The compressed circle

Figure 1: The exhibit in its initial state

As the preceding Figure 1 shows, the experiment at the exhibit “The Compressed Circle” consists of bending up a circular spring by means of two (initially vertical and parallel) rotatable levers in such a way that its shape in the final position (see Figure 2 below) represents a distance from point $A$ to point $B$ .

Figure 2: The exhibit in its final state

Here

$\lvert\overline{AB}\rvert=2\lvert\overline{AC}\rvert=2\lvert\overline{BC}\rvert=2\pi r$

is the length of the circumference of the circle that the spring originally formed. Moreover, the area $F_\Delta$ of the right triangle $\Delta$ with vertices $M$ , $B$ and $C$ (see Figure 1) is equal to half the area of the circle, i.e.

$F_\Delta=\frac{1}{2}\pi r^2.$

Thus, the area $\pi r^2$ of a circle with radius $r$ can be represented by the sum of the areas of two congruent triangles.

And now … the mathematics of it:

If the two levers are turned by the angle $\varphi$ against the vertical axis, the contact point $(x_0,y_0)$ between the right lever arm and the bent-up circle results on the right side. The latter now represents itself as a circular arc with the radius $r^\ast$ and the opening angle $\varphi^\ast$ (in radians!). Thus

$\pi r=r^\ast\varphi^\ast\quad(1)$

und

$x_0^2+(r^\ast-y_0)^2=(r^\ast)^2\quad(2).$

According to Figure 3 above, for the straight lines $y_1$ and $y_2$ ,

$y_1\colon y_1(x)=r+\tan(\pi/2-\varphi)x$

and

$y_2\colon y_2(x)=r^\ast+\tan(\pi/2+\varphi^\ast)x.$

Their intersection $(x_0,y_0)$ is then obtained as the solution of the equation

$y_0\coloneqq y_1(x_0)=y_2(x_0)$

i.e.

$r+\tan(\pi/2-\varphi)x_0=r^\ast+\tan(\pi/2+\varphi^\ast)x_0$

and thus

$x_0(\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast))=r^\ast-r.$

According to equation (1), this results in

$x_0=\frac{r^\ast-r}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}=r\frac{\pi/\varphi^\ast-1}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}$

und somit

$y_0=r+\tan(\pi/2-\varphi)x_0=r+\tan(\pi/2-\varphi)r\frac{\pi/\varphi^\ast-1}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}.$

Equations (1) and (2) then yield

$x_0^2+(r^\ast-y_0)^2=(r^\ast)^2=\frac{\pi^2 r^2}{(\varphi^\ast)^2}$

and thus

$x_0^2+\left(\frac{\pi r}{\varphi^\ast}-y_0\right)^2=\frac{\pi^2 r^2}{(\varphi^\ast)^2},$

which, after multiplication out and truncation, leads to

$x_0^2-\frac{2\pi r y_0}{\varphi^\ast}+y_0^2=0$

leads. This is now called

(1) $\begin{gather*} \left(r\frac{\pi/\varphi^\ast-1}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}\right)^2\ -\frac{2\pi r}{\varphi^\ast}\left(r+\tan(\pi/2-\varphi)r\frac{\pi/\varphi^\ast- 1}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}\right)\+\left(r+\tan(\pi/2-\varphi)r\frac{\pi/\varphi^\ast-1}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}\right)^2=0. \end{gather*}$

Consequently, without restriction, it can be assumed that $r=1$ , so that for a given opening angle $\varphi$ (in radians) of the right lever (see Figure 3), the opening angle $\varphi^\ast$ of the corresponding circular arc (with radius $r^\ast$ ) is obtained as the solution of the following equation:

(2) $\begin{gather*} \left(\frac{\pi/\varphi^\ast-1}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}\right)^2\ -\frac{2\pi}{\varphi^\ast}\left(1+\tan(\pi/2-\varphi)\frac{\pi/\varphi^\ast- 1}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}\right)\+\left(1+\tan(\pi/2-\varphi)\frac{\pi/\varphi^\ast-1}{\tan(\pi/2-\varphi)-\tan(\pi/2+\varphi^\ast)}\right)^2=0. \end{gather*}$

Finally, we give — determined numerically as approximations — for $\varphi_i=\frac{i}{10}(\pi/2+\varphi_0)=\frac{i}{10}(\pi/2+\arctan(1/\pi))$ ( $i=1,\ldots,10$ ) the corresponding angles $\varphi^\ast$ and the radii $r^\ast$ (using the above equation).

$i$	$\varphi_i$	$\varphi_i^\ast$	$\varphi_i^\ast$ (in Radian)	$r^\ast_i$	$x_0^i$	$y_0^i$
$1$	$0,1879$	$2,942$	$168,5$	$1,068$	$0,2148$	$2,1298$
$2$	$0,3760$	$2,716$	$155,6$	$1,157$	$0,4770$	$2,2089$
$3$	$0,5637$	$2,462$	$141,1$	$1,276$	$0,8015$	$2,2680$
$4$	$0,7516$	$2,179$	$124,8$	$1,442$	$1,1819$	$2,2647$
$5$	$0,9395$	$1,869$	$107,1$	$1,681$	$1,6070$	$2,1748$
$6$	$1,1274$	$1,534$	$87,9$	$2,048$	$2,0477$	$1,9726$
$7$	$1,3153$	$1,176$	$67,4$	$2,671$	$2,4656$	$1,6441$
$8$	$1,5031$	$0,800$	$45,8$	$3,927$	$2,8173$	$1,1908$
$9$	$1,6911$	$0,4071$	$23,3$	$7,717$	$3,0555$	$0,6307$
$10$	$1,8789$	$0$	$0$	$\infty$	$\pi$	$0$