How can I mathematically describe the shape of an idealised bean? (In two dimensions and in threes dimensions)

At the moment I'm calling the shape I refer to an ellipse/ellipsoid on a curved major axis.


This seems to work for 2D: $$r \leq \sin^3\theta+\cos^3\theta$$


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    $\begingroup$ I bet you can make a cardoid look like a bean. $\endgroup$ Dec 12, 2012 at 7:32
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    $\begingroup$ For some reason, this made me think of the shape from this question $\endgroup$
    – davidlowryduda
    Dec 12, 2012 at 7:36
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    $\begingroup$ It is bean-shaped. $\endgroup$ Dec 12, 2012 at 7:47
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    $\begingroup$ What kind of bean? There are a number of different types. $\endgroup$ Dec 12, 2012 at 7:50
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    $\begingroup$ this seems to work for 2D $\quad r=\sin^3\theta+\cos^3\theta$ $\endgroup$ Dec 12, 2012 at 8:05

1 Answer 1


The problem is, we do not know exactly what you mean by "idealised bean". Can you describe the shape better? A closed curve shape with two opposite indentations? (To me that seems more like a peanut.)

Then you say in the comments that $r = \sin^3\theta + \cos^3\theta$ seems to work well.

enter image description here

What do you want to do with that curve now? Give it another indentation on the opposite side?

$r = \cos^2(\theta) + 1$

enter image description here

More indented with $r = 2\cos 2\theta + 4$

enter image description here

You also said "an ellipse/ellipsoid on a curved major axis." I'm not sure exactly what that looks like. If you can sketch the picture and show us, we'd have a better idea of the shape you're looking for.

BTW, to make 3D versions, usually we just rotate the 2D shape about some axis.

For reference, here are the equations for an ellipse:

$y = \pm b\sqrt{1 - \frac{x^2}{a^2}} \hspace{1cm}$ or $\hspace{1cm}r = \displaystyle \frac{ab}{\sqrt{a^2\sin^2\theta + b^2\cos^2\theta}}$ in polar coordinates.


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