Back to Course

0% Complete
0/0 Steps

• Do you like this content?

Lesson 2, Topic 3
In Progress

Calculating Distance Along Great Circle

Lesson Progress
0% Complete

Note that the shortest distance between two points on the surface of the earth lies on the arc length of a great circle connecting the two points. Let the arc length be $$\scriptsize \overline{AB}$$, then

$$\scriptsize \overline {AB} = \normalsize \frac{Î¸Â°}{360Â°} \scriptsize \times 2 \pi R$$

Where  Î¸Â° = angular difference

R = radius of the earth = 6,400Km

Â  Ï€ = $$\frac{22}{7}$$

Example 1

A and B are two places on the earth’s surface on the same meridian. B has a Latitude of 20Â° N and A is a point north of B such that the distance AB measured along the meridian is 800Km.
Calculate the Latitude of A correct to the nearest degree,
(Take Ï€ = $$\frac{22}{7}$$ and R = 6,400Km.)

Solution:

$$\scriptsize \overline {AB} = \normalsize \frac{Î¸Â°}{360Â°} \scriptsize \times 2 \pi R$$

$$\scriptsize \overline{AB} = 800$$

800 = Â $$\frac{Î¸}{360} \: \times \: \scriptsize 2 \: \times \: \normalsize \frac{22}{7} \scriptsize \: \times \: 6400$$

Î¸ = $$\frac{800 \: \times \: 360\: \times \: 7}{44 \: \times \: 6400}$$

= $$\frac{45 \: \times \: 7}{44}$$

Î¸ = 7.1590

Î¸ = 7Â° (nearest degrees)

Since the two points are in the same direction

Î¸ = Latitude A – Latitude B

i.e. 7 = Latitude A â€“ 20

Therefore, Latitude A = 7 + 20

i.e. Latitude A = 27Â° N

Example 2

(a) An aeroplane flies from city A (0Â°, 152Â° E) to city B (0Â°, 171Â° W), find the shortest distance travelled to reach city B

(b) How far is city B from the North Pole? (Take Ï€ = $$\frac{22}{7}$$ and R = 6,400Km.)

Solution:

The angular difference between points A and B is 323Â° which gives the major segment â€“ it does not give the shortest distance, rather the minor segment of 37Â° gives the shortest distance.

i.e. $$\scriptsize \bar{AB}\: = \normalsize \frac{Î¸}{360} \scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7} \scriptsize \: \times \: 6400$$

$$= \frac{Î¸}{360} \: \times \: \scriptsize 2 \: \times \: \normalsize \frac{22}{7} \scriptsize \: \times \: 6400$$

= 4,134.60317

i.e.     AB = 4,135Km

Recall that from the equator 0Â° to the North Pole, the angular difference is 90Â°.

Therefore, $$\scriptsize \bar{BN} = \normalsize \frac{Î¸}{360} \: \times \: \scriptsize 2 \: \times \: \normalsize \frac{22}{7} \scriptsize \: \times \: 6400$$

= $$\scriptsize \bar{BN} = \normalsize \frac{90}{360} \: \times \: \scriptsize 2 \: \times \: \normalsize \frac{22}{7} \scriptsize \: \times \: 6400$$

= 10,057.1429

$$\scriptsize \bar{BN} = 10,057Km$$

error: