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    Re: Martelli's Navigational Tables
    From: George Huxtable
    Date: 2005 May 30, 17:35 +0100

    Lu Abel wrote-
    >
    >Sometimes it's easy to forget how different sight reduction was from the
    >earliest days of celestial navigation until about 25 years ago when PCs
    >and pocket calculators arrived.  Navigators had to rely on sight
    >reduction tables and/or longhand calculations of the celestial triangle
    >formulae.   Given that accurate navigation requires 4~5 digit accuracy
    >in answers (dd mm.m), and a rule of thumb is that calculations should be
    >carried out with at least one more digit of accuracy than desired in the
    >final answer, longhand paper calculations must have been daunting indeed!
    >
    >One way to make things easier is to use logarithms for multiplication
    >instead of actually trying to multiply a pair of six-digit numbers.  But
    >there's a problem:  Logarithms are defined only for positive numbers and
    >sines and cosines can be negative as well as positive.  Enter the
    >versine:  Versine (x) = 1 - cos(x).   As you can see, this simply
    >inverts the cosine curve and adds 1 to it, making it range between 0 and
    >2.  It's a bit more convenient to have a function that runs between 0
    >and 1, so it's divided in half, giving the half versine or haversine:
    >hav(x) = (1 - cos(x))/2.
    >
    >The celestial triangle formulae involving sines and cosines can be
    >restated in terms of haversines.  By using a trig function that is
    >always positive, it can be solved with the aid of a table of logarithms.
    >
    >In a quick search I can't find the celestial formulae exactly, but
    >here's a link to the formula for a great circle.  Hc is simply 90
    >degrees minus the great circle distance to the GP of the body.
    >http://www.mathdaily.com/lessons/Haversine_formula
    >
    >By the way, your GPS likely calculates great circle distances using this
    >formula rather than the traditional spherical triangle formula.  That's
    >because calculating short distances using the traditional formula
    >requires taking the difference between two large numbers that are fairly
    >close to one another using the traditional formula.  Tiny differences
    >due to rounding and a limited number of significant digits can result in
    >significant errors.  (Interestingly, errors can creep into the haversine
    >formula with very long distances, but I suspect a one mile error in
    >calculating the distance between New York and Beijing isn't as
    >significant as a one-mile error in a local distance.)  Calculations
    >aren't actually made using haversines, the haversine formulae can be
    >re-expressed in terms of ordinary sines and cosines and that's what's used.
    >
    >Lu Abel
    
    ===============
    
    Lu's reply was spot-on, but I may be able to fill in the missing formulae.
    
    Calculating altitude and azimuth of a celestial body is EXACTLY the same as
    calculating distance and initial-course of a great circle between A and B,
    except that calculated alt. corresponds to (90 - distance in degrees).
    
    You have to remember that haversine (or hav) of an angle X is half(1 - cos
    X), and it varies from 0 at 0 degrees, through 0.5 at 90 degrees, to 1 at
    180 degrees. It's always positive, and has a distinctly different value for
    all angles in the range 0 to 180 deg. These are the qualities required.
    
    Then for a great circle-
    
    hav (distance in degrees) = cos (lat A) * cos (lat B) * hav (difference in
    long) + hav (difference in lat)
    
    For the altitude of a body this would be equivalent to-
    
    hav (90 - Hc) = cos (lat) cos (dec) hav (dlong) + hav (dlat)
    
    It's worked out quite readily from a table which gives, for a range of
    angles, log cos, log hav, and natural hav. I'm not familiar with Martelli,
    but perhaps that's what he provided.
    
    ========================
    
    As well as the calculated alt. of a celestial body, you also need its
    azimuth, measured 0 to 180 deg from North. Whether it's to be measured
    clockwise or anticlockwise is obvious, depending on whether it's to your
    East or your West.
    
    A simple and popular, but VERY faulty, method for calculating az. is-
    
    sin az. = sec Hc sin (dlong) cos dec.
    
    The serious problem with this is that sin 95 deg (say) is exactly the same
    as sin 85 deg, and there is NO SIMPLE WAY to tell whether an azimuth, near
    to East, should be North of East or South of East. This vital matter is
    glossed-over in many texts.
    
    Using haversines instead, one formula is-
    
    hav az. = sec Hc * sec (lat) * hav (90 - dec) - hav (90 ~ (lat + 90 - Hc))
    
    A different approach uses special tables of "half log haversine"
    
    =========================
    
    The above formulae I have lifted from "The principles of Navigation", by E
    W Anderson, with a bit of adaptation, which I hope I've got right...
    
    Of course, all these complication were, as Lu pointed out, the result of
    having to do the arithmetic using logs, which were unable to handle
    negative quantities.
    
    Nowadays calculators and computers can happily deal with trig functions
    which go negative as well as positive, and can use the straightforward
    basic equations. As a result those terrible twistings of the formulae to
    keep everything positive are no longer needed.
    
    Instead of calculating azimuths from their sines, it's just as easy now to
    calculate them from their cosines, or even better, from the tangent, as we
    have discussed more than once before on this list.
    
    George.
    
    
    
    
    ================================================================
    contact George Huxtable by email at george@huxtable.u-net.com, by phone at
    01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
    Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ================================================================
    
    
    

       
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