Advanced coastal navigation techniques

Here are four advanced coastal navigation techniques. The first of the four is of practical use. The last three would require a lot of on-board systems to be broken to make them a real-life use case. On the other hand, some of them can be found in the RYA or Transport Canada Chart and Pilotage exam syllabus.

Below, I detail each problem associated with the techniques, then explain in detail, step by step, how to build the solutions. You’ll need a map, a Breton protractor, a pencil and a compass to make the constructions.

In terms of knowledge, I’m assuming that people are familiar with the concepts of landmark, course to steer, course over the ground, set and drift, and know how to make a running fix. If necessary, the excellent video from Carpe Diem Sailing School can refresh your memory. These prerequisites are part of all intermediate coastal sailing courses offered by the Sail Canada franchise.

Below, I’ve used the International Maritime Organization’s grading conventions (adopted by Sail Canada) to detail the answers.

One last comment: I am much less familiar with the English terms for coastal navigation. I use the following vocabulary to discuss the constructions:

• The surface vector characterizes the movement of the sailboat over the water. Its direction is the course to steer and its magnitude is the surface speed.
• The current vector characterizes the movement of the water over the ground. Its direction is the set and its magnitude is the drift.
• The ground vector characterizes the movement of the sailboat over the ground. Its direction is the course over the ground and its speed is the speed over the ground. I sometimes use the “ground track” as well.
• A line of position (LOP) is a set of points – a line – where the sailboat can plausibly in accordance to a bearing reading of a landmark.
• A circle of position is a set of points – a circle – where the sailboat can plausibly be in accordance to an angle reading.
• I use the verb transpose to describe the movement of an object along a given direction in a manner that its orientation remains unchanged. This is usually applied to a LOP – a line -, which must then remain parallel to the original line.

I think that “ground track” has a different meaning in English (planned vs realized constructs), but here, I use it as a synonym for “ground vector”.

The environment for exercices

I present the exercises in the Cartesian plane. Longitude is replaced by the x-axis and latitude by the y-axis. Units are in nautical miles (scaled). In the image above, point A is 3 nautical miles from the origin in x and 4 nautical miles from the origin in y, which is noted as (3, 4). Points B and C are on the coast at coordinates (8, 3) and (12, 5) respectively.

If you would like to reproduce the exercises below on a sheet of paper, you can easily reproduce these “pseudo-maps” by examining the positions of the landmarks. They are always at integer coordinates and should be easy to read. Reproducing the exercices step by step is the best way to learn.

The Cartesian plane is convenient for making images on the computer. It’s a digital environment that allows you to clearly trace constructions step by step, generating images as needed. This facilitates comprehension. The drawback, of course, is that the person reading the exercises has to make an effort to adapt when faced with real maps. In a Mercator projection, this is not very difficult, since everything is rectangular and only the scales change.

Construction 1: the running fix with current

Setting the scene

At 1000, you pick up the landmark “A” at 135° T (degrees true), then at 190° T at 1045. Your surface speed is 6.0 knots at 080° T and the current is at 1.5 knots towards 110° T.

• What is your position at 1045?
• What was your position at 1000?

The current-sensitive running fix is a useful position-finding tool in practice. It works when there is current and requires only a single landmark to obtain a fix. Because the American east coast is lined with lighthouses, you can use this technique to track your boat’s progress from Nova Scotia to Florida.

Solution

It’s the same idea as an ordinary running fix. However, the first LOP must be transposed by following the ground vector, which must be constructed from the surface vector and the current vector. The steps are as follows:

1. Plot the two line of positions (LOP) at their respective times.
2. Take an arbitrary point on the first LOP and plot the surface vector from this point. The surface vector should have the length appropriate to the elapsed time.
3. At the end of the surface vector, plot the current vector (with length appropriate to the elapsed time).
4. The sum of the two vectors – from the start of the surface vector to the end of the current vector – corresponds to the ground vector.
5. Transpose the first reading to the end of the current vector, keeping it parallel to the first LOP.
6. The intersection of the second LOP with the transposed LOP corresponds to the position at 1045.
7. By then moving the course over the ground and the current vector to this intersection, we can, moving backwards, identify the position at 1000.

These steps are illustrated in the table below. If required, each image can be openend in a new browser tab to make it larger.

