=(L/\pi) \ln(\cosh(\pi(z-a)/L)/\cosh(a \pi/L)) $ while leaving $0$ and $i L$ fixed. The no-hit probability is $${\cal P} = |g'_A(0)|^{5/8}  |g'_A( i L )|^{5/8} = [\tanh(a \pi/L)]^{5/4}$$ Note that this is also the cumulative distribution of $x_{\max}=a$, the [*maximum excursion of a SAW*]{}. The total force exerted on the piston is $f_x = 5 \pi/[2 L \sinh(2 a \pi/L)]$.

Consider now the “door” geometry, i.e. a segment $z=a+ t e^{i \alpha}$ with $t \in [0,h]$, of angle $\alpha=b \pi$. The relevant map $w=g(z)$ has an explicit form in terms of its inverse map $z=f(w)$ with $f(w)=a+(w-x_1)[(w-x_3)/(w-x_1)]^b$, $0<x_1<x_3$ with $a=x_1(x_3/x_1)^b$ and $h=b^b(1-b)^{1-b}(x_3-x_1)$. The no-hit probability is $$\begin{aligned}
\label{21}
 {\cal P} &=&\big[ \mu^{b} (1-b(1-\mu^{-1})) \big]^{-5/8} \\
 h/a &=& b^b(1-b)^{1-b} \mu^{-b} (\mu-1) \ ,
 \label{22}\end{aligned}$$ where $\mu=x_3/x_1>1$ is solution of Eq. (\[22\]). The numerical solution is given on figure \[f:door\]. An interesting limit is represented in Fig. \[f:N1\], where $h=1/\sin(\pi k)$, $a=\ell + \cot(\pi k)$ and $k=1-b$ tends to zero. One finds that $\mu=\frac{1}{w k}  + O(k^0)$ with $w=W(e^{\ell \pi-1})$ the product-log function $W(z)$ solution of $z= W e^W$. This gives the no-hit probability of the horizontal half-line $i + x$ with $x>\ell$, plotted on Fig. \[f:N1\], $$\begin{aligned}
\label{magic}
{\cal P} &=& \Big[1+\frac{1}{W\left(e^{\ell\pi - 