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    Re: Emergency sun declination
    From: Fred Hebard
    Date: 2004 May 21, 22:05 -0400

    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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