skyfield

In the time tests shown below, I found that Skyfield takes several hundred microseconds up to a millisecond to return obj.at(jd).position.km for a single time value in jd , but the incremental cost for longer JulianDate objects (a list of points in time) is only about one microsecond per point. I see similar speeds using Jplephem and with two different ephemerides. My question here is: if I want to random-access points in time, for example as a slave to an external Runge-Kutta routine which uses its own variable stepsize, is there a way I can do this faster within python (without having to

scipy.optimize.minimize using default method is returning the initial value as the result, without any error or warning messages. While using the Nelder-Mead method as suggested by this answer solves the problem, I would like to understand: Why does the default method returns the wrong answer without warning the starting point as the answer - and is there a way I can protect against "wrong answer without warning" avoid this behavior in this case? Note, the function separation uses the python package Skyfield to generate the values to be minimized which is not guaranteed smooth, which may be

The JulianDate object in Skyfield is a handy way to quickly produce and hold a set of time values in Julian Days, and pass them to Skyfield's at() method to calculate astronomical positions in various coordinates. (see an example script ) However, I can't seem to find an add or offset method so that I can add a time offset or an iterable of offsets to a JulianDate object. I always seem to struggle with dates and times. Here is a very simple, abstracted example. I generate jd60 which is offset from an arbitrary jd0 by 60 days. As a simple check I calculate the position of the earth at the two