Opracowanie pionowych łuków kołowych trasy - obliczenia

1) Łuk 1-2-3

Dane :

R₁ = 2400 m

H₁= 110,716 m

H₂ = 119,210 m

H₃ = 112,702 m

ΔH_1-2 = H₂-H₁ = 8,494 m ΔH_2-3 = H₃-H₂ = -6,508 m

d_1-2 = 1452,052 – 700 =752,052 d_2-3= 2320,015- 1452,052 = 867,963 m

i_1-2 =$\frac{{H}_{1 - 2}}{d_{1 - 2}}$= +0,011294 i_2-3 =$\frac{{H}_{2 - 3}}{d_{2 - 3}}$ = -0,007498

t = $\frac{R}{2}$ * ׀(i_1-2 ± i_2-3) ׀=22,550 m

H_p= H₂ ± i_1-2 * t =118,955 m H_k = H₂ ± i_2-3 * t =119,041 m

WS =$\frac{t^{2}}{2R}$ =0,106 m H_S = H₂ –WS =119,104 m

2) Łuk 2-3-4

Dane :

R₂= 4000 m

H₂ = 119,210 m

H₃ = 112,702 m

H₄ = 105,514 m

ΔH_2-3 = H₃-H₂ = -6,508 m ΔH_3-4 = H₄-H₃ = -7,188 m

d_2-3 = 867,963 m d_3-4= 799,995 m

i_2-3 =$\frac{{H}_{2 - 3}}{d_{2 - 3}}$= - 0,007498 i_3-4=$\frac{{H}_{3 - 4}}{d_{3 - 4}}$ = -0,008951

t = $\frac{R}{2}$ *׀(i_2-3 ± i_3-4) ׀= 2,906 m

H_p= H₃ ± i_2-3 * t = 112,723 m H_k = H₃ ± i_3-4 * t = 112,675 m

WS = $\frac{t^{2}}{\ 2R}$ = 0,001 m H_s = H₃- WS= 112,701 m

Opracowanie pionowych łuków kołowych trasy - obliczenia

3) Łuk 3-4-5

Dane:

R =4200 m

H₃ = 112,702 m

H₄ = 105,514 m

H₅ =116,510 m

ΔH_3-4 = H₄-H₃ = -7,188

ΔH_4-5 = H₅-H₄ = 10,966

d_3-4= 799,995m d_4-5 = 2195,290 m

i_3-4=$\frac{{H}_{3 - 4}}{d_{3 - 4}}$ =-0,008951 i_4-5 =$\frac{{H}_{4 - 5}}{d_{4 - 5}}$=0,006994

t = $\frac{R}{2}$ * ׀ i_3-4±i_4-5 ׀= 33,485 m

H_p= H₄ ± i_3-4 * t = 105,814 m H_k = H₄ ± i_4-5 * t = 105,748 m

WS =$\frac{t^{2}}{2R}$ = 0,133 m H_s = H₄+WS = 105,647 m