After applying the correction for elevation and temperature, the runway length is 700 m.
The corrected runway length (in m) for an effective gradient of 1% is __________ (round off to the nearest integer).
Given: Corrected runway length after applying elevation and temperature corrections: \[ l = 700 { m},\quad {Gradient} = 1\% \] According to standards, for every 1% gradient, the runway length increases by 20%. \[ {Correction} = 700 \times \frac{20}{100} = 140 \] \[ {Corrected length} = 700 + 140 = 840 { m} \] \[ {Or simply: } 700 \times 1.2 = 840 { m} \] Hence, the final corrected length is \( \boxed{840 { m}} \).
| Point | Staff Readings Back side | Staff Readings Fore side | Remarks |
|---|---|---|---|
| P | -2.050 | - | 200.000 |
| Q | 1.050 | 0.95 | Change Point |
| R | - | -1.655 | - |