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    Meridional parts
    From: Pete Harmsworth
    Date: 1996 May 02, 12:18 -0600

    I am attempting to set up an Excel spreadsheet in order to input Origin
    & Destination lat & long and output and output rhumb line course and
    distance. This involes calculating meridional parts. Page 552 of
    Bowditch (1995) gives the formula as M=7915.704468 log tan b (45+L/2)-
    23.0133633 sinL-0.051353 sin^3L-0.000206 sin^5L...
    L is the latitude but I can find no mention of b in the text. Does
    anyone know if this is a typo (I wouldn't want to white it out in my
    book if it wasn't). Has anyone coded it into a spreadsheet. I left the b
    out but the result wasn't even close to the value in Table 6. Thanks for
    any input on this.
    
    Pete Harmsworth
    pharms{at}awinc.com
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    To: navigation{at}ronin.com
    Subject: RE meridional parts formula
    Date: Tue, 07 May 1996 20:32:01 GMT
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    Hi,
    
    Following the discussion about meridional parts, I have some
    remarks and questions :
    
    1. The formula comes from integration of  dS = (Earth radius)
        / cos(latitude) between the limits equator and current
        latitude.
    
    2. After Integration, the formula is then
        S = R(Earth) * ln ( tan ( 45d + lat/2) ),
       assuming the Earth to be a SPHERE.
    
    3. What Radius to use :
        We learned : make it 360d * 60' / (2 * pi)  nautical
        miles, giving  R(Earth) = 3437.74677....
    
    4. when using log instead of ln, one has to multiply with
        log (e), yielding the "magic number"  7915.71315....
        When oher numbers do appear,  I think the author uses
       some other definition of Earth radius...
    
    5. Compensating for the ellispoidal form of the earth could
        be done in two ways (hi ho, guessing...)
       a) incorporate a latitude-dependant factor into the number
           for the earth radius  or
       b) subtracting some latitude-dependant number, which will
           have to increase when going to the poles. This seems
           to be  what  the sine - power-series of subtractions is
          about.
    
    However, I do feel uncomfortable when having to apply
    corrections to formulae without knowing the assumptions
    behind. These corrections seem to have something to do with
    the different "Chart Datums", another big secret to all
    navigational textbooks i've seen so far. Does anyone have a
    hint, idea or link about how to find more about this ?
    Many Thanks in advance !!
    
    Until then, I will continue assuming the earth to be a sphere,
    use "ln" and 3437,74677 and forget about all these sinuses ..
    :-))
    
    Best regards,
    and - Don't sink the ship...
    Tom
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    To: navigation{at}ronin.com
    From: puffelej{at}euronet.nl (Jan van Puffelen)
    Subject: Re: RE meridional parts formula
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    Status: RO
    
    Hans Thomas Feuerhelm wrote:
    
    >Hi,
    >
    >Following the discussion about meridional parts, I have some
    >remarks and questions :
    >
    >1. The formula comes from integration of  dS = (Earth radius)
    >    / cos(latitude) between the limits equator and current
    >    latitude.
    >
    >2. After Integration, the formula is then
    >    S = R(Earth) * ln ( tan ( 45d + lat/2) ),
    >   assuming the Earth to be a SPHERE.
    >
    Rather than using the formula LN(TAN(45+lat/2)) the formula LN(TAN lat
    +1/COS lat) may be used as well. This saved a few memory positions in my
    CASIO FX 702P (BASIC) programmable calculator.
    
    I have never known that this method was called the Meridional Parts method.
    Can someone explain the origin of that name? It is called "Vergrotende
    Breedte" (lit. magnifying latitude) in Dutch while while the simple method
    is called Middenbreedte or Middel Breedte method (lit. mid or average latitude).
    
    There is a formula which compensates for the ellipsoidal form of the earth.
    It has a few more terms. I must have it somewhere (probably in the attic). I
    have tested that formula a long time ago and concluded that for all
    practical puposes the simpler formula for the perfect sphere is just as good.
    
