How to determine the distance along a parallel. Determining distances on the map. Tasks for determining distances using a degree grid
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Twitterü Partial area scale (p).
ü Area distortion (vp).
ü Largest scale (a).
ü Smallest scale (b).
ü Maximum distortion angle (w).
ü Shape distortion coefficient (k).
During the course work the following notations were used:
n – parallel scale;
m – scale along the meridian;
e – deviation of angle t from 90°;
t is the angle between the meridian and the tangent to the parallel;
l1 – length of the meridian in the selected trapezoid on the map;
L1 – length of the meridian in the selected trapezoid on the ground;
l2 – length of the parallel in the selected trapezoid on the map;
L2 – length of the parallel in the selected trapezoid on the ground.
The partial scale of the area is determined by the formula:
Where ;
;
Area distortion
.
The largest and smallest scales are determined from the system:
;
where a is the largest scale;
b – smallest scale.
Maximum distortion angle:
Shape distortion coefficient:
1. Let's select point A on the map. Let's limit the area relative to point A from 34° to 36° in longitude, from 58° to 60° in latitude.
Determination of meridian and parallel lengths
2. Determined the scale along the meridian. The scale along the meridian was calculated using the formula:
where l1 is the length of the meridian in mm;
m – map scale denominator;
L1 – arc length of the corresponding meridian along the surface of the ellipsoid.
where Li are the lengths of meridian arcs of 1° latitude
L1 = 222794 m = 222794 ´103 mm
m == 
= 1,000925.
3. Determined the scale by parallel
where l2 is the length of the parallel in mm;
L2 – length of the corresponding parallel on the surface of the ellipsoid (L2 = LjА´Dl)
LjA – parallel length in m corresponds to 1° at latitude jA
Dl – the length of the parallel in degrees is equal to the difference in longitude between the eastern and western meridians.
L2 = 57476 m ´ 2 = 114952 m = 114952 ´103 mm
n == 
= 0,991718.
4. On the map, we measured the angle t (the angle between the meridian and the parallel) with a protractor, and determined the deviation of the angle t from 90° using the formula:
e = 90° – t (3)
e = 90° – 89°59¢ = 0°01¢
5. Calculate the scale of the area:
p = m ´ n ´ cose (4)
where m is the scale along the meridian (1)
n – parallel scale (2)
e – deviation of angle t from 90° (3)
p = 1.000925 ´ 0.991718 ´ cos 0°01¢ = 0.992635
6. We determined the greatest distortion of angles at point A using the formula:
where a – b = 
a+b= 
a – b = = 0.009207
a + b = = 1.992643
7. We calculated the coefficient of distortion of shapes using the formula
For a normal conic projection with one main parallel, the value of m, n partial scales and the area scale p are calculated using the following formula:
where mо= 1000000 (map scale denominator),
r – radii of parallels.
The calculation results are presented in the table in Form 6.
Calculation of length and area scales for a normal conic projection with one main parallel
Based on the found length and area scales, scale change curves m=n, p were constructed.
Graph of length and area scales in normal conformal conic projection
2.4 Contents and purpose of the map
To compile a map at a scale of 1:1000000, topographic maps of different scales are used. It is most convenient to use sheets of a geographical map at a scale of 1:1000000.
When performing this course work, a map of the Vologda region at a scale of 1:1000000 is used as a cartographic source.
The cartographic image includes physical-geographical and socio-economic objects of the map content.
Physiographic objects include:
ü hydrography;
ü relief;
ü vegetation;
Scale is the ratio of the length of a line on a drawing, plan or map to the length of the corresponding line in reality. The scale shows how many times the distance on the map is reduced relative to the actual distance on the ground. If, for example, the scale of a geographic map is 1: 1,000,000, this means that 1 cm on the map corresponds to 1,000,000 cm on the ground, or 10 km. There are 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 scale is 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 are marked on a scale ruler is called the scale base. In our country, the base of the scale is taken to be 1 cm. The number of meters or kilometers corresponding to the base of the scale is called the scale value. When constructing a linear scale, the number 0, from which the divisions begin, is usually placed not at the very end of the scale line, but retreated 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 scale accuracy, and 0.1 mm corresponds to the maximum scale accuracy. A linear scale, compared to a numerical scale, has the advantage that it makes it possible to determine the actual distance on a plan and map without additional calculations.
Named scale– scale expressed in words, for example, 1 cm 75 km. (Fig. 5).
