107 J/m2 divided by 86,400 seconds is... you guessed it, 341.999862 J/sec/m2, or rather 341.999862 W/m2. Happy?
  12. 60, Rosco, No, you are discounting albedo. The energy which is reflected is not absorbed. Note that the maximum temperature on the moon is 120C... that is, the spot on the moon that actually receives 1368 W/m2, for the brief period it does so, can achieve a temperature of 120C (if it hits something black). Really, Rosco, you're tying yourself in knots trying to prove what you misunderstand, when you should be stepping back and saying "whoa, everyone here says something else, and all of science says something else, maybe I better reconsider my position, open my mind, and read and learn instead of posting nonsense."
  13. No - if the earth rotates once in 24 hours and every point on the earths surface is luuminated then the whole area of the arths is illuminated and not the area of the disk which has the same diameter so the factor of 4 used to reduce the solar "constant" of 1368 W/sq m TOA to 342 W/sq m TOA is invalid. I see we will never agree however the Stefan-Boltzman equation is either right or wrong. Climate scientists use it all the time and the way they apply the geometrical manipulations cannot explain the temperature on the moon. We can argue all day about planets where the atmosphere plays a significant role but the simple indisputable fact is if you apply the geometrical reduction to the moon you cannot explain the maximum temperature there. If it fails for the moon why is it valid for Venus, Earth or anywhere else ?
  14. Rosco, The geometric reduction is meant to explain the average temperature over the entire earth/moon, and of course it doesn't explain the maximum "daily" temperature. You are confusing the two concepts. The whole point of the simple radiative model is to provide an understanding of how greenhouse effect increase the average temperature of the planet. If you insist that it should reproduce the maximum temperature in a day/night cycle you are missing the point.
  15. Rosco#63: "cannot explain the temperature on the moon" Your minimalist approach, picking the highest and lowest lunar temperatures and computing an 'average' for use in the SB equation, neglects a few important facts. Are you aware that the moon has negligible axial tilt, so that areas in shadow near the poles (where these ultra cold temperatures were measured) are hardly ever in sunlight? They never warm up, so they do not reach equilibrium with the illuminated portion of the planet. See the images here. The areal extent of these ultracold regions is quite small. What you have done with the lunar temperature range is equivalent to looking at a dataset consisting of {10,10,10,10,10,10,10,0} and concluding the 'average' is 5. Your conclusions about climate science based on that error are thus utterly incorrect. In short, if it doesn't 'fail for the moon,' it is valid for the Earth.
  16. 63, Rosco,
    I see we will never agree however the Stefan-Boltzman equation is either right or wrong.
    This statement is utter nonsense. No one is disputing Stefan-Boltzman, and nothing that is being explained to you is in conflict with it. Stop simply making things up!!!!!!!. The argument is about how to distribute the energy that arrives at the earth on only one side over the entire surface of the earth, a fairly basic bit of geometry that you appear incapable of grasping. You instead would like to pretend that this energy is simply divided among the two hemispheres of the earth, as if it were a flat disk... Oh!!! Now I get it... you're one of those flat-earthers I've heard about. But the earth isn't flat. The energy has to be distributed over the surface of a spherical earth, not a flat earth. Really, I can't believe I'm trying to explain things to someone who can't get past the first page of any introductory climate science text. Please, Rosco, please go do some reading. By the way, throwing the word Venus into your posts doesn't cut it. There is no connection whatsoever to your discussion and the GHE on Venus. It is long past time for this to stop.
