# NavList:

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

**Re: Emergency sun declination**

**From:**Trevor Kenchington

**Date:**2004 May 21, 21:49 +0000

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{at}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