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Solving problems according to topographic plans. Solving problems on topographic plans Technique for measuring and plotting distances on a map

called the scale, which is expressed as a fraction, the numerator of which is equal to one, and the denominator shows how many times the horizontal alignment of the terrain line is reduced when the horizontal alignment of the line is depicted on a plan or map.

Numerical scale- unnamed value. It is written as follows: 1:1000, 1:2000, 1:5000, etc., and in such a record 1000, 2000 and 5000 are called the denominator of the M scale.

The numerical scale means that one unit of line length on the plan (map) contains exactly the same number of units of length on the ground. So, for example, one unit of line length on the 1:5000 plan contains exactly 5000 of the same length units on the ground, namely: one centimeter of the line length on the 1:5000 plan corresponds to 5000 centimeters on the ground (i.e. 50 meters on the ground ); one millimeter of the line length on the 1:5000 plan contains 5000 millimeters on the ground (that is, one millimeter of the line length on the 1:5000 plan contains 500 centimeters or 5 meters on the ground), etc.

When working with a plan, in some cases they use linear scale.

Linear scale

- graphic construction, (Fig. 1) which is an image of a certain numerical scale.
Fig.1

The base of the linear scale called the segment AB of a linear scale (the main share of the scale), usually equal to 2 cm. It is translated into the appropriate length on the ground and signed. The leftmost base of the scale is divided into 10 equal parts.

The smallest division of the base of a linear scale equals 1/10 of the scale base.

Example: for a linear scale (used when working on a 1:2000 scale topographic plan) shown in Figure 1, the base of the AB scale is 2 cm (i.e. 40 meters on the ground), and the smallest division of the base is 2 mm, which in scale 1:2000 corresponds to 4 m on the ground.

Segment cd (Fig. 1), taken from a topographic plan at a scale of 1:2000, consists of two scale bases and two smallest divisions of the base, which, as a result, corresponds to 2x40m + 2x2m = 88 m on the ground.

A more accurate graphical definition and construction of line lengths can be done using another graphical construction - a transverse scale (Fig. 2).

Cross scale

- graphic construction for the most accurate measurement and laying of distances on a topographic plan (map). Scale accuracy is a horizontal segment on the ground, which corresponds to a value of 0.1 mm on a plan of a given scale. This characteristic depends on the resolution of the naked human eye, which (resolution) allows you to see the minimum distance on the topographic plan of 0.1 mm. On the ground, this value will already be equal to 0.1 mm x M, where M is the denominator of the scale

The base AB of the normal transverse scale is, as in the linear scale, also 2 cm. The smallest division of the base is CD = 1/10 AB = 2 mm. The smallest division of the transverse scale is equal to cd \u003d 1/10 CD \u003d 1/100 AB \u003d 0.2 mm (which follows from the similarity of the triangle BCD and the triangle Bcd).

Thus, for a numerical scale of 1:2000, the base of the transverse scale will correspond to 40 m, the smallest division of the base (1/10 of the base) is 4 m, and the smallest division of the scale of 1/100 AB is 0.4 m.

Example: the segment av (Fig. 2), taken from the plan on a scale of 1:2000, corresponds to 137.6 m on the ground (3 bases of the transverse scale (3x40 \u003d 120 m), 4 smallest divisions of the base (4x4 \u003d 16 m) and 4 smallest scale division (0.4x4=1.6 m), i.e. 120+16+1.6=137.6 m) .

Let us dwell on one of the most important characteristics of the concept of "scale".

scale accuracy called a horizontal segment on the ground, which corresponds to a value of 0.1 mm on a plan of a given scale. This characteristic depends on the resolution of the naked human eye, which (resolution) allows you to see the minimum distance on the topographic plan of 0.1mm. On the ground, this value will already be equal to 0.1 mm x M, where M is the denominator of the scale.


Fig.2

The transverse scale, in particular, makes it possible to measure the length of a line on a plan (map) at a scale of 1:2000 precisely with the accuracy of this scale.

Example: 1 mm of the 1:2000 plan contains 2000 mm of terrain, and 0.1 mm, respectively, 0.1 x M (mm) = 0.1 x 2000 mm = 200 mm = 20 cm, i.e. 0.2 m

Therefore, when measuring (constructing) on ​​a line length plan, its value should be rounded with scale accuracy. Example: when measuring (constructing) a line 58.37 m long (Fig. 3), its value on a scale of 1:2000 (with a scale accuracy of 0.2 m) is rounded up to 58.4 m, and on a scale of 1:500 (accuracy scale 0.05 m) - the length of the line is already rounded up to 58.35 m.

The scale is the ratio of the length of a line on a drawing, plan, or map to the actual length of the corresponding line. It shows how many times the distance on the map is reduced relative to the actual distance on the ground. If, for example, the scale geographical map 1: 1,000,000, which means that 1 cm on the map corresponds to 1,000,000 cm on the ground, or 10 km.

