• FEATURED

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.

# Computing volume from marching cubes

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.

14 replies to this topic

### #1scientiphic  Members

Posted 13 January 2009 - 10:51 AM

Hey graphics-gurus, I am using the marching cubes algorithm to construct an isosurface from a 3d field of data. I have the triangular facets, etc. all done, but now I need to compute the volume enclosed in the iso-surface. I can think of one way to do this: using the triangular facets and the vertices of the cube, I create several tetrahedra and then simply calculate the volume of each tet. However, I noticed that as the number of facets in each cube increases, it can become non-trivial to automate the creation of the tets. Could someone point me to a reference that describes the algorithm to automate this process? Or has anybody created a table of vertices for all combinations of tets that are possible corresponding to each of the 256 cases of the MC tritable? Or is there a more efficient way to compute the volume using the MC information? Will greatly appreciate any help!

### #2sjhalayka  Members

Posted 13 January 2009 - 11:03 AM

I've been looking into this recently. You may want to check out:

Llamas I. Real-time voxelization of triangle meshes on the GPU. SIGGRAPH '07: ACM SIGGRAPH 2007 sketches; 2007

The presentation is found at:
http://developer.nvidia.com/object/siggraph-2007.html

### #3Pragma  Members

Posted 13 January 2009 - 01:59 PM

If you already have the triangles, computing the volume is easy. Just loop over all your created triangles, and sum up the quantity dot(v0, cross(v1,v2)) where v0, v1, v2 are the vertices of your triangle.
"Math is hard" -Barbie

### #4sjhalayka  Members

Posted 13 January 2009 - 05:20 PM

Quote:
 Original post by PragmaIf you already have the triangles, computing the volume is easy. Just loop over all your created triangles, and sum up the quantity dot(v0, cross(v1,v2)) where v0, v1, v2 are the vertices of your triangle.

Are there any limitations to the type of mesh one can use this on? e.g.: convex hulls only?

### #5scientiphic  Members

Posted 13 January 2009 - 06:52 PM

Quote:
 Original post by PragmaIf you already have the triangles, computing the volume is easy. Just loop over all your created triangles, and sum up the quantity dot(v0, cross(v1,v2)) where v0, v1, v2 are the vertices of your triangle.

Does your approach work even if the isosurface does not enclose the origin? I guess the expression you have suggested is for determining the volume of tets with one vertex at the origin...

In my case, I can have several closed isosurfaces in one data field (say like several spheres distributed in space - except that the geometry need not be as simple as spheres). So the origin need not necessarily be enclosed within the isosurface.

### #6Pragma  Members

Posted 14 January 2009 - 12:57 AM

The mesh does not need to be convex, and it doesn't need to enclose the origin. To prove to yourself that this is true, try drawing some examples in 2D on paper. You'll see that sometimes dot(v0, cross(v1,v2)) is negative, but that when all contributions are added up the negative terms cancel with even larger positive terms.

The only requirement is that your mesh has to be closed and your triangles have to be consistently oriented. This should happen automatically with data from marching cubes.

EDIT: It should actually be dot(v0, cross(v1,v2)) / 6

[Edited by - Pragma on January 14, 2009 7:57:48 AM]

### #7sjhalayka  Members

Posted 14 January 2009 - 04:17 AM

Quote:
 Original post by PragmaThe mesh does not need to be convex, and it doesn't need to enclose the origin. To prove to yourself that this is true, try drawing some examples in 2D on paper. You'll see that sometimes dot(v0, cross(v1,v2)) is negative, but that when all contributions are added up the negative terms cancel with even larger positive terms.The only requirement is that your mesh has to be closed and your triangles have to be consistently oriented. This should happen automatically with data from marching cubes.EDIT: It should actually be dot(v0, cross(v1,v2)) / 6

Well, this is great. Thank you for this. I'll do some testing today and see how it flies. I am concerned with things like quaternion Julia sets, which have all kinds of interesting surface properties. If anything fails, this will be it. Should have the results within an hour or two.

