Hi Guys,

I'm presently brushing up of some 3D math and am wondering how I would go about working out the final step of where a singular vertex would be positioned in screen space in the -1 to +1 range.

I have done the following calculations based on the following parameters and the results listed below.

At this stage I haven't multiplied anything against the world matrix, so that would effectively mean that the vertex would be located at space 0, 0, 0.

So my question is, what calculation would I need to do against the WVP matrix in order to work out the screen location X & Y on the screen? Expecting the result to be 0, 0 (being direct centre of the screen).

Many thanks!

	float fovY = 60.0f;
	float aspect = 1280.0f / 800.0f;
	float near = 1.0f;
	float far = 1000.0f;

	XMVECTOR from = { 0.0f, 0.0f, -2.5f, 1.0f };
	XMVECTOR to = { 0.0f, 0.0f, 0.0f, 1.0f };
	XMVECTOR up = { 0.0f, 1.0f, 0.0f, 0.0f };
	XMMATRIX matView = XMMatrixLookAtLH(from, to, up);

View Matrix
1       0       0       0
0       1       0       0
0       0       1       0
0       0       2.5     1


Proj Matrix
-0.0975741      0       0       0
0       -0.156119       0       0
0       0       1.001   1
0       0       -1.001  0


View * Proj Matrix
-0.0975741      0       0       0
0       -0.156119       0       0
0       0       3.501   1
0       0       -1.001  0


Identity Matrix
1       0       0       0
0       1       0       0
0       0       1       0
0       0       0       1


WVP Matrix
-0.0975741      0       0       0
0       -0.156119       0       0
0       0       3.501   1
0       0       -1.001  0

Thanks for the reply Pam.

Sorry for my ignorance, but how would you work out the clip co-ordinates, and w co-ordinate of what?

@PAM79 Ah! Awesome. Thanks for that. I'll have a ply around now.