I can suggest some names, which I have learned from others who research these things.
- A 4-simplex, or 4D analogue of a triangle, is called a Pentachoron , describing a regular, 5-sided 4D polytope. Also called a 5-cell. These are the n-simplex.
There are 4 types of ring torus objects in 4D, which can be seen visualized here. The general name of hypertorus works well:
Spheritorus : sphere-bundle over the circle :$S^2$ x $S^1$ $$\left(\sqrt{x^2+y^2} -a\right)^2 +z^2+w^2 = b^2$$
Torisphere : circle-bundle over the sphere : $S^1$ x $S^2$
$$\left(\sqrt{x^2+y^2+z^2} -a\right)^2 +w^2 = b^2$$
3-torus : circle over circle over circle : $T^3$
$$\left(\sqrt{\left(\sqrt{x^2+y^2}-a\right)^2+z^2}-b\right)^2+w^2 = c^2$$
Tiger : circle-bundle over the flat 2-torus (Clifford torus)
$$\left(\sqrt{x^2+y^2} -a\right)^2 +\left(\sqrt{z^2+w^2} -b\right)^2 = c^2$$
As for the bar -> cylinder -> duocylinder, I'm not sure exactly what sequence you are using here. The best fit I can see is describing a specific bisecting rotation around an n-1 plane into n+1 dimensions. In this case, the next 5D shape is called a Cylspherinder , a cartesian product of a $D^2$ and $D^3$ (solid disk times solid sphere). But, you can also make a Spherinder (sphere prism, another type of 4D cylinder) from a rotation of a cylinder into 4D.
Cylinder: $\left|\sqrt{x^2+y^2} -z\right|+\left|\sqrt{x^2+y^2} +z\right| = a$
Duocylinder: $\left|\sqrt{x^2+y^2} -\sqrt{z^2+w^2}\right|+\left|\sqrt{x^2+y^2} +\sqrt{z^2+w^2}\right| = a$
Spherinder : $\left|\sqrt{x^2+y^2+z^2} -w\right|+\left|\sqrt{x^2+y^2+z^2} +w\right| = a$
Cylspherinder : $\left|\sqrt{x^2+y^2+z^2} -\sqrt{w^2+v^2}\right|+\left|\sqrt{x^2+y^2+z^2} +\sqrt{w^2+v^2}\right| = a$
I guess you can call these n-cylinders, but there are even more types of these than just product of n-balls and n-cubes. You can also include the product of n-ball and n-simplex as well. In fact, any shape with both flat and curved cells can fit into this group (product of 2-ball (and higher) with any genus-0 object)
Cyltrianglinder : $\left|\big||x|+2y\big|+|x| -2\sqrt{z^2+w^2}\right|+\left|\big||x|+2y\big|+|x| +2\sqrt{z^2+w^2}\right| = a$
A common way to do this is to find a planar perspective transformation that “warps” the rectangle into the image quadrilateral and then use its inverse to map points on the image back to the rectangle. There are software libraries that include this as standard functionality.
If you want to code this up for yourself, it’s not terribly difficult to work out the necessary mapping. Without going into the detailed derivation, a general planar perspective transformation can be represented (in homogeneous coordinates) by a matrix of the form $$
M=\pmatrix{m_{00} & m_{01} & m_{02} \\ m_{10} & m_{11} & m_{12} \\ m_{20} & m_{21} & 1},
$$ which corresponds to the mapping $$\begin{align}
x' &= {m_{00}x+m_{01}y+m_{02} \over m_{20}+m_{21}+1} \\
y' &= {m_{10}x+m_{11}y+m_{12} \over m_{20}+m_{21}+1}.
\end{align}$$ Given a set of corner-to-corner correspondences between a pair of quadrilaterals, the eight coefficients can by found by solving a system of linear equations.
If one of the quads is a rectangle aligned with the coordinate axes, this transformation can be built up in stages, which can be more convenient for implementation in software. Assuming that we have the matrix $A$ which maps from the unit square to the quadrilateral, the rectangle-to-quadrilateral transformation can be derived by composing it with suitable translation and scaling transformations: $$
M = A\cdot S\left(\frac1w,\frac1h\right)\cdot T(-x_{LL},-y_{LL}).
$$ The inverse map is then $$
M^{-1}=T(x_{LL},y_{LL})\cdot S(w,h)\cdot A^{-1} = \pmatrix{w&0&x_{LL} \\ 0&h&y_{LL} \\ 0&0&1 }\cdot A^{-1},
$$ where $(x_{LL},y_{LL})$ are the coordinates of the rectangle’s lower-left corner. Left-handed coordinate systems can be accommodated by throwing in the appropriate reflections, and skew rectangles can be dealt with by adding a rotation to the transformation cascade.
Now, it’s just a matter of finding the matrix $A$. Let the corners of the quadrilateral be given by the points $q_i'$, and map the corners of the unit square to them as follows: $$\begin{align}
(0,0)&\mapsto q_0' \\
(1,0)&\mapsto q_1' \\
(0,1)&\mapsto q_2' \\
(1,1)&\mapsto q_3'.
\end{align}$$ Solving the resulting system of equations is tedious, but straightforward. One form of the solution is as follows: $$\begin{align}
a_{00} &= a_{20}x_1'+\Delta x_{10} \\
a_{10} &= a_{20}y_1'+\Delta y_{10} \\
a_{20} &= {\omega(\Delta_{10},\Delta_{32})+\omega(\Delta_{20},\Delta_{32})-\omega(\Delta_{30},\Delta_{32}) \over \omega(\Delta_{31},\Delta_{32})} \\
a_{01} &= a_{21}x_2'+\Delta x_{20} \\
a_{11} &= a_{21}y_2'+\Delta y_{20} \\
a_{21} &= -{\omega(\Delta_{10},\Delta_{31})+\omega(\Delta_{20},\Delta_{31})-\omega(\Delta_{30},\Delta_{31}) \over \omega(\Delta_{31},\Delta_{32})} \\
a_{02} &= x_0' \\
a_{12} &= y_0', \\
\end{align}$$ where $\Delta_{ij}=q_i'-q_j'$, $\Delta x_{ij}$ and $\Delta y_{ij}$ are the corresponding coordinate deltas, and $\omega$ is the symplectic form $\omega(\mathbf u, \mathbf v) = u_xv_y-u_yv_x$. (That this looks like a lot of cross products is no coincidence—there’s a geometric interpretation of the solution as the intersection of various planes.)
Since you’re mainly interested in mapping from a quadrilateral to a rectangle, it might be more efficient, and probably more computationally stable, to compute $A^{-1}$ directly from the corner coordinates. It’s similar to the solution for the other direction, but I’ll leave that calculation to you.
Best Answer
there is a shape called a stadium. you could call this shape a half stadium.
Wikipedia link to the stadium shape https://en.wikipedia.org/wiki/Stadium_(geometry)#:~:text=A%20stadium%20is%20a%20two,%2C%20obround%2C%20or%20sausage%20body.