Jeff Molofee - NeHe's OpenGL Tutorials

A plane is represented using its Vector representation as:

Xn dot X = d

Xn, X are vectors and d is a floating point value.

Xn is its normal.

X is a point on its surface.

d is a float representing the distance of the plane along the normal, from the center of the coordinate system.

Essentially a plane represents a half space. So all that we need to define a plane is a 3D point and a normal from that point which is perpendicular to that plane. These two vectors form a plane, ie. if we take for the 3D point the vector (0,0,0) and for the normal (0,1,0) we essentially define a plane across x,z axes. Therefore defining a point and a normal is enough to compute the Vector representation of a plane.

Using the vector equation of the plane the normal is substituted as Xn and the 3D point from which the normal originates is substituted as X. The only value that is missing is d which can easily be computed using a dot product (from the vector equation).

(Note: This Vector representation is equivalent to the widely known parametric form of the plane Ax + By + Cz + D=0 just take the three x,y,z values of the normal as A,B,C and set D=-d).

The two equations we have so far are:

PointOnRay = Raystart + t * Raydirection

Xn dot X = d

If a ray intersects the plane at some point then there must be some point on the ray which satisfies the plane equation as follows:

Xn dot PointOnRay = d or (Xn dot Raystart) + t * (Xn dot Raydirection) = d

solving for t:

t = (d – Xn dot Raystart) / (Xn dot Raydirection)

replacing d:

t= (Xn dot PointOnRay – Xn dot Raystart) / (Xn dot Raydirection)

summing it up:

t= (Xn dot (PointOnRay – Raystart)) / (Xn dot Raydirection)

t represents the distance from the start until the intersection point along the direction of the ray. Therefore substituting t into the ray equation we can get the collision point. There are a few special cases though. If Xn dot Raydirection = 0 then these two vectors are perpendicular (ray runs parallel to plane) and there will be no collision. If t is negative the collision takes place behind the starting point of the ray along the opposite direction and again there is no intersection.

int TestIntersionPlane(const Plane& plane,const TVector& position,const TVector& direction, double& lamda, TVector& pNormal) {

double DotProduct=direction.dot(plane._Normal); // Dot Product Between Plane Normal And Ray Direction

double l2;

// Determine If Ray Parallel To Plane

if ((DotProduct -ZERO)) return 0;

l2 = (plane._Normal.dot(plane._Position-position))/DotProduct; // Find Distance To Collision Point

if (l2<-ZERO) // Test If Collision Behind Start

return 0;

pNormal=plane._Normal;

lamda=l2;

return 1;

}

The code above calculates and returns the intersection. It returns 1 if there is an intersection otherwise it returns 0. The parameters are the plane, the start and direction of the vector, a double (lamda) where the collision distance is stored if there was any, and the returned normal at the collision point.

Computing the intersection between an infinite cylinder and a ray is much more complicated that is why I won't explain it here. There is way too much math involved too easily explain and my goal is primarily to give you tools how to do it without getting into alot of detail (this is not a geometry class). If anyone is interested in the theory behind the intersection code, please look at the Graphic Gems II Book (pp 35, intersection of a with a cylinder). A cylinder is represented as a ray, using a start and direction (here it represents the axis) vector and a radius (radius around the axis of the cylinder). The relevant function is:

int TestIntersionCylinder(const Cylinder& cylinder, const TVector& position, const TVector& direction, double& lamda, TVector& pNormal, TVector& newposition)

Returns 1 if an intersection was found and 0 otherwise.

The parameters are the cylinder structure (look at the code explanation further down), the start, direction vectors of the ray. The values returned through the parameters are the distance, the normal at the intersection point and the intersection point itself.

A sphere is represented using its center and its radius. Determining if two spheres collide is easy. By finding the distance between the two centers (dist method of the TVector class) we can determine if they intersect, if the distance is less than the sum of their two radius.

The problem lies in determining if 2 MOVING spheres collide. Bellow is an example where 2 sphere move during a time step from one point to another. Their paths cross in-between but this is not enough to prove that an intersection occurred (they could pass at a different time) nor can the collision point be determined.

Figure 1

The previous intersection methods were solving the equations of the objects to determine the intersection. When using complex shapes or when these equations are not available or can not be solved, a different method has to be used. The start points, endpoints, time step, velocity (direction of the sphere + speed) of the sphere and a method of how to compute intersections of static spheres is already known. To compute the intersection, the time step has to be sliced up into smaller pieces. Then we move the spheres according to that sliced time step using its velocity, and check for collisions. If at any point collision is found (which means the spheres have already penetrated each other) then we take the previous position as the intersection point (we could start interpolating between these points to find the exact intersection position, but that is mostly not required).

The smaller the time steps, the more slices we use the more accurate the method is. As an example lets say the time step is 1 and our slices are 3. We would check the two balls for collision at time 0 , 0.33, 0.66, 1. Easy !!!!

The code which performs this is:

/*****************************************************************************************/

/*** Find if any of the current balls ***/

/*** intersect with each other in the current timestep ***/

/*** Returns the index of the 2 intersecting balls, the point and time of intersection ***/

/*****************************************************************************************/

int FindBallCol(TVector& point, double& TimePoint, double Time2, int& BallNr1, int& BallNr2) {

TVector RelativeV;

TRay rays;

double MyTime=0.0, Add=Time2/150.0, Timedummy=10000, Timedummy2=-1;

TVector posi; // Test All Balls Against Eachother In 150 Small Steps

for (int i=0;i

for (int j=i+1;j

RelativeV=ArrayVel[i]-ArrayVel[j]; // Find Distance

rays=TRay(OldPos[i],TVector::unit(RelativeV));

MyTime=0.0;

if ( (rays.dist(OldPos[j]))> 40) continue; // If Distance Between Centers Greater Than 2*radius

// An Intersection Occurred

while (MyTime

{

MyTime+=Add;

posi=OldPos[i]+RelativeV*MyTime;

if (posi.dist(OldPos[j])<=40) {

point=posi;

if (Timedummy>(MyTime-Add)) Timedummy=MyTime-Add;