While combing around my notes looking for other possible examples for this question, I chanced upon another one of my unsolved problems:

Cycloidal curves are curves generated by a circle rolling upon a plane or space curve. It's not too hard to derive the required parametric equations if the fixed curve is a plane curve, but I've had some trouble deriving the corresponding expression for space curves.

More specifically, here is the particular problem I was concerned with: consider a (cylindrical) helix:


and imagine a circle of radius $r$ whose plane is always perpendicular to the x-y plane rolling on the helix, starting at the point $(a,0,0)$ ($t=0$). Imagine a point in the plane of the circle at a distance $hr$ from the center. What are the parametric equations for the locus of the point?

The two obvious pieces of information I have are that the center of the circle also traces a helix, whose parametric equation differs from the original by a vertical shift of $r$ (per Tony, that was an erroneous assumption), and that the expression for the arclength of the helix, $s=\sqrt{a^2+c^2}t$, should figure into the final parametric equations. Otherwise, I'm not sure how to start.

How does one derive parametric equations for the "helical cycloid"?

The physical model I had in mind was a screw ramp winding around a cylinder. Supposing that there was a car that needed to go to the top of the cylinder by driving on the ramp, and supposing that a spot is placed on one of the car's wheels, what are the equations for the locus of the spot?

  • $\begingroup$ Where does the point start? $\endgroup$ – Jonas Meyer Dec 24 '10 at 5:06
  • $\begingroup$ @Jonas: I was assuming that in the $h=1$ case, the point of tangency of the circle and the helix is $(a,0,0)$, and that the first cusp of the curve should start from there. $\endgroup$ – J. M. is a poor mathematician Dec 24 '10 at 7:50
  • $\begingroup$ I think you are wrong that the centre of the circle lies directly above a point on the helix. The radius from the centre of the circle to the point of contact is skew, and (except for the point of contact itself) lies outside the bounding cylinder of the helix. $\endgroup$ – TonyK Dec 24 '10 at 8:23
  • $\begingroup$ Aha @Tony, I think that's why my attempts at deriving it looked weird... how would this be fixed? $\endgroup$ – J. M. is a poor mathematician Dec 24 '10 at 8:41
  • $\begingroup$ M.: I think I fixed my answer. $\endgroup$ – Jonas Meyer Dec 24 '10 at 9:04

I like to think of helices on cylinders as images of lines under planar curling, so here's an approach incorporating that idea.

Imagine the circle rolling up a line drawn in the plane, with $P$ the point of tangency along the line and $Q$ a distinguished point on the circumference of the circle. (So, $Q$ traces out a cycloid in the plane.) Now, imagine the plane is made of thin paper, but the circle is made of stiff cardboard. If we curl the plane into a cylinder but the circle remains flat, the plane of the circle will be tangent to the cylinder along the (vertical) line passing through the point $P$. The point $P$ will follow a helix, and the point $Q$ --which lies in the plane of the circle-- traces out the helical cycloid in space. While the path that $P$ (and $Q$) takes is decidedly different after curling than before, the displacement vector between $P$ and $Q$ in the plane of the circle is the same throughout.

Before curling, if the circle of radius $r$ rolls along a horizontal track in the $uv$-plane, then we have $P=(r t, 0)$, and the point $Q$ traces out the standard cycloid:

$$\begin{align} u &= r ( t - \sin t ) \\ v &= r ( 1 - \cos t ) \end{align}$$

The displacement vector from $P$ to $Q$ at time $t$ is given by

$$d := PQ = [m, n] = r[-\sin t, 1 - \cos t ] = - 2 r \sin\frac{t}{2} \; [\cos\frac{t}{2},-\sin\frac{t}{2}]$$

Rotating the plane through an angle, say, $\theta := \rm{atan\frac{c}{a}}$, and writing $b$ for $\sqrt{a^2+c^2}$, the circle's point of tangency along the tilted track is given by ...

$$P_2 = (r t \cos\theta,r t \sin\theta) = \left( \frac{r t a}{b}, \frac{r t c}{b}\right)$$

... and the displacement vector, $d_2$, to the point $Q_2$ tracing out the tilted cycloid is given by ...

$$\begin{align} d_2 :&= [m \cos\theta - n \sin \theta, m \sin\theta + n \cos \theta ] \\ &= \frac{1}{b} [a m - c n, c m + a n] \\ &= -\frac{2r}{b}\sin\frac{t}{2}\left[\cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] \\ \end{align}$$

