Climate Science Glossary

Term Lookup

Enter a term in the search box to find its definition.

Settings

Use the controls in the far right panel to increase or decrease the number of terms automatically displayed (or to completely turn that feature off).

Term Lookup

Settings

All IPCC definitions taken from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the Intergovernmental Panel on Climate Change, Annex I, Glossary, pp. 941-954. Cambridge University Press.

Home Arguments Software Resources Comments The Consensus Project Translations About Support

Bluesky Facebook LinkedIn Mastodon MeWe

Twitter YouTube RSS Posts RSS Comments Email Subscribe


Climate's changed before
It's the sun
It's not bad
There is no consensus
It's cooling
Models are unreliable
Temp record is unreliable
Animals and plants can adapt
It hasn't warmed since 1998
Antarctica is gaining ice
View All Arguments...



Username
Password
New? Register here
Forgot your password?

Latest Posts

Archives

The runaway greenhouse effect on Venus

What the science says...

Select a level... Basic Advanced

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

"I bought off on the “runaway greenhouse” idea on Venus for several decades (without smoking pot) and only very recently have come to understand that the theory is beyond absurd." (Steve Goddard, WUWT)

At a glance

Earth: we take its existence for granted. But when one looks at its early evolution, from around 4.56 billion years ago, the fact that we are here at all starts to look miraculous.

Over billions of years, stars are born and then die. Our modern telescopes can observe such processes across the cosmos. So we have a reasonable idea of what happened when our own Solar System was young. It started out as a vast spinning disc of dust with the young Sun at its centre. What happened next?

Readers who look up a lot at night will be familiar with shooting stars. These are small remnants of the early Solar System, drawn towards Earth's surface by our planet's gravitational pull. Billions of years ago, the same thing happened but on an absolutely massive scale. Fledgeling protoplanets attracted more and more matter to themselves. Lots of them collided. Eventually out of all this violent chaos, a few winners emerged, making up the Solar System as we now know it.

The early Solar system was also extremely hot. Even more heat was generated during the near-constant collisions and through the super-abundance of fiercely radioactive isotopes. Protoplanets became so hot that they went through a completely molten stage, during which heavy elements such as iron sank down through gravity, towards the centre. That's how their cores formed. At the same time, the displaced lighter material rose, to form their silicate mantles. This dramatic process, that affected all juvenile rocky planets, is known as planetary differentiation.

Earth and Venus are the two largest rocky planets. But at some point after differentiation and solidification of their magma-oceans, their paths diverged. Earth ended up becoming habitable to life, but Venus turned into a hellscape. What happened?

There's a lot we don't know about Venus. But what we do know is that the surface temperature is hot enough to melt lead, at 477 °C (890 °F). Atmospheric pressure is akin to that found on Earth - but over a kilometre down in the oceans. The orbit of Venus may be closer to the Sun but a lot of the sunlight bathing the planet is reflected by the thick and permanent cloud cover. Several attempts to land probes on the surface have seen the craft expire during descent or only a short while (~2 hours max.) after landing.

Nevertheless, radar has been used to map the features of the planetary surface and analyses have been made of the Venusian atmosphere. The latter is almost all carbon dioxide, with a bit of nitrogen. Sulphuric acid droplets make up the clouds. Many hypotheses have been put forward for the evidently different evolution of Venus, but the critical bit - testing them - requires fieldwork under the most difficult conditions in the inner Solar System.

One leading hypothesis is that early on, Venus experienced a runaway water vapour-based greenhouse effect. Water vapour built up in the atmosphere and temperatures rose and rose until a point was reached where the oceans had evaporated. In the upper atmosphere, the water (H2O) molecules were split by exposure to high-energy ultraviolet light and the light hydrogen component escaped to space.

With that progressive loss of water, most processes that consume CO2 would eventually grind to a halt, unlike on Earth. Carbon dioxide released by volcanic activity would then simply accumulate in the Venusian atmosphere over billions of years, creating the stable but unfriendly conditions we see there today.

Earth instead managed to hang onto its water, to become the benign, life-supporting place where we live. We should be grateful!

Please use this form to provide feedback about this new "At a glance" section. Read a more technical version below or dig deeper via the tabs above!

Further details

Venus may have experienced a runaway greenhouse effect in the geological past. To use the term 'runaway' is to refer to a highly specific process when discussed by planetary scientists. Simply having a very hot, high-CO2 atmosphere is not it. So let's start with a tutorial on Venus at the present day.

