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    Re: 3-Star Fix - "Canned Survival Problem"
    From: Gary LaPook
    Date: 2008 Jun 14, 10:13 -0400
    Greg R. wrote:
    With the exception of the red numbers on the outside rings, your MB-2A
    computer looks remarkably similar to the Jeppesen CR series (CR-3,
    CR-6, etc.) - is it maybe a military version (or an earlier version)of
    the Jepp one?
    
    



    Gary LaPook responds:



    The MB-2A is very similar to the MB-9 pictured here:

    http://www.rekeninstrumenten.nl/pages%20and%20pictures/12071.jpg

    except the true airspeed goes up to 1800 knots on the MB-9 while it only goes up to 1000 knots on the MB-2A. Since I don't fly planes that can exceed 1000 knots I prefer the MB-2A since its more limited speed range allows an expanded scale for the range it covers.

    It is very similar to the Felesenthal PT computer pictured here:

    http://www.rekeninstrumenten.nl/pages%20and%20pictures/12141.jpg

    in that each of these computers solve the wind triangle with trig, no vector diagram is drawn.

    The Jeppesen CR-3 pictured here:

    http://sliderule.mraiow.com/wiki/Jeppesen_CR-3

    uses a diagram on the back to determine wind factors so its method is completely different than the previous computers.  All of these are similar in that they have the standard time-speed-distance scales and they allow for compressibility in computing true airspeed.

    The original E-6B pictured here:

    http://www.rekeninstrumenten.nl/pages%20and%20pictures/12081.jpg

    uses a wind vector diagram on the back and does not allow for compressibility in TAS computations.

    gl


    Greg R. wrote:
    With the exception of the red numbers on the outside rings, your MB-2A
    computer looks remarkably similar to the Jeppesen CR series (CR-3,
    CR-6, etc.) - is it maybe a military version (or an earlier version)of
    the Jepp one?
    
    And I assume this isn't the other calculator that you alluded to
    earlier (Barger, or something similar?) - it's late and I can't find
    that particular post right now.
    
    --
    GregR
    
    
    
    
    --- "Gary J. LaPook" <glapook@pacbell.net> wrote:
    
      
    Gary J. LaPook wrote:
    
        
    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.
    
    gl
    1
    
    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.
    
    Jeremy
    
    **************Vote for your city's best dining and nightlife.
              
    City's Best
        
    2008.      (http://citysbest.aol.com?ncid=aolacg00050000000102)
    
    gregExerPlot.jpg.JPG
    182KViewDownload
    
    GregExerTools.jpg.JPG
    129KViewDownload
    
    
              
    
    
            
          
        
    
    
    
    
      


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