I haven't been paying due attention here, but you may find some use for this formula for the solid angle at a vertex with three face angles a,b,c: Omega(a,b,c) = acos((cos(c)+cos(b)+cos(a)+1)^2/((cos(a)+1)*(cos(b)+1)*(cos(c)+1))-1) = 2*acos((cos(c)+cos(b)+cos(a)+1)/(4*cos(a/2)*cos(b/2)*cos(c/2))) Mma 7.0 takes several minutes failing to get 0 from the form FullSimplify[ ArcCos[((Cos[c] + Cos[b] + Cos[a] + 1)^2/ ((Cos[a] + 1)*(Cos[b] + 1)*(Cos[c] + 1))) - 1] - 2*ArcCos[((Cos[c] + Cos[b] + Cos[a] + 1)/ (4*Cos[a/2]*Cos[b/2]* Cos[c/2]))], a < Pi/2 && b < Pi/2 && c < Pi/2] , but In:=N[% /. a -> EulerGamma /. b -> 1 /. c -> E, 33] Out= 0.*10^-82 I --rwg (Sorry if you all get two of these. Sqirrelmail is earning its name.)
On 8/14/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
... What's more, using Gaussian curvature (in the form of a "2-D cosine") gives a far easier way to compute the face angle sums than employed by my existing kludgy program (or my dodgy sine formula). WFL
Well, no as it turns out --- computing face angles from polyhedron edges turns out less troublesome. But any algebraic proof of Fred's identity [with appropriate ambiguity concerning which Fred is thus immortalised], to avoid introducing arcsines, must deal with cosines and sines, which are not additive; and to avoid square roots, must deal with squares of sines.
So for two angles [my earlier dodgy identity should have read]
SAB^2 - 2*(SA-2*SA*SB+SB)*SAB + (SA-SB)^2 = 0,
with SA -> sin^2(A), SB -> sin^2(B), SAB -> sin^2(A+B).
But the polytore proof needs three angles, for which the equivalent identity (according to my Maple investigations) has degree 4 in SABC, 8796 terms in SA,SB,SC --- and anyway refuses to check out numerically, let alone formally.
So where does one go from here? WFL
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