Cross-channel fix: set, drift and course to steer
At 0900 you are at 50°30.0′N 001°30.0′W in a 10 m sloop. You steer 120°T at 6.0 kn through the water. At 1100 a GPS fix puts you at 50°24.0′N 001°09.0′W. The log reads 12.0 M. Visibility is good, wind SW 12 kn, sea slight. You are plotting on paper with a hand-bearing compass, GPS, echo sounder and a tidal stream atlas. There is no chartplotter route active. Your next waypoint is 8 nm away and you want to make good 100°T for that leg at the same through-water speed. The tidal stream is expected to remain steady for the next hour. One crew member is an inexperienced helm. You have not yet allowed for leeway or compass deviation; the steering compass deviation is negligible on this heading.
1. Determine the set and rate of the tidal stream experienced between 0900 and 1100. Show your vector calculation.
Read the worked answer
Use latitude and longitude differences to get the actual ground vector. Latitude change is 6.0′ south = 6.0 M south. Longitude change is 21.0′ east at mean latitude about 50°27′, so east distance = 21.0 × cos 50.45° ≈ 13.4 M. Actual displacement over 2 h is therefore 6.0 M south and 13.4 M east.
The boat’s through-water vector over 2 h at 120°T and 6.0 kn is 12.0 M: south component = 12.0 × cos 120° = 6.0 M south; east component = 12.0 × sin 120° = 10.4 M east.
Tide vector = actual displacement − boat vector = (13.4 − 10.4) = 3.0 M east and (6.0 − 6.0) = 0 M north/south. So the stream set 090°T and drifted 3.0 M in 2 h, giving a rate of 1.5 kn.
2. Calculate the course to steer and expected speed made good to make good 100°T at 6.0 kn through water with the same tidal stream. Explain why steering 100°T would not achieve the track.
Read the worked answer
Let heading H be the course to steer. Water vector = 6.0 kn at H, so east component = 6.0 sin H and north component = 6.0 cos H. Tide vector = 1.5 kn at 090°T, i.e. 1.5 east and 0 north. The ground vector must lie on 100°T.
Equations: 6.0 sin H + 1.5 = S sin 100°; 6.0 cos H = S cos 100°. Dividing eliminates S and gives 6.0 sin H + 34.03 cos H = −1.5. Solving gives H ≈ 102.5°T (the other solution, about 277.5°T, makes ground to the west and is rejected). Speed made good S ≈ 7.5 kn.
Steering 100°T would not work because the tide sets 090°T, almost along the desired track. The boat would be pushed east and would make good about 098°T, roughly 2° north of the intended track. Steering 102.5°T gives the required southward component to counter the stream and hold 100°T.
3. What would you check before accepting this course to steer, and how would you use the next fix?
Read the worked answer
Check that the observed stream is consistent with the tidal atlas for position and time; a 1.5 kn stream may be local. Check leeway, compass deviation, depth and charted hazards along the new track, and traffic. Brief the helm that 102.5°T is a steering heading, not a ground track.
Use the next fix to recompute set and drift over the actual interval. If the new vector differs materially, do not keep steering the old heading: replot the ground track, recalculate course to steer and speed made good, and check the ETA and clearance. If the stream has turned, expect the required heading to change.