Bearings and navigation
A boat runs one course, turns, and runs another. Where is it now, and which way is home?
Where you are 5 in Measuring angles
When you reach for it
A navigator does not think in coordinates. Sverrir, skippering a search-and-rescue boat, is told “run 8 nautical miles on a bearing of 040^\circ, then 5 on 145^\circ.” An orienteer reads the same kind of instruction off a map: a direction and a distance, then another, then another. Reach for this skill whenever motion comes as legs, each a direction and a distance, and you want to know where you end up, how far you are from where you began, and which way leads back.
The idea is to break every leg into how far it goes east and how far it goes north. Those amounts simply add, however many legs there are, and at the end a right triangle turns the totals back into a distance and a direction.
The procedure
Two ways to write a direction
A three-figure bearing is the angle measured clockwise from north, from 000^\circ to just under 360^\circ, always written with three digits. Due east is 090^\circ, due south 180^\circ, due west 270^\circ.
A compass bearing starts from north or south, whichever is nearer, and turns an acute angle toward east or west. \text{N}\,35^\circ\,\text{E} reads “35^\circ east of north”: face north, turn 35^\circ toward east. \text{S}\,40^\circ\,\text{E} means face south and turn 40^\circ toward east.
Reading each compass bearing off the figure gives its three-figure form. With \varphi the acute angle:
| Compass | Three-figure |
|---|---|
| \text{N}\,\varphi\,\text{E} | \varphi |
| \text{S}\,\varphi\,\text{E} | 180^\circ - \varphi |
| \text{S}\,\varphi\,\text{W} | 180^\circ + \varphi |
| \text{N}\,\varphi\,\text{W} | 360^\circ - \varphi |
For \text{S}\,40^\circ\,\text{E}, turning clockwise from north you pass east and reach south at 180^\circ; the ray is 40^\circ short of south, so it is 140^\circ.
From a bearing to standard position
An angle in standard position is measured counterclockwise from east. A bearing is measured clockwise from north. Put east along the x-axis and north along the y-axis. North is 90^\circ in standard position, and turning clockwise by \theta subtracts \theta, so a bearing \theta points along the standard angle
with 360^\circ added when the result is negative. Check: due south, \theta = 180^\circ, gives -90^\circ, which is 270^\circ, straight down.
One leg: how far east, how far north
A leg of length d on bearing \theta ends at the point at distance d along the standard angle \alpha, that is, at (d\cos\alpha,\ d\sin\alpha). Since \alpha = 90^\circ - \theta, the cofunction identities give \cos\alpha = \sin\theta and \sin\alpha = \cos\theta. So the leg moves
Sine goes with east because a bearing is measured from north. For an acute bearing you can see it in the right triangle: the north side is adjacent to \theta and the east side is opposite. For any other bearing the signs take care of themselves: \cos\theta < 0 for 90^\circ < \theta < 270^\circ, so y comes out negative, which means south, and \sin\theta < 0 for 180^\circ < \theta < 360^\circ, which means west.
Several legs: add the components
Each leg starts where the last one ended, so the east amounts add and the north amounts add. After all the legs, the total east displacement X and total north displacement Y say exactly where you are relative to the start. The bearing of each leg is measured from north at the point where that leg begins, since north points the same way everywhere on a map.
Distance and bearing, and the way back
From the start, the finishing point is X east and Y north. By the Pythagorean theorem its distance is
For its direction, find the acute angle \varphi that the line makes with the north-south line. The east-west side is opposite \varphi and the north-south side adjacent, so
The signs choose the letters: Y > 0 gives N and Y < 0 gives S; X > 0 gives E and X < 0 gives W. The compass bearing is then converted with the table if a three-figure bearing is wanted.
The back bearing of a direction is the direction exactly opposite it. Reversing a leg negates both components, and since \sin(\theta + 180^\circ) = -\sin\theta and \cos(\theta + 180^\circ) = -\cos\theta, the back bearing is \theta + 180^\circ, with 360^\circ taken off if that passes 360^\circ. In compass form, swap N with S and E with W and keep the angle: the back bearing of \text{N}\,35^\circ\,\text{E} is \text{S}\,35^\circ\,\text{W}. The bearing back to the start is the back bearing of the start-to-finish bearing.
