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» Online Classroom   »   » Public Discussion of Inland and Coastal Navigation   » Question concerning 5-15

   
Author Topic: Question concerning 5-15
Robert Grigsby


 - posted December 06, 2004 05:26 AM      Profile for Robert Grigsby           Edit/Delete Post 
I assumed a 7 degree leeway from a strong northerly wind would affect the course of 085T in a south direction. Making the CMG approx. 92T.
Using this to vector the answers came up with a very close to correct A) 1.75nm & B) 1.14nm but for C) the number was way off. I got a 1.69nm to buoy instead of the 2.25. Where am I going wrong and how does this leeway angle of 7 degrees affect the plotting?

David Burch


 - posted December 06, 2004 02:08 PM      Profile for David Burch           Edit/Delete Post 
Leeway is actual motion through the water. it can be measured, as opposed to current set, which cannot be detected relative to the water. So if the problem says we are sailing 085T with 7° S leeway, it is the same as saying we were traveling 092T.

In Prob 5-15 part C, we have a current of 1.0 kt due S. As in all other parts, we have run for 20 m between sights at 6 kt, which is 2 miles. Hence we start the plot anywhere on the first LOP go 2.0 miles in direction 092T, and then correct that position by moving that point due south by how much the water moved during that time, which is 1 mile in 60 min or 0.333 mile in 20min.

Then we have our DR position corrected for the current and leeway, and we then advance the first LOP to that position, to intersect with the last LOP and then measure the distance from that Running fix to the buoy to check the answer.

The steps are shown in the picture below.

 -

Now, unfortunately, i get 1.93 nmi to the buoy, different from both your answer and the book answer. I am left to either bring out the chart and actually plot it by hand or do a computation. I will do the compuation here as an illustration of using a calculator for such things. The paper plot should in principle not be as accuate as the electronic, but it is often more reassuring.

In any event, for now we compute. And the best way to do this is to use our StarPilot calculator. It can all be done by hand and the formulas, but that is a vastly more difficult task. I will list some alternatives at the end.

=====

i noticed that at one point we said we were going to do this, and put a place hoder for it. Now it is there at this link Numerical running fix.

You will see that the right answer is somewhere between 1.96 and 2.0 (the electronic plot above is slightly off), and the answer in the book (2.25) is definitely wrong.

Hopefully these notes answer the original quesion, although we have not explained the 1.6 result, perhaps looking at this and the e-plot will answer the question.

From: Starpath, Seattle, WA


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