Terrestrial Distance Measurement Help (page 2)

By — McGraw-Hill Professional
Updated on Oct 3, 2011


Stadimetry can be used to measure the distance to an object when the object’s height or width is known. The angular diameter of the object is determined by observation. The distance is calculated using trigonometry. This scheme works in the same way as the base-line method described above, except that the “base line” is at the opposite end of the triangle from the observer.

Figure 8-4 shows an example of stadimetry as it might be used to measure the distance d, in meters, to a distant person. Suppose the person’s height h, in meters, is known. The vision system determines the angle θ that the person subtends in the field of view. From this information, the distance d is calculated according to the following formula:

Surveying, Navigation, and Astronomy Terrestrial Distance Measurement Stadimetry

Fig. 8–4. Stadimetry requires that the height or width of a distant object be known.

d = h/ (tan θ )

In order for stadimetry to be accurate, the linear dimension axis (in this case the axis that depicts the person’s height, h ) must be perpendicular to a line between the observation point and one end of the object. Also, it is important that d and h be expressed in the same units.

Terrestrial Distance Measurement Practice Problems

Practice 1

Suppose we want to determine the distance to an object at the top of a mountain. The base line for the distance measurement is 500.00 meters (which we will call 0.50000 kilometers) long. The angular difference in azimuth is 0.75000° between opposite ends of the base line. How far away is the object?

Solution 1

It helps to draw a diagram of the situation, even though it cannot be conveniently drawn to scale. (The base line must be shown out of proportion to its actual relative length.) Figure 8-3 illustrates this scenario. The base line, which is line segment PQ, is oriented at right angles to the line segment PR connecting one end of the base line and the distant object. The right angle is established approximately, using a hiker’s compass, but for purposes of calculation, it can be assumed exact, so Δ PQR can be considered a right triangle.

Surveying, Navigation, and Astronomy Terrestrial Distance Measurement Accuracy

Fig. 8–3. Illustration for Practice 1 and 2.

We measure the angle θ between a ray parallel to line segment PR and the observation line segment QR, and find this angle to be 0.75000°. The “parallel ray” can be determined either by sighting to an object that is essentially at an infinite distance, or, lacking that, by using an accurate magnetic compass.

One of the fundamental principles of plane geometry states that pairs of alternate interior angles formed by a transversal to parallel lines always have equal measure. In this example, we have line PR and an observation ray parallel to it, while line QR is a transversal to these parallel lines. Because of this, the two angles labeled θ in Fig. 8-3 have equal measure.

We use the triangle model for circular functions to calculate the distance to the object. Let b be the length of the base line (line segment PQ ), and let x be the distance to the object (the length of line segment PR ). Then the following formula holds:

tan θ = b/x

Plugging in known values produces this equation:

Surveying, Navigation, and Astronomy Terrestrial Distance Measurement Accuracy

The object on top of the mountain is 38.195 kilometers away from point P.

Practice 2

Why can’t we use the length of line segment QR as the distance to the object, rather than the length of line segment PR in the above example?

Solution 2

We can! Observation point Q is just as valid, for determining the distance, as is point P. In this case, the base line is short compared to the distance being measured. In Fig. 8-3, let y be the length of line segment QR. Then the following formula holds:

sin θ = b/y

Plugging in known values, we get this:

Surveying, Navigation, and Astronomy Terrestrial Distance Measurement Accuracy

The percentage difference between this result and the previous result is small. In some situations, the absolute difference between these two determinations (approximately 3 meters) could be of concern, and a more precise method of distance measurement, such as laser ranging, would be needed. An example of such an application is precise monitoring of the distance between two points at intervals over a period of time, in order to determine minute movements of the earth’s crust along a geological fault line.

Practice problems for these concepts can be found at:  Surveying, Navigation, and Astronomy Practice Test

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