applied to the azimuthal equidistant projection: The orange spots on the map are all circular on the sphere. The more the circular spots appear stretched, the more the distortion created by the projection. A point on the globe is chosen as "the center" in the sense that mapped distances and
azimuth directions from that point to any other point will be correct. That point, , will project to the center of a circular projection, with referring to latitude and referring to longitude. All points along a given azimuth will project along a straight line from the center, and the angle that the line subtends from the vertical is the azimuth angle. The distance from the center point to another projected point is the
arc length along a
great circle between them on the globe. By this description, then, the point on the plane specified by will be projected to Cartesian coordinates: x = \rho \sin \theta, \qquad y = -\rho \cos \theta The relationship between the coordinates of the point on the globe, and its latitude and longitude coordinates is given by the equations: \begin{align} \cos \frac{\rho}{R} &= \sin \varphi_0 \sin \varphi + \cos \varphi_0 \cos \varphi \cos \left(\lambda - \lambda_0\right) \\ \tan \theta &= \frac{\cos \varphi \sin \left(\lambda - \lambda_0\right)}{\cos \varphi_0 \sin \varphi - \sin \varphi_0 \cos \varphi \cos \left(\lambda - \lambda_0\right)} \end{align} When the center point is the north pole, equals \pi/2 and is arbitrary, so it is most convenient to assign it the value of 0. This assignment significantly simplifies the equations for and to: \rho = R \left( \frac{\pi}{2} - \varphi \right), \qquad \theta = \lambda~~ == Limitation ==