  17. Rosco @54, seriously, "Ask an Astronomer for Kids" is your source of astronomical information? Really? You could have at least tried the moon fact sheet from NASA, where we learn that the blackbody temperature of the moon is 270.7 degrees K. Let me see, 1366/4 * 0.89 (1-lunar albedo) = 303.9 K, or 33.2 K greater than the black body temperature of the moon. The reason for the discrepancy is well known - the divide by 4 approximation is only perfectly accurate for bodies with no temperature variation. Because radiated output varies with the fourth power of temperature, if there is temperature variability, the energy is radiated from the surface more efficiently, resulting in a lower temperature, as can be seen on the moon. That means the Earth's atmosphere and ocean, by redistributing heat do in fact warm the Earth, but they cannot warm the Earth to more than the 255 degrees K indicated by the usual black body approximation. Indeed, as they do not eliminate temperature variation from the surface (as they do on Venus), they warm it to less than that temperature and the total greenhouse effect is more than the normally stated 33 degrees K difference between 255 K 'expected' black body temperature and 288 K average surface temperature. As it happens, the actual black body temperature of the Earth is 254.3 degrees K, only 0.7 degrees K below the expected using the standard approximation, so it is a very good approximation. (I believe Chris Colose discussed this in more detail on this site recently, but cannot remember where.) The question may arise as to whether NASA know lunar temperatures well enough to determine the black body temperature of the moon. Afterall, determining that temperature requires measuring the Outgoing Long wave Radiation integrated across the entire moons surface and over the entire 28 day rotation period. Welcome to the Diviner mission: Lunar Temperatures by latitude and Lunar Hour: (Note, one lunar hour equals 29.53 Earth hours.) Lunar Day Time Temperatures: Lunar Night Time Temperatures:
  18. Muoncounter @65, a minor point - the regions that never see sunlight have temperatures below 35 degrees K. The 120 degrees K is the average night time temperature of the moon, with the poles maintaining a more even temperature during day and night of around 220 K, except at the bottom of craters where the temperatures are much lower. This is probably the best image to see that.
  19. TC#68: Rosco's moon has a low temperature of -233C and a high of 123C, both consistent with the color schemes in the LRO images (purple ~ 40K). It is his average lunar temperature that makes no sense. But consider this: If the lunar temperature is appropriate for its effective (non-black body) radiation balance, as you explain in #67, that means solar input really is driving planetary temperature. But that is in contradiction with Doug Cotton's temperature of the earth's surface is based on the core temperature fantasy. Since we can't have both, which one should we discard? Tom, you've disproved Rosco and Cotton in one shot - all in all, a nice day's work.
  20. Tom Curtis @67 you think NASA would lie to Kids ? If as you say solar insolation is 240 W/ sq m over earth then the Stefan-Boltzman temperature tells you that Earth can never increase in temperature above 255 K. Again - The Stefan-Boltzman equation has 2 variables - radiative flux and temperature. When a surface is irradiated it reaches the temperature determined by that radiative flux. When a surface is not irradiated it will tend towards the temperature of its surroundings. For the moon with no atmosphere to distribute temperature around the sphere it is either irradiated by solar radiation and reaches an equilibrium temperature commensurate with that level of radiation (which has been measured as ~120 C or 393 K) - OR - it is not irradiated and begins cooling to reach thermal equilibrium with the erergy flux of free space which does not have solar radiation incident on it and that is a very low flux commensurate with the temperature of free space which is postulated to be as low as a few K. The extra energy postulated to come from Greenhouse gases to warm the surface of the Earth came from where originally ?
  21. muoncounter @69 (-Snip-)

