派生透视投影平面

Question

Given a perspective projection matrix and Center of Projection how can one derive the projection plane? For example, let us assume that the perspective projection matrix is a 4x4 homogenous matrix:

1 0 0 0
0 1 0 0
0 0 1 0
0 2 0 0

and the center of projection [0, 0, 0]. From this data, it is clear how each point is projected into the projection plane. But how can one use this data to find the exact equation / representation of the projection plane?

Answer 1

Unlike a "center of projection" (which is the effective location of the camera in the 3D scene), there is no need for a concrete "projection plane" in the sense of a notional plane that the image is projected upon.

Instead, the projection matrix determines a transformation into "clip space" which determines the viewport bounds and the depth mapping, from which the x and y coordinates are further scaled and offset to get the target pixel location. However, none of this translates to an actual plane in the scene which somehow corresponds to a physical camera's focal plane.

There are relevant planes you can derive from a camera matrix:

• the w=0 plane, which is the fourth row of the projection matrix. This plane passes through the center of projection, and separates points in front of the camera from those in back; none of the points on the w=0 plane can ever be visible. In your example, this plane is 0 2 0 0 , which is the XZ coordinate plane.

• the z=0 plane, which is the third row of the projection matrix. In conjunction with the w=0 plane, it determines the depth mapping function and the near/far clipping planes. In your example, this plane is 0 0 1 0 , which is the XY coordinate plane.

The remaining rows of the camera matrix also represent planes: the first and second rows are the x=0 and y=0 planes, respectively. In some sense, they are not as fundamentally concrete as the w=0 plane; conventionally, they separate the clip space (and the view frustum) into quadrants. They determine the horizontal and vertical boundary planes of the viewing frustum, by taking the sum and difference with the w=0 plane coordinates.

The "center of projection" is the point at the intersection of the x=0 , y=0 , and w=0 planes (that is, the first, second, and fourth rows of the projection matrix). I should note that this indicates your example is an unconventional configuration: the "center of projection" is not well-defined, and all your points will be mapped with clip-space coordinates such that y/w = 1/2 , which will typically be a horizontal line one quarter way down the screen. Also note that, normally, the z=0 plane should be parallel to the w=0 plane, intersecting only at infinity (instead of along the X-axis, as in your example).

The near and far clip planes are derived from the z=0 and w=0 planes based on the clip space conventions of your graphics system:

• OpenGL's depth clipping is z/w = [near=-1, far=+1]
  • its near clip plane is the sum of the above w=0 and z=0 planes ( 0 2 1 0 in your example)
  • its far clip plane is their difference ( 0 2 -1 0 in the example)
• DirectX's depth clipping is z/w = [near=0, far=-1]
  • its near clip plane is the z=0 plane ( 0 0 1 0 in your example)
  • and its far clip plane is the sum of the two (identical to the OpenGL near clip plane, 0 2 1 0 in your example)