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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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

From: Irv Haworth [mailto:irvhaworth@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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 Hi  Hi  de KA9UHHK--- On Tue, 1/6/09, Irv Haworth wrote: From: Irv Haworth Subject: [NavList 6947] Re: Chronometer SuggestionsTo: NavList@fer3.comDate: 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...andalmostinvariably 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 abeginner...so I took the view that a short simple answer would suffice at this stageof his development.. Hope U all take this as just a bit of sporting...hi...hiIrvin F Haworth (N) VE7CVL & VE0SAO (in case your wondering what hi..histands for)-----Original Message-----From: NavList@fer3.com [mailto:NavList@fer3.com] On BehalfOf glapook@PACBELL.NETSent: January 6, 2009 12:48 PMTo: NavListSubject: [NavList 6945] Re: Chronometer SuggestionsYou don't have that exactly right, the LOP moves westward one minute of arcevery four seconds, not one nautical miles. Since the length of one minuteof longitude varies as the cosine of the latitude the distance the LOP movesalso varies by the cosine of the latitude. At the equator the length of oneminute of longitude is one nautical mile but at 60º latitude it is only onehalf of a nautical mile.In addition to this, the amount of change in altitude also varies with thesine 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=enwhich 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 thebody. Look at the table for four minutes. Since four minutes of time is 60times four seconds of time just divide the tabulated values by 60 to obtainthe change of altitude in four seconds of time. (Or you can just consideerthe tabulated values as seconds of arc.)For example, the first value listedis 60' for latitude zero and azimuth 90º. Go accross the top line to 60ºlatitude and you will find that the change in altitude is 30' exactly onehalf of the change at the equator and which would result in a change of 30NM in the intercept.Go to my July 30, 2008 post on the "Celestial up in the air" threadforfurther explanation at :http://groups.google.com/group/NavList/browse_thread/thread/a270bc3d6aeb66d4/a86eed546f8313d7?hl=en&lnk=gst&q=celestial+up+in+the+air#a86eed546f8313d7glOn 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: 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 fourseconds 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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