Construction	Comment
	The figure shows the two LOPs. We know that the boat is somewhere on the first LOP at 1000 and somewhere on the second LOP at 1045.
	We take any point on the first LOP. From the chosen point, draw the surface vector of length equal to the time interval (45 minutes, i.e. 4.5 nm at 080°). At the end of the surface vector, add the current vector (over 45 minutes, i.e. 1.1 min at 110°). The resulting vector corresponds to the ground vector (not plotted). The ground vector does not reach the other LOP. This means the ship’s course is closer to the coast than that originating from the assumed point. (The opposite is also true).
	The figure shows the transposition of the 1000 LOP to 1045. The transposition is made in such a way as to remain parallel to the original position line, but to pass through the end of the current vector. Note that a running fix with current is the same as an ordinary running fix, but where the drift is included in the boat’s course.
	The intersection between the 1045 LOP and the transposed LOP corresponds to the ship’s position at 1045. (Approximately at position (7.9, 9)). Note that this point is indeed closer to the coast than what the original point suggested. All that remains to be done is to move the construction to identify where the ship was at 1000. This is done so that the end of the current vector arrives at the identified intersection (circled) and the orientation of the vectors does not change.
	The figure on the left shows the displacement of the vectors. The course over the ground is also plotted (original plots are shown dotted) The start of the bottom track corresponds to the ship’s position at 1000.
	The position at 1000 is now indicated by the circle (a fix). This position is approximately at (2.4, 8.6). The entire construction is now consistent: the two positions identified are on the position lines and also correspond to the calculated ground vector.

Construction 2: the inscribed angle

Setting the scene

You measure the horizontal angle between landmark “A” and landmark “B” and obtain a value of 50°. You also measure the horizontal angle between landmark “B” and landmark “C” and obtain a value of 30°.

• Identify your position.

This technique for identifying position readings is based on a classic geometric result about circles, the inscribed angle theorem:

If a circular arc between two points (in this case, landmarks) measures an angle </x></x></x>2\theta from the center of a circle, then any point on the circumference of the circle (the boat’s position) will form an angle of </x></x></x>\theta between the same two points.

This theorem can be used to construct a circle of position. If we measure a given horizontal angle between two landmarks (</x></x></x>\theta), then we know that the boat will necessarily be on a circle whose center measures an angle of </x></x></x>2\theta. The idea is to construct this circle. We will see how below.

Repeating this idea for the two horizontal angle measurements, we can then draw two position circles. The intersection of these two circles gives a positional reading.

Solution

The procedure is as follows:

1. Construct the circle of possible positions consistent with the horizontal angle between “A” and “B”.
   • Draw a straight line between the two landmarks.
   • Draw a perpendicular midway between the two landmarks. The center of the circle lies on this perpendicular.
   • Find the point on this perpendicular such that the angle at “A” is equal to the angle of the survey (here 50°). By the inscribed angle theorem, this point is the center of the circle.
   • Use the compass to draw a circle passing through “A” and “B”.
   • The boat must be on this circle.
2. Construct the circle of possible positions consistent with the horizontal angle between “B” and “C”.
   • Follow the same procedure as above, but adapt the angle to that between “B” and “C” (here 30°).
   • The boat is necessarily on this circle.
3. These two circles intersect at two points. Usually, one of them does not make sense (e.g. the location is on the coast, etc.). The sensible place is kept as a record.

These steps are illustrated in the table below. If required, each image can be openend in a new browser tab to make it larger.

Construction	Comment
	The first step is to draw a straight line between the two landmarks (here “A” and “B”).
	Next, draw a perpendicular to the straight line. This perpendicular must be halfway along the drawn line. The center of the circle of position lies on this perpendicular. We then need to find out how high it is.
	We identify the point on the perpendicular that makes an angle of 50° between the perpendicular and any of the two landmarks (here “A”). The simplest approach is to erect a straight line at an angle equal to the complement from the landmark angle (90° – 50° = 40°). Since the angle from the center is 50° and measures half the angle between the two landmarks, the angle between A and B measured from the center will be double (100°). So we know that by the inscribed angle theorem, any point on the circumference of the circle around the identified center will have an angle of 50°. Now we need to draw this circle.
	The circle can be drawn using a compass. The circle must pass through both landmarks. The ship must be on this position circle.
	Using the same procedure, we construct a second position circle for the other pair of landmarks (in this case “B” and “C”). The ship is necessarily on this second position circle.
	Only two points are simultaneously on both circles. One is circled and the other is the “B” landmark. Here, landmark “B” is on both circles because it was used for both constructions. More generally, the two points can be arbitrary, but one of them will always be meaningless (e.g. on the coast). We therefore retain the point at sea, which gives the ship’s position. Here, the position is about (3.9, 8.2).

Construction 3: three readings with current

Setting the scene

Your course to steer is 270° T. Using your compass, you take three successive readings of the “A” landmark: 210° T at 1000, 180° T at 1100 and 165° T at 1130. The current is 1.0 knot at 210°.

