Hi dharmraj ... The table has four boxes, each with two volume entries. Each box corresponds to a polyhedron of a particular size, and the two entries give the...
5041
Roger Kaufman
vortexswirling
Jul 1, 2010 3:26 pm
Hi Adrian, ... When I was trying examples I went straight to -l 1 (one). However this causes more problems than it solves. A good value to use ended up being...
5042
Roger Kaufman
vortexswirling
Jul 1, 2010 5:13 pm
Hi Adrian, ... This kind of rounding is not necessary. I took that code out and pushed the change to planar.cc Roger...
5043
Steve
stevewaterman51
Jul 1, 2010 5:33 pm
Adrian, ... Don't you mean a tetrahedron of edge length sqrt 2 ? A tet of edge 2 has a volume of 0.942809 [ times 1.06066 = 1 ] A tet of edge sqrt 2 has...
5044
Adrian Rossiter
adrianrossiter
Jul 1, 2010 7:58 pm
Hi Roger ... It is good that it turned out to be something like this, and not something more to do with the algorithm. I'll pick up the change. All being well...
5045
Adrian Rossiter
adrianrossiter
Jul 1, 2010 8:26 pm
Hi Steve ... You are thinking of Cartesian length. The Cartesian unit cube has a face diagonal of length sqrt(2) in the Cartesian system, but this same...
5046
Roger Kaufman
vortexswirling
Jul 1, 2010 8:49 pm
Hi Adrian, ... I don't have any more changes in the near term. I was looking at pyramids and the icosahedron. I took the icosahedron in the default position...
5047
Steve
stevewaterman51
Jul 2, 2010 1:43 pm
Adrian and dharmraj, re - Adrian's S vs C chart/wording So, you are saying that a Cartesian edge length of 1, does not equal a Synergetic edge length of 1 ? A...
5048
Roger Kaufman
vortexswirling
Jul 2, 2010 6:18 pm
Hi Adrian, A couple of years ago I made a web page on this subject. It was a way to produce the size of anti-dipyramid which would be of the correct size that...
5049
Roger Kaufman
vortexswirling
Jul 2, 2010 8:17 pm
Hi Adrian, ... I remember how I did this now. I took the pyramid I wanted to act on and excavated both sides of a regular antiprism of the same N. I would do...
5050
Adrian Rossiter
adrianrossiter
Jul 3, 2010 11:39 am
Hi Roger ... If the polygon is a {n/m} the ratio of heights is antiprism height : pyramid height = (1-cos(PI*m/n))/cos(PI*m/n) = sec(PI*m/n)-1 I have adapted...
5051
swdharmraj
Jul 3, 2010 12:46 pm
Adrian ... Hi Adrian: I agree that a tetrahedron with edge length 1 has a volume of 1/sqrt(72). Here is why. A cube with edge .707 embeds a tetrahedron of edge...
5052
Adrian Rossiter
adrianrossiter
Jul 3, 2010 12:59 pm
Hi Steve ... Yes, in the sense that these wouldn't be length values for the same edge. ... sqrt(2) ... 1/sqrt(2) ... 1 ... It isn't my concept, it is just my...
5053
Adrian Rossiter
adrianrossiter
Jul 3, 2010 1:15 pm
Hi dharmraj ... You have switched to Synergetics length units without applying a conversion. A tetrahedron with Cartesian edge length 1 has a Synergetics edge...
5054
Alan M
a.michelson
Jul 3, 2010 5:14 pm
http://groups.yahoo.com/group/synergeo/message/60714 You are measuring along the mid-face (average between vertex and opposite edge) of the triangular face of...
5055
Alan M
a.michelson
Jul 3, 2010 5:27 pm
Notice that I color-coded <http://groups.yahoo.com/group/synergeo/messages/61376?threaded=1&m=e&va\ ... length? ... see a ... lengths...
5056
Steve
stevewaterman51
Jul 3, 2010 5:28 pm
Adrian, However, having labeled the grids "even" and "odd" you can set up a Cartesian coordinate system so that the even grid has vertices having all even...
5057
Alan M
a.michelson
Jul 3, 2010 5:36 pm
I believe that you're talking about a cube with vertexes at (±1,±1,±1), correct?...
5058
Adrian Rossiter
adrianrossiter
Jul 3, 2010 5:46 pm
Hi Steve ... Having "scaled" the points does that mean that the even grid has vertices with all even coordinates and the odd grid has vertices with all odd...
5059
Roger Kaufman
vortexswirling
Jul 3, 2010 6:05 pm
Hi Adrian, ... I was surprised this turned out to be such a short equation! I've always thought these to be rather important polyhedra as they have one type of...
5060
Steve
stevewaterman51
Jul 3, 2010 6:09 pm
Adrian, Having "scaled" the points does that mean that the even grid has vertices with all even coordinates and the odd grid has vertices with all odd...
5061
swdharmraj
Jul 3, 2010 11:53 pm
dharmraj ... adrian ... Hi: Thanks. I get it. In a way that answers a puzzle going on in my thinking as to how we would know what cube is in which system. I am...
5062
swdharmraj
Jul 4, 2010 12:20 am
dharmraj ... Hi: Just an added thought here. I have always been looking at my graphics and seeing a cube grid which I assumed was the XYZ/Cartesian system. ...
5063
Adrian Rossiter
adrianrossiter
Jul 4, 2010 11:31 am
Hi Roger ... It doesn't seem so easy to obtain particular edge lengths, heights, etc, when making them as the reciprocal of an antiprism. However, it may be...
5064
Adrian Rossiter
adrianrossiter
Jul 4, 2010 12:11 pm
Hi dharmraj ... The general situation is that there is a space, in this case Euclidean space, and systems are set up for addressing objects in that space....
5065
Roger Kaufman
vortexswirling
Jul 4, 2010 2:50 pm
Hi Adrian, ... I was going to look to see what relation there might be between an antiprism and it's reciprocal. But having to calculate the shape of an...
5066
Roger Kaufman
vortexswirling
Jul 5, 2010 1:09 pm
Hi Adrian, I found this page on the Truncated trapezohedron. http://en.wikipedia.org/wiki/Truncated_trapezohedron The pentagonal one is an augmented...
5067
Roger Kaufman
vortexswirling
Jul 5, 2010 1:22 pm
Hi Adrian, ... I forgot to add that conway can make the dodecahedron from the script result, but the edges are not 1/Phi due to canonicalization. However it is...
5068
Roger Kaufman
vortexswirling
Jul 5, 2010 6:03 pm
Hi Adrian, ... I just found this. To complicate the naming even more, the particular type of trapezohedron the script makes is called a "Streptohedron". In the...
5069
Roger Kaufman
vortexswirling
Jul 5, 2010 7:26 pm
Hi Adrian, ... I added a little to the script to see if rotations between 0 and 90 degrees could generate a model. To do that, I multipled the height of the...
