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The Mathologer video Times Tables, Mandelbrot and the Heart of Mathematics discusses several relationships. For the n=2 and 3 cases, the cardiod and catacaustic (or nephroid per @Rahul's comment) curves are shown in reflections in coffee cups.

In the case of the cardiod the cup must be conical but presumably not cylindrical, for the catacaustic it must be a cylinder.

Question: If I wanted to construct these caustics using Blender, what are the specific constraints on the shapes of the coffee cups and directions of illumination? Can the former be any converging or diverging cone, and does the illumination direction need to have the same angle as that of the cone? Must the later be a perfect cylinder and the illumination only oblique?

"Bonus points" for a description of any possible cup and illumination configuration that could work for n=4.

Mathologer Times Tables, Mandelbrot Mathologer Times Tables, Mandelbrot

Mathologer Times Tables, Mandelbrot Mathologer Times Tables, Mandelbrot

Mathologer Times Tables, Mandelbrot Mathologer Times Tables, Mandelbrot

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    $\begingroup$ Cunning combination of classifying concepts in the caption! $\endgroup$
    – joriki
    Jul 23, 2018 at 7:48
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    $\begingroup$ "Catacaustic" is a general term encompassing both the cardioid and the curve in the second image. The latter is, specifically, a nephroid. $\endgroup$
    – user856
    Jul 23, 2018 at 9:40
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    $\begingroup$ If anyone wants the bounty, I found a lead that could be used but I'm not sure how. This webside has information about caustics and the nephroid and cardioid as caustics of the circle.\ $\endgroup$
    – Tbw
    Mar 9, 2020 at 22:38
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    $\begingroup$ @uhoh Thanks for asking such a cool question! Sadly it seems it isn't going to be answered $\endgroup$ May 24, 2021 at 21:13
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    $\begingroup$ @uhoh Yup I was wrong we already have a partial answer! On the lookout for more in the future $\endgroup$ May 25, 2021 at 6:01

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One nuisance is if the cylinder is too tall the rays won't reach the bottom. Similarly for the cone if a side blocks the light. If either cup is too full, the lighting won't trace long enough lines to touch the catacaustic. But tallness and fullness don't change the angle of the lines traced in the coffee.

The allowed height and fullness as a function of the cup angle and lighting is an additional calculation, maybe the more difficult part of the problem.

Neglecting this for now, for a cylinder consider the vertical lines $(x_0, y_0,z)$ on the side of the cup. If you think about it, you can convince yourself for any of these lines the light reflected off points from the line is a straight line in the coffee and independent of the downward angle of the light (sufficient it isn't horizontal or vertical).

So for a cylinder, any downward light angle (except vertical) traces a cardioid.

For a cone centered at the origin with slope $m$, first let the light be parallel to a line on the side of the cup like in the video.

Similarly consider the lines $\Big(t x_0, t y_0, m(t-1)\sqrt{x_0^2+y_0^2} \Big)$, given by varying $t$, on the sides of the cone. The light reflected off these lines also traces straight lines in the coffee because the traced lines are a combination of linear movements.

We claim these are the unique lines on sides which reflect light onto the tangents of the catacaustic. Since these lines partition the sides of the cup, any tangent of the catacaustic is traced by light reflecting off one of these lines. Uniqueness apparently holds since any open neighborhood of a point which reflects light onto a point on the catacaustic is folded upon reflection so all lines through the point except the one we picked out are projected onto a $v$.

Something important to realize is the lines in the coffee are not independent of $m$. Since a cardioid has specific line angles, there should be a unique $m$ which generates a true cardioid.

We will calculate it.

WLOG let the light be in the direction $(-1, 0, -m)$, cone described by $z = m \sqrt{x^2+y^2} - m$, and coffee at level $z=0$.

For $p = (t_0 x_0, t_0 y_0, m(t_0 -1))$ with $x_0^2 + y_0^2 = 1$ on the side, the vectors $(x_0, y_0, m)/\sqrt{1+m^2}, \ (m x_0, m y_0, -1)/\sqrt{1+m^2}$, and $(-y_0, x_0 , 0)$ are an orthogonal basis for $\mathbb{R}^3$. They are parallel, perpendicular, and horizontally tangent to the cone at $p$ respectively.

The light reflected off $p$ has "tangent component" and "parallel component" unchanged. However the perpendicular component is reversed. This give \begin{align} &= \Big((-1, 0, -m)\cdot (x_0, y_0, m)/\sqrt{1+m^2}\Big) (x_0, y_0, m)/\sqrt{1+m^2} - \Big((-1, 0, -m) \cdot (m x_0, m y_0, -1)/\sqrt{1+m^2} \Big) (m x_0, m y_0, -1)/\sqrt{1+m^2} + \Big((-1, 0,-m) \cdot (-y_0, x_0 , 0) \Big) (-y_0, x_0 , 0) \\ &=\left(\frac{(-1 + m^2)x_0^2- 2m^2 x_0}{1+m^2} - y_0^2, \frac{(-1+m^2) x_0 y_0 - 2m^2 y_0^2}{1+m^2} + x_0 y_0 , \frac{-2mx_0 -m^3 + m}{1+m^2} \right):= l \end{align} The reflected light at $p$ can be parameterized $p + t l$. It touches the coffee when \begin{align} &m(t_0 -1) + t \frac{-2mx_0 -m^3 + m}{1+m^2} = 0 \\ &\implies t = \frac{t_0-1}{-2x_0 -m^2 + 1} \end{align} Substituting in $p + tl$ $$ \left(t_0x_0 + \left(\frac{(-1 + m^2)x_0^2- 2m^2 x_0}{1+m^2} - y_0^2\right)\frac{t_0-1}{-2x_0 -m^2 + 1},t_0y_0 + \left(\frac{(-1+m^2) x_0 y_0 - 2m^2 y_0^2}{1+m^2} + x_0 y_0\right)\frac{t_0-1}{-2x_0 -m^2 + 1}, \ 0 \right) \ \ \ \ \ \ \ (1) $$

A necessary condition for a $m$ to give a cardioid is the chord at $(0,1,0)$ described by $(1)$ agrees with the chord for the cardioid at $(0,1,0)$. Wikipedia describes what this should be and it is the line $y = x+1$. (It is seen in the picture, but to use their formula let $\theta = \pi/4$, shift the circle back to the origin, and scale the radius to $1$).

Let $x_0 = 0, y_0 = 1$ in $(1)$ $$ \left(-\frac{t_0-1}{-m^2 + 1},t_0 + \frac{-2m^2}{1+m^2} \frac{t_0-1}{-m^2 + 1}, \ 0 \right) $$ Substitute $s:= \frac{t_0-1}{1-m^2}$ $$ \left(-s, (1-m^2) s + 1 + \frac{-2m^2}{1+m^2} s, \ 0 \right) $$ Set the $y$-component minus the $x$-component equal to $1$

$$ \left(2-m^2 - \frac{2m^2}{1+m^2}\right)s + 1 = 1 $$ i.e. $m^4 + m^2 - 2 = 0$. Since $m>0$ it follows $m=1$ as the unique slope which generates a cardioid.

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  • $\begingroup$ I'm not sure how it happened but it seems I lost track of this question back in May and stopped by here again just now out of curiosity. Wow! I will get right on this. Also, I will add a replacement bounty since I didn't award the first one completely. Thanks! $\endgroup$
    – uhoh
    Jun 28, 2021 at 15:05

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