Back to Course

0% Complete
0/0 Steps

• ## Do you like this content?

Lesson 4, Topic 4
In Progress

# Calculating Distance Along Parallels of Latitude

Lesson Progress
0% Complete

Considering right angle triangle LMO

Cos Î¸ = $$\frac {r}{R}$$

i.e. r = R Cos Î¸

Example 1:

Find the distance between two points A(50oN, 94oW) and B(50oN, 86oE):

(i) Along the parallel of Latitude

(ii) Along a great circle

Solution:

(i) Angular difference = 94o + 86o = 180o

i.e $$\scriptsize \bar{AB} = \normalsize \frac{180}{360}\scriptsize \; \times \; \scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400 \; \times \; cos50$$

= $$\frac{1}{2} \; \times \; \frac{44}{7} \scriptsize \; \times \; 6400 \; \times \; 0.6428$$

i.e $$\scriptsize \bar{AB} = 12,929.213$$

i.e $$\scriptsize \bar{AB} = 13,000km \;(nearest\;kilometre)$$

(ii) Angular difference = 180o – (50 + 50)o

= 180o – 100o

= 80o

i.e $$\scriptsize \bar{AB} = \normalsize \frac{80}{360}\scriptsize \; \times \; \scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400$$

= 8,939.6

i.e. $$\scriptsize \bar{AB}$$ = 8,940Km (nearest Kilometre).

(Recall that it is clear that distance along a great circle is shorter i.e. 8,940Km < 13,000Km.

Example 2

Two points A and B are 800Km apart on the same Latitude with 20o angular difference of longitude. Calculate:

(i) The Latitude

(ii) The speed of an aeroplane that travelled from A to B in 2hours

(iii) The speed of the point B due to the rotation of the earth. (Take $$\scriptsize \pi = \normalsize \frac{22}{7}$$ and R = 6,400Km.)

Solution

$$\scriptsize \bar{AB} = \normalsize \frac{20}{360} \scriptsize\; \times \; \scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400 \; \times \; cos Î¸$$

$$\scriptsize \bar{AB} = \normalsize \frac{1}{18}\scriptsize \; \times \; \normalsize \frac{44}{7} \scriptsize \; \times \; 6400 \; \times \; cos Î¸$$

$$\scriptsize 800 = \normalsize \frac{1}{18} \scriptsize \; \times \; \normalsize \frac{44}{7} \scriptsize \; \times \; 6400 \; \times \; cos Î¸$$

cos Î¸ = $$\frac{18 \; \times \; 800 \; \times \; 7}{44 \; \times \; 6400}$$

Î¸ = cos-1(0.3580)

Î¸ = 69Â°

Let Î¸ Latitude = Î¸Â° N

âˆ´   Î¸ = 69Â° N

(ii) Speed = $$\frac{Distance}{time}$$

Time = 2hours, Distance = 800

i.e.     Speed(s) = $$\frac{800}{2}$$

Therefore, Speed = 400Km/h

(iii) Circumference of Latitude 69o = 2 Ï€r

Where r= RCos Î¸

i.e. Distance = $$\scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400 \; \times \; cos 69$$

= 14,417Km

Therefore, Speed = $$\frac{Distance}{time} = \frac{14,417}{2}$$

= 7,208 Km/h.

Example 3

A plane leaves an airport X, 20.6oE and 36.8oN and flies due South along the same longitude for 8hours at the rate of 1000Km/h to another airport Y, 20.6oE, and Î¸Â°. The plane then flies west to another airport Z for 8hours at the same speed. Calculate, to the nearest degree:

(i) The value of

(ii) The longitude of Z

(Take Ï€ = $$\frac{22}{7}$$ and R = 6,400Km.)

(i) Distance $$\scriptsize \bar{XY}$$ = Speed  x  time

= 1000 x 8

=8,000Km

i.e.  $$\scriptsize \bar{XY} = 8000 = \frac{Î¸}{360} \scriptsize \; \times \; \scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400$$

i.e.      8000 = Î¸ Ã— 117.46032

i.e.     Î¸ = 71.5909

Î¸ = 72Â° nearest degree.

Angular difference = Î¸Â° S + 36.8Â°N

72 = Î¸ + 36.8

Î¸ = 35.2Â°

Î¸ = 35Â°S nearest degrees.

