Thursday, February 22, 2018

A Theorem about Circles and a Volumizing Algorithm

 

A Circle Theorem


Take a circle of radius \(R\). Select a point \(A\) inside it a distance \(a\) from the center, with \(a < R\). From \(A\), construct \(N>1\) line segments starting from A and touching the circle, segment k touching the circle at \(P_k\), such that if \(a-b\equiv \pm 1 \mod N\), then \(\measuredangle P_aAP_b=2\pi/N\), that is all the segments are equally-angularly-spaced. Let \(d_k=\overline{AP_k}\). Then \[ \prod_{k=1}^{N}d_k=\prod_{k=1}^{N}\left ( a\cos \left ( \theta_0+\frac{2 k \pi}{N} \right )+\sqrt{R^2-a^2+a^2\cos^2 \left ( \theta_0+\frac{2 k \pi}{N} \right )} \right ) \\ \prod_{k=1}^{N}d_k=\prod_{k=1}^{N}\left (\sqrt{R^2-a^2}\exp\left ( \sinh^{-1}\left (\frac{a}{\sqrt{R^2-a^2}}\sin \left ( \theta'_0+\frac{2 k \pi}{N} \right ) \right ) \right ) \right ) \] Therefore \[ \sqrt[N]{\prod_{k=1}^{N}d_k}=\sqrt{R^2-a^2}\exp \left (\frac{1}{N}\sum_{k=1}^{N} \sinh^{-1}\left (\frac{a}{\sqrt{R^2-a^2}}\sin \left ( \theta'_0+\frac{2 k \pi}{N} \right ) \right ) \right ) \] It follows that, for N even, as the summation will cancel in every term, \[ \sqrt[N]{\prod_{k=1}^{N}d_k}=\sqrt{R^2-a^2} \] This also holds asymptotically, as the error approaches zero. it is generally not true for N odd.

Note that this much more widely generalizes the well-known geometric mean theorem. This can be seen as a consequence of the power of a point theorem.

A pleasant interpretation of this is that if we take a diametric cross-section of a sphere and choose a point on that disk, the height of the sphere above that point is the geometric mean of the legs of any \(2N>1\) equiangular, planar, stellar net connecting that point to the boundary of the disk.

A Related Volumizing Algorithm


This theorem suggests an algorithm for producing a 3D volume given a closed 2D boundary shape. If we assume the 2D shape is of a diametric cross-section, we simply apply the method detailed above to produce the height above that point. That is, for a given point inside the shape, we take an N-leg equiangular stellar net emanating from that point to the boundary of the shape. The height of the surface at that point is then the geometric mean of the N legs of that net.

This method ensures that circular shapes produce spherical surfaces. However, if N is low, for less regular boundary shapes, the resulting surface may be quite lumpy or sensitive to how the angles of each net are chosen. One solution, then, is simply to make N large enough. However, this may end up being computationally expensive.

In theory, it may be possible to find the asymptotic value: find all parts of the boundary shape visible from the given point, and find the integral of the log of the distance, sweeping over the angle. If the boundary is a polygon, this involves evaluating (or approximating) integrals of the form \[ \int\ln\left ( \sin(x) \right)dx \] Which have no general closed form in terms of elementary functions. However, we can evaluate certain cases. One easy example is that of an infinite corridor formed from two parallel lines. We find that the height profile is double that of a circular cylinder. It may be desirable, then, to determine another function to multiply by which will halve the heights of corridors but leave hemispheres undisturbed.

Below we give some visual examples of the results of the algorithm. The original 2D shapes are shown in red.

In order, an equilateral triangle, an icosagon, a five-pointed star, an almost-donut, an Escherian tesselating lizard, and a tesselating spider.

Sunday, February 18, 2018

Rotating Fluid

 
Suppose we have an infinitely tall cylinder of radius R, filled to a height H with an incompressible fluid. We then set the fluid rotating about the cylindrical axis at angular speed \(\omega\). Suppose we take a differential chunk of fluid on the surface, a radius r from the axis.
The resulting normal force will then be \(N=F_c+W\). This normal force, as the name suggests, will be normal to the fluid surface. It follows by simple geometry, that \[ \frac{dy}{dr}=\frac{F_c}{W}=\frac{r \omega^2}{g} \] From which it follows that the height of the surface at any radius will be given by \[ y=\frac{r^2 \omega^2}{2g}+C \] Let us define \[ \omega_0=2\sqrt{gH}/R \\ u=\omega/\omega_0 \] Given that the fluid is incompressible, we know that the total volume does not change. From this, we can determine that the height of the surface at any radius will be given by: \[ y(r)=2H\left ( ru/R \right )^2+\left\{\begin{matrix} H(1-u^2) \\ 2H(u-u^2) \end{matrix}\right. \, \, \, \, \, \, \, \begin{matrix} u \leq 1\\ u > 1 \end{matrix} \] The highest point on the liquid surface is then given by: \[ y_{\textrm{max}}=\left\{\begin{matrix} H(1+u^2)\\ 2Hu \end{matrix}\right. \, \, \, \, \, \, \, \begin{matrix} u \leq 1\\ u > 1 \end{matrix} \] If \(u > 1\), the center of the base of the cylinder is not covered by fluid. There is a minimum radius at which fluid can be found. This minimum radius is given by: \[ r_{\textrm{min}}=R\sqrt{1-\frac{1}{u}} \] If the fluid is of uniform density and of total mass M, then the moment of inertia of the rotating fluid is given by \[ I=\left\{\begin{matrix} \frac{MR^2}{2}\left ( 1+\frac{u^2}{3} \right )\\ MR^2\left ( 1-\frac{1}{3u} \right ) \end{matrix}\right. \, \, \, \, \, \, \, \begin{matrix} u \leq 1\\ u > 1 \end{matrix} \] Note for each of these piecewise functions, the functions and their first derivatives are continuous.