From csus.edu!csulb.edu!library.ucla.edu!agate!overload.lbl.gov!lll-winken.llnl.gov!uwm.edu!news.alpha.net!news.mathworks.com!uunet!newstf01.news.aol.com!newsbf02.news.aol.com!not-for-mail Sun Feb 26 08:02:28 1995 Path: csus.edu!csulb.edu!library.ucla.edu!agate!overload.lbl.gov!lll-winken.llnl.gov!uwm.edu!news.alpha.net!news.mathworks.com!uunet!newstf01.news.aol.com!newsbf02.news.aol.com!not-for-mail From: hopd@aol.com (HopD) Newsgroups: sci.math Subject: Re: Help with geometrey of dodecahedron! Date: 26 Feb 1995 01:13:33 -0500 Organization: America Online, Inc. (1-800-827-6364) Lines: 46 Sender: root@newsbf02.news.aol.com Message-ID: <3ip66d$hr6@newsbf02.news.aol.com> References: <3hojba$pv0@steele.ohsu.edu> Reply-To: hopd@aol.com (HopD) NNTP-Posting-Host: newsbf02.mail.aol.com :I am trying to make a three dimensional dodecahedron rotate :around the screen and I have a cube working but I can't :figure out the coordinates in 3-space of the vertices of :a dodecahedron. I would like to have the vertives of this :shape centered around the origin, my program can take these :and translate them to screen coordinates. Any help would ge :greatly appreciated. TIA :George Orfanakis OK, start with a pentagon of radius 1. It's hovering a distance of 1.309 units above the xy plane. it's coordinates are (using degrees, not radians): (0, 1, 1.309) (cos18, sin18, 1.309) (cos54, -sin54, 1.309) (-cos54, -sin54, 1.309) (-cos18, sin18, 1.309) Now draw lines from each of the top pentagon's corners to 5 points hovering a distance of .309 units above the xy plane. Their x and y coordinates are further from the center by a factor of 2cos36 (also known as the golden mean) The coordinates of these points are: (0, 2cos36, .309) (2cos36cos18, 2cos36sin18, .309) (2cos36cos54, -2cos36sin54, .309) (-2cos36cos54, -2cos36sin54, .309) (-2cos36cos18, sin18, .309) Now from each of these points draw 2 lines to 5 points submerged .309 units below the xy plane. these points are: (2cos36cos54, 2cos36sin54, -1.309) (2cos36cos18, -2cos36sin18, -1.309) (0, -2cos36, -1.309) (-2cos36cos18, -2cos36sin18, -1.309) (-2cos36cos54, 2cos36sin54, -1.309) Now from these points draw 5 lines to a pentagon submerged 1.309 units below the xy plane. The co-ordinates of these points are: (cos54, sin54, -1.309) (cos18, -sin18, -1.309) (0, -1, -1.309) (-cos18, -sin18, -1.309) (-cos54, sin54, -1.309) The distance of all these vectors from the origin should be the same, approx. 1.64727 Hope this helps. Anyone interested in polyhedra please check out my post on polyhedra models in this newsgroup. I've graphed points of various solids in microsoft excel and then multiplied them by rotation matrices. It's neat to turn a spreadsheet into a 3-D modeling program. HopD