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Diffstat (limited to 'docs/source/operations/projections/merc.rst')
| -rw-r--r-- | docs/source/operations/projections/merc.rst | 30 |
1 files changed, 29 insertions, 1 deletions
diff --git a/docs/source/operations/projections/merc.rst b/docs/source/operations/projections/merc.rst index 7b6e13da..2107b7b2 100644 --- a/docs/source/operations/projections/merc.rst +++ b/docs/source/operations/projections/merc.rst @@ -158,7 +158,35 @@ Inverse projection \lambda = \frac{x}{k_0 a}; \quad \psi = \frac{y}{k_0 a} The latitude :math:`\phi` is found by inverting the equation for -:math:`\psi` iteratively. +:math:`\psi`. This follows the method given by :cite:`Karney2011tm`. +Start by introducing the conformal latitude + +.. math:: + \chi = \tan^{-1}\sinh\psi + +The tangents of the latitudes :math:`\tau = \tan\phi` and :math:`\tau' = +\tan\chi = \sinh\psi` are related by + +.. math:: + \tau' = \tau \sqrt{1 + \sigma^2} - \sigma \sqrt{1 + \tau^2} + +where + +.. math:: + \sigma = \sinh\bigl(e \tanh^{-1}(e \tau/\sqrt{1 + \tau^2}) \bigr) + +This is obtained by taking the :math:`\sinh` of the equation for +:math:`\psi` and using the multiple argument formula. The equation for +:math:`\tau'` can be solved to give :math:`\tau` using Newton's method +using :math:`\tau = \tau'/(1 - e^2)` as an initial guess and with the +needed derivative given by + +..math:: + \frac{d\tau'}{d\tau} = \frac{1 - e^2}{1 + (1 - e^2)\tau^2} + \sqrt{1 + \tau'^2} \sqrt{1 + \tau^2} + +This converges after no more than 2 iterations. Finally set +:math:`\phi=\tan^{-1}\tau`. Further reading ############### |