NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Ix Diagram and Pub 208
From: Gary LaPook
Date: 2012 Dec 11, 13:57 -0800
From: Gary LaPook
Date: 2012 Dec 11, 13:57 -0800
I like your diagram, it is pretty slick. I was trying to figure out how it works and I noticed that your "C" value is the product of sin LHA times cos dec. So this comes from the standard formula for azimuth, sin Z = cos LHA sin dec /cos Hc. You then rearrange this formula to sin Z cos Hc = cos LHA sin dec. Using your diagram, moving from the outside scales to an intersection multiplies the two values so moving out to one scale from an intersection is the same as dividing. So after finding "C" (which is cos LHA sin dec and also sin Z cos Hc) starting with the value of "C" and the intersection with cos Hc and moving out to the azimuth scale your are actually dividing "C" by cos Hc and thereby finding Z. The diagram could also be used for other trig problems involving multiplying and dividing sines and cosines. While trying some different examples, I discovered an interesting set of values to use. Let dec = 30, LHA = 40, Hc = 50 and then the Z comes out to exactly 60. There must be some trig reason for this but I haven't figured out why. gl --- On Mon, 12/10/12, hanno ix <hannoix@att.net> wrote:
|