Oblicz współrzędne punktów 1, 2 i 3 na domiarach prostokątnych.

Dane:
A (212,31m; 588,23m)
B (291,32m; 712,21m)

Pomierzone:
d_11' = 32,32m
d_22' = 5,96m
d_33' = 8,78m
d_AB = 147,05m
d_A1' = -12,40m
d_A2' = 31,14m
d_A3' = 45,11m

ROZWIĄZANIE

Δx_AB = X_B – X_A = 79,01m
Δy_AB = Y_B – Y_A = 123,98m


L_AB = √____________ Δx_AB² + Δy_AB²


L_AB = √________________ 79,01² + 123,989²


D_AB = 147,02m


f = Ιl_AB – L_ABΙ = Ι147,05 – 147,02Ι = 0,03m
f_l = 0,07m

Warunek f ≤ f_l jest spełniony.



Obliczenie współrzędnych punktu 1

X₁ = X_A + Δx_A1' – Δx_11' = 212,31 + (-6,66) – 27,25 = 178,40m
Y₁ = Y_A + Δy_A1' + Δy_11', = 588,23 + (-10,45) + 17,37 = 595,15m

gdzie:

Δx_A1' = l_A1'Δx l_AB


Δx_A1' = -12,40 79,01 147,05


Δx_A1' = -6,66m


Δy_A1' = l_A1' Δy l_AB


Δy_A1' = -12,40 123,98 147,05


Δy_A1' = -10,45m


Δx_11' = d_11'Δy l_AB


Δx_11' = 32,32 123,98 147,05


Δx_11' = 27,25m


Δy_11' = d_11' Δx l_AB


Δy_11' = 32,32 79,01 147,05


Δy_11' = 17,37m




Obliczenie współrzędnych punktu 2

X₂ = X_A + Δx_A2' – Δx_22' = 212,31 + 16,73 – 5,02 = 224,02m
Y₂ = Y_A + Δy_A2' + Δy_22', = 588,23 + 26,25 + 3,20 = 617,68m

gdzie:

Δx_A2' = l_A2'Δx l_AB


Δx_A2' = 31,14 79,01 147,05


Δx_A2' = 16,73m


Δy_A2' = l_A2' Δy l_AB


Δy_A2' = 31,14 123,98 147,05


Δy_A2' = 26,25m


Δx_22' = d_22'Δy l_AB


Δx_22' = 5,96 123,98 147,05


Δx_22' = 5,02m


Δy_22' = d_22' Δx l_AB


Δy_22' = 5,96 79,01 147,05


Δy_22' = 3,20m




Obliczenie współrzędnych punktu 3

X₃ = X_A + Δx_A3' + Δx_33' = 212,31 + 24,24 + 7,40 = 243,95m
Y₃ = Y_A + Δy_A3' – Δy_33', = 588,23 + 38,03 – 4,72 = 617,68m

gdzie:

Δx_A3' = l_A3'Δx l_AB


Δx_A3' = 45,11 79,01 147,05


Δx_A3' = 24,24m


Δy_A3' = l_A3' Δy l_AB


Δy_A3' = 45,11 123,98 147,05


Δy_A3' = 38,03m


Δx_33' = d_33'Δy l_AB


Δx_33' = 8,78 123,98 147,05


Δx_33' = 7,40m


Δy_33' = d_33' Δx l_AB


Δy_33' = 8,78 79,01 147,05


Δy_33' = 4,72m

Kontrola

Numer punktu	Odcięte l	Rzędne d	Różnice odciętych Δl	Różnice rzędnych Δd
A	0,00	0,00	-12,40	32,32
1	-12,40	32,32	43,54	-26,36
2	31,14	5,96	13,97	-14,74
3	45,11	-8,78	101,94	8,78
B	147,05	0,00	147,05	0,00

