AE 429 - Aircraft Performance and
Flight Mechanics
Takeoff and Landing
Takeoff Performance
V=0 (s=0)
V=Vstall
V=Vmcg (min control speed on the ground)
V=Vmca (min control speed in the air
(w/o landing gear in contact with ground))
V=V1 decision speed (or critical engine
failure speed)
– balance speed length
V=VR (takeoff rotation speed) L W
V=VMU (min unstick speed, ground
clearance) α αcl tail
V=VLO=1.1 Vstall (s=sg) (lift off speed)
airborne but before it
clears an obstacle of
height h*
ground
roll
s = sg + sa
50ft
VLO
VMU
V1
VR
Vmca
Vmcg
Vstall
h*
s=sg
s=sa
1
Takeoff Speed and FAR 25
requirements
Speed
FAR 25
Description
Requirement
Vs
stall speed in takeoff configuration
-
Vmc
minimum control speed with one engine inoperative (OEI)
-
V1
OEI decision speed
= or > Vmc
Vr
rotation speed
5% > Vmc
Vmu
minimum unstick speed for safe flight
= or > Vs
Vlof
liftoff speed
10% > Vmu
5% > Vmu
(OEI)
V2
takeoff climb speed at 35 ft
20% > Vs
10% > Vmc
Examples
of Takeoff
Speeds
Aircraft
Takeoff Weight
Takeoff Speed
Boeing 737
100,000 lb
45,360 kg
150 mph
250 km/h
130 kts
Boeing 757
240,000 lb
108,860 kg
160 mph
260 km/h
140 kts
Airbus A320
155,000 lb
70,305 kg
170 mph
275 km/h
150 kts
Airbus A340
571,000 lb
259,000 kg
180 mph
290 km/h
155 kts
Boeing 747
800,000 lb
362,870 kg
180 mph
290 km/h
155 kts
Concorde
400,000 lb
181,435 kg
225 mph
360 km/h
195 kts
2
The critical engine speed defines the point on the runway at which the
distance needed to stop is exactly the same as the that required to
reach takeoff speed. The resulting total takeoff distance is
correspondingly known as the balanced field length.
Definition of critical engine-failure speed and balanced field length
EOM Airplane During Takeoff
This equation gives instantaneous forces during the acceleration
F = T − D − R = T − D − µr (W − L ) = m
P = V∞T
T = const
const
T=
V∞
R ≡ µ r (W − L )
rolling
friction
coeff.
turbojet
T = k1* − k *2V∞ + k *3V∞2
W = const
reciprocating
engine/propeller
dV
dt
turbofan
Weight
r
3
EOM Airplane During Takeoff
–
L=
Both L and D vary with V
1
ρ ∞V∞2 SCL
2
D=
1
ρ∞V∞2 SCD
2
CD = CD 0 + ∆CD 0 + ( k1 + Gk3 ) CL2
∆CD 0 =
W
K uc m −0.219
S
W /S
Drag Polar
[N/m2]
m
k2 = 0
Wave Drag
max mass aircraft [kg]
K uc = 5.81 ⋅10 −5 ; K uc = 3.16 ⋅10−5
zero flap;
–
max flap down
“Ground Effect”
(16h / b )
G=
2
1 + (16h / b )
2
CDi (in ground effect)
=G
CDi (out-of-ground effect)
h = height of wing above the ground, b = wingspan
EOM Airplane During Takeoff
L=
1
ρ ∞V∞2 SCL
2
limited and determined by
features of the geometric design
configuration of the airplane
rolling along the ground
CL ≤ 0.1
F = T − D − R = T − D − µ r (W − L ) = m
r
dV∞
dt
V∞ ( t = 0 ) = 0;V∞ ( t = t LO ) = VLO = 1.1VMU ≈ 1.1Vstall
Ground Roll
Distance
V (t )
ds =
sg
0
solved numerically
ds
dt = V∞ dt
dt
ds =
t LO
0
V∞ dt
ssg =
tLO
0
V∞ dt
4
Average Forces Acting During Takeoff
Feff = T − [D + µr (W − L )]average ≈ CONSTANT
–
Why consider “average” force during the takeoff roll?