• Identify your position at 1000, 1100 and 1130.
• Identify your course over the ground.

The idea of the construction below is to make two transpositions of one line of position. We will take the “middle” LOP (at 1100) and transpose it both into the future (at 1130) and into the past (at 1000). This transposition will enable us to identify the course over the ground.

The problem is that we cannot use the usual transposition technique, as we do not know the boat’s surface speed. So we have to use the principle of proportionality to time to make a transposition. We do not know the speed, but we do know that 60 minutes elapsed between the first and second readings. We also know that 30 minutes elapsed between the second and third readings.

So the distance covered will necessarily be twice as great in the first segment as in the second (because 60 / 30 = 2). All the tracks that satisfy this proportionality ratio have the same angle as the course over the ground: they are parallel lines. All that remains is to identify which one is the ground vector.

Using set and drift, we can then identify the speed over the ground by choosing a line parallel to the one identified, but which is simultaneously consistent with set, drift, course to steer and position readings.

Solution

The procedure is as follows:

1. Plot the three LOPs at the times identified.
2. Calculate the time difference between the first and second LOPs and the second and third LOPs (here 60 minutes and 30 minutes).
3. Draw a perpendicular to the second reading, passing through the landmark.
4. On this perpendicular, identify distances from the Landmark that are proportional to the time elapsed before and after the middle survey. In the solution below, I use 4 nautical miles and 2 nautical miles. Note that their ratio is equal to the elapsed time ratio.
5. Transpose the middle LOP to the two points identified in the previous step. Transposed readings must remain parallel to the original.
6. The straight line identified by the intersection of LOPs 1 and 3 and the transposed LOPs gives the course over the ground.
7. At the start of this hypothetical ground track, draw a straight line following the direction of the surface track (here at 270° T).
8. Over the time interval between the first and last readings, calculate the length of displacement caused by the current (here, the duration is 90 minutes, so the current moves the boat 1.5 nautical miles at 210°).
9. Identify the single straight line that:
   • starts from the course track drawn in step 7.
   • ends on the hypothetical ground track in step 6.
   • Respects the current length and direction identified in step 8.
   • This line identifies the ground vector and the surface vector that are consistent with the set and drift. In particular, it identifies the length of the ground track over the time elapsed between readings.
10. Transpose the ground track identified in step 9 so that the start and end points coincide with the first and third readings.

These steps are illustrated in the table below. If required, each image can be openend in a new browser tab to make it larger.

Construction	Comment
	The figure shows the three LOPs for 1000, 1100 and 1130. The ship must be somewhere on these lines at the times shown.
	Start by building a perpendicular to the LOP (the one between the other two). The perpendicular must pass through the landmark.
	On this perpendicular, I place points whose distance from the landmark is proportional to the time elapsed. 60 minutes have elapsed between the first and second readings, while 30 minutes have elapsed between the second and third readings. The time ratio is thus 2 (60/30). I therefore chose distances of 4 and 2 nautical miles (also a ratio of 2) from the landmark. I could have chosen any other distance with a ratio of 2. It is handy to choose the right size for the plot.
	I transpose the second LOP to the points identified on the perpendicular. Note that these two readings intersect the first and third LOPs.
	By drawing a straight line between the two intersections, we are able to identify the course over the ground, i.e. approximately 257° T. Only tracks with a direction of 257° respect, at constant speed, the fact that twice as much time has elapsed between the first two readings as between the last two. It remains to identify the speed over the ground. If the speed is higher, the ground track will be parallel to the identified straight line, but further away from the coast. Conversely, it will be closer if the speed is lower.
	We now trace the surface route (at 270°) from the hypothetical route. We know that during the time interval (1h30), the current has induced a displacement of 1.5 nautical miles at 210°. We then need to find the only line which, simultaneously: – starts from the surface line at 270°. – arrives at the ground track identified at 257°. – is 1.5 nautical miles long.
	Determining the drift at the required point forms a triangle. This triangle forms the surface vector, the current vector and the ground vector. However, they are not in the right place. We notice that the end of the ground vector, within the 1h30 time interval, does not arrive at the last LOP. This is because our choices of 4 nautical miles and 2 nautical miles were too high. So we have to transpose the construction towards the coast – remaining parallel – so that the start and end coincide with the LOPs.
	The construction is now transposed, with the appropriate notation. Speeds are calculated by measuring the length of each vector, divided by the elapsed time interval. Note that everything is now consistent: – The bottom route starts at the first position line and ends at the last position line. – Is consistent with the surface route. – Is consistent with the route and current amplitude. All that remains is to identify the positions at the appropriate times.
	The three requested positions are now identified on the image. – At 1000, the ship is at (11.6, 9.3). – At 1100, the ship is at (8.0, 8.4). – At 1130, the ship is at (6.2, 8.0).

