# Slope of track

Calculate the average slope (in permille and even in degrees) of the rail tracks between Prievidza (309 m AMSL) and Nitrianske Pravno (354 m AMSL), if the track is 11 km long.

Result

s1 =  4.09
s2 =  0.234 °

#### Solution:

$s=11 \ km \rightarrow m=11 \cdot \ 1000 \ m=11000 \ m \ \\ y=|309 - 354|=45 \ \text{m} \ \\ \ \\ x=\sqrt{ s^2-y^2 }=\sqrt{ 11000^2-45^2 } \doteq 235 \ \sqrt{ 2191 } \ \text{m} \doteq 10999.908 \ \text{m} \ \\ \ \\ s_{1}=1000 \cdot \ \dfrac{ y }{ x }=1000 \cdot \ \dfrac{ 45 }{ 10999.908 } \doteq 4.0909 \doteq 4.09 \ ‰$
$s_{2}=\dfrac{ 180^\circ }{ \pi } \cdot \arctan(\dfrac{ y }{ x } )=\dfrac{ 180^\circ }{ \pi } \cdot \arctan(\dfrac{ 45 }{ 10999.908 } ) \doteq 0.2344 \doteq 0.234 ^\circ \doteq 0^\circ 14'4"$

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