Main scale. You first met the countries of the world in primary school on the map of the hemispheres. In the geographical atlas where this map is placed, its scale is indicated: 900 km in 1 cm. 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. The same distance is actually 20,000 km. This means that the scale of the map will be: 1 cm 1000 km. How can such a discrepancy be explained?
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 contact with surfaces on which a degree grid is projected from the globe to the map. For a hemispheric projection, the point of contact, called the zero distortion point, is at the center of the circle. We will not be able to determine the scale directly at the point, but we can do it over a short distance around this point. To do this, we measure here the length of the arc of the equator at 20 °. It turned out to be equal to 2.5 cm. In nature, this arc is 2220 km (20 ° X 111 km). Divide this distance by 2.5 cm, and we get a scale value approximately equal to that signed on the map (1 cm 900 km).
The question of scale is very important and interesting, and we will consider it in more detail, using the already familiar to us. All three maps shown on it are made in cylindrical projections, and they are characterized by touching the cylinder along the equator line. Therefore, the main scales for our maps will be considered along the equator. It is easy to guess that in this case all the maps have the same main scale, since the gaps between the 10-degree meridians are everywhere equal and equal to 4 mm. It is also easy to determine the value of the main scale. We know that the equatorial arc of 10° on the globe equal to 1110 km. This distance corresponds on the map to a segment 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 as a ratio of 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 for all meridians throughout the entire length is the same. On a map in a conformal 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 along the parallels on all three maps increases sharply as they approach the pole, and it makes no sense to use it at the pole itself, because the point denoting the pole has "stretched" over the entire width of the earth's surface.
Let's define private scales for our maps along the 60th parallel. To solve such a problem, you need to know the lengths of the arcs of parallels 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° will be 558 km.
The partial scale along the 60th parallel on all three maps will be the same, because the segments of the parallels enclosed between the meridians are equal and correspond in the same way as along the equator, 0.4 cm. Let's divide the actual distance by this segment and get the value scale equal to approximately 1390 km in 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 on this segment the average scale is 2 times larger than the main one.
Rice. thirty. Hemispheric maps with the same main scale
Two maps of the same scale. In cartographic practice, the term "medium scale" is not accepted and only the main one is signed on all maps. For those who use the map, the main scale is not always clear, as it often does not express the overall scale of the image. Let's turn to Figure 30, which shows the hemisphere in two projections. According to the type of geometric surface on which the globe grid 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's give proof.
In azimuth transverse projections the cartographic grid is transferred to a plane tangent at a certain point on the equator, which is the point of zero distortion. For her, they sign the main scale on the map. Its value can be determined as follows.
Let's take a cell of the cartographic grid, located in the region of the zero distortion point. In the first approximation, it has the shape of a square and its dimensions in both projections are approximately the same. Let us measure some side of the square, for example, the one that makes up the arc of the equator with a longitude difference 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 (2220:0.5) in 1 cm.
How unsurprising, though. the scale signed on these maps (the main scale) will be the same, despite the different sizes of the hemispheres.
Universal Scale. On maps, not only numerical, but also a linear scale is usually given in the form of a graphical scale. It is clear that for a map of a certain scale, an appropriate scale is built. Is it possible to build one graph that can be used for maps of different scales? Let's try to do it.
Rice. 31. Universal Scale
Let us draw two mutually perpendicular axes and set aside along the vertical axis a segment BC equal to 10 cm, and along the horizontal axis to the left a segment BA equal to 2.5 cm (Fig. 31). (We will consider this last segment as the base of the linear scale for the map 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 plotted), you need to use the ratio. derived 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 scales of other maps.
The chart we have built can be used not only to measure distances on maps of different scales, but also to determine the private or average scale of a map along any meridian and any parallel. The scale of the map along the meridian is determined as follows. Let us take from the map with a measuring compass a segment of the meridian with a latitude difference of 10 °, which will correspond to a distance of 1110 km. This compass opening is carried out according to our schedule along parallel lines until it fits within a distance of 1110 km. In our case, the taken segment MN fit into the distance of 1110 km between the scale lines 1:25,000,000 and 1:30,000,000 (closer to 1:30,000,000). This means that the private scale of the map along this meridian turned out to be 1:28,000,000.