Venus’ orbit is much closer to the sun, which means it receives almost twice the solar radiation at the top of its atmosphere than Earth. Venus also has a very high albedo which ends up over-compensating for the closer distance to the sun. The result is that less than 10% of that incident solar radiation reaches the surface. High albedo can be attributed to sulphur-bearing compounds, along with minor water vapour (around 20 ppm). These substances form globally encircling sulphuric acid-dominated cloud decks (fig. 1). Venus’ atmosphere also has a surface pressure of around 92 bars (or if you like, 92,000 millibars), equivalent to what you’d feel on Earth beneath more than a kilometre of ocean.

Venus in its shroud of clouds.

Fig. 1: Venus in its shroud of clouds - a false colour composite created by combining images taken using orange and ultraviolet spectral filters on the Mariner 10 spacecraft's imaging camera.The images used to create this view were acquired in 1974; the RH one has been enhanced to bring out texture and colour. Image: NASA.

Observations of the water vapour content in the Venusian atmosphere show a high heavy to light hydrogen isotopic ratio (D/H). This is best interpreted as the product of a preferential light hydrogen escape to space: deuterium escapes less easily. Venus is considered to have had at least 100 times its current water content in the past (e.g. Selsis et al. 2007 and references therein).

The greenhouse effect on Venus today is primarily caused by CO2, although water vapour and SO2 are important as well. 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 in infra-red (IR) imagery. In reality, the surface of Venus (Fig. 2) is even hotter than the dayside of Mercury, at a deadly 477 °C (890 °F).

Like Earth, the Venusian clouds also generate a greenhouse effect. However, they are poor IR absorbers and emitters compared to water clouds. The sulfuric acid droplets forming the clouds can also scatter IR radiation back to the surface, producing another form of the greenhouse effect in that way. In the dense Venusian CO2-rich atmosphere, there are IR-handling processes at work that are unimportant on modern Earth.

The Soviet Union's Venera 14 probe.

Fig. 2: The Soviet Union's Venera 14 probe captured two colour panoramas of Venus's surface in 1982. This panorama came from the rear camera. Image: Russian Academy of Sciences. More images can be seen at: https://www.planetary.org/articles/every-picture-from-venus-surface-ever

How to get a Runaway?

To get a true runaway greenhouse effect on Venus, you need a combination of solar radiation and the presence of a greenhouse gas. That gas has two key requirements. It must be condensable and it needs to be in equilibrium with its surface reservoir. In addition, its concentration must increase with temperature, as explained by the Clausius-Clapeyron relation. For Venus to enter a runaway greenhouse state, the greenhouse gas of interest is water vapour, plus its liquid reservoir, the water making up the oceans.

The greenhouse effect on any planet involves impeding the flux of outwards longwave radiation to space. Water vapour is very good at this so can potentially lead to a positive feedback runaway scenario. That works as follows: higher temperatures cause ever more water to evaporate and then drive temperatures even higher and on and on it goes - while there is an available liquid water reservoir.

Through water vapour's effectiveness at blocking IR, the outward longwave radiation flux eventually flatlines. If the incoming Solar flux is constantly greater than that outgoing flatline value, the planet is tipped out of radiative equilibrium and we have that runaway. If you like, it has a fever. The reservoir for water vapour - the oceans - is vast. That means the system may only be able to return to radiative equilibrium once the runaway process has stopped. In the extreme runaway greenhouse effect, that cessation may only happen at the point when the whole ocean has evaporated.

On present-day Earth, there is a strong temperature inversion, called the tropopause. It is situated between the troposphere and stratosphere. You can see it on thundery days when the tops of storm-clouds spread out beneath it to form the familiar anvil-shapes. The tropopause thus forms an effective barrier to moisture getting into the stratosphere. At the height of the tropopause on Earth, in any case, it's already too cold for water to remain in the vapour phase. The wispy clouds making up thunderstorm anvils consist of ice crystals. This impediment to water vapour's ascent is often referred to as a 'cold trap'.

In a runaway scenario, such as that proposed for Venus, no such impediment exists. This means the upper atmosphere would have become moist too. On Venus, the troposphere extends to a much greater height than on Earth. There is no stratosphere - we're talking about a very different situation here. That is critical because water vapour, upon reaching such great heights, has energetic Solar ultraviolet (UV) radiation to contend with. The UV is effective at splitting the H2O molecule into its constituent elements. Once that has happened, the hydrogen in particular is easily lost to space. One can envisage that once a runaway greenhouse effect got going, Venus' water content got steadily depleted in this manner through time. If Venus ever had oceans, they must have evaporated into oblivion. Because of the 'cold trap', this form of water-depletion is of very little significance on Earth - thankfully.

