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    Re: 3-Star Fix - "Canned Survival Problem"
    From: Gary LaPook
    Date: 2008 Jun 14, 03:44 -0400
    Being the conscientious navigator that I am, I followed my usual practice of memorizing some data from the 2008 Nautical Almanac so that I would have it available for emergency use.

    The first thing I memorized was the GHA of Aires at 0000 Z January 1, 2008 which is 100º 01.9' and also remembering that Aires advances 59.139' each day. With this information you can calculate GHA Aires for 0000 Z on June 9, 2008 which is the 161st day of the year but is only 160 days from January 1st. So multiplying 59.139' times 160 days gives  157º 42.2'  to which you add the starting value of 100º 01.9' to come up with the GHA Aires on June 9th at 0000 Z of  257º 44.1'. To this you add the change of GHA for the time since 0000 Z (3 hours 42 minutes 10 seconds for the Vega shot) by multiplying the time interval by the rate of change of 15.041º per hour making 55º 41.6' making the GHA Aires at the time of the Vega shot of 313 º 25.7'.

    I also memorized the SHAs and the Declinations of ten of the navigation stars ( nobody could memorize all 57) which should be enough for emergency use as tabulated for July 1st so that the values will be reasonable for the whole year. Fortunately this included the three stars used in this exercise. So now adding the SHA of Vega, 80º  41' we end up with the GHA of Vega of 34º 06.7' and using the D.R. as the A.P. we get an LHA of 274º 48' and the declination of 38º 47' N. (rounded to the whole minute)

    Using these values on my Bygrave slide rule (see attached work sheet) since I have no tables with me, I computed Hc of 23º 59'.

    The Hs given was 24º 05.5' Computing the dip correction in my head of 4.5' (the square root of 20 must be between 4 and 5 ) and applying the refraction correction of minus 2 gives an Ho of 23º 59' giving and intercept of zero and an azimuth of  58.1 º.  I long ago memorized the refraction table for altitudes above 10º in The Air Almanac and in H.O 249, the cutoff values are 63-33-21-16-12-10º , zero above 63, 1 above 33, 2 above 21, 3 above 16, 4 above 12 and 5 above 10.

    I used the same procedure for Spica and Pollux getting another zero intercept for Pollux, Zn of 290.2º and a  4 NM away for Spica with a Zn of 171.7º.

    Since I am on the road I do not have any of my plotting tools with me so I had to make do with what I found in my briefcase. I used my MB-2A flight computer since it had an azimuth scale and I used a pad of paper  with a right angle at the corner as my straight edge for plotting the LOPs. I used a tape measure from IKEA to measure the length of the intercept (see photo.)  I plotted the LOPs and found the fix by bisecting the three angles giving a fix .4 NM west of the A.P. (D.R.) and 2.8 NM north of it. (Plotting a fix as a distance from the A.P. like this is common in aerial practice and it is often done on an E-6B.) Adding the 2.8 NM north to the D.R. latitude gives a fix latitude of 34º 16' North. To convert the .4 NM west  to a longitude you divide the .4 NM by the cosine of the latitude, .82, to find the difference in longitude of  .5' so the fix longitude is 119º 19.5' West (rounded to either 119º 19' or 20'.)  ( I got the cosine of 34º by finding the sine of 56º on the MB-2A sine scale, used for wind correction calculations.)

    My fix might not be in agreement with others but I used a refraction table tabulated in whole minutes, I only memorized the stars' positions to the nearest minute and I did not have any plotting tools to use but my position is certainly good enough for emergency navigation and done without an almanac, tables or electrons.

    (I will have to send the images when Ii figure out how to make them smaller.)


    m_burkes@msn.com wrote:
    Captain Lecky would be proud of those dividers ha! Speaking of
    interpolation I have found a neat way to get around that pesky DSD and
    interpolation tables by using the aviation E6B computer or the
    equivalent nautical slide rule. Essentially the set up:d-value/60=d-
    correction/declination minutes. Yes the calculator offers the proof.
    Mike Burkes
    On Jun 12, 11:44 pm, Anabasi...@aol.com wrote:
    Thanks for the nice exercise Greg.  I literally had to dust off the  ship's
    Vol III of HO 229 and deflower a Plotting sheet 925 to work this one  out.
    Since I was bereft of electronic gadgets, I did this with a plotting sheet,  
    2 triangles, a pair of dividers, 2 books, a pencil, and small piece of scratch
     paper (wouldn't have reams of paper in the Lifeboat).  I have attached a  
    picture in to this message with the plot and the tools.
    My Lat is a bit lower (plotting or math error?).  I used an assumed  position
    method and HO 229.  I had to assume we were drifting and no  current (didn't
    advance or retard the lines).  I had not done a full HO 229  paper reduction
    of a star in many years, and I had to think a second to remember  how to use
    the interpolation pages on the inside covers for the declination  interpolation.
     I usually whip those off with the calculator.  Still,  I got pretty close to
    the computer solutions with Lat 34deg 11.9' N and  Longitude 119deg 16.0'W.
    As to how you would get an Eastern sight on the west coast, you would have  2
    options in general.  The first would be a back sight.  This would be  
    particularly difficult with a regular sextant at such a low altitude.  The  other
    option would be to use a bubble sight tube or other artificial  horizon.  If you
    were across a bay, you could also use a dip short of  the horizon table.  
    That's all I can think of at the moment.
    **************Vote for your city's best dining and nightlife. City's Best
    2008.      (http://citysbest.aol.com?ncid=aolacg00050000000102)

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