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
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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Thanks for that, Bill --- I'll examine at leisure. I have to admit to haziness about the exact relation between Gaussian curvature (which should zero for an origami polyhedron) and solid angle. But that doesn't matter: the solid angle just needs to be 2\pi. The trickier problem is avoiding square roots in the expressions for the cosines ... My own yard-long program eventully disgorged a quartic in SABC with 504 terms in SA,SB,SC, after some obscure interaction between Maple expand() and simplify() was resolved. WFL On 8/16/09, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
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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On 8/16/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
... My own yard-long program eventully disgorged a quartic in SABC with 504 terms in SA,SB,SC, after some obscure interaction between Maple expand() and simplify() was resolved. WFL
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).
Er, make that a quartic in SABC with 71 terms in SA,SB,SC. To wit, With SA -> sin^2(A), SB -> sin^2(B), SC -> sin^2(C), SABC -> sin^2(A+B+C), 0 = SABC^4 + SABC^3 * ( -16*SA*SB*SC + 8*(SA*SB + SA*SC + SB*SC) - 4*(SA + SB + SC) ) + SABC^2 * ( + 24*SA*SB*SC + 16*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) - 16*(SA^2*SB + SA^2*SC + SA*SB^2 + SA*SC^2 + SB^2*SC + SB*SC^2) + 6*(SA^2 + SB^2 + SC^2) + 4*(SA*SB + SA*SC + SB*SC) ) + SABC * ( - 40*SA*SB*SC - 16*(SA^3*SB*SC + SA*SB^3*SC + SA*SB*SC^3) + 8*(SA^3*SB + SA^3*SC + SA*SB^3 + SA*SC^3 + SB^3*SC + SB*SC^3) + 24*(SA^2*SB*SC + SA*SB^2*SC + SA*SB*SC^2) - 16*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) - 4*(SA^3 + SB^3 + SC^3) + 4*(SA^2*SB + SA^2*SC + SA*SB^2 + SA*SC^2 + SB^2*SC + SB*SC^2) ) + ( 16*SA^2*SB^2*SC^2 + 8*(SA^3*SB*SC + SA*SB^3*SC + SA*SB*SC^3) - 16*(SA^2*SB^2*SC + SA^2*SB*SC^2 + SA*SB^2*SC^2) + (SA^4 + SB^4 + SC^4) - 4*(SA^3*SB + SA^3*SC + SA*SB^3 + SA*SC^3 + SB^3*SC + SB*SC^3) + 6*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) + 4*(SA^2*SB*SC + SA*SB^2*SC + SA*SB*SC^2) ) ; Intriguingly, tidying the above up for publication produces a reduction in numerical rounding error by an order of magnitude! WFL
A postscript --- it took some time to dawn on me that the constant coefficient of the polynomial in SABC is actually the square of 4*SA*SB*SC + SA^2 + SB^2 + SC^2 - 2*(SA*SB + SA*SC + SB*SC), and the vanishing of this polynomial is equivalent to A+B+C = k \pi. This considerably reduces the computational cost of proving the h versus q relations for a (cuboid) polytore; the bad news is that it fatally entices one (me) into attempting to find a similar relation for polytores constructed from prisms based on regular even n-gons, rather than just square cuboids (n = 4). It seems we got lucky with the cuboidal case. For n = 6 (and any n > 4?), the (numerator of square root of) constant coefficient fails to factorise, has degree 16 in q and 12 in h, and numerical coefficients of order 10^8. WFL [21/08/09] On 8/16/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
Er, make that a quartic in SABC with 71 terms in SA,SB,SC. To wit,
With SA -> sin^2(A), SB -> sin^2(B), SC -> sin^2(C), SABC -> sin^2(A+B+C),
0 = SABC^4
+ SABC^3 * ( -16*SA*SB*SC + 8*(SA*SB + SA*SC + SB*SC) - 4*(SA + SB + SC) )
+ SABC^2 * ( + 24*SA*SB*SC + 16*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) - 16*(SA^2*SB + SA^2*SC + SA*SB^2 + SA*SC^2 + SB^2*SC + SB*SC^2) + 6*(SA^2 + SB^2 + SC^2) + 4*(SA*SB + SA*SC + SB*SC) )
