Quiz: Do you understand the greenhouse effect?

Posted on 11 March 2016 by Scottish-Sceptic

There’s also a fun quiz about well known sceptics & similar subjects here: Quiz: Just for fun

[WATU 2]

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25 Responses to Quiz: Do you understand the greenhouse effect?

  1. Scottish-Sceptic says:
    11 March 2016 at 13:56

    Thank you to everyone whose answered the questions
    I would particularly thank those who answered early on as it’s enabled me to improve the wording with a general improvement in score.
    It’s also given me a very good idea of which concepts are most difficult to understand – or indeed which need very clear wording to be understood as intended.

  2. Hugs says:
    11 March 2016 at 16:56

    I got Error occured

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  4. Scottish Sceptic says:
    11 March 2016 at 18:45

    Ooops, it appears I was the only person able to do the survey!
    That should be corrected now.

  5. Bernard Lodge says:
    11 March 2016 at 20:25

    When CO2 emits IR, it does so in all directions, not just down wards to the surface but also upwards into space. Thus, increasing CO2 in the atmosphere will cause more IR to be radiated to space so doesn’t that mean the earth must cool? How can anyone argue with that simple statement? There can be no net warming effect from increasing atmospheric CO2, it must be a net cooling effect.
    Also, low cloud warms the surface at night mostly because the falling temperature causes condensation of water vapor which releases latent heat. High cloud has a much lower warming effect because most of the ‘latent heat’ IR is absorbed and re-radiated to space before it reaches the surface.

  6. Scottish-Sceptic says:
    11 March 2016 at 21:15

    Thanks very much for asking the question. You’re on the right path, but you just need to consider the temperature as well and you will be there.
    The way I try to explain it is to consider a small disk (the CO2 molecule) that is placed in front of a larger disk (the planet). Yes, if you then put two small disks in front of the larger one, then twice as much IR will be received from the two disks – so two disks cause more emission of IR as you suggest. However, in addition they block radiation from the larger disk and the difference in radiation will be the difference between the radiation that was received from the large disk and that now received from the additional new small disk.
    If the small disk is warmer – then we are changing a small part of the area seen of the large disk, by the small warmer disk. Thus as you correctly say, the total radiation increases. However, if the small disk is at a lower temperature, then we have replaced a hotter area with a colder one so that the radiation falls.
    So, what is critical is not only the amount of greenhouse gas, but also the temperature at the region from where it emits. This is something that is usually left out of any explanation – mainly because most of the atmosphere is cooler – so in most of the atmosphere, IR interactive molecules will be cooler than the surface and so reduce the amount of outgoing radiation.

  8. Will Janoschka says:
    12 March 2016 at 09:39

    Hi Mike, try again in parts;
    Where did the questions and multiple guesses originate? I got the same as Hugs plus questions of my email address that were suspicious. The lumping of CO2 and H2O show extreme bias toward the Climatastrophy of all global circulation models that we now know are all complete disasters themselves. How about some real questions?
    1. What is the physical definition of “Greenhouse Effect”?
    2. What evidence is there to demonstrate this fantasy of “Greenhouse Effect”?
    3. What is the physical definition of atmosphere?
    4. What is the the claimed mass and volume of Earth’s atmosphere?
    5. How have such claims of #4 been physically verified?
    6. How would #4 change if all Earth’s temperatures were lower by 20 degrees Celsius?
    7. How does #6 compare to measurements of Venus, Earth, and Mars?
    The Earth’s gravity produces a compressive force upon the atmospheric mass proportional to the 6×10^24 Kg rocky mass. For Earth’s atmosphere at 1 km altitude; the lower local 1km atmospheric mass adds to the compressive force on the higher atmosphere. However that force/kg is 4×10^7 times that for each kg of the rocky planet as the lower atmosphere is 6300 times closer to the above atmosphere than is the COM of earth. Is this significant or not? Just what is the magnitude of Earth’s atmospheric mass? Just how is such verified? Can atmospheric accelerations under known pressure differentials provide some clue? It appears that the whole scam of meteorology has little if any knowledge or understanding of this Earth’s atmosphere.
    All the best! -will-