Worked examples
Converting bearings
(a) Write \text{S}\,40^\circ\,\text{E} as a three-figure bearing and as a standard-position angle. (b) Write 250^\circ as a compass bearing and as a standard-position angle. (c) A direction has standard-position angle 130^\circ. Give it as a three-figure and a compass bearing.
Solution(a) The ray is 40^\circ short of south, so the bearing is 180^\circ - 40^\circ = 140^\circ. Then \alpha = 90^\circ - 140^\circ = -50^\circ, and adding a full turn gives 310^\circ: a ray 50^\circ below the east axis, which is southeast, as it should be.
(b) 250^\circ is 70^\circ past south, turning toward west, so it is \text{S}\,70^\circ\,\text{W}. Then \alpha = 90^\circ - 250^\circ = -160^\circ, which is 200^\circ.
(c) Solve \alpha = 90^\circ - \theta for \theta: \theta = 90^\circ - \alpha = -40^\circ, which is 320^\circ. That is 40^\circ short of north on the west side, \text{N}\,40^\circ\,\text{W}. The figure above shows (a), (b) and (c) as the rays at 140^\circ, 250^\circ and 320^\circ.
One leg, south and east
Sverrir’s boat leaves the harbor H and runs 12 nautical miles on a bearing of 128^\circ. How far east and how far south of the harbor is it, and what bearing would take it straight back?
The boat is about 9.46 nautical miles east and, since y is negative, 7.39 south of the harbor. The figure shows the same thing with a right triangle: the bearing is \text{S}\,52^\circ\,\text{E}, the south side is adjacent to the 52^\circ angle and the east side opposite, so the sides are 12\cos 52^\circ and 12\sin 52^\circ, the same two numbers.
The way back is the back bearing, 128^\circ + 180^\circ = 308^\circ. Check: 9.46^2 + 7.39^2 \approx 144 = 12^2.
Two legs and the way home
Sverrir runs 8.0 nautical miles on a bearing of 040^\circ, then 5.0 nautical miles on a bearing of 145^\circ. Find his distance and bearing from the harbor, and the bearing he must steer to return.
Break each leg into components and add.
| Leg | East, d\sin\theta | North, d\cos\theta |
|---|---|---|
| 8.0 on 040^\circ | 5.142 | 6.128 |
| 5.0 on 145^\circ | 2.868 | -4.096 |
| Total | X = 8.010 | Y = 2.033 |
The distance is \displaystyle r = \sqrt{8.010^2 + 2.033^2} \approx 8.26 nautical miles. Both totals are positive, so the boat is north and east of the harbor, at \displaystyle \varphi = \tan^{-1}\frac{8.010}{2.033} \approx 75.8^\circ from north: bearing \text{N}\,75.8^\circ\,\text{E}, or 075.8^\circ. To return he steers the back bearing, 075.8^\circ + 180^\circ = 255.8^\circ.
Check by another route. At A the course turns from 040^\circ to 145^\circ, a turn of 105^\circ, so the angle inside the triangle at A is 180^\circ - 105^\circ = 75^\circ. The law of cosines gives r^2 = 8^2 + 5^2 - 2(8)(5)\cos 75^\circ \approx 68.29, and r \approx 8.26 again. With two legs either route works; with three or more, components are far shorter.
Three legs on an orienteering course
Asdis runs an orienteering course with three legs: 450 meters on \text{N}\,62^\circ\,\text{E}, then 300 meters on \text{S}\,15^\circ\,\text{E}, then 520 meters on \text{S}\,70^\circ\,\text{W}. How far is she from the start, and on what bearing should she run to get back?