+ SABC * ( - 40*SA*SB*SC - 16*(SA^3*SB*SC + SA*SB^3*SC + SA*SB*SC^3) + 8*(SA^3*SB + SA^3*SC + SA*SB^3 + SA*SC^3 + SB^3*SC + SB*SC^3) + 24*(SA^2*SB*SC + SA*SB^2*SC + SA*SB*SC^2) - 16*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) - 4*(SA^3 + SB^3 + SC^3) + 4*(SA^2*SB + SA^2*SC + SA*SB^2 + SA*SC^2 + SB^2*SC + SB*SC^2) )
+ ( 16*SA^2*SB^2*SC^2 + 8*(SA^3*SB*SC + SA*SB^3*SC + SA*SB*SC^3) - 16*(SA^2*SB^2*SC + SA^2*SB*SC^2 + SA*SB^2*SC^2) + (SA^4 + SB^4 + SC^4) - 4*(SA^3*SB + SA^3*SC + SA*SB^3 + SA*SC^3 + SB^3*SC + SB*SC^3) + 6*(SA^2*SB^2 + SA^2*SC^2 + SB^2*SC^2) + 4*(SA^2*SB*SC + SA*SB^2*SC + SA*SB*SC^2) ) ;
Intriguingly, tidying the above up for publication produces a reduction in numerical rounding error by an order of magnitude! WFL
On 8/21/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
A postscript --- it took some time to dawn on me that the constant coefficient of the polynomial in SABC is actually the square of 4*SA*SB*SC + SA^2 + SB^2 + SC^2 - 2*(SA*SB + SA*SC + SB*SC), and the vanishing of this polynomial is equivalent to A+B+C = k \pi.
This considerably reduces the computational cost of proving the h versus q relations for a (cuboid) polytore; the bad news is that it fatally entices one (me) into attempting to find a similar relation for polytores constructed from prisms based on regular even n-gons, rather than just square cuboids (n = 4).
And in a different --- doubtless equal fatal --- direction, into posing this generalisation: given 4 angles A,B,C,D, what polynomial in the squares of their sines gives a condition (necessary only, since the signs remain unspecified) that A+B+C+D = k \pi ? WFL
On 8/21/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
... And in a different --- doubtless equal fatal --- direction, into posing this generalisation: given 4 angles A,B,C,D, what polynomial in the squares of their sines gives a condition (necessary only, since the signs remain unspecified) that A+B+C+D = k \pi ? WFL
How embarrassing --- the quartic for the sum of 3 angles, with SABC -> SD. At least the result is more symmetric than before: omitting isomorphic terms, - 40*SA*SB*SC*SD - 16*(SA^3*SB*SC*SD + ) + 16*(SA^2*SB^2*SC^2 + ) + 8*(SA^3*SB*SC + ) - 16*(SA^2*SB^2*SC + ) + 24*(SA^2*SB*SC*SD + ) - 4*(SA^3*SB+SA^3*SC + ) + 4*(SA^2*SB*SC + ) + 6*(SA^2*SB^2 + ) + (SA^4 + ) But 5 angles is getting a tad heavy --- can anyone spot a pattern? [From _two_ data items, that is --- 0,1,2 angles are sadly untypical.] WFL
On 8/21/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
... But 5 angles is getting a tad heavy --- can anyone spot a pattern? [From _two_ data items, that is --- 0,1,2 angles are sadly untypical.]
Bolstered only by insane pride, I finally hacked 5 angles. The degree of the terms varies between 8 and 15; there are 2922 terms, falling into 94 symmetry classes. Program/results available to personal enquirers. WFL
Oh dear. Musing on the curious and only almost obvious fact that in the sin^2 criterion f_{n+1} = 0 for n+1 angles the "constant coefficient" [terms not containing the final variable] is just (f_n)^2, I was crushed to discover the following trivial algorithm for generating these polynomials. [Note that they are symmetric in the a_i for n > 2.] Define f_n (a_1, ..., a_n) = 0 just when \sum (+/-) A_i = 0 mod \pi (for some combination of signs), where a_i == sin^2(A_i) ; then f_{n+1} is given in terms of f_n by f_2 (a_1, a_2) = a_1 - a_2 ; f_{n+1} (a_0, a_1, ..., a_n) = f_n (b+c, ..., a_n) f_n (b-c, ..., a_n) , where b = a_0 - 2 a_0 a_1 + a_1 , c^2 = a_0 a_1 (1-a_0) (1-a_1) . Thus f_5 in 3 sec on my ancient Mac G4; though f_6 with an estimated 10^6 terms remains out of reach, leaving (you will be desolate to hear) the direct verification of planarity for the inner corners of any general polytore family likewise unattainable. Fred Lunnon [25/08/09]
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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rwg＠sdf.lonestar.org