= Solution
Write $T_{\rm sid}$ for one sidereal day in minutes and $\Omega_{\rm sid}=2\pi/T_{\rm sid}$. An equatorial <star> at an equatorial site rises along a vertical <great circle>, passes through the <zenith>, and descends on the opposite side. Its <azimuth> changes by $\pi$ at the crossing. The restricted <altitude> travel prevents following it continuously by rotating over the <zenith>.
The maximum <azimuth> rate is $\Omega_{\rm az}=2\pi\omega$ radians per minute, so the shortest half-turn takes $\Delta t=\pi/\Omega_{\rm az}=1/(2\omega)$ minutes. During that lost-track interval the <star> moves through
$$
\boxed{\phi_{\rm full}=\Omega_{\rm sid}\Delta t=\frac{\pi}{\omega T_{\rm sid}}\ \mathrm{radians}=\frac{180^\circ}{\omega T_{\rm sid}}.}
$$
For $T_{\rm sid}\simeq1436.07$ minutes this is approximately $0.12534^\circ/\omega$; using a 24-hour day gives $0.125^\circ/\omega$. If the quoted <zenith blind spot> size means the angular radius of a symmetrically placed crossing gap, it is $\phi_{\rm full}/2$, approximately $0.06267^\circ/\omega$.
This derives the ideal crossing-gap size with instantaneous acceleration and an available <altitude> drive. It does not uniquely define a circular forbidden region for all nearby trajectories. For example, at an equatorial site a <star> of small nonzero <declination> $\delta$ has maximum <azimuth> rate $\Omega_{\rm sid}|\cot\delta|$ at transit: its minimum offset is constrained by $|\tan\delta|\ge\Omega_{\rm sid}/\Omega_{\rm az}$. That rate contour and the half-turn crossing gap are different definitions of a <zenith blind spot>. Actual tracking footprints also depend on acceleration and the reacquisition strategy, as discussed in https://www.astro.uni.torun.pl/~kb/HandbRT32/ChapterII.htm[the telescope designers' tracking discussion].
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