Let
R be the radius of the sphere,
φ the
latitude,
λ the
longitude, and
λ0 the longitude of the central meridian (chosen as desired). Also, define s = \sqrt{1 - \sin \phi} = \sqrt{2} \sin\left(\frac{\pi}{4} - \frac{\phi}{2}\right), where the two forms are equivalent for
φ in the range of possible latitudes. Then the Collignon projection is given by: :\begin{align} x &= \frac{2}{\sqrt{\pi}}R \left( \lambda - \lambda_0 \right) s, \\ y &= \sqrt{\pi} R \left( 1 - s \right). \end{align} This formula gives the projection as pictured above, coming to a point at the
North Pole. For a projection coming to a point at the
South Pole, as in the bottom portion of the HEALPix projection, replace
φ and
y with
-φ and
-y. The standard parallel is 15°51′N. ==See also==