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(See below for update.)

I would like to reverse the following equations. Here, the path is traced out by the polar origin of the wheel. You can put the trace point anywhere on or off of the wheel by translating the wheel relative to the polar origin. To generate a cycloid, $r(t)$ would be a circle which passes through the origin.

Equations of any roulette generated by a wheel given by $r(t)$ rolling on a straight line:

$$x(t) = \int \sqrt{r(t)^2 + r'(t)^2} dt - \frac{r(t) r'(t)}{\sqrt{r(t)^2 + r'(t)^2}}$$ $$y(t) = \frac{r(t)^2}{\sqrt{r(t)^2 + r'(t)^2}}$$

Given such a roulette in $x(t)$ and $y(t)$, how can I determine the original wheel? This is all motivated by a specific example, but I would like to understand the general case.

I worked out the above equations by geometric inspection, more or less. At time $t$, the wheel is touching the road at a point on the road ($y = 0$) and the distance is the arc length around the wheel from the initial angle to $t$ ($x = \int \sqrt{r(t)^2 + r'(t)^2} dt$). Add to that the offset from the tangent point on the wheel to the origin of the wheel. The origin is a distance $r(t)$ from the tangent point at an angle given by the fact that the wheel is also tangent to the road at all times, so doing a bit of trigonometric substitution gives the rest.

Nothing similar is coming to me to work out the reverse equations.

I also tried to work it out by treating the wheel as the limit of a class of polygons with an increasing number of sides where the vertices lie on the wheel. I got as far as:

$\Delta x = (L - x)(1 + \cos(a)) + y \sin(a)$
$\Delta y = -y(1 + \cos(a)) + (L - x) \sin(a)$

Where $L$ is the current arc length traveled so far, $x$ and $y$ are the current trace position and $a$ is the next angle that the polygon approximation of the wheel will pivot on. I'm unable to come up with the angle given simply $t$ and a change in t, since I need three sample points from the wheel to determine the next angle, unless I assume that $\Delta t$ will be constant so the previous $t$ was $t - \Delta t$. I am afraid that might cause problems when I try to reverse it, given who knows what parameterization. It won't be linear with respect to $t$.

So, where should I go from here?

EDIT: New idea. $\dfrac{dy}{dx}(t) = \dfrac{r'(t)}{r(t)}$.
Therefore: $y(t) = \dfrac{r(t)^2}{\sqrt{r(t)^2 + r'(t)^2}} = \dfrac{r(t)}{\sqrt{1 + \frac{r'(t)^2}{r(t)^2}}} = \dfrac{r(t)}{\sqrt{1 + \left(\frac{dy}{dx}(t)\right)^2}}$

$r(t) = y(t) \sqrt{1 + \left(\frac{dy}{dx}(t)\right)^2}$

So given $y(\theta)$ and $x(\theta)$, $r(t) = y(\theta) \sqrt{1 + \left(\frac{dy}{dx}\right)^2}$

Is there any way to find $t$ in terms of $\theta$, the relation between the rate at which the wheel turns and the parameterization of the given $x$ and $y$?

Also, $x(t) + \frac{dy}{dx} y(t) = \int \sqrt{r(t)^2 + r'(t)^2}$ and $\sqrt{r(t)^2 + r'(t)^2}$ = $y(t) \left(1 + \left(\frac{dy}{dx}\right)^2\right)$.

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    $\begingroup$ I think the word roulette is the term you want. en.wikipedia.org/wiki/Roulette_%28curve%29 $\endgroup$ – Matthew Conroy Mar 5 '12 at 23:10
  • $\begingroup$ Yes, a trochoid is a circle on a line roulette (I don't see anything in the wikipedia article which suggests otherwise). $\endgroup$ – Matthew Conroy Mar 6 '12 at 0:26
  • $\begingroup$ Could you clarify: are you only considering rolling circles, or could r(t) represent other curves? $\endgroup$ – Matthew Conroy Mar 6 '12 at 0:44
  • $\begingroup$ let us continue this discussion in chat $\endgroup$ – Matthew Conroy Mar 6 '12 at 6:14
  • $\begingroup$ Since you've restricted to rolling on a line, you can make the assumption that the wheel is convex. Also, if there's a "corner" on the wheel's circumference, then presumably the wheel tips at that point, making the roulette trace out a circular arc, which gives you information about the tracing point relative to the corner. (If the roulette is simply a series of circular arcs, may we conclude that the wheel is a polygon? Hmmm ... Maybe not: for instance, if two adjoining arcs --with differing radii-- meet "smoothly".) Not much, but it's a start! $\endgroup$ – Blue Mar 6 '12 at 15:52
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In fact you can use Steiner-Habich theorem : the pedal of the rolling curve generating the roulette is the Wheel for the roulette (= ground). Or the antipedal of the Wheel is the rolling curve generating the roulette. The problem is just to find pedal or anti-pedal of a plane curve. It is an old problem that goes back to James Gregory 1668 in "Geometriae pars universalis". He invented a transformation between polar and orthonormal coordinates that gives directly with just one inegration a couple ground-wheel :

y=ρ, x=∫ρdθ (=direct Gregory's transformation : GT) if the Wheel is given

ρ=y , θ = ∫ dx/y (= GT^-1) if we know the equations (x,y) of the ground.

There is identity of arc length between the polar curve and (x,y) curve by a rolling motion and the pole runs along x-axis. The theorem of Steiner-Habich is important in the theory (pp 3-4 of the paper I Gregory's transformation). For classes of curves like sinusoidal spirals for which the pedal is in the same class the problem is easier.

You can view examples here http://christophe.masurel.free.fr/#s9 (All papers are open-access).

There are also many informations in "Nouvelles annales de mathematiques" (1842-1927) - but in french language - http://www.numdam.org/numdam-bin/feuilleter?j=nam or on Gallica.fr and also in Mathesis. C. Masurel

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