To determine the scale of the map along the parallel, you must first find the length of the parallel arc of 10 ° at a certain latitude from Table 1, and then the procedure will be the same as when determining the scale of the map along the meridian.
The best option. When a problem has too many solutions, the question always arises as to whether the best one can be chosen from it. In 1856, the Russian mathematician P. L. Chebyshev formulated and solved the following theorem for geographical maps: find the most similar image of a given country so that the scale distortion is minimal. Without proof, he said that for this it is necessary that the scale at all points of the country's border be the same. PL Chebyshev died without publishing his theorem.
For many years, mathematicians around the world have been looking for this proof and, in the end, began to doubt the correctness of the statement. Only in 1896 did the Russian scientist D. A. Grave manage to restore Chebyshev's proof.
A cartographic projection that satisfies the set condition can be created only if the northern and southern borders of the country pass along parallels, and the western and eastern borders - along the meridians. Practically this does not happen. The borders of countries usually pass along curved or broken lines that do not coincide with parallels and meridians. Nevertheless, for each country it is possible to make a projection that is quite close to our condition.
The idea of P. L. Chebyshev found practical implementation in the preparation of maps of the USSR. Such maps are usually compiled in a conic 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 a few 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 °.
Perhaps you have a question: why does the northern parallel of the section, like the southern one, not cross the border of the country, but is located south of it? It's not hard to guess what's going on here. The transfer of the touch parallel to the south is due to the fact that the northern outskirts of our country are poorly inhabited, and therefore preference is given to places that are more populated in the accuracy of the cartographic image.
MAP 2014
1.Concept. MAP - This is a reduced generalized image of a large piece of land built in a cartographic projection in small and medium using conventional signs.
2. map features .
The curvature of the earth is taken into account, - there is a distortion, - there is a degree network - large areas of the earth are depicted
Conventional signs are given in a generalized way (generalization), they do not look like real objects, - medium and small scale
3. map projections - these are mathematical ways of representing a spherical surface on a plane
Types of projection on the auxiliary surface
TYPES OF CARDS
DETERMINATION ON THE MAP OF DISTANCES, HEIGHTS, DEPTH, DIRECTIONS
DEGREE NETWORK
1. Concept- a system of meridians, parallels on maps and globes, used to determine the geographical coordinates of an object
2. reason for existence- rotation of the spherical earth around its axis, as a result of which two fixed points are formed - poles, through which a system of meridians and parallels is drawn.
3. characteristic of poles - these are mathematically calculated points of intersection of an imaginary axis with earth's surface. There is a north and south pole.
4. characteristics of the meridians - this is an imaginary shortest line drawn between the north and south poles.
5 Characteristics of parallels - is an imaginary line drawn at equal distances parallel to the equator
6. latitude characteristic- is the distance from the equator to the given object expressed in degrees
7. longitude characteristic- this is the distance from the zero meridian to the given object expressed in degrees.
8. meaning - determination of coordinates and distances.
TASKS
TASKS FOR DETERMINING DISTANCES ON THE DEGREE GRID
| along the meridians | |
| (After 10°,20…..) | |
| 111 km. | |
| Parallels | |
| (After 10°,20…..) | |
| 3. Find in kilometers the arc length of 1 ° along this 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 on this map | Through20 |
| 2. Calculate the distance in degrees between objects, counting the degree cells or the difference in longitude | 1 cell = 20 degrees T1 lies at 40 W. T2 lies at 20 W. 40-20=20 degrees |
| 3. Remember what the length of an arc of 1 ° along the meridian is in kilometers | 111 km. |
| 4. Multiply this distance in degrees between objects by 111 km | 20 times 111km=2220km |
| Along the parallels between points 1-3 | |
| 1. First, determine how many degrees the parallels are drawn on the maps of the hemispheres | Through 20 Latitude 40 N.S. |
| 2. Calculate the distance in degrees counting, degree cells or difference in latitude | 2 cells=40 degrees |
| 3. Find in kilometers the length of an arc of 1 ° along a given parallel | 20° - 104.6 km |
| 4. Multiply the given distance in degrees between objects by the arc length of 1° along the given parallel | 40 times 104.6 km = |
| | | next lecture ==> | |
scale called the ratio of the length of the line in the 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 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 scale denominator, 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 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, the number 0, from which the counting 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 75 km. (Fig. 5).
