# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**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) >