NavList:

From: Irv Haworth <irvhaworth@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@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...@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