Peeter Joot's (OLD) Blog.

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Archive for May 15th, 2010

Center of mass for a toroidal segment

Posted by peeterjoot on May 15, 2010

Motivation.

[Click here for a PDF of this post with nicer formatting]Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.

After seeing Iron Man II with Lance earlier, a movie with bountiful toroids, and now that the kids are tucked in, finishing up the toroidal center of mass calculation started earlier seems like it is in order. This is a problem I’d been meaning to try since reading this blog post for the center of mass of a toroidal wire segment

Center of mass.

With the prep done, we are ready to move on to the original problem. Given a toroidal segment over angle \theta \in [-\Delta \theta/2, \Delta \theta/2], then the volume of that segment is

\begin{aligned}\Delta V = r^2 R \pi \Delta \theta.\end{aligned} \hspace{\stretch{1}}(3.13)

Our center of mass position vector is then located at

\begin{aligned}\mathbf{R} \Delta V &= \int_{\rho=0}^r \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} \int_{\phi=0}^{2\pi} e^{-j\theta/2} \left( \rho \mathbf{e}_1 e^{ i \phi } + R \mathbf{e}_3 \right) e^{j \theta/2} \rho \left( R + \rho \sin\phi \right) d\rho d\theta d\phi.\end{aligned} \hspace{\stretch{1}}(3.14)

Evaluating the \phi integrals we loose the \int_0^{2\pi} e^{i\phi} and \int_0^{2\pi} \sin\phi terms and are left with \int_0^{2\pi} e^{i\phi} \sin\phi d\phi = i \pi /2 and \int_0^{2\pi} d\phi = 2 \pi. This leaves us with

\begin{aligned}\mathbf{R} \Delta V &= \int_{\rho=0}^r \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} \left( e^{-j\theta/2} \rho^3 \mathbf{e}_3 \frac{\pi}{2} e^{j \theta/2} + 2 \pi \rho R^2 \mathbf{e}_3 e^{j \theta}  \right) d\rho d\theta \\ &= \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} \left( e^{-j\theta/2} r^4 \mathbf{e}_3 \frac{\pi}{8} e^{j \theta/2} + 2\pi \frac{1}{{2}} r^2 R^2 \mathbf{e}_3 e^{j \theta}  \right) d\theta \\ &= \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} \left( e^{-j\theta/2} r^4 \mathbf{e}_3 \frac{\pi}{8} e^{j \theta/2} + \pi r^2 R^2 \mathbf{e}_3 e^{j \theta}  \right) d\theta.\end{aligned} \hspace{\stretch{1}}(3.15)

Since \mathbf{e}_3 j = -j \mathbf{e}_3, we have a conjugate commutation with the e^{-j \theta/2} for just

\begin{aligned}\mathbf{R} \Delta V &= \pi r^2 \left( \frac{r^2}{8} + R^2 \right) \mathbf{e}_3 \int_{\theta=-\Delta \theta/2}^{\Delta \theta/2} e^{j \theta} d\theta \\ &= \pi r^2 \left( \frac{r^2}{8} + R^2 \right) \mathbf{e}_3 2 \sin(\Delta \theta/2).\end{aligned} \hspace{\stretch{1}}(3.18)

A final reassembly, provides the desired final result for the center of mass vector

\begin{aligned}\mathbf{R} &= \mathbf{e}_3 \frac{1}{{R}} \left( \frac{r^2}{8} + R^2 \right) \frac{\sin(\Delta \theta/2)}{ \Delta \theta/2 }.\end{aligned} \hspace{\stretch{1}}(3.20)

Presuming no algebraic errors have been made, how about a couple of sanity checks to see if the correctness of this seems plausible.

We are pointing in the z-axis direction as expected by symmetry. Good. For \Delta \theta = 2 \pi, our center of mass vector is at the origin. Good, that’s also what we expected. If we let r \rightarrow 0, and \Delta \theta \rightarrow 0, we have \mathbf{R} = R \mathbf{e}_3 as also expected for a tiny segment of “wire” at that position. Also good.

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Volume element for a toroidal segment.

Posted by peeterjoot on May 15, 2010

[Click here for a PDF of this post with nicer formatting]Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.

Motivation.

In I love when kids stump me, the center of mass of a toroidal segment is desired, and the simpler problem of a circular ring segment is considered.

