**Spherical Coordinates**

Figure 6-4 shows three systems *of spherical coordinates* for defining points in space. The first two are used by astronomers and aerospace scientists, while the third one is preferred by navigators and surveyors.

In the scheme shown in Fig. 6-4A, the location of a point *P* is defined by the ordered triple ( *θ,ø,r* ) such that *θ* represents the declination of *P,* ø represents the right ascension of *P,* and *r* represents the radius from *P* to the origin, also called the *range.* In this example, angles are specified in degrees (except in the case of the astronomer’s version of right ascension, which is expressed in hours, minutes, and seconds as defined earlier in this chapter). Alternatively, the angles can be expressed in radians. This system is fixed relative to the stars.

Instead of declination and right ascension, the variables *θ* and ø can represent celestial latitude and celestial longitude respectively, as shown in Fig. 6-4B. This system is fixed relative to the earth, rather than relative to the stars.

There’s yet another alternative: *θ* can represent elevation (the angle above the horizon) and ø can represent the azimuth (bearing or heading), measured clockwise from geographic north. In this case, the reference plane corresponds to the horizon, not the equator, and the elevation can cover the span of values between, and including, –90° (the *nadir,* or the point directly underfoot) and +90° (the zenith). This is shown in Fig. 6-4C. In a variant of this system used by mathematicians, the angle *θ* is measured with respect to the zenith (or the positive *z* axis), rather than the plane of the horizon. Then the range for this angle is 0° ≤ *θ* ≤ 180°.

**Spatial Coordinates Practice Problems**

**Practice 1**

Suppose you fly a kite above a perfectly flat, level field. The wind is out of the east-southeast, or azimuth 120°. Thus, the kite flies in a west-northwesterly direction, at azimuth 300°. Suppose the kite flies at an elevation angle of 50° above the horizon, and the kite line is 100 meters long. Imagine that it is a sunny day, and the sun is exactly overhead, so the kite’s shadow falls directly underneath it. How far from you is the shadow of the kite? How high is the kite above the ground? Express your answers to the nearest meter.

**Solution 1**

Let’s work in navigator’s cylindrical coordinates. The important factors are the length of the kite line (100 meters) and the angle at which the kite flies (50°). Figure 6-5 shows the scenario. Let *r* be the distance of the shadow from you, as expressed in meters. Let *h* be the height of the kite above the ground, also in meters.

First, let’s find the ratio of *h* to the length of the kite line, that is, *h* /100. The line segment whose length is *h,* the line segment whose length is r, and the kite line form a right triangle with the hypotenuse corresponding to the kite line. From basic circular trigonometry, we can surmise the following:

sin 50° = h/100

- sing a calculator, we derive
*h*as follows:

We also know, from basic circular trigonometry, this:

cos 50° = *r* /100

- sing a calculator, we derive
*r*as follows:

In this situation, the wind direction is irrelevant. But if the sun were not directly overhead, the wind direction would make a difference. It would also make the problem a lot more complicated. If you like difficult problems, try this one again, but imagine that the sun is shining from the southern sky (azimuth 180°) and is at an angle of 35° above the horizon.

Practice problems for these concepts can be found at: Three-Space and Vectors Practice Test

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