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The gnomonic projection sends the visible open hemisphere of a sphere to a tangent plane by central rays. Work on the unit sphere. Choose a unit vector $\mathbf{b}$ towards the tangent point and a plane $\mathbf{x}\cdot \mathbf{b}=f$, where $f>0$. Choose an orthonormal basis $\mathbf{e},\mathbf{n}$ perpendicular to $\mathbf{b}$. A point $\mathbf{s}$ with $\mathbf{s}\cdot \mathbf{b}>0$ projects toThe positivity condition ensures that the forward ray meets the chosen plane. For $f=1$ the plane is tangent to the unit sphere; changing $f$ rescales the image.

A great circle lies in a plane through the origin. Its projected arc lies in the straight line where that plane meets the image plane. This makes the gnomonic projection useful for drawing shortest spherical routes. For the northern hemisphere, take $\mathbf{b}=(0,0,1)$ and $f=1$: the map is $(x,y,z)\mapsto (x/z,y/z)$. The inverse sends $(\xi ,\eta )$ to $(\xi ,\eta ,1)/\sqrt{1+{\xi }^2+{\eta }^2}$. As a point approaches the bounding great circle $z=0$, its image escapes to infinity. The opposite hemisphere needs a separate chart; using the same formula there would identify antipodal points.

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.