Obliczenie współrzędnych punktu B ze współrzędnych punktu 1

X_B = X₁ + Δx_B1' + Δx_11' = 178,40 + 85,67 + 27,25 = 291,32m
Y_B = Y₁ + Δy_B1' – Δy_11', = 595,15 + 134,43 – 17,37 = 712,21m

gdzie:

Δx_B1' = (l_AB – l_A1')Δx l_AB


Δx_B1' = 159,45 79,01 147,05


Δx_B1' = 85,67m


Δy_B1' = (l_AB – l_A1') Δy l_AB


Δy_B1' = 159,45 123,98 147,05


Δy_B1' = 134,43m


Δx_11' = d_11'Δy l_AB


Δx_11' = 32,32 123,98 147,05


Δx_11' = 27,25m


Δy_11' = d_11' Δx l_AB


Δy_11' = 32,32 79,01 147,05


Δy_11' = 17,37m




Obliczenie współrzędnych punktu B ze współrzędnych punktu 2

X_B = X₂ + Δx_B2' + Δx_22' = 224,02 + 62,28 + 5,02 = 291,32m
Y_B = Y₂ + Δy_B2' – Δy_22', = 617,68 + 97,73 – 3,20 = 712,21m

gdzie:

Δx_B2' = (l_AB – l_A2')Δx l_AB


Δx_B2' = 115,91 79,01 147,05


Δx_B2' = 62,28m


Δy_B2' = (l_AB – l_A2') Δy l_AB


Δy_B2' = 115,91 123,98 147,05


Δy_B2' = 97,73m


Δx_22' = d_22'Δy l_AB


Δx_22' = 5,96 123,98 147,05


Δx_22' = 5,02m


Δy_22' = d_22' Δx l_AB


Δy_22' = 5,96 79,01 147,05


Δy_22' = 3,20m




Obliczenie współrzędnych punktu B ze współrzędnych punktu 3

X_B = X₃ + Δx_B3' – Δx_33' = 243,95 + 54,77 – 7,40 = 291,32m
Y_B = Y₃ + Δy_B3' + Δy_33', = 621,54 + 85,95 + 4,72 = 712,21m

gdzie:

Δx_B3' = (l_AB – l_A3')Δx l_AB


Δx_B3' = 101,94 79,01 147,05


Δx_B3' = 54,77m


Δy_B3' = (l_AB – l_A3') Δy l_AB


Δy_B3' = 101,94 123,98 147,05


Δy_B3' = 85,95m


Δx_33' = d_33'Δy l_AB


Δx_33' = 8,78 123,98 147,05


Δx_33' = 7,40m


Δy_33' = d_33' Δx l_AB


Δy_33' = 8,78 79,01 147,05


Δy_33' = 4,72m

Oblicz współrzędne punktu P metodą wcięcia kątowego w przód.

Dane:
A (150,00m; 500,00m)
B (400,00m; 200,00m)

Pomierzone:
α = 30^g,0000
β = 50^g,0000

ROZWIĄZANIE

Δx_AB = X_B – X_A = 250,00m
Δy_AB = Y_B – Y_A = -300,00m

D_AB = √____________ Δx_AB² + Δy_AB²

D_AB = √________________ 250,00² + (-300,00)²

D_AB = 390,512m


φ_AB = arctg Δy_AB Δx_AB


φ_AB = arctg -300,00 250,00

φ_AB = -55^g,77159


Δx_AB = 250,00m >0
Δy_AB = -300,00m <0
Zatem czwartak leży w IV ćwiartce, więc A_AB = 400^g,0000 + φ_AB

A_AB = 400^g,0000 + (-55^g,77159) = 344^g,22841



d_AP = D_AB•sinβ sin(α + β)

d_AP = 390,512•sin(50^g,0000) sin(80^g,0000)

d_AP = 290,344m


A_AP = A_AB + α = 344^g,22841 + 30^g,0000 = 374^g,22841


Δx_AP = d_APcosA_AP = 290,344•cos(374^g,22841) = 266,877m
Δy_AP = d_APsinA_AP = 290,344•sin(374^g,22841) = -114,353m

