By RUSS SWINNERTON

*Russ offers some ready-reckoning rules of thumb to assist with navigation on the spraydeck*

Let’s face it, calculators get soggy and hard to read when you hang them around your neck on a lanyard and then go paddling. So here’s a collection of handy mental arithmetic tricks and rules-of-thumb to assist with navigating on the spraydeck, and save the need for that calculator.

I know, your GPS can do all this calculating for you. Depending on its map set, it can give you tidal heights and barometric pressure, as well as your position, course and speed. And there’s probably an iPhone app for many of these functions too! But we didn’t buy a kayak so someone else could do the paddling for us, so let’s aim to do our own navigational thinking.

Speed/time/distance

If you know how fast you paddle, then you can figure out your distance travelled by noting the time you’ve been paddling. Using multiples of six-minute intervals is the easy way for working out speed and distance problems.

Six minutes is 1/10 of an hour, so in six minutes, you’ll cover 1/10 of the distance of an hour’s paddling. In six minutes at three knots, you’ll cover 0.3 of a mile. In 24 minutes? That’s four six-minute intervals: four times 0.3, is 1.2nm.

I work in nautical miles (and tenths of miles) because it’s convenient to measure distance off the latitude scale on the right and left hand sides of the chart. One degree of latitude (that’s 60 minutes) is 60 nautical miles. Your fingers make handy dividers, so you can guestimate (sorry, precisely measure with your calibrated fingers) distances quickly and easily, without searching for the kilometres scale (which according to the laws of navigation is always hidden under the folds of the chart in your map case).

The six-minute interval trick works just the same if you’re a kilometre paddler. If you paddle at 7 km/h, in six minutes you’ll cover 0.7 of a kilometre. 48 minutes? That’s eight 6-minute intervals, or 5.6km.

Bearings and Reciprocals

Here’s how to quickly calculate the reciprocal of a bearing or heading – very useful if you’re steering in one direction while keeping track on a transit or back-bearing astern.

Adding 200 and taking away 20 is the same as adding 180, but the math is much easier. Or adding 20 and taking away 200 – it still adds up to 180.

Reciprocal of 135? Why, that’s 335 – 20, or 315. Reciprocal of 057? 257-20, or 237.

Height of Tide – rule of twelfths

The Rule of Twelfths is a simple way of calculating the height of tide for a given time after high or low water. The rule approximates a standard tidal curve, and is as easy as 1,2,3. And 3,2,1 – a rule of thumb for how much the depth changes for each hour of its rise or fall. Here’s how it works:

Time interval – Rise or fall (as a fraction of range)

First hour after HW or LW – 1/12

Second hour – 2/12

Third hour – 3/12

Fourth hour – 3/12

Fifth hour – 2/12

Sixth hour – 1/12

Let’s pick a day, why not Saturday 26 May, when we’ll be (or have been, by the time you read this) at Lake Macquarie for the navigation challenge.

High water Swansea is three minutes before Newcastle, and the data is as follows:

Saturday 26 May

0549 0.53m

1159 1.36m

1727 0.76m

Duration of rise 6h 10min

Range 0.83m

Duration of fall 5h 28min

Range 0.60m

The rise:

We first calculate 1/12 of the range. Range 0.83 divided by 12 = 0.07m approximately (no point working to any more decimal places than two).

At the end of the first hour (0649), the height of tide at Swansea would be LW + 1/12 of range, or 0.53 + 0.07= 0.6m.

And four hours into the rise, it would 9/12 (0.63m) above low water, or 1.16m.

Try it yourself. Range of the falling tide is 0.6m. So 1/12 of the range would be?? And the height of tide at 1400 (Mickey’s little hand on the two, and his big hand on the twelve…), two hours into the fall? [See Answer 1, at the end]

A word of warning on the rule of twelfths. It assumes tides are regular, and semi-diurnal (that is, two full cycles a day, two highs and two lows, with roughly a six hour rise and a six hour fall) as they are on most parts of the NSW coast. But strange things happen in the north of Australia, and even stranger in other parts of the world, such as the south coast of England where you can get things like double low tides. In the north of Australia and parts of the Pacific (such as the Solomon Islands) tides are diurnal, with only one high and one low per day.

Strength of current

Average strength of a current over the period of its run (flood or ebb) is roughly two-thirds of the maximum rate. So a three-knot-maximum flood tidal stream over six hours (between slack waters) would have an average strength of two knots. So if you were paddling with that on your beam, you could expect it to displace you twelve nautical miles in the six hours of the flood (that’s 6 x 2 = 12).

If you’re making a long crossing, where you’ll face the flood and the ebb, then you can ‘boomerang’ it – steer the same course for both tidal streams. You’ll be pushed one way by the flood, and back on the ebb.

