Monday, 16 March 2015

What is the sensitivity?

A new comment on the post "Currygate and wikipedia" is waiting for your approval

Author : PlanetaryPhysicsGroup (but really, its Doug Cotton again)

<b>What is the sensitivity for each 1% of the most prolific "greenhouse gas" (namely water vapor) in Earth's atmosphere?</b>

To help any of you answer the question, here are some facts:

<b>Fact 1:</b>  Water vapor absorbs a significant amount of incident solar radiation as shown <a href=";biw=1920&amp;bih=989&amp;tbm=isch&amp;tbo=u&amp;source=univ&amp;sa=X&amp;ei=Yv8FVYXhAYvo8AXsyoLgAg&amp;ved=0CDEQ7Ak#imgdii=_&amp;" rel="nofollow">here</a>. The atmosphere absorbs about 20% of incident solar radiation and that absorbing is not by nitrogen, oxygen or argon. (Carbon dioxide also absorbs incident photons in the 2.1 micron range which each have about 5 times the energy of 10 micron photons coming up from the surface.  On Venus over 97% of the energy from incident solar radiation is retained in carbon dioxide molecules.)

<b>Fact 2:</b>  The concentration of water vapor varies between about 1% and 4%.  (The concentration of carbon dioxide above Mauna Loa is 0.04% and, as <a href="" rel="nofollow">this</a> graph shows, temperatures there have not increased since 1959.)

<b>Fact 3:</b>  The IPCC claims that water vapor does nearly all of "33 degrees of warming" of Earth's surface.  It must do most of it because it dominates CO2 in concentration and also in the number of frequency bands in which it absorbs and radiates.  But in fact water vapor lowers the "lapse rate" so that the temperature profile rotates downwards at the surface end, making the surface cooler.  (In fact, as per my <a href="" rel="nofollow">paper</a>, there is no 33 degrees of warming being done by any back radiation because it is gravity which props up the surface end of the temperature profile.)

When you have answered the question, work out how much hotter the IPCC conjecture implies a region with 4% water vapor would be than a similar region with 1% water vapor at a similar altitude and latitude. Then look up the study in the Appendix of my <a href="" rel="nofollow">paper</a> and see what real world data tells us about how water vapor cools rather than warms.  And if you don't believe my study, then spend half a day doing your own.

Finally, note that it is quite clear in the energy diagram <a href="" rel="nofollow">here</a> and the text I wrote beneath it that they have certainly added 324W/m^2 of back radiation to 168W/m^2 of solar radiation in order to use this in Stefan Boltzmann calculations to determine the temperature of the surface.  Obviously they worked out by difference what the back radiation figure had to be and made it 66% greater than the 195W/m^2 of upward radiation from the atmosphere to space.  They need not have bothered, because their whole paradigm is wrong, because they ignored the fact that the Second Law of Thermodynamics tells us that gravity forms the temperature and density gradients - which represent the state of thermodynamic equilibrium.

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