    [DB] Off-topic snipped.  Continuing to perpetuate your intransigence in actually taking the time and bother to actually learn something about climate science has become intolerable.

    Please note that posting comments here at SkS is a privilege, not a right.  This privilege can and will be rescinded if the posting individual continues to treat adherence to the Comments Policy as optional, rather than the mandatory condition of participating in this online forum.

    Moderating this site is a tiresome chore, particularly when commentators repeatedly submit off-topic posts. We really appreciate people's cooperation in abiding by the Comments Policy, which is largely responsible for the quality of this site.

    Please take the time to review the policy and ensure future comments are in full compliance with it.  Thanks for your understanding and compliance in this matter.

  22. I give in. I seem to be breaching the commentary policy by arguing that the basis for determining the energy input to a planetary atmosphere is important in determining the topic discussed here. I guess you will always oppose my point of view adn I will continue to oppose the factor of 4 reduction of insolation. But remember - energy cannot be created or destroyed merely transformed. Think about it.

    [DB] The point that you are not grasping is that this thread is about Venus doesn't have a runaway greenhouse effect.

    You have given no indication of talking about that topic, despite able advice from others.  Other threads (nearly 5,000 in number) exist here at SkS on every subject imaginable that pertain to climate science.  Using the seach function in the upper left corner of every page here at SkS, search for that topic you want to hang your hat on and place a comment there (for example Has the greenhouse effect been falsified?).

    Dozens of regular participants here are ready to help you gain a better understanding of climate science.  So the choice stands before you:

    1. Continue in your present path of not listening to others and continuing to be off-topic, with the expected result of forcing the moderators to intervene
    2. Or follow the path outlined above

    Think about it.

  23. The last post. I don't understand why everyone seems to insist I calculate some sort of average - I don't. I think the only relevant thing is the maximum at any time as we all know things will lose energy and cool. I am simply trying to explain why it is possible that Venus when irradiated by the ~2640 W/sq m in the vicinity it occupies in space can have a blackbody temperature of ~464K - much lower than observed but also much higher than usually calculated. I have used the examples of the Earth and the moon to demonstrate because we have some reliable data for these. the graph of the moon above shows varying temperature over various latitudes and agrees with what I posted before the insolation varying as the cosine of the latitude. It also shows a maximum temperature of ~ 380 - 390 K as would be expected using the solar constant. I cannot see any flaw in this logic so I guess we'll have to agree to disagree. The use of the sine and cosine to break a vector down into its normal and tangential component is well established scientific method.

    [DB] "I cannot see any flaw in this logic so I guess we'll have to agree to disagree."

    The flaws in your logic, physics and math have already been pointed out to you.  That you refuse to accept that is telling.

  24. Despite the angst that seems to have developed I have enjoyed the intellectual stimulation. I think people should discuss and argue their beliefs and everyone should have a right to voice theirs. Definately the last post - hopefully not the military version. PS - I don't believe in the flat disk model so I'm definitely not a flat earther.

    [DB] "Despite the angst that seems to have developed"

    Your conduct here is the intellectual version of poking a bee's nest with a stick.  That that activity generates a response you characterize as "angst" should be of no surprise.

    "I think people should discuss and argue their beliefs and everyone should have a right to voice theirs"

    This is a science website.  The conversations and dialogue center on climate science using logic and evidenciary chains.  Not fuzzy terms & beliefs & opinions.  Evidence-based discussions of climate science using peer-reviewed published studies from reputable sources.

    Given your persistence in avoiding using any of the latter some "angst" should be expected.

