The distance between two contour lines in a map with 20m. contour interval is 2.2cm. The scale of the map is 1:50000.
- What is the horizontal distance between the two contour lines?
- What is the slope distance?
- What is the slope in the ground in
- percent
- degrees
- radians
- grads
Let:
$GD = HD$
\[
MS = \frac{MD}{HD}
\]
\[
\frac{1}{50000} = \frac{2.2cm.}{HD}
\]
\[
HD = 110000cm. (\frac{1m.}{100cm.})
\]
\[
HD = 1100m. \leftarrow \text{answer}
\]
\[
SD^2 = HD^2 + VD^2
\]
\[
SD^2 = (1100m.)^2 + (20m.)^2
\]
\[
SD = 1100.18180315801m.
\]
\[
SD \approx 1100.18m. \leftarrow \text{answer}
\]
a. Percent Slope
\[
\%Slope = \frac{VD}{HD} * 100
\]
\[
\%Slope = \frac{20m.}{1100m.} * 100
\]
\[
\%Slope = 1.81818181818
\]
\[
\%Slope \approx 1.82 \leftarrow \text{answer}
\]
b. Degrees Slope
\[
\tan \theta = \frac{VD}{HD}
\]
\[
\tan \theta = \frac{20m.}{1100m.}
\]
\[
\theta = \arctan \frac{20m.}{1100m.}
\]
\[
\theta = 1^{\circ}2’29.86″ \leftarrow \text{answer}
\]
c. Radians Slope
\[
rad = 1^{\circ}2’29.86″ (\frac{2 \pi}{360^{\circ}})
\]
\[
rad = 0.0181798343
\]
\[
rad \approx 0.02 \leftarrow \text{answer}
\]
d. Grads Slope
\[
^g = 1^{\circ}2’29.86″ (\frac{400^g}{360^{\circ}})
\]
\[
^g = 1.15736419753
\]
\[
^g \approx 1.16 \leftarrow \text{answer}
\]