Simple Cube Vertices
Hi,
I''m fairly new to 3D programming and I''m trying to simply get a 3D cube on the screen using Drirect3D Imediate mode (DirextX7)
If I set up some vectors and use draw primitive using D3DPT_TRIANGLELIST. I get some of the sides coming out all funny.
Can some one tell me how to work out what order to draw the trianlges in. Or show me a simple way or drawing a 3d cube.
Thanks,
Jai
newbie
Without knowing more, my first guess would be that you''re probably not specifying the vertices of the triangle list in the correct order.
Oh, and I''m pretty sure you can''t draw a six faced cube with just a single triangle strip. Think about it. If you had a strip of triangles could you fold it to make a cube?
Oh, and I''m pretty sure you can''t draw a six faced cube with just a single triangle strip. Think about it. If you had a strip of triangles could you fold it to make a cube?
A cube can be drawn with a triangle strip, with a little effort.
I''m not a 3d programmer, but I can give a set of co-ordinates which should draw a cube. I''ve listed which triangle each co-ordinate completes.
(0, 0, 0)
(1, 0, 0)
(1, 0, 1) <-- first triangle on bottom of cube
(1, 1, 0) <-- first triangle on right of cube
(1, 1, 1) <-- second triangle on right of cube
(0, 1, 1) <-- first triangle on top of cube
(1, 0, 1) <-- first triangle on back of cube
(0, 0, 1) <-- second triangle on back of cube
(0, 0, 0) <-- second triangle on bottom of cube
(0, 1, 1) <-- first triangle on left of cube
(0, 1, 0) <-- second triangle on left of cube
(1, 1, 0) <-- second triangle on top of cube
(0, 0, 0) <-- first triangle on front of cube
(1, 0, 0) <-- second triangle on front of cube
That should work.
I''m not a 3d programmer, but I can give a set of co-ordinates which should draw a cube. I''ve listed which triangle each co-ordinate completes.
(0, 0, 0)
(1, 0, 0)
(1, 0, 1) <-- first triangle on bottom of cube
(1, 1, 0) <-- first triangle on right of cube
(1, 1, 1) <-- second triangle on right of cube
(0, 1, 1) <-- first triangle on top of cube
(1, 0, 1) <-- first triangle on back of cube
(0, 0, 1) <-- second triangle on back of cube
(0, 0, 0) <-- second triangle on bottom of cube
(0, 1, 1) <-- first triangle on left of cube
(0, 1, 0) <-- second triangle on left of cube
(1, 1, 0) <-- second triangle on top of cube
(0, 0, 0) <-- first triangle on front of cube
(1, 0, 0) <-- second triangle on front of cube
That should work.
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