# Infinite Double Cones in Three Coordinate Systems

Equations for infinite double cones in the Cartesian, cylindrical and spherical coordinate systems.

## Cartesian

We start with the equation for a circle centered at the origin, but instead of treating the radius as a constant, we make it a variable in the third dimension. $$x^2 + y^2 = z^2$$ All points $$(x, y, z)$$ that satisfy this equation are on a double cone. We can picture the double cone as an infinite number of circles parallel to the x-y plane. For each radius $$\sqrt{x^2 + y^2}$$ there are two circles, one at $$z = \sqrt{x^2 + y^2}$$ and one at $$z = -\sqrt{x^2 + y^2}$$. If we make a cut at $$y = 0$$, we get the following shape.

Another way of conceptualizing the double cone is as a line $$z = mx$$ with slope $$m = \tan \alpha$$ rotated about the z-axis. Our above equation uses the radius for z without any scaling, so $$m = 1$$. But in general, we have $$(mx)^2 + (my)^2 = z^2.$$

## Cylindrical

To get the cylindrical coordinates from Cartesian coordinates, we simply replace the x- and y-coordinates with polar coordinates and keep z. Each point then is described with a triple $$(r, \theta, z)$$, where $$r^2 = x^2 + y^2$$ and $$\tan \theta = y/x$$. That's very convenient for our double cone because we only need to describe how r relates to z. $$z = mr$$ As long as this equation holds, points $$(r, \theta, z)$$ with any angle $$\theta$$ are on the double cone.

## Spherical

Spherical coordinates are a triple $$(\rho, \phi, \theta)$$, where $$\rho$$ is the Euclidean distance of the (Cartesian) point $$P = (x, y, z)$$ from the origin, $$\phi$$ is the angle between the positive z-axis and the line through the origin and point $$P$$ (in the graph above $$\phi = \pi / 2 - \alpha$$, in general $$\tan \phi = \sqrt{x^2 + y^2} / z$$), and $$\theta$$ is the angle of the polar coordinates, as in the cylindrical coordinates. We already saw that we don't have any constraint on the angle $$\theta$$ and since the double cone is infinite, we also don't have a constraint on the Euclidean distance. So we can drop another variable from our equation and end up with the spherical equation using only a single coordinate. $$\tan \phi = 1/m$$