coordinate systemsgeometry
I have some polygons I would like to map onto the face of a cone.
I can see from this page that I can convert the points of the polygon to cylindrical coordinates, which is almost what I want.
How do I go about modifying the formulas to work for conical coordinates?
You're misreading the formulas in the Wiki article. As they are written, they only make sense for $n \ge 4$, since the last formula (for $x_n$) breaks the pattern. For $n=3$, the intended meaning is clearly what you have (up to renaming of the variables).
For $n=4$, there are several ways to set up coordinates. For example, as on Wikipedia: $$ \begin{align*} x_1 &= r \cos\theta_1 \\ x_2 &= r \sin\theta_1 \cos\theta_2 \\ x_3 &= r \sin\theta_1 \sin\theta_2 \cos\theta_3 \\ x_4 &= r \sin\theta_1 \sin\theta_2 \sin\theta_3 \end{align*} $$ But the following also works: $$ \begin{align*} x_1 &= r \cos\theta_1 \cos\alpha \\ x_2 &= r \sin\theta_1 \cos\alpha \\ x_3 &= r \cos\theta_2 \sin\alpha \\ x_4 &= r \sin\theta_2 \sin\alpha \end{align*} $$ And as $n$ grows, so does the number of the possibilities. You could classify them by drawing some kind of tree-like diagrams. (This is done in papers by Kalnins and Miller from the 1980s, where they classify coordinate systems in which the Hamilton–Jacobi equation from classical mechanics may be solved by separation of variables, but their general scheme is more complicated, since it involves not only spherical coordinates but also ellipsoidal and paraboloidal coordinates.)
Since $r=\sqrt{x^2+y^2+z^2},$ MathWorld page and your textbook say the same thing. And yes, $z=0$ if and only if $u=v.$
It may help to consider this diagram from the page http://mathworld.wolfram.com/ParabolicCoordinates.html:
This is a cross-section of the system of coordinates showing the curves of constant $u$ and constant $v$ in the $y,z$ plane. One can obtain the complete parabolic coordinate system in $\mathbb R^3$ by rotating the curves in this figure around the $z$ axis, with the result that the points with a given constant $u$ value are all on the surface of a paraboloid, likewise the points with a given constant $v$ value are all on the surface of a paraboloid.
Select an arbitrary positive constant $c$. In the $y,z$ plane, the parabola where $v=c$ is the mirror image (across the $y$ axis) of the parabola where $u=c$. The two parabolas intersect at two points on the $y$ axis. In the full coordinate system, rotating these figures around the $z$ axis, we get a paraboloid where $u=c$ and a mirror-image paraboloid where $v = c$, and the intersection of those paraboloids is a circle in the $x,y$ plane centered at the origin. The radius of that circle is $u^2 = v^2 = c^2$.
If you choose two different constant values of $u$ and $v$, say $u = c_1$ and $v=c_2$ (where $c_1\neq c_2$), the two paraboloids described by $u = c_1$ and $v=c_2$ still intersect in a circle, but the circle is not in the $x,y$ plane. Instead, the circle of intersection is in a plane parallel to the $x,y$ plane at some non-zero value of $z$.
In order to describe a specific point in $\mathbb R^3$ in parabolic coordinates, in addition to the coordinates $u$ and $v$ you need an angle $\theta$, which is an angle of the rotation that turned the parabolas into paraboloids. You can describe a point in the $x,y$ plane in polar coordinates, using $\theta$ as the angle and $u^2$ (or $v^2$, since they are the same in the $x,y$ plane) as the radius.
If you want curvilinear coordinates in two dimensions that are not just a reformulation of polar coordinates, you might want to try parabolic cylindrical coordinates instead (http://mathworld.wolfram.com/ParabolicCylindricalCoordinates.html) and simply delete the $z$ axis. The resulting coordinate system looks like the figure above, except that the vertical axis is labeled $x$ instead of $z$. If you would prefer to have your $x,y$ coordinates in a more conventional orientation (I would), simply reflect the figure through the line $x=y$ (that is, flip it over so the axes are where you want them). Note that there is an ambiguity in the coordinate system because the parabolas for $u=u_1$ and $v=v_1$ (for constants $u_1$ and $v_1$) intersect in two places, one with a positive $y$ coordinate and one with a negative $y$ coordinate. As long as you are looking within a region that is all on one side of the $x$ axis, this is not a problem.
Best Answer
Presumably you already have the formulas for converting from conical to rectangular coordinates as listed on the Wikipedia page for conical coordinates. You'll need to solve for r, μ, and ν in terms of x, y, and z to get your answer. I can't see offhand the easiest way to find a general formula, but if you're trying to find it for particular values of r, μ, and ν, it shouldn't be too hard.