Ah, from 3.5 yrs ago,
From gosper@alum.mit.edu Mon Jul 16 08:47:39 2007 Date: Mon, 16 Jul 2007 10:39:48 -0400 (EDT) Message-ID: <23173122.9411184596788302.JavaMail.gbourne@brunch.mit.edu> From: Bill Gosper <gosper@alum.mit.edu> To: math-fun@mailman.xmission.com Lemma: The apical solid angle of an isosceles trapezoidal pyramid with vertex angles a,b,a,c is
2 c b c 2 b - sin (-) + (cos(a) - 1) sin(-) sin(-) - sin (-) + cos(a) + 1 2 2 2 2 (d79) 2 acos(-------------------------------------------------------------) b c (cos(a) + 1) cos(-) cos(-) 2 2 I derived this by splitting the general a,b,a,c hedral vertex into trihedrals a,b,d and d,a,c, then maximizing over d, getting b c (d38) cos(d) = cos(a) - 2 sin(-) sin(-) 2 2 Clearer might have been the difference between two isosceles trihedrals: solid_angle(d,d,b)-solid_angle(d-a,d-a,c) with d chosen to equalize the dihedrals, using dihedral(a,b,c):=acos(csc(a)*csc(b)*cos(c)-cot(a)*cot(b)), which gives sin(a) d = atan(----------------------), b c cos(a) - csc(-) sin(-) 2 2 and eventually the same solid angle formula. But the max technique generalizes from trapezoidal pyramids to arbitrary quadrilateral pyramids with apex angles a,b,c,d, whose maximal solid angle is cos(d) + cos(c) + cos(b) + cos(a) a b c d --------------------------------- - tan(-) tan(-) tan(-) tan(-) a b c d 2 2 2 2 4 cos(-) cos(-) cos(-) cos(-) 2 2 2 2 Note we get the old formula when d=0. Note also the symmetry, giving the same result for a,b,d,c. In this maximal case, the "diagonal angles", i.e. the apex angles exposed by splitting into two triangular pyramids, are a d b c b d a c 2 (sin(-) sin(-) + sin(-) sin(-)) (sin(-) sin(-) + sin(-) sin(-)) 2 2 2 2 2 2 2 2 acos(1 - -----------------------------------------------------------------) c d a b sin(-) sin(-) + sin(-) sin(-) 2 2 2 2 and same(a<->c). This is analogous to the maximal area of a quadrilateral with sides a,b,c,d (or a,b,d,c): sqrt((c + b + a - d) (d - c + b + a) (d + c - b + a) (d + c + b - a)) ---------------------------------------------------------------------, 4 with diagonals (a d + b c) (b d + a c) (b d + a c) (c d + a b) sqrt(-----------------------) and sqrt(-----------------------). c d + a b a d + b c Whose formula is this? (Superhero of Alexandria?) And what about arbitrary pentagons, etc? Are the maximal areas fixed w.r.t. permuting the sides? The formulas must be doozies. --rwg IMPORTUNATE PERMUTATION ------=_Part_509_2287320.1184596788300-- On Sun, Feb 6, 2011 at 7:13 PM, Bill Gosper <billgosper@gmail.com> wrote:
David Wilson>On the unit sphere, measuring angles in radians.
Place an arc with measure A on the equator. Place arcs with measures B and C at each end of A running north along meridians (perpendicular to A). Join the northern
endpoints of B and C with a great circle arc. What is the area of the spherical quadrilateral thus formed?
rwg>If you know the angles between those arc at those endpoints to be a, isn't it just the angular excess 2 a + 2 pi - 2 pi = 2a? If you know all four arcs,
max_solid_angle(a,b,c,d):=2*acos((cos(d)+cos(c)+cos(b)+cos(a))/(4*cos(a/2)*cos(b/2)*cos(c/2)*cos(d/2))-tan(a/2)*tan(b/2)*tan(c/2)*tan(d/2)) --rwg