Examination

[ftkl]

Should be correct answer, just extend thr ground track and then fix the drift distance, join a line from A point, then join White Rock with an 90° junction on the leeway course.

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[Ex-Calif]

Originally Posted by ftkl

Should be correct answer, just extend thr ground track and then fix the drift distance, join a line from A point, then join White Rock with an 90° junction on the leeway course.

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Please don't take my advice if you think this is wrong but I think you may have it backwards.

The current at 175 comes from 155* off the port bow. 25* off the port stern. I apologize above saying you will drift to port. I was doing math in my head backwards. Becuase the current comes from port the set is to starboard - This also makes sense for the question becuase they want you to be aware of hitting the Rock (WHITE ROCK)

The diagram attached should be what you see.

I used the speed and drift calculator to get the raw numbers so I did not do any math.

You still would have to work out time/distance to derive set but with the tool I used you get angle - 4.09 degress - If you plot 334.09 on the chart you should also get the math answer "time at drift" and arrive at the same point.

BTW - If the distance to abeam WR is 8 miles the time is about 81 minutes @5.92 kts

In regards to drift maybe this rule of thumb is right (I am not sure). (Current/90) / (X/Angle)

If the current (1 knot) were applied on the beam (90) the boat would drift 1 NM each hour. But the vector of the current is 25 degrees(angle) off the stern

So - (1/90) / (X/25) = .27 - So the boat drifts .27 miles per hour or .00462 per minute

.00462 X 81 minutes = .375 NM.

[ftkl]

Geometry and Pythagorean Theorem can find the answers by an advanced calculator, however, I believe the exam would like us to answer by a raw simple equipment.

Last but not least, drift 175° is from north to south, it is a different with your diagram.

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[Ex-Calif]

You are probably right about the current. Hard to tell from your original translation of the question.

[jackdale]

I have used Open CPN and my photo editor to construct the set and drift solution.



The next step is to draw the actual CMG/SMG line from the fix



Continue down that line until White rock is abeam.

[ftkl]

1) SMG should be in the middle of the ground track,
2）Course starting point is at the first fix point not after sideway one.
In the beginning, there are two unknown info that are time for travel and distance.
So, I just estimate the distance that just pass abeam WR after fighting the current. My fix is not an arithmetic solution, it is still a question on me, I will ask sailing fans who study the education ctr if I know them one day.

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[jackdale]

Originally Posted by ftkl

1) SMG should be in the middle of the ground track,
2）Course starting point is at the first fix point not after sideway one.
In the beginning, there are two unknown info that are time for travel and distance.
So, I just estimate the distance that just pass abeam WR after fighting the current. My fix is not an arithmetic solution, it is still a question on me, I will ask sailing fans who study the education ctr if I know them one day.

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SMG and CNG are the same line / vector

I am confused about what you are getting at with you second comment.

[ftkl]

Yes, you right smg and cng are the same. How to determine the drift distance, if 4.15 kn, how long boat will be abeam WR. I believe the question has incomplete info provide or wrong questioning, it had better asked the compass course or when will be abeam WR. The position abeam WR cam be varied by distance and travelling time. I have other exercises, if meet problem, share here again, thx

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[jackdale]

Originally Posted by ftkl

Yes, you right smg and cng are the same. How to determine the drift distance, if 4.15 kn, how long boat will be abeam WR.

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Extend the CMG/SMG line. Find a point at which White Rock is 90 degrees to the CMG (234T if to port, 054T if to starboard)

[nigel1]

I took the liberty to have a go at this as well.

Not sure where "White Rock" is. On the chart I have, the nearest object is Baipai Reef (see the attached screen shot), so I have used this a White Rock.

The Red line is the course steered and speed through water vector. In this case, 330 True at 5 knots.

From the end of this vector, draw the tidal vector (shown in green), 175 True at 1 knot.

A liner from your initial position to the end of the tidal vector is the course and speed over the ground vector (shown in black)

This gives a course over ground of 324 True and 4.16 knots.

From White Rock (Baipai) draw a bearing line in direction 060 --240 True.This is the abeam bearing line, at 90 degrees to the boats heading.

On the chart, I have used a square to mark where the beam bearing line intersects the Course Made Good line.

From the initial fix to the intersect, the distance is 3.71 miles.
Time to reach this position will be 3.71/4.16 = 53.5 minutes.

Please note, the answer can vary depending on where this White Rock is.

[ftkl]

I have done the question aldy. Firstly, I find an ep and the smg, and then extend the set line, after that, use a triangle right angle ruler to hit white rock and the starting position finding a new position on the set line, so, i can find the abeam white rock and the ground track (may adjust the set line by parallel ruler).Measure the ground distance over smg, then I can get the time for the distance.

Many thx all of you. I have another one question in the this forum, take a look and discuss again.

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