TAKE-OFF REQUIREMENTS AND THRUST Flashcards
What factors have to be taken into accounct when considering take-off requirements?
- TOFL
- Climb requirements
- Brake Energy limits/Tyre speed limits
- Structural limits
ACN/PCN
Obstacle clearances
Define TOFL
The greater of
- ASD
- ASG
- TOD x 1,15
Define the term “Regulated Take-off weight”
The most restrictive of all relevant weight limitations
What two parts can Climb requirements be divided into?
- Climb limit mass: Air gradient that doesn’t take obstacles into account
- Obstacle limit mass: Ground gradient
What effect does flaps have on FLTOM and climb limit mass?
More flaps increases FLTOM but reduces climb limit mass
What are the obstacle clearance requirements for Take-off?
Obstacles have to be cleared by:
- 35 ft vertically or by
- 90 m + (0,125 x D)
How big is the domain widths for horizontal obstacle clearances?
- Turns less than 15 deg:
Up to 300 m in VMC by day or good nav accuarcy in IMC/Night
Up to 600 m if nav accuracy is not good enough - Turns more than 15 deg:
600 m in VMC by day
900 m in IMC or by night
Turn limitations up to 1’500 ft?
No turns up to 50 ft
Less than 15 deg up to 400 ft
Less than 25 deg up between 400 ft - 1’500 ft
What is “Reference zero”?
35 ft above runway is reference zero and that’s where the 1 st climb segment starts
Describe the 4 climb segments
- Starts at 35 ft. Ends when gear is up
Speed is V2(+10)
Gradient must be positive - Starts when gear is up and ends at 400 ft
Speed is V2(+10)
Gradient must be min 2,4% - Starts at 400 ft and ends when flaps are up
Speed is accelerating to Venr
Gradient is 0% - Starts when flaps are up and ends at 1’500 ft
In the take-off flight path, what are the gross gradients reduced by?
0,8% for 2-engines
0,9% for 3-engines
1,0% for 4-engines
Calculate climb gradient from the following:
Aircraft must be at 4’500 ft by 15 nm
What’s the required gradient?
Horizontal dist : 15 nm = 91’200 ft
Vertical gain: 4’500 ft
4’500/91’200 x 100 = 4,9%
Calculate the following:
Required gradient is 4%
TAS = 110, GS = 100
10’336/100 x 4 = 413 fpm
In 1 min @ 100 kts the a/c will travel 1,7 nm = 10’336 ft (1nm=6080ft)
10’336 ft is 100% so therefore 4% becomes 413 fpm
Describe the effect of flaps on take-off
Flaps increases CLmax
Flaps decreases the TODR
Flaps decreases the climb gradient
Describe the forces “Thrust” and “Power” when engines are spooled up just before brake release
There’s a lot of thrust but TAS is zero so there’s no power