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


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

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:

Figure 3: Mechanism of the exhibit

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

    \[\pi r=r^\ast\varphi^\ast\quad(1)\]



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

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


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

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

    \[y_0\coloneqq y_1(x_0)=y_2(x_0)\]



and thus


According to equation (1), this results in


und somit


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_ix_0^iy_0^i
Table 1: The values \varphi^\ast and r^\ast as a function of \varphi.

Figure 4 below summarizes this in a diagram.

Figure 4: Diagram showing the points (x_0,y_0) for the different opening angles.

Note: This exhibit is closely related to the exhibits “What is Pi?”, “What is the area of a circle?”, and “Twelve Corners”.