Basic geodetic calculations position of points is defined

Basic geodetic calculations • position of points is defined by rectangular plane coordinates Y, X in given coordinate system (reference frame) • geodetic coordinate system S-JTSK is clockwise 1

coordinate differences x 12 = x 2 – x 1 y 12 = y 2 – y 1 x 21 = x 1 – x 2 y 21 = y 1 – y 2 distance s 12 = s 21 s 12 = y 12 /sin 12 s 12 = x 12 /cos 12 2

Bearing • oriented angle between parallel to the axis +X and the join of the points 21 = 12 + 180 21 = 12 + 200 gon = 12 + 200 g 3

Bearing Quadrant I II IV y 12 + + - - x 12 + - - + 12= 12 12= 200 g - 12 12= 200 g + 12 = 400 g - 12 4

Bearing – examples P. N. Y [m] X [m] 1 2000 7000 2 2300 7200 3 2300 6800 4 1700 6800 5 1700 7200

Bearing – examples

Determination of a point defined by polar coordinates (bearing and distance) Given: rectangular coordinates of points P 1 [y 1, x 1] and P 2 [y 2, x 2], distance d 13 horizontal angle 1 Calculated: P 3 [y 3, x 3] 7

according to the table 12 13 = 12 + 1 Coordinate differences: y 13 = d 13. sin 13 x 13 = d 13. cos 13 y 3 = y 1 + y 13 = y 1 + d 13. sin 13 x 3 = x 1 + x 13 = x 1 + d 13. cos 13 8

Calculation of the coordinates by intersection from angles Given: rectangular coordinates of points P 1 [y 1, x 1] and P 2 [y 2, x 2], horizontal angles 1 a 2 Calculated: P 3 [y 3, x 3] 9

according to the table 12 21 = 12 + 200 gon s 13 = s 12. sin 2 / sin (200 gon – ( 1 + 2)) = s 12. sin 2 / sin ( 1 + 2) , s 23 = s 12. sin 1 / sin (200 gon – ( 1 + 2)) = s 12. sin 1 / sin ( 1 + 2) (law of sines) 10

13 = 12 + 1 23 = 21 – 2 y 3 = y 1 + s 13. sin 13 = y 2 + s 23. sin 23 x 3 = x 1 + s 13. cos 13 = x 2 + s 23. cos 23 Coordinates of the point P 3 are determined twice using bearings and distances to check the calculation. 11

Intersection from distances Given: rectangular coordinates of points P 1 [y 1, x 1] and P 2 [y 2, x 2], measured horizontal distances d 13 a d 23 Calculated: rectangular coordinates of P 3 [y 3, x 3] 12

21 = 12 + 200 gon 13

13 = 12 + 1 23 = 21 – 2 y 3 = y 1 + s 13. sin 13 = y 2 + s 23. sin 23 x 3 = x 1 + s 13. cos 13 = x 2 + s 23. cos 23 Coordinates of the point P 3 are determined twice using bearings and distances to check the calculation. 14

Resection Given: rectangular coordinates of points P 1 [y 1, x 1], P 2 [y 2, x 2], P 3 [y 3, x 3] measured horizontal angles 1 a 2 Calculated: rectangular coordinates of P 4 [y 4, x 4] 15

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Traverse (polygon) • • a broken line connecting two survey points traverse points = vertexes of the broken line traverse legs = joins of nearby traverse points horizontal angles at all traverse points and lengths of traverse legs are measured • coordinates Y, X of the traverse points are calculated 17

Traverse • connected (at one or both ends) – the traverse is connected to the survey points whose coordinates are known • disconnected – the traverse is connected to the survey points whose coordinates are not known Dividing traverses according to a shape: • traverse line • closed traverse – the start point = the end point Orientation of a traverse = measurement of the horizontal angle at the start (or the end) point. 18

Traverse connected and oriented on both ends 19

Given: coordinates of the start and the end points 1 [y 1, x 1], n [yn, xn] (here n = 5) coordinates of the orientation points A [y. A, x. A], B [y. B, x. B] measured horizontal distances d 12, d 23, d 34, d 45 measured horizontal angles ω1, ω2, ω3, ω4, ω5 Calculated: coordinates of points 2 [y 2, x 2], 3 [y 3, x 3], …, n-1 [yn-1, xn-1] 20

