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    Re: Certaine Errors in Navigation Corrected
    From: Gary LaPook
    Date: 2007 Dec 04, 16:55 -0800
    Gary LaPook writes:

    You can look at his explanation yourself and you will see that is no allowance for an elliptical earth so it uses the round earth assumption used throughout celestial navigation.

    I would think his method could produce better accuracy with either modern printing of the form to use, larger scale or precision machining of a mechanical device to do the computation. One tenth minute precision is not needed for flight navigation and many methods and devices were used that produced accuracy that was attainable by the Wright method. I suspect that the method was just forgotten in the mists of time.

    gl


    Fred Hebard wrote:
    Some naive comments/questions:
    
    First, how much of the discrepency between Wright's calculated
    distance and the modern digital calculator is due to the elliptical
    shape of the earth, or were you using the same assumptions?
    
    Second, one could guess that a graphical method would be good to 3
    decimal places (about what you got for question 1).  Five-decimal-
    place precision is needed to get 0.1 arcminute accuracy, more or
    less, so a graphical method would only be good to 10 arcminutes, more
    or less.  Perhaps it's the lack of precision that led to Wright's
    method not being adapted to standard sight reduction.  Certainly back
    in his time, simple reduction of noon sights for altitude was easy
    enough.  By the period when time sights for longitude became
    prevalent, and especially by the point when intercept methods took
    over, 3 decimal places wasn't close enough anymore.
    
    Fred
    
    On Dec 4, 2007, at 4:18 AM, Gary J. LaPook wrote:
    
      
    Gary J. LaPook wrote:
    
    
    It is not surprising that nobody ever noticed this before
    (considering that Wright published in 1599 almost 300 years prior
    to Marc St. Hilaire) that Wright's method of calculating the great
    circle distance on the earth using only a strait edge and a compass
    could just as easily be used to calculate the altitude of a
    celestial body. The great circle distance is simply 60 NM times the
    number of degrees of the great circle between two points and this
    is exactly the same as the zenith distance to a body having the
    geographical position represented by the second point.     The
    formula is 90º minus zenith distance equals altitude.
    
    Wright's example of calculation of the great circle distance
    between London and Jerusalem resulted in his calculated distance of
    2325 NM and a modern digital calculator comes up with 2316.8 NM a
    difference of  only 8.2 NM or minutes of zenith distance or of
    computed altitude for those coordinates! Using his method Wright
    could compute altitudes to a precision of 8.2'. It is surprising in
    light of the many devices invented later in an attempt to find a
    mechanical method for this calculation that none (that I am aware
    of) attempted to use Wright's method, a method that would seem
    easily adapted to a mechanical device and that could provide much
    greater accuracy using a larger scale and precise machining of the
    parts.
    
    I would really like it if someone could explain why Wright's method
    works since I have not been able to find such an explanation
    anywhere. I am attaching pages 45-52 of "Certaine Errors" in which
    he lays out his method. I am also including the errata sheet
    showing that the corrections of typos I identified in my previous
    posts were correct.
    
    gl
    
    
        
    <Wright Pages 45-52.pdf>
    <Errata.pdf>
        
    
    
    
    
    
      


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