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Calculation of coordinates of a point by a linear intersection

Figure 1. Linear intersection in a coordinate system

Search:
P (X_P, Y_P)


Data:
A (X_A, Y_A)
B (X_B, Y_B)

Measured:
d_AP, d_BP

To calculate the coordinates of point P on the basis of the coordinates of two points A and B, and measured the length of the sides of the AP and BP use a linear intersection method. First, we sketch by placing a point in a coordinate system (Figure 1.).
We calculate the northing and the easting difference Δx_AB = X_B – X_A, Δy_AB = Y_B – Y_A, which are necessary for determining bearing* A_AB oraz długości* boku AB:
D_AB = √____________ Δx_AB² + Δy_AB²

*If you do not remember
- method of calculating the length of the coordinates, click
- how to calculate the bearing from coordinates of two points, click



To calculate the angles α, β and γ in a triangle ABP use the cosine theorem:

D_AB² = d_AP² + d_BP² –2d_APd_BPcosγ ⇒ γ = arctg D_AB² –(d_AP² + d_BP²) –2d_APd_BP


d_AP² = D_AB² + d_BP² –2D_ABd_BPcosβ ⇒ β = arctg d_AP² – (D_AB² + d_BP²) –2D_ABd_BP


d_BP² = d_AP² + D_AB² – 2d_APD_ABcosα ⇒ α = arctg d_BP² – (d_AP² + D_AB²) –2d_APD_AB

Ensure you have calculated the sum of the angles is equal to 200^g,0000.


In the remainder of the calculations proceed as though the angular intersection.

Calculate the bearing angle of the side AP:

A_AP = A_AB –α

and northing and the easting difference based on the the length and bearing of the side AP:

Δx_AP = d_APcosA_AP
Δy_AP = d_APsinA_AP

The final coordinates of point P:

X_P = X_A + Δx_AP
Y_P = Y_A + Δy_AP

Control calculations are re-enumeration of coordinates of a point P on the basis of point B and compare them with the coordinates of point P calculated on the basis of point A.

Calculate the bearing angle of the side BP:
A_BP = A_AB + β

and northing and the easting difference based on the the length and bearing of the side BP:
Δx_BP = d_BPcosA_BP
Δy_BP = d_BPsinA_BP


The final coordinates of point P:
X_P = X_B + Δx_AB
Y_P = Y_B + Δy_AB

Example

Calculate the coordinates of point P by a linear intersection.

Data:
A (100,00m; 100,00m)
B (100,00m; 300,00m)

Measured:
d_AP = 100,00m
d_BP = 173,21m

SOLUTION

Δx_AB = X_B – X_A = 0,00m
Δy_AB = Y_B – Y_A = 200,00m

D_AB = √____________ Δx_AB² + Δy_AB²

D_AB = √________________ 0,00² + 200,00²

D_AB = 200,00m


If Δx_AB = 0 oraz Δy_AB > 0, then A_AB = 100^g,0000



γ = arctg D_AB² –(d_AP² + d_BP²) –2d_APd_BP



γ = arctg 200,00² –(100,00² + 173,21²) –2•100,00•173,21



γ = 99^g,99687



β = arctg d_AP² –(D_AB² + d_BP²) –2D_ABd_BP



β = arctg 100,00² –(200,00² + 173,21²) –2•200,00•173,21



β = 33^g,33333



α = arctg d_BP² –(d_AP² + D_AB²) –2d_APD_AB


α = arctg 173,21² –(100,00² + 200,00²) –2•100,00•200,00


α = 66^g,66980



α + β + γ = 66^g,66980 + 33^g,33333 + 99^g,99687 = 200^g,0000



A_AP = A_AB –α = 100^g,00000 –66^g,66980 = 33^g,33020


Δx_AP = d_APcosA_AP = 100,00•cos(33^g,33020) = 86,605m
Δy_AP = d_APsinA_AP = 100,00•sin(33^g,33020) = 49,996m


The final coordinates of point P:
X_P = X_A + Δx_AP = 100,00 + 86,605 = 186,605m = 186,60m
Y_P = Y_A + Δy_AP = 100,00 + 49,996 = 149,996m = 150,00m



Control
A_BA = A_AB + 200^g,00000 = 300^g,00000


A_BP = A_BA + β = 300^g,00000 + 33^g,33333 = 333^g,33333


Δx_BP = d_BPcosA_BP = 173,21•cos(333^g,33333) = 86,605m
Δy_BP = d_BPsinA_BP = 173,21•sin(333^g,33333) = -150,004m


The final coordinates of point P:
X_P = X_B + Δx_BP = 100,00 + 86,605 = 186,605m = 186,60m
Y_P = Y_B + Δy_BP = 300,00 + (-150,004) = 149,996m = 150,00m