SolutionThe three-figure bearings are 062^\circ, 180^\circ - 15^\circ = 165^\circ and 180^\circ + 70^\circ = 250^\circ.
| Leg | East | North |
|---|---|---|
| 450 on 062^\circ | 397.33 | 211.26 |
| 300 on 165^\circ | 77.65 | -289.78 |
| 520 on 250^\circ | -488.64 | -177.85 |
| Total | X = -13.67 | Y = -256.37 |
She is \displaystyle r = \sqrt{13.67^2 + 256.37^2} \approx 256.7 meters from the start. Both totals are negative, so she is south and west of it, and \displaystyle \varphi = \tan^{-1}\frac{13.67}{256.37} \approx 3.1^\circ: from the start she is on \text{S}\,3.1^\circ\,\text{W}, or 183.1^\circ. The way back swaps both letters, \text{N}\,3.1^\circ\,\text{E}, which is 003.1^\circ: almost due north.
A new destination
Still at the end of Example 3, Sverrir gets a call: a vessel in trouble lies 15 nautical miles from the harbor on a bearing of 300^\circ. What course and distance take the boat straight to it?
SolutionGive both points coordinates from the harbor. The boat is at (8.010,\ 2.033). The vessel is at
The leg the boat needs is the difference, vessel minus boat:
Its length is \displaystyle \sqrt{21.001^2 + 5.467^2} \approx 21.70 nautical miles. It runs north and west, at \displaystyle \varphi = \tan^{-1}\frac{21.001}{5.467} \approx 75.4^\circ from north: \text{N}\,75.4^\circ\,\text{W}, which is 360^\circ - 75.4^\circ = 284.6^\circ. The bearing of the vessel from the harbor (300^\circ) is not the course from the boat; the course depends on where the boat now is.
Choosing a bearing
Einar runs 400 meters on a bearing of 330^\circ. His second leg is 650 meters, and he wants it to finish due east of the start. What bearing should he run, and how far east of the start does he finish?
SolutionThe first leg gives 400\sin 330^\circ = -200 and \displaystyle 400\cos 330^\circ = 200\sqrt{3} \approx 346.41: 200 meters west and 346.41 north. Due east of the start means the total north displacement is 0, so the second leg, on an unknown bearing \theta, must satisfy
Cosine is negative for bearings between 90^\circ and 270^\circ, and \cos^{-1}(-0.5329) \approx 122.2^\circ. Cosine takes the same value at 360^\circ - 122.2^\circ = 237.8^\circ, so there are two candidates, and the east total decides between them. Since \displaystyle 650^2 - (200\sqrt{3})^2 = 302{,}500 = 550^2, the second leg’s east part is \pm 550:
- on 122.2^\circ, \sin\theta > 0, so X = -200 + 550 = 350: due east, 350 meters out;
- on 237.8^\circ, \sin\theta < 0, so X = -200 - 550 = -750: due west.
He runs on 122.2^\circ and finishes 350 meters due east of the start.
Boundaries
A flat map. Adding east and north components treats the ground as a plane, with north pointing the same way everywhere. Over a few miles that is exact enough for any boat or runner. Over hundreds of miles the earth’s curve matters, and long routes use spherical trigonometry instead.
The inverse tangent alone cannot find a bearing. \tan^{-1} returns only angles between -90^\circ and 90^\circ, and (X, Y) and (-X, -Y) give the same ratio, though they point opposite ways. That is why the method takes \tan^{-1} of the sizes |X| and |Y| and lets the signs choose the quarter. When Y = 0 the ratio is undefined: the point is due east (X > 0) or due west (X < 0), bearing 090^\circ or 270^\circ.
Back at the start, there is no bearing. If X = Y = 0, the distance is 0 and no direction leads back.
Bearings are measured from true north here. A magnetic compass points to magnetic north, which is generally not the same direction. The difference depends on where you are and slowly changes over the years, so a compass reading must be corrected to true north before using this method.
Each bearing belongs to its own leg. A bearing measured from one point says nothing directly about the direction from any other point, as Example 5 shows.
