# 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:**Stan K

**Date:**2017 Feb 11, 11:39 -0500

Frank,

That was quite a ramble, but I think the situation is simpler than that. The location of the problem is at 33º12'W, and the full quote is, "

*Remember that at LAN the LHA of the sun and the GHA are the same. Therefore, the longitude should be the same as the GHA."*

I think that Berson simply "mis-spoke", and intended to say something more like, "Remember that at LAN the LHA is zero. Therefore, the longitude should be the same as the GHA."

And who am I to complain about the lack of proofreading? I make a mistake in every email I write every email. ;-)

Stan

-----Original Message-----

From: Frank Reed <NoReply_FrankReed@fer3.com>

To: slk1000 <slk1000@aol.com>

Sent: Fri, Feb 10, 2017 11:19 pm

Subject: [NavList] Re: Ocean Navigator March/April 2017 Nav Problem

From: Frank Reed <NoReply_FrankReed@fer3.com>

To: slk1000 <slk1000@aol.com>

Sent: Fri, Feb 10, 2017 11:19 pm

Subject: [NavList] Re: Ocean Navigator March/April 2017 Nav Problem

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

and then the inverse cosine yields the distance:

gcdist = cos

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

sum = cos(Lat

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

gcdist = cos

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

alt = 90° - gcdist

is identical to

gcdist = sin

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(*)·cos(***dec***) + 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