  9. Will Janoschka says:
    12 March 2016 at 09:49

    The claimed mass of Earth 6×10^24 Kg while claimed mass of atmosphere 5×10^18 kg via the meteorologic fantasy hydrostatic equilibrium equation, stolen without understanding from Navier-Stokes equation:
    Partial(P)/partial(h); = -density x g. where:
    P is pressure in Pascalls,
    h is altitude in meters,
    density in kg/m^3
    g is gravitational force in Newtons/kg!
    The Climatastrophy folk with no understanding of normalization also claim:
    https://www.grc.nasa.gov/www/k-12/airplane/pressure.html
    gas pressure = force/area; which is nonsense as pressure would go to infinity as area goes to zero. Gas pressure is a Laplacian residue and remains constant.
    Properly: vector force is proportional to scalar pressure normal to integrated over some area.
    All the best! -will-

  10. Scottish-Sceptic says:
    12 March 2016 at 10:40

    Will, thanks. The aim of the questions were partly to check whether at least some people out there had the same understanding of the greenhouse effect – and the answer is apparently they do, but also to try to understand which concepts are most difficult to understand.
    One of the key points about the greenhouse effect is that it has two components:
    1. IR interactive surfaces (like layers with “greenhouse” gases)
    2. The temperature of that layer.
    You keep going on at quite some length about the various mechanisms that set the temperature of the various layers. And as you rightly infer, atmospheric pressure and various thermal effects are involved. But you are missing the key ingredient in your own model of how the incoming solar is matched with the outgoing IR.
    And, even you would have to admit, that if the earth were allowed to cool to near absolute zero (say just enough for some ‘gas’) in the coolness of space, then without any incoming radiation, there could not be a greenhouse effect.
    Any may I draw your attention to question 10:
    10. If the pressure of the atmosphere doubled what would happen to surface temperature? (Note this question is about the total greenhouse effect? This is around 33C which is the total increased warming that is caused by the atmosphere over what it would be without an atmosphere)
    And the correct answer is “massive increase”
    Do you agree with that?

  11. Will Janoschka says:
    12 March 2016 at 10:59

    “So, what is critical is not only the amount of greenhouse gas, but also the temperature at the region from where it emits. This is something that is usually left out of any explanation – mainly because most of the atmosphere is cooler – so in most of the atmosphere, IR interactive molecules will be cooler than the surface and so reduce the amount of outgoing radiation.”
    Sort of, There is no evidence that a lower temperature radiative atmospheric volume with optical depth, need have lower total radiative exitance to space than higher temperature surface with low emissivity at angles far from normal! ‘For example’ the whole ocean, and anything covered with ice or snow! …. For CO2 especially, at any co2 level above 100ppmv the 15 micron optical depth is 2 meters at the surface and 900 meters at the tropopause. This means that at the surface all surface exitance is inhibited by the opposing CO2 radiance within 8 meters of the surface in any direction. While at the tropopause the temperature remains at 220 Kelvin for 5 Km in any direction.
    From that altitude and temperature exitance is outward into 2 PI steradians,the whole outward hemisphere. Twice that even possible from a surface. As CO2 concentration increases above 100ppmv the geometry of this radiative situation changes not in the slightest. Unless CO2 has some atmospheric property other than radiative emissivity in the 15 micron band increasing CO2 levels cannot affect the atmosphere, including the radiative effects..
    It appears that the whole scam of meteorology has little if any knowledge or understanding of this Earth’s atmosphere. The meteorological scammers have been scammed by the AlGoreistas, in fine fashion, using the exact same techniques.
    All the best! -will-

