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    Re: Beginner Meridian Passage Question
    From: Bill B
    Date: 2004 Sep 2, 20:36 -0500

    Bill, Zorcec, and Jim,
    
    Than you very much for your help.  A couple of questions on your
    information:
    
    > b) when you think Polaris is directly below or above the pole, add
    > or subtract its difference (which you have memorized as about 45');
    
    It looks to me that the average declination for Polaris in 2004 was close to
    89 d 17'.  90 - 89 d 17' = 43'.  Am I missing some 2nd and 3rd order
    variables when calculating the distance of Polaris from PN?  Id so, what?
    
    > And the misconceptions...
    >
    >>> -"I am guessing the same is true for the Southern Hemisphere"
    > Do not speak about southern or northern hemisphere.
    > The pole in the same hemisphere of the observer zenith is called : elevated
    > pole.
    
    Understood.  I was relating to the upper transit as being south of the PN in
    the Northern Hemisphere, and vice versa in the Southern Hemisphere. After
    your explanation I see there are better ways to state it (between elevated
    pole and zenith, or LHA = 0).
    
    This presents me with a conundrum. If I am moving along the equator, what
    becomes of the elevated pole?
    1. There are two
    2. There are none
    3. It depends on whether I am facing north or south
    4. It doesn't matter.  Even if I get far enough above the water to see
    Polaris its low angle and refraction make it a lousy choice. And there is
    currently no southern pole star.
    5. I am like a dog between four trees, not a leg to stand on 
    6. Other
    
    Now what started me down this path in the first place.  Not sure it is very
    productive or useful, but my self assignments force me to get a better grasp
    on the coordinate systems and relationships, as well as grinding the rust
    off my algebra and trig skills (or lack thereof).
    ============================================================================
    Following is my best guess on determining latitude directly from meridian
    passages of other circumpolar stars (and possibly rough longitude if
    observations are made before and after predicted transit) from other. Does
    it hold water?
    
    Use star finder of planisphere to locate a suitable candidate
    
    Distance from pole = 90 - Declination
    
    Given:  LHA = SHA + GHA Aries - Longitude
    Given:  Meridian passage will occur when LHA = 0
    Therefore:  GHA Aries = Longitude - SHA (+/- 360 if necessary)
    NOTE:  If only info is RA in hh:mm:ss,  SHA = 360 - (RA * 15)
    
    Determine meridian passage in a manner similar to finding LAN.
    * From the daily page in almanac locate the GHA Aries that bracket the LHA
    * Subtract the lesser GHA Aries from the LHA
    * Convert the remainder from arc to time*
    * Add that time to the GHA Aries hour used above
    * Convert the sum from GMT to local time
    
    *NOTE:  Standard arc-to-time table cannot be used as it is based on a
    24-hour solar day and a star is based on a 23:56:04 sidereal day.  Two
    approaches may be used:
    
    1.Work the Increments and Corrections almanac table in reverse:
    * Find the degrees dd:mm!m difference from above in the Aries column
      and read off the time.
    
    2.
    * Divide the solar day by the sidereal day   (24/23:56:04) using
      their decimal equivalents
    * Multiply the above quotient by 15 degrees = 15d 02' 27.9"
    * Divide above arc (LHA-GHA Aries) by 15d 02' 27.9" (both in decimal form)
    * Convert the quotient to mm:ss
    
    The same procedure can be used for a lower branch transit where LHA = 180.
    
    Lower branch meridian passage will occur when LHA = 180
    Therefore:  GHA Aries = 180 + Longitude - SHA (+/- 360 if necessary)
    
    To determine latitude:
    
    * Correct from Hs to Ho as per usual
    * Latitude = Declination of upper branch passage - Ho
    * Latitude = Declination of upper branch passage + Ho
    
    Thanks again
    
    Bill
    
    
    

       
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