Now the fun part: Curl the $(u,v)$-plane around a cylinder of radius $a$, such that the $uv$-origin aligns with the $xyz$-point $(a,0,0)$ and the $v$-axis runs parallel to the $z$ axis. The curled, tilted track will coincide with a helix. The horizontal distance travelled by $P_2$ in the plane becomes the length of horizontal circular arc travelled by $P_3$ around the cylinder; upon dividing by the radius, $a$, of the cylinder, this length becomes the angular "distance" --$s := \frac{rt}{b}$-- travelled by $P_3$; the vertical distances match. Therefore:

$$P_3 = \left(a\cos s, a\sin s, \frac{rtc}{b} \right)=(a\cos s,a\sin s,c s)$$

As for the displacement vector: The $u$ direction of the tangent plane coincides with the horizontal vector tangent to the cylinder at $P_3$; the $v$ direction coincides with the $z$ direction. Thus, the transformation from $uv$-coordinates to $xyz$-coordinates is given by

$$[1,0]\to[-\sin s, \cos s, 0]\hspace{0.5in}[0,1]\to[0,0,1]$$

The image, $d_3$, of the displacement vector $d_2$, then, is

$$d_3 = -\frac{2r}{b}\sin\frac{t}{2}\left[-\sin s \cos\left(\theta-\frac{t}{2}\right), \cos s \cos\left(\theta-\frac{t}{2}\right), \sin\left(\theta - \frac{t}{2} \right) \right] $$

and the path of $Q_3$, which traces the helical cycloid, is given by

$$ Q_3 = P_3 + d_3 = \left\{ \begin{align} x &= a \cos s &+ \frac{2 r}{b} \sin s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\ y &= a \sin s &- \frac{2 r}{b} \cos s \sin\frac{bs}{2r} \cos\left(\theta-\frac{bs}{2r}\right)\\ z &= c s &- \frac{2 r}{b} \sin\frac{b s}{2r}\sin\left(\theta-\frac{bs}{2r}\right) \end{align} \right. $$

Here's a picture with $a=c=1$ and $r=1/2$:

Helical Cycloid

Note: The above does not simply curl the planar cycloid around the cylinder. Since

$$x^2 + y^2 = a^2 + \frac{4r^2}{b^2} \sin^2\frac{bs}{2r} \cos^2\left(\theta-\frac{bs}{2r}\right) \ge a^2$$

we see that most of the helical cycloid lies outside the surface of the cylinder.


The tangent to the helix at $P = (a \cos t, a \sin t, ct)$ is

$T = (-a \sin t, a \cos t, c)$.

The vector

$H = (\cos t, \sin t, 0)$

is horizontal, and perpendicular to $T$. The radius vector $R$ from $P$ to the centre $C$ of the circle is perpendicular to both of these, so it is proportional to

$H \wedge T = (c \sin t, -c \cos t, a)$,

which has length $\sqrt{a^2 + c^2}$. So the centre of the circle is at $P + h H \wedge T$, with $h = r/\sqrt{a^2 + c^2}$.

  • $\begingroup$ Should that be $-c\cos t$ in the middle component of $H\wedge T$? $\endgroup$ – Jonas Meyer Dec 24 '10 at 9:10
  • 1
    $\begingroup$ Yes, you're right. I've fixed it. $\endgroup$ – TonyK Dec 24 '10 at 9:22

I'm adjusting my answer, which as TonyK points out was based on an incorrect assumption. Hopefully it's correct now.

The center of the circle is not directly above the point of contact with the helix as my incorrect answer had asserted. To see where it is, first picture the starting point at $(a,0,0)$, where the circle is parallel to the $y$-$z$ plane. The tangent vector to the helix is in the direction of $(0,a,c)$, so the radius is perpendicular to this in the direction of $(0,-c,a)$. So you have to add $\frac{r}{\sqrt{a^2+c^2}}(0,-c,a)$ to get to the center. As the circle rolls along, the vector you add to get to the center will rotate around the $z$ axis to $\frac{r}{\sqrt{a^2+c^2}}(c\sin(t),-c\cos(t),a)$. So the center of the circle is parametrized by $(a\cos(t),a\sin(t),ct)+\frac{r}{\sqrt{a^2+c^2}}(c\sin(t),-c\cos(t),a)$.

The arclength formula allows you to write down the rotation angle of the point about the center of the circle in terms of $t$ as $\frac{\sqrt{a^2+c^2}}{r}t$. Let $b=\frac{\sqrt{a^2+c^2}}{r}$ from now on.

If we pretend for a minute that the point was just going to rotate about the center of the circle at this rate while the circle remains fixed, and simplify the problem for now by assuming that the point starts above the center of the circle, then the vector we would add to the center of the circle to parametrize the point would be $(0,hr\sin(bt),hr\cos(bt))$. When looking toward the $y$-$z$ plane from the positive $x$ axis, this motion would appear clockwise, as it should because the circle will be starting off to the right from this perspective.

But the vector we actually add to the center will also be rotated by $t$ about the $z$-axis, so putting it all together (with minimal simplification) we get

$\begin{align*} x&=a\cos(t)+\frac{c}{b}\sin(t)-hr\sin(t)\sin(bt)\\ y&=a\sin(t)-\frac{c}{b}\cos(t)+hr\cos(t)\sin(bt)\\ z&=ct+\frac{a}{b}+hr\cos(bt). \end{align*}$

Now to adjust the starting point, you can change $bt$ to $b(t-t_0)$ for suitable $t_0$.

  • $\begingroup$ See my comment to the OP's question ("I think you are wrong that the centre of the circle lies above a point on the helix"). $\endgroup$ – TonyK Dec 24 '10 at 8:25
  • $\begingroup$ @TonyK: You're right, thank you. I'll try to fix it. $\endgroup$ – Jonas Meyer Dec 24 '10 at 8:29

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