NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Amplitudes
From: Hewitt Schlereth
Date: 2010 Jun 3, 19:03 -0400
From: Hewitt Schlereth
Date: 2010 Jun 3, 19:03 -0400
Thank you for the math, John. One way to visualize this to imagine you are facing more or less East waiting for the sun to rise on the morning of an equinox. This means that you are standing near the terminator with the dark side behind you. Since the sun's rays are parallel, it's rays strike the terminator at a right angle and because on that day the terminator is aligned N-S, looking 90� to it, you are looking East exactly. Now assume this is the March equinox. As the sun moves North, the terminator tilts, but the sun's rays still strike it at a right angle. So, you still have to look at right angles to it to see the sun rise. But now you will be looking somewhat to the left (North) of East. The other thing about the terminator is, it's a great circle, and so will cross a North latitude and a South latitude of the same number (34N & 34S,say) at the same angle. Applying 90� to the terminator will give the same bearing for both 34N and 34S. Similarly observers any other N-S pair of latitudes - 10N-10S, 40N-40S, 60N-60S, etc. - have to look in the same direction to see the sun rise. Hewitt On 6/3/10, John Karlwrote: > > > Oops, scratch that comment in my last post, that the azimuth angle A is > symmetric in d, the declination, for Hc = 0. (Cos A is, not A.) > > JK > > > > ---------------------------------------------------------------- > NavList message boards and member settings: www.fer3.com/NavList > Members may optionally receive posts by email. > To cancel email delivery, send a message to NoMail[at]fer3.com > ----------------------------------------------------------------