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The straight sine method can be "refined" by dividing the days since
spring by 93 instead of 90, in which case,
sine(61/93*90)=.8575*23.5=20*9.1'.  Not much of a refinement, but good
enough for lifeboat work.

On May 21, 2004, at 7:18 PM, Royer, Doug wrote:

> You betcha.It's 23.5* though as Frank pointed out.
> Here's what I got in a few minutes useing a rose and the "graphic"
> method:
> The date's(05-21)number of inches on the vertical axis = 3.3 in.
> The vertical radius length = 3.81 in.
> 3.3/3.81 = 0.86614
> 23.50 x 0.86614 = 20.3543* = 20* 21.3'
> dec. = 20* 21.3'N
> 05-21-04 1601zt sun dec = 20*24.1'N
> Not to bad for 3 minutes from start to finish.That includes marking the
> rose.Good enough for an emergency.
> Let's try the below:
> 61 days since the equinox. 05-21 - 03-20
> sin(61) = 0.8746
> 23.5 x 0.8746 = 20.5531 = 20* 33.2'
> dec = 20* 33.2'N
> In the ballpark.
> Good enough to use if one had no idea otherwise.
>
>
>
>
>
> 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.
>
> Fred
>
>> 1.Take a parrallel rule and from the horizontal axis,makeing sure it
>> stays
>> parrallel with that axis,move the rule up to the date's point on the
>> arc.
>> 2.Draw a line from that point to the vertical axis.
>> 3.Measure the distance from the line thus drawn to the point where the
>> vertical and horizontal axis's meet in the center of the rose.
>> 4.Measure the distance from the center of the rose to the edge of the
>> arc.
>> 5.Divide the "date measurement"(#3) by the total radius
>> measurement(#4)to
>> get the ratio of the two.
>> 6.Multiply that ratio by 22.50*(degree)
>

   

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