Let’s try the solid torus problem for fun using the geometric algebra toolbox. To setup the problem, it seems reasonable to introduce two angle, plus radius, toroidal parametrization as shown in the figure

toriodal parameterization

Our position vector to a point within the torus is then

\begin{aligned}\mathbf{r}(\rho, \theta, \phi) &= e^{-j\theta/2} \left( \rho \mathbf{e}_1 e^{ i \phi } + R \mathbf{e}_3 \right) e^{j \theta/2} \\ i &= \mathbf{e}_1 \mathbf{e}_3 \\ j &= \mathbf{e}_3 \mathbf{e}_2 \end{aligned} \hspace{\stretch{1}}(1.1)

Here i and j for the bivectors are labels picked at random. They happen to have the quaternion-ic property i j = -j i, which can be verified easily.

Volume element

Before we can calculate the center of mass, we’ll need the volume element. I don’t recall having ever seen such a volume element, so let’s calculate it from scratch.

We want

\begin{aligned}dV = \pm \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \left( \frac{\partial {\mathbf{r}}}{\partial {\rho}} \wedge \frac{\partial {\mathbf{r}}}{\partial {\theta}} \wedge \frac{\partial {\mathbf{r}}}{\partial {\phi}} \right) d\rho d\theta d\phi,\end{aligned} \hspace{\stretch{1}}(2.4)

so the first order of business is calculation of the partials. After some regrouping those are

\begin{aligned}\frac{\partial {\mathbf{r}}}{\partial {\rho}} &= e^{-j\theta/2} \mathbf{e}_1 e^{ i \phi } e^{j \theta/2} \\ \frac{\partial {\mathbf{r}}}{\partial {\theta}} &= e^{-j\theta/2} \left( R + \rho \sin\phi \right) \mathbf{e}_2 e^{j \theta/2} \\ \frac{\partial {\mathbf{r}}}{\partial {\phi}} &= e^{-j\theta/2} \rho \mathbf{e}_3 e^{ i \phi } e^{j \theta/2}.\end{aligned} \hspace{\stretch{1}}(2.5)

For the volume element we want the wedge of each of these, and can instead select the trivector grades of the products, which conveniently wipes out a number of the interior exponentials

\begin{aligned}\frac{\partial {\mathbf{r}}}{\partial {\rho}} \wedge \frac{\partial {\mathbf{r}}}{\partial {\theta}} \wedge \frac{\partial {\mathbf{r}}}{\partial {\phi}} &= \rho \left( R + \rho \sin\phi \right) {\left\langle{{ e^{-j\theta/2} \mathbf{e}_1 e^{ i \phi } \mathbf{e}_2 \mathbf{e}_3 e^{ i \phi } e^{j \theta/2} }}\right\rangle}_{3} \end{aligned} \hspace{\stretch{1}}(2.8)

Note that \mathbf{e}_1 commutes with j = \mathbf{e}_3 \mathbf{e}_2, so also with e^{-j\theta/2}. Also \mathbf{e}_2 \mathbf{e}_3 = -j anticommutes with i, so we have a conjugate commutation effect e^{i\phi} j = j e^{-i\phi}. Together the trivector grade selection reduces almost magically to just

\begin{aligned}\frac{\partial {\mathbf{r}}}{\partial {\rho}} \wedge \frac{\partial {\mathbf{r}}}{\partial {\theta}} \wedge \frac{\partial {\mathbf{r}}}{\partial {\phi}} &= \rho \left( R + \rho \sin\phi \right) \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \end{aligned} \hspace{\stretch{1}}(2.9)

Thus the volume element, after taking the positive sign, is

\begin{aligned}dV = \rho \left( R + \rho \sin\phi \right) d\rho d\theta d\phi.\end{aligned} \hspace{\stretch{1}}(2.10)

As a check we should find that we can use this to calculate the volume of the complete torus, and obtain the expected V = (2 \pi R) (\pi r^2) result. That volume is

\begin{aligned}V = \int_{\rho=0}^r \int_{\theta=0}^{2\pi} \int_{\phi=0}^{2\pi} \rho \left( R + \rho \sin\phi \right) d\rho d\theta d\phi.\end{aligned} \hspace{\stretch{1}}(2.11)

The sine term conveniently vanishes over the 2\pi interval, leaving just

\begin{aligned}V = \frac{1}{{2}} r^2 R (2 \pi)(2 \pi),\end{aligned} \hspace{\stretch{1}}(2.12)

as expected.

Center of mass.

With the prep done, we are ready to move on to the original problem … but math playtime has run out for the day, so that will come later.

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