Where this goes next
A leg is an arrow with a length and a direction, which is what a vector is: the east and north amounts are its components, and adding legs is adding Vectors. Turning a distance and a direction into east and north amounts is the same move as converting polar coordinates to rectangular ones, with the angle measured from north instead of east. For two legs, the law of cosines and the law of sines work directly on the triangle. And sighting one point from the two ends of a known baseline, rather than walking to it, is Applications of oblique triangles.
Exercises
Round distances to two decimal places for nautical miles and one for meters, and angles to the nearest tenth of a degree.
Reading and converting bearings
- Write each as a three-figure bearing: (a) \text{N}\,25^\circ\,\text{W} (b) \text{S}\,47^\circ\,\text{E} (c) \text{S}\,8^\circ\,\text{W}.
- Write each as a compass bearing: (a) 075^\circ (b) 200^\circ (c) 345^\circ.
- Give the standard-position angle, between 0^\circ and 360^\circ, for the bearings (a) 030^\circ (b) 165^\circ (c) 290^\circ.
- Give the three-figure and compass bearings for the standard-position angles (a) 15^\circ (b) 240^\circ.
- Give the back bearing of (a) 064^\circ (b) 213^\circ (c) \text{N}\,18^\circ\,\text{W}.
One leg
- The rescue boat runs 9.0 nautical miles on a bearing of 057^\circ. How far east and how far north does it go?
- Einar runs 280 meters on a bearing of 235^\circ. How far west and how far south does he go?
- The boat runs 6.5 nautical miles on \text{N}\,72^\circ\,\text{W}. How far west and how far north does it go?
Several legs and the way back
For each, find the distance and bearing of the finish from the start, and the bearing back to the start.
- The boat runs 10 nautical miles due east, then 7 nautical miles due south.
- Asdis runs 350 meters on 020^\circ, then 500 meters on 110^\circ.
- The boat runs 4.0 nautical miles on 300^\circ, then 9.0 nautical miles on 235^\circ.
- Einar runs 600 meters on \text{N}\,40^\circ\,\text{E}, then 450 meters on \text{S}\,80^\circ\,\text{E}, then 700 meters on \text{S}\,25^\circ\,\text{W}.
Planning a course
- From the harbor, a lighthouse is 11 nautical miles away on a bearing of 075^\circ and a buoy is 6 nautical miles away on 160^\circ. What course and distance take the boat from the lighthouse to the buoy?
- Asdis runs 500 meters on 030^\circ. Her next leg is 300 meters, and it must end due north of the start. Find every bearing that works, and how far north of the start each one finishes.
- A knot is a speed of one nautical mile per hour. The boat runs for 1.5 hours at 12 knots on 250^\circ, then for 0.5 hours at 16 knots on 340^\circ. Find its distance and bearing from the harbor.
Reasoning
- In standard position the x-coordinate uses cosine, yet the east component of a leg is d\sin\theta. Explain why there is no contradiction.
- Explain why a back bearing differs from the bearing by exactly 180^\circ, and why in compass form both letters swap while the angle stays the same.
- A runner goes d meters on bearing \theta, then d meters on bearing \theta + 180^\circ. Use components to show where she finishes.
- Two runners do the same three legs in different orders. Explain why they finish at the same point, and whether their paths are the same.
Answers
- (a) 335^\circ, since 360^\circ - 25^\circ = 335^\circ. (b) 133^\circ, since 180^\circ - 47^\circ = 133^\circ. (c) 188^\circ, since 180^\circ + 8^\circ = 188^\circ.
- (a) \text{N}\,75^\circ\,\text{E}. (b) \text{S}\,20^\circ\,\text{W}, since 200^\circ is 20^\circ past south. (c) \text{N}\,15^\circ\,\text{W}, since 345^\circ is 15^\circ short of north.
- (a) 60^\circ, from 90^\circ - 30^\circ. (b) 285^\circ, from 90^\circ - 165^\circ = -75^\circ. (c) 160^\circ, from 90^\circ - 290^\circ = -200^\circ.
