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I'm thinking about a circle rolling along a parabola. Would this be a parametric representation?

$(t + A\sin (Bt) , Ct^2 + A\cos (Bt) )$

A gives us the radius of the circle, B changes the frequency of the rotations, C, of course, varies the parabola. Now, if I want the circle to "match up" with the parabola as if they were both made of non-stretchy rope, what should I choose for B?

My first guess is 1. But, the the arc length of a parabola from 0 to 1 is much less than the length from 1 to 2. And, as I examine the graphs, it seems like I might need to vary B in order to get the graph that I want. Take a look:

I played with the constants until it looked ALMOST like what I had in mind.

This makes me think that the graph my equation produces will always be wrong no matter what constants I choose. It should look like a cycloid:

Cycloid

But bent to fit on a parabola. [I started this becuase I wanted to know if such a curve could be self-intersecting. (I think yes.) When I was a child my mom asked me to draw what would happen if a circle rolled along the tray of the blackboard with a point on the rim tracing a line ... like most young people, I drew self-intersecting loops and my young mind was amazed to see that they did not intersect!]

So, other than checking to see if this is even going in the right direction, I would like to know if there is a point where the curve shown (or any curve in the family I described) is most like a cycloid--

Thanks.

"It would be really really hard to tell" is a totally acceptable answer, though it's my current answer, and I wonder if the folks here can make it a little better.

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2  
I think you want the osculating circle ... en.wikipedia.org/wiki/Osculating_circle –  deoxygerbe Apr 12 '11 at 22:08
    
are you suggesting that as a way to compare the curvatures?.. but what if it is near one of the cusps where the "true" cycloid is not differentiable? –  a little don Apr 12 '11 at 22:33
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A question reserved for the legendary mathematician: "How do you do that with a plain old $f(x)$?" –  muntoo Apr 13 '11 at 0:09
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I did a few animations of the "parabolic trochoid" back in the day (both for the case of the circle rolling inside and outside the parabola); when I get to see my Mathematica notebooks again, I'll get back to you. –  J. M. Apr 13 '11 at 1:25
    
Thanks J.M. that would be awesome. –  a little don Apr 13 '11 at 11:37

4 Answers 4

up vote 42 down vote accepted

(I had been meaning to blog about roulettes a while back, but since this question came up, I'll write about this topic here.)

I'll use the parametric representation

$$\begin{pmatrix}2at\\at^2\end{pmatrix}$$

for a parabola opening upwards, where $a$ is the focal length, or the length of the segment joining the parabola's vertex and focus. The arclength function corresponding to this parametrization is $s(t)=a(t\sqrt{1+t^2}+\mathrm{arsinh}(t))$.

user8268 gave a derivation for the "cycloidal" case, and Willie used unit-speed machinery, so I'll handle the generalization to the "trochoidal case", where the tracing point is not necessarily on the rolling circle's circumference.

Willie's comment shows how you should consider the notion of "rolling" in deriving the parametric equations: a rotation (about the wheel's center) followed by a rotation/translation. The first key is to consider that the amount of rotation needed for your "wheel" to roll should be equivalent to the arclength along the "base curve" (in your case, the parabola).

I'll start with a parametrization of a circle of radius $r$ tangent to the horizontal axis at the origin:

$$\begin{pmatrix}-r\sin\;u\\r-r\cos\;u\end{pmatrix}$$

This parametrization of the circle was designed such that a positive value of the parameter $u$ corresponds to a clockwise rotation of the wheel, and the origin corresponds to the parameter value $u=0$.

The arclength function for this circle is $ru$; for rolling this circle, we obtain the equivalence

$$ru=s(t)-s(c)$$

where $c$ is the parameter value corresponding to the point on the base curve where the rolling starts. Solving for $u$ and substituting the resulting expression into the circle equations yields

$$\begin{pmatrix}-r\sin\left(\frac{s(t)-s(c)}{r}\right)\\r-r\cos\left(\frac{s(t)-s(c)}{r}\right)\end{pmatrix}$$

So far, this is for the "cycloidal" case, where the tracing point is on the circumference. To obtain the "trochoidal" case, what is needed is to replace the $r$ multiplying the trigonometric functions with the quantity $hr$, the distance of the tracing point from the center of the rolling circle:

$$\begin{pmatrix}-hr\sin\left(\frac{s(t)-s(c)}{r}\right)\\r-hr\cos\left(\frac{s(t)-s(c)}{r}\right)\end{pmatrix}$$

At this point, I note that $r$ here can be a positive or a negative quantity. For your "parabolic trochoid", negative $r$ corresponds to the circle rolling outside the parabola and positive $r$ corresponds to rolling inside the parabola. $h=1$ is the "cycloidal" case; $h > 1$ is the "prolate" case (tracing point outside the rolling circle), and $0 < h < 1$ is the "curtate" case (tracing point within the rolling circle).

