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VprMatrix89

Ray-Plane Intersection

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I understood line intersection on the wiki, and I found one for ray collision. They are essentially the same,but I'd like to know what one of the variables (d) represents in this picture. Can anyone explain what d is in this equation? http://www.cs.princeton.edu/courses/archive/fall00/cs426/lectures/raycast/sld017.htm

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Quote:
Original post by VprMatrix89
I understood line intersection on the wiki, and I found one for ray collision. They are essentially the same,but I'd like to know what one of the variables (d) represents in this picture.
Can anyone explain what d is in this equation?
http://www.cs.princeton.edu/courses/archive/fall00/cs426/lectures/raycast/sld017.htm
d is the distance from the origin to the plane as measured along the plane normal N, in units of |N|. If N is unit-length, d represents the 'actual' distance to the plane.

The value of d can be computed by taking the dot product of the plane normal and any point known to lie on the plane.

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Ok, so you say d is the distance, in that case, from what?
If I assume that you mean distance to Po, then I dont see how thats possible,
because the plane normal and a point in the plane tells you nothing about Po.
Please explain?

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Quote:
Original post by VprMatrix89
Ok, so you say d is the distance, in that case, from what?
If I assume that you mean distance to Po, then I dont see how thats possible,
because the plane normal and a point in the plane tells you nothing about Po.
Please explain?
It has nothing to do with the ray - we're just talking about N and d here.

One of the ways you can represent a plane is with a normal (usually made unit-length for convenience) and a distance. The distance we're speaking of here is the distance from the origin to the plane, measured along a line that is perpendicular to the plane and passes through the origin.

Think of it this way. If all we had was the normal, we wouldn't have enough information to fully describe the plane, correct? We would know its orientation, but we would not know its location in space.

We can imagine this 'incomplete' plane as passing through the origin. Now, let's 'push' the plane some distance, in the direction of its normal. How far we 'push' it becomes the 'distance' that I spoke of earlier.

If you think about it, you'll see that by choosing an orientation for the plane (defined by the normal) and then 'pushing' it some distance along the normal, we can specify any plane in 3-space.

Does that help at all?

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I sort of figured that, when I plugged in some coordinates. That cleared it up, thanks. It makes sense for an intersection, since your finding the scale to the vector, I just didn't realize it at first.

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