Ostateczne współrzędne szukanego punktu P:
X_P = X_A + Δx_AP = 150,00 + 266,877 = 416,877m = 416,88m
Y_P = Y_A + Δy_AP = 500,00 + (-114,353) = 385,647m = 385,65


Kontrola
A_BA = A_AB - 200^g,0000 = 144^g,22841

d_BP = D_AB•sinα sin(α + β)

d_BP = 390,512•sin(30^g,0000) sin(80^g,0000)

d_BP = 186,412m


A_BP = A_BA − β = 144^g,22841 - 50^g,0000 = 94^g,22841


Δx_BP = d_BPcosA_BP = 186,412•cos(94^g,22841) = 16,877m
Δy_BP = d_BPsinA_BP = 186,412•sin(94^g,22841) = 185,646m



Ostateczne współrzędne szukanego punktu P:
X_P = X_B + Δx_BP = 400,00 + 16,877 = 416,877m = 416,88m
Y_P = Y_B + Δy_BP = 200,00 + 185,646 = 385,646m = 385,65m




Dodatkowa kontrola

γ = 200^g,0000 − (α + β) = 120^g,0000

Δx_PB = X_B – X_P = 400,00 – 416,877 = -16,877m
Δy_PB = Y_B – Y_P = 200,00 – 385,646 = -185,647m


φ_PB = arctg Δy_PB Δx_PB


φ_PB = arctg -185,647 -16,877


φ_PB = 94^g,22841

A_PB = 200^g,0000 + φ_PB = 294^g,22841


Δx_PA = X_A – X_P = 150,00 – 416,877 = -266,877m
Δy_PA = Y_A – Y_P = 500,00 – 385,646 = 114,353m


φ_PA = arctg Δy_PA Δx_PA


φ_PA = arctg 114,353 -266,877


φ_PA = -25^g,77161

A_PA = 200^g,0000 + φ_PA = 174^g,22839


γ = A_PB – A_PA = 294^g,22841 – 174^g,22839 = 120^g,0000

Oblicz współrzędne punktu P metodą wcięcia liniowego.

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

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

ROZWIĄZANIE

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


Jeśli Δx_AB = 0 oraz Δy_AB > 0, wówczas 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


Ostateczne współrzędne szukanego punktu 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



Kontrola
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


Ostateczne współrzędne szukanego punktu 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

Oblicz współrzędne punktów 2 i 3 mając dane współrzędne punktów A, B, C i 1.

Dane:
A (100,00m, 100,00m),
B (300,00m, 600,00m),
C (250,00m, 750,00m),
1 (50,00m, 200,00m);

Pomierzone
- kąty:
α₁ = 110^g,0600,
α₂ = 230^g,4000,
α₃ = 210^g,5006,
α₄ = 240^g,0000;

- długości:
d₁₂=155,00m,
d₂₃=150,00m,
d_3B=185,00m;

ROZWIĄZANIE

Δx_A1 = X₁ – X_A = -50,00m
Δy_A1 = Y₁ – Y_A = 100,00m


φ_A1 = arctg Δy_AB Δx_AB


φ_A1 = arctg 100,00 -50,00

φ_A1 = -70^g,48328


Δx_A1 = -50,00m <0
Δy_A1 = 100,00m >0
Zatem czwartak leży w II ćwiartce, więc A_A1 = 200^g,0000 + φ_A1

A_A1 = 200^g,0000 + (-70^g,48328) = 129^g,51672



Δx_BC = X_C – X_B = -50,00m
Δy_BC = Y_C – Y_B = 150,00m


φ_BC = arctg Δy_BC Δx_BC


φ_BC = arctg 150,00 -50,00

φ_BC = -79^g,51672


Δx_BC = -50,00m <0
Δy_BC = 150,00m >0
Zatem czwartak leży w II ćwiartce, więc A_BC = 200^g,0000 + φ_BC