- (a) 075^\circ, \text{N}\,75^\circ\,\text{E}, from \theta = 90^\circ - 15^\circ. (b) 210^\circ, \text{S}\,30^\circ\,\text{W}, from \theta = 90^\circ - 240^\circ = -150^\circ.
- (a) 244^\circ. (b) 033^\circ, from 213^\circ - 180^\circ. (c) \text{S}\,18^\circ\,\text{E}.
- 7.55 nautical miles east and 4.90 north: 9\sin 57^\circ \approx 7.55, 9\cos 57^\circ \approx 4.90.
- 229.4 meters west and 160.6 meters south: 280\sin 235^\circ \approx -229.4, 280\cos 235^\circ \approx -160.6.
- 6.18 nautical miles west and 2.01 north. The bearing is 288^\circ: 6.5\sin 288^\circ \approx -6.18, 6.5\cos 288^\circ \approx 2.01.
- 12.21 nautical miles on 125.0^\circ (\text{S}\,55.0^\circ\,\text{E}); back on 305.0^\circ. X = 10, Y = -7, \displaystyle r = \sqrt{149}, \displaystyle \varphi = \tan^{-1}\frac{10}{7} \approx 55.0^\circ.
- 610.3 meters on 075.0^\circ; back on 255.0^\circ. X \approx 589.55, Y \approx 157.88. The legs are perpendicular, so \displaystyle r = \sqrt{350^2 + 500^2} also works, and the bearing is \displaystyle 20^\circ + \tan^{-1}\frac{500}{350} \approx 75.0^\circ.
- 11.29 nautical miles on 253.7^\circ (\text{S}\,73.7^\circ\,\text{W}); back on 073.7^\circ. X \approx -10.84, Y \approx -3.16.
- 590.0 meters on 115.4^\circ (\text{S}\,64.6^\circ\,\text{E}); back on 295.4^\circ (\text{N}\,64.6^\circ\,\text{W}). Bearings 040^\circ, 100^\circ, 205^\circ give X \approx 533.00, Y \approx -252.93.
- 12.06 nautical miles on 225.3^\circ (\text{S}\,45.3^\circ\,\text{W}). Lighthouse (10.625,\ 2.847), buoy (2.052,\ -5.638); buoy minus lighthouse is (-8.573,\ -8.485).
- 236.4^\circ, finishing 267.2 meters north, or 303.6^\circ, finishing 598.8 meters north. The first leg gives X = 250, so 300\sin\theta = -250, \displaystyle \sin\theta = -\frac{5}{6}, and both bearings have that sine. The north totals are 433.01 + 300\cos\theta with \displaystyle \cos\theta = \mp\frac{\sqrt{11}}{6}, and both are positive.
- 19.70 nautical miles on 274.0^\circ (\text{N}\,86.0^\circ\,\text{W}). The legs are 18 and 8 nautical miles; X \approx -19.65, Y \approx 1.36.
- Standard position measures from east, a bearing from north. The standard angle is \alpha = 90^\circ - \theta, and the east coordinate is d\cos\alpha = d\cos(90^\circ - \theta) = d\sin\theta. Both statements say the same thing with angles measured from different starting lines.
- The opposite direction is a half turn, 180^\circ, whichever way you turn. With components, \sin(\theta \pm 180^\circ) = -\sin\theta and \cos(\theta \pm 180^\circ) = -\cos\theta, so both east and north parts flip. In compass form, flipping the north part swaps N and S, flipping the east part swaps E and W, and the acute angle from the north-south line is unchanged because the ray and its opposite lie on one straight line through the center.
- Back at the start. The second leg has east part d\sin(\theta + 180^\circ) = -d\sin\theta and north part d\cos(\theta + 180^\circ) = -d\cos\theta, which cancel the first leg’s parts, so X = Y = 0.
- The final position is the sum of the east parts and the sum of the north parts, and addition is commutative and associative, so the sums, and the finish, do not depend on the order. The paths differ: the points visited in between depend on the order, as does the shape of the track.