|
Author
|
Topic: USCG Cel Nav -- ETA problems
|
JimG
|
posted February 08, 2006 06:00 PM
First of all I like to thank you for answering my previous questions. Your answers where very informative, & I've been able to successfully work through numerous problems of that type. I am currently having problems correctly working out the ETA problems, that start on page 85 of the USCG Notes. I was hoping you could show me how to solve those, or a source showing me how. Thanks!
From: Steamboat Sp, Co.
|
|
David Burch
|
posted February 08, 2006 08:37 PM
Here is a sample problem to look at.
00602. At 1845 zone time on 24 October 1981, you depart Bimini Island, LAT 25° 50.0' N, LONG 77° 00.0' W (ZD +5). You are bound for Bishop Rock, LAT 49° 40.0' N, LONG 6° 4.0' W, and you estimate your speed of advance at 13.6 knots. The distance is 3,491 miles. What is your estimated zone time of arrival at Bishop Rock?
A. 0627, 3 November
B. 1642, 3 November
C. 0939, 4 November
D. 1627,.4 November
------------------
The key to these problems is work everything in GMT and then convert to the ZT of the destination.
(step 1) Find duration of trip: 3491 mile at 13.6 kts = 3491/13.6 = 256.69h --> /24 = 10.695 day (-10 x 24) = 10d + 16.6911 hr (-16 x60) = 10d 16h 41m.
(step 2) find gmt of arrival. they tell us we are in ZD+5 when leaving else, 77/15 = 5.133 -->5, so GMT of departure = 1845 oct 24 + 5h = 2345 oct 24. then we travel 10d + 16h 41m = oct 34 2345 + 1641 = oct 34 29h 86m oct 34 40h 26m = oct 35 16h 26m = nov 4 1626 GMT.
(step 3) convert to local ZT. 6° 40' round to nearest degree (7°) then divide by 15, then round to nearest hour, 7/15 = 0.467 which rounds to 0. So in this case ZT = GMT and the answer is D.
note if the destination longitude had been 25° 25' W, the ZD = 25/15 = 1.666 which rounds to 2, and the ZD would be +2h, so the answer would then be 1426, etc.
remember the definition, GMT = ZT + ZD, where ZD for west lon is + and for east lon is -.
From: Starpath, Seattle, WA
|
|
|
