Degree network and its elements. Degree network and its elements 1 degree of the length of the meridian arc is equal to

12.08.2020

Arc length ( X ) meridian from the equator ( V =0 0) to a point (or to a parallel) with latitude ( V ) is calculated by the formula:

Task 4.2 Calculate the length of the meridian arcs from the equator to points with latitudesB 1 = 31°00" (the width of the bottom frame of the trapezoid) andB 2 \u003d 31 ° 20 "(the width of the upper frame of the trapezoid).

X o B1 = 3431035.2629

X o B2 = 3467993.3550

To control the length of meridian arcs from the equator to points with latitudes B 1 , and B 2 can also be calculated using the formula:

For the example under consideration, we have:

X o B1 = 3431035.2689

X o B2 = 3467993.3605

Laboratory work No. 5 Calculation of the dimensions of the shooting trapezoid.

Arc length ( ΔX ) meridian between parallels with latitudes V 1 and V 2 calculated by the formula:

(5.1)

where ΔB=B 2 -V 1 – increment of latitude (in arcseconds);

- average latitude; ρ” = 206264.8” is the number of seconds in radians; M 1 ,M 2 and M m – radii of curvature of the meridian at points with latitudes V 1 ,V 2 and V m .

Task 5.1 Calculate the radii of curvature of the meridian, the first vertical and the average radius of curvature for points with latitudes B 1 = B 2 = 31°20" (trapezoid upper frame width) and and B m ,= (B 1 + B 2 )/2 (mean latitude of the trapezoid)

For the example under consideration, we have:

Task 5.2 Calculate the length of the meridian arc between points with latitudes B 1 = 31°00" (width of the lower frame of the trapezoid),B 2 = 31 ° 20 "(latitude of the upper frame of the trapezoid) on the ground and on a map at a scale of 1: 100,000.

Solution.

Calculation of the length of the meridian arc between points with geodetic latitudes B 1 , and B 2 according to formula 5.1 gives the result on the ground:

ΔХ = 36958.092 m.,

on a map at a scale of 1:100,000:

ΔХ = 36958.09210m. : 100000 = 0.3695809210m. ≈ 369.58mm.

To control the length of the meridian arc ΔX between points with geodetic latitudes B 1 , and B 2 can be calculated using the formula:

ΔX \u003d X o B 2 -X o B 1 (5.2)

where X 0 B1 and X 0 B2 are the lengths of the meridian arc from the equator to parallels with latitudes V 1 and V 2 which gives the result on the ground:

ΔX \u003d 3467993.3550 - 3431035.2629 \u003d 36958.0921 m.,

on a map at a scale of 1:100000:

ΔХ = 36957.6715 m.m. : 100000 = 0.369575715m. ≈ 369.58mm.

Parallel arc length

The length of the parallel arc is calculated by the formula:

(5.3)

where N is the radius of curvature of the first vertical at a point with latitude V ;

Δ L= L 2 - L 1 – the difference between the longitudes of the two meridians (in arcseconds);

ρ” = 206264.8” is the number of seconds in a radian.

Task 5.3Calculate the arc lengths of parallels ongeodetic latitudesB 1 =31°00"andB 2 =31°20"between meridians with longitudesL 1 = 66°00"andL 2 =66°30".

Solution.

Calculating the length of the parallel arc at geodetic latitudes B 1 and B 2 between points with longitudes L 1 "and L 2 using formula 5.3 gives the result on the ground:

ΔU H = 47 752.934 m., ΔU V = 47 586.020 m.

on a map at a scale of 1:100,000:

ΔU H = 47,752.934m. : 100000 = 0.47752934 m. ≈ 477.53mm.

ΔУ В = 47 586.020m. : 100000 = 0.47586020m m. ≈ 475.86mm.

Calculation of the area of the shooting trapezoid.

The area of the shooting trapezoid is calculated by the formula:

(5.4)

Task 5.4Calculate the area of the survey trapezoid bounded by parallels with latitudes B 1 =31°00"andB 2 =31°20"and meridians with longitudesL 1 = 66°00"andL 2 =66°30".

Solution

Calculating the area of the shooting trapezoid according to formula 5.4 gives the result:

P \u003d 1761777864.9 m 2. = 176177.7865 ha. \u003d 1761.778 km 2.

For rough control the area of the shooting trapezoid can be calculated using the approximate formula:

(5.5)

Calculation of the diagonal of the shooting trapezoid.

The diagonal of the shooting trapezoid is calculated by the formula:

(5.6)

d is the length of the diagonal of the trapezoid,

ΔY H is the length of the arc of the parallel of the lower frame, ΔY B is the length of the arc of the parallel of the upper frame of the trapezoid,

ΔХ is the length of the arc of the meridian of the left (right) frame.

