On chart 3: course to steer with a diamond
How to think it through
- 1You must go from A to B on chart 3; the tide at the nearby diamond pushes you sideways.
- 2One hour of tide from A, then from its end an arc as long as your speed to the A–B line.
- 3That side is the course to steer; from A to the arc is your speed over the ground, which gives the arrival time.
Worked example
Every step says why you do it, which formula you use, the working and the answer.
Course to steer on chart 3, with diamond A
At 20:10 you leave A (46°21.4′N 005°50.6′W) for B (46°16.9′N 005°58.3′W), 7.0 miles on 230°T. You make 6.0 kn through the water. HW Victoria 14:10, range 4.8 m: use diamond A.
- A
- 46°21.4′N 005°50.6′W
- B
- 46°16.9′N 005°58.3′W · 230°T · 7.0 M
- Speed through the water
- 6.0 kn
- HW Victoria · range
- 14:10 · 4.8 m
- Diamond A, row HW+6
- 110° · 1.8 / 0.9 kn
At 20:10, which row of diamond A do you use? (hours from HW Victoria, −6 to +6)
Why: The chart's tidal stream table is referred to HW Victoria: each row covers the hour centred on it.
Working: 20:10 − 14:10 = +360 min → row HW+6: 110°, 1.8 kn springs, 0.9 kn neaps
Answer: 6 h
What is today's rate, between neaps and springs?
Why: The diamond gives two rates: today's is interpolated with the range at Victoria (springs 4.9 m, neaps 2.4 m).
Working: (4.8 − 2.4) ÷ 2.5 = 0.96; 0.9 + 0.96 × 0.9 = 1.8 kn towards 110°
Answer: 1.8 kn
What true course must you steer (CTS) to stay on the line from A to B?
Why: Lay off one hour of tide first, then from its end swing an arc of the boat's speed to the A–B line: that side is the course to steer.
Working: CTS 245°T
Answer: 245°
What is your speed over the ground (SOG)?
Why: The distance from A to where the arc cuts the line is how far you go over the ground in an hour.
Working: SOG 4.9 kn
Answer: 4.9 kn
Leaving at 20:10, what time do you reach B?
Why: Time is distance over speed over the ground, not through the water.
Working: 7.0 ÷ 4.9 × 60 = 85 min → 21:35
Answer: 21:35