I have a Helicoid described by the following parametric equations:
$$x = u\cos(v)$$ $$y = cv$$ $$z = u\sin(v)$$
The helicoid revolves around the y-axis:
Eliminating $u$ and $v$, we obtain the Cartesian form:
$$z = x\tan{\frac{y}{c}}$$
Rearranging for $y$ yields:
$$y = c \arctan{\frac{z}{x}} \qquad(1)$$
I have a line, described by a point and a direction:
$$p + \alpha n$$
I want to find the intersection distance $\alpha$ from the point to the helicoid along the direction $n$.
We know that at the intersection we have:
$$p_x + \alpha n_x = x$$ $$p_y + \alpha n_y = y$$ $$p_z + \alpha n_z = z$$
Substituting into $(1)$ we obtain:
$$p_y + \alpha n_y = c \arctan{\frac{p_z + \alpha n_z}{p_x + \alpha n_x}} \qquad(2)$$
As a quick check, I tried $p = (-2, 0, 1)$ and $n = (0, 1, 0)$ and obtained $\alpha \approx -0.464$. The point $p + \alpha n$ is shown on the image above, which looks good.
The equation $(2)$ is trivial to solve for cases where $n_x = 0$ and $n_z = 0$. How can I most efficiently solve this equation for the non-trivial case? Must I resort to the bisection or Newton methods?
You need to eliminate $u$ and $v$ from
$$\begin{align}p_x + \alpha n_x &= u\cos(v)\\ p_y + \alpha n_y &= cv\\ p_z + \alpha n_z &= u\sin(v).\end{align}$$
You have
$$\tan(v)=\frac{p_z + \alpha n_z}{p_x + \alpha n_x}=\tan\left(\frac{p_y + \alpha n_y}c\right).$$
This is a transcendental equation in $\alpha$, for which no closed formula can be found.
In general it has an infinity of solutions, and for large $\alpha$,
$$\frac{n_z}{n_x}\approx \tan\left(\frac{p_y + \alpha n_y+k\pi}c\right),$$ or
$$\alpha\approx\frac1{n_y}\left(c\arctan\left(\frac{n_z}{n_x}\right)-p_y+k\pi\right).$$