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  • Members: 82
  • Category: Geometry
  • Founded: Nov 2, 2006
  • Language: English
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Name Creator
Folder David Fontaine
Some Polytope Stuff Here
 
Folder Modelling Programs
 
"The Archimedean Honeycomb Duals" by Guy Inchbald
You need to subscribe to JSTOR: The Mathematical Gazette Vol. 81, No. 491 (Jul., 1997), pp. 213-219 (article consists of 7 pages)
http://www.jstor.org/stable/pdfplus/3619198.pdf 		a.michelson
*IN CASE OF DELAYED MESSAGES:*
http://help.yahoo.com/l/us/yahoo/groups/original/members/contact/ygpostform1.html 		a.michelson
*Yahoo! Customer Care Help*
http://help.yahoo.com/l/us/yahoo/groups/original/members/contact/forms_index.html 		a.michelson
17 Types of Symmetry
A polyhedron can have one of 17 types of symmetry.
http://AntiPrism.com/album/855_symmetry/ 		a.michelson
Antiprism
Programs for viewing and modelling polyhedra, and working with OFF format files
http://www.antiprism.com 		adrianrossiter
Geometry Study
When I heard "Boadicea" by Enia on the radio, I was reminded of watching this Geometry video.
http://EnergyVision.BlogSpot.com/2006/02/geometry-study.html 		a.michelson
Gregg Fleishman
3850 Main Street, Culver City, California 90232. Notice his use of the various zonohedra. Gregg Fleishman also brought you Cluster Structures! See also Play Mountain Place.
http://GreggFleishman.com/ 		a.michelson
Hedrondude's Home Page
http://www.polytope.net/hedrondude/home.htm 		a.michelson
Natural Color System — Wikipedia, the free encyclopedia
JOVO Toys <http://www.jovotoys.com/> has 6 colors: red, yellow, blue, green, black, white!
http://en.wikipedia.org/wiki/Natural_Color_System 		a.michelson
Opponent process — Wikipedia, the free encyclopedia
red-green. yellow-blue.
http://en.wikipedia.org/wiki/Opponent_process 		a.michelson
Prove that it is EXACTLY 1/9.
First of all, let me define ACOS(1/3) as the dihedral angle of that oct-tet IVM. Then cos(dihedral) = 1/3 and sin(dihedral) = (2/3)*SQRT(2). For sin(dihedral/2) = sqrt((1-cos(dihedral))/2) = sqrt((1-(1/3))/2) = sqrt((2/3)/2) = sqrt(1/3). And sin(2*dihedral) = 2*(sin(dihedral))*(cos(dihedral)) = 2*((2/3)*SQRT(2))*(1/3) = (4/9)*sqrt(2). Therefore sin(5*dihedral/2) = sin(2*dihedral+dihedral/2) = (sin(2*dihedral))*(cos(dihedral/2)) + (cos(2*dihedral))*(sin(dihedral/2)) = (4/9)*sqrt(2)*sqrt(1-(1/3))-sqrt(1-2*16/81)*sqrt(1/3) = (4/9)*sqrt(2)*sqrt(2/3)-sqrt(1-32/81)*sqrt(1/3) = (4/9)*sqrt(2)*sqrt(2/3)-sqrt(49/81)*sqrt(1/3) = (4/9)*sqrt(2)*sqrt(2/3)-(7/9)*sqrt(1/3) = (8/9)/sqrt(3)-(7/9)*sqrt(1/3) = (8/9)/sqrt(3)-(7/9)/sqrt(3) = (1/9)/sqrt(3). Now we know that sin(5/2*acos(1/3)) = (1/9)/sqrt(3). But then Adrian would calculate sin(5/2*acos(1/3))*sqrt(3) = ((1/9)/sqrt(3))*sqrt(3) and all we got is 1/9. So we all computed it to INFINITY decimal places, and we all got all ones!!!!!!
http://groups.yahoo.com/group/synergeo/messages/10632?threaded=1&m=e&var=1&tidx=1 		a.michelson
Waterman Polyhedron
Steve Waterman's NEW Polyhedron Site!
http://WatermanPolyhedron.Com/ 		a.michelson
Waterman polyhedron — Wikipedia, the free encyclopedia
Steve Waterman, the American inventor of Waterman polyhedron and Waterman "Butterfly" World Projection
http://en.wikipedia.org/wiki/Waterman_polyhedron 		a.michelson
what is the difference?
Numbers of points on the various polyhedrons up to and including the dodecahedron: x = 8F^2 + 2. A108100 (2*n-1)^2+(2*n+1)^2 = 4*n^2+4*n^2+1+1 = 8*n^2+2 is the count for an elongated or gyroelongated square dipyramid.
http://groups.yahoo.com/group/synergeo/message/35898 		a.michelson
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