 Solution

Write $T_{\mathrm{sid}}$ for one sidereal day in minutes and ${\Omega }_{\mathrm{sid}}=2\pi /T_{\mathrm{sid}}$. 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 ${\Omega }_{\mathrm{az}}=2\pi \omega$ radians per minute, so the shortest half-turn takes $\Delta t=\pi /{\Omega }_{\mathrm{az}}=1/(2\omega )$ minutes. During that lost-track interval the star moves throughFor $T_{\mathrm{sid}}\simeq 1436.07$ 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 ${\varphi }_{\mathrm{full}}/2$, 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 ${\Omega }_{\mathrm{sid}}|\cot \delta |$ at transit: its minimum offset is constrained by $|\tan \delta |\ge {\Omega }_{\mathrm{sid}}/{\Omega }_{\mathrm{az}}$. 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 the telescope designers' tracking discussion.