Shortest path on the surface of a cone

Points on the cone can written as

$p = ( s \sin \theta_c cos \phi, s \sin \theta_c sin \phi, H - s \cos \theta_c )$

where $\tan \theta_c = R/H = 5/20 = 1/4$ , from which $\cos \theta_c = \dfrac{1}{\sqrt{1+\frac{1}{16}}} = \dfrac{4}{\sqrt{17}}$

and $\sin \theta_c = \dfrac{1}{\sqrt{17}}$

Now if we unfold the surface of the cone onto a plane, the surface of the cone

is mapped into a circular sector, which can be described by polar

coordinates $(r, \psi)$ , that can obtained by the identities:

$r = s$

$r \psi = s (\sin \theta_c ) \phi$ (equal arc length)

hence, $\psi = ( \sin \theta_c ) \hspace{4pt} \phi$

So, let's compute the polar coordinates of the unfolded points A and B,

For point A, we have , $s_A \sin \theta_c = 5$ , from which $s_A = 5 \sqrt{17}$

and $\phi_A = 0$ , hence $\psi_A = 0$ .

For point B, a similar computation, reveals that $s_B = 2.5 \sqrt{17}$ , and $\phi_B = \frac{3}{4} \pi$ , hence $\psi_B = \dfrac{3}{4 \sqrt{17}} \pi$

Now the shortest path is the straight line segment connecting the unfolded A and B, and its length is given by

the distance fromula

$d = \sqrt{ (s_A \cos \psi_A - s_B \cos \psi_B )^2 + (s_A \sin \psi_A - s_B \sin \psi_B )^2 } \approx 13.18$