NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Basics of computing sunrise/sunset
From: Frank Reed
Date: 2009 Jun 30, 19:51 -0700
From: Frank Reed
Date: 2009 Jun 30, 19:51 -0700
Bill B, you wrote: "My questions. If I recall the GP would be approximately 5400 nautical miles (nm) away when I see Sun set (using appropriate eye protection of course). In my mental picture, I am not wrapping my vision along the surface of the Earth. I have a line of sight from my eye position to the Sun, which will run out of atmosphere *long* before 5400 nm " Aha. A good little technical point there, Bill! As Christian noted in a later message, I wasn't taking his wording too literally. I was merely agreeing that if we had detailed observational information on the state of the atmosphere between us and the object we're viewing, then the problem is straight-forward physics: a numerical integration of the differential equations for the law of refraction in a continuous medium. It's solvable with enough data. But you're right in pointing out that the points we would require for this integration do not extend to the GP of the Sun but rather they extend along a line inclining up through the atmosphere. The details in the high atmosphere don't much affect the refraction so let's suppose we stop when we reach an altitude of 40,000 feet. So if I am standing on the roof of a building 100 feet above sea level, and I look out and see the lower limb of the Sun just on the horizon (sea horizon), how many "weather balloons" (observation points) would I need? Just for fun, let's suppose I place a detector every 500 feet between me and the top of the troposphere (around 40,000 feet) all the way out towards the lower limb of the Sun. From my location to the horizon is a distance of about 10n.m. (sqrt(ht in feet)). From that spot, continuing on a straight ray to an altitude of 40,000 feet is about 200 n.m. (sqrt again). So that's 210 miles or, doing the math, around 2500 detectors or "weather balloons". We're very lucky that all of that detail is irrelevant to most refraction. For altitudes above 3 or at worst 5 degrees, nothing matters except the temperature and pressure right at the observer's location at the bottom of the atmosphere. It's worth remembering that this fortunate circumstance is what made celestial navigation possible. If we were dependent on the observation of altitudes very close to the horizon, the accuracy of celestial navigation would be ten times worse and sometimes considerably worse than that. -FER --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---