Cone projected tangent angle

Question

A cone has a slope of 45 degrees.
The cone is projected on a plane that is inclined to the axis of the cone by x degrees. If x was 0, the projection would be 2 lines converging at 90(45 + 45) degrees to each other.

cone projection parallel to cone's axis

If x was 90 degrees, the projection would cover the infinite plane in all directions.

cone projection perpendicular to cone's axis

In fact, if x is anything greater than the cone's slope(45 degrees), the projection will completely cover the plane.

cone projection at roughly 80 degrees between projected plane and cone's axis

If x was 45 degrees, the projection would be a straight line because one side of the cone would be perfectly perpendicular to the projected plane.

cone projection with 45 degrees

If x is between 0 and 45, the projection will be 2 converging lines at a point representing the vertex of the cone.

cone projection around 20 degrees

Question 1: If the cone is projected at an angle of x between 0 and 45 degrees, what expression represents the angle between these converging lines?

illustration of projected tangent lines from the cone and angle in question

Question 2: If the cone's slope was represented by y degrees instead of being a constant 45, what expression would represent the angle between the converging lines for x between 0 and (90 - y) degrees?

Answer 1

Let $\alpha$ be the semi-aperture of the cone and $\theta$ (named $x$ in the question) the angle its axis forms with a given plane (see figure below). The projection of the cone onto the plane, for some values of $\theta$, is an angle whose sides are tangent to the projections of all the circular sections of the cone on the plane (twice angle $\delta$ in the figure).

If $ABCD$ is one such circular section of radius $h\tan\alpha$, its projection $A'B'C'D'$ is an ellipse with semi-axes $a=O'C'=OC=h\tan\alpha$ and $b=O'B'=OB\sin\theta=h\tan\alpha\sin\theta$, while the distance from (the projection of) the vertex to the centre of the ellipse is $y_0=VO'=h\cos\theta$.

It is then a straightforward exercise to find the slope of the tangents: $$ \tan\delta={a\over\sqrt{y_0^2-b^2}}= {\tan\alpha\over\sqrt{\cos^2\theta-\tan^2\alpha\sin^2\theta}}. $$

This formula works as long as the expression inside the square root is not negative, that is for $0\le\theta\le90{°}\!-\alpha$. For $\theta=90°\!-\alpha$ a generatrix of the cone is perpendicular to the plane and $\delta=90°$.

For $\alpha=45°$, in particular: $$ \tan\delta ={1\over\sqrt{\cos^2\theta-\sin^2\theta}} ={1\over\sqrt{\cos2\theta}}. $$

Answer 2

When axis is tilted by $\beta$ and considering reduction of projected length in the denominator (simple figure, not drawn) we have new vertical angle for $90^{\circ}$ vertex angle:

$$2 \tan^{-1}\sec \beta$$

For the general case (using $\alpha $ in place of $x$). Let the tan of vertical angle at cone ( base radius $r$, height $h$ ) vertex be $T$. Considering projections comparing tan of semi-vertical angles $$ \tan \alpha=\dfrac{r}{h}$$ when the axis of symmetry is tilted by $ \beta$ $$ \tan \alpha_1=\dfrac{r \cos \beta}{h}$$ Dividing $$ \sec \beta= \dfrac{\tan \alpha_1}{\tan \alpha}$$ Expressing half angle tangent $t$ in terms of the full angle tangent $T$ which is the projected angle between generators

$$\dfrac{{\dfrac{\sqrt{1+T1^2}-1}{T1}}}{{\dfrac{\sqrt{1+T^2}-1}{T}}} = \sec \beta, $$

an implicit equation between $ \tan^{-1} T_1 $ and $\tan^{-1} T. $