geometry
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.
EDIT
This seems to work for 2D:
$$r \leq \sin^3\theta+\cos^3\theta$$
This exercise can be understood as an application of a general result about perimeter bisectors of triangles.
Proposition. Given $\triangle ABC$ with incircle $\bigcirc I$ meeting the edges at $D$, $E$, $F$ as shown. If $F^\prime$ is the point opposite $F$ in $\bigcirc I$, and if $F^{\prime\prime}$ is the point where $\overleftrightarrow{CF^\prime}$ meets $\overline{AB}$, then $$|\overline{CA}|+|\overline{AF^{\prime\prime}}| = |\overline{CB}|+|\overline{BF^{\prime\prime}}| \tag{$\star$}$$ so that $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$.
Proof of Proposition. Let the perpendicular to $\overline{FF^\prime}$ at $F^\prime$ meet the edges of the triangle at $A^\prime$ and $B^\prime$. By tangent properties of circles, we have $$\overline{CE}\cong\overline{CD} \qquad \overline{A^\prime E}\cong\overline{A^\prime F^\prime} \qquad \overline{B^\prime D}\cong\overline{B^\prime F^\prime}$$ Consequently, $|\overline{CA^\prime}| + |\overline{A^\prime F^\prime}| = |\overline{CB^\prime}| + |\overline{B^\prime F^\prime}|$, so that $\overline{CF}$ is a perimeter bisector of $\triangle A^\prime B^\prime C$. The Proposition holds by the similarity of $\triangle ABC$ and $\triangle A^\prime B^\prime C$. $\square$
The Proposition has a helpful corollary.
Corollary. Given $\triangle ABC$ with incenter $I$ and perimeter bisector $\overline{CF^{\prime\prime}}$, if $M$ is on $\overline{AB}$ such that $\overline{IM} \parallel \overline{CF^{\prime\prime}}$, then $M$ is the midpoint of $\overline{AB}$.
Proof of Corollary. The points of tangency of the triangle with its incircle separate the perimeter into three pairs of congruent segments, marked $a$, $b$, $c$. Thus, the semi-perimeter of $\triangle ABC$ is $a+b+c$, and since $|\overline{BC}| = b+c$, it follows that $|\overline{BF^{\prime\prime}}| = a = |\overline{AF}|$. Thus, $\overline{FF^{\prime\prime}}$ lies between congruent segments. In $\triangle FF^\prime F^{\prime\prime}$, segment $\overline{IM}$ passes through the midpoint of one side ($\overline{FF^\prime}$) and is parallel to another ($\overline{F^\prime F^{\prime\prime}}$); it necessarily meets the third side ($\overline{FF^{\prime\prime}}$) at its midpoint, which must also be the midpoint of $\overline{AB}$. $\square$
To solve the original problem, it basically suffices to embed the above triangle into an ellipse:
In the above, the ellipse's foci are $C$ and $F^{\prime\prime}$, and $\overline{AB}$ is a chord through the latter. The fundamental nature of ellipses implies that $(\star)$ holds; therefore, $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$. Moreover, the reflection property of ellipses implies that normals at $A$ and $B$ bisect angles $\angle CAF^{\prime\prime}$ and $\angle CBF^{\prime\prime}$; therefore, the intersection of these normals is the incenter of $\triangle ABC$. The result follows by the Corollary. $\square$
Euler angles (and Tait-Bryan angles) are evil. They can be defined in a number of ways, and even when people remember to specify the axes and the order they are rotated, it all gets too hairy too fast. Instead, I show how to easily deal with this using only elementary vector algebra.
Let's say you have an ellipsoid centered at $\vec{c}$ with semi-principal axes $\vec{r}_1 $, $\vec{r}_2$, and $\vec{r}_3$: $$\vec{r}_1 = \left [ \begin{array}{c} r_{11} \\ r_{21} \\ r_{31} \end{array} \right ], \qquad \vec{r}_2 = \left [ \begin{array}{c} r_{12} \\ r_{22} \\ r_{32} \end{array} \right ], \qquad \vec{r}_3 = \left [ \begin{array}{c} r_{13} \\ r_{23} \\ r_{33} \end{array} \right ]$$
The surface of the ellipsoid passes through points $\vec{c} \pm \vec{r}_1$, $\vec{c} \pm \vec{r}_2$, and $\vec{c} \pm \vec{r}_3$.
The semi-principal axes are all perpendicular: $\vec{r}_1 \cdot \vec{r}_2 = 0$, $\vec{r}_1 \cdot \vec{r}_3 = 0$, and $\vec{r}_2 \cdot \vec{r}_3 = 0$.
