The WordViewProjection matrix is what your looking for. that is the objects position in screen space, where x is -1 to 1, y is -1 to 1 and z is (assuming it's actually on the screen) 0 to 1, which is the depth.

to get the WVP, you just have to multiply the objects world space matrix with the view matrix, then with the projection matrix (in that order, World * View * Projection)

The world matrix stores the objects position (translation), orientation (rotation), and size (scale) in world space, or in other words, all the transformations you do on the model.

The View matrix stores the cameras position, target (what its looking at, or direction it's facing), and up (which is the direction that should be orthogonal to the target, so they make a right angle. it's the cameras "up" direction, and not the worlds up direction) You can get the view matrix from 3 vectors describing it like this:


D3DXMATRIX View;
D3DXVECTOR3 Position = D3DXVECTOR3(0.0f, 0.0f, 0.0f); // Position of your camera
D3DXVECTOR3 Target = D3DXVECTOR3(0.0f, 0.0f, 1.0f); // Cameras target
D3DXVECTOR3 Up = D3DXVECTOR3(0.0f, 1.0f, 0.0f); // Cameras up, should be a unit vector, and orthogonal to Target

D3DXMatrixLookAtLH( &View, &Position, &Target, &Up );

The projection matrix is like a pyramid, where the camera is the point or top of the pyramid, and everything inside the pyramid is what the camera sees. The near and far planes decide how close to the camera and how far away from the camera objects should be clipped, or not rendered. You could look at the far plane as the base of the pyramid and the near plane like cutting the tip off of the pyramid, so that anything that's left inside the pyramid is rendered to the screen. The projection is defined by two floats that describe the near and far planes, an aspect ratio (width/height) and a vertical field of view (FOV, in radians). We can get the projection matrix like this: (notice the (float) next to Width/Height. This is because Width and Height are most likely integers, so without the (float) this would be an integer division, and the result would most likely be a 1 instead of a real aspect ratio)

D3DXMATRIX Projection;
D3DXMatrixPerspectiveFovLH(&Projection, 0.4f*3.14f, (float)Width/Height, 1.0f, 1000.0f);

Finally, after you have all three matrices, you can create the WVP, or world view projection matrix:

D3DXMATRIX WVP;
WVP = World * View * Projection;

The WVP matrix is the objects position in screen space. As i mentioned,above, screen space's x and y are between -1 and 1, and z is 0 to 1. If your looking to turn this into 0 to the width and height of your screen, you will have to do an extra calculation, which i will show you in a minute. Now that we have the WVP, we can get the objects screen space position like this:

D3DXVECTOR3 screenCoord = D3DXVECTOR3(0.0f, 0.0f, 0.0f); // Position of object in screen space
D3DVec3TransformCoord(&screenCoord, &screenCoord, &WVP);

So now how do we turn the screen space from -1 to 1, to 0 to the width and height of your window? easy, just add 1 to x and y, divide it by two, and multiply it by the screen width or height, like this:

screenCoord.x = ((screenCoord.x + 1.0f) / 2.0f) * Width;
screenCoord.y = ((screenCoord.y + 1.0f) / 2.0f) * Height;

Maybe you already knew how to get the world, view, and projection matrices, but i thought i would be thorough and make sure you understand exactly how to do what you want ;)

hope this helps! good luck!

oh my bad! thanks for mentioning the flipping of y in screenspace!

i'm not going to lie, i wasn't even aware of D3DXVec3Project, haha, so thanks for that too, i'm pretty sure that's exactly what he's looking for, no sense recreating the wheel ;)

try setting the world matrix to the indentity