An AxisAligned Bounding Box (AABB) Sweep Test
Just like the name says, the faces of an axisaligned bounding box are aligned with the coordinate axes of its parent frame (see Figure 3). In most cases AABBs can fit an object more tightly than a sphere, and their overlap test is extremely fast.
To see if A and B overlap, a separating axis test is used along the x, y, and zaxes. If the two boxes are disjoint, then at least one of these will form a separating axis. Figure 4 illustrates an overlap test in one dimension.
In this example the xaxis forms a separating axis because
Note that the separating axis test will return true even if one box fully contains the other. A more general separating axis test is given in the section below on oriented bounding boxes (OBB’s). Listing 3 defines an AABB class that implements this overlap test.
Listing 3. An AABB class.
#include "vector.h"
// An axisaligned bounding box
class
AABB
{
public:
VECTOR P; //position
VECTOR E; //x,y,z extents
AABB( const VECTOR& p,
const VECTOR& e ): P(p), E(e){}
//returns
true if this is overlapping b
const bool
overlaps( const AABB& b ) const
{
const VECTOR T = b.P  P;//vector from A to B
return fabs(T.x) <= (E.x + b.E.x)
&&
fabs(T.y) <= (E.y + b.E.y)
&&
fabs(T.z) <= (E.z + b.E.z);
}
//NOTE:
since the vector indexing operator is not const,
//we must cast
away the const of the this pointer in the
//following
min() and max() functions
//min x,
y, or z
const SCALAR min( long i ) const
{
return ((AABB*)this)>P[i]  ((AABB*)this)>E[i];
}
//max
x, y, or z
const SCALAR
max( long i ) const
{
return ((AABB*)this)>P[i] + ((AABB*)this)>E[i];
}
};
For more information on AABBs and their applications, please see [8].
Just like spheres, AABBs can be swept to find the first and last occurrence of overlap. In Figure 5(a), A and B experienced displacements v_{a} and v_{b}, respectively, while Figure 5(b) shows B's displacement as observed by A.
Figure 6 shows B's displacement as observed by A.
In this example, the normalized times it took for the xextents and yextents to overlap are given by
and it can be seen that the xextents will cross before the yextents. The two boxes cannot overlap until all the extents are overlapping, and the boxes will cease to overlap when any one of these extents becomes disjoint. If u_{0,x}, u_{0,y}, and u_{0,z} were the times at which the x, y, and zextents began to overlap, then the earliest time at which the boxes could have begun to overlap was
Likewise, if u_{1,x}, u_{1,y}, and u_{1,z} are the times at which the x, y, and zextents become disjoint, then the earliest time at which the boxes could have become disjoint was
In order for the two boxes to have overlapped during their displacement, the condition
must have been met. Just like in the sphere sweep test, the positions of first and last overlap can be linearly interpolated with u. Listing 4 gives an implementation of this AABB sweep algorithm.
Listing 4. An AABB sweep algorithm.
#include "aabb.h"
//Sweep
two AABB's to see if and when they first
//and last
were overlapping
const
bool AABBSweep
(
const VECTOR& Ea, //extents of AABB A
const VECTOR& A0, //its previous position
const VECTOR& A1, //its current position
const VECTOR& Eb, //extents of AABB B
const VECTOR& B0, //its previous position
const VECTOR& B1, //its current position
SCALAR& u0, //normalized time of first collision
SCALAR& u1 //normalized time of second collision
)
{
const AABB A( A0, Ea );//previous state of AABB A
const AABB B( B0, Eb );//previous state of AABB B
const VECTOR va = A1  A0;//displacement of A
const VECTOR vb = B1  B0;//displacement of B
//the problem is solved in A's frame of reference
VECTOR v = vb  va;
//relative velocity (in normalized time)VECTOR u_0(0,0,0);
//first times of overlap along each axisVECTOR u_1(1,1,1);
//last times of overlap along each axis//check if they were overlapping
// on the previous frame
if( A.overlaps(B) )
{
u0 = u1 = 0;
return true;}
//find the possible first and last times
//of overlap along each axis
for( long i=0 ; i<3 ; i++ )
{
if( A.max(i)<B.min(i) && v[i]<0 )
{
u_0[i] = (A.max(i)  B.min(i)) / v[i];}
else if( B.max(i)<A.min(i) && v[i]>0 )
{
u_0[i] = (A.min(i)  B.max(i)) / v[i];}
if( B.max(i)>A.min(i) && v[i]<0 )
{u_1[i] = (A.min(i)  B.max(i)) / v[i];}
else if( A.max(i)>B.min(i) && v[i]>0 )
{
u_1[i] = (A.max(i)  B.min(i)) / v[i];}
}
//possible first time of overlap
u0 = MAX( u_0.x, MAX(u_0.y, u_0.z) );//possible last time of overlap
u1 = MIN( u_1.x, MIN(u_1.y, u_1.z) );//they could have only collided if
//the first time of overlap occurred
//before the last time of overlap
return u0 <= u1;
}
Dave Moss 
20 Mar 2012 at 11:23 am PST

The scaler u0 and u1 represent the time the objects will collide...



Petter Oberg 
Shouldn't SphereSphereSweep() return true only if 0 <= u0 <= 1, since else the collision does not occur within this frame?