Measuring distances on the map and plan. Measuring distances using 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 map with a scale of 1: 100,000 (in 1 cm 1 km), the distance is 5 cm, i.e. on the ground this distance is 1x5 = 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, the following quantities can be used: the length of an arc of 1° of the meridian and 1° of the 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.
ü Private area scale (p).
ü Area distortion (vp).
ü Largest scale (a).
ü Smallest scale (b).
ü Maximum distortion angle (w).
ü Form distortion coefficient (k).
During term paper the following designations are used:
n is the scale along the parallel;
m is the scale along the meridian;
e is the deviation of the angle t from 90°;
t is the angle between the meridian and the tangent to the parallel;
l1 is the length of the meridian in the selected trapezoid on the map;
L1 - the length of the meridian in the selected trapezoid on the ground;
l2 is the length of the parallel in the selected trapezoid on the map;
L2 is the length of the parallel in the selected trapezoid on the ground.
The private 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 is the smallest scale.
Maximum distortion angle:
Shape Distortion Factor:
1. Let's choose point A on the map. Let's limit the area relative to point A in longitude from 34 ° to 36 °, in latitude from 58 ° to 60 °.
Determining Meridian and Parallel Lengths
2. Determine the scale along the meridian. The scale along the meridian was calculated by the formula:
where l1 is the meridian length in mm;
m is the map scale denominator;
L1 is the length of the arc of the corresponding meridian along the surface of the ellipsoid.
where Li are the lengths of meridian arcs in 1° latitude
L1 = 222794 m = 222794 ´103 mm
m == 
= 1,000925.
3. Determine the scale along the parallel
where l2 is the length of the parallel in mm;
L2 is the length of the corresponding parallel on the surface of the ellipsoid (L2 = LjА´Dl)
LjA - the length of the parallel in m corresponds to 1° at latitude jA
Dl is the length of the parallel in degree measure equal to the difference in longitude between the east and west meridians.
L2 = 57476 m ´ 2 = 114952 m = 114952 ´103 mm
n == 
= 0,991718.
4. On the map, the protractor measured the angle t (the angle between the meridian and the parallel), determined the deviation of the angle t from 90 ° by 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 - scale along the parallel (2)
e – angle deviation t from 90° (3)
p = 1.000925 ´ 0.991718 ´ cos 0°01¢ = 0.992635
6. We determined the greatest distortion of the angles at point A by the formula:
where a – b = 
a+b= 
a – b = = 0.009207
a + b == 1.992643
7. We calculated the coefficient of distortion of forms by the formula
For a normal conic projection with one principal parallel, the value m, n of the partial scales and the area scale p are calculated by the following formula:
where mo = 1000000 (map scale denominator),
r are the radii of the parallels.
the calculation results are presented in the table in form 6.
Calculation of length and area scales for normal conic projection with one main parallel
On the basis of the found scales of lengths and areas, the curves of change of scales m=n, p are constructed.
Plot of length and area scales in normal conformal conic projection
2.4 Content and purpose of the map
To draw up a map at a scale of 1:1000000, topographic maps different scales. 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 Oblast at a scale of 1: 1,000,000 is used as a cartographic source.
The cartographic image includes physical-geographical and socio-economic objects of the map content.
The physical-geographical objects include:
ü hydrography;
ü relief;
the vegetation;
how to determine the distance by parallels? how to determine the distance by parallels in the atlas? and got the best answer
Answer from Nat f[newbie]
With the help of 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 measured line. Multiply the number of permutations of the compass by the distance taken between the needles. Then this number is multiplied by the scale.
For example, the distance between Kyiv and St. Petersburg, located approximately on the 30° meridian, is 111 km *9.5° = 1054 km; the distance between Kyiv and Kharkov (approximately 50° parallel) is 71 km * 6° = 426 km.
Source:
Answer from Marina Cherentseva[active]
What have the greats come to!
Answer from Beykut Balgysheva[active]
The meridians of the Earth are semicircles or arcs that contain 180 degrees, (the whole circumference 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; on 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 the 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 an arc of 1 ° parallel, indicated on the map or determined from the tables.
The length of the arcs of parallels and meridians on the Krasovsky ellipsoid
Answer from Alexander Silin[newbie]
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Answer from 3 answers[guru]
Hello! Here is a selection of topics with answers to your question: how to determine the distance by parallels? how to determine the distance by parallels in the atlas?