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: Hanno Ix
Date: 2012 Dec 12, 10:26 -0800
From: Gary LaPook <garylapook@pacbell.net>
To: NavList@fer3.com
Sent: Tuesday, December 11, 2012 9:35 PM
Subject: [NavList] Re: Ix Diagram and Pub 208
From: Hanno Ix
Date: 2012 Dec 12, 10:26 -0800
Greg,
you are correct.
I believe there is at least one practical way to deal with this.
The question then is, what is the most practical way?
I have to ponder this - unfortunately I am pressed for time.
h.
From: Gary LaPook <garylapook@pacbell.net>
To: NavList@fer3.com
Sent: Tuesday, December 11, 2012 9:35 PM
Subject: [NavList] Re: Ix Diagram and Pub 208
"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. " --------------------------------------------------------------------------------- The answer just came to me in a flash. The diagram solves this equation: sin Z cos Hc = sin LHA cos dec = C ( I wrote it wrong in my first post) In my example: sin 60 cos 50 = sin 40 cos 30 = C since cos 50 = sin 40 and also sin 60 = cos 30 so Z ends up coming out exactly as 60 because it is arc sin (cos 30) gl --- On Tue, 12/11/12, hanno ix <hannoix@att.net> wrote:
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