  12. Scottish-Sceptic says:
    12 March 2016 at 11:23

    Will, you highlight the very issue that makes the normal “heat trapping” model so ridiculous. That is, you start looking at the effect from the ground where as you say IR travels almost negligible distance. That is why you need to consider not what IR is trapped, but the average height at which it is irradiated from the atmosphere.
    The best analogy I have is looking into a wood in summer and winter. There’s a couple of good pictures here:
    Greenhouse warming
    As you will see, the change in the foliage, has negligible effect on whether we can see through the wood to the other side. So, as you rightly say adding CO2 doesn’t fundamentally affect the fact that IR doesn’t travel far.
    However, if instead of taking our viewpoint as being the planet surface, or indeed within the atmosphere, and instead we view the planet from outside (or using the analogy – view the wood from outside), it is immediately clear that a small change in density of trees/gas, has a significant effect on how far we can see into the wood/atmosphere.
    This is as you say, a small different at the surface, of perhaps 8-10m because the atmosphere is dense at the surface. But, because the atmosphere is much thinner at the level from which IR finally escapes to space, the mean distance between hitting CO2 molecules is larger, so there is a much higher change in relative distance. At around the average radiation height, the change in height at which of the last molecule to emit IR into space, for a doubling of CO2 is around 100-200m.
    That doesn’t sound much … it’s your same 8-10m change but scaled to the expanded atmosphere at height … and it isn’t much at all compared to the effects you are concerned with which is the total pressure of the atmosphere. So, in no way does it conflict with your own ideas about atmospheric pressure.
    However, if you do the calculations, with a 6.5C/km lapse that 100-200m is equivalent to the 0.5 or 1C of warming. (the higher one is IPCC, the lower figure is the one that one seems to get with more recent HITRAN data).
    So, there is actually no conflict between your own statements on atmospheric pressure and how that causes warming, and the small perturbation to this “effective radiation height” and thus small perturbation to total greenhouse effect caused by CO2.

  13. Will Janoschka says:
    12 March 2016 at 12:08

    “But you are missing the key ingredient in your own model of how the incoming solar is matched with the outgoing IR.”
    I don’t have a model, nor want one! I admit I am not that good at understanding Navier-Stokes equation, not even enough to guess at an appropriate form to use. It appears that no one else has done so either! I have made measurements of the generation and transmission of thermal EMR in and about this atmosphere. Likely overall temperature and emissivity will spontaneously adjust until all forms of power to the planet equilibrate with all forms of power exiting the planet. I would never call such a greenhouse anything.
    “10. If the pressure of the atmosphere doubled what would happen to surface temperature? (Note this question is about the total greenhouse effect? This is around 33C which is the total increased warming that is caused by the atmosphere over what it would be without an atmosphere). And the correct answer is “massive increase””
    Possible theoretical increase in surface temperature for a hypothetical question. By your ‘warming’ do you mean increase in temperature, or actually adding thermal energy independent of your work of doubling?
    “Do you agree with that?”
    No.
    If the rocky mass stays at 6×10^24kg, doubling the ‘same’ atmosphere would double the surface pressure and surface density but increase the atmospheric volume (height) and total temperature lapse somewhat. Adding sufficient SF6 to the atmosphere to double surface pressure may actually decrease atmospheric volume somewhat. I do not know.
    If the mass of the rocky planet doubles cooling is likely.
    Question: How ya-all gonna keep the added atmosphere ‘heer’ and/or where ya-all gonna get the extra rocks.
    All the best! -will-

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  15. Scottish-Sceptic says:
    12 March 2016 at 12:46

    OK, if you don’t the like idea of doubling, will you accept that halving the atmosphere would decrease the greenhouse effect massively?
    Otherwise, you don’t seem to be starting from any actual scientific basis at all, and unless you can agree on at least one physical principle that is based in science, it is going to be really difficult to have a conversation.

  16. Will Janoschka says:
    12 March 2016 at 13:15

    “However, if you do the calculations, with a 6.5C/km lapse that 100-200m is equivalent to the 0.5 or 1C of warming. (the higher one is IPCC, the lower figure is the one that one seems to get with more recent HITRAN data).”
    There is no effective radiating altitude. As per the Miskolczi calculation from measurement, there is always a continuing accumulation of radiative exit flux from surface all the way to 200km at every location. As far as some average temperature lapse, tis another red herring from either a latitude or time of day point of view. Where the atmosphere is rapidly ascending, surface pressure is low, lapse is low, tropopause is higher. Where the atmosphere is rapidly descending, surface pressure is high, lapse is high, tropopause is lower. all the way from equator to poles. most of this are the requirements of continuum mechanics of this atmosphere in motion. We don need no steenkin equilibrium heer!!
    Just past noon the whole atmosphere has its greatest volume, greatest insolation, greatest temperature, and greatest exit flux. The minimums are just prior to sunrise. This atmosphere refuses to get stuffed unto some average meteorological value, that has no meaning! Please give the lat, long, time, for average weather/climate.
    All the best! -will-

