Lemma: Bounds and resting collar

Uniformly for $\mu \in I$,

Consequently,

For $s>0$, the region $s+k_{\mu }\ge h_{\mu }$ has largest $\rho$-coordinate $s/c<r$ and is not cut by the cap. For $\rho \ge r$, the graph is resting:

In particular, $7/2\le \rho \le 4$ is a fixed resting collar.

Proof. Write $\alpha /2=\pi /4+t$, with $|t|\le \pi /120<27/1000$. Then

The inequalities $\cos t\ge 1-t^2/2$, $|\sin t|\le |t|$, and $\cos t\le 1$ give the stated strict bounds on $a$. Since $0<\alpha <\pi$, $b=(3/5)\sin \alpha \in (0,3/5]$, whence $c>a-3/10>69/200$.

The comparisons in (eq:loop-clearance) follow from $r=2/a$ and $R_{\ast }=1/c$, together with direct rational comparisons. In particular, $200/69<29/10$ is equivalent to $2000<2001$.

Since $\min \{\xi ,\zeta \}\le \rho /2$,

Thus $s+k_{\mu }\ge h_{\mu }$ implies $\rho \le s/c$. Equality occurs on the bisector. The inequality $b<a$ gives $a<2a-b=2c$, hence $1/c<2/a=r$. Finally, $h_{\mu }\ge 2$ when $\rho \ge r$, so the outer minimum in (eq:family) equals $2$.