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    Re: Emergency sun declination
    From: Trevor Kenchington
    Date: 2004 May 21, 21:49 +0000

    Fred,
    
    I think you are missing the point. Yes, the sine of an angle is equal to
    the perpendicular divided by the hypotenuse. However, a sine curve is
    also a shape drawn out by a point on the circumference of a revolving
    circle: Set up a bicycle wheel with its axle horizontal, mark one point
    on the tyre, rotate the wheel and measure the vertical height of the
    marked spot above or below the plane of the axle. If you plot those
    heights against the angle through which the wheel has been rotated (or
    against time, if the wheel is rotating at a constant rate), then what
    you draw out will be a sine curve.
    
    What Doug, Jim and others have described are ways of finding the
    "vertical height" above the "axle" by using a circle drawn on paper. In
    Doug's method, that "height" is his measurement #3. To turn it from a
    point on a sine curve into the sine of the angle, it is necessary to
    scale the curve to a maximum of one (and a minimum of minus one). To do
    that, Doug divides by his #4, which is the radius of the drawn circle
    and which equals his #3 when the angle in question is 90 degrees -- the
    sine of 90 being, of course, one.
    
    You could draw a right-angle triangle with the perpendicular and
    hypotenuse having lengths equal to Doug's #3 and #4 but that is a bit of
    a different way of looking at the same geometry. The two things do line
    up though:
    
    Consider the case where the angle of interest is 60 degrees (or the date
    of interest is 61 days after the last equinox). Look at the circle laid
    out as a compass card. Doug's #4 is the dimension from the centre of the
    circle to the north cardinal point. However, that is equal to the
    distance from the centre of the circle to the 270+060=330 degree mark.
    Draw a line from the 330 mark to the 030 mark and it will be parallel to
      the line from the centre to the 330 mark. So the angle centre-330-030
    will be 60 degrees. Meanwhile, the 330-030 line will cut the radius
    leading to the north cardinal point at a distance from the centre which
    Doug has defined as his #3. You now have a right-angle triangle with one
    angle of 60, an opposite side (the perpendicular) of #3 and the
    hypotenuse equal in length to #4.
    
    Not Doug's #4 but equal in length to it.
    
    
    And don't forget that the multiplier should be 23.44 -- 23.5 as a close
    approximation. 22.5 was a typo.
    
    
    Trevor Kenchington
    
    
    Fred Hebard wrote:
    
     > Ahha,
     >
     > A graphical method of computing the sine of a function.  Sine is
     > opposite over the hypotenuse. #3 =opposite, #4=hypotenuse.  Then dec =
     > 22.5*sine(days past vernal equanox), etc for other seasons.
    
    
    
    
    --
    Trevor J. Kenchington PhD                         Gadus{at}iStar.ca
    Gadus Associates,                                 Office(902) 889-9250
    R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
    Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555
    
                          Science Serving the Fisheries
                           http://home.istar.ca/~gadus
    
    
    

       
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