Hello,
Can I solve a geodesic problem with lon1, lat1, s12, and az2 given, using GeographicLib?
I would like to make an tangential polygon inscribing an equidistant area (a pseudo-circle) on the ellipsoidal earth.
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it's Problem 7 in Section 10. Here I only give an iterative method to
improve an approximate solution. Perhaps the approximate solution can
be found by considering the spherical case. For this, see
The solution of the spherical problem is given in https://en.wikipedia.org/wiki/Spherical_trigonometry#Solution_of_triangles It's case 3 in the first subsection (oblique triangles), and the next subsection outlines its solution with right-angle triangles. You'll have to watch out for multiple solutions, obviously; but this falls out naturally from the right-angle method.
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I'm working at implementing the iterative solution following your advice. When \cos \alpha_2 vanishes, the derivative will be as follows:
\frac{\textup{d}s_{12}}{\textup{d}\alpha_1} \Biggl|{\phi_1,\alpha_2} \xrightarrow[\cos \alpha_2 \rightarrow 0]{} \ -a w_2 \cot \beta_2 \cos \sigma{12} \qquad \textrm{Eq.}\ (79^\prime)
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Hello,
Can I solve a geodesic problem with lon1, lat1, s12, and az2 given, using GeographicLib?
I would like to make an tangential polygon inscribing an equidistant area (a pseudo-circle) on the ellipsoidal earth.
I discuss the solution of this problem in
it's Problem 7 in Section 10. Here I only give an iterative method to
improve an approximate solution. Perhaps the approximate solution can
be found by considering the spherical case. For this, see
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Thank you for your answer and suggenstion. Hinks' paper, however, didn't describe equations of the calculation. I begin writing my program. Thank you.
The solution of the spherical problem is given in https://en.wikipedia.org/wiki/Spherical_trigonometry#Solution_of_triangles It's case 3 in the first subsection (oblique triangles), and the next subsection outlines its solution with right-angle triangles. You'll have to watch out for multiple solutions, obviously; but this falls out naturally from the right-angle method.
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I'm working at implementing the iterative solution following your advice. When \cos \alpha_2 vanishes, the derivative will be as follows:
\frac{\textup{d}s_{12}}{\textup{d}\alpha_1} \Biggl|{\phi_1,\alpha_2} \xrightarrow[\cos \alpha_2 \rightarrow 0]{} \ -a w_2 \cot \beta_2 \cos \sigma{12} \qquad \textrm{Eq.}\ (79^\prime)
You're nearly right... The correct limit is (I believe!)
−a w2 cotβ2 M21
(i.e., M21 in place of cosσ12).
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Great! It revealed that my solution was f=0 limiting one.