Surveying I– coordinate account

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Calculation method of polar coordinates

The method of poles is one of the fundamental methods of measuring location of detail points. Geodetic connection to the two control points A and B and the measured directions to these points and to the point P, we can set the coordinates of point P.

Figure 1. Distribution of points in a coordinate system

Search:
P (X_P, Y_P)


Data:
A (X_A, Y_A)
B (X_B, Y_B)
S (X_S, Y_S)

Measured:
directions: k_A,k_B, k_P,
length: d_SP


To tie in the two control points A and B, calculate the orientation angle of 0˚ distribution limbus lines as follows:

γ' = A_SA − k_A
γ" = A_SB − k_B

where: A_SA, A_SB - bearings* of the sides of the SA and SB.
*If you do not remember how to calculate the azimuth from coordinates of two points, click

We check whether the deviation between the angles of γ' and γ" is situated at the limit permissible deviation f_γ = 2m₀√_ 2:
Ιγ' – γ"Ι ≤ f_γ

If this condition is met, then:

γ = γ' + γ" 2

We calculate the bearing of the SP side of the formula:

A_SP = k_P + γ

and northing and the easting difference based on the measured length and bearing side SP:

Δx_SP = d_SPcosA_SP
Δy_SP = d_SPsinA_SP

The final coordinates of a point P

X_P = X_S + Δx_SP
Y_P = Y_S + Δy_SP

Control calculation:
Comparison of the measured length d_SP with calculated length D_SP = √__________________ (X_P − X_S)² + (Y_P − Y_S)²

Example

Calculate the coordinates of P point of the polar method.

Data:
A (1400,34m; 1398,22m)
B (534,15m; 1729,04m)
S (535,40m; 681,72m)

Measured
directions:
k_A = 23^g,2719 ± 2^cc
k_B = 79^g,3064 ± 2^cc
k_P = 64^g,7378 ± 2^cc

length:
d_SP = 341,43m

SOLUTION

Δx_SA = X_A – X_S = 864,94m
Δy_SA = Y_A – Y_S = 716,50m


d_SA = √____________ Δx_SA² + Δy_SA²


d_SA = √________________ 864,94² + 716,50²


d_SA = 1123,16m



φ_SA = arctg Δy_SA Δx_SA


φ_SA = arctg 716,50 864,94

φ_SA = 44^g,04190


Δx_SA = 864,94m >0
Δy_SA = 716,50m >0
So the quadrant bearing is in first quadrant: A_SA = φ_SA

A_SA = 44^g,04190



Δx_SB = X_B – X_S = -1,25m
Δy_SB = Y_B – Y_S = 1047,32m


d_SB = √____________ Δx_SB² + Δy_SB²


d_SB = √________________ (-1,25)² + 1047,32²


d_SB = 1047,32m



φ_SB = arctg Δy_SB Δx_SB



φ_SB = arctg 1047,32 -1,25

φ_SB = -99^g,92402


Δx_SB = -1,25m <0
Δy_SB = 1047,32m >0
So the quadrant bearing is in second quadrant A_SB=200^g,0000 + φ_SB

A_SB = 200^g,0000 + (-99^g,92402) = 100^g,07598


γ' = A_SA − k_A = 44^g,04190 – 23^g,2719 = 20^g,77000
γ" = A_SB − k_B = 100^g,07598 – 79^g,3064 = 20^g,76958

f_γ = 2m₀√_ 2= 5^cc,7
Ι 20^g,77000 – 20^g,76958Ι = 4^cc,2 ≤ f_γ

This condition is met, then:

γ = γ' + γ" 2

γ = 20^g,77000 + 20^g,76958 2

γ = 20^g,76979


A_SP = k_P + γ = 64^g,7378 + 20^g,76979 = 85^g,50759

Δx_SP = d_SPcosA_SP = 341,43• cos(85^g,50759) = 77,056m
Δy_SP = d_SPsinA_SP = 341,43• sin(85^g,50759) = 332,621m


The final coordinates:
X_P = X_S + Δx_SP = 535,40 + 77,056 = 612,456m = 612,46
Y_P = Y_S + Δy_SP = 681,72 + 332,621 = 1014,341m = 1014,34


Control

D_SP = √__________________ (X_P − X_S)² + (Y_P − Y_S)²

D_SP = 341,43m = d_SP