How many points are required to make a black box

I have a black box with an arbitrary mass distribution inside it. I want to replace that object with n point masses without changing any mechanical properties of the box (center of mass, total mass, moment of inertia tensor). What is the fewest number of points I need and how can I find them?

I count 10 constraints. 3 from center of mass, 1 from total mass, and 6 from the moment of inertia tensor. 3 point masses give me 12 degrees of freedom so at first glance it seems the answer should be 3. In practice however it does not seem possible to represent some objects (a sphere for instance) with only coplanar points. At least 4 are required.

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Related: physics.stackexchange.com/q/64949. – dmckee Jun 5 '13 at 16:48
I tried a simple check using 3 point masses distributed on the apexes of an equilateral triangle and calculated the moment of inertia about 3 orthogonal axes through the centroid of the triangle. Two of the MoIs are the same but one differs, so it would seem that you're right about not being able to model a sphere in a black box with 3 point masses. But on the other hand at a glance your number of degrees of freedom and number of constraints look right to me... I'm stumped. +1 :) – Kyle Oman Jun 5 '13 at 19:30
Not sure why this is attracting close votes. The numbers are a necessary, concrete part of counting degrees of freedom, and this question actually gets at reasonably deep, general principles in classical mechanics. – Chris White Jun 6 '13 at 3:55
@ChrisWhite l feel that this might well be my fault because I first misread the question and wrongly flagged it. Happily it has 2 leave open statements now so that it will ptobably be ok. I apologize to Hammer and anybody else for my wrong flag. – Dilaton Jun 6 '13 at 4:43
Maybe there's some redundancy in those constraints. E.g. if you know the total mass, centre of gravity, and five of the inertia tensor's degrees of freedom, perhaps there's a way to deduce the sixth. If this is the case there would be only nine constraints and the answer would be four. However, if there is redundancy in those constraints, I haven't been able to see it yet. – Nathaniel Jun 11 '13 at 2:50

I have an argument which suggests that 4 points are sufficient, Please verify its correctness. The constraints that come from center of mass are easy to handle, by a simple translation.

The argument critically relies on the fact that scaling distances along any given principal axis does not change the direction of the principal axis. Then we proceed by scaling the distances in the system to fix the Momentum of Inertia.

Scaling x->a will leave $I_{xy} =0$ invariant, where x,y,z are chosen to be the principal axis. This can be verified, by writing down the off-diagonal terms of the moment of inertia matrix. $I_{xy}= -\Sigma m_i r_{xi} r_{yi} = 0$ and clearly scaling does not change $I_{xy}$

Now consider a regular Tetrahedron with each mass m = M/4(M is total mass) with arbitrary size. As it is a regular tetrahedron, moment of inertia, I is diagonal in any basis, so one is free to choose co-ordinates in any convenient way.

Now Consider the terms of the form $\Sigma m_i r_{xi}^2$,$\Sigma m_i r_{yi}^2$ and $\Sigma m_i r_{zi}^2$

We scale the system along its principal axis as x->ax, y->by,z->bz, The following equation determines a, $a^2 \Sigma m_i r_{xi}^2 = (I_{xx}+I_{yy}-I_{zz})/2$ Similarly for y,z

We now have a system consisting of 4 particles, and the desired moment on inertia. Lastly translate to fix the COM location.

As the OP demonstrated that 3 is too few degrees of freedom, 4 is necessary and sufficient.

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This seems to be the simplest way to show that four masses are sufficient. (+1) – user26872 Jun 11 '13 at 20:37
Good argument between you and oen you answered everything. – Hammer Jun 12 '13 at 17:02

Three masses are too few. Suppose we find three masses that satisfy the constraints. The three masses are coplanar. Go to a coordinate system in which the masses are in the $xy$ plane. Then the moment of inertia of the three masses is $$I = \left( \begin{array}{ccc} I_{xx} & I_{xy} & 0 \\ I_{xy} & I_{yy} & 0 \\ 0 & 0 & I_{xx}+I_{yy} \end{array} \right).$$ In particular, notice that $I_{zz}$ is not independent of $I_{xx}$ and $I_{yy}$.

