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Greg, thank you, too, for the math. Hewitt

On 6/3/10, Hewitt Schlereth  wrote:
> 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 Karl  wrote:
>  >
>  >
>  > 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
>  >
>  >
>  >
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