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I am trying to find the intersection between the surface of revolution described by rotating a 3D line $\vec{r}_1(s)$ an angle $\theta$ around the $z$ axis and another 3D line $\vec{r}_2$. In parametric form, I know that the solution to this problem is described by the solutions to

\begin{equation} \begin{pmatrix} x_1(s)\cos{\theta} + y_1(s)\sin{\theta} \\ -x_1(s)\sin{\theta} + y_1(s)\cos{\theta}\\ z_1(s) \end{pmatrix} = \begin{pmatrix} x_2(t) \\ y_2(t) \\ z_2(t) \end{pmatrix} \end{equation}

I don't know how to manipulate the equations to find the values possible values of $s,t,\theta$.

I am sure that if $\vec{r}_1$ pases through the z-axis, rotating it describes a cone with the z-axis as axis and finding the intersection is straight forward. However, in the case where $\vec{r}_1(s)$ doesn't intersect the $z$ axis I am no entirely sure of the surface that it describes. It is not necesarrily a cone for example, if $\vec{r}_2$ is parallel to the z-axis, the surface is a circular cylinder.

Using cylindrical coordinates, I have found that I have found the following equation for the surface: let $\vec{q}_1 = \vec{r}_1(0)= (x_1,y_1,z_1)$ and $\vec{vq}_1 =\vec{r}_1(1) - \vec{r}_1(0) = (vqx_1,vqy_1,vqz_1)$:

\begin{equation} \rho^2 = (x_{1}^2 + y_{1}^2) + f(z)^2 (vqx_1^2 + vqy_1^2) \end{equation}

With $f(z) = (z - z_1)/(vqz_1)$

I am not sure how to use that equation. I guess I could substitute $\rho^2 = x_2^2(t) + y_2^2(t)$ and $z = z_2(t)$ and try to solve for $t$ in order to find the intersection but I am not sure this would work.

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  • $\begingroup$ These equations are of little use as long as you don't write the functions of $s$ and $t$ explicitly. $\endgroup$
    – user65203
    Commented Mar 29, 2021 at 20:59

2 Answers 2

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This is a comment, just too long to fit in the comment box.

Rotating a line segment $(x_0, y_0, z_0) - (x_1, y_1, z_1)$ (where $z_0 \ne z_1$) around the $z$ axis yields a surface of revolution $$r(z) = \sqrt{ Z_2 z^2 + 2 Z_1 z + Z_0 } \tag{1a}\label{G1a}$$ where $$\begin{aligned} Z_2 &= \displaystyle \frac{(x_1 - x_0)^2 + (y_1 - y_0)^2}{(z_1 - z_0)^2} \\ Z_1 &= \displaystyle \frac{(x_1 - x_0)(x_0 z_1 - x_1 z_0) + (y_1 - y_0)(y_0 z_1 - y_1 z_0)}{(z_1 - z_0)^2} \\ Z_0 &= \displaystyle \frac{(x_0 z_1 - x_1 z_0)^2 + (y_0 z_1 - y_1 z_0)^2}{(z_1 - z_0)^2} \end{aligned} \tag{1b}\label{G1b}$$


This is because each point on the line defines the radius for that $z$. Parametrising the line using $0 \le t \le 1$, $$\begin{aligned} x(t) &= (1-t) x_0 + t x_1 \\ y(t) &= (1-t) y_0 + t y_1 \\ z(t) &= (1-t) z_0 + t z_1 \\ r(t) &= \sqrt{x(t)^2 + y(t)^2} \\ \end{aligned}$$ and solving $z(t) = z$ for $t$ yields $$t = \frac{z - z_0}{z_1 - z_0}$$ Substituting this $t$ into $r(t)$ yields $\eqref{G1a}$ and $\eqref{G1b}$.

Note that you will be evaluating $r^2(z) = Z_2 z^2 + 2 Z_1 + Z_0$, really.


