LikeLike

]]>Still, the explanation of how it really works is quite simple and the equations quite simple too so I was always biased towards that as Feynman says we should be !

Thanks for putting it all down properly for all of us. I am trying to come up with some ideas for relatively simple experiments which could be used to assist students in understanding the truth here. Have you got any thoughts on experiments ?

LikeLike

]]>Good blog and website!

LikeLike

]]>LikeLike

]]>You don’t define how areas are categorized into either category, but let’s take three end-member examples.

[snip. Yes, the study in the Appendix explains the methodology and the selection process for the 15 locations. Your cherry-picked “three” locations and your complete disregard for appropriate methodology, lack of consideration of the angle of the Sun and lack of adjustments for altitude or proximity of large water masses render your data as being far from an appropriate counter study. To you and others, do NOT submit comments here without spending at least an hour reading the paper and the study in the Appendix thereof.]

LikeLike

]]>“The Stefan-Boltzmann Law in physics confirms that the solar radiation impinging on Earth’s surface (168 watts per square meter according to NASA) is nowhere near sufficient to explain the mean surface temperature. In fact, the flux is like that from an iceberg at -40°C.”

[Yes my statement (which you quoted) is correct. See NASA energy diagrams here: http://whyitsnotco2.com/PSI.html showing the 168 figure and (below the NASA diagram) the Stefan-Boltzmann calculator showing 168W/m^2 flux and the resulting temperature of 233K which is -40°C. Please read that whole page.]

LikeLike

]]>When you stop looking at an atmosphere as having “A temperature” and realise that the temperature is a distribution of individual molecule temperatures ranging from absolute zero to some thousands of degrees the rest becomes obvious. Because rising air must convert KE to PE when rising it MUST cool. Yes there is no physical container but there is in fact a gravitational container.

So there must be a temperature gradient from bottom to top. [snip – the rest of the comment was incoherent and would be confusing for silent readers.]

LikeLike

]]>Latest book and documentary.

‘The Deliberate Corruption of Climate Science’.

LikeLike

]]>Hence, don’t start out with the assumption I’m wrong, because that is not the case. Instead, try to understand the physics I have correctly explained using Kinetic Theory. Gravity acts on molecules and forms both a density gradient and a temperature gradient, as seen in every planetary troposphere. The pressure gradient is a mere corollary. As I said, in an “ideal” column of a planet’s troposphere in the state of maximum entropy, we know that, for an imaginary horizontal plane (with thickness, say, 1 nanometer or less) the pressure from above and below is equal, so we also know that the numbers of molecules crossing each way is equal (the density) and so the temperature (mean kinetic energy at the time of crossing) must also be equal. Hence, because of the acceleration due to gravity, those crossing from above must have had less kinetic energy just after their previous collision, and so the mean kinetic energy at that higher level must have been less – that is, the temperature was cooler there when we consider the macro scale. If I were wrong, then no vortex cooling tube would work, and Earth would be frozen solid because radiation from a colder atmosphere cannot heat the warmer surface, and nor can a mean of 168W/m^2 of solar radiation reaching that surface account for a mean temperature above the black body temperature of 233K which is about -40°C.

As a footnote, let me add that we don’t always have to differentiate some expression for entropy to determine its potential maximum. We just have to understand that it is maximized when all unbalanced energy potentials have dissipated, and those potentials must also consider gravitational potential energy. The latter is ignored in deriving the Clausius corollary of the Second Law, which thus only applies in a horizontal plane. If we were to have isothermal conditions in, say, a perfectly insulated, sealed tall cylinder of non-radiating argon, then there would be unbalanced energy potentials due to the extra gravitational potential energy at the top, and so entropy could and would increase until the mean sum of molecular (kinetic energy + gravitational potential energy) was homogeneous. Then we would have a temperature gradient with d(PE) = -d(KE) and so m.g.dH = – m.cp.dT and thus the gradient is dT/dH = -g/cp. I suggest you also read this guy’s website and the pages there on the Second Law: http://entropylaw.com and perhaps some of the cited references.

LikeLike

]]>You are correct when you say: A gas in thermodynamic equilibrium has as many molecules crossing an arbitrary horizontal plane upward and downward. However, the pressure of a gas in a gravitational field varies the weight of the gas above, creating a pressure gradient dP/dz. According to the ideal gas law (density = rho = n/V = RT/P), the density gradient in the gas (drho/dz), depends on both dP/dz and dT/dz. The latter is a controversial subject. (:)) The flux in either direction depend on both density and velocity, but the downward velocity is increased by the acceleration of gravity between collisions while the upward velocity is decreased. To get the physics right, one needs to deal with all of these complications using differential calculus, not handwaving approximations.

LikeLike

]]>