# Parametric curve resembling a bean.

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.

• Cardiods can look a bit like Lima beans Feb 19, 2023 at 7:08
• while this surely an odd question, one of the related questions from the sidebare might be useful to you: math.stackexchange.com/questions/4202315/… Feb 19, 2023 at 7:20
• Use splines : it is the most flexible tool for drawing curves of any shape. Feb 19, 2023 at 8:14
• @JeanMarie Splines are annoying in that you must manually place the control points for you to get anything. An explicit parametrization without the need for control points like the suggested answer is preferable. Feb 19, 2023 at 8:16
• This is quite frankly one of the best questions on this site. I am astounded by the sheer - elegance - of the bean drawing. Feb 19, 2023 at 9:39

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: $$(\cos(t)-\frac{1}{1+4t^2},\sin(t))$$

• For future convenience, here's the curve in Desmos with adjustable variables. Feb 19, 2023 at 23:10
• +1 for this nice answer Mar 10, 2023 at 23:45

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).

• Excellent idea. Feb 20, 2023 at 6:50
• @Jean Marie: Thank you for the kind words! I wonder if we can take a 3D shape and try to write a certain mapping of the sphere ( preferably polynomial) that resembles the image. It's not easy since we need to manipulate so many parameters and we do not understand well the effect on the image. Neural networks maybe? Feb 20, 2023 at 7:06
• If I understand it correctly this solution is both periodic and analytic, two features the currently accepted solution is lacking. So if these are required this is the way to go. Feb 20, 2023 at 14:19

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.

\begin{align*}x&=3+2\sin t+\cos 2t\\y&=4\cos t-\sin 2t\end{align*}\quad t\in[0,2\pi]
• Nice! Could be equivalently written as $3ie^\vartheta + ie^{-\vartheta} + e^{2\vartheta}$, with $x$ and $y$ as real and imaginary parts. This is of course a (very short) Fourier series. And any shape could be represented arbitrarily close as a Fourier series. Feb 20, 2023 at 12:13