Distinguish between numerical, linear and named scales .

Numerical scale is depicted as a fraction, in which the numerator is equal to one, and the denominator is a number showing how many times the lines on the map (plan) are reduced relative to the lines on the ground. For example, a scale of 1:100,000 shows that all linear dimensions on the map are reduced by 100,000 times. Obviously, the larger the denominator of the scale, the smaller the scale, with a smaller denominator, the larger. The numerical scale is a fraction, so the numerator and denominator are given in the same measurements (centimeters).

Linear scale is a straight line divided into equal segments. These segments correspond to a certain distance on the depicted terrain; divisions are indicated by numbers. The measure of length along which the divisions on the scale bar are marked is called the base of the scale. In our country, the scale base is taken equal to 1 cm. The number of meters or kilometers corresponding to the scale base is called the scale value. When constructing a linear scale, a figure 0 , from which the countdown of divisions begins, is usually placed not at the very end of the scale line, but retreating one division (base) to the right; on the first segment to the left of 0, the smallest divisions of the linear scale are applied - millimeters. The distance on the ground corresponding to one smallest division of the linear scale corresponds to the accuracy of the scale, and 0.1 mm corresponds to the maximum accuracy of the scale. The linear scale compared to the numerical one has the advantage that it makes it possible to determine the actual distance on the plan and map without additional calculations.

Named Scale - the scale expressed in words, for example, in 1 cm 32 km.

Measuring distances on the map and plan.

Measuring distances with a scale. You need to draw a straight line (if you need to know the distance in a straight line) between two points and use a ruler to measure this distance in centimeters, and then you should multiply the resulting number by the scale value. For example, on a scale map 1: 100 000 (in 1 cm 1 km) the distance is 5 cm, i.e. on the ground this distance is 1 * 5 = 5 (km). You can also measure the distance on the map using a measuring compass. In this case, it is convenient to use a linear scale.

Measuring distances using a degree network. To calculate distances on a map or globe, you can use the following quantities: arc length 1° meridian and 1° equator is approximately 111 km. For meridians, this is always true, and the length of an arc of 1 ° along the parallels decreases towards the poles. At the equator, it can also be taken equal to 111 km. And at the poles 0 (because the pole is a point). Therefore, it is necessary to know the number of kilometers corresponding to the length of 1 ° of the arc of each particular parallel. To determine the distance in kilometers between two points lying on the same meridian, calculate the distance between them in degrees, and then multiply the number of degrees by 111 km. To determine the distance between two points on the equator, you also need to determine the distance between them in degrees, and then multiply by 111 km.

INTRODUCTION

The topographic map is reduced a generalized image of the area, showing the elements using a system of conventional signs.
In accordance with the requirements, topographic maps are highly geometric accuracy and geographic fit. This is provided by their scale, geodetic basis, map projections and a symbolic system.
The geometric properties of a cartographic image: the size and shape of areas occupied by geographical objects, the distances between individual points, directions from one to another - are determined by its mathematical basis. Mathematical basis maps include as components scale, a geodesic base, and a map projection.
What is the scale of the map, what types of scales are there, how to build a graphical scale and how to use the scales will be considered in the lecture.

6.1. TYPES OF SCALE OF TOPOGRAPHIC MAP

When compiling maps and plans, horizontal projections of segments are depicted on paper in a reduced form. The degree of such a decrease is characterized by scale.

map scale (plan) - the ratio of the length of the line on the map (plan) to the length of the horizontal laying of the corresponding terrain line

m = l K : d M

The scale of the image of small areas on the entire topographic map is almost constant. At small angles of inclination physical surface(on the plain) the length of the horizontal projection of the line differs very little from the length of the oblique line. In these cases, the length scale can be considered as the ratio of the length of the line on the map to the length of the corresponding line on the ground.

The scale is indicated on the maps in different options

6.1.1. Numerical scale

Numerical scale expressed as a fraction with a numerator equal to 1(aliquot fraction).

Or

Denominator M the numerical scale shows the degree of reduction in the lengths of the lines on the map (plan) in relation to the lengths of the corresponding lines on the ground. Comparing numerical scales, the largest is the one whose denominator is smaller.
Using numerical scale maps (plan), you can determine the horizontal distance dm lines on the ground

Example.
Map scale 1:50 000. The length of the segment on the map lk\u003d 4.0 cm. Determine the horizontal location of the line on the ground.