That only takes care of the rotation corresponding to "rolling"; to get the circle into the proper position, a further rotation and a translation has to be done. The further rotation needed is a rotation by the tangential angle $\phi$, where for a parametrically-represented curve $(f(t)\quad g(t))^T$, $\tan\;\phi=\frac{g^\prime(t)}{f^\prime(t)}$. (In words: $\phi$ is the angle the tangent of the curve at a given $t$ value makes with the horizontal axis.)

We then substitute the expression for $\phi$ into the anticlockwise rotation matrix

$$\begin{pmatrix}\cos\;\phi&-\sin\;\phi\\\sin\;\phi&\cos\;\phi\end{pmatrix}$$

which yields

$$\begin{pmatrix}\frac{f^\prime(t)}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}&-\frac{g^\prime(t)}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}\\\frac{g^\prime(t)}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}&\frac{f^\prime(t)}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}\end{pmatrix}$$

For the parabola as I had parametrized it, the tangential angle rotation matrix is

$$\begin{pmatrix}\frac1{\sqrt{1+t^2}}&-\frac{t}{\sqrt{1+t^2}}\\\frac{t}{\sqrt{1+t^2}}&\frac1{\sqrt{1+t^2}}\end{pmatrix}$$

This rotation matrix can be multiplied with the "transformed circle" and then translated by the vector $(f(t)\quad g(t))^T$, finally resulting in the expression

$$\begin{pmatrix}f(t)\\g(t)\end{pmatrix}+\frac1{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}\begin{pmatrix}f^\prime(t)&-g^\prime(t)\\g^\prime(t)&f^\prime(t)\end{pmatrix}\begin{pmatrix}-hr\sin\left(\frac{s(t)-s(c)}{r}\right)\\r-hr\cos\left(\frac{s(t)-s(c)}{r}\right)\end{pmatrix}$$

for a trochoidal curve. (What those last two transformations do, in words, is to rotate and shift the rolling circle appropriately such that the rolling circle touches an appropriate point on the base curve.)

Using this formula, the parametric equations for the "parabolic trochoid" (with starting point at the vertex, $c=0$) are

$$\begin{align*}x&=2at+\frac{r}{\sqrt{1+t^2}}\left(ht\cos\left(\frac{a}{r}\left(t\sqrt{1+t^2}+\mathrm{arsinh}(t)\right)\right)-t-h\sin\left(\frac{a}{r}\left(t\sqrt{1+t^2}+\mathrm{arsinh}(t)\right)\right)\right)\\y&=at^2-\frac{r}{\sqrt{1+t^2}}\left(h\cos\left(\frac{a}{r}\left(t\sqrt{1+t^2}+\mathrm{arsinh}(t)\right)\right)+ht\sin\left(\frac{a}{r}\left(t\sqrt{1+t^2}+\mathrm{arsinh}(t)\right)\right)-1\right)\end{align*}$$

A further generalization to a space curve can be made if the rolling circle is not coplanar to the parabola; I'll leave the derivation to the interested reader (hint: rotate the "transformed" rolling circle equation about the x-axis before applying the other transformations).

Now, for some plots:

parabolic trochoids

For this picture, I used a focal length $a=1$ and a radius $r=\frac34$ (negative for the "outer" ones and positive for the "inner" ones). The curtate, cycloidal, and prolate cases correspond to $h=\frac12,1,\frac32$.


(added 5/2/2011)

I did promise to include animations and code, so here's a bunch of GIFs I had previously made in Mathematica 5.2:

Inner parabolic cycloid, $a=1,\;r=\frac34\;h=1$

inner parabolic cycloid

Curtate inner parabolic trochoid, $a=1,\;r=\frac34\;h=\frac12$

curtate inner parabolic trochoid

Prolate inner parabolic trochoid, $a=1,\;r=\frac34\;h=\frac32$

prolate inner parabolic trochoid

Outer parabolic cycloid, $a=1,\;r=-\frac34\;h=1$

outer parabolic cycloid

Curtate outer parabolic trochoid, $a=1,\;r=-\frac34\;h=\frac12$

curtate outer parabolic trochoid

Prolate outer parabolic trochoid, $a=1,\;r=-\frac34\;h=\frac32$

prolate outer parabolic trochoid

The Mathematica code (unoptimized, sorry) is a bit too long to reproduce; those who want to experiment with parabolic trochoids can obtain a notebook from me upon request.

