Question

Determine the turning radius of a taxiway for super-sonic jet, with following data:
Wheel base = 18 m
Tread of main landing gear = 6.5 m
Coefficient of friction between tyre and pavement = 0.15
Turning speed = 50 kmph

• A. 133.33 m
• B. 148.42 m
• C. 120 m
• D. 180 m

Answer 1

The correct option is D 180 m

(i) Turning radius, $R=\frac{V^2}{125\times f}=\frac{{50}^2}{125\times 0.15}=133.33m$

(ii) From Horonjeff's equation,

$R=\frac{0.388W^2}{[\frac{T}{2}-S]}=\frac{0.388\times {18}^2}{\frac{22.5}{2}-(6+\frac{6.5}{2})}=62.86m$

(iii) For super-sonic jet,
R = 180 m
$\therefore$ $R_{min}$ = [Minimum of (i), (ii) and (iii)] = 180 m