Solution.
Multiplying the value of the segment on the map in centimeters by the denominator of the numerical scale, we get the horizontal distance in centimeters.
d\u003d 4.0 cm × 50,000 \u003d 200,000 cm, or 2,000 m, or 2 km.

note to the fact that the numerical scale is an abstract quantity that does not have specific units of measurement. If the numerator of a fraction is expressed in centimeters, then the denominator will have the same units of measurement, i.e. centimeters.

for instance, a scale of 1:25,000 means that 1 centimeter of the map corresponds to 25,000 centimeters of terrain, or 1 inch of the map corresponds to 25,000 inches of terrain.

To meet the needs of the economy, science and defense of the country, maps of various scales are needed. For government topographic maps, forest management plans, forestry plans and forest plantations, standard scales are defined - scale range(Tables 6.1, 6.2).

Scale series of topographic maps

Table 6.1.

Numerical scale

Map name

1 cm card corresponds
on the ground distance

1 cm2 card corresponds
on the territory of the square

five thousandth

0.25 hectare

ten thousandth

twenty-five thousandth

6.25 hectares

fifty thousandth

hundred thousandth

two hundred thousandth

five hundred thousandth

millionth

Previously, this series included scales of 1:300,000 and 1:2,000.

6.1.2. Named Scale

named scale called the verbal expression of the numerical scale. Under the numerical scale on the topographic map there is an inscription explaining how many meters or kilometers on the ground corresponds to one centimeter of the map.

for instance, on the map under a numerical scale of 1:50,000 it is written: "in 1 centimeter 500 meters." The number 500 in this example is named scale value .
Using a named map scale, you can determine the horizontal distance dm lines on the ground. To do this, it is necessary to multiply the value of the segment, measured on the map in centimeters, by the value of the named scale.

Example. The named scale of the map is "2 kilometers in 1 centimeter". The length of the segment on the map lk\u003d 6.3 cm. Determine the horizontal location of the line on the ground.
Solution. Multiplying the value of the segment measured on the map in centimeters by the value of the named scale, we obtain the horizontal distance in kilometers on the ground.
d= 6.3 cm × 2 = 12.6 km.

6.1.3. Graphic scales

To avoid mathematical calculations and speed up work on the map, use graphic scales . There are two such scales: linear and transverse .

Linear scale

To build a linear scale, choose an initial segment that is convenient for a given scale. This original segment ( a) are called scale base (Fig. 6.1).



Rice. 6.1. Linear scale. Measured segment on the ground
will CD = ED + CE = 1000 m + 200 m = 1200 m.

The base is laid on a straight line the required number of times, the leftmost base is divided into parts (segment b), to be the smallest divisions of the linear scale . The distance on the ground that corresponds to the smallest division of the linear scale is called linear scale accuracy .

How to use a linear scale:

  • put the right leg of the compass on one of the divisions to the right of zero, and the left leg on the left base;
  • the length of the line consists of two counts: a count of whole bases and a count of divisions of the left base (Fig. 6.1).
  • If the segment on the map is longer than the constructed linear scale, then it is measured in parts.

Cross scale

For more accurate measurements, use transverse scale (Fig. 6.2, b).



Fig 6.2. Cross scale. Measured distance
PK = TK + PS + ST = 1 00 +10 + 7 = 117 m.

To build it on a straight line segment, several scale bases are laid ( a). Usually the length of the base is 2 cm or 1 cm. Perpendiculars to the line are set at the points obtained. AB and draw through them ten parallel lines at regular intervals. The leftmost base from above and below is divided into 10 equal segments and connected by oblique lines. The zero point of the lower base is connected to the first point WITH top base and so on. Get a series of parallel inclined lines, which are called transversals.
The smallest division of the transverse scale is equal to the segment C 1 D 1 , (fig. 6. 2, a). The adjacent parallel segment differs by this length when moving up the transversal 0C and vertical line 0D.
A transverse scale with a base of 2 cm is called normal . If the base of the transverse scale is divided into ten parts, then it is called hundreds . On a hundredth scale, the price of the smallest division is equal to one hundredth of the base.
The transverse scale is engraved on metal rulers, which are called scale.

How to use the transverse scale:

  • fix the length of the line on the map with a measuring compass;
  • put the right leg of the compass on an integer division of the base, and the left leg on any transversal, while both legs of the compass should be located on a line parallel to the line AB;
  • the length of the line consists of three counts: a count of integer bases, plus a count of divisions of the left base, plus a count of divisions up the transversal.

The accuracy of measuring the length of a line using a transverse scale is estimated at half the price of its smallest division.

6.2. VARIETY OF GRAPHIC SCALE

6.2.1. transitional scale

Sometimes in practice it is necessary to use a map or an aerial photograph, the scale of which is not standard. For example, 1:17 500, i.e. 1 cm on the map corresponds to 175 m on the ground. If you build a linear scale with a base of 2 cm, then the smallest division of the linear scale will be 35 m. Digitization of such a scale causes difficulties in the production of practical work.
To simplify the determination of distances on a topographic map, proceed as follows. The base of a linear scale is not taken to be 2 cm, but calculated so that it corresponds to a round number of meters - 100, 200, etc.