- average 0.7 VLO
Takeoff Roll
We are no longer considering a “Statics”
problem
–
–
Finite (even large) accelerations are present
we apply Newton’s second law to any body initially
at rest,
V>0
V=0
Feff = constant
s=0
t=0
F = ma = m
–
Integrating
V
0
s
0
t
F
dV = eff dt
0
m
ds =
t
0
Vdt =
Feff = constant
s
t
dV
dt
dV =
F
V = eff t
m
F
F
tdt = eff
0 m
m
t
OR
t
0
tdt
F
dt
m
t=
Vm
Feff
s=
Feff t 2
m 2
5
Approximate Analysis Ground Roll
ds
dt
dt = V∞ dt = V∞
dV∞
dt
dV∞
ds =
ds = V∞
2
dV∞
1 d (V∞ )
=
( dV∞ / dt ) 2 ( dV∞ / dt )
dV∞ 1
g
= (T − D − µr (W − L ) ) = (T − D − µr (W − L ) )
dt
m
W
VLO
sg =
0
KT =
d (V∞2 )
2 g ( KT − K AV∞2 )
ρ∞
0
d (V∞2 )
W
+ NVLO
2 g (T − D − µ r (W − L ) )
2 (W / S )
CD 0 + ∆CD 0 + k1 +
N = 3 large aircraft;
NVLO
VLO
approx
T
− µr
W
KA = −
=
~ constant , value at V = 0.7 VLO
G
C 2 − µr CL
π eAR L
N = 1 small aircraft;
distance covered during rotation
Takeoff Roll
Lift-off distance for a jet airplane
–
Using the average Force we have postulated
The sum
Feff = T − D + µ r (W − L )
average
Is fairly constant
Thrust is also fairly
constant
–
s=
V 2m
2 Feff
sLO =
Substituting into
D + µ r (W − L )
sLO =
{
2
VLO
W
2 gFeff
2
VLO
W
2 g T − D − µ r (W − L )
average
}
+ NVLO
6
Takeoff Roll
Lift-off distance (continued)
–
Generally we use a safety margin at lift-off by 10%
VLO = 1.1Vstall = 1.1
–
sLO =
2W
ρ∞ SCLmax
Substituting into the expression for sLO
{
1.21(W / S )
g ρ ∞CLmax T / W − D / W + µr (1 − L / W )
–
D + µ r (W − L )
average
average
}
+ 1.1N
≈ D + µ r (W − L )
1 W
ρ ∞ CLmax S
2
0.7VLO
One method of estimating quickly the average drag and rolling
resistance force is to use 0.7VLO in calculating the aerodynamic forces
Takeoff Ground Roll Distance –
Simplified Equation
T >> D + µ r (W − L )
sLO =
0.7VLO
1.21(W / S )
g ρ ∞CLmax {T / W }
T ∝ ρ∞
sLO ∝ W 2
sLO ∝ 1/ ρ ∞2
hot days, low air density
cold days, high air density
sLO at altitude is > sLO at sea level
7
Distance While Airborne to Clear an
Obstacle
V ≈ 1.15Vstall
2
1
ρ S 1.15Vstall ) 0.9CL max
L 2 ∞ (
n= =
W
W
2
1
ρ∞ S (1.15Vstall ) 0.9CL max
=2
= 1.19
2
1
ρ∞ S (Vstall ) CL max
2
R=
2
2
1.15Vstall
6.95Vstall
=
g (1.19 − 1)
g
cos θ OB =
R − hOB
h
= 1 − OB
R
R
sa = R sin θOB
R-hOB
CL ≈ 0.9CL max
R=
R
θOB
Flight
Path
V∞2
g ( n − 1)
hOB
sa
V ≈ 1.1Vstall
ground
roll
V ≈ 1.2Vstall
airborne but before it
clears an obstacle of
height h*
s = sg + sa
8
L = W cos θ a
Landing Performance
V∞ = Va
V∞ = VTD
D
L = W cos θ a
D −T
D T
θa = −
W
W W
o
θa ≤ 3
cos θ a ≈ 1 L ≈ W
L
Flight
Path
sa
Approach
Distance
sin θ a =
Touch Down
R
θa
hf
W
θf
R
T
hOB=50ft
D = T + W sin θ a
sf
θa ≈
sfr
Flare
Distance
V∞ = VTD
V∞ = 0
Free Roll
Distance
sg
sa =
( 50 − h )
f
tan θ a
1
T
−
L /W W
Ground
Roll Distance
h f = R − R cos θ f = R (1 − cos θ f ) = R (1 − cos θ a )
since
θ f = θa
Landing Performance
R=
V∞2
g ( n − 1)
from Va = 1.3Vstall
to
VTD = 1.15Vstall
commercial airplanes
from Va = 1.2Vstall
to
VTD = 1.1Vstall
military airplanes
Vaverage = V flare = 1.23Vstall
commercial airplanes
Vaverage = V flare = 1.15Vstall
military airplanes
Load factor n = 1.2
sa =
( 50 − h )
f
tan θ a
R=
2
V flare
0.2 g
θa = θ f
Radius of turn
s f = s flare = R sin θ f
9
Landing data
The figure shows the landing data for two different flap angles for some
airplanes.
