can someone please help me, I need to know how to do line and plane collision detection.

I dont need anything fancy, I just need help!!

can someone please help me, I need to know how to do line and plane collision detection.

I dont need anything fancy, I just need help!!

Are you wanting to simply determine if a line intersects a plane? Or where a line segment intersects a plane if it does?

just a line intersecting a plane

Ok, given a plane with normal N and displacement d, and a line parameterized by r(t)=r(0)+(r(1)-r(0))*t:

Take the dot product of N and r(1)-r(0). If the result is (or is close to) zero then assume no intersection. Otherwise there is a intersection. Do you want to know where that intersection is?

Thanks for your help, how do I get the d? Also, I would like to know where the intersection occurs

Any plane can be defined by a normal N and displacement d, such that for any point P in the plane, the following must be true:

N â€¢ P - d = 0

Sometimes people will use +d rather than -d but that doesnâ€™t change anything significant (only a sign convention). So you see to get d, you just pick a point in the plane and stick it in that equation and solve for d.

To find where the intersection of the line and plane is, substitute r(t) for P and solve the system of equations for t (see Cramerâ€™s rule). Then solve for r(t). If you want to see if r(t) is a point on the line segment between r(0) and r(1), you easily check to see if 0<=t<=1, if so then r(t) is a point on the line segment, and therefore the line segment intersects the plane (this is good to know for collision detection since each frame you can only move a step at a time, the path of which is a line segment).

[This message has been edited by DFrey (edited 02-15-2001).]

I once had nothing to do and solved these equations:

a + r*b + s*c = d + t*e

That results in very LONG termsâ€¦ And itâ€™s still longer than any other intersection equation. I wonder why we donâ€™t get teached the easiest equations in school.

But, DFrey, I wonder why d is called displacement. Displacement sounds like that the plane is translated in some direction about d units.

I wonder why there are so many representations of planes. How do you come to your representation, DFrey. I somehow remebers me of the normal represenation of a plane. But that again looked differently.

It is called displacement because it is a displacement from the origin. See, the shortest line from the origin to the plane will have length |d|. And Iâ€™m not doing anything special to come up with these plane definitions either, they are just what I learned in Calculus and Analytical Geometry with all the other first year students.

[This message has been edited by DFrey (edited 02-15-2001).]

I start with the normal representation

(x-q)*n = 0

(x1-q1)*n1 + (x2-q2)*n2 + (x3-q3)*n3 = 0

x1*n1 - q1*n1 + x2*n2 - q2*n2 + x3*n3 - q3*n3 = 0

x1*n1 + x2*n2 + x3*n3 = q1*n1 + q2*n2 + q3*n3

x*n = q*n

where q*n is your N*P So:

q*n - x*n = 0

But x*n is no distance.

Or is your operator another product?

You see, DFrey, my problem is that we donâ€™t learn enough here. We always calculate to long with the equations we get. We already stopped with linear algebra. It will come the next time when I study on the universityâ€¦ So I try to follow. Iâ€™m sorry if I sound a bit idioticâ€¦

Any light on it? Thanks!!!

Iâ€™m wondering if your computer is not using a font with the dot symbol. Thatâ€™s what I wrote.

(N dot P) - d = 0

N dot P has units of length since N is a unitless vector and the components of P have units of length.

Well, we didnâ€™t discuss the dot product in school nor did we discuss the cross product. Are there any articles where I could learn about them? Thanks!

PS: The dot is correct here, just my interpretation isnâ€™tâ€¦

Hi!

Michael and DFrey: I think you guys are talking about the same. I mean n=N: normal vector with unit length, q=P: a point of the plane and x*n=d: the distance of the plane from the origin. Or am I wrong?

dav