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    FW: Re: Chronometer Suggestions
    From: Irv Haworth
    Date: 2009 Jan 9, 14:35 -0800
     


    From: Irv Haworth [mailto:irvhaworth{at}shaw.ca]
    Sent: January 8, 2009 3:11 PM
    To:  . Subject: RE: [NavList 6952] Re: Chronometer Suggestions

     I don't think this forum was intended to be used as a  teaching platform , but in deference to a novice I'll take a chance.
     
     
    Re G.L . comments...my humble reply is really directed to Bill Sellar as a novice celestial navigator who simply wanted to know the effects of time error on distance (fix). U.K. George is quite right in that we have probably confused him with somewhat irrelevant information-errors. Let me explain ,in simple terms, what  I think is the information you need to get on with it..
     
    A pound (mass) is not a pound the world around but a 1' of arc is (currently) 6076.11549...... feet  on the surface of the earth regardless of location. (consider a Great Circle).
     
     Simply stated one  either from spherical trig formulae or sight reduction tables one obtains  Hc (height computed) . From this he applies his measured etc Ho (sextant height observed and corrected) . .If Ho is  equal to Hc (unlikely unless his DR is bang on( consider  use of  HO211 ),then his ultimate 3 body fix is his DR. Using the modern short methods tables (HO214 /HO229 etc) he would then use Ho-Hc to obtain an altitude intercept = a. (If Ho is greater than Hc then he is closer to the body (T) if les than he is further away from the body (A). Since both Ho and Hc are angular measurements the difference  will be in minutes ( ' )   of arc. and thus in nautical mile(s).
    Turning to GL 's statement , there is no question that the length of 1 ' of arc varies as the cos. of the Latitude and if he used a spheroid (oblate spheroid for earth) he could plot his fix   ( position )  directly on a globe . (Our simple statement would now become Ho-Hc= a x cos L). Since this technique , having been tried  -and dismissed- in WW11 and  much  earlier , is not feasible the we must fall back on Marc Saint Hilaire's technique (altitude difference or altitude intercept) and take it that Bowditch is right in asserting "for practical navigation purposes 1 NM is considered the length of 1' of latitude on any great circle on earth regardless of location.
     
    Given the G.L.'s statement is true how do we manage to obtain a fix using Marc St. Hilaire's technique...Gerhard Mercator solved this for us by developing the Mercator chart. Thus after obtaining the results from a 3 body fix we construct ( or  purchase) a Plotting sheet and use the mid (horizontal) line as our mid latitude and plot the LOP's . This fix is then transferred (plotted) on to our Mercator chart.   
    I will now sit back and wait for the flack -hopefully - flavoured with some construction criticism as well as any correction(s) .
     
    Irvin F. Haworth
     
     

    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Gary LaPook
    Sent: January 6, 2009 3:25 PM
    To: NavList@fer3.com
    Subject: [NavList 6952] Re: Chronometer Suggestions

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    --- On Tue, 1/6/09, Irv Haworth <irvhaworth{at}shaw.ca> wrote:
    From: Irv Haworth <irvhaworth{at}shaw.ca>
    Subject: [NavList 6947] Re: Chronometer Suggestions
    To: NavList@fer3.com
    Date: Tuesday, January 6, 2009, 1:23 PM

    Quite right notwithstanding my comment. Hate to be a dog with a bone, but my
    "quick ans" took into consideration that a LOP is not a fix...and
    almost
    invariably 3 LOP's produce a cocked hat..with each LOP subject to an error.
    (in fact one may not even be within the cocked hat ! )
    Finally I "factored" my reply based on the assumption he was a
    beginner
    ...so I took the view that a short simple answer would suffice at this stage
    of his development..

    Hope U all take this as just a bit of sporting...hi...hi

    Irvin F Haworth (N) VE7CVL & VE0SAO (in case your wondering what hi..hi
    stands for)

    -----Original Message-----
    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf
    Of glapook{at}PACBELL.NET
    Sent: January 6, 2009 12:48 PM
    To: NavList
    Subject: [NavList 6945] Re: Chronometer Suggestions


    You don't have that exactly right, the LOP moves westward one minute of arc
    every four seconds, not one nautical miles. Since the length of one minute
    of longitude varies as the cosine of the latitude the distance the LOP moves
    also varies by the cosine of the latitude. At the equator the length of one
    minute of longitude is one nautical mile but at 60º latitude it is only one
    half of a nautical mile.

    In addition to this, the amount of change in altitude also varies with the
    sine of the azimuth so you have to combine these two factors. Go to :

    http://navlist.googlegroups.com/attach/c09c132c9a92fad1/HO+249+extracts+.pdf
    ?view=1&part=4&hl=en

    which contains two tables called "motion of the body" or
    "M.O.B."
    tables that are used in flight navigation to allow for the motion of the
    body. Look at the table for four minutes. Since four minutes of time is 60
    times four seconds of time just divide the tabulated values by 60 to obtain
    the change of altitude in four seconds of time. (Or you can just consideer
    the tabulated values as seconds of arc.)For example, the first value listed
    is 60' for latitude zero and azimuth 90º. Go accross the top line to 60º
    latitude and you will find that the change in altitude is 30' exactly one
    half of the change at the equator and which would result in a change of 30
    NM in the intercept.

    Go to my July 30, 2008 post on the "Celestial up in the air" thread
    for
    further explanation at :


    http://groups.google.com/group/NavList/browse_thread/thread/a270bc3d6aeb66d4
    /a86eed546f8313d7?hl=en&lnk=gst&q=celestial+up+in+the+air#a86eed546f8313d7


    gl

    On Jan 6, 12:20 pm, "Federico Rossi"
    <fedeastro.ro...{at}libero.it>
    wrote:
    > Lu,
    >
    > If I’ve understood well, this error doesn’t depend on your latitude on

    > earth, i.e. it’s a maximum of 1 nm for every 4 seconds (for bodies due
    > east or west) whether you are on the equator or far from it, does it?
    >
    > Federico
    >
    > Da: NavList@fer3.com [mailto:NavList@fer3.com] Per
    > conto di Lu Abel
    > Inviato: martedì 6 gennaio 2009 20.05
    > A: NavList@fer3.com
    > Oggetto: [NavList 6941] Re: Chronometer Suggestions
    >
    > Irv and Bill:
    >
    > It's a MAXIMUM of 1 NM for every 4 seconds, not a minimum.
    >
    > If the body you're sighting is directly north or south of you, even a
    > fairly significant time error would result in a very minimal shift in
    > the LOP produced by the body (the extreme example is Polaris).   On
    > the other hand, if the body you're sighting is directly east or west,
    > then it's Geographic Position is moving by 1 NM every four seconds and

    > any LOP developed from that sight would be off by 1 NM for every four
    seconds of clock error.
    >
    > Lu Abel
    >
    > Irv Haworth wrote:
    >
    > Minimum of 1 NM for every 4 seconds..( a quick answer)..
    >
    > Irvin F Haworth
    >
    > W, Van BC Canada
    >
    >   _____
    >
    > From: NavList@fer3.com [mailto:NavList@fer3.com] On
    > Behalf Of William Sellar
    > Sent: January 6, 2009 5:05 AM
    > To: NavList@fer3.com
    > Subject: [NavList 6931] Re: Chronometer Suggestions
    >
    > As a beginning celestial navigator, I am wondering how much time and
    > watch accuracy is actually required for practical navigation.  Can we
    > predict how many miles off one would be for every second of time error?
    >
    > Bill






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