# 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?

• I suppose one could assume this gif is centered at the origin in Euclidian 3 space and parametrically describe the paths of the white and black points, then compare the distance between them for different values of t. Feb 9, 2017 at 4:18
• Most of the drawing actually obscures the question. Feb 9, 2017 at 12:34
• If you put the yin and yang into orthogonal planes in $\mathbf{R}^{4}$ and let them rotate in the analogous way (or, indeed, in an arbitrary way), it's another matter: If $C_1$ and $C_2$ are circles of radius $r$ lying in orthogonal planes and having a common center, the distance from each point of $C_1$ to each point of $C_2$ is $r\sqrt{2}$. I have a delightful physical model which unfortunately this universe is too small to contain.... Feb 10, 2017 at 11:44

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 $\sqrt{2} \,r$.

• Best answer. Nice and simple, no parametrizations involved. Feb 9, 2017 at 8:25

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 $\sqrt2$.

• There is no real need to think of a sphere, just two circles.
– user65203
Feb 9, 2017 at 7:52

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).

• Sorry but this ellipsoidal model a nonsense. You have two (incomplete) disks that are seen as ellipses as they are viewed using an isometric projection. They are inscribed in a perfect sphere. IMO it is nearly impossible to tell visually if the distance is constant or not as we are missing depth clues.
– user65203
Feb 9, 2017 at 8:00
• @YvesDaoust Yes you are right, it is total nonsense if you assume that the two rotating objects are rigid (inflexible) bodies. My mistake. What I meant is that if we take the rigid motion and map it under $(x,y,z) \mapsto (x, y/\sqrt 2,z/\sqrt 2)$ then you get equidistant motion. But the two rotating objects will not be rigid in such a model, they will bend and flex. Feb 9, 2017 at 8:12

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.

• Nothing new here...
– user65203
Feb 10, 2017 at 11:30
• Hi, saw it this afternoon only, did not see or compare with any other post. Feb 10, 2017 at 11:45

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

• Could you elaborate on why this is? It may be clear to you, but for the majority of people posting on this question it is not. Feb 9, 2017 at 15:57
• @Zxyrra: you can refer to Toffomat's answer for a little more details. But it is essentially a matter of observation. Ignore the "tadpoles", imagine the center point and see how the three points evolve. I want my answer to be as short as possible.
– user65203
Feb 9, 2017 at 16:24
• I understand you want it to be as short as possible but without reasoning - just saying "clearly" you don't really answer the question. Feb 9, 2017 at 16:27
• @Zxyrra: thanks, you helped me shorten it !
– user65203
Feb 9, 2017 at 16:28

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.

• I am puzzled. Your solution implies that the distance varies in a ratio $\sqrt3:1$, while mine says $\sqrt2:1$. I can't make up my mind about which is correct. Update: On second thoughts, your equations can't be right because they don't allow the points in opposition.
– user65203
Feb 9, 2017 at 8:28
• Ach, yes. I forgot the coefficient of 2 in the binomial expansion of (x + y)^2. D'oh. Feb 9, 2017 at 8:34
• Oh, wait. No, I didn't. Um... I'm not sure what's up. Um, the points never are in opposition, are they? Feb 9, 2017 at 8:37
• Of course they are, when they pass through the holes.
– user65203
Feb 9, 2017 at 8:38
• but it's still not in agreement and I still don't have opposition. So I must be wrong. Feb 9, 2017 at 8:47