Normal vector of a cone pointing outward Consider a cone with height $h$ and maximum radius $R$.
Assume that the vertex of the cone is in $(0,0,0)$ and its axis is the $z$-axis.
For $(x,y,z)$ lying in the outside of the cone I want to determine the normal vector pointing away from the cone.
My plan is to use adjusted cylindrical coordinates:
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}z\frac{R}{h}\cos(\phi)\\z\frac{R}{h}\sin(\phi)\\z\end{pmatrix}=:\psi(\phi, z)$
and then calculating $\frac{\psi_\phi\times\psi_z}{\|\psi_\phi\times\psi_z\|_2}$ where $\times$ is the cross product:
$\psi_\phi\times\psi_z=(-z\frac{R}{h}\sin(\phi),z\frac{R}{h}\cos(\phi),0)\times(\frac{R}{h}\cos(\phi),\frac{R}{h}\sin(\phi),1)$.
And then using $x\times y:=\begin{pmatrix}x_2y_3-x_3y_2\\x_3y_1-x_1y_3\\x_1y_2-x_2y_1\end{pmatrix}$
Does this work this way?

Cylindrical coordinates are hopefuly not the most appropriate in this case.
It looks to me simpler to work with cartesian coordinates.
For this, we need the cartesian equation of this cone, which is :
$$f(x,y,z) := x^2+y^2-(Rz/h)^2=0$$
(why that ? If such a cone is cut by plane $z=z_0$, its horizontal "slices" are circles centered in $(0,0)$ with equation $x^2+y^2=(Rz/h)^2$ ; in particular, when $z_0=0$ we get a radius equal to $0$ ;  when $z_0=h$ we get a radius equal to $R$).
Therefore, the outward normal vector is 

$$(\partial f/\partial x, \ \partial f/\partial y, \ \partial f/\partial z)=\left(2x, \ 2y, \ \frac{2R^2z}{h^2}\right)$$

