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Simple Intersection Tests For Games
by miguel gomez []

October 18, 1999 Article Start Previous Page 3 of 7 Next

An Axis-Aligned Bounding Box (AABB) Sweep Test

Just like the name says, the faces of an axis-aligned 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 z-axes. 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 x-axis 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 axis-aligned bounding box

class AABB



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 va and vb, 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 x-extents and y-extents to overlap are given by

and it can be seen that the x-extents will cross before the y-extents. 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 u0,x, u0,y, and u0,z were the times at which the x, y, and z-extents began to overlap, then the earliest time at which the boxes could have begun to overlap was

Likewise, if u1,x, u1,y, and u1,z are the times at which the x, y, and z-extents 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 axis

VECTOR 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;


Article Start Previous Page 3 of 7 Next

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