Measuring distances on a map and plan. Measuring distances using a scale. You need to draw a straight line (if you need to find out the distance in a straight line) between two points and use a ruler to measure this distance in centimeters, and then multiply the resulting number by the scale value. For example, on a map of scale 1: 100,000 (1 cm in 1 km) the distance is 5 cm, i.e. on the ground this distance is 1х5 = 5 (km). You can also measure distance on a 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 values: the arc length of 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 (since a pole is a point). Therefore, it is necessary to know the number of kilometers corresponding to the length of 1° arc of each specific 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.
Main scale. You first became acquainted with the countries of the world in elementary school using a map of the hemispheres. In the geographical atlas where this map is placed, its scale is indicated: 1 cm is 900 km. Let's check it out. On one of the hemispheres, we measure the distance along the equator or along the middle meridian. It is 20 cm. This same distance is actually 20,000 km. This means that the scale of the map will be: 1 cm 1000 km. How can we explain this discrepancy?
For the convenience of the cartographer, the concept of “main scale” was introduced, which refers to certain projection locations. Such places can be points or lines of tangency of surfaces onto which a degree grid is projected from the globe onto the map. For a hemispheric projection, the tangent point, called the point of zero distortion, is at the center of the circle. We will not be able to determine the scale directly at a point, but we can do this over a short distance in the area of this point. To do this, we measure here the length of the equatorial arc of 20°. It turned out to be equal to 2.5 cm. In reality, this arc is 2220 km (20° X 111 km). Let's divide this distance by 2.5 cm, and we get a scale value approximately equal to that indicated on the map (1 cm is 900 km).
The question of scale is very important and interesting, and we will look at it in more detail, using the one we are already familiar with. All three maps shown on it are drawn up in cylindrical projections, and they are characterized by the cylinder touching the equator. Consequently, the equator will be the main scale for our maps. It is not difficult to guess that in this case all maps have the same main scale, since the intervals between 10-degree meridians are equal everywhere and amount to 4 mm. It is also easy to determine the magnitude of the main scale. We know that a 10° arc of the equator on the globe is 1110 km. This distance corresponds to a segment on the map equal to 0.4 cm. This means that 1 cm of the map contains 2780 km (1110: 0.4) and the numerical scale will be expressed by the ratio 1:278,000,000.
In addition to the main scale, each map has private scales. On the map in a square projection (Fig. 27, b), the partial scale along all meridians is the same throughout. On a map in an equiangular projection (Fig. 27, c), it will gradually increase from the equator to the pole, and on a map in an equal-area projection (Fig. 27, a), on the contrary, it will decrease. The partial scale of parallels on all three maps increases sharply as they approach the pole, and at the pole itself it is pointless to use it, because the point denoting the pole has “stretched” over the entire width of the earth’s surface.
Let's determine the private scales for our maps along the 60th parallel. To solve such a problem, you need to know the lengths of parallel arcs at different latitudes. We take their values in 1° from . The length of an arc of 10° will be 10 times greater and at a latitude of 60° it will be 558 km.
The partial scale along the 60th parallel on all three maps will be the same, because the segments of parallels concluded between the meridians are equal and correspond in the same way as along the equator, 0.4 cm. Let us divide the actual distance by this segment and get the value scale equal to approximately 1390 km per 1 cm (558:0.4), i.e. the scale will be 2 times larger than the main one. This way you can determine the partial scale when it remains constant along the entire line. If the scale is constantly changing, then we will get only its average value. For example, on a map in a conformal projection (Fig. 27, c) the segment between the 60th and 70th parallels is 2 times larger than that of the equator. This means that in this segment the average scale is 2 times larger than the main one.
Rice. thirty. Hemisphere maps with the same major scale
Two maps of the same scale. In cartographic practice, the term “medium scale” is not accepted and only the main one is labeled on all maps. For those who use a map, the main scale is not always clear, since it often does not express the overall scale of the image. Let us turn to Figure 30, which shows the hemisphere in two projections. According to the type of geometric surface on which the globe mesh is projected, both projections are transverse azimuthal, and according to the type of distortion, one of them is equiangular, and the second is arbitrary. The diameter of the hemisphere in the first projection is twice as large as in the second. And yet their main scale is the same. It's hard to believe, but it's true. Let us provide evidence.
In azimuthal transverse projections, the map grid is transferred to a plane tangent to a certain point on the equator, which is the point of zero distortion. It is for this reason that the main scale is written on the map. Its value can be determined as follows.