To calculate the intersection between a surface of revolution described by $r(z)$ and an (unrelated to above) line or line segment between points $(x_A, y_A, z_A)$ and $(x_B, y_B, z_B)$, we parametrise the line (segment) the same way as above, say using $\lambda$, $$\begin{aligned} x(\lambda) = (1 - \lambda) x_A + \lambda x_B \\ y(\lambda) = (1 - \lambda) y_A + \lambda y_B \\ z(\lambda) = (1 - \lambda) z_A + \lambda z_B \\ \end{aligned}$$ and the intersection(s) occur when $z(\lambda)$ is within the $z_0$ and $z_1$, and $$x(\lambda)^2 + y(\lambda)^2 = r^2\bigl(z(\lambda)\bigr)$$ It turns out this has zero, one, or two real algebraic solutions.

First, calculate $$\begin{aligned} D &= Z_2 (z_B - z_A)^2 - (x_B - x_A)^2 - (y_B - y_A)^2 \\ R &= Z_2 ( (x_A z_B - x_B z_A)^2 + (y_A z_B - y_B z_A)^2 - Z_0 (z_B - z_A)^2 ) \\ ~ &+ Z_1^2 (z_B - z_A)^2 - (x_A y_B - x_B y_A)^2 \\ ~ &+ 2 Z_1 ( (x_B z_A - z_B x_A) (x_B - x_A) + (y_B z_A - y_A z_B)(y_B - y_A) ) \\ ~ &+ Z_0 ( (x_B - x_A)^2 + (y_B - y_A)^2 ) \\ \end{aligned}$$ If $D = 0$ or $R \lt 0$, there is no intersection. Otherwise, there is one or two real solutions $$\lambda = \frac{S \pm \sqrt{R}}{D}$$ where $$S = x_A (x_B - x_A) + y_A (y_B - y_A) - (Z_1 + Z_2 z_A) (z_B - z_A)$$ Note that $0 \le \lambda \le 1$ is on the line segment between the points, $\lambda \lt 0$ is before point $(x_A, y_A, z_A)$, and $\lambda \gt 1$ is after point $(x_B, y_B, z_B)$.

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  • $\begingroup$ In the case when $z_0 = z_1$ and $z_A = Z_B$, the problem becomes 2D, but the equations are not reduced. Any idea what I could do in that case? $\endgroup$
    – sayeg84
    Commented Apr 2, 2021 at 1:16
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Hint.

Solving

$$ \cases{ \cos\theta x_1+\sin\theta y_1 = x_2\\ -\sin\theta x_1 + \cos\theta y_1 = y_2 } $$

for $\{\sin\theta,\cos\theta\}$ we have

$$ \cases{ \sin\theta = \frac{x_2y_1-x_1 y_2}{x_1^2+y_1^2}\\ \cos\theta = \frac{x_1x_2+y_1y_2}{x_1^2+y_1^2} } $$

so

$$ \left(\frac{x_2y_1-x_1 y_2}{x_1^2+y_1^2}\right)^2+\left(\frac{x_1x_2+y_1y_2}{x_1^2+y_1^2}\right)^2 = 1 $$

but $$ \cases{ (x_1,y_1,z_1) = p_{01}+\lambda_1\vec v_1\\ (x_2,y_2,z_2) = p_{02}+\lambda_2\vec v_2 } $$

so making the pertinent substitutions we get two equations and two unknowns

$$ \cases{ \left(\frac{x_2y_1-x_1 y_2}{x_1^2+y_1^2}\right)_{\lambda_1,\lambda_2}^2+\left(\frac{x_1x_2+y_1y_2}{x_1^2+y_1^2}\right)_{\lambda_1,\lambda_2}^2 = 1\\ z_1(\lambda_1) =z_2(\lambda_2) } $$

now depending on $p_{01},p_{02},\vec v_1,\vec v_2$ the ruled surface and the line can or cannot intersect. If them do not intersect the system of a quadratic an a linear equation will have not real solutions for $\lambda_1,\lambda_2$

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