Example. It is required to calculate the length of the base corresponding to 400 m for a map at a scale of 1:17,500 (175 meters in one centimeter).
To determine what dimensions a segment of 400 m long will have on a 1:17,500 scale map, we draw up the proportions:
on the ground on the plan
175 m 1 cm
400 m X cm
X cm = 400 m × 1 cm / 175 m = 2.29 cm.

Having solved the proportion, we conclude: the base of the transitional scale in centimeters is equal to the value of the segment on the ground in meters divided by the value of the named scale in meters. The length of the base in our case
a= 400 / 175 = 2.29 cm.

If we now construct a transverse scale with a base length a\u003d 2.29 cm, then one division of the left base will correspond to 40 m (Fig. 6.3).


Rice. 6.3. Transitional linear scale.
Measured distance AC \u003d BC + AB \u003d 800 +160 \u003d 960 m.

For more accurate measurements on maps and plans, a transverse transitional scale is built.

6.2.2. Step scale

Use this scale to determine the distances measured in steps during eye survey. The principle of constructing and using the scale of steps is similar to the transitional scale. The base of the scale of steps is calculated so that it corresponds to the round number of steps (pairs, triplets) - 10, 50, 100, 500.
To calculate the value of the base of the steps scale, it is necessary to determine the survey scale and calculate the average step length Shsr.
The average step length (pairs of steps) is calculated from the known distance traveled in the forward and backward directions. By dividing the known distance by the number of steps taken, the average length of one step is obtained. When the earth's surface is tilted, the number of steps taken in the forward and reverse directions will be different. When moving in the direction of increasing relief, the step will be shorter, and in the opposite direction - longer.

Example. A known distance of 100 m is measured in steps. There are 137 steps in the forward direction and 139 steps in the reverse direction. Calculate the average length of one step.
Solution. Total covered: Σ m = 100 m + 100 m = 200 m. The sum of the steps is: Σ w = 137 w + 139 w = 276 w. The average length of one step is:

Shsr= 200 / 276 = 0.72 m.

It is convenient to work with a linear scale when the scale line is marked every 1 - 3 cm, and the divisions are signed with a round number (10, 20, 50, 100). Obviously, the value of one step of 0.72 m on any scale will have extremely small values. For a scale of 1: 2,000, the segment on the plan will be 0.72 / 2,000 \u003d 0.00036 m or 0.036 cm. Ten steps, on the appropriate scale, will be expressed as a segment of 0.36 cm. The most convenient basis for these conditions, according to the author, there will be a value of 50 steps: 0.036 × 50 = 1.8 cm.
For those who count steps in pairs, a convenient base would be 20 pairs of steps (40 steps) 0.036 × 40 = 1.44 cm.
The length of the base of the steps scale can also be calculated from proportions or by the formula
a = (Shsr × KSh) / M
where: Shsr - average value of one step in centimeters,
KSh - number of steps at the base of the scale ,
M - scale denominator.

The length of the base for 50 steps on a scale of 1:2,000 with a step length of 72 cm will be:
a= 72 × 50 / 2000 = 1.8 cm.
To build the scale of steps for the above example, it is necessary to divide the horizontal line into segments equal to 1.8 cm, and divide the left base into 5 or 10 equal parts.


Rice. 6.4. Step scale.
Measured distance AC \u003d BC + AB \u003d 100 + 20 \u003d 120 sh.

6.3. SCALE ACCURACY

Scale Accuracy (maximum scale accuracy) is a segment of the horizontal line, corresponding to 0.1 mm on the plan. The value of 0.1 mm for determining the accuracy of the scale is adopted due to the fact that this is the minimum segment that a person can distinguish with the naked eye.
for instance, for a scale of 1:10,000, the scale accuracy will be 1 m. In this scale, 1 cm on the plan corresponds to 10,000 cm (100 m) on the ground, 1 mm - 1,000 cm (10 m), 0.1 mm - 100 cm (1m). From the above example, it follows that if the denominator of the numerical scale is divided by 10,000, then we get the maximum scale accuracy in meters.
for instance, for a numerical scale of 1:5,000, the maximum scale accuracy will be 5,000 / 10,000 = 0.5 m

Scale accuracy allows you to solve two important problems:

  • determination of the minimum sizes of objects and terrain objects that are depicted at a given scale, and the sizes of objects that cannot be depicted at a given scale;
  • setting the scale at which the map should be created so that it depicts objects and terrain objects with predetermined minimum sizes.

