Surveying I– coordinate account

Page takes part in the competition "Notes on the Internet" organized by the Rector of Academy of Science and Technology in Cracow.
V o t e !

Back>>

Calculation of bearing on the basis of the coordinates of several control points

The geodetic measurements is very important to the accuracy and control of the results obtained. Therefore, if we do so, we refer to as many control points. This involves the setting of a fixed orienting φ.

Figure 1. Sketch the direction is determined to establish control points to directions

Search:
A_AB


Data:
A (X_A, Y_A)
1 (X₁, Y₁)
2 (X₂, Y₂)
n (X_n, Y_n)

Measured:
k₁, k₂, ..., k_n,
k_B

Having measured the directions k₁, k₂, ..., k_n related to the 0° in a limbus and the calculated bearings A_A1, A_A2, ..., A_An we are setting a fixed orienting for each direction:

φ₁ = A_A1 − k₁,
φ₂ = A_A2 − k₂,
...
φ _n = A_An − k_n,

and then the average value:

φ = φ₁ + φ₂ + ... + φ_n n

Calculating the bearing angle AB:

A_AB = φ + k_B

Example

Calculate the bearing angle of the direction AB.

Data:
A ( 202,74m; 436,77m)
1 ( 191,74m; 431,78m)
2 ( 213,45m; 431,14m)

Measured:
k₁=122^g,3009
k₂=264^g,3623
k_B=93^g,2654

SOLUTION

Δx_A1 = X₁ – X_A = –11,00m
Δy_A1 = Y₁ – Y_A = –4,99m


φ_A1 = arctg Δy_A1 Δx_A1

φ_A1 = arctg -4,99 -11,00= 27^g,11197


Δx_A1 = –11,00m <0
Δy_A1 = –4,99m <0
So quadrant bearing is in the third quarter: A_A1=200^g,0000 + φ_A1

A_A1=200^g,0000 + 27^g,11197 = 227^g,1120

Δx_A2 = X₂ – X_A = 10,71m
Δy_A2 = Y₂ – Y_A = –5,63m


φ_A2 = arctg Δy_A2 Δx_A2

φ_A2 = arctg -5,63 10,71= –30^g,81097


Δx_A2 = 10,71m >0
Δy_A2 = –5,63m <0
So quadrant bearing is in the fourth quarter: A_A2=400^g,0000 + φ_A2

A_A2=400^g,0000 + (–30^g,81097) = 369^g,1890

φ₁ = A_A1 − k₁ = 227^g,1120 − 122^g,3009 = 104^g,8111
φ₂ = A_A2 − k₂ = 369^g,18097 − 264^g,3623 = 104^g,8267


φ = φ₁ + φ₂ 2=104^g,8111 + 104^g,8267 2= 104^g,8189

A_AB = φ + k_B = 104^g,8189 + 93^g,2654 = 198^g,0843