Predicting the tide
Tonight's idea
The sea rises and falls about twice a day. The moon does most of the pulling, and one cycle, high water to high water, takes about 12.4 hours. Near many coasts the depth is close to a sine wave, so four numbers describe it (see Reading a sine wave):
\(d(t) = a \sin\!\left(\dfrac{2\pi}{P}(t - c)\right) + m\), with \(t\) in hours after midnight and \(d\) the depth in metres.
- Midline \(m\): halfway between high and low water.
- Amplitude \(a\): half the difference between high and low water.
- Period \(P\): the time from one high water to the next.
- Phase \(c\): a time when the water is at the midline and rising. That is a quarter period before high water.
Then "when is the water deep enough?" is a trig equation. Write \(\theta\) for the angle inside the sine, in radians. In each cycle, \(\sin\theta \ge k\) holds from \(\theta = \arcsin k\) to \(\theta = \pi - \arcsin k\). Turn those angles back into times.
Watch one: crossing the bar
An invented tide table for a sandbar: high 5.2 m at 04:06, low 1.2 m at 10:18, high 5.2 m at 16:30, low 1.2 m at 22:42.
Midline \(\tfrac{5.2 + 1.2}{2} = 3.2\). Amplitude \(\tfrac{5.2 - 1.2}{2} = 2\). Period \(16.5 - 4.1 = 12.4\) hours. Phase \(4.1 - \tfrac{12.4}{4} = 1.0\).
\[d(t) = 2\sin\!\left(\frac{2\pi}{12.4}(t - 1)\right) + 3.2.\]
A boat needs 4.2 m, so \(\sin\theta \ge 0.5\), so \(\tfrac{\pi}{6} \le \theta \le \tfrac{5\pi}{6}\). Since \(t = 1 + \tfrac{12.4}{2\pi}\theta\), the window is \(t = 2.03\) to \(6.17\): 02:02 to 06:10. Check: \(d(2.03) = 4.2\).
See what each of the four numbers does to a wave: mathdetours.com/demos/sinewave.html.
Your turn
- Another invented table, for a harbour: high water 7.5 m at 05:00, low water 1.5 m at 11:12, high water 7.5 m at 17:24. Find the midline, the amplitude and the period.
- Find the phase \(c\) for the harbour and write down its formula. Use it to find the depth at 08:00.
- A ship needs 5.5 m of water in the harbour. Find the first window after midnight, to the nearest minute.
- Back at the sandbar. A week later the tides are weaker: the amplitude has fallen to 1.25 m, with the same midline and period. How long is each 4.2 m window now? What is the smallest amplitude at which the boat can cross at all?
Stretch: why the tides weaken and strengthen
A real tide is a sum of sine waves. The two largest have periods of about 12.42 hours (the moon's) and 12.00 hours (the sun's). When their high waters coincide, the tide is large. How many days pass before they coincide again?
Answers
- Midline \(\tfrac{7.5 + 1.5}{2} = 4.5\) m. Amplitude \(\tfrac{7.5 - 1.5}{2} = 3\) m. Period \(17.4 - 5 = 12.4\) hours.
- \(c = 5 - 3.1 = 1.9\), so \(d(t) = 3\sin\!\left(\tfrac{2\pi}{12.4}(t - 1.9)\right) + 4.5\). At \(t = 8\) the angle is \(\tfrac{2\pi}{12.4} \times 6.1 = 3.091\) radians, \(\sin 3.091 = 0.0506\), and \(d = 4.65\) m. Sensible: 08:06 is halfway from high to low, when the depth is the midline 4.5 m.
- \(3\sin\theta + 4.5 \ge 5.5\) means \(\sin\theta \ge \tfrac13\), so \(0.3398 \le \theta \le \pi - 0.3398 = 2.8018\). With \(t = 1.9 + \tfrac{12.4}{2\pi}\theta\): from \(t = 2.571\) to \(t = 7.429\), which is 02:34 to 07:26. Check: \(d(2.571) = 5.50\).
- \(1.25\sin\theta + 3.2 \ge 4.2\) means \(\sin\theta \ge 0.8\), so \(0.9273 \le \theta \le 2.2143\), a width of 1.2870 radians. In time that is \(\tfrac{12.4}{2\pi} \times 1.2870 = 2.54\) hours, about 2 h 32 min, down from 4 h 8 min. The boat can cross only if high water reaches 4.2 m: \(3.2 + a \ge 4.2\), so the smallest amplitude is 1 m, and then only for a moment at high water.
Stretch. In a time \(T\) hours the sun's wave completes \(\tfrac{T}{12}\) cycles and the moon's \(\tfrac{T}{12.42}\). They are back in step when the sun's wave has gained exactly one cycle: \(\tfrac{T}{12} - \tfrac{T}{12.42} = 1\), so \(T = \tfrac{12 \times 12.42}{0.42} = 354.9\) hours, about 14.8 days. Large tides come round about once a fortnight, with weak ones in between.