Some details

In this coordinate system $z_i = 0$. Therefore, $$\begin{eqnarray*} I_{xx} &=& \sum m_i (y_i^2 + z_i^2) = \sum m_i y_i^2 \\ I_{yy} &=& \sum m_i (x_i^2 + z_i^2) = \sum m_i x_i^2 \\ I_{zz} &=& \sum m_i (x_i^2 + y_i^2) = I_{xx}+I_{yy} \\ I_{xy} &=& I_{yx} = -\sum m_i x_i y_i \\ I_{yz} &=& I_{zy} = -\sum m_i y_i z_i = 0 \\ I_{xz} &=& I_{zx} = -\sum m_i x_i z_i = 0. \end{eqnarray*}$$

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This is a really succinct way to show that three is impossible thank you (+1). Unfortunately I can't split the bounty between you and Prathyush and I think his answer solves the larger question – Hammer Jun 12 '13 at 17:02
@Hammer: Glad to help. No worries, Prathyush deserves the bounty. – user26872 Jun 12 '13 at 18:00

Yes it can be done with 6 point masses. 3 pairs of masses, placed along x-axis, y-axis and z-axis from the center of gravity and along the principal moment directions.

A mass $m$ with center of gravity $\vec{c} = (c_x,c_y,c_z)$ expressed along the principal inertia axes and mass moment of inertia tensor $I_c = {\rm diag}(I_{xx}, I_{yy}, I_{zz})$ is equivalent to the following arrangement:

1. Mass of $m/6$ placed at $\vec{r}_1 = (c_x+r_x, c_y, c_z)$
2. Mass of $m/6$ placed at $\vec{r}_2 = (c_x-r_x, c_y, c_z)$
3. Mass of $m/6$ placed at $\vec{r}_3 = (c_x, c_y+r_y, c_z)$
4. Mass of $m/6$ placed at $\vec{r}_4 = (c_x, c_y-r_y, c_z)$
5. Mass of $m/6$ placed at $\vec{r}_5 = (c_x, c_y, c_z+r_z)$
6. Mass of $m/6$ placed at $\vec{r}_6 = (c_x, c_y, c_z-r_z)$

with

$$r_x = \sqrt{ \frac{3 (I_{yy}+I_{zz}-I_{xx})}{2\;m} }$$ $$r_y = \sqrt{ \frac{3 (I_{zz}+I_{xx}-I_{yy})}{2\;m} }$$ $$r_z = \sqrt{ \frac{3 (I_{xx}+I_{yy}-I_{zz})}{2\;m} }$$

Proof

From the summation of masses and position it is easy to deduce that the center of gravity and mass is preserved.

To show that the inertia matrix matches we much show that

$$I_c = (\mbox{Inertia of 1}+\mbox{Inertia of 2})+(\mbox{Inertia of 3}+\mbox{Inertia of 4})+(\mbox{Inertia of 5}+\mbox{Inertia of 6})$$

$$I_c = (-\frac{m}{6} \begin{pmatrix} 0&0&0\\0&0&-r_x\\0&r_x&0\end{pmatrix}\begin{pmatrix} 0&0&0\\0&0&-r_x\\0&r_x&0\end{pmatrix}-\frac{m}{6} \begin{pmatrix} 0&0&0\\0&0&r_x\\0&-r_x&0\end{pmatrix}\begin{pmatrix} 0&0&0\\0&0&r_x\\0&-r_x&0\end{pmatrix})+(\ldots)+(\ldots)$$

$$\begin{pmatrix} I_{xx} & 0 & \\ 0 & I_{yy} & 0 \\ 0 & 0 & I_{zz} \end{pmatrix} = \begin{pmatrix} 0 & 0 & \\ 0 & \frac{m}{3} r_x^2 & 0 \\ 0 & 0 & \frac{m}{3} r_x^2 \end{pmatrix}+\begin{pmatrix} \frac{m}{3} r_y^2 & 0 & \\ 0 & 0 & 0 \\ 0 & 0 & \frac{m}{3} r_y^2 \end{pmatrix} + \begin{pmatrix} \frac{m}{3} r_z^2 & 0 & \\ 0 & \frac{m}{3} r_z^2 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

$$\begin{pmatrix} I_{xx} & 0 & \\ 0 & I_{yy} & 0 \\ 0 & 0 & I_{zz} \end{pmatrix} = \begin{pmatrix} \frac{m}{3} (r_y^2+r_z^2) & 0 & \\ 0 & \frac{m}{3} (r_z^2+r_x^2) & 0 \\ 0 & 0 & \frac{m}{3} (r_x^2+r_y^2) \end{pmatrix}$$

Which is solved when the relationships for $r_x$, $r_y$ and $r_z$ are used.

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This shows that you can do it with 6 points masses, but it doesn't prove that it is impossible with less than 6, does it? – Michiel Jun 6 '13 at 5:18
Very nice explanation. I actually got this far myself. I wrote myself some python code that can calculate the locations of those 6 points. I don't think it is an optimal solution though. – Hammer Jun 6 '13 at 5:51