Severe help with ray-triangle intersection
Author
I've been working many many hours trying to do collision detection. I'd like to have a setup where I can make everything collidible without labeling it a wall or floor and such.
I've been trying ray-trangle collision but I can't seem to get it to work. I have a feeling that if I get this all my problems will be solved but I've never done anything like this. I'm completely lost! If someone could explain that from the top or recommend another method that'd be very very swell! Thanks in advance!
Note: If you find something out of place ehh sorry! I'm over half sleep while I'm typing this.
A ray/triangle hit computation could be done as follows (perhaps there is a more efficient way):
Let the ray be originate at a point R0 and travel along a track r, then
R(t) := R0 + t * r
denotes the ray's current point at a travel distance of t.
Let the triangle be defined by 3 points
{ P0, P1, P2 }
so that a plane is spanned out as
P(u,v) := P0 + u * ( P1 - P0 ) + v * ( P2 - P0 )
and the triangle in this plane is given by the constraints
0 <= (u,v) <= 1 and u+v<=1
The plane's perpendicular vector
p := ( P1 - P0 ) x ( P2 - P0 )
where x denotes the cross product, can then be used to define the plane's unit length normal
n := p / |p|
We are searching for the point on the ray that is 0 length units away from the plane, so that from the Hessian formula of planes the equation
( R(t') - P0 ) . n = 0
has to be solved for t'
t' = ( ( P0 - R0 ) . n ) / ( r . n )
what could obviously done only if
r . n != 0
or else the ray is parallel to the plane and has none or infinite many hit points.
Now, solve
P(u',v') = R(t')
for u' and v' and check for the condition
0 <= (u',v') <= 1 and u'+v'<=1
If true, then P(u',v') = R(t') is the hit point, and comparing the "current" point's R(k) parameter k with t' tells whether it is located before, onto, or behind the triangle w.r.t. the ray's track.
[Edited by - haegarr on April 2, 2006 8:33:21 AM]
Let the ray be originate at a point R0 and travel along a track r, then
R(t) := R0 + t * r
denotes the ray's current point at a travel distance of t.
Let the triangle be defined by 3 points
{ P0, P1, P2 }
so that a plane is spanned out as
P(u,v) := P0 + u * ( P1 - P0 ) + v * ( P2 - P0 )
and the triangle in this plane is given by the constraints
0 <= (u,v) <= 1 and u+v<=1
The plane's perpendicular vector
p := ( P1 - P0 ) x ( P2 - P0 )
where x denotes the cross product, can then be used to define the plane's unit length normal
n := p / |p|
We are searching for the point on the ray that is 0 length units away from the plane, so that from the Hessian formula of planes the equation
( R(t') - P0 ) . n = 0
has to be solved for t'
t' = ( ( P0 - R0 ) . n ) / ( r . n )
what could obviously done only if
r . n != 0
or else the ray is parallel to the plane and has none or infinite many hit points.
Now, solve
P(u',v') = R(t')
for u' and v' and check for the condition
0 <= (u',v') <= 1 and u'+v'<=1
If true, then P(u',v') = R(t') is the hit point, and comparing the "current" point's R(k) parameter k with t' tells whether it is located before, onto, or behind the triangle w.r.t. the ray's track.
[Edited by - haegarr on April 2, 2006 8:33:21 AM]
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