Whether the surface is regular or irregular, mathematical area measurements basically require knowledge of the linear magnitude of two dimensions. However, beside accurate mathematical formulas, several practical techniques and procedures are available to estimate surface areas.
Area determination of the earth's surface usually considers the area of the orthogonal projection of the irregular land surface to a datum-level plane. Area determination for portions of the earth's surface can be expressed as the square of any of the linear units. The most common area units are the international metric system units: square centimeters (cm2), square meters (m2), hectares (ha), and square kilometers (km2). However, in various countries, other units have become standard through common usage. For example, in the United States, the acre (43,560 square feet or 10 square chains) is the most common area unit (see Appendix, Table 2 for converting factors).
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If the coordinates are designated as x1, x2, x3, ....., xn and the y coordinates are y1, y2, y3, ...., yn, then the area of the polygon whose vertices are (y1, y1), (y2, y2), (y3, y3), ..., (yn, yn) is the absolute value of:
if the ground coordinates of vertices are known the ground area may be directly determined by substituting these coordinates in equation 7.25. Otherwise, the scale of the photograph may be used to determine the surface area on the ground as follows: after determining the polygon area using the photo coordinates, this surface is then multiplied by the square of the photo scale reciprocal:
Photo area ´ PSR2 = Ground area
Example 7.19
The photo coordinates of the 5 vertices of a polygon delineated on a 1:12,000 vertical aerial photograph were measured to be x1 = 5.12 cm, y1 = 1.54 cm, x2 = 5.61 cm, y2 = 4.37 cm, x3 = 7.13 cm, y3 = 6.20 cm, x4 = 9.83, y4 = 6.81 cm, x5 = 8.08 cm, y5 = 3.44 cm. Find the area of the polygon.
Solution:
Using equation 7.25, the photographic area of the polygon is:
½(|[(5.12)(4.37) + (5.61)(6.20) + (7.13)(6.81) + (9.83)(3.44) + (8.08)(1.54)] -
[(5.61)(1.54) + (7.13)(4.37) + (9.83)(6.20) + (8.08)(6.81) + (5.12)(3.44)]|)
= |151.97 - 173.38| = 21.41 cm2
The ground area of the polygon is then:
21.41 cm2 ´ (12000)2 = 308304 104 cm2 or 30.83 hectares
Example 7.20
Using the same photograph in example 7.19, another polygon was delineated and the ground coordinates of its 5 vertices were found to be X1 = 514852, Y1 = 5178256, X2 = 515495, Y2 = 5180687, X3 = 518168, Y3 = 5181112, X4 = 518738, Y4 = 5179680, X5 = 517766, Y5 = 5178184.
Find the ground area of the polygon.
Solution:
Using equation 7.25, the ground area of the polygon can directly be computed as:
½|[(514852)(5180687) + (515495)(5181112) + (518168)(5179680) + (518738)(5178184) + (517766)(5178256)] - [(515495)(5178256) + (518168)(5180687) + (518738)(5181112) + (517766)(5179680) + (514852)(5178184)]| = 851.50 ha
If the polygon is delineated by arcs instead of straight segments, the
above formula can still be used to estimate the area of the polygon. In
this case, the boundaries of the polygon should be divided into small arcs
that could approach a straight segment (figure 7.19). The line segments
shown in figure 7.19 are purposely drawn large to better illustrate the
concept of this technique. Usually, the curved boundary is divided into
small arcs that approach straight segments. The shorter the arcs, the more
accurate the area estimation.
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However, a more accurate method to determine the area of a "non-straight-line-boundary-polygon" is the use of integration, planimeters or computers. All of these techniques are based on the coordinate method described above to determine the area of a polygon.
Where dx and dy are the small portions of the curve that
approach a straight segment. This technique is rarely used in practical
photography area measurement and, therefore, will not be discussed further.
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Ground Area = Photo Area ´ (PSR)2
For example, if a 23 cm by 23 cm aerial photo weighs 20 grams, then the ratio of area to unit weight is 529 cm2 ¸ 20 g = 26.45 cm2/g. In other words, 1 g of paper would be equivalent to an area of 26.45 cm2 on the photo or 26.45 cm2 ´ PSR2 on the ground. Areas of polygons of any configuration can then be cut out and weighed. Each gram will be equivalent to 26.45 cm2 ´ PSR2 ground area.
Suppose we have interpreted and delineated the units of interest on a set of photographs. The units can then be cut out and sorted by category of information (e.g., all the polygons that represent bare soil are cut and put in the same pile, all the polygons that represent forested areas are cut out and put in a separate pile, etc.). Then, by weighing the piles of paper for each category, the total area of each category can be determined and the percentage area may be compared to the entire project area. The total project area will also serve for checking against the total weights of the individual categories.
Example 7.21
suppose we interpreted an aerial photograph at a nominal scale of 1:20,000 for a land use project. If the aerial photograph is 23 cm by 23 cm and weighs 20 grams, find the ratio of area to the unit weight and the area of a lake that was cut out and found to weigh 4 grams.
Solution:
20 g is equivalent to an area of 23 cm ´ 23 cm = 529 cm2 on the photograph or 529 ´ (20,000)2 cm2 = 2116 ´ 108 cm2 on the ground = 2116 ´ 104 m2 on the ground (because 1 m2 = 10,000 cm2) = 2116 ha on the ground (because 1 ha = 10,000 m2).
