Course to steer across a tidal stream

How to think it through

1You want to go from A to B, but the tide pushes you sideways.
2Lay off one hour of tide from A; from its end, with a radius equal to your speed, cut the track.
3That line is the course to steer; where it meets the track gives your speed over the ground, which you use for the time of arrival.

Worked example

Every step says why you do it, which formula you use, the working and the answer.

Course to steer across the tide

You want to reach a harbour 13.5 miles away bearing 238°T. Boat speed 5.0 knots; the tide sets 350° at 1.5 knots. What course do you steer, what is your speed over the ground and how long will it take?

Track required: 238°T
Boat speed: 5.0 kn
Tide (set, rate): 350° · 1.5 kn
Distance: 13.5 NM

What is the true course to steer (CTS)?
Why: If you point straight, the tide carries you off: aim into it by an angle that depends on how hard it pushes sideways.
$sin (WCA) = \frac{Rate \times sin ( set - track )}{V _{boat}}$
Working: The triangle gives CTS 222°T: you point the bow a little into the tide.
Answer: 222°
What is the speed over the ground (SOG)?
Why: Over the ground you go faster or slower depending on whether the tide pushes you along or against you.
$SOG = V_{boat} cos (WCA) + Rate cos (set - track)$
Working: SOG = 4.2 knots
Answer: 4.2 kn
How many minutes to cover 13.5 miles?
Why: You close the destination at your speed over the ground, not through the water.
$T_{min} = \frac{D}{SOG} \times 60$
Working: 13.5 ÷ 4.2 × 60 = 191 min
Answer: 191 min

Now you try

With the examiner (even from a photo)