mean of maximum daily temperature is obtained as $36^o$ c and $42^o$ C respectively . Assume the effective gradient of 2% on the runway.

• Basic Runway length $=800 \mathrm{~m}$.
  • Elevation correction $=\frac{7}{100} \times 800 \times \frac{180}{300}$ $=33.6 \mathrm{~m}$
  • Correction Runway length after Elevation correction $=833.6 \mathrm{~m}$
  • $A R T$ (Airport Ref. $\operatorname{Temp} ) =T_{a}+\left(\frac{T_{m}-T_{a}}{3}\right)$ $=36+\left(\frac{42-36}{3}\right)=38^{\circ} \mathrm{C}$
  • Stand. Atm Temp. $=15^{\circ}-(0.0065 \times 180)=13.83^{\circ} \mathrm{C}$
  • Rise in temp $=38-13.83=24.17^{\circ} \mathrm{C}$
  • $\begin{aligned} \text { Temp.Correction } &=\frac{1}{100} \times 833.6 \times \frac{24.17^{\circ} \mathrm{C}}{1^{\circ} \mathrm{C}} \\ &=201.48 \mathrm{~m} \end{aligned}$
  • Corrected Runway length after Temp $=833.6+201.48$ $=1035.08 \mathrm{~m}$
  • Total% correction (Temp. & Elevation) $=\frac{(1035 \cdot 08-800)}{800}$ $=29.38\lt35 \% \mathrm{OK}$
  • Correction For gredient $=\frac{0.2 \%}{1 \%} \times 1035.08 \times 20$=414.032m;
  • corrected length $=(1035.08+414.032 )\mathrm{~m. }$
  • Hence, Final Runway Length $=1449.11 \mathrm{~m}$