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# Optimization on collision detection between oriented bounding box

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6 replies to this topic

### #1sobeit  Members

Posted 27 September 2013 - 05:22 PM

Hi there,

I want to check out with your guys see if my thought is right on the optimization of collision detection of OBB.

Since I'm using separating axis theorem, so the axes I need to test on include normals of very faces of each OBB, and vectors generated by a cross product of two edges, one from each OBB. My thought is that since every box has 3 pairs of parallel faces, so I can reduce the number of separating axes from each box from 6 to 3. And all edges of a box are only in 2 directions, can I reduce the number of the second type separating axis to 4?

Am I right? Is there any other optimization?

### #2Dirk Gregorius  Members

Posted 27 September 2013 - 07:48 PM

In 2D:

Here 4 axes are correct. You only need to test the two distinct face normals of each box.

In 3D:

You have three distinct face directions one ach box and you need to test each of these -> 2 x 3 = 6 axes

You have three distinct edge directions on each box and you need to test each combination -> 3 x 3 = 9 axes

So for a SAT between two OBB you need to test a total of 15 axes

I recommend looking at Christer Ericson and Gino v.d. Bergen books. They deal with this topic in detail and also explain numerical issues.

HTH,

-Dirk

Edited by Dirk Gregorius, 27 September 2013 - 07:50 PM.

### #3sobeit  Members

Posted 27 September 2013 - 09:49 PM

In 2D:

Here 4 axes are correct. You only need to test the two distinct face normals of each box.

In 3D:

You have three distinct face directions one ach box and you need to test each of these -> 2 x 3 = 6 axes

You have three distinct edge directions on each box and you need to test each combination -> 3 x 3 = 9 axes

So for a SAT between two OBB you need to test a total of 15 axes

I recommend looking at Christer Ericson and Gino v.d. Bergen books. They deal with this topic in detail and also explain numerical issues.

HTH,

-Dirk

Thanks, I'm afraid I can't afford the book right now. Do you know any method could generate the contact information after detecting collision?

### #4Dirk Gregorius  Members

Posted 28 September 2013 - 01:02 PM

You test all axes and keep track of the maximum separation. If the maximum is negative (e.g. the objects are penetrating) this is called the axis of minimum penetration.

1) The axis of minimum penetration is associated with a face: Let's call the face of this axis the reference face. Now find the most anti-parallel face on the other box. We call this face the incident face. Finally clip the incident face against the side planes of the reference face and keep all points below the reference face.

2) The axis of minimum penetration is associated with two edges: Compute the closest points between the two edges and use the average point

If you look for a good implementation of this I recommend looking at dBoxBox() in the ODE.

You will find good presentations about collision detection and constraint solving.

HTH,

-Dirk

### #5sobeit  Members

Posted 07 October 2013 - 03:05 AM

If you look for a good implementation of this I recommend looking at dBoxBox() in the ODE.

as you suggested, I looked into dboxbox.cpp. But i have trouble understanding it. could you help me out, thanks.

dMultiply1_331 (pp,R1,p);		// get pp = p relative to body 1


what is pp?

// Rij is R1'*R2, i.e. the relative rotation between R1 and R2
R11 = dCalcVectorDot3_44(R1+0,R2+0); R12 = dCalcVectorDot3_44(R1+0,R2+1); R13 = dCalcVectorDot3_44(R1+0,R2+2);
R21 = dCalcVectorDot3_44(R1+1,R2+0); R22 = dCalcVectorDot3_44(R1+1,R2+1); R23 = dCalcVectorDot3_44(R1+1,R2+2);
R31 = dCalcVectorDot3_44(R1+2,R2+0); R32 = dCalcVectorDot3_44(R1+2,R2+1); R33 = dCalcVectorDot3_44(R1+2,R2+2);


what is Rij? does R1' mean the transposition of R1? if it is, why it uses R1 directly?

  Q11 = dFabs(R11); Q12 = dFabs(R12); Q13 = dFabs(R13);
Q21 = dFabs(R21); Q22 = dFabs(R22); Q23 = dFabs(R23);
Q31 = dFabs(R31); Q32 = dFabs(R32); Q33 = dFabs(R33);


Why do we need absolute value of matrix?

### #6Dirk Gregorius  Members

Posted 07 October 2013 - 11:16 AM

p is the position and R is the orientation. Rij is the matrix element at row i and column j. You essentially transform the position and orientation of one box into the local space of the other. This makes one box axis aligned and simplifies the math.

I found this tutorial. Maybe this is helpful:

http://www.jkh.me/files/tutorials/Separating%20Axis%20Theorem%20for%20Oriented%20Bounding%20Boxes.pdf

### #7sobeit  Members

Posted 08 October 2013 - 01:39 AM

Thanks, the tutorial explains a lot.

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