giuliano-macedo/reverse-jhs.md

## reverse-jhs.md

      
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              reverse-jhs.md
            
          
    JHS Reverse formula

More comprehensive reverse JHS formula for computing latitude,longitude given northing, easting of a projection.
reference: Geomatics guidance Note number 7 part 2 - 3.5.3.1
(EPSG Dataset coordinate operation method code 9807)
Constants

spheroid constants:


$\text{f}$ flattening

$a$ semi-major axis

$e^2= 2\text{f} - \text{f}^2$ eccentricity

projection constants:


$\varphi_0$ Latitude of orgin

$\lambda_0$ Longitude of origin (sometimes called central meridian)

$k_0$ scale factor

$FE$ false easting

$FN$ false northing

Derived constants

$n=\text{f}/(2-\text{f})$
$B = [\frac{a}{1+n}](1+\frac{n^2}{4}+\frac{n^4}{64})$


$h_1 = \frac{n}{2} - \frac{2}{3}n^2 + \frac{5}{16}n^3 + \frac{41}{180}n^4$


$h_2 = \frac{13}{48}n^2 - \frac{3}{5}n^3 + \frac{557}{1440}n^4$


$h_3 = \frac{61}{240}n^3 - \frac{103}{140}n^4$


$h_4 = \frac{49561}{161280}n^4$


Obs: $h_1$ - $h_4$ are only used in the reverse formulas for calculating $M_0$ when $\varphi_o \notin \set{0, \frac{\pi}{2}, -\frac{\pi}{2}}$


$h'_1 = \frac{n}{2} - \frac{2}{3}  n^2 + \frac{37}{96}  n^3 - \frac{1}{360}  n^4$


$h'_2 = \frac{1}{48}  n^2 + \frac{1}{15}  n^3 - \frac{437}{1440}  n^4$


$h'_3 = \frac{17}{480}  n^3 - \frac{37}{840}  n^4$


$h'_4 = \frac{4397}{161280}  n^4$


$M_0$

meridional arc distance from equator to the projection origin $M_0$ is computed from:
if $\varphi_0=0$ then $M_0=0$
else if $\varphi_0=90^{\circ}\text{N}=\frac{\pi}{2}\text{ radians}$ then $M_0=B\frac{\pi}{2}$
else if $\varphi_0=90^{\circ}\text{S}=-\frac{\pi}{2}\text{ radians}$ then $M_0=-B\frac{\pi}{2}$
else:
$Q_0 = \text{asinh}(\tan{\varphi_0}) - [e\text{ atanh}(e\sin{\varphi_0})]$
$\beta_O=\text{atan}(\text{sinh } Q_0)$
$\xi_{O0}=\text{asin}(\sin B_0)$


$\xi_{O1} = h_1 \sin(2\xi_{O0})$


$\xi_{O2} = h_2 \sin(4\xi_{O0})$


$\xi_{O3} = h_3 \sin(6\xi_{O0})$


$\xi_{O4} = h_4 \sin(8\xi_{O0})$


$\xi_O= \xi_{O1}+\xi_{O2}+\xi_{O3}+\xi_{O4}$
$M_0=B\xi_O$
end
Formula

Given easting $E$ and northing $N$ in meters, and $\varphi$, $\lambda$  as resulting longitute, latitude in radians:
$\eta' = \frac {E-FE}{B k_0} $
$\xi' = \frac { (N-FN) + k_0 M_0 }{B k_0} $

$\xi'_1 = h'_1 \text{sin}(2\xi')\text{ cosh}(2\eta')$
$\xi'_2 = h'_2 \text{sin}(4\xi')\text{ cosh}(4\eta')$
$\xi'_3 = h'_3 \text{sin}(6\xi')\text{ cosh}(6\eta')$
$\xi'_4 = h'_4 \text{sin}(8\xi')\text{ cosh}(8\eta')$
$\xi'_0 = \xi' - (\xi'_1 + \xi'_2 + \xi'_3 + \xi'_4)$


$\eta'_1 = h'_1 \text{cos}(2\xi')\text{ sinh}(2\eta')$
$\eta'_2 = h'_2 \text{cos}(4\xi')\text{ sinh}(4\eta')$
$\eta'_3 = h'_3 \text{cos}(6\xi')\text{ sinh}(6\eta')$
$\eta'_4 = h'_4 \text{cos}(8\xi')\text{ sinh}(8\eta')$
$\eta'_0 = \eta' - (\eta'_1 + \eta'_2 + \eta'_3 + \eta'_4)$


$\beta' = \text{asin}( \frac{\sin{\xi'_0}}{\text{cosh }\eta'_0} ) $
$Q' = \text{asinh }(\tan{\beta'})$
$Q'' = Q' + [e \text{ atanh}(e \text{ tanh }Q')]$
$Q'' = Q' + [e \text{ atanh}(e \text{ tanh }Q'')]$ which should be iterated until the change in Q" is insignificant
Finally:
$\varphi = \text{atan }(\text{sinh } Q'')$
$\lambda = \lambda_0 + \text{asin }(\frac{\text{tanh}(\eta'_0)} {\cos \beta'} )$