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    Re: FW: Re: Chronometer Suggestions
    From: Gary LaPook
    Date: 2009 Jan 9, 23:31 -0800

    And here is the simple answer. If, where he is sailing, he can
    tolerate a one minute of longitude error all he needs is a watch that
    is accurate within 4 seconds. If sailing in the middle of the ocean
    more than fifteen degrees of longitude away from any danger you can
    use a watch with a one hour error.
    On Jan 9, 2:35�pm, "Irv Haworth"  wrote:
    > � _____ �
    > From: Irv Haworth [mailto:irvhawo...@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
    > Hi �Hi �
    > de KA9UHH
    > K
    > --- On Tue, 1/6/09, Irv Haworth  wrote:
    > From: Irv Haworth 
    > 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 glap...@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 60� 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+extrac...
    > ?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 90�. Go accross the top line to 60�
    > 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/a270bc3d6...
    > gl
    > On Jan 6, 12:20 pm, "Federico Rossi"
    > 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: marted� 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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