First, please note that the topic title isn't what you want. Per the article you reference, it's about how to "Calculate position as 0 to 1 *between *planes."

The equation ( which is poorly written and even more poorly described IMHO ) appears to represent the player's position "value" as a ratio of the distance from the player to a plane along a predefined direction to the full distance between the planes along that same direction, passing through the player's position. E.g., if a string is stretched(*) from one plane to the other through the player, and the player is two-thirds of the way along that string, the "value" is 0.333 for camera1 and 0.667 for camera 2.

(*) *always *in the same direction.

[In the following discussion, if you want to have a ratio = 1 when the player reaches the far plane, switch 1's and 2's around. It's moot however, because you'll need the **ratio** and **1-ratio** in either case to weight the two "cameras"]

The algorithm he mentions is to calculate the intersection point in both plane 1 (back) and plane 2 (forward), calling them point1 and point2, defined by a vector from the player's position along the direction vector to each plane. The length (absolute value) of a vector defined with those two points ( p2 minus p1 ) is the "full" distance between the planes. Define a vector from the player's position to point 2 ( p2 minus playerPos ). Now calculate the dot product of the player vector with the normal to plane 2 (player-distance). Calculate the dot product of the p2-p1 vector with the normal to plane 2 (full-distance). The ratio player-distance / full-distance is what (I believe) the author is trying to represent as the "value" for weighting. That **ratio** will vary from 1 to 0 as the player hits plane 1 and progresses to plane 2. So, as the player progresses from plane 1 to plane 2, camera 1 is weighted by **ratio**; camera 2 is weighted by 1 - **ratio**.

He shows two parallel planes in his example, but that algorithm will work for any two planes. He (apparently) also assumes there's a predefined direction vector that the user will have to come up with. You can pick whatever you'd like, provided that vector is not parallel with either plane. It's tough to calculate the intersection of a vector with a plane if it never intersects that plane.

The part of the math he (conveniently) skips over is how to determine the points in the planes. For that you'll have to have the predefined vector (best to use a unit vector) and solve two equations using the two equations for the planes, one for the distance from the player to plane1; the other for the distance to plane 2. N.B., those are not dot product distances, but distances along a direction!

**Edited by Buckeye, 12 February 2014 - 08:52 PM.**