Once that water was lost, the chemical processes that lock up carbon in rocks on Earth could not operate. All of them involve water somewhere. Thereafter, every addition of carbon to the atmosphere, large or small, stayed up there. Most CO2 was probably of volcanic origin. The result was the 96.5% CO2 atmosphere and hellish surface temperature of Venus today.

Earth and the Runaway: Past and Future

Currently, Earth is well under the absorbed solar radiation threshold for a runaway greenhouse effect to occur. Its water condenses and is recycled back to the surface as rain, rather than accumulating indefinitely throughout the atmosphere. The opposite is true for CO2, which builds up and up through our emissions, only checked by natural removal processes. Note here that the runaway greenhouse threshold is largely independent of CO2 since the IR opacity is swamped by the water vapour effect. This makes it difficult to justify concerns over a CO2-induced runaway on Earth.

However, this immunity to a runaway greenhouse effect will not last forever. The most realistic scenario for Earth entering a runaway occurs a few billion years in the future, when the sun's brightness has substantially increased. Earth will then receive more sunlight than the outgoing longwave radiation escaping to space. A true runaway greenhouse effect is then able to kick in. Caveats apply, though. For example, greater cloud cover could increase planetary albedo and delay this process.

Interestingly, 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 of its existence. Geothermal heat and the heat flow from the moon-forming impact would have made up for the difference between the net solar insolation and the runaway greenhouse threshold. But if this happened it could only have lasted for a relatively short period of time - since we still have plenty of water on Earth.

For further reading, a recent review paper (Gillmann et al. 2022) explores the various hypotheses concerning the evolution of the Venusian atmosphere over geological time. There's also an excellent book chapter (Arney & Kane. 2020, currently available as a PDF at arXiv). As might be expected, difficulties in fieldwork are plural on Venus and designing a probe that survives touchdown and can go on to do the required data-collecting is still some time away. The key piece of evidence we need to confirm the existence of a runaway greenhouse effect in deep time would be for free water having once been present. But it is apparent that large parts of the surface were covered with lava flows from monster volcanoes at some point. Is that evidence nowhere to be seen, or is it just hiding? Time will tell.

Last updated on 21 January 2024 by John Mason. View Archives

Printable Version  |  Offline PDF Version  |  Link to this page

Argument Feedback

Please use this form to let us know about suggested updates to this rebuttal.

Comments

Prev  1  2  3  4  5  6  7  8  9  Next

Comments 76 to 100 out of 205:

  1. 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.
  2. 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 ?
  3. 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.
  4. 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 ?
  5. 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.
  6. 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.
  7. 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.
  8. 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.
  9. 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.
  10. 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'.
  11. muoncounter - the 51 % was after albedo of ~30% and atmospheric absorbtion of ~19 %. Please read this from the IPCC http://www.ipcc.ch/publications_and_data/ar4/wg1/en/ch1s1-4-3.html 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. http://www.physicalgeography.net/fundamentals/7f.html
  12. 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.
    Response:

    [DB] An open-pdf can be found here:

    http://people.psych.cornell.edu/~dunning/publications/pdf/unskilledandunaware.pdf

  13. 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.
  14. Rosco, Enough. Go study, then come back.
  15. Sphaerica a question. http://www.ipcc.ch/publications_and_data/ar4/wg1/en/ch1s1-4-3.html 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 ?
    Response:

    [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.

  16. http://www.atmos.ucla.edu/~liougst/Lecture/Lecture_3.pdf 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 !
  17. 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).
  18. 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).
  19. But isn't the area of the atmosphere radiating ~240 W/sq m actually at ~5 km above the Earth's surface?
  20. 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.
  21. 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.
  22. 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.
  23. 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.
  24. 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. (shades@iburst.co.za)
  25. 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.

Prev  1  2  3  4  5  6  7  8  9  Next

Post a Comment

Political, off-topic or ad hominem comments will be deleted. Comments Policy...

You need to be logged in to post a comment. Login via the left margin or if you're new, register here.

Link to this page



The Consensus Project Website

THE ESCALATOR

(free to republish)


© Copyright 2024 John Cook
Home | Translations | About Us | Privacy | Contact Us