During the calculations of the triangles with marching cubes, you should add the normal of each triangle to the used vertex normals. When all triangles has been defined the normals are normalized. (If you use OpenGL or Direct3D the API can do the normalization for you, probably faster than you yourself can do it.)

You of course need some way of storing the vertices and vertex normals so that they can be shared among the triangles. To figure out the correct way of storing the vertices you should think of that the cubes has 12 edges that are all shared among four cubes.

- WitchLord

The solution to that particular problem is that you add a fraction of the triangle normal depending on the angle that the two edges sharing the vertex make.

Let's say you want to compute the normal for vertex V0 in a triangle V0,V1,V2

You can compute the normal like this:

E1 = V1-V0
E2 = V2-V0
Normal = CrossProduct(E1,E2);

The length of this normal is dependant on the angle between the two edges E1 and E2. And also the length of the edges like this:

LengthOf(Normal) = LengthOf(E1)*LengthOf(E2)*sin(SmallestAngleBetween(E1,E2))

If you add this normal to the vertex directly without normalizing it first you'll almost have what you want. The larger triangles have more influence than than smaller. I suggest that you try this way as I believe it gives an acceptable approximation.

If you want it to be perfect then you need to compute the actual angle between the edges. Which is done with either

angle = asin(LengthOf(Normal)/LengthOf(E1)/LengthOf(E2))

or

angle = acos(DotProduct(E1,E2)/LengthOf(E1)/LenghtOf(E2))

The second is derived from the formula for the dot product:

DotProduct(E1,E2) = LengthOf(E1)*LengthOf(E2)*cos(SmallestAngleBetween(E1,E2))

In either case the determination of the angle requires a lot of computations (as you can see).

Then the normals should be summed together like this:

V0Normal += angle/2/pi*Normalize(Normal);

Does this make me a marching-cube-guru?

- WitchLord

Edited by - WitchLord on May 19, 2000 8:39:44 PM

whoa whoa wait.. cant you just use the first derivative? i havent looked into metaballs a whole lot, but as far as i understand, they're implicit surfaces and as such can be differentiated..


--
Float like a butterfly, bite like a crocodile.



Edited by - goltrpoat on May 21, 2000 10:42:00 AM

The metaballs can be described mathematically with implicit surfaces, but it is my belief that it would require a lot more computations to find the normal from the first derivative than the method I describe. At least if you are using marching cubes to produce the surfaces.

But then again I haven''t worked with metaballs myself either so I could be completely wrong.

- WitchLord

Why, thank you!

I can''t take credit for the solution though as I read it somewhere. I just can''t remember where.

- WitchLord