Geodesic on a right circular cone problem.

Question

Is there a solution available to solve geodesic on a right circular cone problem?

We are given a cone with diameter $D$ and height $H$. The center of the base of the cone is at $(x=0,y=0,z=0)$ and the cone point is at $(0,0,H)$, The geodesic goes through $(0,R,0)$ and $(0,-R,0)$ on the cone base.

• a. Determine the arc length formula for the geodesic.
• b. Determine the equation for the geodesic.

Let $ L= \sqrt{R^2 + H^2}$ be the radius of the circle or slant radius of cone

and let $(\theta, c,s) $ be the (central angle in radians , chord length, arc length) respectively. Then, are the following OK?

$$s = \theta * L, s = \pi * R, $$

$$ \theta = \pi R / L,\quad c = 2 L* \sin(\pi * R / 2L ).$$

Answer 1

We just have to consider that a right circular cone is isometric with a circle sector. If we assume that the depicted cone has base radius $R$ and height $h$, by "unfolding" it we get a circle sector for a circle having radius $\sqrt{R^2+h^2}$, with arc length $2\pi R$. By "unfolding" the depicted geodesic between two opposite points on the base of the cone, we have a chord in a circle with radius $\sqrt{R^2+h^2}$ and the length of the corresponding arc is $\pi R$. It easily follows that the length of the depicted geodesic is

$$ 2\sqrt{R^2+h^2}\sin\left(\frac{\pi R}{2\sqrt{R^2+h^2}}\right). $$

When $h\to 0^+$, the length of such path tends to $2R$ (not to $\pi R$) and there is nothing strange in that: if the height of the cone is small, it is faster to reach the opposite point by going through the vertex of the cone, instead of making a half-turn around the base.

Answer 2

If you cut the cone surface along a ray emanating from the apex, the cone is isometric to a circle sector, so the length of the geodesic is the length of a chord in a circle.

The geodesic between (x=0, y=R, z=0) and (x=0, y=-R, z=0) on the cone base appears as a straight red line in the attached picture.

Attached is a picture of the "unfolding" of the right circular cone.

Answer 3

Of course there is a solution without Maple. You just have to use geodesic polar coordinates on the surface of the cone. They are like usual polar coordinates in a disk, just here the disc has a cone singularity at its center and its angle is $\Theta = \frac{2\pi R}{\sqrt{R^2+H^2}}$ is radians, assuming that $2R = D$ in your post. According to my brief (and maybe not too precise) calculations, the geodesic polar coordinate representation of the surface of the cone is \begin{align} x &= \frac{R \, \rho}{\sqrt{R^2+H^2}}\cos\left(\frac{\theta \, \sqrt{R^2+H^2}}{R}\right)\\ y &= \frac{R \, \rho}{\sqrt{R^2+H^2}}\sin\left(\frac{\theta\, \sqrt{R^2+H^2}}{R}\right)\\ z &= H - \frac{H \, \rho}{\sqrt{R^2+H^2}} \end{align}
where the coordinate parameters are $(\theta,\rho) \, \in \, [0, \Theta]\times(0, \sqrt{R^2+H^2}]$ As the straight lines in polar coordinates are $$\rho = \frac{c_0}{a_0 \cos{\theta} + b_0 \sin{\theta}},$$ the geodesics on the cone are \begin{align} x(\theta) &= \frac{R \, c_0}{(a_0 \cos{\theta} + b_0 \sin{\theta})\sqrt{R^2+H^2}}\,\cos\left(\frac{\theta \, \sqrt{R^2+H^2}}{R}\right)\\ y(\theta) &= \frac{R \, c_0}{(a_0 \cos{\theta} + b_0 \sin{\theta})\sqrt{R^2+H^2}} \,\sin\left(\frac{\theta\, \sqrt{R^2+H^2}}{R}\right)\\ z(\theta) &= H - \frac{H \, c_0}{(a_0 \cos{\theta} + b_0 \sin{\theta})\sqrt{R^2+H^2}}. \end{align}

Answer 4

An exact way is .... if radius at closest point to axis of symmetry near cone vertex is $r_{min}$ and cone semi-vertical angle

$$ \alpha = \sin ^{-1} \frac{R}{\sqrt { R^2+H^2}}, $$

then

$$ r_{min} = r \cos ( \theta\cdot \sin \alpha) $$

Arc lengths and geodesics are obtained by integration using :

$$ ds^2 = dr^2 + dz^2 + ( r d \theta)^2 $$ and using Clairaut's Law:

$$ r_{min}= r^2 \cdot \frac{d \theta }{ ds } = r \cdot \sin \psi $$

EDIT 1:

which gives

$$\frac{d \psi}{ds} = - \sin \alpha = -\frac12, $$

say as an example of $30^0$ semi-vertical angle cone. Then we must have

$$ \psi= (\pi/4, \pi/2, 3 \pi/4 ) $$

respectively at points of (entry, $r_{min}$, exit) of geodesic line on the truncated cone. If R = 1, these radii are accordingly $( 1, 1/\sqrt 2, 1 ). $

Calculation of radius and arc length of cone /geodesic from middle position:

$$ dr / \sin \alpha = r d \theta \cot \psi$$

$$ dr / \sin \alpha = r d \theta \sqrt{(r/r_{min} )^2-1 }$$ Integrating , integration constant =0 with central boundary condition

$$ r = r_{min} \sec ( \theta \sin \alpha ); z = r \cot \alpha = r_{min}\cdot \cot\alpha\, \sec ( \theta \sin \alpha ) \tag {1} $$

which is the required equation of polar projection of geodesic.

Considering differential lengths/angles

$$ d \theta /ds = \sin \psi /r = r _{min} /r^2 $$

plug in from above and integrate

$$ s/r_{min} = \tan (\theta \, \sin \alpha ) / \sin \, \alpha \tag{2} $$

which is general equation of (semi) arc length of geodesic arc.

For $ \alpha =0 $, or $ \pi/2 $ we have $ s = r _{min} \tan \theta $ as we can expect.

Again plugin from (1) for semi-arc lengthFor

$$ s /r = \sin ( \theta \sin \alpha )/\sin \alpha $$

At the requiredparticular diametrical opposite position $ \theta = \pi/2 $ the full arclength $$ = \sqrt{R^2+H^2}\sin ( \pi R/(2 \sqrt{R^2+H^2})) \tag {3} $$

However it is more convenient to retain $(r-\theta, s- \theta)$ forms.