# NavList:

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

**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. gl On Jan 9, 2:35�pm, "Irv Haworth"wrote: > � _____ � > > From: Irv Haworth [mailto:irvhawo...{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 > > 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...{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 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 > > --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To unsubscribe, email NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---