There was something posted recently about how to calculate the Solar ecliptic-longitude, & thereby the Equation of Time (EqT).
. (I should emphasize, that, when the Solar ecliptic-longitude is determined as described below, of course the Solar declination can be gotten from it by the method that we discussed earlier.) . It used a formula derived by solving our Solar orbit. . With matters like that, it’s of interest how the formulas are derived. . I once solved the orbital problem. It wasn’t for planetary-orbits. I wanted to find out how far the fastest rifle bullet (4110 fps, from something I’d read 20 years previous) could go, on the moon. . If the ground were flat, with uniform gravitational-field, I get about 600 miles. But, for such a long range, those assumptions aren’t good enough. It’s necessary to do it as an orbital problem. …an orbit about the moon’s center, that intersects the moon’s surface. . (The answer that I got was just a bit more than 800 miles. It might have been about 820 miles.) . For me, by far the most straightforward solution-method was by conservation-laws. …as opposed to the solution-method that uses dynamics. . Well one of the conservation-laws—conservation of angular-momentum, is proved via dynamics. By far the simplest & most straightforward way to do that is in Lagrangian dynamics. . The conservation-laws solution involves an integral that can be solved in closed form. (Numerical-integration isn’t necessary.) . And, as is so often the case, it’s one of those integrations that requires trial-&-error to find the Integrand’s antiderivative. . There are several methods for converting the problem of integrating one function, to a problem of integrating a different one. So you apply whichever of those methods seems most promising, & if it seems to give you a new integration problem that looks simpler or more promising, then you apply one of the conversion methods to that new integral…& keep doing so till it leads to an expression whose integral is known. . So the integration involved a bit of trial & error, but was solvable in closed form. (…as opposed to requiring a numerical approximation.) . But, if you’re on a desert island, & need the EqT for position-determination, or for some reason you need mean time or standard time from your sundial, then to use a solution of the Earth’s orbit, you’d have to: . 1. Solve our orbit to derive the formulas. . OR . 2. Have a piece of paper on which the formula is written . OR . 3. Have been solving orbits so regularly & recently that you don’t need to look up the needed formulas. . And another problem is that you’d need the initial conditions at some recent epoch. That too would have to be looked-up. …again, unless you’ve been doing the problem so much lately that you know the initial-conditions. . ….&, if you’re going to carry around a piece of paper with the orbital-solution formula & the initial-conditions…well then, why not just carry a piece of paper with the EqT & Solar declination for each day of the current year (…& maybe the next few years if you might be on your desert island or at sea for a few years)? . So it would be desirable to have an easier approximation for the EqT. . I’ll suggest one. . The month-lengths of the ecliptic-month approximations in the Indian National Calendar can give you an estimate of the Solar ecliptic-longitude for any day, if you know the date of the nearest equinox (or even roughly if not exactly). . Start with a day known to have an EqT of zero. September 1st & Christmas are such days. . The Indian National Calendar has 30-day & 31-day months. . Taurus thru Virgo have 31 days. The other months have 30 days. . Taurus starts in April. Virgo starts in August. . So, Taurus & Virgo are the ecliptic months that start in a month that starts with “A”. . In a 31 day month, the average motion rate along the ecliptic is 30/31 degrees per day. . In a 30 day month, the average motion rate along the ecliptic is 1 degree per day. . Start with a day known to have an EqT of zero. September 1st & Christmas are two such days. . (There are four of them, but you might not need all four.) . So, based on that, determine the Solar ecliptic longitude on (say) September 1st, & on the date whose EqT you want. . For each of those two EqT values, use spherical-trigonometry’s Cosine Formula to determine the equatorial-longitude--the Right-Ascencion (R.A.), in degrees--that corresponds to that ecliptic-longitude for motion along the ecliptic. . Subtract the two R.A. values to find out how much the actua Sun increases its R.A. between September 1st & the date whose EqT you want. . Next you want to know how far the mean-Sun moves in R.A. during that same period, from September 1st, to the date whose EqT you want. . Well, it takes 1 year for the Sun to go 360 degrees around the ecliptic, to return to some initial point. That’s called a tropical-year. The length of a tropical-year differs for different initial points on the ecliptic. That’s because of the precession of the equinoxes, & our orbit’s ellipticity. . Because this is about the mean-Sun, then it seems reasonable to use the mean tropical year, an average tropical year length, averaged over the ecliptic. . That’s about 365.2422 mean solar days. . The fictitious mean-Sun moves at a uniform rate, in terms of R.A. per day. When the Sun goes around the ecliptic, that alsoconsists of one completre circuit in R.A. So the 365.2422 day mean tropical year is also the time it takes for the mean-Sun to complete a circuit in R.A. . Alright, then the mean Sun goes 360 degrees in 365.2422 days. . That rate of motion allows you to determine how far, in R.A. the mean sun goes, from September 1st to the day whose EqT you want. . So, from the distance in R.A. that the mean Sun goes during that period, you subtract the distance in R.A. that the actual Sun goes during that period. . That gives you the amount by which the mean Sun is ahead of the actual Sun on that day whose EqT you want. . Multiply that by 4, to convert degrees to minutes. . Then you have the mean minus true EqT for that day. The amount of clock-fast for that day. . If it’s negative, then it’s clock-slow. . When you do that for today, October 29th, you get a result for EqT that differs only by a minute from the value that you find when you look it up. . As you know, a minute is about the best accuracy that you can expect for a shadow-sundial. . (Well alright, maybe a half-minute?) . But, for a sundial, a minute’s error isn’t bad at all. . For navigation? Well, a minute of time corresponds to a quarter of a degree of longitude. …about 15 nautical-miles. You can’t expect much better than that with any observational-instrument improvised on a desert-island or boat. . (BTW, above, throughout, I spoke of R.A. in degrees, rather than the customary hours.)
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