# NavList:

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

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Re: Question re. A and B factors (corner cosines) lunars
From: Magnus Sjoquist
Date: 2012 Jan 24, 11:44 +0100

Jaap vd Heide:

My mistake in thinking was: "Since azimuths do not appear in the formula it would not matter what they were. With altitudes and distance known (all three sides in the triangle) the observed bodies could be placed anywhere and the corner cosine at the moon/sun would not be affected".

Of course, this was nonsense and far out from trigonometry. With three sides given, there is only one triangle (or rather two: If you cut out the spherical triangle from the sphere, the rest of the sphere could be seen as a second triangle).

Frank?s explanation put my attention  to this very elementary fact, embarrrising enough for me.

Furthermore: Please correct me if I am wrong again:

Distance Chicago ? Waukegan corresponds to  the difference between apparent altitude and true altitude.

(The other case (factor) is if I start from Chicago but travel to a place not too far from but outside Paris, as I understand it (which does not say very much).

Best ?

/Magnus

----Ursprungligt meddelande----
Från: Jaap.vdHeide@xs4all.nl
Datum: 2012-01-24 10:25
Till: <NavList@fer3.com>
Ärende: [NavList] Re: Question re. A and B factors (corner cosines) lunars

Franks example obviously has its limits, as in case Paris would have been due east from Chicago, Chicago is on a great "circle of equal distance to Paris" and Waukegan - being due North - isn't. The difference obviously depends on the ratio between the distance North of Chicago considered in respect to the distance (tied to the radius of the mentioned circle) to the far reference point. (in the example Paris)

Just my two cents,

Jaap
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