In practice, it is accepted that the length of a segment on a plan or map can be estimated with an accuracy of 0.2 mm. The horizontal distance on the ground, corresponding to a given scale of 0.2 mm (0.02 cm) on the plan, is called graphic accuracy of scale . Graphical accuracy of determining distances on a plan or map can only be achieved using a transverse scale..
It should be borne in mind that when measuring the relative position of the contours on the map, the accuracy is determined not by the graphical accuracy, but by the accuracy of the map itself, where errors can average 0.5 mm due to the influence of errors other than graphical ones.
If we take into account the error of the map itself and the measurement error on the map, then we can conclude that the graphical accuracy of determining distances on the map is 5–7 worse than the maximum scale accuracy, i.e., it is 0.5–0.7 mm on the map scale.

6.4. DETERMINATION OF UNKNOWN MAP SCALE

In cases where for some reason the scale on the map is missing (for example, cut off when gluing), it can be determined in one of the following ways.

  • On the grid . It is necessary to measure the distance on the map between the lines of the coordinate grid and determine how many kilometers these lines are drawn through; This will determine the scale of the map.

For example, the coordinate lines are indicated by the numbers 28, 30, 32, etc. (along the western frame) and 06, 08, 10 (along the southern frame). It is clear that the lines are drawn through 2 km. The distance on the map between adjacent lines is 2 cm. It follows that 2 cm on the map corresponds to 2 km on the ground, and 1 cm on the map corresponds to 1 km on the ground (named scale). This means that the scale of the map will be 1:100,000 (1 kilometer in 1 centimeter).

  • According to the nomenclature of the map sheet. The notation system (nomenclature) of map sheets for each scale is quite definite, therefore, knowing the notation system, it is easy to find out the scale of the map.

A sheet of a map at a scale of 1:1,000,000 (millionth) is indicated by one of the letters of the Latin alphabet and one of the numbers from 1 to 60. The map notation system is more large scale is based on the nomenclature of sheets of a millionth map and can be represented by the following scheme:

1:1 000 000 - N-37
1:500 000 - N-37-B
1:200 000 - N-37-X
1:100 000 - N-37-117
1:50 000 - N-37-117-A
1:25 000 - N-37-117-A-g

Depending on the location of the map sheet, the letters and numbers that make up its nomenclature will be different, but the order and number of letters and numbers in the nomenclature of a map sheet of a given scale will always be the same.
Thus, if a map has the M-35-96 nomenclature, then by comparing it with the above diagram, we can immediately say that the scale of this map will be 1:100,000.
See Chapter 8 for details on card nomenclature.

  • By distances between local objects. If there are two objects on the map, the distance between which on the ground is known or can be measured, then to determine the scale, you need to divide the number of meters between these objects on the ground by the number of centimeters between the images of these objects on the map. As a result, we get the number of meters in 1 cm of this map (named scale).

For example, it is known that the distance from n.p. Kuvechino to the lake. Deep 5 km. Having measured this distance on the map, we got 4.8 cm. Then
5000 m / 4.8 cm = 1042 m in one centimeter.
Maps on a scale of 1:104 200 are not published, so we make rounding. After rounding, we will have: 1 cm of the map corresponds to 1,000 m of terrain, i.e., the map scale is 1:100,000.
If there is a road with kilometer posts on the map, then it is most convenient to determine the scale by the distance between them.

  • According to the length of the arc of one minute of the meridian . Frames of topographic maps along the meridians and parallels have divisions in minutes of the meridian and parallel arcs.

One minute of the meridian arc (along the eastern or western frame) corresponds to a distance of 1852 m (nautical mile) on the ground. Knowing this, it is possible to determine the scale of the map in the same way as by the known distance between two terrain objects.
for instance, the minute segment along the meridian on the map is 1.8 cm. Therefore, 1 cm on the map will be 1852: 1.8 = 1,030 m. After rounding, we get a map scale of 1:100,000.
In our calculations, approximate values ​​of the scales were obtained. This happened due to the approximation of the distances taken and the inaccuracy of their measurement on the map.

6.5. TECHNIQUE FOR MEASURING AND PUTTING DISTANCES ON A MAP

To measure distances on a map, a millimeter or scale ruler, a compass-meter is used, and a curvimeter is used to measure curved lines.

6.5.1. Measuring distances with a millimeter ruler

With a millimeter ruler, measure the distance between the given points on the map with an accuracy of 0.1 cm. Multiply the resulting number of centimeters by the value of the named scale. For flat terrain, the result will correspond to the distance on the ground in meters or kilometers.
Example. On a map of scale 1: 50,000 (in 1 cm - 500 m) the distance between two points is 3.4 cm. Determine the distance between these points.
Solution. Named scale: in 1 cm 500 m. The distance on the ground between the points will be 3.4 × 500 = 1700 m.
At angles of inclination of the earth's surface more than 10º, it is necessary to introduce an appropriate correction (see below).

