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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@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

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