# NavList:

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

**Re: Ocean Navigator March/April 2017 Nav Problem**

**From:**Frank Reed

**Date:**2017 Feb 10, 20:09 -0800

Stan, you quoted the Ocean Navigator column:*"Remember that at LAN the LHA of the sun and the GHA are the same." *

Heh. Yeah, that is *bad*. I think part of the problem with this is that the math is taught backwards. The fundamental mathematical problem is the great circle distance. If you understand that, then you can understand that LHA is nothing more than the difference in longitude between the observer's location and the spot where the Sun is at the zenith (sometimes known as the GP). If we focus on the idea that LHA is a dLon, then the concept is much simpler, and these sorts of mistakes are much harder to make. At LAN, of course, the dLon between the observer and the Sun (Sun's GP) is zero. Therefore the LHA is zero. At Greenwich, or anywhere on the prime meridian, the difference in longitude from the Sun's longitude is identical to the GHA. Therefore on the prime meridian, the LHA of the Sun and the GHA are the same.

When a navigator knows that the altitude problem is really identical to the great circle problem, it's also easy to see that the great circle formula directly yields a calculated altitude. In the great circle problem, we multiply together the cosines of the two latitudes and the cosine of the difference in longitude and we add to that the product of the sines of the two latitudes:

sum = cos(Lat_{1})·cos(Lat_{2})·cos(dLon) + sin(Lat_{1})·sin(Lat_{2}),

and then the inverse cosine yields the distance:

gcdist = cos^{-1}(sum).

Actually it yields the distance as a pure angle or as an angle in degrees. If it's in degrees, we have to multiply by 60 to get distance in nautical miles, but that much is fairly obvious. For the celestial problem, we replace the second location by the Sun or other celestial body (or more plainly, we replace the second location by the lat, lon of the location where the celestial body is straight up). The latitude is the declination, of course. And the dLon is still a difference in longitude, but we *name it* LHA. It is simply a difference in longitude. We have then:

sum = cos(Lat_{1})·cos(* dec*)·cos(

*) + sin(Lat*

**LHA**_{1})·sin(

*),*

**dec**and then the inverse cosine yields the distance between these points:

gcdist = cos

^{-1}(sum).

This distance across the globe is identical to the zenith distance from the observer's zenith to the celestial body's position in the sky. That's the fundamental principle of celestial navigation: distance on the globe equals zenith distance. Finally, we can get altitude by subtracting this zenith distance from 90°:

alt = 90° - gcdist.

Although it slightly breaks the nice analogy between the great circle problem and the celestial navigation problem, it is customary in this calculation to save this last step. Rather than subtracting from 90°, we get the same result by using an inverse sine in the second to last step. That is,

gcdist = cos

^{-1}(sum),

alt = 90° - gcdist

is identical to

gcdist = sin

^{-1}(sum).

Well, I have rambled enough. With any luck, this perspective will be useful to someone out there in NavListLand.

Frank Reed