6.5.2. Measuring distances with a compass

When measuring distance in a straight line, the needles of the compass are set at the end points, then, without changing the solution of the compass, the distance is read off on a linear or transverse scale. In the case when the opening of the compass exceeds the length of the linear or transverse scale, the integer number of kilometers is determined by the squares of the coordinate grid, and the remainder - by the usual scale order.


Rice. 6.5. Measuring distances with a compass-meter on a linear scale.

To get the length broken line sequentially measure the length of each of its links, and then summarize their values. Such lines are also measured by increasing the compass solution.
Example. To measure the length of a polyline ABCD(Fig. 6.6, a), the legs of the compass are first placed at points A and V. Then, rotating the compass around the point V. move the back leg from the point A exactly V" lying on the continuation of the line sun.
Front leg from point V transferred to a point WITH. The result is a solution of the compass B "C"=AB+sun. Moving the back leg of the compass in the same way from the point V" exactly WITH", and the front of WITH v D. get a solution of the compass
C "D \u003d B" C + CD, the length of which is determined using a transverse or linear scale.


Rice. 6.6. Line length measurement: a - broken line ABCD; b - curve A 1 B 1 C 1;
B"C" - auxiliary points

Long curves measured along the chords with compass steps (see Fig. 6.6, b). The step of the compass, equal to an integer number of hundreds or tens of meters, is set using a transverse or linear scale. When rearranging the legs of the compass along the measured line in the directions shown in fig. 6.6, b arrows, count the steps. The total length of the line A 1 C 1 is the sum of the segment A 1 B 1 equal to the step value multiplied by the number of steps, and the remainder B 1 C 1 measured on a transverse or linear scale.

6.5.3. Measuring distances with a curvimeter

Curved segments are measured with a mechanical (Fig. 6.7) or electronic (Fig. 6.8) curvimeter.


Rice. 6.7. Curvimeter mechanical

First, turning the wheel by hand, set the arrow to zero division, then roll the wheel along the measured line. The reading on the dial against the end of the arrow (in centimeters) is multiplied by the scale of the map and the distance on the ground is obtained. A digital curvimeter (Fig. 6.7.) is a high-precision, easy-to-use device. Curvimeter includes architectural and engineering functions and has a convenient display for reading information. This unit can process metric and Anglo-American (feet, inches, etc.) values, allowing you to work with any maps and drawings. You can enter the most commonly used type of measurement and the instrument will automatically translate scale measurements.


Rice. 6.8. Curvimeter digital (electronic)

To improve the accuracy and reliability of the results, it is recommended that all measurements be carried out twice - in the forward and reverse directions. In case of insignificant differences in the measured data, the arithmetic mean of the measured values ​​is taken as the final result.
The accuracy of measuring distances by these methods using a linear scale is 0.5 - 1.0 mm on a map scale. The same, but using a transverse scale is 0.2 - 0.3 mm per 10 cm of line length.

6.5.4. Converting horizontal distance to slant range

 It should be remembered that as a result of measuring distances on maps, the lengths of horizontal projections of lines (d) are obtained, and not the lengths of lines on the earth's surface (S) (Fig. 6.9).



Rice. 6.9. Slant Range ( S) and horizontal spacing ( d)

The actual distance on an inclined surface can be calculated using the formula:


where d is the length of the horizontal projection of the line S;
v - the angle of inclination of the earth's surface.

The length of the line on the topographic surface can be determined using the table (Table 6.3) of the relative values ​​of the corrections to the length of the horizontal distance (in%).

Table 6.3

Tilt angle

Rules for using the table

1. The first line of the table (0 tens) shows the relative values ​​of the corrections at angles of inclination from 0° to 9°, the second - from 10° to 19°, the third - from 20° to 29°, the fourth - from 30° up to 39°.
2. To determine the absolute value of the correction, you must:
a) in the table, by the angle of inclination, find the relative value of the correction (if the angle of inclination of the topographic surface is not given by an integer number of degrees, then the relative value of the correction must be found by interpolation between the tabular values);
b) calculate the absolute value of the correction to the length of the horizontal span (i.e., multiply this length by the relative value of the correction and divide the resulting product by 100).
3. To determine the length of a line on a topographic surface, the calculated absolute value of the correction must be added to the length of the horizontal distance.

