Forwarded conversation Subject: [Jprogramming] Szilassi's toroidal heptahedron: interfacial angles? Last try. ------------------------
From: *PMA* <[email protected]> Date: 2009/8/21 To: [email protected] Hi All. Here is a final crack at posing my question -- this is as whittled as I can whittle: Given a 3-D vertex of 3 faces with all edge-to-edge angles known, how does one calculate its face-to-face angles (all, one per edge)? If there come to mind other sites especially interested in this sort of query, I'd appreciate hearing. Thanks for your consideration. P.A. ------------------------------------------------------------------------------------------------ Dear J Folks, From this shape's coordinates as given in Stewart's _Adventures Among The Toroids_, I've calculated planar angles (internal angles successively around the perimeter of each face) and wish now to compute the (12 unique of 21) interfacial angles as well. There are 14 vertices, each the meeting of 3 edges. Am I correct, that at any vertex an appropriate function should be able to calculate the I-F angles, given just the planar angles among the edge pairs? If so, what would be a workable J version? If not, what am I not seeing? Would appreciate any comment. Thanks in advance! P.A. ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm ---------- From: *Sherlock, Ric* <[email protected]> Date: 2009/8/21 To: Programming forum <[email protected]> Hi Peter, You could try providing some sample data and show what you expect the result to be. That might elicit a better response. > From: PMA ---------- From: *Bo Jacoby* <[email protected]> Date: 2009/8/21 To: Programming forum <[email protected]> Isn't it just plain linear algebra? You have a 3-corner and knows the 3 cosines of angles between 3 vectors. Compute the cosines of angles between pairwise normals to the vectors. Venlig hilsen, Bo. --- Den fre 21/8/09 skrev Sherlock, Ric <[email protected]>: > Fra: Sherlock, Ric <[email protected]> > Emne: Re: [Jprogramming] Szilassi's toroidal heptahedron: interfacial angles? Last try. > Til: "Programming forum" <[email protected]> > Dato: fredag 21. august 2009 03.07 Trænger du til at se det store billede? Kelkoo giver dig gode tilbud på LCD TV! Se her http://dk.yahoo.com/r/pat/lcd ---------- From: *Bo Jacoby* <[email protected]> Date: 2009/8/21 To: Programming forum <[email protected]> Alternatively: Draw a sphere with its center in the corner. The edge to edge angles of the corner are sides in the spherical triangle, and the face to face angles of the corner are angles in the spherical triangle. Use spherical trigonometry formulas. - Bo --- Den fre 21/8/09 skrev Bo Jacoby <[email protected]>: > Fra: Bo Jacoby <[email protected]> > Dato: fredag 21. august 2009 04.58 ___________________________________________________________ Skal du købe ny bil? Sammenlign priser på brugte biler med Kelkoo og find et godt tilbud! - Se mere her http://dk.yahoo.com/r/pat/mmb ---------- From: *PMA* <[email protected]> Date: 2009/8/22 To: Programming forum <[email protected]> Bo, Thank you for these leads! I felt the algebra would be just plain, but hadn't the grasp to specify it. Will pursue.... Best regards, Pete ---------- From: *Istvan Kadar* <[email protected]> Date: 2009/8/28 To: Programming forum <[email protected]> Hi Pete, A NUMERICAL EXAMPLE NB. A B known NB. \ / edge-angles NB. \ / a=:57 01 45.2 NB. \ c / b=:87 42 41.1 NB. \ / c=:55 37 19.6 NB. \ / NB. D unknown NB. a | b face-angles NB. | C=: ? NB. | A=: ? NB. | B=: ? NB. | NB. C a + b + c << 360 (57 01 45.2+87 42 41.1+55 37 19.6=200 21 45.9<<360) A formula for the solution cos_C =: ((cos c)-~(cos a)*cos b)%(-sin a)*sin b C =: acos cos_C ]sin_a=:1&o.rfd((%60)p.~a) 0.838948 9j7":sin_a 0.8389482 ]cos_a=:2&o.rfd((%60)p.~a) 0.544211 9j7":cos_a 0.5442112 ]sin_b=:1&o.rfd((%60)p.~b) 0.999202 9j7":sin_b 0.9992024 ]cos_b=:2&o.rfd((%60)p.~b) 0.0399327 9j7":cos_b 0.0399327 ]cos_c=:2&o.rfd((%60)p.~c) 0.564649 9j7":cos_c 0.5646485 ]cos_C =: ((cos_c)-~cos_a*cos_b)%-sin_a*sin_b 0.647656 9j7":cos_C =: ((cos_c)-~cos_a*cos_b)%-sin_a*sin_b 0.6476563 _2&o.0.6476563 0.866292 dfr=:rfd^:_1 9j7":_2&o.0.6476563 0.8662919 radian dfr 0.8662919 49.6349 C=:49.6349 degree (49° 38' 05.5" ) Best regards Pista 2009/8/22 PMA <[email protected]> ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