### #8sjhalayka  Members

Posted 14 January 2009 - 06:43 AM

Yeah, I'm a bonehead! Took me a second to make the link with determinants... Thanks for pointing this out.

[Edited by - taby on January 14, 2009 1:43:56 PM]

### #9scientiphic  Members

Posted 14 January 2009 - 07:30 AM

Quote:
 Original post by PragmaThe mesh does not need to be convex, and it doesn't need to enclose the origin. To prove to yourself that this is true, try drawing some examples in 2D on paper. You'll see that sometimes dot(v0, cross(v1,v2)) is negative, but that when all contributions are added up the negative terms cancel with even larger positive terms.The only requirement is that your mesh has to be closed and your triangles have to be consistently oriented. This should happen automatically with data from marching cubes.EDIT: It should actually be dot(v0, cross(v1,v2)) / 6

Thanks for clarifying this. However, I missed a detail in my original question that could complicate matters. The way my code works is I compute the volume corresponding to each closed isosurface at the level of the individual cells and then sum up the individual volumes by cycling through all the cells (this way it is just easier to run on parallel machines; each processor computes its portion of the volume. Furthermore, a closed isosurface may be shared by several processors.). In other words, if I take a marching cube, then I really need the volume formed by the intersection of the triangular facets with the cube. Once this is done, I store the contribution of each cube to the total volume in each cell (cube).

### #10Pragma  Members

Posted 14 January 2009 - 01:47 PM

Do you care about the volume contained in each cell? Or do you just need to know the total volume?

If it's the former then you'll need to use a different trick. But if it's the latter, you can compute dot(v0, cross(v1,v2)) / 6 for all the triangles in each cell and store them in the cell. Then you can add this up for all the cells. You can easily do this in parallel over multiple machines.

### #11scientiphic  Members

Posted 14 January 2009 - 02:07 PM

Quote:
 Original post by PragmaDo you care about the volume contained in each cell? Or do you just need to know the total volume?If it's the former then you'll need to use a different trick. But if it's the latter, you can compute dot(v0, cross(v1,v2)) / 6 for all the triangles in each cell and store them in the cell. Then you can add this up for all the cells. You can easily do this in parallel over multiple machines.

Yes, I see that your suggested approach will work regardless of whether the isosurface is on one processor or shared across several processors, and it should work perfectly fine if I only needed the total volume.

However, I do need the volume in each cell. One primary reason is that I do not a priori know the total number of closed isosurfaces in the data field. Once I am done calculating the "volume" present in each cell, I will then need to search through the entire data field for connected "volumes". To be more precise, I will need to compute the volume distribution (say for instance, the pdf of sphere volumes) of the closed isosurfaces in the data field.

I can think of one **laborious** procedure at the moment: construct a table (similar to the tritable in the mc-algorithm) that contains all the triangle facets that enclose the required volume in each cube (this would mean breaking up the relevant part of the cube faces into triangular facets). I could then use your suggested approach for the tabulated triangles in each cube. So essentially I will have a "closed" volume in each cube.

Do you have other suggestions?

### #12Pragma  Members

Posted 15 January 2009 - 02:05 AM

Nope, I'm out of ideas.

### #13XBTC  Members

Posted 28 February 2012 - 10:53 AM

Sorry for resurrecting this age old thread. Did you find a easy solution for calculating the volume per marching cube?

### #14FLeBlanc  Members

Posted 28 February 2012 - 01:06 PM

Considering the OP was last active in 2009, it is highly likely they will not see your 3 year later response asking them for clarification.

### #15Álvaro  Members

Posted 28 February 2012 - 01:15 PM

Isn't this basically trivial? In order to implement marching cubes you need to have some classification of cases, and in each case you generate triangles in some specific way. In each of those cases, you can compute what the volume contained in this cube is without too much trouble. Am I missing anything?

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.