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    Re: Rejecting outliers
    From: Peter Hakel
    Date: 2011 Jan 2, 13:11 -0800
    Thank you, George, for reminding me that this data from Peter Fogg was available.  I added more output to my spreadsheet in order to provide further information as to what is going on with the weighted least squares procedure in this case.

    The yellow column D now displays the weight of each data point relative to the maximum, marked by the value of 1.000, that is,
    col D = weight / maximum(all weights), i.e.,
    D = BL / MAX(BL)

    The output ("best" UT and H) is in column G.  The "Scatter" parameter (F17, earlier called "Resolution") should be selected "within reason" and this data set provides a good illustration of that.  If this parameter is too low, all but two data points will be removed from consideration.  If it is too high, you end up with ordinary (non-weighted) least squares, where every data point is equally important.  The optimum is somewhere in the middle, around 2.5' for this data set, which reflects the inherent scatter in the original data.  (Cell G15 can serve as a rough guide for the magnitude of this random spread.)  The value of 2.5' is not too different from the error bars in your plot.

    Varying F17 I get the following results in the yellow and orange column G:
    Scatter       UT              Hfit
    0.1'          5:28:50         66d 28.5'      (line through #4 and #8, like your solid line, crossing with your dashed)
    1.5'          5:28:09         66d 24.1'
    2.0'          5:28:14         66d 24.1'
    2.5'          5:28:12         66d 24.0'      (this looks like the best result)
    3.0'          5:28:08         66d 23.9'
    7.0'          5:28:10         66d 25.6'      (non-weighted fit, all D=1.000, your mean)

    What I called the "best result" is just about at the lower end of the error bar on the mean from your plot.  In column D you can see the various (and non-negligible) weights with which the individual data points affect the final result.  I notice that observations #1 and #3, rejected by Peter Fogg, have low weights; #9 has the lowest weight of all of  only 9.1% of maximum.  Based on the fitted slope in cell BI28 (a=0.3710 degrees over the 4min 51sec of observation duration) I calculate the slope of 22.9' per 5 minutes, with altitude of 66d 13.9' at the initial UT=5:26:00.

    One can certainly plot the data and look at trends in order to average. Alternatively, looking at the numbers as the "Scatter" parameter is adjusted to "optimum" is also an option.  One could automatize the search for this optimum Scatter but that would be a bigger project. :-)

    I also, belatedly, thank Peter Fogg for providing this data that enabled further understanding.


    Peter Hakel



    From: George Huxtable <george{at}hux.me.uk>
    To: NavList@fer3.com
    Sent: Sun, January 2, 2011 9:04:07 AM
    Subject: [NavList] Re: Rejecting outliers: was: Kurtosis.

    Fred Hebard wrote-
    "All of this discussion could be informed immensely by some data and
    associated analyses.  Data talk."

    And in a later posting, objected to the use of simulated data, rather than
    real data.

    In my view, both have their place. With simulated data, it's possible to
    know the exact value of the original quantity, before it's subjected to
    deliberate perturbations, so you can discover how well an analysis
    procedure recovers that original quantity from the perturbed data.

    But in general, I agree with Fred. I can't think of any more appropriate
    set of data to investigate than that set of 9 observations which has been
    proffered by Peter Fogg on many occasions over the last 3 years, most
    recently on 13 Dec 2010, attached as "102278.example slope.jpg" under
    threadname "[NavList] A 'real-life' example of slope", and attached here
    under the same label. That plots and also tabulates the 9 observations,
    together with values for latitude and azimuth, which if correct result in a
    calculated slope of 32 arc-minutes in the 5 minite period of observation.

    I have replotted those points as an Excel chart, attached as "slope fit
    3.xls", and for those who don't have Excel, as a simple picture, "slope fit
    3.gif", which shows exactly the same thing. 5 hours should be added to the
    time in minutes on the bottom scale to correspond to time of day, and 66º
    to the altitudes shown in arc-minutes.

    What I have done is to plot all 9 points of that series, discarding none.
    Error-bars have been estimated, based on the observed scatter, of +/- 4.45
    arc minutes. They have then been analysed in a number of different ways.

    First, they have been treated as most navigators would do, by simple
    averaging. This boils them down to a single mean point we will call P, at
    which the averaged time is 5h 28m 10s, and the averaged altitude is 66º
    25.9'. The standard deviation of that mean is reduced, compared with that
    of each individual observation, by a factor of 3 (= root 9) to +/- 1.48
    arc-minutes, as shown by its error-bar. This single point is then chosen to
    represent the altitude in subsequent calculations.

