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» Online Classroom   »   » Public Discussion of Cel Nav   » longitude by LAN

Author Topic: longitude by LAN

 - posted January 22, 2016 11:58 AM      Profile for David           Edit/Delete Post 
I have digested your Celestial Navigation Home Study Course. I have a question regarding Longitude by LAN. For example if LAN is at 12:10 local time and I am in zone 5 and the equation of time is +1 min. is the formula : (A)17:10 minus 12:00 or is the formula (B) 17:10 minus 12:01 or is the is it (C) 17:10 minus 11:59 ????

From: Minnesota
Capt Steve Miller

 - posted January 22, 2016 12:19 PM      Profile for Capt Steve Miller           Edit/Delete Post 
The GMT of LAN would be 1710 if you are at the center of your Time Zone i.e. 75°. The Equation of Time is not used anymore. The Equation of Time was used before time was reckoned by our clocks of today (it was based on Solar time).
Now if you are not in the center but West of the center of the Time Zone you will have to add your Longitude minus 75° and converted to time (suppose you are at 80°W, you are 5°W of the center of the Time Zone & 5° equals 20 minutes of time thus in this case Local LAN would be at 1210 + 20 = 1230 or 1730 GMT).
On the other hand if you are say at 70°W you are 5° East of the center of the Time Zone so again 5° = 20 minutes and Local LAN would be 1210 - 20 = 1150 or 1650 GMT.

From: Starpath

 - posted January 23, 2016 06:33 PM      Profile for David           Edit/Delete Post 
Now I am really confused. Since I am determining LAN by sextant, I do NOT know if I am east or west of the center of my time zone. I am trying to determine where I am. Since the equation of time no longer is used, why would I not just subtract 12 from 17:10.......for a result of 5.166.......and a longitude of 77.5 or 77 degrees 30 min??????

From: Minnesota
David Burch

 - posted January 23, 2016 06:39 PM      Profile for David Burch           Edit/Delete Post 
Lon by LAN is not a recommended procedure, but to carry this out you do not need to know anything at all about time zones.

You just need to know what the GMT was at the time of the peak height of the sun. Then look up the GHA of the sun at that time. That is all there is to it.

From: Starpath, Seattle, WA

 - posted January 23, 2016 06:51 PM      Profile for David           Edit/Delete Post 
Yes I am aware of that method. However if one does not have access to the nautical almanac to look up the GHA of the sun, one should be able to calculate longitude by subtracting GMT from LAN....is that not correct???
From: Minnesota
David Burch

 - posted January 23, 2016 07:01 PM      Profile for David Burch           Edit/Delete Post 
if you know the GMT of mer pass at greenwich on the same date, then you can subtract the GMT you observe for LAN from that time and convert the time diff to angle and that will be your Lon.

You can figure the former from the Eq of Time if you have that.

For doing the most with the least, I would recommend the book Emergency Navigation. we have a good price at starpath on this book, and there are ebooks also available.


.... again, however, time zones never enter the process at all. They are not part of cel nav.

From: Starpath, Seattle, WA

 - posted January 24, 2016 09:03 AM      Profile for David           Edit/Delete Post 
Eureka!!!! Refer to page 15 of the User's Guide for the Davis Master Sextant. In this example they added the 33' W AFTER they did the calculation. It seemed to me that one cold add (if the Equation of time was "W") to the local time of LAN and then do the calculation. But that did not work. Where I got screwed up was not realizing that the 33'W in the problem is 33' arc not sec in time. If one converts the 33' arc to time seconds (ie 1' arc = 4 sec in time) ie 132" in time and adds the 132" of time to the local time of LAN, the problem works out the same as if one adds the 33' or arc to the longitude position of mean sun. Do you agree? This was driving me nuts since, intuitively it should work out....and it does!
From: Minnesota

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