# NavList:

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

**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 Polaris. 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, Isonomiawrote: > 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 --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To unsubscribe, send email to NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---