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I am confused about the triangle strip representation of closed mesh .The vertex buffer for triangle strip representation of the a figure is shown below:

A(0,1) B(0,0) C(1,1) D(1,0)E(2,1)

vertex buffer=[A,B,C,D,E] where, T1=(A,B,C) T2=(B,C,D) T3=(C,D,E)

Now, if I have a tetrahedron with four vertex A(0,0,0),B(1,0,0),C(0,1,0) and D(0,0,1) then what would be its vertex buffer representation?

Thank you very much.

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up vote 2 down vote accepted

short answer:[A,B,C,D,A,B]

you'll need 4 triangles to create a Tetrahedron. so you'll need 6 vertices in your buffer. not that for the first three vertices you'll only get one triangle but for every other vertex you'll have one extra triangle. now back to your problem, as I said you'll need 4 triangles represented as [A,B,C], [A,B,D], [A,C,D] and [B,C,D] with a little bit sorting you can easily arrange them into a form that each two consecutive triangle share two vertices [A,B,C], [B,C,D], [C,D,A] and finally [D,A,B]. then by combining all those together you'll have your full vertices arrangement which is exactly as I mentioned above.

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Thanks a lot. Yes I also think so. But here the writer represent it by 2 strips: [0,1,2,3] and [2,3,0,1]. I will be glad if you describe it. – gangcil Aug 25 '12 at 4:46
@gangcil as far as I know it doesn't really matter which representation you choose. They both will result in creation of same shape, except I used only one draw call and passed 6 vertices, while they used 2 calls and passed 8 vertices. – Ali.S Aug 26 '12 at 21:17

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