Can You Squeeze the Sheets?

Two identical flat sheets carry parallel semicircular ridges of radius $1$, with adjacent ridges a distance $L\ge2$ apart. The sheets face each other, ridges toward ridges, and cannot slide sideways. Which $L$ leaves the most empty space between them, measured as cross-sectional area per unit length?

The Fiddler, Zach Wissner-Gross, January 3, 2025(original post)

Solution

To pack as tightly as possible the sheets are offset by half a period, so each ridge nestles between two opposing ridges. A top ridge centre and the nearest bottom ridge centre are then separated horizontally by $L/2$ and vertically by the gap $h$; the ridges (radius $1$ each) touch when the centres are $2$ apart: $$\Bigl(\tfrac{L}{2}\Bigr)^2+h^2=4 \implies h=\sqrt{4-\tfrac{L^2}{4}}.$$ In one period of width $L$ the slab between the two centre-lines has area $hL$, from which the two half-disks (one per sheet, total area $\pi$) are solid. The empty area per unit length is therefore $$E(L)=\frac{hL-\pi}{L}=\sqrt{4-\frac{L^2}{4}}-\frac{\pi}{L}.$$ Setting $E'(L)=0$ gives $4\pi\sqrt{4-L^2/4}=L^3$, so $$L\approx\boxed{2.658},\qquad E\approx 0.3126.$$

The computation

Rather than evaluate $E(L)$, encode the geometry directly: for each spacing $L$, find the gap $h$ at which the offset ridges just touch, then lay a fine grid over one period of the cross-section and measure the fraction of it that lies inside neither sheet’s semicircles. Scanning $L$ locates the most open spacing.

import numpy as np
from scipy.optimize import brentq, minimize_scalar
def empty_per_length(L):
    h = brentq(lambda h: (L / 2)**2 + h**2 - 4, 0, 2)   # ridges just touch
    n = 1500
    xs = (np.arange(n) + 0.5) * L / n
    ys = (np.arange(n) + 0.5) * h / n
    X, Y = np.meshgrid(xs, ys)
    bottom = (X**2 + Y**2 < 1) | ((X - L)**2 + Y**2 < 1)  # bumps up at x=0, L
    top = (X - L / 2)**2 + (Y - h)**2 < 1                 # bump down at x=L/2
    return (~(bottom | top)).mean() * h                   # empty area / period / L
res = minimize_scalar(lambda L: -empty_per_length(L),
                      bounds=(2.001, 3.999), method='bounded')
print(round(res.x, 3), round(-res.fun, 4))   # 2.658  0.3126

Extra Credit

With hemispherical dimples of radius $1$ instead of ridges (centres at least $2$ apart, any number, sheets not necessarily identical), what is the minimum empty space, as volume per unit area of one flat sheet?

The two-dimensional trick (offset the bumps and let them touch) becomes a sphere-packing question in the gap, and the optimum trades the gap height against how densely the hemispheres tile each sheet. The densest planar arrangement of centres at spacing $2$ is the triangular lattice. The minimum is the source’s paywalled extra credit; this is a packing optimisation rather than a single derivative, so I leave it as a computational target rather than assert a number I cannot check.