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Take a right prism whose bases are regular $n$ polygons and twist it uniformly by some angle $\phi<\pi$. This surface can be seen as the trajectories of the sides of the polygon as it translates with velocity $v_z$ and twists with constant angular frequency $\omega$. After that its sides will no longer be planes, but rather (I think) ruled surfaces. I have two questions then:

1) What's the area of the lateral faces of the prism?

2) How to prove or disprove the lateral faces are ruled surfaces?

Edit: as suggested below, the lateral surface area should be the same. One possible way to prove it is something like: by analysing the twist deformation on a small patch of the surface with the top edge twisted in relation the bottom edge. One gets the top edge by evolving the bottom edge along helicoidal paths. Here's an image with one sketch of a such twisted prism: enter image description here

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    $\begingroup$ Note that this is a cube twisted around the z-axis, this does not affect the area of the sides. $\endgroup$ – User123456789 Sep 25 '18 at 14:23
  • $\begingroup$ Just see it as cutting up a box up in pieces of size $\Delta z$ and rotating each slab over a small angle proportional to the total rotation. If $\Delta z \rightarrow 0$, you get the obtained shape, but never change the total area of all slabs. $\endgroup$ – User123456789 Sep 25 '18 at 15:20
  • $\begingroup$ The rulings are horizontal (i.e., they're the edges you get it you slice the twisted shape with a plane orthogonal to the axis of twisting). $\endgroup$ – John Hughes Sep 25 '18 at 18:44
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    $\begingroup$ Nice sketch. The twisting does not change the volume (Cavalieri's principle) , but it does change the surface area. $\endgroup$ – Ethan Bolker Sep 28 '18 at 21:34
  • $\begingroup$ Thanks, but I got the sketch off the web. Yes I know it doesn't change volume. See two answers below for the change in area. $\endgroup$ – minmax Sep 28 '18 at 22:00
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We need a clean parametrization of one of these twisted sides. To this end put ${\pi\over n}=:\alpha$. Then one edge of the polygon in its initial position is given by $$u\mapsto(r\cos\alpha, r\sin\alpha\> u,0)\qquad(-1\leq u\leq1)\ .$$ The twisting with angular velocity $\omega=1$ turns this edge counterclockwise and at the same time moves it in the vertical direction with a certain speed $\lambda$. It follows that we obtain a helical band with the parametric representation $${\bf r}(u,t)=\bigl(r\cos\alpha\cos t-r\sin\alpha\>u\,\sin t,\>r\cos\alpha\sin t+r\sin\alpha\>u\,\cos t,\lambda\, t\bigr)$$ whereby $-1\leq u\leq 1$ and $t\geq0$. One then has to compute $$d\pmb\omega(u,t)={\bf r}_u(u,t)\times{\bf r}_t(u,t)$$ and in the sequel the scalar surface element $${\rm d}\omega(u,t)=\bigl|d\pmb\omega(u,t)\bigr|\ .$$ This is the quantity that has to be integrated over $-1\leq u\leq1$ and $0\leq t\leq T$, whereby $T$ has to be chosen according to your specifications. The resulting integral is elementary; but in any case: The claim that the area of the side is equal to the untwisted area is wrong.

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  • $\begingroup$ I agree with you. I added my calculation below. I didn't claim that fortunately. I said it was plausible though. I have to check the elementary integral below. I think I also proved below that the surface is ruled. $\endgroup$ – minmax Sep 28 '18 at 21:30
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Here is my attempt:

Let's take the edge between points $\vec r_1(0)=(1/2, -1/2, 0)$ and $\vec r_2(0)=(1/2, 1/2,0)$. This edge can be parametrized as $$ \vec r(s, 0)= (1-s)\vec r_1(0)+s\vec r_2(0)=\vec r_1(0)+s(\vec r_2(0)-\vec r_1(0))=(1/2, s-1/2, 0), $$ where $s\in[0,1]$. As this edge rotates on the $xy$ plane with angular velocity $\omega$ and translates along the $z$-axis, it is mapped on $$ \vec r(s, t)= \left[ \begin{array}{ccc} \cos(\omega t) & -\sin(\omega t) & 0\\ \sin(\omega t) & \cos(\omega t) & 0\\ 0 & 0 & 1 \end{array} \right]\left[ \begin{array}{c} 1/2\\ s-1/2\\ 0 \end{array} \right] + \left[ \begin{array}{c} 0\\ 0\\ v_z t \end{array} \right] $$ In cartesian coordinates we can write this as $$ \vec r(s, t)=\left(\frac{1}{2}\cos(\omega t)-(s-\frac{1}{2})\sin(\omega t), \frac{1}{2}\sin(\omega t)+(s-\frac{1}{2})\cos(\omega t), v_z t\right)=\vec r_1(t)+s(\vec r_2(t)-\vec r_1(t)). $$ We can show that \begin{align} \vec r_1(t) &= \left(\frac{\cos(\omega t)+\sin(\omega t)}{2},\frac{\sin(\omega t)-\cos(\omega t)}{2}, v_z t\right),\nonumber\\ \vec r_2(t) &=\vec r_1(t)+(-\sin(\omega t), \cos(\omega t), 0).\nonumber \end{align} Hence this proves the twisted surface is ruled. Let us now take two orthogonal differential displacement vectors tangent to the twisted prism surface: \begin{align} d\vec r_s &= \frac{\partial \vec r(s,t)}{\partial s}ds=\left(-\sin(\omega t), \cos(\omega t), 0\right)ds\\ d\vec r_t &= \frac{\partial \vec r(s,t)}{\partial t}dt=\left(-\frac{\omega}{2}\sin(\omega t)-\omega(s-\frac{1}{2})\cos(\omega t), \frac{\omega}{2}\cos(\omega t)-\omega(s-\frac{1}{2})\sin(\omega t), v_z \right)dt. \end{align} The differential area vector is given by the cross product between these two vectors $$ d\vec r_s\times d\vec r_t=\left(v_z\cos(\omega t), v_z\sin(\omega t), \omega(s-\frac{1}{2})\right)dsdt. $$ Taking the modulus of this $$ ||d\vec r_s\times d\vec r_t||=\sqrt{v_z^2+\omega^2(s-\frac{1}{2})^2}dsdt. $$ The area of the twisted face is then given by $$ A=\Delta t\int_0^1\sqrt{v_z^2+\omega^2(s-\frac{1}{2})^2}ds=\frac{\Delta t v_z^2}{\omega}\int_{-\omega/(2v_z)}^{\omega/(2v_z)}\sqrt{1+u^2}du= \frac{\Delta t v_z^2}{\omega}\left[\frac{\omega}{2v_z}\sqrt{1+\frac{\omega^2}{4v_z^2}}+\sinh^{-1}\left(\frac{\omega}{2v_z}\right)\right] $$ This shows though that the surface is actually stretched when it is twisted.

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