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Go Back   SailNet Community > General Interest Forums > Seamanship & Navigation
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Seamanship & Navigation Forum devoted to seamanship and navigation topics, including paper and electronic charting tools.


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  #41  
Old 03-01-2010
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It would be nice if the people posting the questions/problems would confirm the answers to their problems/questions and correct them if necessary, with an explanation of the correct answer and why it is correct.

Otherwise, this is a pretty useless thread for the newbies.
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  #42  
Old 03-01-2010
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The questions I have posted so far have been answered correctly but for the last one.
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  #43  
Old 03-01-2010
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Bump

This one is not that tough - or did it get missed in the shuffle.

Navigation Questions. Can you do them?
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  #44  
Old 03-01-2010
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Quote:
Originally Posted by Boasun View Post
Your vessel is on a course of 297° T at 11 ktnots. At 0019 a light bears 274.5°T and at 0048 the light bears 252°T. at what time and at what distance off will your vessel be when abeam of the light?
New answers (used chart rather than graph paper)

Distance off = 3.9

Time abeam = 0109
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Last edited by jackdale; 03-01-2010 at 06:19 PM.
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  #45  
Old 03-02-2010
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0108 @ 3.7 nm. But on a sail boat with a tiny compass you are close enough. This was also a 7/10th rule and could have been done via math.

You are steering 246°T, and a light is picked up dead ahead at a distance of 14 miles at 1037. You change course to pass the light 2.5 miles off abeam to port. If you are making 12 kts, what is your ETA at the position 2.5 miles off the light??
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  #46  
Old 03-02-2010
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11:46 Dan S/V Marian Claire
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Very good Dan.
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  #48  
Old 03-02-2010
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Looking ahead a light pops up over the horizon.

The light list shows a height of 121 feet.
The tide tables show a tidal range of 9 feet and a tide of 6 feet.
Your eyes are 8 feet above the waterline.

What is the distance to the light?
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I used the Pythagorean Theorem to get my answer. a squared plus b squared = c squared. c is the hypotenuse. The 14 nm is the hypotenuse because the new course and the light being abeam form the 90 degree angle required. The 2.5 nm is one of the sides so I solved for the other side. 14 squared = 2.5 squared + ? squared or 14 squared -2.5 squared = ? squared. 196-6.25 = 189.75. square root of 189.75 is 13.77 nm. Divide by 12nm/hr = 1.148 hrs. .148hrs x 60min/hr = 9 min. So 10:37 plus 1 hr and 9 min = 11:46 Hope that makes sense. Dan S/V Marian Claire
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Old 03-02-2010
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Uh Dan your idea is fine but I think you got it sideways. The 14 mile distance from the boat to the mark and the 2.5 miles to the new position of the boat when its abeam of the mark are the two legs of the right triangle. the line between the orginal position and new position 2.5 miles abeam of the mark is the hypotenuse. The actual sum is 202.25 and the distance from the original position to the new position 2.5 miles off the mark is the square root of that.
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