Example. On the topographic map, the length of the horizontal laying is 1735 m, the angle of inclination of the topographic surface is 7°15′. In the table, the relative values ​​of the corrections are given for whole degrees. Therefore, for 7°15" it is necessary to determine the nearest larger and nearest smaller multiples of one degree - 8º and 7º:
for 8° relative correction value 0.98%;
for 7° 0.75%;
difference in tabular values ​​in 1º (60') 0.23%;
the difference between the specified angle of inclination of the earth's surface 7 ° 15 "and the nearest smaller tabular value of 7º is 15".
We make proportions and find the relative amount of the correction for 15 ":

For 60' the correction is 0.23%;
For 15′ the correction is x%
x% = = 0.0575 ≈ 0.06%

Relative correction value for tilt angle 7°15"
0,75%+0,06% = 0,81%
Then you need to determine the absolute value of the correction:
= 14.05 m approximately 14 m.
The length of the inclined line on the topographic surface will be:
1735 m + 14 m = 1749 m.

At small angles of inclination (less than 4° - 5°), the difference in the length of the inclined line and its horizontal projection is very small and may not be taken into account.

6.6. MEASUREMENT OF AREA BY MAP

The determination of the areas of plots from topographic maps is based on the geometric relationship between the area of ​​the figure and its linear elements. The area scale is equal to the square of the linear scale.
If the sides of a rectangle on the map are reduced by n times, then the area of ​​this figure will decrease by n 2 times.
For a map with a scale of 1:10,000 (in 1 cm 100 m), the area scale will be (1: 10,000) 2, or in 1 cm 2 there will be 100 m × 100 m = 10,000 m 2 or 1 ha, and on a map of scale 1 : 1,000,000 in 1 cm 2 - 100 km 2.

To measure areas on maps, graphic, analytical and instrumental methods are used. The use of one or another measurement method is determined by the shape of the measured area, the given accuracy of the measurement results, the required speed of obtaining data, and the availability of the necessary instruments.

6.6.1. Measuring the area of ​​a parcel with straight boundaries

When measuring the area of ​​a plot with rectilinear boundaries, the plot is divided into simple geometric figures, measure the area of ​​each of them in a geometric way and, summing up the areas of individual sections calculated taking into account the scale of the map, get the total area of ​​the object.

6.6.2. Measuring the area of ​​a plot with a curved contour

Object with curvilinear contour they are divided into geometric shapes, having previously straightened the boundaries in such a way that the sum of the cut-off sections and the sum of the excesses mutually compensate each other (Fig. 6.10). The measurement results will be approximate to some extent.

Rice. 6.10. Straightening curvilinear site boundaries and
breakdown of its area into simple geometric shapes

6.6.3. Measurement of the area of ​​a plot with a complex configuration

Measurement of plot areas, having a complex irregular configuration, more often produced using pallets and planimeters, which gives the most accurate results. grid palette is a transparent plate with a grid of squares (Fig. 6.11).


Rice. 6.11. Square Mesh Palette

The palette is placed on the measured contour and the number of cells and their parts inside the contour is counted. The proportions of incomplete squares are estimated by eye, therefore, to improve the accuracy of measurements, palettes with small squares (with a side of 2 - 5 mm) are used. Before working on this map, determine the area of ​​​​one cell.
The area of ​​the plot is calculated by the formula:

P \u003d a 2 n,

Where: a - the side of the square, expressed on the scale of the map;
n- the number of squares that fall within the contour of the measured area

To improve accuracy, the area is determined several times with an arbitrary permutation of the palette used in any position, including rotation relative to its original position. The arithmetic mean of the measurement results is taken as the final value of the area.

In addition to grid palettes, dot and parallel palettes are used, which are transparent plates with engraved dots or lines. Points are placed in one of the corners of the cells of the grid palette with a known division value, then the grid lines are removed (Fig. 6.12).


Rice. 6.12. dot palette

The weight of each point is equal to the price of the division of the palette. The area of ​​the measured area is determined by counting the number of points inside the contour, and multiplying this number by the weight of the point.
Equidistant parallel lines are engraved on the parallel palette (Fig. 6.13). The measured area, when applied to it with a palette, will be divided into a series of trapezoids with the same height h. Segments of parallel lines inside the contour (in the middle between the lines) are the middle lines of the trapezoid. To determine the area of ​​​​a plot using this palette, it is necessary to multiply the sum of all measured middle lines by the distance between the parallel lines of the palette h(taking into account the scale).

P = h∑l

Figure 6.13. Palette consisting of a system
parallel lines

Measurement areas of significant plots made on cards with the help of planimeter.


Rice. 6.14. polar planimeter

The planimeter is used to determine areas mechanically. The polar planimeter is widely used (Fig. 6.14). It consists of two levers - pole and bypass. Determining the contour area with a planimeter comes down to the following steps. After fixing the pole and setting the needle of the bypass lever at the starting point of the circuit, a reading is taken. Then the bypass spire is carefully guided along the contour to the starting point and a second reading is taken. The difference in readings will give the area of ​​the contour in divisions of the planimeter. Knowing the absolute value of the division of the planimeter, determine the area of ​​the contour.
The development of technology contributes to the creation of new devices that increase labor productivity in calculating areas, in particular, the use of modern devices, among which are electronic planimeters.


