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    Re: running fix + lunar
    From: Paul Hirose
    Date: 2010 Nov 22, 21:31 -0800

    Andres Ruiz wrote:
    > The "practical circumstance", is: one is sailing with known course and
    > speed and to obtain a fix uses two altitudes with a sextant and a
    > lunar distance at different times ant the chronometer or on-board
    > watch has an error. Please Paul, correct me if I have not
    > misunderstood your example.
    
    You understand me perfectly. The idea of celestial sight reduction with
    a set of linear equations is nothing new. An especially elaborate
    example is the Kaplan paper, referenced in your other message. The only
    new idea I've introduced is that time is only approximately known, and
    must be found by a lunar. In reality this is an improbable scenario, but
    I thought a few readers might find it interesting.
    
    
    Here is another example. This time we'll shoot an afternoon lunar with
    the Sun, followed by three altitudes at evening twilight. Normally, Sun
    and Moon altitudes would be observed with the lunar distance, but to
    make the problem more difficult let's say the horizon is bad at that 
    time. So the lunar distance must be combined with three altitudes 
    several hours later.
    
    Here are the observations on 2010-12-01 and -02, with UTC from the
    chronometer, and estimated positions.
    21:36:02 at N25°57.7' W142°00.5' Moon-Sun 49°53.1'
    02:27:11 at N26°35.3' W140°05.7' Jupiter 51°28.1'
    02:32:56 at N26°36.1' W140°03.5' Schedar 51°11.7'
    02:38:24 at N26°36.8' W140°01.4' Altair 50°21.1'
    
    Those times and positions have considerable error. As before, assume the
    same error affects all observations. The observed angles are accurate.
    The lunar is near limb to near limb. Once again I'll try to describe the
    method with simple language.
    
    Begin by computing altitude (or lunar distance) at the given times and
    locations. Compare these to the observed angles.
    
    Moon - Sun computed separation angle is 1.7' greater than observed. One
    degree north changes computed angle -.7'. One degree east changes it
    +.6'. One minute later changes it -.4'.
    
    I don't know a way to gauge the effect of latitude and longitude on
    lunar distance, except to recompute from scratch for a different
    position. However, I have a program which outputs the rate of change per 
    minute.
    
    The "partial derivatives", as these are called, are much easier to
    compute for the altitude observations. The increase in altitude equals
    the (north) increase in latitude, times the cosine of azimuth. It also
    equals the (east) increase in longitude, times the sine of azimuth,
    times the cosine of latitude.
    
    And since the passage of time has the effect of moving east in
    longitude, its partial derivative is simply some fraction of the
    longitude partial derivative. E.g., one minute of time is equivalent to
    1/4 degree of longitude, so if 1 degree of longitude changes the
    computed altitude by 40', 1 minute of time changes it 10'.
    
    Getting back to our observations, Jupiter's computed altitude is 1°41'
    less than observed. Moving one degree north changes computed altitude
    -42'. Moving one degree east changes it +38'. One minute later changes
    it +9.5'.
    
    Schedar computed altitude is 3°09' less than observed. Moving one degree
    north changes computed altitude +50'. One degree east changes it +29'.
    One minute later changes it +7.3'.
    
    Altair computed altitude is 4°05' greater than observed. One degree
    north changes computed altitude -24'. One degree east changes it -49'.
    One minute later changes it -12.3'.
    
    Set up the equations as described in my previous message:
    -1.7 = -.7∆φ + .6∆λ - .4∆t
    101 = -42∆φ + 38∆λ + 9.5∆t
    189 = 50∆φ + 29∆λ + 7.3∆t
    -245 = -24∆φ - 49∆λ - 12.3∆t
    
    Unlike before, we have three unknowns and four (not three) equations. A
    solution may be found by the method of least squares. Results:
    ∆φ = +1.38° = N 1°23'
    ∆λ = +2.75° = E 2°45'
    ∆t = +5.99 m = +5m 59s
    
    Add those corrections to the times and positions. The observed angles
    are the same as before.
    21:42:01 at N27 20.7 W139 15.5 Moon 49 53.1
    02:33:10 at N27 58.3 W137 20.7 Jupiter 51 28.1
    02:38:55 at N27 59.1 W137 18.5 Schedar 51 11.7
    02:44:23 at N27 59.8 W137 16.4 Altair 50 21.1
    
    Now repeat the process of determining (computed - observed), and the
    partial derivatives. Set up new equations:
    .6 = -.7∆φ + .4∆λ - .4∆t
    11.4 = -46.6∆φ + 33.3∆λ + 8.3∆t
    -2.8 = 50.9∆φ + 28.1∆λ + 7.0∆t
    -4.3 = -22.4∆φ - 49.1∆λ - 12.3∆t
    
    Note the values on the left side are much smaller than before. It shows 
    the first set of corrections were a big improvement. New solution:
    ∆φ = -.138° = S 8.2'
    ∆λ = +.369° = E 22.2'
    ∆t = -.878 m = -53 s
    
    In the upper line of each pair below I show the corrected time and
    position, and the corresponding computed lunar distance or altitude. The
    lower line has the true values, which are unknown to the navigator
    (except for the observed angles).
    
    21:41:08 at N27 12.5 W138 53.3 Moon-Sun 49 53.1
    21:41:18 at N27 11.8 W138 53.5 Moon-Sun 49 53.1
    
    02:32:17 at N27 50.1 W136 58.5 Jupiter 51 28.1
    02:32:27 at N27 50.0 W137 01.0 Jupiter 51 28.1
    
    02:38:02 at N27 50.9 W136 56.3 Schedar 51 11.7
    02:38:12 at N27 50.8 W136 58.8 Schedar 51 11.7
    
    02:43:30 at N27 51.6 W136 54.2 Altair 50 21.1
    02:43:40 at N27 51.5 W136 56.7 Altair 50 21.1
    
    We now have a perfect match between computed and observed angles.
    However, some error remains in the time and position. I believe this is
    mainly due to error in the dead reckoning between the afternoon lunar
    and the evening sights. The navigator used 22 knots on true course 069
    to come up with the estimated positions at dusk. I introduced an error
    in both course and distance made good, equivalent to 3% of the distance
    run. Thus, the same set of time and position corrections cannot be valid
    for both the afternoon and evening observations.
    
    The results were still pretty good. I could have injected errors in the
    observed angles too, but there are limits to how hard I want to work!
    Fortunately my HP 49G calculator makes it easy to solve simultaneous
    linear equations. It automatically uses the method of least squares if
    the solution is "overdetermined" (the number of equations exceeds the
    number of unknowns). This is not a very modern calculator, either. It's
    about 10 years old.
    
    The problem in my first message, with three equations in three unknowns,
    could actually be solved with a simple calculator, pencil, and paper.
    Some readers may have solved similar sets of equations that way in high
    school. Unfortunately, a least squares solution is not so simple. I'd
    hate to work it with nothing but a simple calculator. It wouldn't be too
    bad with something like an early 1980s HP-15C, which can invert and
    multiply matrices. That's the hard part.
    
    One refinement of the least squares solution is to put additional weight
    on a lunar distance, because usually that angle is more accurate than an
    altitude. In my example, that would be accomplished by multiplying the
    coefficients of the first equation by say 1.5 or 2.0.
    
    The same thing can be accomplished by expressing the residuals (observed
    - computed value) and the partial derivatives in "sigmas", where one
    sigma is the standard deviation of the measurement. This could simply be
    the navigator's estimate.
    
    -- 
    I filter out messages with attachments or HTML.
    
    
    
    

       
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