Intersection of a line with an elliptical torus

The most general torus is best described by an ellipse with semi-axes $a$ and $b$, centred at $(d,0)$ in the r-z plane ( $r = \sqrt{x^2 + y^2}$), which is then rotated about the central axis of the torus (along the z-direction). The equations describing the torus then become:

$$\frac{z^2}{b^2} + \frac{r'^2}{a^2} = 1$$ (13.6)

$$x^2 + y^2 = r^2 = (r' + d)^2$$ (13.7)

where $r'$ is a dummy variable. Finding the intersections with the line will then naturally give a quartic in $t$ of the form

$$c_4t^4 + c_3t^3 + c_2t^2 + c_1t + c_0 = 0$$ (13.8)

where

$$c_4 = (1 + \beta^2w^2)^2$$

$$c_3 = 4(1+\beta^2w^2)((\underline{u}.\underline{R})+ z_0w\beta^2)$$

$$c_2 = 4((\underline{u}.\underline{R})+ z_0w\beta^2)^2 + 2(1 + \beta^2w^2)(R^2+z_0^2\beta^2-a^2-d^2)$$

$$+4\left(\frac{ad}{b}\right)^2w^2$$

$$c_1 = 4((\underline{u}.\underline{R})+ z_0w\beta^2)(R^2 + z_0^2\beta^2 - a^2 - d^2) +8\left(\frac{ad}{b}\right)^2wz_0$$

$$c_0 = (R^2 + z_0^2\beta^2 - a^2 -d^2)^2 - 4a^2d^2 +4\left(\frac{ad}{b}\right)^2z_0^2$$ (13.9)

and

$$\beta^2 = \frac{a^2 - b^2}{b^2}$$ (13.10)

Using the method of Cashwell and Everett [1], equation 13.8 may be solved to find all real solutions. The normal to a surface described by the equation $F(x,y,z) = 0$ can be shown to be $\nabla F$. From this, the (unnormalised) normal to an elliptical torus is found to be:

$$\underline{n} = \underline{i}x$$

$$+ \underline{j}y$$

$$+ \underline{k}\left(a\sqrt{1-\frac{z^2}{b^2}} + d\right) \frac{az}{b\sqrt{b^2-z^2}}$$ (13.11)

In the limit of $z \longrightarrow \pm b$, $\hat{\underline{n}} \longrightarrow \pm \underline{k}$.