Distances along small circles

Distances along Small Circles

Suppose P and Q are places west or east of each other, i.e. they lie on the same circle of latitude. Then when you travel due east or west from P to Q you travel along an arc of the circle of latitude.

The situation here is slightly different from that of the previous section. While circles of longitude all have the same length, circles of latitude get smaller as they get nearer the poles.

Consider the circle of latitude 50°S. Let its radius be $r$ km.

Nautical miles

Then taking a point P on the circle, and letting $R$ be the radius of the Earth:

$\cos(50°) = \frac{r}{R} \quad \text{or} \quad r = R \times \cos(50°)$

In general, the circle of latitude ($\alpha$) has radius $R \times \cos(\alpha)$. This is true for latitudes both north and south of the equator.

Now the distance along a circle of latitude can be found. Suppose the difference between longitudes is $\theta$. Then the length of the arc going from P to Q is:

$\text{Arc length} = \frac{2\pi r\theta}{360} = \frac{2\pi(R \times \cos(\alpha))\theta}{360}$

But in kilometers, $\frac{2\pi R}{360} = 111.7$, so the arc length is:

$\text{Arc length} = 111.7 \times \cos(\alpha) \times \theta \text{ km}$

Example 1

Find the distance in km and nm along a circle of latitude between (20°N, 30°E) and (20°N, 40°W).

Solution:

Both places are on latitude 20°N. The difference in longitude is 70°. Use the formula for distance.

Distance = $111.7 \times \cos20° \times 70°$. Hence the distance in nautical miles is $60 \times 70 \times \cos20°$

The distance is 3,950 nm.

Example 2

A ship starts at (40°S, 30°W) and sails due west for 1,000 km. Find its new latitude and longitude.

Solution:

Because it sails due west, the latitude remains unchanged. Suppose it has sailed through $x°$ of longitude. Use the formula:

$1,000 = 111.7 \times \cos(40°) \times x$

Hence, $x = \frac{1,000}{111.7 \times \cos(40°)} = 12$

Add 12° to the longitude, obtaining 42°. Therefore, the new position is (40°S, 42°W).

Example 3

A ship sails west from (20°S, 15°E) to (20°S, 23°E), taking 37 hours. Find speed, in knots and in km per hr.

Solution:

The difference in longitude between the two points is 8°. Hence, the distance, in nautical miles, is:

$60 \times 8 \times \cos(20°) = 451 \text{ nm}$

Divide by 37 to obtain the speed. The speed is 12.2 knots.

To obtain the speed in kilometers per hour, multiply by 1.862. The speed is 22.7 km/hr.

Navigation

Suppose a ship is sailing in a sea current, or that a plane is flying in a wind. Then the course set the ship or plane is not the direction that it will move in. The actual direction and speed can be found either by scale or by the use of Pythagoras's theorem and trigonometry.

Draw the line representing the motion of the ship relative to the water. At the end of this line draw a line representing the current. Draw the third side of the triangle. This side, shown with a double-headed arrow, is the actual course of ship.

Example 4

A ship sets course due east. In still water the ship can sail at 15 km/hr. There is a current flowing due south of 4 km/hr. Use a scale drawing to find:

1. The speed of the ship
2. The bearing of the ship.

Solution:

In one hour the ship sails 15 km east relative to the water. Draw a horizontal line of length 15 cm. In one hour the current pulls the ship 4 km south. At the end of the horizontal line, draw a vertical line of length 4 cm.

(a) Measure the third side of the triangle as 15.5 cm. The speed of the ship is 15.5 km/hr.

(b) Measure the angle between the course of the ship and east as 13°. Add this to 90°. The ship is sailing along a bearing of 105°.

Note: These results can also be found by Pythagoras' theorem and trigonometry. Thus, the speed is

$\text{Speed} = \sqrt{15^2 + 4^2} = \sqrt{225 + 16} = \sqrt{241} \approx 15.5 \text{ km/hr}$

The angle with east is $\theta = \tan^{-1}\left(\frac{4}{15}\right) \approx \tan^{-1}(0.2667) \approx 14.9°$

Example 5

The ship of Example 4 needs to travel due east. Calculate the following:

1. What course should be set?
2. How long will the ship take to cover 120 km?

Solution

The ship needs to set a course slightly north of east, consider the following diagram.

(a) By trigonometry: $\sin \theta = \frac{4}{15}$, hence $\theta = 15.945°$. Subtract this from 90°. A course of 074.5° should be set.

(b) Using Pythagoras' theorem in the triangle, actual speed of the ship = $\sqrt{15^2 - 4^2} = 14.46 \text{ km/hr}$.

Now divide 120 by this speed. The ship will take 8.3 hrs.

Note: With no current, the journey would take 8 hrs. The journey takes slightly longer when there is a current.

Suppose a ship or a plane does not directly reach a position. We can still find how close the ship or plane is to the position.

In the diagram above, the ship S (or plane) is travelling on a straight line AB. The shortest distance to point P is when SP is perpendicular to AB.

Example 6

A small island is 200 km away on a bearing of 075°. A ship sails on a bearing of 070°. Find the closest that the ship is to the island.

Solution:

Consider the diagram below.

The angle between the directions of the island and the path of the ship is: $75° - 70° = 5°$

When a ship is closest to the island, the line from the island to the ship is at 90° to the path of the ship, hence, by trigonometry:

$d = 200 \times \sin 5° = 17.4 \text{ km}$

The closest distance is 17.4 km.