Task 5.4Calculate the diagonal of the survey trapezoid bounded by parallels with latitudes B 1 =31°00"andB 2 =31°20"and meridians with longitudesL 1 = 66°00"andL 2 =66°30".

The meridian of the earth's ellipsoid is an ellipse, the radius of curvature of which is determined by the value M latitude dependent. The arc length of any curve of variable radius can be calculated by the well-known formula of differential geometry, which, as applied to the meridian, has the expression

Here IN 1 and IN 2 latitudes for which the length of the meridian is determined. The integral is not taken in closed form in elementary functions. Only approximate methods of integration are possible for its calculation. When choosing the method of approximate integration, we pay attention to the fact that the value of the eccentricity of the meridian ellipse is a small value, so here it is possible to apply a method based on the expansion in a series in powers of a small value ( e /2 cos 2 B < 7*10 -3) биномиального выражения, стоящего под знаком интеграла. Число членов разложения будет зависеть от необходимой точности вычисления длины дуги меридиана, а также от разности широт ее конечных точек.

In geodetic practice, various cases may arise, more often it is necessary to perform calculations for small lengths (up to 60 km), but for special purposes it may be necessary to calculate arcs of long meridians: from the equator to the current point (up to 10,000 km), between the poles (up to 20,000 km). The required accuracy of calculations can reach a value of 0.001 m. Therefore, we will first consider the general case, when the difference in latitudes can reach 180 0, and the length of the arc is 20,000 km.

To expand a binomial expression into a series, we use a formula known from mathematics.

Hold Calculation Error m it is sufficient here to determine the terms of the expansion using the remainder term in the Lagrange form, which is not less in absolute value than the sum of all the discarded terms of the expansion and is calculated by the formula

, (4. 27)

as the first of the discarded terms of the expansion, calculated at the maximum possible value of the quantity x.

In our case we have

Substituting the resulting expression into equation (4. 25), we obtain

, (4. 28)

which allows term-by-term integration with retention of the required number of expansion terms. Let us assume that the length of the meridian arc can reach a value of 10,000 km (from the equator to the pole), which corresponds to the difference in latitudes DB = p / 2, while it is required to calculate it with an accuracy of 0.001 m, which will correspond to a relative value of 10–10. The value of cosB will not exceed one in any case. If in the calculations we keep the third degrees of expansion, then the remainder term in the Lagrange form has the expression

As you can see, to achieve the required accuracy, such a number of expansion terms is not enough, it is necessary to keep four expansion terms and the residual term in the Lagrange form will have the expression

Therefore, when integrating, it is necessary to keep in this case four degrees of decomposition.

Term-by-term integration (4.28) is easy if you convert even powers to multiple arcs ( cos 2 n B v Cos(2nB)) using the well-known double argument cosine formula

; cos2 B = (1 + cos2B)/2,

successively applying which, we get

Acting in this way until cos 8 B, we obtain after simple transformations and integration

Here, the latitude difference is taken in radian measure and the following designations are used for coefficients that have constant values for an ellipsoid with given parameters.

;

It is useful to remember that the length of the meridian arc with a latitude difference of one degree is approximately equal to 111 km, one minute - 1.8 km, one second - 0.031 km.

In geodetic practice, very often there is a need to calculate the meridian arc of small length (on the order of the length of the side of the triangulation triangle), in the conditions of Belarus this value will not exceed 30 km. In this case, there is no need to apply the cumbersome formula (4.29), but you can get a simpler one, but providing the same accuracy of calculations (up to 0.001 m).

Let the latitudes of the end points on the meridian be B1 and B2 respectively. For distances up to 30 km, this will correspond to the difference in latitude in radian measure, not more than 0. 27. Calculating the average latitude Bm meridian arcs according to the formula B m = (B 1 + B 2) / 2, we take the arc of the meridian for the arc of a circle with a radius

(4. 30)

and its length is calculated by the formula for the length of the arc of a circle

, (4. 31)

where the difference in latitude is taken in radians.

The length of the arc of the meridian and parallel. Sizes of trapezium frames for topographic maps

Kherson-2005

Meridian arc length S M between latitudes B1 and B2 is determined from the solution of an elliptic integral of the form:

(1.1)

which, as is known, is not taken in elementary functions. Numerical integration is used to solve this integral. According to Simpson's formula, we have:

(1.2)

(1.3)

where B1 and B2 are the latitudes of the ends of the meridian arc; M 1, M 2, Msr are the values of the radii of curvature of the meridian at points with latitudes B1 and B2 and Bcp=(B 1 +B 2)/2; a is the semi-major axis of the ellipsoid, e 2 is the first eccentricity.

Parallel arc length S P is the length of a part of the circle, so it is obtained directly as the product of the radius of the given parallel r=NcosB for the difference in longitude l extreme points desired arc, i.e.

where l \u003d L 2 -L 1

The value of the radius of curvature of the first vertical N calculated by the formula

(1.5)

Filming trapezoid is a part of the surface of an ellipsoid bounded by meridians and parallels. Therefore, the sides of the trapezoid are equal to the lengths of the arcs of the meridians and parallels. Moreover, the northern and southern frames are arcs of parallels a 1 and a 2, and eastern and western - arcs of meridians With, equal to each other. Trapezium Diagonal d. To obtain specific dimensions of the trapezoid, it is necessary to divide the mentioned arcs by the scale denominator m and, to obtain dimensions in centimeters, multiply by 100. Thus, the working formulas are:

(1.6)

where m- denominator of the survey scale; N 1, N 2, are the radii of curvature of the first vertical at points with latitudes B1 and B2; M m- radius of curvature of the meridian at a point with latitude Bm=(B1+B2)/2; ΔB \u003d (B 2 -B 1).

Task and initial data

1) Calculate the length of the meridian arc between two points with latitudes B 1 =30°00"00.000"" and B 2 \u003d 35 ° 00 "12.345" "+1" No., where № is the number of the variant.

2) Calculate the length of the arc of the parallel between the points lying on this parallel, with longitudes L1 = 0°00"00.000"" and L 2 \u003d 0 ° 45 "00.123" "+ 1" "No., where № is the number of the variant. Latitude of the parallel B=52°00"00.000""

3) Calculate the dimensions of the trapezoid frame at a scale of 1:100,000 for the N-35-№ map sheet, where № is the trapezoid number given by the teacher.

Solution scheme

Meridian arc length		Parallel arc length
Formulas	results		Formulas	results
a	6 378 245,0		a	6 378 245,0
e 2	0,0066934216		e 2	0,0066934216
a(1-e 2)	6335552,717		L1	0°00"00.000""
B1	30°00"00.000""		L2	0°45"00.123""
IN 2	35°00"12.345""		l \u003d L 2 -L 1	0°45"00.123""
bcp	32°30"06.173""		l(rad)	0,013090566
sinB 1	0,500000000		V	52°00"00.000""
sinB 2	0,573625462		sinB	0,788010754
sinBcp	0,537324847		cosB	0,615661475
1+0.25e 2 sin 2 B 1	1,000418339		1-0.25e 2 sin 2 B	0,998960912
1+0.25e 2 sin 2 B 2	1,000550611		1-0.75e 2 sin 2 B	0,996882735
1+0.25e 2 sin 2 Bcp	1,000483128		N	6 391 541,569
1-1.25e 2 sin 2 B 1	0,997908306		NcosB	3 935 025,912
1-1.25e 2 sin 2 B 2	0,997246944		S P	51 511,715
1-1.25e 2 sin 2 Bcp	0,997584361
M1	6 351 488,497
M2	6 356 541,056
Mcp	6 353 962,479
M1+4Mcp+M2	38 123 879,468
(M 1 +4Mcp+M 2)/6	6 353 979,911
B2-B1	5°00"12.345""
(B 2 -B 1) glad	0,087326313
S M	554 869,638

Trapeze frame sizes
Formulas	results	Formulas	results
a	6 378 245,0	1-0.25e 2 sin 2 B 1	0,998960912
e 2	0,0066934216	1-0.75e 2 sin 2 B 1	0,996882735
a(1-e 2)	6 335 552,717	1-0.25e 2 sin 2 B 2	0,998951480
0.25e2	0,001673355	1-0.75e 2 sin 2 B 2	0,996854439
0.75e2	0,005020066	1+0.25e 2 sin 2 Bm	1,001043808
1.25e2	0,008366777	1-1.25e 2 sin 2 Bm	0,994780960
B1	52°00"00""	N 1	6 391 541,569
IN 2	52°20"00""	N 2	6 391 662,647
bm	52°10"00""	mm	6 375 439,488
sinB 1	0,788010754	l	0°30"00""
sinB 2	0,791579171	l(rad)	0,008726646
sinBm	0,789798304	∆B	0°20"00""
cosB 1	0,615661475	∆B(rad)	0,005817764
cosB2	0,611066622	a 1	34,340
m	100 000	a 2	34,084
100/m	0,001	c	37,091
		d	50,459

The length of the arc of parallels and meridians on the Krasovsky ellipsoid,
taking into account distortions from the polar compression of the Earth

To determine the distance on a tourist map, in kilometers between points, the number of degrees is multiplied by the length of the arc 1 ° parallel and meridian (in longitude and latitude, in the system geographical coordinates), the exact calculated values of which are taken from the tables. Approximately, with a certain error, they can be calculated by the formula on the calculator.

