NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Time of Sun & Moon rise and set
From: Gary LaPook
Date: 2016 Feb 6, 07:12 +0000
From: Brad Morris <NoReply_Morris@fer3.com>
To: garylapook@pacbell.net
Sent: Friday, February 5, 2016 10:59 PM
Subject: [NavList] Re: Time of Sun & Moon rise and set
From: Gary LaPook
Date: 2016 Feb 6, 07:12 +0000
Years ago we discussed the problem of identifying "noon" from a fast moving airplane since its speed in a north or south direction will swamp out the change in altitude of the sun which is what is used in the traditional noon sight to determine "noon." (Frank, I searched for those posts and couldn't find them, can you?)
gl
From: Brad Morris <NoReply_Morris@fer3.com>
To: garylapook@pacbell.net
Sent: Friday, February 5, 2016 10:59 PM
Subject: [NavList] Re: Time of Sun & Moon rise and set
Frank
Whoops and correct. I moved the decimal point by mistake. Where's my sword, I need to fall on it! The correct value is indeed 1.0 minutes of arc change per hour.
But the declination of the sun whilst experiencing that 'rapid' declination rate of change is near 0°, placing the sun near the celestial equator. Meaning that at either pole, the sun would be either just above or below the horizon. That may make measuring the altitude of the sun just a bit challenging.
Further I agree, Steve's question was explicit. He wanted to know if the time of meridian transit (his words) was affected by latitude, specifically at the poles. It is not, clearly.
Herbert made the assertion that culmination time would not necessarily correspond to meridian transit time AND that this was due to the change in declination of the body in question. I can see that's true, I still don't understand the relevance.
The LOP of the sun at either meridian transit or culmination will differ by an amount small enough to be negligible for CN purposes. It's noise. Even if you are attempting to prove you are at the pole using CN, then the error is still less than 1 mile.
Finding longitude by noon sun has been well debated here. I think the issues will be exacerbated when near the poles, as the curvature of altitude vs time curve is very shallow (nearly flat).
Further, all of the manuals of polar navigation explicitly state that longitude is fairly meaningless near the poles, due to the convergence of the meridians. So why use a shakey method to determine longitude, when longitude is not useful?
So perhaps Herbert would care to explain more about what he meant. Still so I may learn!
Brad
On Feb 6, 2016 12:10 AM, "Frank Reed" <NoReply_FrankReed@fer3.com> wrote:
Brad, you wrote:
"The sun has a maximum rate of change in declination of 0.1 minutes of arc per hour."Looks like you moved a decimal point somewhere. I like to remember that the Sun's "speed" in latitude is very nearly one knot on the equinoxes --or 1.0 minutes of arc per hour (in longitude on the equinoxes, it's 15° per hour, which is nearly 900 knots, by the way). We're still only talking about one knot rate of change in latitude (=declination) so if you were trying to determine longitude by symmetry of sights about local noon, the correction would still be relatively small. In other words, not much difference between time of meridian passage and time of culmination. The nominal correction to longitude gets larger rapidly as we approach the poles, but if you convert that into miles in the fix, the problem is much reduced since longitude lines are closer together at the poles. Convenient that.By the way, I'm pretty sure that this was not what Steve Bryant was asking about. Time of meridian passage does not depend on latitude. That answer still stands.Frank Reed
Conanicut Island USA
