camera transformation

In most rendering APIs, the concept of a camera is more of a convenience than an integral part of the renderer. Basically, you have a chain of transformations -- model-to-world, world-to-view, view-to-clip, and something like clip-to-screen, and the renderer just deals in terms of these transformations. Luckily, the camera fits into this scheme very easily -- it generates the world-to-view transformation.

To generate the world-to-view transformation, you take the position and orientation of the camera in world space and compute a transformation (view-to-world) from that, and finally invert that to get the world-to-view transformation.

That's the basics. I left out a lot of details.

Thanks for your response, but I knew all that... what I need IS that world to view transformation, given the sort of data I have about the camera.

How to Compute the View Matrix from the Camera's Look-At and Eye Positions, and an Up Vector

The view matrix is simply the inverse of the camera's orientation/position matrix.

To generate the camera's orientation/position matrix, combine the local x, y, and z axes, and eye position. You compute the local axes like this:

    z = normalize( lookAt - eye );    x = normalize( cross( up, z ) );    y = cross( z, x ); 

The x, y, and z vectors go into the first, second, and third columns (if you are using column vectors) of the matrix. The eye position goes into the 4th column (if you are using column vectors). If you are using row vectors, then they go into the rows of the matrix. Now you have your camera's orientation/position matrix, which conveniently happens to be the view-to-world matrix. To make a world-to-view matrix, you simply invert it.

Now, rather than doing a general 4x4 matrix inversion, you can take advantage of some of the properties of the matrix in order to invert it quickly. First, let's simplify the notation. The matrix can be represented like this:

    R  T    0  1 

where R is the upper-left 3x3 part of the 4x4 matrix, T is the upper-right 3x1 part of the 4x4 matix, 0 is the 1x3 row of zeros on the bottom, and 1 is the 1 in the lower-right corner. Note that R and T conveniently correspond to the rotation and translation portions of the matrix. If you are using row vectors, then the matrices here are shown transposed. That doesn't affect the math, it just affects the notation.

The inverse of this 2x2 matrix is:

    R<sup>-1</sup> T'    0  1 

where T' = -R^-1T. In addition, since R is orthonormal (or orthogonal, or whatever the correct terminology is), R^-1 = R^T. The result is:

    R<sup>T</sup> T'    0  1 

where T' = -R^TT, and so no inversion is actually necessary. This is your world-to-view matrix.

One more possible optimization... T' = -R^TT can also be computed like this:

    T'<sub>X</sub> = -(R<sub>X</sub> dot T)    T'<sub>Y</sub> = -(R<sub>Y</sub> dot T)    T'<sub>Z</sub> = -(R<sub>Z</sub> dot T) 

where R_X, R_Y, and R_Z are the first, second, and third columns (or rows) of R -- the x, y, and z axes you computed at the beginning!

Note: It doesn't matter how the camera's orientation/position matrix is computed. The matrix inversion presented above will work for any transformation matrix that involves only rotation and translation.

Thanks a lot for the detailed reply! I will implement this and hopefully it works fine.

I don't know a ton about transformation matrices, but I think that the top 3x3, or R, is just a rotation matrix, and T is a translation matrix, just combined into the 4x4 matrix that way. Could I simply perform the translation of -T, then multiply R * [point] to rotate the point? The inverse of a 3x3 matrix isn't hard to calculate, although your optimizations make a 4x4's easy also.

Quote:

Original post by deej21 Could I simply perform the translation of -T, then multiply R * [point] to rotate the point?

Not R, R^T. Transforming a point by the view matrix looks like this:

    | v' |  =  | R<sup>T</sup>  -R<sup>T</sup>T |  | v |    | 1  |  =  | 0<sup> </sup>   1<sup> </sup>  |  | 1 | 

which simplifies to v' = R^Tv - R^TT, or v' = R^T(v - T).

Quote:

Original post by deej21 The inverse of a 3x3 matrix isn't hard to calculate, although your optimizations make a 4x4's easy also.

Don't forget that the inverse of a rotation matrix is simply its transpose.

Sweet, that worked out great. Thanks a lot, johnbolton.
Just a note, it also worked just using the 3x3 matrix. I just subtracted the eyepoint, then multiplied by the transposed rotation matrix. This eliminates the need to calculate T'