An exit taxiway for an airport is to be designed. If the turnoff speed is 80 kmph and the coefficient of friction between tire and pavement surface is 0·13, the radius of the central curve will be nearly
- A354 m
- B372 m
- C394 mCorrect
- D422 m
Solution
Concept
Radius of the exit taxiway curve
Formula
R = V² / (125 f) (V in km/h)
Step-by-step solution
1. R = 80² / (125 × 0.13) = 6400 / 16.25 = 393.8 m ≈ 394 m.
Explanation
The radius of the central curve of a taxiway is obtained from the speed and the coefficient of friction, using the formula used for airfield curves.
Why the other options are wrong
354, 372 and 422 m do not equal 80²/(125 × 0.13).
Shortcut
R = V²/125f.
Common mistake
Using 127 (highway formula): gives 387.6 m.
Correct answer: C) 394 m