Question

The longitudinal section of a runway provides the following data: \[ \begin{array}{|c|c|} \hline \text{End-to-end runway (m)} & \text{Gradient (\%)} \\ \hline 0 \text{ to } 300 & + 1.2 \\ 300 \text{ to } 600 & - 0.7 \\ 600 \text{ to } 1100 & + 0.6 \\ 1100 \text{ to } 1400 & - 0.8 \\ 1400 \text{ to } 1700 & - 1.0 \\ \hline \end{array} \] The effective gradient of the runway (in %, rounded off to two decimal places) is \(\underline{\hspace{2cm}}\). 
 

Hint:

To calculate the effective gradient, use the weighted average of the gradients for each section of the runway, considering their lengths.

Answer 1

Correct Answer: 0.3 - 0.34

The effective gradient of a runway is determined by finding the net change in elevation over the total length of the runway. Given the data, we'll calculate the total elevation change and then derive the effective gradient.

The runway is divided into segments, each with its gradient. We calculate the elevation change for each segment as follows:

Segment (m)	Gradient (%)	Length (m)	Elevation Change (m)
0 to 300	+1.2	300	300×(1.2/100)=3.6
300 to 600	-0.7	300	300×(-0.7/100)=-2.1
600 to 1100	+0.6	500	500×(0.6/100)=3.0
1100 to 1400	-0.8	300	300×(-0.8/100)=-2.4
1400 to 1700	-1.0	300	300×(-1.0/100)=-3.0

Sum the elevation changes: \(3.6 - 2.1 + 3.0 - 2.4 - 3.0 = -0.9\) m.

Total runway length: 1700 m.

Effective gradient (%) = \(\frac{\text{Total Elevation Change (m)}}{\text{Total Runway Length (m)}}\times 100 = \frac{-0.9}{1700}\times 100 \approx -0.053\%\).

Thus, the effective gradient of the runway is approximately \(0.05\%\). Since gradients are typically represented as positive values when concerning effective gradient calculations, the absolute value will be used.