3

Consider a circle of radius $r_s$ that is tangent to two curves $r(\theta)$ and $R(\theta)$ at points $E_1, E_2$ respectively, defined in polar coordinates. Knowing the function $r(\theta)$, find the function $R(\theta)$, so that when rotating both function in opposite directions by some angle $d\phi$ around the center of the coordinate system, the circle remains on the line OS (where O is the center of the coordinate system and S being the center of the circle) and is tangent to both curves at one point respectively. Can be thought of as if the circle rolls on both surfaces without slipping.

In case where curve $r(\theta)$ is a circle with radius $r$, the other curve $R(\theta)$ is also a circle with the radius $R = r+2r_s$.

I can generate the points of function $R(\theta)$ numerically, but I am struggling to find a way to get an analytic solution for the curve $R(\theta)$. Any thoughts?

A drawing for (maybe) better understanding of the problem: problem

tragus
  • 31
  • 4
  • 1
    Ignore the function $R(\theta)$ for the moment and consider just $r(\theta)$ and the circle centered at $S.$ What kind of function $r(\theta)$ that is not a circle can roll continuously against the circle at $S$ without losing contact or requiring the circle at $S$ to move? – David K Feb 11 '24 at 20:08
  • True, in this case there's none. If point $S$ is fixed then that means both $r(\theta)$ and $r_s$ must be constant.

    What if we allow the movement of the point $S$ along the $r$ axis, on the line $OS$, without allowing its angular displacement? Then we could have a function $r(\theta)$ and I assume a possible corresponding curve $R(\theta)$ that allows the rolling motion of the circle.

     – tragus Feb 12 '24 at 05:46
  • 1
    See comments under Jean-Marie's question: I think we have a consensus that it would be good to allow the circle's center to move along the line $OS$ as you propose. You can edit this into the question above in order to encourage people to answer it that way. – David K Feb 12 '24 at 21:40
  • ... I meant Jean-Marie's answer (below), not "question", sorry. – David K Feb 12 '24 at 23:36

1 Answers1

2

The fact that your curves are given in polar coordinates is less important than the fact that you need curves having a good representation in terms of curvilinear abscissa.

There are very few of them, set apart circles (the trivial case you have mentionned) and "Involutes of a Circle" (IC in short ; disclaimer : non standard abbreviation).

Precisely, these curves are used in the profile of gears : see the animation here displaying two opposite ICs rolling without slipping one on the other (see as well the graphical representation below).

Now, if you separate them by a distance $2R$, you can introduce a disk (circular galley) with radius $R$ between them : this disk will communicate the desired rolling-without-slipping movement from one IC to the other while staying in-place.

enter image description here

Fig. 1 : see explanation below.

One can have a good picture of an IC as a limit curve (like for a circle which is the limit of inscribed regular polygons). Consider the "spiral-looking" red curve in fig. 1 ; imagine a tethered goat in point A of an octogonal pillar discovering that its rope has in fact been rolled around the pillar ; the goat, by progressively unrolling the rope will be able to graze more and more land ; The frontier curve is an approximate IC ; replace now the octogon by any n-gon with $n \to \infty$ (always inscribed in the same circle) : you have now a correct idea of what an IC is.

Now, if we place a second similar IC in the position represented on fig. 1, the two ICs can roll one on the other because the addition of the resp. distances to their centers is a constant.

I haven't made a figure representing the circle that can be placed between the two ICs but it is not difficult to imagine it.

See many inspiring animations on this site.

Jean Marie
  • 81,803
  • The part about the circle "not moving in any direction" seems to rule out any non-trivial solution, doesn't it? – David K Feb 11 '24 at 20:11
  • @David K I don't understand your remark. Where you have found "not moving in any direction" : the circle has its axis fixed but turns under the action of the first CI, then communicates its movement to the second CI. – Jean Marie Feb 11 '24 at 21:21
  • 1
    I was quoting the last sentence of the first paragraph of the question. Earlier in that paragraph it says "the circle remains in place (with its center at point S)", which is a better but more wordy way to say this. I should have mentioned that in the question, both curves rotate about the origin, so their axes are fixed as well. – David K Feb 11 '24 at 21:30
  • 1
    So in order to construct the curves as requested in the question, one must first show that the curve $r(\theta)$, while rotating about a fixed center, rolls against the circle while the circle's center also remains fixed. Once that problem has been solved then we can consider the curve $R(\theta)$, where the question requires to two curves to rotate about the same center (not necessarily at the same rate). But I think just solving the first part dictates that $r(\theta)$ is constant. – David K Feb 11 '24 at 21:35
  • I understand now a little better what you mean. Let me take time to see why my proposal isn't a solution. – Jean Marie Feb 12 '24 at 19:49
  • In a comment under the question, OP is considering whether to allow the circle's center to move along a line in order to allow the circle to stay in contact with a shape of non-constant radius. Would this improve the question or be too much of a difference (requiring a new question), do you think? – David K Feb 12 '24 at 21:25
  • @David K I think it would improve the question, without needing a new question because it would remain in the same spirit. – Jean Marie Feb 12 '24 at 21:29
  • 1
    Agreed. So recommended to OP. – David K Feb 12 '24 at 21:41
  • 1
    I edited my question. Still, your answer gave me an idea to construct an equation with regards to curve lenght.

    Considering that the arc lenght traveled be a circle is the same on both curves, then $L_r$ = $L_R$ where $L$ is the curve lenght of each function. Using equation for curve lenght gives an equality

    $\int_0^{\phi_1} \sqrt{r(\theta)^2 + \left ( \frac{\mathrm d r(\theta)}{\mathrm d\theta} \right)^2} \mathrm d \theta= \int_0^{\phi_2} \sqrt{R(\theta)^2 + \left ( \frac{\mathrm d R(\theta)}{\mathrm d\theta} \right)^2} \mathrm d \theta$

    Which gives an integral function.

     – tragus Feb 13 '24 at 07:41