As a final bonus, here is an animation of a three-dimensional generalization of the prolate parabolic trochoid:

3D prolate parabolic trochoid

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I might edit this post much later to include animations and Mathematica code; watch this space! –  J. M. Apr 14 '11 at 5:33
    
Please do! Very nice answer. Do you have a link to the blog you mentioned? –  Andres Caicedo Apr 14 '11 at 6:09
    
Hi @Andres; the last entry in my blog was intended to transition into a discussion of roulettes, but unfortunately real life got in the way. When I saw this question, I decided to post a portion of the supposed follow-up to that blog entry here as an answer. As I said, the animations would have to wait since the routines I wrote are in a computer currently far away from me. –  J. M. Apr 14 '11 at 7:37
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Very nice answer. It may also be interesting to say a few words about the generic behaviour, that for the curtate and prolate cases the curves are smooth, while for the cycloidal cases there is a cusp of the form $x^2 = y^3$ –  Willie Wong Apr 14 '11 at 10:43
    
Thanks! –  Andres Caicedo Apr 14 '11 at 14:22

If I understand the question correctly:

Your parabola is $p(t)=(t,Ct^2)$. Its speed is $(1,2Ct)$, after normalization it is $v(t)=(1,2Ct)//\sqrt{1+(2Ct)^2)}$, hence the unit normal vector is $n(t)=(-2Ct,1)/\sqrt{1+(2Ct)^2)}$. The center of the circle is at $p(t)+An(t)$. The arc length of the parabola is $\int\sqrt{1+(2Ct)^2}dt= (2 C t \sqrt{4 C^2 t^2+1}+\sinh^{-1}(2 C t))/(4 C)=:a(t)$. The position of a marked point on the circle is $p(t)+An(t)+A\cos(a(t)-a(t_0))\,n(t)+A\sin(a(t)-a(t_0))\,v(t)$ -that's the (rather complicated) curve you're looking for.

edit: corrected a mistake found by Willie Wong

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hum, for the trigonometric terms inside the parenthesis in your final answer, there should also be a correction coming from the current slope of the curve, due to the fact that the point of contact between the circle and the parabola is not at the "bottom" of the circle. –  Willie Wong Apr 13 '11 at 15:35
    
@Willie Wong: thanks a lot (I knew I would have there some mistake) –  user8268 Apr 13 '11 at 19:42

User8268 already gave you the case for the parabola. Here is the formula for the general case: let $\gamma(s)$ be a $C^2$ curve in $\mathbb{R}^2$ parametrised by arc-length $s$ (so $\frac{d}{ds}\gamma$ has norm 1). Let $\epsilon:\mathbb{R}^2\to\mathbb{R}^2$ be the linear transformation representing counterclockwise rotation by $\pi/2$ radians. The cycloid-like curve you form from rolling a circle of radius $R$ on the "left hand side" of this curve is given by

$$ s \to \gamma(s) - R\sin \frac{s}{R} \dot{\gamma}(s) + R\left(1-\cos \frac{s}{R}\right)\epsilon\dot{\gamma}(s) $$

where $\dot{\gamma} = \frac{d}{ds}\gamma$, whenever the signed curvature of $\gamma$ is at most $R^{-1}$. (If the radius of curvature is less than $R$, it is impossible to roll a circle of radius $R$ without slipping or jumping along the curve). The signed curvature is defined to be

$$ \kappa = \ddot{\gamma} \cdot \epsilon \dot{\gamma} $$

where the $\cdot$ is the usual dot product on $\mathbb{R}^2$. Note positive curvature corresponds to the curve bending toward the side which the circle is on, and negative curvature is bending away. So negative curvature does not present an obstruction to rolling, while large positive curvatures can represent something like a pot-hole.)

If you want the circle to be on the "right hand side" of the curve, replace $\epsilon$ by the clockwise rotation matrix.

In general, however, it is not necessarily possible to solve for the arclength parametrisation. On the other hand, computing the arclength when you have a known parametrisation is easier. So if we write $\tau(s)$ to be an arbitrarily parametrised curve and $|\tau(s)|$ to be the arclength between $s$ and $0$ along $\tau$, the equation can be re-written as

$$ s \to \tau(s) - R \sin \frac{|\tau(s)|}{R} \frac{\dot{\tau}(s)}{|\dot{\tau}(s)|} + R\left(1-\cos \frac{|\tau(s)|}{R}\right) \frac{\epsilon \dot{\tau}(s)}{|\dot{\tau}(s)|} $$

So for the parabola, you can plug-in the well-known formula for the arc length and try to graph it.

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If you know about Frenet frames, then the expressions above is clearer: In the (moving) Frenet frame, the position of a point on the boundary of the circle relative to its center is just $-\sin \theta \mathbf{t} - \cos\theta \mathbf{n}$, where $\theta$ measures the angle turned. The no-slip condition connects the angle turned to the arclength. And you add the motion of the center to transform back to rectilinear coordinates. –  Willie Wong Apr 13 '11 at 15:39

It sounds like you are interested in roulettes. The wikipedia page does not mention the circle-rolling-on-a-parabola, however.

There is a Wolfram Demonstration Project visualizing your problem.

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