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I am given a tetrahedron with the following points:

$$\begin{align} P_1 &= (2,0,1)\\ P_2 &= (-1,1,1)\\ P_3 &= (1,0,2)\\ P_4 &= (3,1,4) \end{align}$$

and I am tasked with finding its center of mass, M.

My attempt: $$\vec{OM} = \frac{1}{3}(\vec{OP_1} + \vec{OP_2} + \vec{OP_3})$$

$\vec{OM_1}$ is the coordinate for the center of mass of the base and $\vec{OM_2}$ is the coordinate of the center of mass of the tetrahedron. To find the center of mass of the tetrahedron, I do the following:

$$\begin{align} \vec{OM_2} &= \frac{\vec{OM} - \vec{OP_4}}{2} \\ \vec{OM_2} &= \frac{\frac{1}{3}(2,1,4) - (3,1,4)}{2} \\ \vec{OM_2} &= (-\frac{1}{3}, -\frac{1}{6}, -\frac{8}{6}) \end{align}$$

This is wrong since the answer is: $$(\frac{5}{4}, \frac{1}{2}, 2)$$

What have I done wrong? I'd really appreciate tips / explanations exactly where I am wrong.

Hosein Rahnama
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user262493
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    Note that there is more "stuff" near $M$ than near $P_4$. The center of mass is going to be closer to $M$ than to $P_4$. I suggest just computing $\vec{OM} = \frac{1}{4}(\vec{OP_1} + \vec{OP_2} + \vec{OP_3} + \vec{OP_4})$. – Darth Geek Dec 28 '15 at 22:39
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    You want $\frac14(\vec{OP_1}+\vec{OP_2}+\vec{OP_3}+\vec{OP_4})$ – Empy2 Dec 28 '15 at 22:40
  • @Michael Oh alright thanks. Am I supposed to derive this formula myself?Cause I have never seen it. And why is my method not correct? – user262493 Dec 28 '15 at 22:41
  • You have $\frac16\vec{OP_1}+\frac16\vec{OP_2}+\frac16\vec{OP_3}-\frac12\vec{OP_4}$. This makes $P_4$ different from the other points, but they all play a similar role. Also, for an average, the coefficients should add to 1; your coefficients add to $\frac16+\frac16+\frac16-\frac12=0$ – Empy2 Dec 28 '15 at 22:44
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    You are working in three dimensions not two. It just happens that the centroid (centre of mass) of a simplex made of uniform material is the same as the centroid of a figure with equal masses at the vertices of the simplex. A simplex in three dimensions is a tetrahedron, which has four vertices. NB In your question you have a tetrahedron, but you say nothing about how the mass is distributed. Is it made of uniform material? – Mark Bennet Dec 28 '15 at 22:48
  • I have a suggestion for you. First try to find the centroid of a triangle with vertices $P_1$, $P_2$, and $P_3$ which will be $C=\frac 13 (P_1+P_2+P_3)$. Then generalize your method to obtain the centroid of the tetrahedron. It is suggestive that you obtain something like $C=\frac 14 (P_1+P_2+P_3+P_4)$ – Hosein Rahnama Dec 28 '15 at 23:13

2 Answers2

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Fact. Let $\{v_1,v_2,v_3,v_4\}$ be the vertices of a tetrahedron $T$. Then the center of mass of $T$ is $$ \overline x=\frac{1}{4}\bigl(v_1+v_2+v_3+v_4\bigr) $$ Details of this proof can be found here.

Applying the result to your tetrahedron gives \begin{align*} \overline x &= \frac{1}{4}\bigl(v_1+v_2+v_3+v_4\bigr) \\ &= \frac{1}{4}\bigl(2-1+1+3,0+1+0+1,1+1+2+4\bigr) \\ &= \frac{1}{4}(5,2,8) \end{align*} which matches your desired result.

Essentially, you need to weight each vertex evenly and you seem to have taken a weighted sum.

Community
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Brian Fitzpatrick
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suppose first we locate the origin of coordinates at the centroid. assume the position of the centroid is a linear function of the vertices. by symmetry such a function must be

$$ \tau(x_1+x_2+x_3+x_4) $$

if we uniformly expand the tetrahedron by a factor $\lambda$ by linearity the coordinates of the centroid are also multiplied by $\lambda$. however the centroid remains at the origin. hence in this system of coordinates we must have $$ \sum_1^4 x_k=0 \tag{1} $$ now shift the origin by $a$. in the new system of coordinates the centroid is $-a$, and the vertices are $x_k-a$

hence $$ \tau\sum_1^4 (x_k-a) =-a \tag{2} $$ (1) and (2) imply $\tau=\frac14$

David Holden
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