    Regards,
    puffelej{at}euronet.nl (Jan van Puffelen)
    
    
    
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    To: navigation{at}ronin.com
    Subject: RE: meridional parts again + Datum
    Date: Sat, 11 May 1996 15:37:10 GMT
    Message-Id: <3194b420.62945203{at}mailto.btx.dtag.de>
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    Status: RO
    
    Hi Crew,
    
    Meridional parts (MP) are still bothering us over here...
    This is what I've collected so far :
    
    1) Here in Old Germany, all the books say that MP's are ok
       for practical navigation assumming the Earth being a
       sphere (e.g. M?ller Krauss, the "bible" for merchant navy
       education, and also Fulst, Nautical Tables). M?ller Krauss
       mentions that correcting for the ellipsoid is done in chart
       making, but doesnt bother to tell the dumb guys how this is
       done...
    
      So, to stay friends with my european sailing buddies, I'd
      have to stick to the following :
      - Earth is a shpere
      -1 nm = 1852,000. m,
      - the equator has 360 * 60  = 21600 nm
      - Earth radius from this is 21600 / (2pi) = 3437,74677078 nm
     giving the equivalent formulae :
    
     (USING NATURAL LOGARITHMS)
    
      MP(LAT) = 3437,74677 * ln ( tan ( 45 + LAT/2) )
     or, many thanks to Jan van Puffelen  (!!):
      MP(LAT) = 3437,74677 * ln ( tan (LAT) -1/cos(LAT) )
    
    ( USING LOG to Base 10 )
    
     MP(LAT) =  7915,70447 * LOG ( tan ( 45 + LAT/2) )
     MP(LAT) =  7915,70447 * LOG ( tan(LAT) - 1/cos(LAT) )
    
     I checked Jan's Formula (using the cos..) and it gave exactly
    the same results, except for LATs above 85d, but who'd want to
    go there, anyway. It looks like tan (45 + x/2)  is the same as
     tan(x) -1/cos(x). Seems worthwhile to refresh all that trig
    stuff from high school....
    
    2.) Assuming Earth an ellispoid, as many Silicon sea
    navigators do (and honestly, nothing wrong with that!!)  gives
    need for correction with the sine power series from the
    postings. Checking this, it looks sufficient to me to just use
    the first term, as DAN already said.
    
    Redoing the last DR from Silicon Sea leg 12, I have :
    
    111,3 d / 1167,8 nm  ( MP "elliptical")
    111,4 d / 1164,5 nm  ( MP "sphere" )
    111,4 d / 1163,3 nm  ("mid Latitude")
    
    At first sight, MP "sphere" is closer to "mid Latitude", which
    until now, I've always used as a parallel check. Second
    thought : "mid-Latitude" also assumes a sphere and not an
    ellipsoid (sigh !) This is a hard life for beginning
    navigators !!!
    
    So what do I do, stick with my old german buddies, ignore the
    ellipoid and make a bad impression on the "elliptical" people,
    
    or become "ellipsoidal", resulting to fail german exams on
    navigation ?
    
    BTW : the "sphere" .model is 3,3 nm "faster", an equivalent of
    25 minutes at 7,8 knots, but what's 25 minutes on a leg of 6,5
    days with all possible surprises to come.....
    
    3. Referencing the ellipsoid
    Reading about the different ellipsoids (there's 20 I've
    collected so far), the books say that apart from using the
    equator and polar radius, one has to define a well established
    reference ("DATUM"), so that the ellipsoidal grid of LATs +
    LONs is adjusted to a certain wellknown point. The ones I've
    found :
    
    Potsdamer Datum :
      uses Bessel ellipsoid and references to
      "Helmert Turm"  52d 22'' 53,954'' N,  013d 04' 01,153'' E
    
    European Datum :
      uses "international Hayford" and references to
      "Helmert Turm"  52d 22' 51,45'' N,  013d 03'  58,74'' E
      (now used for German Maps of Europe)
    
    What "Datums" are used in the US ?
    
    This would mean, that you'd have different LATs and LONs even
    when reading exactly the same object from different chart
    designs. Well, I do hate to think about these consequences.
    What about all these waypoint tabulations ?  GPS ?  (ha,ha),
    the integrated navigation programs ?
    
    Almost being as lost as Odysseus,  I'll go for some sherry..
    
    bye for now,
    TOM
    
    
    
    
    
    
    
    
    
    
    
    
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