(I’d post this as a “reply” to my existing topic, but that thread doesn’t show up in my display, & so it’s necessary to post this as a new topic.)
. Yesterday I described a way of estimating EqT, which depended on the fact that the EqT is zero on some known day. But EqT can be also estimated without knowing when it is zero. …by use of Sidereal Time. . First in summary, & then with details for the summarized elements: . For some given time of day, in Standard-Time, convert it to local mean time, by applying your longitude-correction. . 1. Determine ecliptic-longitude as before. Even if not reliably, estimate it as well as possible for a particular time of day. (Each half-day in error for when an ecliptic-month starts only makes a 2 minute error in the EqT estimate.) . 2. As before, use Spherical Trigonometry’s Cosine Formula to determine the Sun’s R.A., in degrees, from its ecliptic-longitude. . 3. Determine the Sidereal Time (…the hour-angle of the Vernal-Equinox. Hour-angle is measured westward around the Earth’s axis, from the meridian, in the south.) . 4. Subtract the Sun’s R.A. from 360, & add the result to the Sidereal Time in degrees. . 5. From the local mean time, determine the hour-angle of the mean Sun, in degrees. . 6. Subtract the hour-angle of the Sun from the hour-angle of the mean-Sun. Multiply by 4, & that gives EqT. . Details, where needed: . 2. The Spherical-Trigonometry Cosine Formula, for ecliptic-longitude & R.A: . To get the Sun’s R.A.: . Multiply the tangent of its ecliptic-longitude by the cosine of the obliquity, & then take the inverse tangent of the result. . 3. Sidereal-Time: . Sidereal Time, like Mean Time advances uniformly. . It takes us roughly 365.25 days to orbit the Sun. That’s slightly longer than a mean tropical year. That’s because the equinoxes come about 1/72 of a degree to meet us partway, making our tropical year about 20 minutes shorter than the sidereal year. (Of course those year-lengths really depend on from what part of our orbit they’re measured.) . Because of the Sun’s apparent eastward movement with respect to the fixed stars, due to our orbital motion around it, in a sidereal year the fixed stars will have been perceived to go around us one more time than the Sun has. In a sidereal year, the stars will have gained 360 degrees over the mean Sun. . So, in a day they’ll gain about (360/365.25) degrees, or about .9856 degree. . On the day of the Autumnal Equinox, if it occurs at Local Mean Time (LMT) midnight, the Vernal Equinox will be on the meridian at midnight, because the equinoxes are opposite. . So, how many days since the Autumnal Equinox, & how many hours since midnight? Each hour since midnight advances the Vernal-Equinox 15 degrees westward. Each day since the Autumnal Equinox avances the Vernal Equinox about .9856 degree westward. . Of course if the Autumnal Equinox occurred some hours later than midnight LMT, then at any later date & time the Vernal Equinox’s hour-angle will be advanced 15 degrees less for each of those hours later. …& of course the opposite if it’s earlier. . In that way, you can determine hour-angle of the Vernal Equinox at any time & date. . 5. Hour angle of the local mean sun: . It’s 15 degrees for every hour past noon local mean time. , I haven’t tried this method yet, but its error is probably about the same as the other method I described. . I guess most of the error comes from the approximation of the Solar Ecliptic Longitude via the Indian National Calendar. . That calendar seeks to make each of its ecliptic-months start as close as possible to the actual astronomical ecliptic-month, & so I wouldn’t expect large cumulative errors to accumulate. . The length of an Indian National Calendar ecliptic-month could be off by up to half a day at most. Each half-day of error makes a 2 minute EqT error. But the calendar’s errors are intended to not accumulate. As I described for October 29th, the EqT error was only 1 minute.
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