# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Zenith distance and Ho question.**

**From:**Frank Reed

**Date:**2020 Dec 19, 22:17 -0800

Robert, you wrote:*"I don't understand how the sun can stay on the 23.4 N declination and be 90 degrees away from me and this equal 5400 nm. I'm thinking the 23.4 declination or latitude is a small circle, not a great circle and that there's fewer miles in a circle the further you go from the equator." *

Yes, it can be confusing. You're trying to imagine that the Sun would be east of you in the scenario. But it's not. It's northeast. And it's not separated from your position by 90° of longitude. It's separated by 90° (5400 n.m.) of *distance*. Where would that be?

Start on the ground. Suppose I tell you that I have two cities, A and B, both located at 23.45° N latitude. They're both on the Tropic of Cancer, just like the observer and the subSun point in the example you have been pondering. Suppose in addition that I tell you that the distance between A and B is exactly 5400 nautical miles. So what is the difference in longitude between them? You can figure this out by playing with a globe, either a physical globe or a virtual one like Google Earth. Or you can do it by calculation. If you're interested, you can code up the great circle distance equation yourself, or you can try out various online great circle distance calculators. What you should find is that the difference in longitude between these two places is a bit less than 101°. And by the way, for this specific case, where the two points have identical latitudes and the distance is specified to be exactly 90°, the great circle math shakes out and simplifies to something short and sweet:

*cos(dLon) = - tan ^{2}(Lat)*

That short equation, which is a direct consequence of the standard

*law of cosines*for this special case, will directly yield the difference in longitude between A and B if you feed it a latitude of 23.45° [fun with math: try this equation for a latitude of 46°... unexpected?? what is that result telling us?].

You added:*"But, I assume you will say the sun travels in a great circle. I'm having trouble comprehending this."*

The GP (subStar point) of a celestial object necessarily follows a circle very close to a latitude small circle since, after all, that motion is due to the rotation of the Earth, almost entirely. They're never quite perfect, and if you take into account the changing declination of the Sun, its ground track is like a tightly-wound spiral running down the globe to the Tropic of Capricorn from June 21 to December 21 and then reversing and coming back to the Tropic of Cancer six months later, that cycle repeating over and over again. But if you stick with a star, the ground track is very nearly a perfect circle, and that's a circle of latitude. But of course this motion has no bearing on the question you were asking. Leave the Sun where it is! Puzzle this geometry out for some fixed instant of time, which is what you would have for a celestial navigation sight anyway. A celestial sight consists of a sextant altitude at some instant of UT. We process that altitude to get a zenith distance, and the zenith distance is identical to the "distance off" from the GHA and Dec for that celestial body at the exact instant of UT as recorded (and we must get the GHA and Dec from some astronomical database, like the Nautical Almanac).

Frank Reed