Slope of track

Calculate the average gradient (in per mille and in degrees) of the railway tracks between Prievidza (309 m above sea level) and Horná štubňa (624 m above sea level), given that the track is 37 km long.

Final Answer:

s₁ =  8.51 ‰
s₂ =  0.4878 °

Step-by-step explanation:

$$\begin{gathered} s=37\text{km}\to \text{m}=37\cdot 1000\text{m}=37000\text{m} \\ y=|309-624|=315\text{m} \\ {s}^2=x^2+y^2 \\ x=\sqrt{s^2-y^2}=\sqrt{37000^2-315^2}\doteq 36998.6591\text{m} \\ {s}_1=1000\cdot \frac{y}{x}=1000\cdot \frac{315}{36998.6591}=8.51‰ \end{gathered}$$

$$\begin{gathered} r=\arctan (\frac{y}{x})=\arctan (\frac{315}{36998.6591})\doteq 0.0085\text{rad} \\ {s}_2=r\to \text{°}=r\cdot \frac{180}{\pi }\text{°}=0.0085\cdot \frac{180}{\pi }\text{°}=0.488\text{°}=0.4878^{\circ }=0°29^{\prime }16" \end{gathered}$$

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