Angle off – radian or clock-face rule

Here’s a problem. You’re planning to maintain a three-knot average speed with that three-knot flood stream at 90 degrees to your direction of travel for the full six hours of its flow. How far do you need to aim off to counter the expected two nautical miles per hour displacement? Welcome to the radian rule!! The answer is 40 degrees. And here’s how we calculate it.

You’ll remember from school days (or your PhD in nuclear physics) that a radian is the angle at the centre of a circle that subtends an arc (a segment of the circumference) equal in length to the radius. A radian works out to be about 57.3 degrees – so that’s the angle of the pointy end of the pie-shaped segment with its three sides of equal length. Half a radian – subtends half the distance. And that relationship between angle and distance subtended holds right down to small angles.

To simplify the sums, we approximate the radian at 60 degrees, same as the number of minutes on a clock face, giving us a powerful and flexible tool for navigation. (And even digital watches have 60 minutes to the hour!!)

One radian, or roughly 60 degrees, at one nautical mile subtends an arc of 1 nautical mile. 30 degrees subtends half a mile (30 minutes on a clock-face equals half an hour). 15 degrees subtends a quarter of a mile, and so on until we get to 1 degree. It subtends 1/60 of a mile, or 30m.

(I nm is roughly 1800m, and divided by a radian (roughly 60 degrees) gives 30m. If you fancy more complex arithmetic, 1852/57.3=32.32m, which is near enough to 30m for kayaking… it’s best not to get too hung up on accuracy.)

Here’s a table to assist visualising – but the essence of this is to do it in your head!!

Degrees 1nm 5nm 10nm

1 30m 150m 300m

6 180m 900m (0.5nm) 1.8km (1nm)

12 360m 1.8km (1nm) 3.6km (2nm)

18 540m 2.7km (1.5nm) 5.4km (3nm)

24 720m 3.6km (2nm) 7.2km (4nm)

30 900m 4.5km (2.5nm) 9km (5nm)

45 1350m 6.75km (3.75nm) 13.5km (7.5nm)

Back to our aiming off problem. We know that our tidal stream will displace us a total of two miles per hour over the six hours of its run, that’s 12 nm. Paddling at 3 kts, we’d cover 18 miles in that time. So what angle do we need to steer to displace ourselves two-thirds of the distance we’d travel? The clock-face rule tells us that two-thirds of 60 is 40, so we’d aim off 40 degrees.

That’s how you calculate your ferry-glide angle. Try it yourself. You’re planning to cross a two-knot stream, while paddling at four knots. Angle off to steer?? [See Answer 2, at the end.]

Of course, if you’ve got land ahead or astern when ferry-gliding across a current, you can use a natural transit on shore to see how well you’re doing holding the planned line. The calculation of course to steer to offset the current is just a rough estimate to get you heading in the right direction.

Clock-face rule to calculate distance off

We can also use the clock-face rule to estimate our distance off-shore, by measuring the horizontal angle subtended by a known distance off the chart (such as the width of an island, or the gap between headlands, and so on).

If you say ‘measuring the angle’ really quickly, it sounds simple! And really it’s not all that tough – using a kamal, a flash name for a stick on the end of a piece of string of known length.

Here’s the theory: if we make our string 57.3cm long, about an arm’s length if your arms are short like mine, then one-centimetre graduation marks on our stick will be worth one degree (the radian rule again). Hang the expense, we could even buy a metric ruler, and use the scale engraved!

Let’s assume we’ve made landfall on an island we see from our chart to be 1.5nm wide. We measure the angle between its left and right edges as 10 degrees. So how far off are we? Ten degrees is 1/6 of our clock face, so the width of the island is 1/6 of our distance off. So the object must be 6 x 1.5nm away, or 9nm.

We can also use the kamal and the radian rule for distance by vertical angle, providing the object’s close (there’s a problem of the earth not being flat, if the object’s a long way off).

And we all carry natural kamals on the ends of our hands: our fingers’ width is worth about two degrees (you can calibrate them with a kamal). So three fingers is six degrees – and the clock-face rule tells us that, for an object three fingers high (3 x 2 = 6), our distance off must be 10 times the height.

So what might look like a paddler giving the Maritime NSW inspectors two fingers could be an attempt to measure four degrees. That’s my story, anyway, and I’m sticking to it!

[Answer 1: 1/12 of the range is 0.05m. Two hours of fall is 3 x 0.05 = 0.15m, subtracted from 1.36, gives 1.21m]

[Answer 2: For every hour, you need to displace yourself across track by half the distance you’ll paddle, for as long as you and the tidal stream can maintain the pace!! So aim 30 degrees off.]