TITLE: Are the points moving around a sphere in this manner always equidistant?
QUESTION [34 upvotes]: I recently encountered this gif:

Pretend that there are visible circles constructed along the paths of the smaller black and white "discs", tracing how their individual centers move as they revolve around the center of the whole design. These circles together form an imaginary sphere inside the design.
Assuming that the exact centers of each smaller "disc" are points moving along the sphere, and that they move in perfect circles at the same rate, how does the distance between the points change over a single revolution?
Are they equally distant from each other at all times, or is there a period in which they grow closer, which appears to happen when both "discs" enter the holes of the opposite color?

REPLY [2 votes]: Ignore the YinYang pointed/holed plate (very nice demo !) as we want distance between white/black discs only. Watch demo for a short time. Motion is along equatorial and (prime) meridional circles.We see how longitude $\theta $ is independent of latitude $\phi.$ If radius of holes from origin $=a$,
White disc
$$ (x_w,y_w,z_w)= a( \cos \omega t ,\sin \omega t ,0)\tag1$$
Black disc
$$ (x_b,y_b,z_b)= a( -\cos \omega t,0 ,\sin \omega t) \tag2$$
Let $$ \omega t =\theta \tag3 $$
Distance between white and black discs $=d$
$$ d^2 = (x_w-x_b)^2 +(y_w-y_b)^2+ (z_w-z_b)^2 \tag4 $$
Plugging in
$$ d/a= \sqrt{2 (1+ \cos^2 \theta)} \tag5 $$
Maximum /minimum values are $(2a , \sqrt{2}a)$ at $ \theta = 0, \pi,...\, \theta= \pi/2, 3 \pi/2,... $ respectively.