(ii) $$\scriptsize \bar{ZY}$$ = Speed  x  time

$$\scriptsize \bar{ZY}$$ = 1000 x 8

i.e.  $$\scriptsize \bar{ZY}$$ = 8,000Km

i.e.    $$\scriptsize \bar{ZY} = 8000 = \frac{Î¸}{360} \scriptsize \; \times \; \scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400 \; \times \; cos 35.2$$

i.e.      8000 = Î¸ Ã— 91.312699

i.e.  Î¸ = 87.6110335

i.e. Angular difference = 20.6oE + Î¸Â°W

i.e. 87.611 = 20.6 + Î¸Â°W

i.e. Î¸Â°W = 67.011Â°W

Example 4:

A and B are two points on latitude 55oS and their longitudes 23oW and 33oE respectively. Calculate the distance between A and B along:

(i) A great circle

(ii) The parallel of Latitude

(Take Ï€ = $$\frac{22}{7}$$ and R = 6,400Km.)

:- $$\scriptsize \bar{AM} = r sin \normalsize \frac {Î¸}{2}$$Â Â Â

$$\scriptsize \bar{AM} = Rcos(Latitude) sin \normalsize \frac{\theta}{2}$$ ………….â€¦.(a)Â

$$\scriptsize \bar{AM} = R$$

i.e.   $$\scriptsize \bar{AM} = R sin x$$         ………………… (b)

Equate equation (a) and (b)

R cos (latitude) $$\scriptsize sin \normalsize \frac {Î¸}{2}\scriptsize = R sin x$$

Therefore $$\scriptsize sin x = cos \; (latitude) \; sin \normalsize \frac {Î¸}{2}$$

i.e.    sin x   = Cos 55 Sin 28

i.e. x = sin -1 (0.5736Ã—0.4695) Â

= sin -1 (0.2693 )Â Â Â

x = 15.6219

2x = 31.24Â° =31Â°

Therefore, $$\scriptsize \bar{AB}$$ along a great circle = $$\frac{2x}{360} \; \times \; \scriptsize 2 \pi r$$

i.e $$\scriptsize \bar{AB}= \normalsize \frac{31.24}{360}\scriptsize \; \times \; \scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400$$

i.e $$\scriptsize \bar{AB}$$ = 3,491.38

i.e $$\scriptsize \bar{AB}$$ = 3,491 km

(b) $$\scriptsize \bar{AB}$$ along parallel of Latitude = $$\frac{Î¸}{360}\scriptsize \; \times \; \scriptsize 2 \pi R cos \; (latitude)$$

i.e $$\scriptsize \bar{AB} =\normalsize \frac{56}{360}\scriptsize \; \times \; \scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400 \; \times \; cos 55$$

= 3,589.31

i.e. $$\scriptsize \bar{AB}$$ = 3,589Km.

Example 5

The position of P is (60oS, 60oW) and that of Q is (60oS, 32oE). Calculate the distance between P and Q along:

(i) The parallel of Latitude

(ii) A great circle

(i) $$\scriptsize \bar{PQ} = \normalsize \frac{Î¸}{360}\scriptsize \; \times \; \scriptsize 2 \pi R cos \; (latitude)$$

i.e $$\scriptsize \bar{PQ} = \normalsize \frac{Î¸}{360} \scriptsize \; \times \; 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400 \; \times \; cos 60$$

$$\scriptsize \bar{PQ}$$ = 5,140.3174603

$$\scriptsize \bar{PQ}$$ = 5,140Km

(ii)Sin x = Cos 60 Sin46

= 0.5000 x 0.7193

Sin x = 0.3600

x = Sin-1(0.3600)

x = 21.1o

2x = 42.2o

2x = 42o

$$\scriptsize \bar{PQ} = \normalsize \frac{2x}{360}\scriptsize \; \times \; \scriptsize 2 \pi R$$

i.e $$\scriptsize \bar{PQ}= \normalsize \frac{42}{360}\scriptsize \; \times \; \scriptsize 2 \; \times \; \normalsize \frac{22}{7} \scriptsize \; \times \; 6400$$

= 4,693.333

PQ = 4,693Km.

error: