Author

Topic: Problem 514b

Michael Mitgang

posted January 25, 2005 11:55 PM
Hi. I seem to have gotten parts a and c correct but seem to have gotten b incorrect. I find b to be 1.15 nmi rather than .8 nmi. please advise.
Thanks.
From: Menlo Park, CA


David Burch

posted January 27, 2005 02:11 PM
Are you sure you put the current in? If you got the first part right, which has answer 1.5, then if you put current in, it seems the answer must be different?
Also, you can solve this numerically for precise answers as explained in Numerical running fixes
From: Starpath, Seattle, WA


Michael Mitgang

posted January 28, 2005 05:00 PM
I put the current vector in at .666 miles (amount it traveled in 20 minutes) at 295T, however I did not get the same answer as you show in the answers. I realize the answer will be less than the 1.5 found in part A, but I did not get 0.8 rather I got 1.15. Please help clarify difference.
From: Menlo Park, CA


David Burch

posted February 04, 2005 09:03 PM
Michael, i apologize again for the delay in getting this redone. You are right. Here is what i get (i compute rather than plot as i do not have a chart at hand, but it is actually easier to just plot it than to compute it):
Sailing 6kt at 085T in a current of 2.0 toward 295T yields an SMG = 4.38 kts in direction CMG = 071.8T, and therefore we have a distance run between sights (20 min at 4.38 kt) of 1.46 nmi.
Taking into account the actual way we are moving, the bearing of sight one is 071.8043 = 28.8° and the bearing at sight two is 360341+71.8 = 90.8°.
Using tables or calulators to find distance off from two bearings, we get that the distance off at the time of sight two is 0.8 x 1.46 = 1.17 nmi, which should be the correct ans to 514b. I will add this to the tech support section.
Thanks for finding this.
See computers for distance by two bearings
See related note and more examples of this technique Numerical running fixes
From: Starpath, Seattle, WA


