Are the points moving around a sphere in this manner always equidistant?

Question

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?

Answer 1

They are not, as the other answers point out. The simplest way (I think) to see it is the following:

1. The discs move on circles. We can think of the circles as the equator and a meridian of a sphere. Call the white-disc-circle the equator.
2. At the moment the discs cross the holes, they are at opposite points of the sphere, and their distance is thus maximal.
3. When the black disc is at its highest point ("north pole"), the white one is still on the equator. Hence, they are not at opposite points of the sphere, and their distance is clearly less than the maximum value. This is actually the point of closest approach.

The (spatial) distance oscillates between twice the radius ($2 r$) and $\sqrt{2} \,r$.

Answer 2

The two points are revolving on circles in the $xy$ and $yz$ planes, in a synchronized way, on the trajectories

$$x=\sin t,y=\cos t,z=0$$ and $$x=0,y=-\cos t,z=\sin t.$$

The phases are such that the points are on opposite locations on the $y$ axis at $t=0$.

The distance is thus

$$\sqrt{\sin^2t+4\cos^2t+\sin^2t}=\sqrt{2(1+\cos^2t)}.$$

It is maximum on every half-turn (from the position at $t=0$), forming a straight line with the center, and minimum every quarter turn later, forming a right angle.

The ratio of the long distance over the short one is $\sqrt2$.

Answer 3

Let's say that the sphere has radius 1 centered at the origin in $\Bbb R^3$ and the disks are moving with speed 1. And let's also say that in the picture the $z$-axis points upward, and the $y$ axis points normal to the plane of motion of the black thing. Let's also use the convention that at time zero the black disk has center at $(0,0,1)$ and the white one at $(0,1,0)$ Then You can literally parametrize the two curves which describe the center of the disks in a simple way:

$\gamma_{black}(t) = (\sin t, 0, \cos t)$

$\gamma_{white}(t) = (-\sin t, \cos t, 0)$

So if you compute distances (by subtracting and taking the norm squared), it is clear that they are not always equidistant.

So why does it seem like they are always equidistant in the picture?

Well, if you take a good look, the motion of the two disks is not along an exact sphere but more like some kind of ellipsoid.

So instead of using the spherical model $x^2+y^2+z^2=1$, try the ellipsoidal model $x^2+y^2/2+z^2/2=1$. In other words, let's consider the image of this rigid motion under the map $(x,y,z) \mapsto (x, \sqrt 2 y, \sqrt 2 z)$. So the new parametrizations will be

$\gamma_{black}(t) = (\sin t, 0, \sqrt 2 \cos t)$

$\gamma_{white}(t) = (-\sin t, \sqrt 2 \cos t, 0)$

And in this model we see that they are always equidistant! (Well those are my two cents at least. Might be total BS. Really it depends on how you want to interpret this two dimensional projection of a three dimensional motion - for all we know the planes of motion may not even be orthogonal, which would make the formulas more complicated).

Answer 4

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.

Answer 5

The two disks and the center of the sphere alternate bewteen alignment (distance $2r$) and a right angle configuration (distance $\sqrt2r$).

Answer 6

At time $t$ the black disc is at $(\cos t, \sin t, 0)$ and the white at $(-\cos t, 0, -\sin t)$ (up to orientation: Black spins along the unit circle in the $x-y$ plane where $z = 0$. White spins along a unit circle in the $x-z$ plane where $y=0$. White is so oriented that it starts at $z = 0 = -\sin t$, $x = -1, = -\cos 0$ and so forth.)

The distance is $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 } =$

$ \sqrt {(2\cos t )^2 + \sin^2 t + \sin^2 t} = \sqrt {4\cos^2 t + 2 \sin^2 t}=\sqrt{2 + 2\cos^2 t}$ which is not constant.