Let's take a map grid cell located in the area of the point of zero distortion. To a first approximation, it has the shape of a square and its dimensions in both projections are approximately the same. Let's measure some side of the square, for example, the one that makes up the arc of the equator with a difference in longitude of 20°. It turned out to be equal to 0.5 cm in both projections. Its actual distance along the equator is 2220 km. This means that the scale in the central part of both projections will be equal to 1:444,000,000, or 4440 km in 1 cm (2220:0.5).
It's not surprising, though. the scale labeled on these maps (the main scale) will be the same, despite the different sizes of the hemispheres.
Universal scale. Maps usually show not only a numerical scale, but also a linear scale in the form of a graphic scale. It is clear that for a map of a certain scale a corresponding scale is built. Is it possible to build one graph that can be used for maps of different scales? Let's try to do this.
Rice. 31. Universal scale
Let us draw two mutually perpendicular axes and plot a segment BC equal to 10 cm along the vertical axis upward, and a segment BA equal to 2.5 cm along the horizontal axis to the left (Fig. 31). (We will consider this last segment to be the base of a linear scale for a map of 1:20,000,000. On this scale, it will correspond to 500 km. To find the distance CE from which the base of the next scale (1:25,000,000) needs to be set aside, you need to use the relationship. obtained from the similarity of triangles ABC and DEC: CB/AB = CE/DE; CE = (CB x DE)/AB.
The value DE - the base of the linear scale - for a map scale of 1:25,000,000 will be equal to 2 cm (500 km: 25,000,000), and CE - 8 cm. In the same way, the distances from point C to the lines where the bases of the linear lines will be built are calculated scales of other maps.
The graph we constructed can be used not only to measure distances on maps of different scales, but also to determine the partial or average scale of the map along any meridian and any parallel. The scale of the map along the meridian is determined as follows. Using a measuring compass, let’s take from the map a segment of the meridian with a latitude difference of 10°, which will correspond to a distance of 1110 km. We draw this compass solution according to our graph along parallel lines until it fits within a distance of 1110 km. In our case, the taken segment MN fell within the distance of 1110 km between the lines of scales 1:25,000,000 and 1:30,000,000 (closer to 1:30,000,000). This means that the partial scale of the map along this meridian is equal to 1:28,000,000.
To determine the map scale by parallel, you must first find from Table 1 the length of the parallel arc of 10° at a certain latitude, and then the procedure will be the same as when determining the map scale by meridian.
The best option. When a problem has too many solutions, the question always arises whether it is possible to choose the best one. In 1856, the Russian mathematician P. L. Chebyshev posed and solved the following theorem for geographic maps: find the most similar image of a given country so that the scale distortion is minimal. Without proof, he said that this requires that the scale at all points of the country's border be the same. P. L. Chebyshev died without publishing his theorem.
For many years, mathematicians around the world searched for this proof and, in the end, began to doubt the correctness of the statement. Only in 1896, the Russian scientist D. A. Grave was able to restore Chebyshev’s proof.
A cartographic projection that satisfies the stated condition can be created only in the case when the northern and southern borders of the country run along parallels, and the western and eastern borders along meridians. In practice this does not happen. The borders of countries usually follow curves, or broken lines, that do not coincide with parallels and meridians. Nevertheless, for each country it is possible to create a projection that comes quite close to our condition.
The idea of P. L. Chebyshev found practical implementation in the compilation of maps of the USSR. Such maps are usually drawn up in a conical projection with the condition of maintaining scale along all meridians and two parallels, one of which crosses the southern border of the country, and the second passes several degrees south of the coast of the Arctic Ocean. It turns out that the cone does not touch the globe, but cuts it along two given parallels: 47 and 62°.
You may have a question: why does the northern parallel of the section, like the southern parallel, not cross the country’s border, but is located south of it? It's not hard to guess what's going on here. The transfer of the parallel of tangency to the south is due to the fact that the northern outskirts of our country are poorly populated, and therefore preference for the accuracy of the cartographic image is given to places that are more populated.
how to determine distance by parallels? how to determine the distance from parallels in the atlas? and got the best answer
Answer from Nat f[newbie]
Using a ruler, the distance from point “A” to point “B” is measured, the resulting distance is multiplied by the scale and the distance on the ground is obtained,
Using a compass, install a small solution between the legs of the measuring compass, then move the compass along the line being measured. Multiply the number of permutations of the compass by the distance taken between the needles. Then multiply this number by the scale.
