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Re: Questions ko0ky runs away from really fast

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In article <n919ml$ps1$1@gioia.aioe.org>, kkensington01@gmail.invalid 
says...


> 
> On 04/02/2016 12:16 PM, Friendly Neighborhood Vote Wrangler Emeritus wrote:
> >> 1. On a winter's night, with snow on the ground, is it typically colder
> >>      or warmer when there is cloud cover versus when it is clear?
> >
> > clouds over water, snow or ice will have negligible effect on temperature.
> 
> BZZZT!
> 
> http://littleshop.physics.colostate.edu/activities/atmos1/ColderOnClearThanCloudy.pdf
> 
> You bit my shiny hook at last, ko0ky, and thus exposed your l0on climate 
> change denial, and yourself, as intellectually bankrupt.
> 
> Final victory on the climate change question goes to kensi.
> 
> Thank you for playing, goodbye.
> 
> >> 2. What might NASA have meant by "an active Sun" *other* than sunspot
> >>      activity?
> >
> > There's no mention of "sunspot activity" in the *three* referenced
> 
> Non-answer. What I said was mentioned word-for-word was "an active Sun". 
> You have k'lamed that they meant something different from "sunspot 
> activity" by this. You have been asked to state exactly what.
> 
> Our ruling on question #2 is that you answered with (approximately 200 
> screedy lines of) *crickets*.
> 
> Thus, Question 2 remains unanswered.
> 
> > And you call yourself an astrophysicist? Pshaw, you're not even smart
> > enough to be an astrologer.
> 
> To get angles summing to other than 180 degrees the triangle has to be 
> in a curved manifold. The angle sum will be less the more negative the 
> curvature's manifold, and greater the more positive, being 180 when the 
> manifold is flat. A good example: take the meridians at 0 degrees and 90 
> degrees longitude on Earth, and take the segments of these meridians 
> from north pole to equator. Add in the arc of the equator joining these 
> endpoints by the shortest distance. This is a triangle in Earth's 
> surface as a curved manifold, as the sides are segments of great 
> circles, which are geodesics of that manifold. The interior angles are 
> all 90 degrees, so the angle sum is 270 degrees.
> 
>  From this can be intuited that the relative scale of the triangle and 
> the (possibly negative) radius of curvature of the manifold determines 
> the angle excess/deficit, as making the sphere and triangle smaller by 
> the same factor preserves the angle sum of 270. (Earth is actually 
> (approximately) an oblate spheroid, which doesn't alter the angles, 
> though it makes the triangle a not-quite-equilateral isosceles 
> triangle.) Make only the triangle smaller and the angle excess shrinks 
> toward zero, the angle sum tending to 180, and ditto making only the 
> sphere bigger. The exact quantitative relationship is given by the 
> Gauss-Bonnet theorem, which states:
> 
> integral_M K dA + integral_dM k_g ds = 2*pi*chi(M)
> 
> where M is the portion of the manifold enclosed by the triangle (or, 
> actually, any figure) and dA differential units of the enclosed area, K 
> is the Gaussian curvature over M, dM is the triangle itself (or other 
> figure), k_g is the geodesic curvature of this boundary, ds is the 
> differential on the perimeter (so integral_M dA = area A and integral_dM 
> ds = perimeter s), and chi(M) is the Euler characteristic of M (1 for a 
> simply-connected, bounded area with no holes such as our triangle). For 
> a piecewise-geodesic boundary like our triangle, integral_dm k_g ds is 
> the sum of the angles by which the geodesic segments turn at the corners 
> of the boundary, which is pi minus the interior angles. So for the 
> triangle, it's 3*pi minus the sum of the interior angles whose sum with 
> integral_M K dA equals 2*pi*chi(M) = 2*pi, i.e. 3*pi - angle_sum + 
> curvature_integral = 2*pi, or angle_sum - pi = curvature_integral -- the 
> deviation of the angle sum from 180 degrees (but expressed in radians) 
> equals the integral of the curvature of the manifold over the portion 
> enclosed by the triangle. For a manifold of constant curvature (e.g. a 
> sphere) that integral equals the triangle's area times the constant 
> curvature (so, the angle deviation grows as the triangle's area grows, 
> and as the curvature grows); the curvature in question is 1/r^2 where r 
> is the radius of curvature (for a sphere, this is the radius of the 
> sphere; for a manifold of constant negative curvature, it's actually 
> imaginary and of smaller magnitude as the curvature becomes stronger).
> 
> The importance of the foregoing to an astrophysicist is, of course, 
> general relativity. Space itself (and indeed, space-time) is a curved 
> manifold, and in fact this phenomenon affects not only "plane" figures 
> but solids, and thus spherical angles, in curved space(-time). For 
> instance, a tetrahedron's four interior solid angles' sum will change 
> near, say, a black hole or other dense object. (There's no nice tidy 
> constant in steradians here analogous to angle sum = pi radians for a 
> plane triangle; the sum of the solid angles of a tetrahedron depends on 
> the six dihedral angles of the tetrahedron, being twice their sum minus 
> 4*pi. For the regular, Platonic solid tetrahedron with 
> equilaterial-triangle faces, in Euclidean space, it's 6*cos^-1(1/3) - 
> 4*pi, and will exceed this in positively-curved space and be deficient 
> of it in negatively-curved space.)
> 
> > <sn<SMACKAK00K!>
> 
> Since you have now been thoroughly and utterly *destroyed* on every one 
> of your kOoky contentions, our ruling is: Snicker denied.
> 
> *snicker*

Well done, Paul!  You get to join the "I Kicked Fakey's Ass On Usenet" 
club.  Free 'shrooms for everyone!

<snicker>

-- 
Checkmate, AUK DoW #1
AUK Hammer of Thor award, Feb. 2012 (Pre-Burnore)
Destroyer of the AUK Ko0k Vote  (Post-Burnore)
Originator of the "Dance for me" (tm) lame
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Re: Questions ko0ky runs away from really fast "Checkmate, DoW #1" <Lunatic.Fringe@The.Edge> - 2016-02-04 21:30 -0800

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