It would also be possible to have a similar problem, where we give the speed over the bottom and the direction of the current without giving the speed of the current. The construction is then virtually the same, with the exception of the last step, where the magnitude of the current must be deduced from the surface vector.

Construction 4: four readings

Setting the scene

You take three successive readings of landmark “A”: at 150° T at 1000, at 180° T at 1100 and at 210° T at 1200. At 1230, you take a reading on landmark “B” at 200° T.

• Identify your position at 1000, 1100, 1130 and 1200.
• Identify your course on the bottom.

Start this process in the same way as construction 3. This will give us the direction of the course over the ground. The last measurement of the bittern will enable us to identify the speed over the ground. The crucial idea to understand is the same as in the previous construction, i.e. that parallel vectors, but of different sizes, respect the proportionality of time lapses. Only one of these vectors constitutes the ground vector.

The procedure is as follows:

1. Identify the course over the ground by repeating steps 1 to 6 of the previous construction. This produces a straight line with the same orientation as the ground vector.
2. On the perpendicular erected at the second transposed LOP, identify the additional point corresponding to the time difference between the second reading and landmark “B”. Transpose the position line of the second measurement at this point (keeping it parallel).
3. Draw the LOP for landmark “B”.
4. Draw a straight line starting at landmark “A” and going to the intersection of the straight line with the same orientation as the ground vector and the transposed straight line from step 2. Extend this straight line until it intersects the LOP of landmark “B”. This point of intersection corresponds to the ship’s position at the time of the landmark “B” survey.
5. Transpose the line with the same orientation as the ground vector to make it pass through this point. This gives the ground vector.
6. By working backwards, we can identify all the positions at the time of the past surveys.

These steps are illustrated in the table below. If required, each image can be openend in a new browser tab to make it larger.

Construction	Comment
	The LOPs are shown in this figure. Note that the time elapsed between the first and second readings is identical to that between the second and third readings. Similarly, the angle variations are identical (30° in both cases).
	We can find the course over the ground by reproducing steps 1 to 6 of the previous construction. But in this particular case, we can use the fact that the only line that is consistent with identical angle and time variations is the line parallel to the coast. We therefore know that the course on the bottom is at 090°. In the figure on the left, a straight line parallel to the coast is drawn, with the perpendicular required for construction (passing through the sea wall).
	We now transpose the 1100 LOP to the moment of observation of the landmark “B”. If we had the right ground vector, the transposed position line would intersect the LOP of the landmark “B”. This is not the case, meaning that the straight line assumes the speed over the ground is too small. It needs to be moved away from the coast.
	To obtain the ground vector, extend the straight line from the landmark “A” to the intersection between the transposed straight line just drawn and the hypothetical ground vector (dotted line on the map). The intersection between this dotted line and the position line of landmark B gives us the position at 1230. All we need to do now is to transpose the hypothetical ground track through this point to obtain the actual ground vector.
	The ground vector is now correctly identified. The speed is identified simply by measuring the length of the vector divided by the time period. Note that the route simultaneously respects the time differences and places the boat on the position lines at the appropriate times. All that remains is to identify the positions correctly.
	The position at 1000 is approximately at (2.3, 9.5). Position 1100 is at (6, 9.5). The position at 1200 is approximately at (9.7, 9.5). The position at 1230 is about (11.6, 9.5).

Conclusion

The first problem is important enough to keep in mind when navigating. The other three are more in the range of clever constructions, showing the ingenuity of navigators from another era… and the importance of lighthouses in the days before GPS (or LORAN systems). Old-schoolers will however argue the importance of being able to identify one’s position and speed even when on-board electronics are dead.

In such a case, you can still take old-school readings with two landmarks and two sets of readings at different times. It is much simpler. And personally, I cannot imagine doing constructions 3 and 4 on a typical sailboat navigation table, measuring no more than half a meter by one meter, and whose stability depends solely on the absence of swell. In short, these constructions are mainly useful… for examinations.

If you want to understand why the recipes identified above work, it can be very useful to do some Euclidean geometry. A course in classical geometry can be of great help. Greek navigators only made these constructions using a compass and a ruler without graduation. Times change!

If you want to develop this intuition without taking a formal course, the Euclidea application, available on both Google Apps and the Apple store, can help you develop the necessary intuitions in geometry. It could be just the thing to keep you busy during a long passage at sea.