Global Navigation Help

Global Navigation

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

Practice 1

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?

Solution 1

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:

Global Trigonometry Global Navigation

Fig. 11–11. Illustration for Practice 1-4.

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°

= 109.50°

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°

= 102.00°

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°

= 40.00°

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

Practice 2

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.

Solution 2

We multiply the angular distance, 1.95 rad, by 6371 kilometers. This gives us 12,400 kilometers, accurate to three significant figures.

Practice 3

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 should the initial azimuth heading be as the aircraft approaches cruising altitude after taking off from point R ? Express the answer to the nearest degree.

Solution 3

In order to determine this, we must figure out the measure of angle ψ _r in degrees. The initial azimuth heading is 360° – ψ _r . This is because side s of Δ _sph QRS runs directly northward, or toward azimuth 0°, from point R . The spherical law of sines tells us this about Δ _sph QRS :

(sin q )/(sin ψ _q ) = (sin r )/(sin ψ _r )

We can solve this equation for _r by manipulating the above expression and then finding the arcsine:

sin ψ _r = [(sin r )(sin ψ _r )]/(sin q )

ψ _r = arcsin {[(sin r )(sin ψ _r ])/(sin q )}

We already have the following information, having solved Problem 11-7:

q = 1.95 rad

ψ q = 109.5°

r = 102.00°

Plugging in the numbers gives us this:

ψ _r = arcsin {[sin 102.00°)(sin 109.5°)]/(sin 1095 rad)}

= arcsin [(0.97815 × 0.94264)/0.92896]

= arcsin 0.99255

= 83.00°

This means that the pilot’s initial heading, enroute on a great circle from point R to point S , should be 360° – 83.00°, or 277° to the nearest degree. This is 7° north of west.

Practice 4

Suppose an aircraft pilot wants to fly a great-circle route from point S to point R in the scenario of Problem 11-7. What should the initial azimuth heading be as the aircraft approaches cruising altitude after taking off from point S ? Express the answer to the nearest degree.

Solution 4

This is similar to the previous problem. We must figure out the measure of angle ψ _s in degrees. The initial azimuth heading is equal to ψ _s because side r of Δ _sph QRS runs directly northward, or toward azimuth 0°, from point S. According to the spherical law of sines:

(sin q )/(sin ψ _q ) = (sin s )/(sin ψ _s )

We can solve this equation for ψ _s by manipulating the above expression and then finding the arcsine:

sin ψ _s = [(sin s )(sin ψ _q )]/(sin q )

ψ _s = arcsin {[(sin s )(sin ψ _q ])/(sin q )}

We already have the following information, having solved Problem 11-7:

q = 1.95 rad

ψ _q = 109.5°

s = 40.00°

Plugging in the numbers gives us this:

ψ _s = arcsin {[sin 40.00°)(sin 109.5°)]/(sin 1.95 rad)}

= arcsin [(0.64279 × 0.94264)/0.92896]

= arcsin 0.65226

= 40.71°

This means that the pilot’s initial heading, enroute on a great circle from point S to point R, should be 41° to the nearest degree. This is 41° east of north.

Practice problems for these concepts can be found at: Global Trigonometry Practice Test

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