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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. View Archives

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Comments 51 to 75 out of 267:

  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.

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