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Trevor,

I'm not sure what point it is I'm supposed to be missing, but Doug is
finding the sine of 61 degrees by a graphical method while I was
finding it using sine tables or a calculator or a Taylor series.  A
Taylor series is how the tables are calculated, and, I'm not sure of
this one, but believe it to be correct, how the calculator evaluates
it.  The radius in Doug's method is the hypotenuse of the triangle and
Doug's height is the opposite side, except he translates it to the
y-axis to measure it.

Off to Maine, so I'll be off-list for 5 days.

Fred

On May 21, 2004, at 5:49 PM, Trevor J. Kenchington wrote:

> 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
>
>                      Science Serving the Fisheries
>                       http://home.istar.ca/~gadus
>

   

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