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On the unit sphere with one point deleted, a distinct nonantipodal pair has a length minimizer exactly when its shorter great circle arc avoids the deleted point. Antipodal pairs admit an avoiding semicircle. If the unique shorter arc crosses the puncture, smooth detours approach its length without attaining it.

The gnomonic projection intersects rays from a sphere’s centre with a tangent plane. If $\mathbf{b}$ is its unit normal and $\mathbf{e},\mathbf{n}$ are tangent basis vectors, a direction $\mathbf{s}$ maps to $f(\mathbf{s}\cdot \mathbf{e},\mathbf{s}\cdot \mathbf{n})/(\mathbf{s}\cdot \mathbf{b})$. It maps great circle arcs to straight lines and is singular at the tangent-plane horizon.

The length of a curve is $L(\alpha )={\int }_a^b\Vert {\alpha }^{\prime }(t)\Vert dt$. Its arc length parameter is $s(t)={\int }_a^t\Vert {\alpha }^{\prime }(u)\Vert du$. Regularity gives $s^{\prime }(t)>0$, so the inverse function theorem gives a smooth inverse and $\beta (s)=\alpha (t(s))$ has $\Vert {\beta }^{\prime }(s)\Vert =1$.

Reparametrize a length minimizer to constant speed on $[a,b]$ and use the energy functional $E=\frac{1}{2}{\int }_a^b\Vert {\alpha }^{\prime }{\Vert }^{2}dt$. By Cauchy-Schwarz inequality, every competitor has $E\ge L^2/(2(b-a))$, with equality for constant speed, so this parametrization also minimizes $E$. For a fixed-endpoint variation with tangential vector field $V$, differentiation and integration by parts giveThe boundary term is zero. Arbitrary smooth tangential $V$ of compact interior support are realizable by the stated variation fact. Taking $V$ to be a nonnegative cutoff times ${\nabla }_t{\alpha }^{\prime }$ proves ${\nabla }_t{\alpha }^{\prime }=0$. Hence a length minimizer is a geodesic after constant-speed reparametrization. The original parameter need not be affine: arbitrary varying-speed parametrizations preserve length but have nonzero covariant acceleration. This qualification is necessary for the literal wording.

For distinct nonantipodal $p,q$ on the punctured unit sphere, there is a unique shorter great circle arc, of length $d=\arccos (p\cdot q)<\pi$. A minimizing curve exists exactly when that shorter arc avoids the removed north pole. If it avoids the pole it attains the spherical lower bound. If it passes through the pole, arbitrarily small smooth detours have lengths tending to $d$, but equality would force the unique shorter arc, which is unavailable. Thus the infimum is not attained; the longer great circle arc is not a substitute minimizer.

For antipodal $p,q$ in the punctured sphere, one can choose a great circle semicircle avoiding the north pole; it attains length $\pi$. Thus every admissible antipodal pair has a minimizer. If identical endpoints are included despite the earlier distinctness stipulation, the infimum is zero, but no smooth regular curve attains it; only a nonregular constant curve does.

The great circle containing the pointing and celestial pole is the intersection of the unit sphere with the meridian plane $y=0$. The orthogonal projection of $\mathbf{s}$ onto that plane, normalized to unit length, isWithin the local meridian chart $\cos \Delta >0$, its declination is therefore $q$. Outside that chart, the same construction gives an oriented great circle parameter; the foot can lie on the opposite right-ascension half of the meridian, and $q$ need not be its ordinary declination.

To verify the right angle geometrically, the tangent to the great circle from $L$ toward $S$ is proportional to $\mathbf{s}-(\mathbf{s}\cdot \mathbf{\ell })\mathbf{\ell }$. Since $\mathbf{s}\cdot \mathbf{\ell }=\rho$, that tangent has only a $y$ component. The tangent from $L$ toward the pole, $\mathbf{P}-(\mathbf{P}\cdot \mathbf{\ell })\mathbf{\ell }$, lies in $y=0$. Their orthogonality proves the claimed spherical right angle. Thus within the local meridian chart, $q$ is the declination of the perpendicular foot on the pointing meridian.

There are antipodal perpendicular feet on the complete great circle. On the local meridian chart, the telescope field selects the foot near the pointing; the cotangent equation alone determines $q$ only modulo $\pi$. At a coincident foot or pole, the corresponding angle is understood by continuity rather than by a nonzero tangent vector.

Use the gnomonic projection associated with an undistorted focal plane. Write $\Delta =\alpha -A$ and rotate the celestial Cartesian coordinate system so that the pointing meridian has longitude zero. The stellar direction and an orthonormal basis adapted to the optical axis areHere $\mathbf{e}$ points east and $\mathbf{n}$ points north. Intersect the ray $t\mathbf{s}$ with the plane $\mathbf{x}\cdot \mathbf{b}=f$. It gives $t=f/(\mathbf{s}\cdot \mathbf{b})$, so the detector coordinates areThese equations also avoid spurious singularities caused by writing individual tangents or cotangents.

To put this gnomonic projection in the desired form, let $\rho =({\cos }^2\delta {\cos }^2\Delta +{\sin }^2\delta {)}^{1/2}$ and choose $q$ locally byThus $\cot q=\cot \delta \cos \Delta$. The denominator becomes $\rho \cos (q-D)$, the numerator for $\eta$ becomes $\rho \sin (q-D)$, and $\cos \delta \sin \Delta =\rho \cos q\tan \Delta$. ConsequentlyUse the local meridian chart $\cos \Delta >0$, with $q\in [-\pi /2,\pi /2]$ chosen continuously near $D$ and $q=D$ at the image centre. A visible gnomonic projection requires $\mathbf{s}\cdot \mathbf{b}>0$, but this front-hemisphere condition alone does not select that meridian chart. A narrow field near a celestial pole can cross the opposite meridian; for such fields use the Cartesian expressions above and distinguish the oriented great circle parameter from ordinary declination. In particular, the small-field limit is $\xi \simeq f\cos D(\alpha -A)$ and $\eta \simeq f(\delta -D)$, with angles in radians. The factor $\cos D$ is the shrinking angular distance per unit right ascension near the pole.