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An equatorial coordinate system locates directions on the celestial sphere using right ascension and declination, referred to a specified equator, equinox and epoch.

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.