A_BC = 200^g,0000 + (-79^g,51672) = 120^g,48328



∑α_p = α₁ + α₂ + α₃ + α₄ = 790^g,96060

∑α_t = A_BC – A_A1 + 4•200^g,0000 = 790^g,96656
f_α = ∑α_p – ∑α_t = -59^cc,6
f_dop = ± 360^cc
Warunek f ≤ f_dop jest spełniony.



v_αi = – f_α n


v_αi = – -59^cc,6 4

v_αi = 14^cc,9


v_α1 = 14^cc,9
v_α2 = 14^cc,9
v_α3 = 14^cc,9
v_α4 = 14^cc,9
∑v_αi = -59^cc,6 = -f_α


_ α₁ = α₁ + v_α1 = 110^g,06149
_ α₂ = α₂ + v_α2 = 230^g,40149
_ α₃ = α₃ + v_α3 = 210^g,50209
_ α₄ = α₄ + v_α4 = 240^g,00149


A₁ = A_A1 + _ α₁ – 200^g,0000 = 39^g,57821
A₂ = A₁ + _ α₂ – 200^g,0000 = 69^g,97970
A₃ = A₂ + _ α₃ – 200^g,0000 = 80^g,48179
A_BC = A₃ + _ α₄ – 200^g,0000 = 120^g,48328


ΔX₁₂ = d₁₂cosA₁ = 125,999m
ΔY₁₂ = d₁₂sinA₁ = 90,274m
ΔX₂₃ = d₂₃cosA₂ = 68,141m
ΔY₂₃ = d₂₃sinA₂ = 133,629m
ΔX_3B = d_3BcosA₃ = 55,835m
ΔY_3B = d_3BsinA₃ = 176,373m


∑ΔX_p = 249,975m
∑ΔY_p = 400,276m


∑ΔX_t = X_B – X₁ = 250,000m
∑ΔY_t = Y_B – Y₁ = 400,000m


f_x = ∑ΔX_p – ∑ΔX_t = -0,025m
f_y = ∑ΔY_p – ∑ΔY_t = 0,276m


f_L = √_________ f_x² + f_y²

f_L = √________________ (-0,025)² + 0,276²

f_L = 0,277m


f_Ldop = 0,170m

Za odchyłkę liniową przyjmujemy podwójną wartość dopuszczalnej odchyłki liniowej stabelaryzowanej w Instrukcji technicznej G-4, zał.3.


d = d₁₂ + d₂₃ + d_3B = 490,000m



v_x12 = – f_x d₁₂ d


v_y12 = – f_yd₁₂ d


v_x23 = – f_xd₂₃ d


v_y23 = – f_y d₂₃ d


v_x3B = – f_x d_3B d


v_y3B = – f_y d_3B d


∑v_xi = 0,025m = -f_x
∑v_y;i = -0,276m = -f_y



__ ΔX₁₂ = X₁₂ + v_x12 = 126,007m

__ ΔY₁₂ = Y₁₂ + v_y12 = 90,187m

__ ΔX₂₃ = X₂₃ + v_x23 = 68,149m

__ ΔY₂₃ = Y₂₃ + v_y23 = 133,544m

__ ΔX_3B = X_3B + v_x3B = 55,844m

__ ΔY_3B = Y_3B + v_y3B = 176,269m



_ X₂ = X₁ + __ ΔX₁₂= 176,007m

_ Y₂ = Y₁ + __ ΔY₁₂= 290,187m

_ X₃ = X₂ + __ ΔX₂₃= 244,156m

_ Y₃ = Y₂ + __ ΔY₂₃= 423,731m

_ X_B = X₃ + __ ΔX_3B= 300,000m = X_B

_ Y_B = Y₃ + __ ΔY_3B= 600,000m = Y_B