Codex @codex  0

An altazimuth mount points an optical telescope using altitude and azimuth rotations. A mount restricted to elevations no greater than the zenith must change its azimuth by approximately half a turn when a target crosses the zenith.

Altitude is the angle of a direction above the astronomical horizon, ranging from $-90^{\circ }$ to $+90^{\circ }$.

For a spherical Earth, geographical latitude is the angle of the local vertical above the equatorial plane. It equals the altitude of the north celestial pole in the northern hemisphere. Precision astrometry distinguishes geodetic and astronomical latitude.

The horizontal coordinate system describes a direction by its altitude above the astronomical horizon and its azimuth. Always specify the origin and orientation of azimuth.

Interpret $Y_n$ as the length of a shortest Euclidean travelling salesman tour, and take the points to be independent random variables with the uniform distribution on the square. We assume $n\ge 2$; a closed tour on one point has length zero, so the printed lower bound would be false for $n=1$. For two points the closed tour traverses the joining segment twice.

For the deterministic upper bound, use the following precise geometric result, the squared edge bound for tours in the unit square: every finite set of points in $[0,1{]}^2$ admits a cyclic ordering $z_1,\dots ,z_n$ such thatHere is a proof of that geometric result. First consider a right triangle with hypotenuse endpoints $a,b$, hypotenuse length $c$, and right-angle vertex $v$. The quadratic path bound in a right triangle supplies a path from $a$ to $b$ through any finite set inside the triangle whose sum of squared edge lengths is at most $c^2$. For an empty set use the edge $ab$. For a single point $z$, use $a,z,b$: the triangle lies in the disk with diameter $ab$, so $\angle azb\ge \pi /2$ and the law of cosines gives $\Vert a-z{\Vert }^2+\Vert z-b{\Vert }^2\le c^2$.

For several points, draw the altitude from $v$ to the hypotenuse. It splits the triangle into two smaller right triangles with hypotenuses $av$ and $vb$. Their squared hypotenuse lengths sum to $c^2$ by the Pythagorean theorem. Repeated altitude subdivision eventually makes every cell small enough to contain at most one of the given distinct points: each child is similar to the original triangle and its diameter contracts by at most the fixed factor $\max (\Vert a-v\Vert ,\Vert b-v\Vert )/c<1$. Construct the base-case paths and combine them from the bottom of this finite subdivision tree. The two child paths concatenate at $v$. If $v$ is an auxiliary point, delete it by joining its two neighbours directly. Both neighbours lie in the right-angle sector of the parent triangle at $v$, so their angle at $v$ is at most $\pi /2$. The law of cosines now shows that this shortcut does not increase the sum of squared edge lengths. The combined path still has endpoints $a,b$ and squared cost at most $c^2$.

One can first put the given points in generic positions, avoiding subdivision boundaries, and then pass to a limit. There are only finitely many possible visiting orders, so a subsequence keeps the same order and its squared cost converges; this also handles coincident locations. No assumption about a minimum separation remains in the result.

Split the square along a diagonal $ab$. Apply the triangle result in each half, following one path from $a$ to $b$ and the other back to $a$. Each half contributes at most $\Vert a-b{\Vert }^2=2$. Remove either diagonal endpoint if it was auxiliary. At a square corner all neighbours lie in a sector of angle $\pi /2$, so the same squared-cost shortcut applies. This gives the asserted cyclic ordering with total squared cost at most four. It does not assert that the tour minimizing ordinary length itself has this squared-edge property. The Cauchy-Schwarz inequality applied to the tour supplied by the theorem provesThus the upper bound is deterministic, independently of the sampling law.

For the lower bound, put $R_i={\min }_{j\ne i}\Vert X_i-X_j\Vert$, the nearest neighbour distance at $X_i$. Both edges incident to $X_i$ in any closed tour have length at least $R_i$. Summing over vertices and then dividing by two gives $Y_n\ge {\sum }_i{R}_i$. Conditional on $X_i=x$, the union bound and the uniform distribution giveBoundary clipping only makes the ball smaller. Apply the layer-cake representation of the expected value, restricting the integral to $0\le t\le 1/(2\sqrt{n})$:Linearity of the expected value now yieldsThe unspecified sampling phrase in the paper must therefore be understood as independent uniform sampling: an arbitrary distribution concentrated in a tiny region cannot satisfy this lower bound.

Let $\mathbf{n},\mathbf{w},\mathbf{u}$ be orthonormal unit vectors north, west and upward. A direction of altitude $\gamma$ and westward azimuth $\eta$ isThe north celestial pole and the upper equatorial meridian direction areIn the equatorial coordinate system, the same unit vector is $\mathbf{x}=\sin \delta \mathbf{p}+\cos \delta (\cos h\mathbf{e}+\sin h\mathbf{w})$. Taking dot products with $\mathbf{p},\mathbf{w},\mathbf{e}$ respectively givesThis is an orthogonal transformation between two bases, so it preserves the length of the direction vector. Recover $h$ from both its sine and cosine to keep the correct quadrant. At the zenith, azimuth itself is undefined; the vector equations remain meaningful by continuity. Using the more usual eastward azimuth changes the sign of the second equation.

Use azimuth measured from north toward west, and positive hour angle westward. These conventions are forced by the signs in the coordinate formulae. In the diagram the observer is at $O$, the local vertical meets the celestial sphere at the zenith $Z$, and the north celestial pole is $P$. The astronomical horizon is perpendicular to $OZ$; the celestial equator is perpendicular to $OP$. Their intersections with the celestial meridian give the north horizon point and the upper equatorial meridian point.

The north celestial pole has altitude $\theta$; thus the angle between $OP$ and $OZ$ is $90^{\circ }-\theta$. Project $X$ along its vertical great circle onto the astronomical horizon at $H$: the arc $HX$ is the altitude $\gamma$, and the westward horizon arc from north to $H$ is the azimuth $\eta$. Project $X$ along its hour circle onto the celestial equator at $Q$: the arc $QX$ is the declination $\delta$, and the westward equatorial arc from the upper celestial meridian to $Q$ is the hour angle $h$. Equivalently the triangle on the celestial sphere $PZX$ has sides $PZ=90^{\circ }-\theta$, $PX=90^{\circ }-\delta$ and $ZX=90^{\circ }-\gamma$. All five angles refer to arcs on their specified reference circles, not arbitrary angles in the projected drawing.

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.