Aircraft Stability and Performance
2nd Year, Aerospace Engineering
Dr. M. Turner
March 5, 2015
Turning Aircraft
V
R
To execute a turn, aircraft can
either
◮
“Skid to turn” - alter yaw of
aircraft to create side-slip
(basically using rudder)
◮
β = angle of side-slip
Centripetal acceleration: α =
2
Centripetal force: m VR
β
V
(NB - slightly
non-standard illustration)
V2
R
Turning Aircraft
φ
Alternatively
◮ “Bank-to-turn” - roll the
aircraft to alter the direction of
the lift vector
(using ailerons)
◮ φ = roll angle
Lt
α
Aircraft nose
points into
slide
mg
◮
Bank-to-turn is typically more efficient (easier) in most aircraft
◮
Correctly banked turn: centripetal force provided solely by inward
component of lift vector due to banking of aircraft
NB: Rudder input necessary to prevent aircraft side-slipping
Turning aircraft - fundamental relationships
Resolving horizontally and vertically:
Lt cos φ =
φ
mg
Lt
2
Lt sin φ =
V
mα = m
R
α
φ = arctan
V2
gR
Aircraft nose
points into
slide
mg
This implies
◮
◮
For given radius of turn, R, bank angle required to execute turn
must increase
⇒ the faster the aircraft is travelling, the more it must bank to
adhere to a turn of a given radius
For a fixed bank angle, the radius of the turn must increase as the
square of the velocity
Load Factor
◮
An important parameter indicating the “stress” an aircraft (and its
pilot/passengers) can be subjected to.
◮
Load Factor, N defined as
N
=
=
=
lift
in turn
weight
Lt
mg
1
mg /(cos φ)
=
= sec φ
mg
cos φ
◮
N is a ratio of forces (dimensionless) although often given in “g”.
◮
When bank angle is zero (φ = 0), N = 1
Load factor a significant parameter in aircraft turning.
◮
Load Factor - illustration
φ=0
N=1
◮
φ=60
N=1.2
N=2
Load factor indicates “weight gain” of passengers/pilots.
◮
◮
◮
φ=30
Straight, level flight: normal sensation of weight
Correctly banked turn at 60o : sensation of twice normal weight
Typical load factors: commercial aircraft ≈ 2; fighter aircraft
≈ 8 − 10.
Relationship between load factor and turning
Bank angle
sec2 φ
= 1 + tan2 φ
tan2 φ
= sec 2 φ − 1 = N 2 − 1
p
N2 − 1
=
tan φ
Turn radius
tan φ =
p
N2 − 1 =
R
Centripetal
acceleration
α
=
V2
gR
V2
√
g N2 − 1
mV 2
mV 2
=
2
R
√V
g N 2 −1
p
= mg N 2 − 1
=
Bank angle increases with
load factor (for small φ)
For a given load factor,
radius of turn increases
as square of speed
Centripetal acceleration
increases with load
factor.
Alternative expressions for turn radius
Two expressions for lift:
◮
N = Lt /(mg )
◮
Lt = 12 ρSCL,t V 2
⇔




Lt = Nmg



⇒
V2
=
=
Using this value in the expression for turn radius gives
R=
V2
√
g N2 − 1
=
=
Nmg
√
N2 − 1
Nm
√
1
2
2 ρSCL,t N − 1
1
2 ρSCL,t g
Minimum radius turn, Rmin , attained at CL,max .
Rmin =
Nm
√
N2 − 1
1
2 ρSCL,max
Lt
1
ρSC
L,t
2
Nmg
1
2 ρSCL,t
Rate of turn
Rate of turn ω given by
ω=
V
=
R
V
2
√V
g N 2 −1
=
gp 2
N −1
V
Using
V =
s
Nmg
1
2 ρSCL,t
in the expression for ω gives
ω
=
s
=
s
1
2 ρSCL,t
Nmg
g
1
2 ρSCL,t g
m
p
N2 − 1
r
N2 − 1
N
Implication: turn rate increases as a function of CL,t
Turning flight vs Straight level flight
When entering a turn, lift vector direction is altered. Thus to keep
same CL during turn, velocity (throttle) must be adjusted
How does this compare to straight level flight?
Straight Level
Vs2
Turn
mg
= 1
2 ρSCL,s
Vt2 =
mg
= 1
2
2 ρSVs
CL,t =
CL,s
Nmg
1
2 ρSCL,t
Nmg
1
2
2 ρSVt
Thus for same lift coefficient, CL,s = CL,t :
√
NVs
√
i.e speed needs to be increased by N to keep CL the same in turn
as in straight level flight
Vt =
Turning flight vs Straight level flight
Assume now that change in lift coefficient (incidence angle) is
acceptable during turning flight.
How can straight-level airspeed be maintained during turning flight?
Turn
Straight Level
mg
= 1
ρSC
L,s
2
Hence if Vs = Vt :
Vs2
CL,s = CL,t /N
Vt2 =
Nmg
1
2 ρSCL,t
or CL,t = NCL,s
which implies
◮
◮
◮
For same speed in turn, lift coefficient must increase by a factor of N
If aircraft is flying close to stalling speed, attempts to turn the
aircraft while keeping velocity constant could stall the aircraft
In practice, for gentle turns (φ ≤ 25o ) effects of turning are small
(sin φ ≈ φ). For |φ| ≤ 30o , pilots tend not to adjust throttle and
accept a slight loss in speed.
Turning vs straight level flight: remarks
Constant CL
Constant V
⇒ Velocity
√ must be increased:
Vt = NVs
⇒ Increase in thrust required (or
height loss)
⇒ Turn radius increased:
R
=
NV 2
√ s
g N2 − 1
⇒ Lift coefficient must be
increased CL,t = NCL,s
⇒ Increased CL ⇒ increased α
⇒ Turn radius decreased (danger
of stall!)
Height loss during turning
Assumptions:
◮
◮
Thrust (throttle) is constant and independent of speed
Turn takes place at same lift coefficient for initial level flight
Straight Level Resolving normal to
flight path
LT
=
mg cos γ
⇒ L cos φ =
mg cos γ
DT
γ
Assuming γ small, cos γ ≈ 1 and
hence
L cos φ = mg
LT
(N = sec φ)
γ − flight path angle
TT
mg
Height loss during turning
LT
Resolving along flight path
DT
TT + mg sin γ = DT
γ
γ − flight path angle
TT
mg
Since thrust = drag in straight level flight
T = D = TT
Also,
DT = ND
Hence
D + mg sin γ
=
sin γ
=
ND
D(N − 1)
mg
=
1
(N − 1)
CL /CD
Implication: minimum angle occurs at minimum drag condition (also