Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 4 down vote favorite

I'm trying to improve a site that shows the region of the Earth currently under daylight, and I need a formula that, given the current time, tells where (latitude/longitude) the sun and moon are overhead, accurate to 1 mile. Can anyone come up with one?

Ideally I'm hoping to get something JavaScript can calculate at a "reasonable speed" without recursion, loops, or extra libraries.

I've tried several things (eg, Fourier series on the Sun/Moon's RA/DEC), but nothing seems quite accurate enough.

Note: I realize my calculations for sun/moon rise/set ignore refraction and lunar parallax: for now, I'm focusing on finding the overhead positions. I realize I can pull data from a non-JavaScript CGI program (and have tried that), but it seems like a lot of unnecessary network access.

share|improve this question
Have you looked at ngc7000.org/astrotools/ephemtool.html ? – Larry OBrien Sep 29 '11 at 21:50
2  
It might be easier to retrieve the data from an online database than to calculate it. – JoeHobbit Oct 2 '11 at 4:18

1 Answer

up vote 0 down vote

No one else has answered so I'll do so, but I'm going to ignore details and speak to the geometry. The latitude of the subsolar point is simply Sun's declination. You should easily be able to convince yourself of that. The longitude is a bit more challenging. You know that when Sun is at the zenith, it must also be on the local celestial meridian. That means its local hour angle (LHA) must be zero. You know that LHA is the difference between local sidereal time (LST) and Sun's right ascension (RA). For LHA to be zero, you must have LHA = RA. If you know LHA and the sidereal time at Greenwich (GST), the difference between the two is the required longitude. I neglected the difference between mean and apparent sidereal time and Sun's mean, apparent, and topocentric coordinates. You can account for these.

A similar process can be applied to Moon to find the sublunar point.

share|improve this answer
Thanks, @JoeH. The difficulty I'm having is finding a simple formula for the sun's (and especially the moon's) RA and DEC that's easy to calculate and accurate enough to find the subsolar/sublunar point to within 1 mile. There are several ways to estimate this, but none are super-accurate. – barrycarter Jan 3 at 15:04
@JoeH, Yeah! He's right! This is the method I thought when I first saw the question, but this is not accurate enough in the real world. The celestial motion is far more complex than they do it on paper – Cheeku Mar 3 at 23:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.