Parametric curve resembling a bean.

Question

I am looking for a parametric closed curve that roughly resembles a bean.

I am looking for something with an explicit parametrization of the form $C(t) = (X(t), Y(t))$

I tried searching online but "parametric bean" is not yielding much of use.

Answer 1

A bit of experimentation got me to this:

It has the formula $$(\cos(t)-\frac{0.6}{1+9t^2},\sin(t))$$

Taken over the interval $[-\pi,\pi]$.

It has the disadvantage of not being periodic, but I don't know if that's a requirement, and it should be a relatively easy fix. It's also not easy to work with analytically, but I don't know what you're planning to do with it, so I don't know if that's important.

If you want it more curved you can go with: $$(\cos(t)-\frac{1}{1+4t^2},\sin(t))$$

Answer 2

HINT:

The domain inside is equivalent to the inside of a disk ( Riemann theorem)-- so try using a conformal map. Take a disk centered at the origin, dilate on the vertical and shift to the right. Now apply the complex map $z \mapsto z^2$, (even $z\mapsto z^n$). Because the square map doubles the argument the tall straight bean will wrap around the origin. Example:

$$t \mapsto (x^2-y^2, 2 x y)$$ where $(x,y)= (2 + \cos t , 3 \sin t)$ gets us a cashew nut.

$\bf{Added:}$ Should we want a filliform shape, like a snake, consider its spine ( a curve ), write as is the image of $P(t)$ over a segment ( $P$ analytic), then extend $P$ around the segment and consider the image of a flat ellipse approximating the segment).

$\bf{Added:}$ We can also do surfaces in $3D$, by deforming some ellipsoids. Here is an example of a surface, image of an ellipsoid under inversion. (Looks like a cashew nut, in 3D).

Answer 3

Here is a simple approach using connected circular arcs with (set apart parameter $t$ which is implicit), uses two parameters $a,b$ (which are the circles radii). For an interactive experience, move the sliders $a,b$ in this Geogebra animation here ; click on the "algebra" icon on the left if you want to see details.

The circular arcs are $P'P$ centered in $O$, $PR$ centered in $Q$, $RR'$ centered again in $O$, $R'P'$ centered in $Q'$.

Please note that any connecting point of two circular arcs is aligned with the centers of these arcs, warranting "smoothness" (differentiability) at the connecting point.

Answer 4

This interesting question about curve synthesis made me de-lurk. Here's my modest attempt:

$$\begin{align*}x&=3+2\sin t+\cos 2t\\y&=4\cos t-\sin 2t\end{align*}\quad t\in[0,2\pi]$$

I don't want to say too much about where it came from, except that it was part of a series of experiments on fudging around with hypotrochoids.