This looks suspiciously like a quaternion product in drag... (Quaternions are very nice for spherical geometry.) At 11:22 AM 12/24/02 -0800, R. William Gosper wrote:
In my version of Macsyma's Geofuncs package is now (c1) solid_angle(a,b,c) 2 (cos(c) + cos(b) + cos(a) + 1) (d1) acos(-------------------------------------- - 1) (cos(a) + 1) (cos(b) + 1) (cos(c) + 1)
where a, b, and c are the ordinary angles at the trihedral vertex. It seems too nice to be new, but an hour or so of (low bandwidth) Googling failed to turn it up. Here are some playings with it.
A sector with wedge angle v will have solid angle v/(2 pi) 4 pi = 2 v. But we can split it perpendicular to its straight edge to make two trihedrals with angles pi/2, pi/2, and v:
(c2) 2*solid_angle(%pi/2,%pi/2,v)
(d2) 2 v
We can dissect a regular n-gonal pyramid of height h and base circumradius r into n congruent tetrahedra with trihedral components v, v, and 2 asin(sin(v) sin(pi/n)), where v = atan(r/h), i.e., half the apex angle. This gives, after much simplification, two expressions for the solid angle of the apex:
2 %pi 2 %pi 2 %pi 2 cos (---) (2 sin (---) cos(v) + cos(-----)) n n n (d3) n acos(--------------------------------------------- 2 %pi 2 2 %pi sin (---) cos (v) + cos (---) n n
2 %pi - cos(-----)) n
n = acos((- 1) n - 1 /===\ 2 %pi 2 | | %pi %pi k 2 2 cot (---) n | | (cos(v) + cot(---) cot(-----)) n | | n n k = 1 (1 - ------------------------------------------------------)) 2 2 %pi n (cos (v) + cot (---)) n
Letting n -> oo, the solid angle of a cone is 2 pi (1 - cos v), where, again, tan v = (base radius)/height.
Exercise: what is the black/white (pentagons/hexagons) area ratio of a soccer ball (spherical truncated icosahedron)?
Thanks to Ed Hudson for the loan of the laptop on which this was derived, amid numerous calls to the formerly great technical support organization of COMPAQ ("taken from you by HP").