The FAR landing field length is defined as the actual demonstrated distance
from a 50 ft. height to a full stop increased by the factor 1/0.60, a 67% increase.
Landing Roll – Ground Roll
The landing roll is very similar to the takeoff ground roll,
except:
–
–
Thrust = 0
(or near zero)
The sign of the
acceleration is
negative
− [D + µ r (W − L )]average = m
–
dV
dt
We seek an approximate
expression like
STO for landing
using an average
Force again
10
Landing Roll
Following this logic
F ≈ −[D + µ r (W − L )]average
–
–
≈ −[D + µ r (W − L )]0.7V
T
Notice that this assumption is less accurate for landing than
for takeoff
Nonetheless, let’s do the integration again
0
sL
ds =
F
m
t
0
sL = −
dt
F t2
m 2
or sL = −
V 2m
2F
Landing Roll
Combining the previous expression for sL and using
the approximately constant retarding force
sL ≈
–
VTD2 W
2 g D + µ r (W − L )
again, adding a safety factor (10% in this case) to touch
down at a speed above the stall speed
VTD ≥ 1.15Vstall commercial airplanes
Vstall =
r
VTD ≥ 1.10Vstall military airplanes
–
0.7VTD
2W
ρ∞ SCL
max
Substituting into the expression above for sL
sL ≈
2.645W 2
g ρ ∞ SCLmax D + µr (W − L )
(commercial airplane)
0.7VTD
11
Landing Roll
Landing roll can be reduced if a thrust reverser is installed
Thrust reverser: 40-50% max T for Jet airplane;
40% max T for Reciprocating engine/propeller; 60% max T for Turbofan
−TR − D − µr (W − L ) = m
dV
dt
where TR = Reverse Trust
D can be increased with: Spoilers, speed brakes, drogue chutes
–
If the thrust reverser produces constant thrust,
W2
sL ≈
g ρ ∞ SC Lmax
–
{ T+
D + µ r (W − L )
0.7VT
}
Lift and drag forces are calculated, accounting for ground effect as
we did for takeoff roll
L=
1
ρ ∞V∞2 SCL
2
D=
(16h / b )
G=
2
1 + (16h / b )
1
C2
ρ ∞V∞2 S CD ,0 + G L
2
π eAR
2
Where
Landing Roll - sg
m
dV
<0
dt
ds =
d (V∞2 )
m
2 −TR − D − µr (W − L )
sg − s fr =
W
2g
sg = NVTD +
d (V∞2 )
0
VTD
W
2g
−TR − D − µr (W − L )
VTD
0
d (V∞2 )
s fr = NVTD
;
TR + D + µr (W − L )
= NVTD +
VTD ≥ 1.15Vstall = jVstall
commercial airplanes
VTD ≥ 1.10Vstall = jVstall
military airplanes
sg = jN
N=1 (mil); 3 (comm)
WVTD2
1
2 g TR + D + µ r (W − L )
Vstall =
0.7VTD
2W
ρ∞ SCLmax
2 W 1
W /S
1
+ j2
ρ ∞ S CL max
ρ ∞ gCL max TR / W + D / W + µr (1 − L / W )
0.7VTD
12