Nancy, I'm afraid I don't understand how to do this. How can I convert a position (r,phi=0,z) to a circular line in (x,y,z)? Right, to display cylindrically symmetric data originally described in the (r,z) plane as a full 3-D cylindrically symmetric object, I would need to map the position say (r,z) = (1,1) to the line segment x^2 + y^2 = 1, z=1. The reason I don't understand this is that now my point is a line segment. Can it be?
Or, are you saying that I should comput a 3-D [x,y,z] grid from my 2-D [r,z] data and display that? I didn't think this is what you meant since you said to convert the positions. Tom On Wed, 21 Jun 2000, nancy collins wrote: >there's no problem with having 2d surfaces in 3d. you can have >quads in 3 space, and/or you can stack sheets of quads into volumes >consisting of cubes/hexahedra. they can be in [r,phi,z] space, and >you can slice, stack, cut, whatever in your cylindrical coordinates, >and then right before you display the results, you convert just >the positions to [x,y,z] because we live in a euclidean world.
