Global Navigation Help
Spherical trigonometry, when done on the surface of the earth, is of practical use for mariners, aviators, aerospace engineers, and military people. It is the basis for determining great-circle distances and headings. Here are four problems relating to global navigation. These problems can be solved almost instantly by computer programs nowadays, but you can get familiar with the principles of global trigonometry by performing the calculations manually.
Global Navigation Practice Problems
Consider two points R and S on the earth’s surface. Suppose the points have the following coordinates:
Lat R = +50.00° Lon R = +42.00° Lat S = –12.00° Lon S = –67.50°
What is the angular distance between points R and S , expressed to the nearest hundredth of a radian?
To solve this problem, a spherical triangle can be constructed with R and S at two of the vertices, and the third vertex at one of the geographic poles. Let’s use the north pole, and call it point Q. (The south pole will work too, but its use is more awkward because the sides of the spherical triangle extend over greater portions of the globe.) We label the sides opposite each vertex q, r, and s. Therefore, q is the angular distance we seek (Fig. 11-11).
We can use the spherical law of cosines to determine q, provided we can figure out three things:
- The measure of ψ q , the spherical angle at vertex Q
- The angular length of side r
- The angular length of side s
The measure of ψ q is the absolute value of the difference in the longitudes of the two points R and S :
ψ q = |Lon R − Lon S |
= | + 42.00° – (–67.50°)|
= 42.00° + 67.50°
The angular length of side r is the absolute value of the difference in the latitudes of points Q and S :
r = |Lat Q − Lat S ]
= | + 90.00° – (–12.00°)|
= 90.00° + 12.00°
The angular length of side s is the absolute value of the difference in the latitudes of points Q and R:
s = |Lat Q – Lat R |
= | +90.00° – (+50.00°)|
= 90.00° – 50.00°
Now that we know r, s, and ψ q , the spherical law of cosines tells us that:
cos q = cos r cos s + sin r sin s cos ψ q
and therefore the following holds:
q = arccos (cos r cos s + sin r sin s cos ψ q )
= arccos (cos 102.00° × cos 40.00° + sin 102.00° × sin 40.00° × cos 109.50°)
= arccos [(–0.20791) × 0.76604 + 0.97815 × 0.64279 × (–0.33381)]
= arccos (–0.36915)
= 1.95 rad
Suppose an aircraft pilot wants to fly a great-circle route from point R to point S in the scenario of Problem 11-7. What is the distance, in kilometers, the aircraft will have to fly? Express the answer to three significant figures.
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