NavList:

The equinox occurs tomorrow at 15:33:28 UT (for those near me, that's 11:33:28 Eastern US time). If the sky isn't too cloudy tomorrow, I'm going to try to get a Sun sight right at the time of the equinox. Anyone else in for the game?

At the time of the equinox, the Sun's SHA will be 0° 00.0' (and GHA 51° 30.7'), and its Dec will also be 0° 00.0'. The Sun will be directly above the Earth's equator at a point in Brazil (GHA is longitude, so 51° 30.7' W) close to the mouth of the Amazon River. At this time of the year the Sun's Dec increases northbound at very nearly 1 knot. That is, its Dec is increasing northbound at 1' per hour. Meanwhile the Sun's GHA is increasing westbound at 900' per hour (or 15' per minute of time, or 1' in four seconds of time).

Let's start with a local noon sight, not taken at the exact time of the equinox. Suppose you shoot the Noon Sun from your location tomorrow, and let's say the UT at the time of your sight is 17:09 UT, and you get a corrected altitude for the Sun's center of 52° 25'. We know that
   Latitude = Z.D. + Dec.
For the Z.D., we just subtract the corrected altitude from 90°. In this case, I get 37° 35'. If we don't care about a couple of miles, this zenith distance on the equinox is telling us our distance from the equator. The angular distance of the Sun away from the zenith (straight up) is identical to the distance over the globe, as an angle, from the observer to the subSun point. And since we measured the altitude at local noon, the direction to the Sun is due South. Obviously then the Z.D. is identical to the latitude in this case. So as a good approximation, our observer's latitude is 37° 35' N. We can do better by adjusting the Dec. The time of this sight was 1h 36m after the equinox. That means that the Sun was actually north of the equator. Speed northbound is 1 knot. Sun's Dec therefore would be 0° 01.6'. And the observer's latitude is pushed north by the same amount: Lat = 37° 36.6'.

For the real "fun" tomorrow, suppose you shoot the Sun at the exact time of the equinox, 15:33:28 UT, but it's nowhere near local noon. Instead, maybe you're sailing in San Francisco Bay, and you find the Sun, about 10° south of east (azimuth 100° true), and its corrected altitude is 14° 57'. You would like to compare that with an Hc calculated from an "Assumed Position", and in this case I'm picking an arbitrary AP of 37° 48' N, 122° 32' W (just outside the Golden Gate). As always, we use the great circle distance equation where point 1 is the subSun location near the mouth of the Amazon and point 2 is the AP near the Golden Gate:
   cos(dist) = sin(Lat1)·sin(Lat2) + cos(Lat1)·cos(Lat2)·cos(Lon2-Lon1),
and remember that the distance away, measured over the globe as an angle, is identical to the zenith distance. For this carefully timed equinox sight, Lat1 is exactly zero degrees. That implies that sin(Lat1) is zero, knocking out the first term on the right, and cos(Lat1) is exactly one, shortening the second term. We're left with
   cos(ZD) = cos(Lat)·cos(Lon-GHA),   [equinox only!].
And we can make one more change and swap out ZD for altitude (or this can done later). If we do it now, since ZD = 90° - Alt, then cos(ZD) becomes sin(Alt). And finally we can drop in the specific values for the Lat, Lon, and GHA:
   sin(Alt) = cos(37°48')·cos(122°32' - 51° 30.7')
               = 0.25697,
So Alt = 14.890° or 14° 53.4'. This is an Hc in the jargon of the intercept method. Its what the true altitude of the Sun would be from our selected AP. Since our observed (corrected) altitude, Ho, was 14° 57', which is 3.6' higher in the sky, the intercept vector is 3.6 nautical miles towards the observed Sun azimuth, which is 100° roughly. The key "fun" in all of this is the nice short equation for the altitude (Hc) at the time of the equinox:
   sin(Hc) = cos(APLat)·cos(APLon - GHASun).
And note that if your timing is a little off, you can easily adjust GHASun at the rate of 15' further west for every minute of time after the exact time of the equinox.

Frank Reed
Clockwork Mapping / ReedNavigation.com
Conanicut Island USA