  17. Scottish-Sceptic says:
    12 March 2016 at 13:34

    This is a reply to your last post
    The effective altitude is just the average altitude of the molecules that radiate IR into space.
    You are telling me what you don’t accept, but most of them are irrelevant to what I’m saying, it is very difficult to know what you will accept.
    So, can I just check a few details so I know what you actual:
    1. Do you accept there is incoming solar energy
    2. Do you accept that if incoming solar energy not matched by outgoing there will be heating
    3. Do you accept the earth loses heat via IR
    4. Do you accept there is a lapse rate
    5. Do you accept the lapse rate is proportional to gravity over heat capacity (so 9.8C for dry air and less for moist so 6.5C)
    6. Do you accept that CO2 emits IR radiation
    7. Do you accept that CO2 IR radiation increases with temprature?
    8. Do you accept that given the above, that IR radiation will decrease if the CO2 from which it originates is higher in the atmosphere (where it is colder).
    And, as I’ve got this far, I may as well finish it off …
    a) Do you accept that an increase in CO2 in the atmosphere, means there is more higher uP
    b) Do you therefore accept (using the wood analogy) – and your own statement about a 8-10m distance that CO2 travels) that it will be the highest CO2 molecules from which IR escapes into space.
    c) Do you accept that with increasing CO2, because there are more at higher altitude, that the average height of the CO2 from which IR escapes will increase
    d) Do you accept that that CO2 will be cooler, and therefore emit less IR?
    Thus do you accept that the concentration of CO2, makes a marginal change in the average height of IR emissions and that because that because of the lapse rate the temperature of that CO2 will be lower so emitting less IR and thus tending toward warming the planet.
    Or to put it another way … if the average radiation height increases by about 100m, the average temperature from the surface to that average radiation height, given a lapse rate of 6.5K/m will be 0.65K

  18. Will Janoschka says:
    12 March 2016 at 13:58

    There is no physical ‘greenhouse effect’! Why not admit that you to have been scammed by professionals. Meteorology comes not within a barge pole length of any science. You will have lots of company eventually!
    “Otherwise, you don’t seem to be starting from any actual scientific basis at all, and unless you can agree on at least one physical principle that is based in science, it is going to be really difficult to have a conversation.”
    Fine! please demonstrate your knowledge or guess at physical meaning of even one term in any form of the Navier-Stokes partial differential equations describing compressible fluid motion. They are a true BITCH, and likely still incomplete, and certainly unprovable for any atmosphere. They are the very best available from a scientific, engineering, chemical, mathematical standpoint. Nothing else comes close. Cal tech (JPL) is having some success!
    It seems your form of science is but some sort of policy statement prepared for high ranking government officials leading to the “new world order”!
    All the best! -will-

  19. Bergstrom says:
    12 March 2016 at 15:32

    I also completed the test, but then there were a page asking me to connect with my e-mail address and the result got lost after that.
    VB

  20. Scottish-Sceptic says:
    12 March 2016 at 16:56

    Thanks for telling me that. I had to install the quiz modular specially so I’m new to it and so every bit of feedback is useful.

