Intersection of a line with a frustum of cone

The equation describing a right circular cone of radius $r_{1}$ and height $h$ is given by $x^{2} + y^{2} = k^{2}z^{2}$, where $k = r_{1}/h$. The intersection of a line with this cone lead to a quartic in $t$ with solutions of the form:

$$t = \frac{- ((\underline{u}.\underline{R})- wz_0(1 + k^{2}))... ...)w^2)(x_0^2 + y_0^2 - k{^2}z_{0}^{2})}} {1 - (1 + k^{2})w^2}.$$ (13.3)

In fact, this solution is valid only for an unbounded, infinite cone, where the radius to height ratio at the point $z = h$ is given by $k$. For the case of the frustum of cone, the cone is bounded and if the intersection(s) is outside the planes defining the top and bottom of the frustum of cone then the intersection(s) with the plane(s) is given by $t=(z_{plane} - z_0)/w$. Provided $t$ is positive, this can be used to form a proposed solution $(x_0 + tu,y_0+tv, z_0+tw)$ where the condition for intersection is that $(x_0 + tu)^2+(y_0+tv)^2 \le a^2$. In this case, $a$ is equal to $r_{1}$ or $r_{2}$ for the case of the top or bottom plane respectively.

The normal to the curved sides of the frustum of cone is given by $\nabla F$, where $F(x,y,z) = 0$ is the equation of a cone. The (unormalized) normal to the cone at the point $x$, $y$ and $z$ is then:

$$\underline{n} = x \underline{i} + y \underline{j} - \frac{r_{1}}{h}\sqrt{x^{2}+y^{2}} \underline{k}.$$ (13.4)

Hence, the (signed) distance to the point $\underline{R} = (x_{0}, y_{0}, z_{0})$ is just

$$D = \frac{\underline{R} \cdot \underline{n}}{\vert\underline{n}\vert}$$ (13.5)

and is the nearest distance from the point $\underline{R}$ to the curved sides of the frustum of cone for the case were $z_{0}$ is between the planes defining the top and bottom of the frustum of cone.