1. calculation of bearings according to the table 1 A and n. B 21

2. angular adjustment Angular error Oω is calculated (error = „it should be“ minus „it is“. „It should be“ is the bearing n. B calculated from coordinates, „it is“ is the bearing αn. B calculated using measured horizontal angles). i = 1, … , n n … number of the traverse points (here n = 5) 22

Clause for the angular adjustment: The angular error is divided equally to the measured horizontal angles: = O / n ´ 1 = 1 + , . . . , ´n = n + . 23

3. calculation of bearings 12 = 1 A + ´ 1 23 = 12 + ´ 2 ± 200 g … n-1, n = n-2, n-1 + ´n-1 ± 200 g n. B = n-1, n + ´n ± 200 g = n. B Check! 24

4. calculation of coordinate differences y 12 = d 12. sin 12 … yn-1, n = dn-1, n. sin n-1, n x 12 = d 12. cos 12 … xn-1, n = dn-1, n. cos n-1, n 25

5. calculation of coordinate deviations y 1 n = yn – y 1 x 1 n = xn – x 1 y 1 ncal = y 12 + y 23+ y 34 + y 4 n = y x 1 ncal = x 12 + x 23+ x 34 + x 4 n = x Oy = y 1 n – y Ox = x 1 n – x 26

Positional difference Clause for the adjustment: 27

Corrections of coordinate differences The corrections of coordinate differences are not equal, they depend on values of coordinate differences. 28

6. corrected coordinate differences … Check! 29

7. calculation of adjusted coordinates y 1 = given x 1 = given y 2 = y 1 + y´ 12 x 2 = x 1 + x´ 12 …. yn = yn - 1 + y´n – 1, n = given Check! xn = xn - 1 + x´n – 1, n = given Check! 30

Closed traverse without orientation 31

Given: measured horizontal distances d 12, d 23, d 34, d 41 measured horizontal angles ω1, ω2, ω3, ω4 Calculated: coordinates of points P 1 [y 1, x 1], P 2 [y 2, x 2], P 3 [y 3, x 3], P 4 [y 4, x 4] 32

1. choice of a local coordinate system One of the traverse points is chosen as a beginning of a local coordinate system (here P 1) and one axis is put in the traverse leg from this point (here axis +Y is put in P 1 P 2). Coordinates of the beginning are chosen, usually: y 1 = 0, 00, x 1 = 0, 00 Result from this choice: x 2 = 0, 00, 12 = 100 g 33

The calculation is the same as previous one, the start point = the end point = P 1. 2. angular adjustments i = 1, … , n n … number of the traverse points (here n = 4) 34

Clause for the angular adjustment: Angular error is divided equally to the measured horizontal angles: = O / n ´ 1 = 1 + , . . . , ´n = n + . 35

3. calculation of bearings 12 = 100 g 23 = 12 + ´ 2 ± 200 g … 41 = 34 + ´ 4 ± 200 g 12 = 41 + ´ 1 ± 200 g = 12 Check! 36

4. calculation of coordinate differences y 12 = d 12. sin 12 … y 41 = d 41. sin 41 x 12 = d 12. cos 12 … x 41 = d 41. cos 41 37

5. calculation of coordinate deviations y 1 n = yn – y 1 = 0 x 1 n = xn – x 1 = 0 y 1 ncal = y 12 + y 23+ y 34 + y 4 n = y x 1 ncal = x 12 + x 23+ x 34 + x 4 n = x Oy = – y Ox = – x 38

Positional difference Clause for the adjustment: 39

Corrections of coordinate differences 40

6. corrected coordinate differences …. Check! 41

7. calculation of adjusted coordinates y 1 = given x 1 = given y 2 = y 1 + y´ 12 x 2 = x 1 + x´ 12 …. y 1 = y 4 + y´ 41 = given Check! x 1 = x 4 + x´ 41 = given Check! 42