Rice. 6.15. Electronic planimeter

6.6.4. Calculating the area of ​​a polygon from the coordinates of its vertices
(analytical way)

This method allows you to determine the area of ​​a plot of any configuration, i.e. with any number of vertices whose coordinates (x, y) are known. In this case, the numbering of the vertices should be done in a clockwise direction.
As can be seen from fig. 6.16, the area S of the polygon 1-2-3-4 can be considered as the difference between the areas S "of the figure 1y-1-2-3-3y and S" of the figure 1y-1-4-3-3y
S = S" - S".



Rice. 6.16. To the calculation of the area of ​​a polygon by coordinates.

In turn, each of the areas S "and S" is the sum of the areas of trapezoids, the parallel sides of which are the abscissas of the corresponding vertices of the polygon, and the heights are the differences in the ordinates of the same vertices, i.e.

S "\u003d pl. 1u-1-2-2u + pl. 2u-2-3-3u,
S" \u003d pl 1y-1-4-4y + pl. 4y-4-3-3y
or: 2S " \u003d (x 1 + x 2) (y 2 - y 1) + (x 2 + x 3) (y 3 - y 2)
2S " \u003d (x 1 + x 4) (y 4 - y 1) + (x 4 + x 3) (y 3 - y 4).

In this way,
2S= (x 1 + x 2) (y 2 - y 1) + (x 2 + x 3) (y 3 - y 2) - (x 1 + x 4) (y 4 - y 1) - (x 4 + x 3) (y 3 - y 4). Expanding the brackets, we get
2S \u003d x 1 y 2 - x 1 y 4 + x 2 y 3 - x 2 y 1 + x 3 y 4 - x 3 y 2 + x 4 y 1 - x 4 y 3

From here
2S = x 1 (y 2 - y 4) + x 2 (y 3 - y 1) + x 3 (y 4 - y 2) + x 4 (y 1 - y 3) (6.1)
2S \u003d y 1 (x 4 - x 2) + y 2 (x 1 - x 3) + y 3 (x 2 - x 4) + y 4 (x 3 - x 1) (6.2)

Let us represent expressions (6.1) and (6.2) in general form, denoting by i the ordinal number (i = 1, 2, ..., n) of the vertices of the polygon:
(6.3)
(6.4)
Therefore, twice the area of ​​the polygon is equal to either the sum of the products of each abscissa and the difference between the ordinates of the next and previous vertices of the polygon, or the sum of the products of each ordinate and the difference of the abscissas of the previous and subsequent vertices of the polygon.
An intermediate control of calculations is the satisfaction of the following conditions:

0 or = 0
Coordinate values ​​and their differences are usually rounded to tenths of a meter, and products to whole square meters.
Complex formulas for calculating the area of ​​\u200b\u200bthe plot can be easily solved using spreadsheets MicrosoftXL. An example for a polygon (polygon) of 5 points is given in tables 6.4, 6.5.
In table 6.4 we enter the initial data and formulas.

Table 6.4.

y i (x i-1 - x i+1)

Double area in m2

SUM(D2:D6)

Area in hectares

In table 6.5 we see the results of the calculations.

Table 6.5.

y i (x i-1 -x i+1)

Double area in m2

Area in hectares

6.7. EYE MEASUREMENTS ON THE MAP

In the practice of cartometric work, eye measurements are widely used, which give approximate results. However, the ability to visually determine distances, directions, areas, steepness of the slope and other characteristics of objects on the map contributes to mastering the skills of correctly understanding the cartographic image. The accuracy of eye measurements increases with experience. Eye skills prevent gross miscalculations in instrument measurements.
To determine the length of linear objects on the map, one should visually compare the size of these objects with segments of a kilometer grid or divisions of a linear scale.
To determine the areas of objects, squares of a kilometer grid are used as a kind of palette. Each square of the grid of maps of scales 1:10,000 - 1:50,000 on the ground corresponds to 1 km 2 (100 ha), scale 1:100,000 - 4 km 2, 1:200,000 - 16 km 2.
The accuracy of quantitative determinations on the map, with the development of the eye, is 10-15% of the measured value.

Video

Scaling tasks
Tasks and questions for self-control
  1. What elements does it include mathematical basis kart?
  2. Expand the concepts: "scale", "horizontal distance", "numerical scale", "linear scale", "scale accuracy", "scale bases".
  3. What is a named map scale and how do you use it?
  4. What is the transverse scale of the map, for what purpose is it intended?
  5. What transverse map scale is considered normal?
  6. What scales of topographic maps and forest management tablets are used in Ukraine?
  7. What is a transitional map scale?
  8. How is the base of the transitional scale calculated?
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