Trying to optimize line vs cylinder intersection - c++

Question

My brain has been melting over a line segment-vs-cylinder intersection routine I've been working on.
/// Line segment VS <cylinder>
// - cylinder (A, B, r) (start point, end point, radius)
// - line has starting point (x0, y0, z0) and ending point (x0+ux, y0+uy, z0+uz) ((ux, uy, uz) is "direction")
// => start = (x0, y0, z0)
// dir = (ux, uy, uz)
// A
// B
// r
// optimize? (= don't care for t > 1)
// <= t = "time" of intersection
// norm = surface normal of intersection point
void CollisionExecuter::cylinderVSline(const Ogre::Vector3& start, const Ogre::Vector3& dir, const Ogre::Vector3& A, const Ogre::Vector3& B, const double r,
const bool optimize, double& t, Ogre::Vector3& normal) {
t = NaN;
// Solution : http://www.gamedev.net/community/forums/topic.asp?topic_id=467789
double cxmin, cymin, czmin, cxmax, cymax, czmax;
if (A.z < B.z) { czmin = A.z - r; czmax = B.z + r; } else { czmin = B.z - r; czmax = A.z + r; }
if (A.y < B.y) { cymin = A.y - r; cymax = B.y + r; } else { cymin = B.y - r; cymax = A.y + r; }
if (A.x < B.x) { cxmin = A.x - r; cxmax = B.x + r; } else { cxmin = B.x - r; cxmax = A.x + r; }
if (optimize) {
if (start.z >= czmax && (start.z + dir.z) > czmax) return;
if (start.z <= czmin && (start.z + dir.z) < czmin) return;
if (start.y >= cymax && (start.y + dir.y) > cymax) return;
if (start.y <= cymin && (start.y + dir.y) < cymin) return;
if (start.x >= cxmax && (start.x + dir.x) > cxmax) return;
if (start.x <= cxmin && (start.x + dir.x) < cxmin) return;
}
Ogre::Vector3 AB = B - A;
Ogre::Vector3 AO = start - A;
Ogre::Vector3 AOxAB = AO.crossProduct(AB);
Ogre::Vector3 VxAB = dir.crossProduct(AB);
double ab2 = AB.dotProduct(AB);
double a = VxAB.dotProduct(VxAB);
double b = 2 * VxAB.dotProduct(AOxAB);
double c = AOxAB.dotProduct(AOxAB) - (r*r * ab2);
double d = b * b - 4 * a * c;
if (d < 0) return;
double time = (-b - sqrt(d)) / (2 * a);
if (time < 0) return;
Ogre::Vector3 intersection = start + dir * time; /// intersection point
Ogre::Vector3 projection = A + (AB.dotProduct(intersection - A) / ab2) * AB; /// intersection projected onto cylinder axis
if ((projection - A).length() + (B - projection).length() > AB.length()) return; /// THIS IS THE SLOW SAFE WAY
//if (projection.z > czmax - r || projection.z < czmin + r ||
// projection.y > cymax - r || projection.y < cymin + r ||
// projection.x > cxmax - r || projection.x < cxmin + r ) return; /// THIS IS THE FASTER BUGGY WAY
normal = (intersection - projection);
normal.normalise();
t = time; /// at last
}
I have thought of this way to speed up the discovery of whether the projection of the intersection point lies inside the cylinder's length. However, it doesn't work and I can't really get it because it seems so logical :
if the projected point's x, y or z co-ordinates are not within the cylinder's limits, it should be considered outside. It seems though that this doesn't work in practice.
Any help would be greatly appreciated!
Cheers,
Bill Kotsias
Edit : It seems that the problems rise with boundary-cases, i.e when the cylinder is parallel to one of the axis. Rounding errors come into the equation and the "optimization" stops working correctly.
Maybe, if the logic is correct, the problems will go away by inserting a bit of tolerance like :
if (projection.z > czmax - r + 0.001 || projection.z < czmin + r - 0.001 || ... etc...

Answer 1

The cylinder is circular, right? You could transform coordinates so that the center line of the cylinder functions as the Z axis. Then you have a 2D problem of intersecting a line with a circle. The intersection points will be in terms of a parameter going from 0 to 1 along the length of the line, so you can calculate their positions in that coordinate system and compare to the top and bottom of the cylinder.
You should be able to do it all in closed form. No tolerances. And sure, you will get singularities and imaginary solutions. You seem to have thought of all this, so I guess I'm not sure what the question is.

