Ruffles have maximally aligned ridges

Two large planar sheets have parallel semicircular cylindrical ridges with radius $1.$ Neighboring ridges are separated by a distance $L \geq 2$. The sheets are placed so that the ridges extrude toward each other, and so that the sheets cannot shift relative to each other in the horizontal direction, as shown in the cross-section below:

Which value of $L$ maximizes the empty space between the sheets?

Firstly, let's establish some parameters in order to make sure we can turn this into a nice(r) optimization problem. Let $h$ be the vertical distance between the two planar sheets for a given value of $L \geq 2$. Let's assume that one cylindrical ridge on the lower sheet is centered at the origin, that is, $x = 0.$ By construction, the next cyliindrical ridge on the lower sheet (to the right) is centered at $x = L$ and the center of the semicircular cylindrical ridges on the upper sheet that is tangent to these two lower ridges should be centered at $x = \frac{L}{2}.$ Adding in the vertical axis, we have semicircles of radius $1$ centered at $(0,0)$, $(L, 0)$ and $(\frac{L}{2}, h).$

Let's next determine the objective function for optimization, in terms of $h$ and $L$. That is, we are trying to determine the maximal value of the cross-sectional area of empty space per unit length of one sheet. That is, we have $f(h, L) = \frac{A(h,L)}{L},$ where $A(h,L)$ is the cross-sectional area of the empty space bounded by the rectangle $0 \leq x \leq L$ and $0 \leq y \leq h.$ Now, the non-empty space in this rectangle is half of the unit semicircle centered at (0,0), the entire unit semicircle centered at (L/2, h) and half of the unit semicricle centered at (L, 0), so we need to subtract out the total non-empty area of $\pi,$ that is, $A(h,L) = hL - \pi,$ so we arrive at $$f(h, L) = \frac{A(h, L)}{L} = \frac{hL - \pi}{L} = h - \frac{\pi}{L}.$$

Now, let's determine the height $h$ as a function of $L,$ that is, $h = h(L).$ Since there is no ability to shift the sheets horizontally, it must be that the semicircle centered at $(\frac{L}{2}, h(L))$ is tangent to both of the lower semicircles. Thus, we can draw a straight line from $(0,0)$ to $(\frac{L}{2}, h(L))$ which must have length $2$, since each of the semicircles have unit radii, thus we have the Pythagorean theorem $$2^2 = \left(\frac{L}{2}\right)^2 + h(L)^2,$$ or equivalently $$h(L) = \frac{1}{2} \sqrt{16 - L^2}.$$ In actuality, we should be careful to make sure that the space between the sheets is at least the unit radius of the semicircular ridges, that is, $$h(L) = \max \left\{ 1, \frac{1}{2} \sqrt{16 - L^2} \right\}.$$ However, we can disregard the case where $L \gt 2\sqrt{3}$ where $h(L) = 1 \gt \frac{1}{2} \sqrt{16-L^2}$, since otherwise, barring some extreme static friction, we could horizontally shift the sheets relative to one another which is a no-no of the setup.

Having decided to ignore the other case, we have $h(L) = \frac{1}{2} \sqrt{16-L^2}$ for all $2 \leq L \leq 2\sqrt{3},$ so we now have $$f(L) = f(h(L), L) = \frac{1}{2} \sqrt{16 - L^2} - \frac{\pi}{L}.$$ Taking derivatives, we $$f^\prime(L) = \frac{-L}{2 \sqrt{16-L^2}} + \frac{\pi}{L^2},$$ which if we set equal to zero and solve for $L$ results in the sextic polynomial $$L^6 - 4\pi^2 L^2 + 64 \pi^2 = 0.$$ If we set $\ell = L^2,$ then the condition for optimality becomes a depressed cubic in $\ell$ with a single real zero, that can be handled by Cardano's formula, that is, $$\ell = \sqrt[3]{ 32 \pi^2 + \sqrt{ 1024 \pi^4 + \frac{64}{27} \pi^6} } + \sqrt[3]{ 32\pi^2 - \sqrt{1024 \pi^4 + \frac{64}{27} \pi^6}} \approx 7.06550552638\dots$$ Therefore, the optimal spacing of the ridges is $$L^* = \sqrt{\ell} \approx 2.65810186531\dots.$$