How to calculate the point on the globe with maximum magnitude with given Besselian elements and given time for a partial solar eclipse? I am currently working on improvement of an Android app that calculates solar eclipses. For many days now I have not been able to solve a mathematical problem concerning solar eclipses, so I decided to try to ask for help.
I am working with the book &quot;Elements of solar eclipses 1951-2200&quot; by Jean Meeus. Calculating the central line of an eclipse from given Besselian elements already works good, so I am able to calculate the path of the central line of total and annular eclipses and the curves of equal magnitude.
My problem now is that Meeus only describes the calculation of the central line, so for a given time, that means for given Besselian elements I can calculate the corresponding point of the central line on the globe. Using these algorithms I can not calculate the line of maximum magnitude for partial phases of total and annular eclipses or for partial eclipses. So my question is:
How can I calculate latitude and longitude of the point on the globe with maximum magnitude with given Besselian elements for a time when eclipse is partial, that means, when no central line exists. As far as I understand the problem, this should be the point on the globe that is nearest to the shadow axis, which in the case of a partial eclipse or partial phase of an eclipse does not intersect the globe.
I would very much appreciate any help.
Jean Meeus, Morsels 2, chapter 69 talks about the Nelder-Mead simplex algorithm for minimising a function of two variables - in this instance magnitude as a function of latitude and longitude. Maximising the magnitude equates to minimising the negative of the magnitude. Apparently the method first appeared in magazine Byte in the 1960s, there are various references on-line.
Maximum magnitude always occurs near the polar circle, so try starting with three points equally spaced round the 60th parallel. They would have to be converted to Cartesian X Y with centre at the pole (giving an equilateral triangle), and the result converted back to latitude/longitude at the end. The sign of the element Y will make it obvious which polar circle to use.
Because the shadow is conic, not cylindrical, the resulting point will not (in general) be exactly on the horizon.