  25. 73, Rosco,
    I think the only relevant thing is the maximum at any time...
    You are wrong about this. Also, please notice how often you use the words "I think" and "I believe." These are fuzzy broadcasts of the fact that you don't know, but refuse to learn.
    I think people should discuss and argue their beliefs...
    If you want to do so, visit a site about religion. This site is about science, and as such it is about facts, not beliefs. You are entitled to your own beliefs, but not your own facts.
    ...and everyone should have a right to voice theirs.
    No. You do not have a right to broadcast misinformation here, any more than I have a right to walk into a classroom and teach children that mathematics is evil and the language of the Satan.
    PS - I don't believe in the flat disk model so I'm definitely not a flat earther.
    Yes, you do, and if you understood the math you'd recognize this. My suggestion... leave this site, stop posting, and surf the Internet under the assumption that there is something you really, really do not understand. Try to figure it out so that you can come back, apologize for your recalcitrance, and discuss things on the level of understanding that is appropriate to this site.
  26. Rosco#73: "I don't understand why everyone seems to insist I calculate some sort of average - I don't." The other day I measured 109F air temperature in my backyard; at the same time the ground surface (in direct sunlight) read 128F. You would use the maximum local transient (the area in direct sun changes during the day) temperature as the day's high? 109F was bad enough, but 128F? Why not record the temperature inside a car as the day's high? Or inside an abandoned refrigerator? Everyone's maximum temperature would vary depending on sun exposure and materials. That approach is just flat wrong; it is unbelievable that you don't understand something so straightforward. Recognizing that you may continue reading after your 'last post' and your 'definitely the last post,' there's no need of a reply. It is a shame, however, to encounter such rigid opposition to learning. In education, we advocate for 'life-long learning;' that requires a willingness to abandon preconceived beliefs. To be so clearly incorrect and yet remain unwilling to learn strongly suggests it is by deliberate choice. Sad, very sad.
  27. Sorry - I didn't intend to post again. muoncounter measured 109 F air temperature - (~43 C or ~316 K) and 128 F ( 53.3 C or 326 K) on the ground. The energy flux associated with this temperature is ~640 W/sq m. You have just supplied some real data - the maximum temperature at your place that day required ~640 W/sq m. And presumably it built up during the day to a maximum in the afternoon and began cooling after the peak. Surely we can agree on that ?
  28. Rosco#77: This demonstrates exactly why you are wrong about using these peak temperatures. Put aside the fact that the ground was not at equilibrium (adjacent areas in shade had lower temperatures), hence there must be heat flow along this very local thermal gradient. At 30N latitude during August, I am basking in the glow of a mere 440 W/m^2. -- source To paraphrase, it would indeed be a travesty if you believed something had to account for the missing 200 W/m^2.
  29. Interesting graph and site - I liked the fact they use the sine and cosine of angles to break down non normal incident vectors into normal components to determine the value incident on a surface - I think I've seen that somewhere - I'll research the material provided on that site but according to Stefan-Boltzman 440 W/sq m is the radiative flux associated with a temperature of ~ 297 K or ~24 C or ~76 F ?
  30. Sorry I pressed submit accidentally I should have said the sine of the angle to the sun in the sky at any time is the magnitude of the vector of importance for insolation. The cosine of the latitude of where on earth you are situated is the magnitude of the vector of importance for insolation. The combination of the two gives a vector relating insolation to a point on earth but the problem is the sine value changes daily whilst the cosine changes seasonally.
  31. As I understand it, Rosco, you keep trying to apply theory based on equilibrium temperatures resulting from radiation balance to scales of observation where equilibrium clearly doesn't apply and where a large number of variables besides solar radiation are important - incuding downwelling LW radiation, which you seem to believe does not exist. You then wonder why your simple approach gives you startling numbers for incoming radiation. What you are doing is trying to predict weather - using too few variables. At the planetary scale, things are simpler. The only way heat can be exchanged between a planet and the surrounding space is via radiation. That simplifies the problem to one of radiation balance. Simlicity is a good thing! Because solar inputs and LW outputs are not necessarily constant across the globe, a proper balance requires integrating gains and losses over the globe through proper averaging. Certainly, as Tom has pointed out, spatial variation in surface temp can influence the efficiency by which heat is shed to space, and therefore the equilibrium planetary temp. But, luckily, one need not understand what is happening at every point on earth to get a reasonable radiation balance for Venus.
