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    Re: Figure out LAN?
    From: Brad Morris
    Date: 2014 Sep 27, 13:50 -0400


    I do so hope you caught the adjustment due to delta longitude from the central meridian of your timezone. 

    The Nautical Almanac produces the time of meridian passage at the central meridian of our time zone (yes, we are in the same time zone). 

    But as neither of us is located on that central meridian, an adjustment to the time published must be made.  Why is that?  Its because of the steady rotation of the earth!  In your case, it takes the earth just a bit more time such that the sun crosses your meridian.  In my case, the published time comes after my meridian passage, as I am before the central meridian.

    The time of meridian crossing can thus be directly calculated.  For us, the central meridian of our time zone is 75°.  Your longitude is 75° 22'.5. Subtracting 75° from 75° 22'.5 leaves 22'.5. 
    Now the earth rotates at 15°/hour; or INVERTING that (60 minutes/15°) yields 4 minutes of time per degree.
    Your delta is 22'.5 (22'.5/60minutes of arc per degree) which is 0.375 degrees.
    Multiply the 4 minutes per degree by 0.375 degrees yields 1.5 minutes of time or 1 minute 30 seconds.
    (I urge you to write this out on a piece of paper, including the units, so that it becomes clearer)

    We're almost there.  To the time published in the Nautical Almanac, simply add the 1m30s we just found for your longitude. 

    As Greg points out, you would want to be shooting the altitude minutes before until minutes after LAN.  It is impractical to try to obtain the altitude just at the very moment of meridian crossing.  Further, there is no need to do so, as the altitude changes very slowly around meridian crossing.

    My longitude is 72° 47'.6.  What is my adjustment for LAN? 


    On Sep 27, 2014 1:07 PM, "Samuel L" <NoReply_SamuelL@fer3.com> wrote:


    Thanks. I usually calculate LAN from 12 UT and subtract the 1 hour for DST.



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