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David

Hardly a spherical trig expert, but I will propose a KISS thought
experiment which you can easily do by determining the Hc's of a
theoretical body keeping the SHA (hence LHA) and latitude constant and
varying the declination. Then solve for the azimuths.

You should see the azimuth from your AP to the body decrease as its
declination increases. This makes sense as the meridians (not YOUR
meridian, but the meridians of longitude) converge at your elevated
pole. Clearly with an LHA of zero a change in declination will not
affect the azimuth.

Now the thought experiment. Imagine two bodies with identical
declinations, one to the east and one to the west of YOUR meridian.
Consider their respective azimuths with a declination of 15 degrees. Now
consider their respective azimuths with a declination of 75 degrees.

If I recall Frank has a special case of separation given roughly equal
declinations, but I cannot recall it.

Perhaps oversimplified, but hope it helps.

Bill


On 9/25/2017 12:33 PM, David Pike wrote:
> Last night in a post, I speculated that you could get a rough idea of
> the difference in azimuth of stars by examining their difference in
> SHA.I wasn’t entirely happy with that statement, hence my use of the
> word ‘roughly’.This was not a good problem to go to bed with.After much
> thought, I concluded that it might only be true close to the poles.Study
> of AP3270/HO249 Vol 1 this morning shows this is true at the poles but
> gets progressively more untrue the further from the poles you travel.By
> the time you get to the latitude of Belgiumit’s not much use.Even the
> difference of the differences isn’t constant.
>
> Perhaps one of our spherical geometry experts might explain the true
> relationship, if any.DaveP
>
> 
> 

>
>

   

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