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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.

gl

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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