    Second, a line, constrained to have a slope of 32' in 5 min of time, has
    been fitted to those points as well as possible, minimising the squared
    deviations from it, and plotted as a dotted line. That slope was chosen  to
    accord with Peter Fogg's own estimate. It's on the basis of those
    deviations from that line, that the standard deviation of the data-points
    has been assessed. Whatever its slope, every such line has to pass through
    the point P, as Lars Bergman pointed out in a posting on 9th December. It
    will be clear to most navigators that the observed points appear perfectly
    compatible with that line, in the way they scatter around it. The largest
    departure is that of point 1, which differs by 1.85 standard deviations.
    Just according to regular Gaussian statistics, we would expect
    one-fifteenth of the points (on average) to differ by that much or more, so
    it should cause no great surprise to find one such, in a sample of 9. It
    provides no grounds whatever for rejecting that point as any sort of
    deviant outlier.

    Third, we have to consider how Peter Fogg has analysed this observation,
    which presents some problems. He has discarded not just observation 1, but
    also no. 3. His grounds for doing so have nowhere been stated clearly,
    despite numerous requests. Statements about his procedures have used the
    word "intuition", more than once. Through the remaining 7 points he has
    attempted to fit a straight line, as shown here by dashes. Unfortunately,
    though he has specified that line to have a slope of 32' over 5 minutes of
    time, his own plot (and therefore mine as well) has actually been drawn
    with a slope of 34. (This is the second example in which he has drawn an
    erroneous calculated slope.) Whether that error has contributed to his
    rejection of points 1 and 3, only he can tell us. After all this, his
    slope-fit passes about 1.4' away from point P. Which result is most
    true-to-life is impossible to say.

    Fourthly, we can ask for a best-fit straight-line to the data, allowing the
    best-slope to be freely chosen instead of being constrained to 32. Just a
    glance at the data is enough to indicate that the chosen slope of 32' per 5
    min does not accord particularly well with the observed data, and a reduced
    slope would fit it better. When we do so, the resulting continuous line,
    showing a significantly better fit to the 9 observations, has a slope of
    only 24. That's by no means conclusive; no more than suggestive, that the
    calculated slope of 32 may be somewhat suspect. It would be worth checking
    it out once again, to be sure. The quoted latitude of 34º corresponds with
    the (South) lat. of his home port of Sydney, so is unlikely to be wrong,
    but what about the calculated azimuth of 149º? We haven't been given
    sufficient information to check that for ourselves; perhaps we can be
    provided with the missing details.

    ================

    I hope this has provided Fred, and maybe others, with enough data to be
    able to assess whether Peter Fogg's data-rejection is justified, whether
    his procedure (whatever it may be) offers any improvement over standard
    statistics, and whether all the prolonged resulting hoo-hah has been
    worthwhile.

    George.

    contact George Huxtable, at george{at}hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ----- Original Message -----
    From: "Peter Fogg" <piterr11---.com>
    To: <NavList@fer3.com>
    Sent: Monday, December 13, 2010 7:49 AM
    Subject: [NavList] A 'real-life' example of slope


    | The attached file, an example of slope in action, comes from a post I
    made
    | on 10 March 2007 [NavList 2278].  It may seem like a poor round of sights
    | compared with Antoine's, but remember the crucial difference often
    ignored
    | by our armchair navigators: the relative stability of the platform used.
    | Unless the sea conditions are abnormally calm, the near-perfection of
    | Antoine's sights is in practice unachievable from the deck of a smallish
    | sailing boat, in my experience.
    |
    | The analysis below comes from [NavList 2455] of 22 March 2007:
    |
    | Sights 1 and 3 have been discarded, as they cannot be matched to the
    | slope. This slope, a fact, is then best matched to the pattern of the
    | other sights that exhibit random error.
    |
    | What is the alternative to this technique? In this example, taking
    | just the one sight could have been equivalent to choosing any one of
    | these sights at random. What were the odds of obtaining as good an
    | observation as the slope will produce with just the one sight?
    |
    | Of a poor sight (#1&3):                      2 out of 9;        22%
    | Of a mediocre sight (#2,4,6,7,8,9):          6 out of 9;        67%
    | Of poor or mediocre:                        8 out of 9;        89%
    | Of an excellent sight (#5):                  1 out of 9;        11%
    |
    | So at the cost of a little extra calculation and the drawing up of a
    | simple graph this 11% chance has been converted to a 100% chance of a
    | similar result to what appears to be the best sight of the bunch,
    | together with all the other advantages of KNOWING a lot more about
    | this round of sights, and being able to derive extra information (eg;
    | standard deviation) at will.
    |
    | Is this a typical example? No. Typically there are fewer sights in the
    | 5 minutes, and NONE of the individual sights is as good as the derived
    | slope; confirmed by comparing the resulting position lines to a known
    | position.
    |

    File:


    File: 115117.average2.xls
       
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