  32. All I am pointing out is in muoncounter's back yard the earth in the sun was emitting ~640 W/sq m at 123 F, the air around him was emitting ~565 W/ sq m at 109 F - these calculations are from Stefan-Boltzman's law used correctly. I don't know what temperatures you get in winter but if you convert them to Kelvin and calculate the radiative flux you'll see a huge difference - for example at 60 F (~288K)the radiative flux is ~390 W/sq m and for say 40 F (~277.6 K) the radiative flux is ~336 W/sq m. There is an enormous difference that cannot be explained by a "constant irradiation" model.
  33. The IPCC state in their scientific discussion that approximately 51 % of incoming solar radiation warms the surface of earth - it is not front and centre like Figure 1 but it is there. If you calculate 51 % of the solar constant and multiply that by the cosine of the latitude of where you live you arrive at the maximum irradiation for your home on a clear day at the summer solstice for where you live. You will see that this maximum energy corresponds to a temperature - calculated by Stefan-Boltzman's law - which is much more than recorded values. We all know heated air rises and is replaced with cooler air which is why (no matter how you believe the air is heated) the temperature never achieves the theoretical maximum. If you had a perfectly sealed greenhouse you would record a temperature inside it near to the theoretical maximum for the insolation minus some losses because there is never perfect transmission - there are always losses. That is what I have been saying.
  34. Rosco, you still haven't gone and done the reading, have you? As someone said earlier, you're disagreeing with the stuff on just about the first page of a climatology or atmospheric physics textbook, yet you're utterly convinced you're right and everyone else is wrong. It's sad to see, really.
  35. Rosco#76 and beyond, Please stop insisting you are the only person who understands radiative physics. I'm willing to bet that everyone who has commented on this thread (except you) has seen the equations that go into those insolation curves. You did not discover that sine / cosine of the solar azimuth angle and latitude are involved. But in #80, you refer to sine and cosine of angles as 'magnitudes of vectors.' That would be a -5 on any physics test where I teach. As far as my backyard is concerned, reduce the emissivity to ~0.7 and see what radiative flux you obtain at an absurdly hot 326K. In winter, when my backyard air temp could go as low as 270K and the ground temp is closer to 288K, an emissivity of ~0.7 again matches the insolation of 270 W/m^2. Guess the sand and clay of my backyard aren't black body emitters; I'll have to sell the place. Please do some reading before you post your next 'definitely definitely the last post'.
  36. muoncounter - the 51 % was after albedo of ~30% and atmospheric absorbtion of ~19 %. Please read this from the IPCC 1.4.3 Solar Variability and the Total Solar Irradiance 3rd paragraph "Between 1902 and 1957, Charles Abbot and a number of other scientists around the globe made thousands of measurements of TSI from mountain sites. Values ranged from 1,322 to 1,465 W m–2, which encompasses the current estimate of 1,365 W m–2. Foukal et al. (1977) deduced from Abbot’s daily observations that higher values of TSI were associated with more solar faculae (e.g., Abbot, 1910)." the solar insolation is certainly a vector quantity This link shows ~51 % solar irradiance reaches Earth.
  37. Rosco, I would like to suggest a paper for you to read. I think it might help you understand what so many in this thread have been trying to communicate you. Kruger, Justin; David Dunning (1999). "Unskilled and Unaware of It: How Difficulties in Recognizing One's Own Incompetence Lead to Inflated Self-Assessments". Journal of Personality and Social Psychology 77 (6): 1121–34.
  38. 82, Rosco,
    There is an enormous difference that cannot be explained by a "constant irradiation" model.
    How the heck do you get to this conclusion? Please rephrase it more accurately as follows: There is an enormous difference that cannot be explained by Rosco's horribly insufficient understanding of radiative physics.
  39. Rosco, Enough. Go study, then come back.
  40. Sphaerica a question. 1.4.3 Solar Variability and the Total Solar Irradiance 3rd paragraph "Between 1902 and 1957, Charles Abbot and a number of other scientists around the globe made thousands of measurements of TSI from mountain sites. Values ranged from 1,322 to 1,465 W m–2, which encompasses the current estimate of 1,365 W m–2. Foukal et al. (1977) deduced from Abbot’s daily observations that higher values of TSI were associated with more solar faculae (e.g., Abbot, 1910)." How did they measure this ?

