Welcome to the NavList Message Boards.


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

Compose Your Message

Add Images & Files
    Re: Averaging
    From: Alexandre Eremenko
    Date: 2004 Oct 21, 16:00 -0500

    Thanks to all who participated in the
    Averging discussion.
    Now I can formulate the following Rules:
    Suppose you are averaging altitudes over
    an intevral of 5 minutes.
    1. If the body is 20 degrees away from the meridian,
    the error due to nin-linearity does not exceed 0.25'.
    2. If the body is 10 degrees away from the meridian,
    the error due to non-linearity does not exceed 0.5'.
    3. If the altitude of the body is at most 70d,
    the error due to non-linearity does not exceed 0.5',
    no matter what the hour angle is.
    4. If the altitude is at most 60d, the error does not exceed
    0.3', no matter what the hour angle is.
    Thus the "worst case scenario" mentioned by Herbert Prinz
    is indeed the worst. Moreover, this is essentially the
    ONLY bad scenario for averaging: measuring a large altitude
    close to the meridian.
    Once more: DO NOT average if the altitude is more than 60d
    AND the body bearing is within 2 compass points of N or S.
    In ALL OTHER cases, averaging helps to increase precision.
    Few fords for "mathematically inclined" on how this was obtained.
    The problem is to estimate the non-linearity of
    altitude that comes from the formula
    (1)  sinA=sinDsinL+cosDcosLcost,
    where A,D,L and t are altitude, declination, latitude and LHA
    It is easily seen that L and D change slowly and approximately linearly
    (I assume we are on a boat, not an airplane!), so t is the fastest
    variable in the formula. Moreover, t changes linearly for all practical
    So I estimate the non-linearity of A(t), assuming D and L fixed.
    Let t_0 be the middle of our 5-minute interval.
    By Taylor formula,
    A(t_0+h)=A(t_0)+hA'(t_0)+(h^2/2)A''(t_0)+...=linear part + the error term.
    We need an estimate for the error term.
    We have |h|<2.5 min=0.0109 rad, so
    h^2<0.000118 rad=0.4'.
    It remains to estimate the second derivative A''.
    To do this, one differentiates the main formula twice.
    (sorry, I omit the expression for the second derivative,
    but it is easy to derive).
    Now for |A|<60 deg, the expression is easy to estimate
    by hand. But in general, i just did it on computer, (running Matlab,
    but any other soft will do the job).
    The results are:
    |A"|<2.4 if 10deg
    Browse Files

    Drop Files


    What is NavList?

    Join NavList

    (please, no nicknames or handles)
    Do you want to receive all group messages by email?
    Yes No

    You can also join by posting. Your first on-topic post automatically makes you a member.

    Posting Code

    Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.

    Email Settings

    Posting Code:

    Custom Index

    Start date: (yyyymm dd)
    End date: (yyyymm dd)

    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site