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Intersection distance of Ray and Plane


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#1 AODBAMF   Members   -  Reputation: 102

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Posted 06 June 2014 - 06:16 PM

Hi, I'm really confused about the returned values, for example: 

      Ray ray = new Ray(new Vector3(0, 12, 0), -Vector3.UnitY);
      Plane planeLow = new Plane(Vector3.UnitY, 1);
      Plane planeHigh = new Plane(Vector3.UnitY, 2);

      float distanceLow;
      float distanceHigh;

      Ray.Intersects(ray, planeLow, out distanceLow);
      Ray.Intersects(ray, planeHigh, out distanceHigh);

this returns 13 for distanceLow and 14 for distanceHigh, while I am expecting values of 11 and 10 respectively. The ray is at Y = 12 facing downwards, the planeLow is facing upwards with a distance of 1 to the origin, which is to me a distance of 11 along the ray where the ray and the plane intersects. When I lower the plane to -1 (so farther away from the ray origin), the value actually decreases to 11 (which I initially expected) instead of increasing to 13.

 

What am I missing? Thanks!

 

Edit: The solution appears to be that I simply just need to flip the D (distance) value of the plane temporarily while doing Intersection calculations, I just don't get why this is how it works, there must be a situation where it makes more sense this way?


Edited by AODBAMF, 06 June 2014 - 06:38 PM.


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#2 Buckeye   Crossbones+   -  Reputation: 5745

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Posted 06 June 2014 - 07:16 PM

Your constructor for the plane gives you planes NOT at +1 and +2, but -1 and -2. The second parameter of the constructor is the distance along the normal to the origin. The direction of the normal is important. That is, a plane with the normal facing +y points to the origin at a distance of +1 only if the plane is at y == -1.

 

Change your constructors to (UnitY, -1) and (UnitY, -2) and see if that changes things.

 

EDIT: for clarity, if you want a plane at y = +1, the distance to the origin along the normal = (0,1,0) is -1 (i.e., 1 unit in the negative Y direction). To put it another way, if you travel along the direction of the normal by the signed distance, you will arrive at the origin.

 

I think, perhaps, you assumed the second parameter was the distance from the origin to the plane. The second parameter is, in fact, the distance from the plane to the origin. Again, the direction is important.


Edited by Buckeye, 06 June 2014 - 08:12 PM.

Please don't PM me with questions. Post them in the forums for everyone's benefit, and I can embarrass myself publicly.


#3 AODBAMF   Members   -  Reputation: 102

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Posted 07 June 2014 - 04:14 AM

Alright thanks, I assumed it was the other way around smile.png


Edited by AODBAMF, 07 June 2014 - 04:18 AM.


#4 unbird   Crossbones+   -  Reputation: 5631

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Posted 07 June 2014 - 05:57 AM

I've seen people stumble upon this occasionally. I guess it's implemented this way because the evaluation of a (homogenous) point will now be a 4D dot product, i.e.

vx * nx + vy * ny + vz * nz + vw * d == dot(v, plane)

instead of

vx * nx + vy * ny + vz * nz- vw * d (this is not a dot product)

This makes a plane "shader compatible", e.g. manually clipping in a pixel shader is just:

clip(dot(position, plane));





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