    [DB] "How did they measure this ?"

    Please read the links given in the IPCC chapter you refer to.  And at least try to be on-topic here.  You were previously given advice on using the Search function to find better threads for this type of discussion and advice to read first and then ask questions: try to follow that sage advice. 

    Further off-topic comments will be deleted.

  41. Radiative Equilibrium Energy Conservation Principle at the top of the atmosphere (TOA) Incoming solar energy absorbed = Outgoing infrared energy emitted S (1 – r) x π ae2 = σTe4 x 4 π ae2 σ = Stefan-Boltzmann constant S = Solar constant (total solar energy/time/area at TOA) r = Global albedo ≅ 0.3 ae = Radius of the Earth Te = Equilibrium temperature of the Earth-atmosphere-system Leading to Te = ∜(S(1-r)/4σ) But if we restate the “equality” Incoming solar energy absorbed = Outgoing infrared energy emitted S (1 – r) x π ae2 = Ro x 4 π ae2 (where Ro = σTe4 is the outgoing radiation) we see that this simply says that the outgoing radiation is a quarter of the incoming radiation. Alternatively, taking the original “equality” and dividing both sides by π ae2 you arrive at the interesting postulation that :- S (1 – r) = σTe4 x 4 and thus proving the Stefan-Boltzman equation is wrong. So what has gone wrong here? Clearly the original statement is not an equality !
  42. 91, Rosco, This goes back to the problem you were having with the geometry. The energy absorbed is proportional to S(1-r) over the area of a disk. The energy emitted is the same, but it is distributed over 4 times the area because the earth is a sphere and not a flat disk. Perhaps this discussion will make it more clear, but really it comes down to two simple statements: The earth absorbs energy as a disk (πr2). The earth emits energy as a sphere (4πr2). The temperature of the earth reflects this, which is why we divide 1368 by 4 to get 342, which in turn is cut to 240 by the earth's albedo (1-r).
  43. Rosco @91: The equation you misquote applies only for surfaces with no variation in temperature, and with an emissivity of 1. Neither is true of the Earth. In particular, the effective emissivity of the Surface of the Earth significantly less than 1 due to the effects of the Greenhouse gases in the atmosphere. In contrast, the effective brightness temperature of the Earth as observed from space has, by definition, an emissivity of 1. Therefore, for planets with greenhouse gases, this equation only applies at the Top of the Atmosphere. Further, the Earth does not have equal temperatures across the globe, but it has near equal temperatures - sufficiently close that this equation represents a fair approximation. So, let's check out the approximation at the TOA for the Earth. According to NASA, the Black Body temperature of the Earth as observed from space (aka, the brightness temperature) is 254.3, and the radius of the Earth is 6378100 meters. Therefore the total energy radiated to space by the Earth is approximately: 5.6704*10^-8 * 254.3^4 * 4 * pi * 6378100^2 = 1.2122 * 10^17 Watts The TSI equals 1366 Watts/m^2, and the Bond albedo equals 0.306 so the total energy absorbed by the Earth is approximately: 1366 * (1 - 0.0306) * pi * 6378100^2 = 1.2116 *10^17 Watts. That is just a 0.06% difference. Not bad for an approximation. Of course, I understand that you have been so blinded by your need for AGW to be wrong that you will not be able to acknowledge this clear result. Consequently, I suggest you do as climate scientists do and ignore the approximation. Instead, work out the total energy absorbed from the sun using in each cell of a 5 degree latitude by 5 degree longitude cell of the Earth's surface over a one year period. Remember to compensate for the different areas of each cell depending on their latitude. Sum the total energy absorbed. Then determine the total energy emitted from each cell over the year based on the variation of temperature and surface type (which effects emissivity), and sum the results. If you are correct, the sums should be approximately equal. But in fact they will not be, for as has been shown in the General Circulation Models the equality is at the top of the atmosphere, not the surface. IF you are going to dispute the climate scientists who have gone to the effort of just that sort of detailed calculation (and much more), you should yourself make that same level of detailed calculation instead of disputing an approximation used for teaching purposes (as you have been doing).
  44. But isn't the area of the atmosphere radiating ~240 W/sq m actually at ~5 km above the Earth's surface?
  45. Rosco @94, the level of the atmosphere with a temperature of 255 degrees K is on average approximately 5 km, and if the GHG in the atmosphere absorbed and emitted equally at all wavelengths, it would radiate approximately 240 W/m^2 at that level. But in fact the GHG's only radiate at a limited number of wavelengths, so much of the radiation comes from the much warmer surface. Consequently the it is only when you get to the TOA that the sum of energies across all wavelengths of outgoing radiation sums to ~240 W/m^2.
