[vtkusers] Centroid of a surface

[Boettcher, Prof. Dr. Peter]

I recently asked for assistance for the calculation of the center of
mass of a triangle mesh. I found some information on the net in the
meantime, but still need assistance. 

 

As far as I understood, I am looking for the centroid being located at
the surface of my 3D mesh. In 2004 there was a similar discussion, but I
am not sure whether it will fix my problem.

 

Do you mean the center of mass of the object whose boundary is the
surface

you have triangulated? I am not sure what the centroid of a _surface_
would

be, unless you are thinking of a mass distribution concentrated on the

surface.

 

To compute the center of mass of the volume, you could use the fact that
the

volume integral can be written as a surface integral, and the surface

integral written as a line integral, and the line integral evaluated as
a

difference. 

 

For example for the x-coordinate of the center of mass, you want to

integrate x over the volume, ie the form x dx dy dz, and divide the
result

by the volume. 

 

To get the numerator, you could instead integrate x^2/2 dy dz over the

surface, because d(x^2/2 dy dz) = x dx dy dz, and to integrate
d(something)

over a set V is the same as integrating 'something' over the boundary of
V. 

 

Now use your triangulation and view this surface integral as a sum over

triangles of the integrals over each triangle. If your surface is

polyhedral, the triangles are flat and lie in an affine plane described
by a

linear equation, say ax + by + cz = d. If a = 0 then dy dz = 0 in the
plane

so this triangle contributes 0 to the sum; otherwise you can write x =
py +

qz + r (p = -b/a, q = -c/a, r = d/a) and you need to integrate (py + qz
+

r)^2/2 dy dz over the triangle. This is the same as integrating (py + qz
+

r)^3/(3*2*p) dz over the oriented edges of the triangle. 

 

Now use the same argument to lower the dimension again: each edge lies
in a

line described by another linear equation relating y and z; if the

coefficient of y is zero then dz = 0 on this line and the edge integral
is

0, otherwise y = mz + n say and the edge integral is the integral of

(p(mz+n) + qz + r)^3/(3*2*p) = ((pm+q)z + pn + r)^3/(3*2*p) over the

interval from z0 (the z coordinate of the start of the edge) to z1 (the
z

coordinate of the terminal point of the edge). 

 

This integral is ((pm+q)z + pn + r)^4/(4*3*2*p*(pm+q)) evaluated from z1
to

z0, from calculus. 

 

So for each triangle compute p,q,r; then for each edge compute m,n; then

accumulate the difference of the 4th degree polynomial above across

endpoints. 

 

You get the numerator for the y and z coordinates of the center of mass

similarly. You could just permute coordinates and use the above verbatim
but

perhaps it's more efficient to use d(yx dy dz) = y dx dy dz and follow
the

above recipe with appropriate changes, since you use the same data

p,q,r,m,n, z0, z1 then.  

 

The denominator is the volume, and you can follow the same steps (and do
the

computation simultaneously) with the form dx dy dz. You integrate x dy
dz

over the boundary, view it as a sum over triangles, get p, q, r, the for

each oriented edge get m,n and evaluate the resulting 3rd degree
polynomial

over the start and end z coordinates of the edge.

 

If you mean the center of mass of a uniform mass distribution
concentrated

on the surface, that would be the sum over triangles T of

area(T)*centroid(T) divided by the total surface area, ie the sum over T
of

area(T). [This is different from the center of mass of the volume
enclosed,

for example for the tetrahedron whose vertices are (0,0,0), (1,0,0),

(0,1,0), (0,0,1).]   

 

Reid Huntsinger

 

 

With kind regards, Peter.

 

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