Re: 3D Projection

*To*: mathgroup at smc.vnet.net*Subject*: [mg19742] Re: 3D Projection*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Wed, 15 Sep 1999 03:52:59 -0400*Organization*: Universitaet Leipzig*References*: <7ri605$8it@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi, it is not the right place, comp.graphics.algorithms is the correct one. You have to say where your camera is placed, what is your viewing depth and how is the camera oriented in space. Any standard book of 3d graphics will give you the transformation between the different coordinate systems .. In homogeneous coordinates with a projection plane: t ^ | | l<-----+-----> r | | V b and a near plane position at n (perpendicular to the axis above) and a far plane position f You have the matrix {{2 n/(r-l), 0, (r+l)/(r-l), 0}, { 0 , 2 n/(t-b), (t+b)/(t-b), 0}, { 0, 0, -(f+n)/(f-n),-2 f n/(f-n)}, { 0 0, -1, 0}} where your point has the coordinates {x,y,z,1} Hope that helps Jens Daniel Kolb wrote: > > Hi > I don't know if this is the right place for this question. > I'm searching for a formula for 3D central-projection > thanx