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    Re: Amplitudes
    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 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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