Given the following information, what is the distance from the end of the TODR to 1500 ft above reference zero for the purpose of obstacle clearance for a performance class B aeroplane?
Cloud base: 300 ft above reference zero
Tas: 101 kt
No wind
All engines rate of climb at take off power: 1830 ft/min
Single engine rate of climb at take off power: 400 ft /min
Refer to figures.
- "Above Reference zero", reference zero is the point on the ground where the aircraft reaches 50 ft during the take-off phase.
- "The gradient from 50 ft to the assumed engine failure height is the average all-engine gradient x 0.77"
- "If visual reference for obstacle avoidance is lost, it is assumed that the critical power unit becomes inoperative at this point. All obstacles encountered in the accountability area must be cleared by a vertical interval of 50 ft"
- Wind Calm => TAS = GS = 101 kt
The exercise can be solved in 2 steps:
1) AEO gradient (with fatorization of 0.77) from 50 ft to cloud base (300 ft):
Net gradient = 1830 x 0.77 = 1410 ft/min
Climb = 300 (cloud base) - 50 ft (regulatory height) = 250 ft
250 ÷ 1410 ft/min = 10.63 sec
(10.63 sec x 101 kt) ÷ 3 600 = 0.3 NM
2) OEI gradient from cloud base to 1500 ft:
1500 ft - 300 (cloud base) = 1200 ft
1200 ft ÷ 400 ft/min = 3 min
3 min x 101 kt / 60 s = 5.05 nm
- Total Ground Distance to climb to 1500 ft = 0.30 + 5.05 = 5.35 nm
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