An example of converting numerical values of geographical coordinates from tenths to degrees and minutes.

The approximate longitude of the city of Sverdlovsk is 60.8° (sixty point and eight tenths of a degree) east longitude.
8 / 10 = X / 60
X \u003d (8 * 60) / 10 \u003d 48 (from the proportion we find the numerator of the right fraction).
Result: 60.8° = 60° 48" (sixty degrees and forty-eight minutes).

To add a degree symbol (°) - press Alt + 248 (with numbers in the right numeric keypad of the keyboard; in a laptop - with the special Fn button pressed or by turning on NumLk). This is done in Windows and Linux operating systems, and in Mac OS - using the Shift + Option + 8 keys

Latitude coordinates are always indicated before longitude coordinates (whether printed on a computer or written down on paper).

In the maps.google.ru service, supported formats are determined by the rules

Examples of how it would be correct:

Full form angle records (degrees, minutes, seconds with fractions):
41° 24" 12.1674", 2° 10" 26.508"

Abbreviated forms of writing an angle:
Degrees and minutes with decimals - 41 24.2028, 2 10.4418
Decimal degrees (DDD) - 41.40338, 2.17403

The Google map service has an online converter for converting coordinates and converting them to the desired format.

As a decimal separator of numerical values, on Internet sites and in computer programs, it is recommended to use a dot.

tables

The length of the parallel arc in 1°, 1" and 1" in longitude, meters

Latitude, degree	The length of the parallel arc in 1° longitude, m	Parallel arc length in 1", m	Arc length par. h1",m

A simplified formula for calculating the arcs of parallels (without taking into account distortions from polar compression):

L pairs \u003d l equiv * cos (Latitude).

The length of the meridian arc in 1 °, 1 "and 1" in latitude, meters

Latitude, degree	The length of the meridian arc in 1° latitude, m

Drawing. 1-second arcs of meridians and parallels (simplified formula).

A practical example of using tables. For example, if the map does not indicate a numerical scale and there is no scale bar, but there are lines of a degree cartographic grid, you can graphically determine the distances, based on the fact that one degree of the arc corresponds to the numerical value obtained from the table. In the "north-south" directions (between the horizontal lines of the geographical grid on the map) - the values of the lengths of the arcs change, from the equator to the poles of the Earth, insignificantly and amount to approximately 111 kilometers.

Andreev N.V. Topography and Cartography: Optional course. M., Enlightenment, 1985

Mathematics textbook.

Http://ru.wikipedia.org/wiki/Geographic_coordinates

The length of the arc of parallels and meridians, taking into account the polar compression of the Earth

To determine the distance on a tourist map, in kilometers between points, the number of degrees is multiplied by the arc length of 1 ° of the parallel and meridian (in longitude and latitude, in the geographic coordinate system), the exact calculated values of which are taken from the tables. Approximately, with a certain error, they can be calculated by the formula on the calculator.

An example from a school geography lesson (according to an old textbook and from study guide for optional course)

Determine the partial scale of a small-scale (1:1,000,000, 1:6,000,000, 1:20,000,000 and smaller) maps earth's surface(atlas for class VI) in the area of Kazan and Sverdlovsk (now Yekaterinburg, see the list of renamed cities). Both of these cities are located approximately at latitude 56°N.
The longitude of Kazan is 49°E, Yekaterinburg is 60°E.
The distance between them on the map is 1.1 cm (determined using a measuring compass and a ruler with millimeter divisions).
The length of the arc of a parallel in 1 ° for a latitude of 56 ° N is equal to 62394 meters.

60 - 49 = 11° (longitude difference).
L \u003d 62394 * 11 \u003d 686,334 meters \u003d 68,633,400 cm (distance between points in centimeters).

m = 1 / (68 633 400 / 1.1) ~ 1 / 62 400 000

Answer: private scale (m) - 1 cm 624 km.

The main scale (signed in the marginal
registration of this card) - 1 / 75,000,000 (1 cm 750 km).

Private m-b may be more or less than the main one, depending on the location of the selected area on the map.

An example of converting numerical values of geographical coordinates from tenths to degrees and minutes.

To add a degree symbol (°) - press Alt + 248 (numbers in the right numeric keypad of the keyboard; in a laptop - with the special Fn button pressed or by turning on NumLk)). This is done in Windows and Linux operating systems, and in Mac OS - using the Shift + Option + 8 keys