Wyznaczanie trasy o zadanym spadku

i = 2,6%

Punkt A

d₁= 5 m d₂=11 m

H₁=190 m H₂=190,5 m

d=16 m ∆H= 0,5 m

$$i = \ \frac{H}{d} = \frac{0,5}{16} = 0,03$$

H_A= H₁ + d₁∙i = 190 + 5 ∙ 0,03 = 190,15 m

Punkt B

d₁= 2 d₂=5 m

H₁=190,5 m H₂=190,2 m

d=7 m ∆H= 0,3 m

$$i = \ \frac{H}{d} = \frac{0,3}{7} = 0,03$$

H_A= H₁ - d₁∙i = 190,5 - 2 ∙ 0,04 = 190,42 m

$$d_{w} = \frac{h_{w}}{i_{p}} = \frac{0,5}{0,026} = \mathbf{19,23\ m}$$

$$d_{A} = \frac{h_{A}}{i_{p}} = \frac{H_{2} - H_{A}}{i_{p}} = \frac{190,5 - 190,15}{0,026} = \mathbf{13,46}\mathbf{m}$$

$$d_{B} = \frac{h_{B}}{i_{p}} = \frac{H_{1} - H_{B}}{i_{p}} = \frac{190,5 - 190,42}{0,026} = \mathbf{3,08}\mathbf{m}$$