Answer 2

This is what I use, it may help:
bool d3RayCylinderIntersection(const DCylinder &cylinder,const DVector3 &org,const DVector3 &dir,float &lambda,DVector3 &normal,DVector3 &newPosition)
// Ray and cylinder intersection
// If hit, returns true and the intersection point in 'newPosition' with a normal and distance along
// the ray ('lambda')
{
DVector3 RC;
float d;
float t,s;
DVector3 n,D,O;
float ln;
float in,out;
RC=org; RC.Subtract(&cylinder.position);
n.Cross(&dir,&cylinder.axis);
ln=n.Length();
// Parallel? (?)
if((ln<D3_EPSILON)&&(ln>-D3_EPSILON))
return false;
n.Normalize();
d=fabs(RC.Dot(n));
if (d<=cylinder.radius)
{
O.Cross(&RC,&cylinder.axis);
//TVector::cross(RC,cylinder._Axis,O);
t=-O.Dot(n)/ln;
//TVector::cross(n,cylinder._Axis,O);
O.Cross(&n,&cylinder.axis);
O.Normalize();
s=fabs( sqrtf(cylinder.radius*cylinder.radius-d*d) / dir.Dot(O) );
in=t-s;
out=t+s;
if (in<-D3_EPSILON)
{
if(out<-D3_EPSILON)
return false;
else lambda=out;
} else if(out<-D3_EPSILON)
{
lambda=in;
} else if(in<out)
{
lambda=in;
} else
{
lambda=out;
}
// Calculate intersection point
newPosition=org;
newPosition.x+=dir.x*lambda;
newPosition.y+=dir.y*lambda;
newPosition.z+=dir.z*lambda;
DVector3 HB;
HB=newPosition;
HB.Subtract(&cylinder.position);
float scale=HB.Dot(&cylinder.axis);
normal.x=HB.x-cylinder.axis.x*scale;
normal.y=HB.y-cylinder.axis.y*scale;
normal.z=HB.z-cylinder.axis.z*scale;
normal.Normalize();
return true;
}
return false;
}

Answer 3

Have you thought about it this way?
A cylinder is essentially a "fat" line segment so a way to do this would be to find the closest point on line segment (the cylinder's center line) to line segment (the line segment you are testing for intersection).
From there, you check the distance between this closest point and the other line segment, and compare it to the radius.
At this point, you have a "Pill vs Line Segment" test, but you could probably do some plane tests to "chop off" the caps on the pill to make a cylinder.
Shooting from the hip a bit though so hope it helps (:

Answer 4

Mike's answer is good. For any tricky shape you're best off finding the transformation matrix T that maps it into a nice upright version, in this case an outright cylinder with radius 1. height 1, would do the job nicely. Figure out your new line in this new space, perform the calculation, convert back.
However, if you are looking to optimise (and it sounds like you are), there is probably loads you can do.
For example, you can calculate the shortest distance between two lines -- probably using the dot product rule -- imagine joining two lines by a thread. Then if this thread is the shortest of all possible threads, then it will be perpendicular to both lines, so Thread.LineA = Thread.LineB = 0
If the shortest distance is greater than the radius of the cylinder, it is a miss.
You could define the locus of the cylinder using x,y,z, and thrash the whole thing out as some horrible quadratic equation, and optimise by calculating the discriminant first, and returning no-hit if this is negative.
To define the locus, take any point P=(x,y,z). drop it as a perpendicular on to the centre line of your cylinder, and look at its magnitude squared. if that equals R^2 that point is in.
Then you throw your line {s = U + lamda*V} into that mess, and you would end up with some butt ugly quadratic in lamda. but that would probably be faster than fiddling matrices, unless you can get the hardware to do it (I'm guessing OpenGL has some function to get the hardware to do this superfast).
It all depends on how much optimisation you want; personally I would go with Mike's answer unless there was a really good reason not to.
PS You might get more help if you explain the technique you use rather than just dumping code, leaving it to the reader to figure out what you're doing.