# Cassini (Cassini-Soldner)¶

Although the Cassini projection has been largely replaced by the Transverse Mercator, it is still in limited use outside the United States and was one of the major topographic mapping projections until the early 20th century.

 Classification Transverse and oblique cylindrical Available forms Forward and inverse, Spherical and Elliptical Defined area Global, but best used near the central meridian with long, narrow areas Implemented by Gerald I. Evenden Options +lat_0 Latitude of origin (Default to 0)

## Usage¶

There has been little usage of the spherical version of the Cassini, but the ellipsoidal Cassini-Soldner version was adopted by the Ordnance Survey for the official survey of Great Britain during the second half of the 19th century [Steers1970]. Many of these maps were prepared at a scale of 1:2,500. The Cassini-Soldner was also used for the detailed mapping of many German states during the same period.

$echo 0.17453293 -1.08210414 | proj +proj=cass +lat_0=10.44166666666667 +lon_0=-61.33333333333334 +x_0=86501.46392051999 +y_0=65379.0134283 +a=6378293.645208759 +b=6356617.987679838 +to_meter=0.201166195164 +no_defs 66644.94 82536.22  Example using EPSG 3068 (Soldner Berlin): $ echo 13.5 52.4 | proj +proj=cass +lat_0=52.41864827777778 +lon_0=13.62720366666667 +x_0=40000 +y_0=10000 +ellps=bessel +datum=potsdam +units=m +no_defs
31343.05    7932.76


## Mathematical definition¶

The formulas describing the Cassini projection are taken from Snyder’s [Snyder1987].

$$\phi_0$$ is the latitude of origin that match the center of the map (default to 0). It can be set with +lat_0.

### Spherical form¶

#### Forward projection¶

$x = \arcsin(\cos(\phi) \sin(\lambda))$
$\DeclareMathOperator{\arctantwo}{arctan2} y = \arctantwo(\tan(\phi), \cos(\lambda)) - \phi_0$

#### Inverse projection¶

$\phi = \arcsin(\sin(y+\phi_0) \cos(x))$
$\DeclareMathOperator{\arctantwo}{arctan2} \lambda = \arctantwo(\tan(x), \cos(y+\phi_0))$

### Elliptical form¶

#### Forward projection¶

$N = (1 - e^2 \sin^2(\phi))^{-1/2}$
$T = \tan^2(\phi)$
$A = \lambda \cos(\phi)$
$C = \frac{e^2}{1-e^2} cos^2(\phi)$
$x = N ( A - T \frac{A^3}{6} - (8-T+8C)T\frac{A^5}{120})$
$y = M(\phi) - M(\phi_0) + N \tan(\phi)(\frac{A^2}{2} + (5-T+6C)\frac{A^4}{24})$

and M() is the meridionial distance function.

#### Inverse projection¶

$\phi' = M^{-1}(M(\phi_0)+y)$

if $$\phi' = \frac{\pi}{2}$$ then $$\phi=\phi'$$ and $$\lambda=0$$

otherwise evaluate T and N above using $$\phi'$$ and

$R = (1 - e^2)(1 - e^2 sin^2 \phi')^{-3/2}$
$D = x/N$
$\phi = \phi' - \tan \phi' \frac{N}{R}(\frac{D^2}{2}-(1+3T)\frac{D^4}{24})$
$\lambda = \frac{(D - T\frac{D^3}{3} + (1+3T)T\frac{D^5}{15})}{\cos \phi'}$