## Friday, January 20, 2012

### Spirograph !

From simplicity: complexity, and beautiful symmetry.

Born in 1956, I definitely OCD'd on Spirograph as a kid. First, the brief Wikipedia biography on its inventor, then the entry on Spirograph itself.

Denys Fisher
(11 May 1918, LeedsEngland – 17 September 2002, Barrow-in-Furness, England) was an English engineer who invented the spirograph toy.
He left Leeds University to join the family firm, Kingfisher (Lubrication) Ltd. In 1960 he left the firm to set up his own company, Denys Fisher Engineering, in Leeds. In 1961 the company won a contract with NATO to supply springs and precision component for its 20 mm cannon. Between 1962 and 1964 he developed various drawing machines from Meccano pieces, eventually producing a prototype Spirograph. Patented in 16 countries, it went on sale in Schofields department store in Leeds in 1965. A year later, Fisher licensed Spirograph to Kenner Products in the United States. In 1967 Spirograph was chosen as the UK Toy of the Year.
In 1970, Fisher sold his company, which, as Denys Fisher Toys, produced other toys and board games, before being bought by Hasbro. Through the 1980s & 1990's he continued to work with Hasbro in developing new toys and refining Spirograph.
Spirograph is a geometric drawing toy that produces mathematical curves of the variety technically known as hypotrochoids and epitrochoids. The term has also been used to describe a variety of software applications that display similar curves, and applied to the class of curves that can be produced with the drawing equipment (so in this sense it may be regarded as a synonym of hypotrochoid). The name is a registered trademark of Hasbro, Inc.

## History

Drawing toys based on gears have been around since at least 1908, when "The Marvelous Wondergraph" was advertised in the Sears catalog.[1][2] The Boys Mechanicpublication of 1913 had an article describing how to make a Wondergraph drawing machine.[3] An instrument called a spirograph was invented by the mathematicianBruno Abakanowicz between 1881 and 1900 for calculating an area delimited by curves.[4] The Spirograph itself was developed by the British engineer Denys Fisher, who exhibited it in 1965 at the Nuremberg International Toy Fair. It was subsequently produced by his company. US distribution rights were acquired by Kenner, Inc., which introduced it to the United States market in 1966, promoting it as a creative children's toy.
In 1968, Kenner introduced Spirotot, a less complex version of Spirograph, for preschool-age children, too young for Spirograph.

## Operation

Several Spirograph designs drawn with a Spirograph set
A Spirograph consists of a set of plastic gears and other shapes such as rings, triangles, or straight bars. There are several sizes of gears and shapes, and all edges have teeth to engage any other piece. For instance, smaller gears fit inside the larger rings, but also can engage the outside of the rings in such a fashion that they rotate around the inside or along the outside edge of the rings.
To use it, a sheet of paper is placed on a heavy cardboard backing, and one of the plastic pieces—known as a stator—is pinned to the paper and cardboard. Another plastic piece—called the rotor—is placed so that its teeth engage with those of the pinned piece. For example, a ring may be pinned to the paper and a small gear placed inside the ring: the actual number of arrangements possible by combining different gears is very large. The point of a pen is placed in one of the holes of the rotor. As the rotor is moved, the pen traces out a curve. The pen is used both to draw and to provide locomotive force; some practice is required before the Spirograph can be operated without disengaging the stator and rotor. More intricate and unusual-shaped patterns may be made through the use of both hands, one to draw and one to guide the pieces. It is possible to move several pieces in relation to each other (say, the triangle around the ring, with a circle "climbing" from the ring onto the triangle), but this requires concentration or even additional assistance from other artists.

