SIAM News Blog
SIAM News

# Measuring Areas with a Shopping Cart

Figure 1. Trajectories of the two wheels as the front one repeatedly circumscribes a closed curve.
One late night in a deserted supermarket, I was waiting in a check-out aisle for a cashier. With no one around and nothing better to do, I began rolling the shopping cart’s front wheel around the outline of a floor tile. The cart ended up rotated after one traversal, as Figure 1 illustrates (for a round “tile” and with a “bike” instead of a shopping cart). It then dawned on me that the angle $$\theta$$, by which the cart turns in one cycle (see Figure 1), is proportional to the area $$A$$ of the tile, up to a small relative error if the diameter $$\varepsilon$$ of the curve (not necessarily a circle) is small:

$A= \theta L ^2 + O( \epsilon ^3 ), \tag1$

where $$L$$ is the length of the "bike."

Figure 2. The hatchet planimeter; (1) gives the area.
As happens with nearly every observation, someone noticed this before. Holger Prytz (a Danish cavalry officer) proposed the idea of calculating areas more than 100 years ago, and without the advantage of a shopping cart. One can find a beautiful description of this in [2], along with the discovery that the bike’s direction angle changes after a cycle according to the Möbius transformation restricted to a circle,

$z\mapsto e^{i \alpha } \frac{z-a}{1-\overline a z},$

Figure 3. RF is a directed segment with R’s velocity constrained to the line RF.
where $$\alpha$$ and $$a$$ are functionals of the front path. The Prytz planimeter, also called the hatchet planimeter, is sketched in Figure 2. The hatchet moves like a rear wheel, not sliding sideways. The needle is guided around the region’s boundary, and the angle $$\theta$$ gives the area according to $$(1)$$.

A geometrical explanation of $$(1)$$ rests on two observations of independent interest.

Figure 4. The doubly-swept area contributes zero, leaving AQ-AP as the net swept area.

### Observation 1

The signed area1 swept by a moving segment (as described in Figure 3) remains unchanged if the longitudinal velocity is altered (and in particular made to vanish).

This statement is intuitively plausible since the longitudinal motion has no effect on the rate at which $$RF$$ sweeps the area.

### Observation 2

A segment $$PQ$$ executing a cyclic motion in the plane sweeps the signed area $$A_Q-A_P$$ (see Figure 4).

Figure 5. Proof of Prytz’s formula (1).

### Proof of Prytz’s Formula (1)

Figure 5 shows the motion of $$RF$$ over one zig-zag, with $$F$$ returning to its starting position; the rotation around the start/stop point through $$\theta$$ completes the cyclic motion, bringing $$RF$$ to its initial position (the rotation violates the no-slip condition).

During the “sliding” stage, $$RF$$ sweeps area $$\frac{1}{2} \theta L ^2$$, according to Observation 1. During the “rotating” stage, $$RF$$ sweeps the sector of area $$\frac{1}{2} \theta L ^2$$. The total swept area is thus $$\theta L ^2$$.

But this area equals $$A -A_R$$ by Observation 2, so that

$A -A_R= \theta L ^2.$

Figure 6. πc2=πa2+ring=πa2+πb2.
It is not hard to show that $$A_R = O ( \varepsilon ^3 )$$; this completes the outline of the proof of $$(1)$$.

It is worth noting that if one extends the “bike” to $${\mathbb R} ^3$$, Prytz’s formula $$(1)$$ admits an eye-opening explanation entirely different from the one I just described. This explanation (given in [1]) is similar in spirit to the explanation of the finiteness of the radius of convergence of $$1/(1+ x ^2 )$$ by extending to the complex domain.

As a concluding remark, choosing a circular annulus in Figure 3 yields another proof of the Pythagorean theorem (modulo the proof of Observation 1), as Figure 6 illustrates.

1 The area counts with the positive sign if the motion is to the left of $$RF$$, as it is in Figure 3.