  21. Will Janoschka says:
    12 March 2016 at 18:04

    Scottish-Sceptic says:
    12th March 2016 at 1:34 pm
    “1. Do you accept there is incoming solar energy”
    Yes! no one knows the absorptance of such flux by atmosphere or surface.
    “2. Do you accept that if incoming solar energy not matched by outgoing there will be heating.
    Mabe! If the much lesser ‘absorbed flux’ is not converted to other forms of accumulated power (energy), mechanical, chemical, electrical, not involving sensible heat (temperature of mass).
    “3. Do you accept the earth loses heat via IR”
    Yes! depends on your meaning of heat! from both the Sun and Earth thermal EMR dispatch to space is mostly of waste energy (entropy) from the lowest temperature region. Other higher temperature energy is still being converted to useful “work” such an ‘hurricanes’.
    “4. Do you accept there is a lapse rate”:
    Yes! Three measurable atmospheric lapse; pressure, density, temperature. The gravitationally maintained replacement for atmospheric weight or atmospheric downward acceleration via the ‘force of gravity’. Atmosphere like orbiting satellites is ‘weightless’, including all airborne bugs, bees, birds, aircraft, aerosols; including; smoke, fly ash, volcanic ash, and the H20 related smog, solid and liquid water and the two airborne water colloids! Do not get me started!
    “5. Do you accept the lapse rate is proportional to gravity over heat capacity (so 9.8C for dry air and less for moist so 6.5C)”
    Never ever ever! with this atmospheric mass mixture of N2 and O2, and gravitational compressive force, dry thermal lapse is slightly greater than -10 degree Celsius/km. That meteorological scam is only because the Cp specific heat of this dry atmosphere is so very very close to 1.00000000 J/(kg x degree Celsius) The lesser water induced rate is not any sort of gravitational lapse it is the balance ‘tween conversion of H2O WV ‘latent heat of evaporation’ back to sensible heat (temperature) and the rate that local atmosphere can dispatch such to elsewhere/when by any means permissible. The thermal lapse in those monster equatorial columnobullshit clouds may well be zero. No one that survived such a a ride have any notes, or remembrances, except “I want my mommy”! You have been scammed again by professionals!
    “6. Do you accept that CO2 emits IR radiation”
    I accept that CO2, as with any matter, not only can, but ‘must’ thermally emit in every direction of lower radiance proportional to the difference in opposing radiance times the emissivity of that matter at that wavelength in that direction.
    “7. Do you accept that CO2 IR radiation increases with temprature?”
    I accept that such can increase but only if the surround radiance temperature increases in a lesser amount.
    “8. Do you accept that given the above, that IR radiation will decrease if the CO2 from which it originates is higher in the atmosphere (where it is colder).’
    To damned many ifs! Have you any physical evidence that CO2 exitance to space originates at any higher (colder) atmospheric location than the tropopause?
    “And, as I’ve got this far, I may as well finish it off …”
    OK! perhaps later, I need sleep and sober
    All the best! -will-

  22. James McGinn says:
    12 March 2016 at 22:54

    Will (to Mike):
    Please demonstrate your knowledge or guess at physical meaning of even one term in any form of the Navier-Stokes partial differential equations describing compressible fluid motion. They are a true BITCH, and likely still incomplete, and certainly unprovable for any atmosphere. They are the very best available from a scientific, engineering, chemical, mathematical standpoint. Nothing else comes close. Cal tech (JPL) is having some success!
    Paul Aubrin:
    – hide quoted text –
    Source: https://groups.google.com/d/msg/alt.global-warming/PwjztcEeC8A/N4DzJIbCBwAJ
    The Clay Mathematics Institute will give a one million dollars prize for anyone who will /significantly/ improve the understanding of the equations which rule the weather and the climate.
    http://www.claymath.org/millennium-problems/navier%E2%80%93stokes-equation
    “Although these equations were written down in the 19th Century, our
    understanding of them remains minimal. The challenge is to make substantial progress toward a mathematical theory which will unlock the secrets hidden in the Navier-Stokes equations.”
    Official problem statement:
    http://www.claymath.org/sites/default/files/navierstokes.pdf
    James McGinn;
    According to my own theory of atmospheric flow, there is a significant increase in viscosity that occurs at wind shear boundaries between moist bodies of air and dry bodies of air. This increase in viscosity (think of it as a plasma) is a consequence of H2O surface tension. (We can think of it in accordance with the principle that if you maximize H2O surface area you maximize H2O surface tension. [see * below]) Whether or not this is something that could be included in a Navier-Stokes equation is not something I’m qualified to to speculate on. However, I do believe I can explain the mechanism that underlies the significant changes in viscosity associated with H2O in our atmosphere:
    BREAKTHROUGH: Hydrogen Bonding as The Mechanism That Neutralizes H2O Polarity
    https://groups.google.com/d/msg/sci.physics/Cin1MQ4ZyFU/QmNEM9mnDgAJ
    https://zenodo.org/record/37224
    * There is an alternate explanation for this notion that maximization of surface area maximized surface tension. It involves a direct analogy to non-Newtonian fluids: Go to Youtube and do a search using the following: How Non-Newtonian Fluids Reveal the Mechanism Underlying Ice

  23. Scottish-Sceptic says:
    13 March 2016 at 00:47

    I was hoping to find a point where your views diverged from my own.

  24. markstoval says:
    13 March 2016 at 18:50

    Well I got a “Professor”, but does that mean I am as dumb as “Dr.” Mikey Mann?

  25. Scottish Sceptic says:
    13 March 2016 at 22:40

    Mickey would never get that high!

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