  46. I always had a problem with the factor of 4 thingie being used to reduce insolation. I have actually learnt some things from this discussion. And, whilst I am a sceptic I wasn't always and I really do not have an agenda. The equation always caused me to question. Whilst I could appreciate that it seemed right by all the science I had ever read it still seemed at odds with calculating the effective temperature of earth - heck I know I never bought it. But if I look at as a radiative balance statement- that to maintain radiative balance for earth the OLR over the surface of a sphere only has to be one quarter of the insolation over the disk - well heck that makes some sense - although why we need to simplify insolation to a constant irradiated disk when we have computer programs which ought to deal with the complex equations seems unnecessary. So we have some common ground - there is an area where the insolation is balanced by the OLR and the temperature of this "interface" - for lack of a better word at the moment - is ~255 K or -18 C. I still think all of this discussion has been relevant and there is still some way to go. I still have doubts about the relevance to the surface temperature but that can wait till another day.
  47. 96, Rosco,
    ...although why we need to simplify insolation to a constant irradiated disk when we have computer programs which ought to deal with the complex equations seems unnecessary.
    This is exactly true, in that many computer models, which run on super computers and take months to churn through a hundred years of simulation, do in fact do things at that level. The Stefan-Boltzman, disk/area approximation is purely a very simple model for 50,000 foot arguments (meaning, for introductory teaching, trivial blog discussions and explanations like this one). There's an important lesson to be learned there. Everything that is discussed is a simplification of what is really going on (to make it manageable by human minds, or in discussions). In particular, when people get around to "visualizing" the actual interactions of the greenhouse effect they get it all wrong. They try to picture simple "back radiation" and other things that are useful gross oversimplifications in some situations, but are really totally and completely wrong. When you find something that doesn't look right, dig deeper! The answer lies in the as yet unknown complexity, not in the idea that thousands of scientists over hundreds of years have made some obvious, stupid blunder.
    I still have doubts about the relevance to the surface temperature but that can wait till another day.
    That's fine. Keep learning. But really... all of this discussion belongs on another thread. In the future, please be more attentive to the appropriate discussion for the thread in question. Feel free to find a better thread, and leave behind a comment explaining where you've posted your question (although many of us will find it anyway by looking at the "recent comments" page of this site). I'm glad you've come to some terms with the factor of 4 aspect of the equations and can now move ahead.
  48. Aye, Bob--good grief, when I first came to the climate situation a few years ago, I knew that complaints about the use of "greenhouse" were misplaced, because it was actually stratospheric CO2 radiating heat to the surface. Sigh . . . well, live and learn, and whenever an absolute pops into the head, beat it with a hammer until it breaks down into mere probability.
  49. I've read every comment here. Rosco could be one of three things: an idiot savant, an agent provocateur, or a fifth-columnist. Whichever way, it's resulted in an interesting exposition of the basics of climate science. Please can anyone help? The CO2 forcing equation F=5.35ln(C/Co); how is the parameter 5.35 calculated? How would I calculate the appropriate parameter for Venus with its higher atmospheric pressure, partial pressure, and depth of atmosphere? Myhre et al quoted in the article is behind a paywall, and an hour of searching for radiative transfer models hasn't yielded results. (
  50. Mike, The equation you present simply says that temperatures will increase 5.35˚F for every doubling of CO2. It's a restatement of the equation in Celsius:
    F = 3.0ln(C/C0)
    From this perhaps it is easy for you to see that this equation simply assumes a climate sensitivity of 3˚C per doubling of CO2, something I'm sure you've seen before. Search for "climate sensitivity" to see the various ways in which that is estimated.
    Response: [Sph] Ignore this. It was wrong.

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