If 20 grams = 2116 ha, therefore, 1 gram = 105.8 ha, which represents the ratio of area to the unit weight. If the polygon corresponding to the lake weighs 4 grams, its surface is:
4 ´ 105.8 = 423.2 ha or 20 % ([423.2 ¸ 2116] ´ 100) of the entire area covered by the photo.
The same procedure can be applied for the rest of the categories interpreted on the photograph or a set of photographs or a map.
This simple technique was developed by the U.S. Soil Conservation Service and has been found to be both efficient and accurate. The accuracy is of course dependent, among others, upon the homogeneity of the thickness of the paper, the precision of cutting the polygons, and the scale used to weigh the paper.
where a is the area of the category to be determined, A
is the total area of all the categories (i.e., the project area), l
is the sum of all the segments (transects) passing through the category,
and L is the sum of all the transects inside the project area considered.
For example, in figure 7.21, L = S Li
= L1 + L2 + ... +L11, l = S
li = l1 + l2 + l3 + l4,
A
= x ´ y, and a = (A ´
l) ¸ L.
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The number of and spacing between transects used depend upon the size
and arrangement of the polygons to be measured; the higher the number,
the more accurate the measurements. Generally, at least 10 transects should
be used and a higher number for accuracy, depending on the size of the
area to be measured. The transect lines may randomly be laid across the
area to be measured. However, for regularly shaped units (e.g., rectangular
or linear), the transects lines should be laid across the short dimensions
to obtain the best sample (figure 7.22). The transect lines must not
be laid parallel to the boundaries of long polygons.
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One of the procedures of using the transect method may be as follows:
Starting about half a centimeter from an appropriate boundary (so that
the transect lines are not parallel to the sides of long polygons as described
in the previous paragraph) of the area to be measured, transect lines are
drawn 1 cm apart until the opposite boundary of the area (figure 7.23).
The same process may be performed by overlaying a transparency on which
the transects lines are already drawn. Next, the line segments intercepting
the boundaries of the polygons are measured and the transect line distances
are summarized for each polygon and each category (Table 7.1). The sum
of proportions of all categories can then be verified to make sure
they total one hundred percent (or 1.00) and the sum of the areas of individual
categories totals the total area of the entire project area considered.
| Polygon type
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Total distance in a category
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Proportion to total distance in
the entire area
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Area of a category
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Correction factor
(5) |
Adjusted area
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Measurements from paper may be subject to errors introduced from shrinkage or dilatation. When making measurements from a grided map, errors may also be introduced from defective griding. In either case, these errors must be known prior to area measurement. The dimensions of the grid on a map are usually known and errors due to paper deformation or defective griding may be determined by measuring the dimensions of the grid on the map and comparing them to the actual dimensions. For example, if a 100 mm by 100 mm grid were measured to be 100.2 mm and 100.3 mm, its area is 10050.06 mm2. Therefore, a correction factor of -50 mm2 per 10000 mm2 (or -5/1000) should be applied to the measured surfaces.
Determining areas using a planimeter
Before making measurement with a planimeter, the following issues must
be taken into account.
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PA = n ´ k (7.40)
The value of n is a function of the radius of the wheel and the polar arm of the planimeter and the scale factor, k, is a function of the scale of the plane containing the polygon being measured. The scale factor also varies with the polar arm of the planimeter and is usually provided with the planimeter. A table for unit area may be presented as indicated in table 7.3.
Table 7.3. Table unit for area in metric scale for two tracer arm lengths
at different scales.
| Tracer Arm Length | 149.5 | 116.2 |
| Scale | Unit Area | Unit Area |
| 1:100 | 0.1 m2 | 0.08 m2 |
| 1:200 | 0.4 m2 | 0.32 m2 |
| 1:300 | 0.9 m2 | 0.72 m2 |
| 1:400 | 1.6 m2 | 1.28 m2 |
| 1:500 | 2.5 m2 | 2 m2 |
| 1:600 | 3.6 m2 | 2.88 m2 |
| 1:1000 | 10 m2 | 8 m2 |
| 1:1500 | 22.5 m2 | 18 m2 |
| 1:2000 | 40 m2 | 32 m2 |
| 1:3000 | 90 m2 | 72 m2 |
| 1:5000 | 250 m2 | 200 m2 |
| 1:6000 | 360 m2 | 288 m2 |
| 1:10000 | 1000 m2 | 800 m2 |
| 1:20000 | 4000 m2 | 3200 m2 |
| 1:24000 | 5760 m2 | 4608 m2 |
| 1:25000 | 6250 m2 | 5000 m2 |
| 1:30000 | 9000 m2 | 7200 m2 |
| 1:50000 | 25000 m2 | 20000 m2 |
Assume in figure 7.27, the scale is 1:24,000. After tracing the perimeter of the polygon the difference between the final and the initial readings of the planimeter was found to be 1265. What is the ground area of the polygon?
Solution:
From the table 7.3, if the scale coefficient is 5760 m2 (tracer arm length = 149.5) the area is: 1265 ´ 5760 m2 = 7286400 m2 or 728.64 ha if the scale coefficient is 4608 m2 (tracer arm length = 149.5) the area is:
1265 ´ 4608 m2 = 5829120 m2 or 582.91 ha
7.9
Factors Affected by Scale | 7.10 Determining Areas from
Aerial Photographs | References![]()