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    Re: How far is polaris?
    From: Gary LaPook
    Date: 2007 Nov 21, 13:21 -0800

    Gary writes:
    Polaris is now at declination 89� 18.1' North meaning it circles the
    true pole at a radius of 41.9'.  When its LHA is 0� it is 41.9' above
    the pole; when LHA is 180� it is 41.9' below the pole; when LHA is
    either 90� or 270� it is at the same height as the pole. You can
    figure this correction by multiplying 41.9' times the cosine of LHA
    I posted the following on September 9th about the Polaris correction
    table in the Nautical Almanac and the equivelent "Q" correction tables
    in the American Air Almanac and in H.O 249:
    "A0 accounts for the the cosine of  LHA of Polaris times the distance
    Polaris is from the pole (90� - decl. Polaris).
    A1 accounts for "the tilt of the diurnal circle of Polaris with
    respect to the vertical." (Bowditch, article 2105, 1962 ed.)
    A2 accounts for the movement of Polaris from its nominal position
    during the year.
    To make all the factors positive, constants were added to all three
    and the sum of all the constants is 1� so 1� is subtracted at the end
    of the computation.
    The "Q" correction table found in H.O. 249 only corrects for A0 as A1
    is small and irrelevant to the precision possible in aerial celestial
    navigation. The "Q" table strictly is only accurate for the year of
    the  "epoch" (2005) of the table as it is based on the coordinates of
    Polaris on the date of the epoch. Since H.O. 249 is used for a 10 year
    period  a "Precession and Nutation" correction table is provided to
    correct positions obtained for any year 2001 through 2009. This
    correction should also be applied to a Polaris LOP calculated with the
    "Q" table as this will correct for the movement of Polaris from its
    nominal position similar to factor A2.
    Since A0 is the largest factor you can compare the N.A. Polaris table
    with the "Q" table by subtracting 1� from A0 and it will be
    approximately the same as the "Q" factor for the same LHA. If you do
    this with a copy of the 2005 N.A. it should end up with the same value
    to the precision of the "Q" table.
    Another easy way to derive the correction for a Polaris sight is to
    plot the position of Polaris on a rotating circular plotting board
    such as the Navy Mk 5A or Mk 6A; or the Air Force Polhemus Celestial
    Computer, CPU-41/P; or even on the 2101-D Star Finder. Subtract
    Polaris' current GHA from 360� and then plot Polaris on the plotting
    board on that azimuth and at a distance out from the center of the
    board equivalent to its distance from the pole, currently about 42'.
    (Plotting it this way allows you to use LHA Aries instead of LHA
    Polaris, saving you one step in the computation.) You then only have
    to set the plotting board to LHA Aries and the distance Polaris is
    above or below the center gives you the "Q" correction. You can update
    the position of Polaris from time to time. "
    On Nov 21, 6:12 am, Isonomia  wrote:
    > OK, I've now determined that this latitude error can't be due to the
    > wabble of polaris, because it is an orders of magnitude too large (the
    > monthly table provides this correction based on the relative movement
    > of the earth around the sun).
    > It still leaves me perplexed. At LHA Aries of 120-129 it is 0.2' (at
    > 0deg lat) and 1.0' at (68deg latt) whilst with aries at 230-239, it is
    > unvarying at 0.6'.
    > Oh ***k! -and how stupid of me! Finally got it - the latitude
    > correction on the polaris table is the same latitude correction as is
    > needed for any other star but because polaris' wobble is only 1deg
    > this correction can fit on a table a lot lot lot smaller than any
    > other star!
    > Mike
    > On Nov 21, 10:34 am, Isonomia  wrote:
    > > I was looking at the navigation table for Polaris table and couldn't
    > > work out what the correction of up to 1' with latitude (and LHA) on
    > > the polaris table represented.
    > > It seems to be constant at one LHA going to a maximum at 180degrees
    > > and therefore the only only thing I can imagine it would be is a
    > > correction for the none infinite distant from earth to the star - is
    > > this right?
    > > If so, do other stars have similar correction tables which I've
    > > missed?
    > > Mike
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