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:

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

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