Center of mass from shape boundary

Question

It is possible to find a shape center of mass by only its boundary? Would the average coordinates of X and Y would approximate my center of mass? (If it would work how good the approximation is going to be)

using average coordonates:

http://www.researchgate.net/post/How_can_I_find_the_centroid_of_the_shape_radius_signature_and_implement_DFT_equation_for_the_shape_in_MATLAB

Answer 1

For constant density and no holes, knowing $d\Omega$ tells everything: you can still exactly compute the center of mass with $\int\int\int_{P \rm{~inside~} d\Omega} \rho P ~dP $ (or the equivalent in polar form - with Jacobian - using $r <R(\theta,\phi)$ ).

Now you might want to relate the center of mass of the volume with the center of mass of the boundary itself (e.g., considered as a shell). But this is a different question ;-) . Here it would be approximative. How much really depends on the kind of shape and variations around mean R you allow. For moderate slopes and moderate variations, it's pretty good.

A very bad case would be to have infinitly thin very high walls on your planet: as a volume they wouldn't count, but as a surface they would.

Answer 2

$$\vec R_{CM}={{\iiint (x\hat {i}+y\hat {j}+z\hat {k})\rho(x,y,z)\,dV}\over{{\iiint \rho(x,y,z)\,dV}}}=\left({{\iiint x\rho(x,y,z)\,dV}\over{{\iiint \rho(x,y,z)\,dV}}},{{\iiint y\rho(x,y,z)\,dV}\over{{\iiint \rho(x,y,z)\,dV}}},{{\iiint z\rho(x,y,z)\,dV}\over{{\iiint \rho(x,y,z)\,dV}}}\right)$$ If you can find vector functions that: $$div(\vec f_1)=x\rho(x,y,z)\\div(\vec f_2)=y\rho(x,y,z)\\div(\vec f_3)=z\rho(x,y,z)\\div(\vec g)=\rho(x,y,z)$$ You get: $$\vec R_{CM}=\left({{\iint \vec f_1.d\vec S}\over{{\iint \vec g.d\vec S}}},{{\iint \vec f_2.d\vec S}\over{{\iint \vec g.d\vec S}}},{{\iint \vec f_3.d\vec S}\over{{\iint \vec g.d\vec S}}}\right)$$ From Stokes theorem. for example if density was constant (=2): $$\vec f_1=x^2\hat i,\,\,\vec f_2=y^2\hat j,\,\,\vec f_3=z^2\hat k,\,\,\vec g=2x\hat i$$