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## Tracking edges in time and detect orientation flip of normal

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### #1Dirk Gregorius  Members

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Posted 08 February 2013 - 06:37 PM

Given two vectors in a 3D space. Both are simulated independently. You can think of them as edges on a collision shape of a rigid body. I need to detect when they cross over each other such that the orientation (sign) of the cross product changes. To make this less abstract here is an example:

Given two edges with vertices u1,v1 and u2,v2. At each frame I build the edges e1 = v1 - u1 and e2 = v2 - u2 and the normal n = e1 x e2. Over the time interval the edges can cross over each other creating a discontinuity in the normal direction. I am looking for a way to detect this discontinuity.

This is in the context of rigid bodies, but it might be a pretty common problem in cloth continuous collision detection. All ideas are very welcome. There is also a singularity when the edges become parallel, but this is not important initially.

Thanks,

-Dirk

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Posted 08 February 2013 - 06:50 PM

If the normal flips the direction the dot product from this frame to the previous frame of the normal will be negative...
"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

### #3Dirk Gregorius  Members

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Posted 08 February 2013 - 07:03 PM

That might be one solution, but I am not keeping track of this. I am evaluating the cross product at arbitrary times. Mostly at t = 0 and t = 1. You can rotate around one of the edges and between these extreme frames the dot product between n(0) and n(1) might have flipped.

Edited by Dirk Gregorius, 08 February 2013 - 07:10 PM.

### #4DT....  Members

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Posted 08 February 2013 - 10:21 PM

Given two edges with vertices u1,v1 and u2,v2. At each frame I build the edges e1 = v1 - u1 and e2 = v2 - u2 and the normal n = e1 x e2. Over the time interval the edges can cross over each other creating a discontinuity in the normal direction. I am looking for a way to detect this discontinuity.

Instead of taking the cross product, maybe something along the lines of Gram-Schmidt would help...

float a           =  Dot( e1, e2 );

Vector3  e3  =  a * e2;

Vector3  e4  = e2 - e3;  // or e3 - e1

...and interpolate using e4( 0.0 ... 1.0 ) ?

### #5deekr  Members

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Posted 11 February 2013 - 01:52 AM

I don't think this problem is well-defined, or at least the language isn't all there. In general, the result of a cross-product is a vector. The "sign" doesn't make much sense in arbitrary 3D, without a reference. You mentioned a "normal direction", but what does this mean? The edges could be on two different faces, maybe on two different rigid bodies, maybe the normal direction is the screen, ...

However, it does sound like you are trying to compare the "orientation" or "handedness" of these two vectors with respect to some direction, which makes sense to me. In this case, cross the two edges and then dot the result with the direction (do all of this in some consistent order). Positive means one way, negative means the other, zero is indeterminate.

edit: (assuming the two vectors are not degenerate)

Edited by deekr, 11 February 2013 - 01:54 AM.

### #6Dirk Gregorius  Members

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Posted 11 February 2013 - 10:24 AM

Thanks deekr. This is what I am thinking as well, but I didn't come up with some reference frame. I'd like to come up with some handness/orientation so I could run a quick determinant test.

Here is some context. I am trying to come up with a distance function between two edges. Essentially I use s = dot( SupportB( -n( t )) - SupportA( n( t ) ), n( t ) ) where n( t ) = cross( e1( t ), e2( t ) ) and e( t ) = R( t ) * e'. Then for continuous collision detection I want to find the root of this function.

This works great, but the discontinuity makes the root finder fail. When I plot the function you can get something like s(0.50) = 7, s(5.1) = 7.5, s( 5.2 ) = -8.0. T. I use a mix of bisection and Regula Falsi.

Ideas to get this better behaved?

Edited by Dirk Gregorius, 11 February 2013 - 10:27 AM.

### #7deekr  Members

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Posted 11 February 2013 - 10:58 AM

Sorry dirk, I missed the part in your original post where you were talking about the direction being something like n = e1 x e2.

If your difficulty is with giving a negative value to your root-finder algorithm, you could look for the roots of s^2. That function is always positive and has the same roots as s.

### #8Dirk Gregorius  Members

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Posted 11 February 2013 - 12:08 PM

Interesting idea, but I guess that will not work since the root finder expect the root to be bracketed.

Edited by Dirk Gregorius, 11 February 2013 - 12:11 PM.

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