## Mathematical basis

Consider a fixed circle C1 of radius R centered at the origin. A smaller circle C2 of radius r < R is rolling inside C1 and it is tangent to C1. The inner circle cannot slip since teeth are present in a real Spirograph. Now assume that a point A that corresponds to hole in the inner circle of the Spirograph is located at the distance ρ < r from the center of C2. Without loss of generality it can be assumed that at the initial moment the point A was on the X-axis. In order to find the trajectory created by a Spirograph, follow A as the inner circle is set in motion.
Now mark two points T on C1 and B on C2. The point T indicates where two circles are tangent all the time. Point B however will travel on C2 and its initial location coincides with T. After setting C2 in motion counterclockwise, there is a clockwise rotation with respect to its center. The distances that point B traverses on the small circle is the same as the distance that the tangent point T travels on the large circle due to absence of any slipping effects.
Now the new (relative) system of coordinates $(\hat{X},\hat{Y})$ with its origin at the center of C2 and its axes parallel to X and Y is obserbable. If the parameter t is defined as the angle by which the tangent point T rotates on C1 and $\hat{t}$ is the angle by which C2 rotates (i.e. by which B travels) in the relative system of coordinates, then the distances traveled by B and T along their respective circles must be the same (no slipping). Therefore
$tR=(t-\hat{t})r$
or equivalently
$\hat{t}=-\frac{R-r}{r}t.$
It is common to assume that a counterclockwise motion results in a positive change of angle and a clockwise one will correspond to a negative change of angle. A minus sign in the above formula. ($\hat{t}<0$)to accommodate this convention.
Let (xc,yc) be the coordinates of the center of C2 in the absolute system of coordinates. Then R − r represents the radius of the trajectory of the center of the inner circle, and
$\begin{array}{rcl} x_c&=&(R-r)\cos t,\\ y_c&=&(R-r)\sin t. \end{array}$
The coordinates of A in the new system are $(\hat{x},\hat{y})$ and they obey the regular law of circular motion (the angle of rotation in the relative system is $\hat{t}$):
$\begin{array}{rcl} \hat{x}&=&\rho\cos \hat{t},\\ \hat{y}&=&\rho\sin \hat{t}. \end{array}$
In order to obtain the trajectory of A in the absolute (old) system of coordinates, add these two motions:
$\begin{array}{rcrcl} x&=&\hat{x}+x_c&=&(R-r)\cos t+\rho\cos \hat{t},\\ y&=&\hat{y}+y_c&=&(R-r)\sin t+\rho\sin \hat{t},\\ \end{array}$
where ρ is defined above.
Now, use the relation between t and $\hat{t}$ as discussed above to obtain equations describing the trajectory of point A in terms of one parameter t:
$\begin{array}{rcrcl} x&=&\hat{x}+x_c&=&(R-r)\cos t+\rho\cos \frac{R-r}{r}t,\\[4pt] y&=&\hat{y}+y_c&=&(R-r)\sin t-\rho\sin \frac{R-r}{r}t.\\ \end{array}$
(using the fact that function sin is odd)
It is convenient to represent the equation above in terms the radius R of the largest circle and dimensionless parameters describing the structure of the Spirograph. Namely, let
$l=\frac{\rho}{r}$
and
$k=\frac{r}{R}.$
The parameter $0\le l \le 1$ represents how far the point A is located from the center of the inner circle. At the same time, $0\le k \le 1$ represents how big the inner circle is with respect to the large one.
It is now observed that
$\frac{\rho}{R}=lk,$
and therefore the trajectory equations take form of
$\begin{array}{rcl} x(t)&=&R\left[(1-k)\cos t+lk\cos \frac{1-k}{k}t\right],\\[4pt] y(t)&=&R\left[(1-k)\sin t-lk\sin \frac{1-k}{k}t\right].\\ \end{array}$
Parameter R is a scaling parameter and will not affect the structure of the Spirograph. It is interesting to note that two extreme cases of k = 0 and k = 1 will result in degenerate trajectories of the Spirograph. Namely when k = 0 we will have a simple circle of radius R. And indeed this case corresponds to the case when the inner circle is shrunk into a point. (Division by k = 0 in the formula is not a problem since bothsin  and cos  are bounded functions).
The other extreme case k = 1 corresponds to the inner circle matching the large circle. In this case the trajectory is a single point since the inner circle is too large to roll without slipping.
If l = 1 then it is the case when the point A is on the circumference of the inner circle. In this case the trajectories are called hypocycloids and the equations will match the one describing a hypocycloid.