For example, the distance between Kiev and St. Petersburg, located approximately on the 30° meridian, is 111 km * 9.5° = 1054 km; distance between Kiev and Kharkov (approximately parallel 50°) – 71 km * 6° = 426 km.
Source:
Answer from Marina Cherentseva[active]
what have the excellent students come to!
Answer from Beykut Balgysheva[active]
The Earth's meridians are semicircles or arcs that contain 180 degrees (the entire circle is 360) or 20,000 km. (the circumference of the Earth is 40,000 km), then 1 degree of the meridian is approximately 111 km. (40,000 km divided by 360 degrees) - knowing the distance in meridian degrees, you can calculate the distance in kilometers by multiplying this distance by 111 km.
Parallels are circles whose radii decrease towards the poles; at different parallels the value of 1 degree in kilometers is not the same. To determine the distance in kilometers on a map or globe between two points located on the same meridian, the number of degrees between points is multiplied by 111 km. To determine the distance in kilometers between points lying on the same parallel, the number of degrees is multiplied by the length of the arc of 1° parallel, indicated on the map or determined from tables.
Length of arcs of parallels and meridians on Krasovsky's ellipsoid
Answer from Alexander Silin[newbie]
A
Answer from 3 answers[guru]
Hello! Here is a selection of topics with answers to your question: how to determine distance from parallels? how to determine the distance from parallels in the atlas?
MAP 2014
1.Concept. MAP - This is a reduced generalized image of a large area of land constructed in a cartographic projection in small and medium using conventional symbols.
2. map signs .
The curvature of the earth is taken into account, there is distortion, there is a degree network - large areas of the earth are depicted
Conventional signs are given in a generalized manner (generalization), do not resemble real objects, medium and small scale
3. map projections - these are mathematical methods for depicting a spherical Surface on a plane
Types of projection along an auxiliary surface
TYPES OF CARDS
DETERMINATION OF DISTANCES, HEIGHTS, DEPTH, DIRECTIONS BY MAPS
DEGREE NETWORK
1.Concept- a system of meridians, parallels on maps and globes, used to determine the geographic coordinates of an object
2. reason for existence- rotation of a spherical earth around its axis, resulting in the formation of two fixed points - poles, through which a system of meridians and parallels is drawn.
3. pole characteristics - these are mathematically calculated points of intersection of an imaginary axis with the earth's surface. There is a north and a south pole.
4. characteristics of meridians - this is the imaginary shortest line drawn between the north and south poles.
5 Characteristics of parallels - this is an imaginary line drawn at the same distance parallel to the equator
6. latitude characteristic- this is the distance from the equator to a given object expressed in degrees
7. longitude characteristic- this is the distance from the prime meridian to a given object expressed in degrees.
8. meaning - determination of coordinates and distances.
TASKS
TASKS FOR DETERMINING DISTANCES ON A DEGREE GRID
| Along the meridians | |
| (After 10°,20…..) | |
| 111 km. | |
| By parallels | |
| (After 10°,20…..) | |
| 3. Find in kilometers the length of an arc of 1° along a given parallel 0° – 111.3 km 10° – 109.6 km 20° – 104.6 km 30° – 96.5 km 40° – 85.3 km | 50° – 71.1 km 60° – 55.8 km 70° – 38.2 km 80° – 19.8 km 90° – 0 km |
| Along the meridians between points 1-2 | |
| 1. First, determine how many degrees the meridians are drawn through on a given map | In 20 |
| 2. Calculate the distance in degrees between objects, counting degree cells or the difference in longitude | 1 cell = 20 degrees T1 lies at 40 west. T2 lies at 20 west. 40-20=20 degrees |
| 3. Remember what the length of an arc of 1° along the meridian is equal to in kilometers | 111 km. |
| 4.Multiply the given distance in degrees between objects by 111 km | 20 times 111km=2220km |
| Along parallels between points 1-3 | |
| 1. First, determine how many degrees the parallels are drawn on the maps of the hemispheres | After 20 Latitude 40 N. |
| 2. Calculate the distance in degrees by counting degree cells or the difference in latitude | 2 cell=40 degrees |
| 3. Find the length of an arc of 1° along a given parallel in kilometers | 20° – 104.6 km |
| 4. Multiply the given distance in degrees between objects by the length of an arc of 1° along a given parallel | 40 times 104.6 km= |
| | | next lecture ==> | |
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