# Cone projected tangent angle

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?

• Welcome to mse. What have you tried? Where are you stuck? Jun 21, 2020 at 4:07
• Andrei, I created a little web page to visualize the relationship between the angle x and the tangent projection angle and approximate the answers inefficiently. I used that to create the illustrations in my question. An answer to the following question should help: Given that a cone was cut at angle x, what would be the ratio of radii in the elliptical conic section? A cylindrical section should have ratio abs(cos x) but a cone is more complex. If I could get properties of an elliptical conic section at any distance away from the cone's vertex, would help.
– Josh
Jun 21, 2020 at 13:13

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: $$\tan\delta ={1\over\sqrt{\cos^2\theta-\sin^2\theta}} ={1\over\sqrt{\cos2\theta}}.$$

• Thanks a lot! That worked very well. I tested that with my interactive visualization and the angles line up perfectly through various cone slopes and various projection angles.
– Josh
Jun 21, 2020 at 22:22
• Thank you for this solution @intelligenti pauca. I'm wondering whether it is possible to generalise a solution including the cases, where the angle between the cone axis and the projected plane is equal or greater than 90 degrees. Any thoughts ? thanks a lot ! Nov 20, 2022 at 22:22
• @mysterium If $\theta=90°$ then the projection is the whole plane. The case $\theta>90°$ doesn't make sense: $\theta$ is the angle between the axis of the cone and the plane (i.e. with the projection of that axis on the plane) and cannot be obtuse. Nov 21, 2022 at 14:01
• Thank you for the reply @Intelligenti pauca. However, if one can imagine an ice-cream cone held upright with an extended arm, let's say at a geometry similar to the statue of liberty, and seen by their eyes, wouldn't the plane containing the eyes and the ice-cream cone axis (vertical in this case) make an obtuse angle ? in addition, if one solves for tan 𝛼 the above equation (going from a given projected half angle 𝛿 to the original half angle 𝛼 ), it seems that 𝜃 can be defined for angles >90°. Or I'm mistaken somewhere ? Nov 21, 2022 at 20:27
• @mysterium In that case angle $\theta$ is the supplementary of your angle. Formally, you could use your angle as well, because $\cos(\pi-\theta)=-\cos\theta$ and the formula for $\tan\delta$ contains only $\cos^2\theta$. Nov 21, 2022 at 21:03

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 $$\tan^{-1} T.$$