When can we launch? Modelling the tide at our sailing club's slipway

Our sailing club launches its dinghies down a concrete slipway that only reaches the water for part of each tide. The notice by the boat park says “Launching possible from 2 hours before to 2 hours after high water.” Last autumn I followed that rule to the minute and still ran the trolley's wheels into mud about an hour and fifty minutes after high water, with the boat not yet floating. The instructor's reply was “it's a neap tide, it's always tight at neaps”. I wanted to know whether the rule is actually wrong, and if so, how often.

Aim. I will (i) build a sinusoidal model of the tide at our harbour from a day of tide-gauge readings, (ii) use it to find the length of the launch window and how sensitive that length is to the depth the trolley needs, (iii) test the sailors' “rule of twelfths” against the model, and (iv) use a month of the tide table to decide on how many days the club's two-hour rule is not safe.

Sinusoidal functions were the part of our trigonometry unit I found most convincing, because the graphs in the textbook looked like tide curves. This exploration was my chance to see whether they are good enough to answer a real question on the water.

The harbour authority's tide gauge shows the height of the water above chart datum (a fixed reference level close to the lowest tides). On the day my trolley stuck (day 9 of the tide table I use in Section 6), I downloaded the gauge's record for that day from the harbour authority's website at 30-minute intervals from 00:00 to 25:00 — 51 readings. The club's survey puts the toe of the slipway at 2.8 m above chart datum, and I measured that the trolley needs 0.5 m of water over the toe before the dinghy floats off, so launching is possible while the height is at least $H=3.3$ m.

Data note: the gauge readings and the tide table in this exemplar are illustrative, generated to behave like a real tide; a student must use their own (for example from their national tide-gauge network, cited).

I use $t$ = time in hours after 00:00 on the log day and $h(t)$ = height of the water above chart datum in metres. The data rise and fall about twice a day with a fairly constant range, so I chose the model

$$h(t)=a\cos (b(t-c))+d,(1)$$

with cosine rather than sine so that $c$ is simply the time of high water. First estimates came straight from the readings: $a$ = half the difference between the highest and lowest readings, $d$ = their mean, $c$ = the time of the highest reading (08:30), and the period from twice the time between that high water and the next low water (08:30 to 15:00), giving $\frac{2\pi }{b}\approx 13.0$ hours. These are in the first column of Table 1.

Launching is possible while $h(t)\ge H$. Around the high water at $t=c$, with $\theta =b(t-c)$, this is

$$a\cos \theta +d\ge H⟺\cos \theta \ge \frac{H-d}{a}⟺-{\theta }_0\le \theta \le {\theta }_0,\text{where }{\theta }_0={\cos }^{-1}\left(\frac{H-d}{a}\right).(2)$$

This needs $-1\le \frac{H-d}{a}\le 1$; if $H-d>a$, high water never reaches $H$ and there is no window at all. Converting back to time, the window runs from $c-\frac{{\theta }_0}{b}$ to $c+\frac{{\theta }_0}{b}$, so its length is

$$W=\frac{2}{b}{\cos }^{-1}\left(\frac{H-d}{a}\right)=\frac{T}{\pi }{\cos }^{-1}\left(\frac{2(H-d)}{R}\right),(3)$$

where $T=\frac{2\pi }{b}$ is the period and $R=2a$ the range. On my log day $\frac{H-d}{a}=\frac{3.3-2.596}{1.090}\approx 0.6456$, so ${\theta }_0\approx 0.8689$ rad and $W\approx 3.49$ hours: from 07:07 to 10:37 in the morning and 19:46 to 23:15 in the evening. The club's rule allows 4 hours. It reaches 15 minutes past the end of the window, when the model gives a height of only 3.19 m — about 11 cm less water over the toe than the trolley needs. That matches what happened to me: I was stuck about 110 minutes after high water — after the window had closed (105 minutes) but inside the time the rule allows (120 minutes).

How much does the depth matter? I only measured the depth the trolley needs roughly, so I wanted to know how much an error in $H$ would change the answer. Differentiating (1) with the chain rule, $h^{\prime }(t)=-ab\sin (b(t-c))$. At the end of the window, $t=c+\frac{{\theta }_0}{b}$, the water is falling at $|h^{\prime }|=ab\sin {\theta }_0\approx 0.414$ m per hour. Near that moment the height changes by about $h^{\prime }\mathrm{\Delta }t$ in a short time $\mathrm{\Delta }t$, so an error of $\mathrm{\Delta }H=0.1$ m in the required height moves each end of the window by about $\frac{\mathrm{\Delta }H}{|h^{\prime }|}\approx 14$ minutes — 29 minutes on the whole window. Even if the trolley needs 10 cm less water than I measured, the window is still under 4 hours (solving (2) again with $H=3.2$ m gives 3.96 hours, close to the estimate 3.98 hours), so the conclusion for this day does not depend on that measurement — although it is closer than I expected. The same derivative shows the tide is fastest at mid-tide, $|h^{\prime }{|}_{max}=ab\approx 0.54$ m per hour, when $\sin \theta =\pm 1$ — which is why the window closes quickly once the ebb gets going.

On my log day the tide was modelled well by $h(t)\approx 1.09\cos (0.497(t-8.87))+2.60$, with residuals of about 4 cm. Solving $h(t)\ge H$ gave a launch window of 3.49 hours, and in general $W=\frac{T}{\pi }{\cos }^{-1}\left(\frac{2(H-d)}{R}\right)$. The rule of twelfths is a straight-line version of the cosine model, exact at four of its six steps and never more than about 2.6% of the range out, so it is not the problem. The problem is the club's fixed rule of two hours either side of high water: it is only safe when the range is at least about 2.65 m, and on about 36% of days — around every neap tide — it sends people down the slipway after the water has gone. I will suggest to the club that the notice gives the rule for springs and a shorter one (about 1.6 hours either side) for neaps, or better, a small table of window length against range from equation (3).

Limitations. The most important one is the mean level $d$. I estimated it from one day, but the whole answer depends on $H-d$, which is only 0.70 m: raising $d$ by 0.1 m (a storm surge or low air pressure raises the sea; high pressure lowers it) lengthens every window, and lowering it shortens them. Repeating the calculation of Section 6 with $d$ 0.1 m higher or lower moves $R^{\ast }$ to 2.28 m or 3.03 m, and the share of days on which the rule fails to between 20% and 49%. Weather can shift the level by that much, so on any particular day the model is a guide, not a guarantee — but in every case the rule fails around the neaps. Second, I used a single cosine for each day, while the real tide contains more than two components; the small residual pattern I could not see in 25 hours might appear over a longer log. Third, the tide table gives heights only to 0.1 m, which limits how precisely $p$, $q$ and $P$ are known — but the conclusion (the rule fails around neaps) holds for any values close to mine, because $R^{\ast }$ sits well inside the range of the cycle. My sensitivity estimate in Section 4 showed that my rough trolley measurement is not a serious limitation.

What I will remember is that the formula I used most in class, $a\cos (b(t-c))+d$, answered a question that the experienced sailors at my club got wrong — not because their rule of thumb about the rise was poor, but because the rule of thumb about the window ignored one parameter.