The axis-aligned bounding box for the above ellipsoid is defined by $$\begin{array}{c} x = c_x \pm \sqrt{r_{11}^2 + r_{12}^2 + r_{13}^2} \\ y = c_y \pm \sqrt{r_{21}^2 + r_{22}^2 + r_{23}^2} \\ z = c_z \pm \sqrt{r_{31}^2 + r_{32}^2 + r_{33}^2} \end{array}$$
Note that this applies also to the case where the homogenous transformation matrix $\mathbf{M}$ that transforms the unit sphere to the ellipsoid is $$\mathbf{M} = \left [ \begin{array}{cccc} r_ {11} & r_{12} & r_{13} & c_x \\ r_{21} & r_{22} & r_{23} & c_y \\ r_{31} & r_{32} & r_{33} & c_z \\ 0 & 0 & 0 & 1 \end{array} \right ]$$
We can describe the transformation of the unit sphere to the above ellipsoid using $$\vec{p}(u, v, w) = \vec{p}(\vec{t}) = \vec{c} + u \vec{r}_1 + v \vec{r}_2 + w \vec{r}_3 = \vec{c} + \mathbf{R} \vec{t} \tag{1}\label{1}$$ where $\vec{t} = ( u , v , w )$ fulfills $$\left\lVert \vec{t} \right\rVert = 1 \qquad \iff \qquad u^2 + v^2 + w^2 = 1$$ on the surface of the ellipsoid, and $$\left\lVert \vec{t} \right\rVert \lt 1 \qquad \iff \qquad u^2 + v^2 + w^2 \lt 1$$ within the ellipsoid, and matrix $\mathbf{R}$ has the semi-principal axes as column vectors, $$\mathbf{R} = \left [ \begin{array}{ccc} \vec{r}_1 & \vec{r}_2 & \vec{r}_3 \end{array} \right ] = \left [ \begin{array}{ccc} r_{11} & r_{12} & r_{13} \\ r_{21} & r_{22} & r_{23} \\ r_{31} & r_{32} & r_{33} \end{array} \right ]$$
Essentially, we scale the unit sphere by the semi-principal axes, rotate it to the new orientation, and translate the center.
The above operation is invertible, if the ellipsoid is not degenerate; that is, if $\left\lVert\vec{r}_1\right\rVert \gt 0$, $\left\lVert\vec{r}_2\right\rVert \gt 0$, and $\left\lVert\vec{r}_3\right\rVert \gt 0$.
Since rotation matrices are orthonormal, its inverse is its transpose. To counteract the scaling, we scale by the inverses of the semi-principal axes.
The scaling vectors are now row vectors of the transformation matrix, with lengths inverse to the corresponding semi-principal axis length; i.e. $$\vec{s}_n = \left ( \frac{\vec{r}_n}{\left\lVert \vec{r}_n \right\rVert^2} \right)^{T}, \qquad n = 1, 2, 3$$ so that $$\left\lVert \vec{s}_n \right\rVert = \left\lVert \vec{r}_n \right\rVert^{-1}, \qquad n = 1, 2, 3$$ i.e. $$\vec{s}_1 = \frac{\vec{r}_1}{\left\lVert\vec{r}_1\right\rVert^2} = \left [ \begin{array}{ccc} s_{11} & s_{12} & s_{13} \end{array} \right ]$$ $$\vec{s}_2 = \frac{\vec{r}_2}{\left\lVert\vec{r}_2\right\rVert^2} = \left [ \begin{array}{ccc} s_{21} & s_{22} & s_{23} \end{array} \right ]$$ $$\vec{s}_3 = \frac{\vec{r}_3}{\left\lVert\vec{r}_3\right\rVert^2} = \left [ \begin{array}{ccc} s_{31} & s_{32} & s_{33} \end{array} \right ]$$ $$s_{i j} = \frac{r_{j i}}{ r_{1 i}^2 + r_{2 i}^2 + r_{3 i}^2 }, \qquad i = 1, 2, 3; \; j = 1, 2, 3$$ $$\mathbf{S} = \left [ \begin{array}{ccc} s_{11} & s_{12} & s_{13} \\ s_{21} & s_{22} & s_{23} \\ s_{31} & s_{32} & s_{33} \end{array} \right ]$$
Now, $$\vec{t}(\vec{p}) = \vec{s}_1 \cdot ( \vec{p} - \vec{c} ) + \vec{s}_2 \cdot ( \vec{p} - \vec{c} ) + \vec{s}_3 \cdot ( \vec{p} - \vec{c} ) = \mathbf{S}\left(\vec{p} - \vec{c}\right) \tag{2}\label{2}$$
Essentially, $\eqref{1}$ and $\eqref{2}$ are the inverse of each other.
If you insist on using Euler or Tait-Bryan angles or other similar angular convention, I recommend you use them to calculate the semi-principal axes $\vec{r}_1$, $\vec{r}_2$, and $\vec{r}_3$ only, and then apply the above vector solution.
Best Answer
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.
What do you want to do with that curve now? Give it another indentation on the opposite side?
$r = \cos^2(\theta) + 1$
More indented with $r = 2\cos 2\theta + 4$
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.