wfl>On 2/7/11, Bill Gosper <billgosper@gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote: rwg> Ah, from 3.5 yrs ago,
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From gosper@alum.mit.edu <http://gosper.org/webmail/src/compose.php?send_to=gosper%40alum.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 <http://gosper.org/webmail/src/compose.php?send_to=23173122.9411184596788302.JavaMail.gbourne%40brunch.mit.edu>> From: Bill Gosper <gosper@alum.mit.edu <http://gosper.org/webmail/src/compose.php?send_to=gosper%40alum.mit.edu>> To: math-fun@mailman.xmission.com <http://gosper.org/webmail/src/compose.php?send_to=math-fun%40mailman.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). ------------ WFL>If there is a connection with the original thread, I'm afraid it has escaped me. Then I'm confused--I thought this was exactly what David requested. For the unit sphere, solid angle := area. If you place the apex of a pyramid at the center, the vertex angles will equal the great arcs on the surface. The base of an isosceles trapezoidal pyramid will project onto David's trapezoidal area. Is it clear that the area of a hinged quadrilateral abac is maximal when flexed into the isosceles configuration? Is this all wrong? Irrelevant? DW>When I worked at ComputerVision, I implemented an algorithm for polygonal area. For each edge of the polygon, I computed the area of the quadrilateral between the edge and its shadow on the x-axis via orthogonal projection. The edges were directed counterclockwise around the polygon, allowing computation of directed normals, the areas were added or subtracted according to whether the normal pointed toward or away from the x-axis. Adding up all these areas gave the area of the polygon. (rwg)>I sent this to the kids a few months ago. Probably shoulda cc'ed math-fun: -------------- Subj: That Im[z w*] trick is really handy. Here's the area of an arbitrary polygon: In[100]:=area[L_List]:=Total[MapThread[Im[#1\[Conjugate]*#2]&,{L,RotateLeft[L]}]]/2 E.g., the b x h triangle in general position: In[104]:= FullSimplify[ ComplexExpand[area[z+E^(I*t)*{0,b,h*I+x}]],{b,h,x,t}\[Element]Reals] Out[104]= (b h)/2 In[99]:= Area[Pentagon] Out[99]= 5/(4 Sqrt[5 - 2 Sqrt[5]]) In[106]:= area[Table[Exp[2*k*I*\[Pi]/5]/2/Sin[\[Pi]/5], {k, 5}]] Out[106]= (5 Sqrt[5/8 + Sqrt[5]/8])/(8 (5/8 - Sqrt[5]/8)) In[107]:= FullSimplify[% - %99] Out[107]= 0 -------------- --rwg