If the plane is tangent at some point other than a pole, the concentric circles represent distances from the point of tangency. In this case, meridians and parallels appear as curves.

The projection can be used to portray the entire earth, the point 180° from the point of tangency appearing as the largest of the concentric circles. The projection is not conformal, equal area, or perspective. Near the point of tangency distortion is small, increasing with distance until shapes near the opposite side of the earth are unrecognizable (Figure 1).

The projection is useful because it combines the three features of being azimuthal, having a constant distance scale from the point of tangency, and permitting the entire earth to be shown on one map. Thus, if an important harbor or airport is selected as the point of tangency, the great-circle course, distance, and track from that point to any other point on the earth are quickly and accurately determined. For communication work with the station at the point of tangency, the path of an incoming signal is at once apparent if the direction of arrival has been determined and the direction to train a directional antenna can be determined easily. The projection is also used for polar charts and for the star finder.

Mathematics

Let $\phi_1$ and $\lambda_0$ be the latitude and longitude of the center of the projection, then the transformation equations are given by:

\begin{align*} x &= k \cos \phi \sin \left( \lambda-\lambda_0 \right) \\ y&=k \left[ \cos \phi_1 \sin\phi - \sin \phi_1 \cos\phi\cos\left(\lambda-\lambda_0 \right)\right]\end{align*}

Here,

$k = \frac{c}{\sin c}$

and

$\cos c = \sin\phi_1\sin\phi+\cos\phi_1\cos\phi\cos\left(\lambda-\lambda_0\right)$

where $c$ is the angular distance from the center. The inverse formulas are:

$\phi=\sin^{-1}\left(\cos c \sin\phi_1 + \frac{y\sin c \cos\phi_1}{c}\right)$

and

$\lambda = \left\{\begin{matrix} \lambda_0 + \tan^{-1} \left(\frac{x \sin c}{c \cos\phi_1 \cos c - y \sin\phi_1 \sin c} \right) & for \phi_1 \neq \pm 90^o\\ \lambda_0 + \tan^{-1} \left(- \frac{x}{y}\right) & for \phi_1 = 90^o\\ \lambda_0 + \tan^{-1} \left( \frac{x}{y}\right) & for \phi_1 = -90^o, \end{matrix}\right$

with the angular distance from the center given by:

$c = \sqrt{x^2 + y^2}$