View more

View more

View more

### Image of the Day Submit

IOTD | Top Screenshots

### The latest, straight to your Inbox.

Subscribe to GameDev.net Direct to receive the latest updates and exclusive content.

# Ray-Box Intersection

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

6 replies to this topic

### #1Agwan  Members

Posted 21 July 2007 - 02:11 PM

Hey, i'm trying to write a Ray-Box Intersection test, i've spent some time searching through google andf this site for a way to implement this. I found two pappers on the topic in google but have been unable to get them to work correctly. (Fast Ray Axis Aligned Bounding Box Overlap Tests with Plucker Coordinates, An Effcient and Robust Ray Box Intersection Algorithm). I'm looking for somthing like this.. Using C# DirectX 9.0c
bool RayBoxIntersection(Vector3 BoxMin, Vector3 BoxMax, Vector3 Origin, Vector3 End, out Vector3 hit ){}


Does anyone have a working solution that I could look at or be able to provide any information on the topic..

### #2scgames  Members

Posted 21 July 2007 - 03:40 PM

This has been discussed plenty as of late, so I'm going to be lazy and just post some code I have lying around:

template <class T> bool Math3D<T>::IntersectLineAABB(
const Vector3<T>& O,
const Vector3<T>& D,
const Vector3<T>& C,
const Vector3<T>& e,
T t[],
T epsilon)
{
int parallel = 0;
bool found = false;
Vector3<T> d = C - O;
for (int i = 0; i < 3; ++i)
{
if (Math<T>::Fabs(D[i]) < epsilon)
parallel |= 1 << i;
else
{
T es = (D[i] > (T)0.0) ? e[i] : -e[i];
T invDi = (T)1.0 / D[i];

if (!found)
{
t[0] = (d[i] - es) * invDi;
t[1] = (d[i] + es) * invDi;
found = true;
}
else
{
T s = (d[i] - es) * invDi;
if (s > t[0])
t[0] = s;
s = (d[i] + es) * invDi;
if (s < t[1])
t[1] = s;
if (t[0] > t[1])
return false;
}
}
}

if (parallel)
for (int i = 0; i < 3; ++i)
if (parallel & (1 << i))
if (Math<T>::Fabs(d[i] - t[0] * D[i]) > e[i] || Math<T>::Fabs(d[i] - t[1] * D[i]) > e[i])
return false;

return true;
}



This is old code, so no style comments please. Also, most people would probably arrange the logic a little differently (I can't really remember why I did it this way).

Beyond that though, the code should be correct, and can perhaps serve as a point of reference for you. There was another thread on this topic recently to which oliii posted some code as well, so you might look through the last few pages to see if you can find it.

Note that this is a line-AABB test, but it can easily be adapted for use with rays.

### #3Agwan  Members

Posted 21 July 2007 - 11:26 PM

Thanks but i'm also trying to understand the solution, and how it works.. any comments?

### #4Gage64  Members

Posted 21 July 2007 - 11:44 PM

This page has some info on various intersection tests.

### #5Agwan  Members

Posted 21 July 2007 - 11:49 PM

Also what's with the T's

T t[],
T epsilon)

T es = (D[i] > (T)0.0) ? e[i] : -e[i];
T invDi = (T)1.0 / D[i];

### #6scgames  Members

Posted 22 July 2007 - 04:50 AM

Quote:
 Original post by AgwanAlso what's with the T's...
It's a function template. If you're not familiar with templates (or don't need to parameterize the function) just replace the T's with 'float' (or whatever).

As for how it works, it goes something like this. You clip the line, in sequence, against each of the 'slabs' that make up the box (a 'slab' is, generally speaking, the space bounded by two parallel hyperplanes). So we start by clipping the line against the 'x axis' slab:
Before clipping:   |   | /   |   |/   |   |   |  /|-----/-----   |/  |-----------  /|   | / |   |/  |   |After clipping:   |   |     |   |    |   *   |  /|-----/-----   |/  |---*-------   |   |   |   |   |   |
Next we clip against the 'y axis' slab:
Before clipping:   |   |     |   |    |   *   |  /|-----/-----   |/  |---*-------   |   |   |   |   |   |After clipping:   |   |     |   |    |   |   |   |-----*-----   |/  |---*-------   |   |   |   |   |   |Leaving us with:   --*--   |/  |   *----
If the line hasn't been completely clipped away at this point, the endpoints of whatever remains (indicated by the *'s in this case) are the intersection points.

If you carefully step through the code I posted (and perhaps draw a diagram or two), I think you'll begin to see how the algorithm works.

### #7DeeFar  Members

Posted 22 July 2007 - 02:38 PM

Quote:
 Original post by Gage64This page has some info on various intersection tests.

Thanks for this huge information.

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.