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G. Leng, Flight Dynamics, Stability & Control
Lecture 3 : Static Longitudinal Stability
Or the balancing of lift forces and pitching moments
G. Leng, Flight Dynamics, Stability & Control
1.0 Trim condition
An aircraft is trimmed if there are no nett forces and momentsacting on it
E.g. The nett pitching moment coefficient is zero
M = 1/2 V2 S c Cm = 0
pitching moment coefficient
G. Leng, Flight Dynamics, Stability & Control
1.1 Concept of static stability
Static stability refers to the tendency of the aircraft todevelop forces or moments to return to its trimcondition when disturbed
Question : What does that mean ?
G. Leng, Flight Dynamics, Stability & Control
Figure 1.1 : Static stability
G. Leng, Flight Dynamics, Stability & Control
1.2 Longitudinal static stability
Longitudinal static stability refers to the tendency of the aircraft toreturn to its trim condition after a nose up or nose down disturbance
This implies that Cm must vary with AOA in a certain manner !!
Question : What manner ?
G. Leng, Flight Dynamics, Stability & Control
Figure 1.2 : Cm variation with AOA
Cm
Cm
G. Leng, Flight Dynamics, Stability & Control
Figure 1.3 : F4 wind tunnel data - pitching moment
Source : NASA Technical Note D 6425
G. Leng, Flight Dynamics, Stability & Control
Moral
A statically stable aircraft must have Cm / < 0
Question : Is static stability always a good thing ?
Note : Aeronautical engineers usually write Cm / as Cm
G. Leng, Flight Dynamics, Stability & Control
1.3 Tail configuration - statically stable
W
c.g. wing
Lw
Ltlw
lt
tail
G. Leng, Flight Dynamics, Stability & Control
1.4 Tail configuration - statically unstable
W
c.g.wing
Lw
Lt
lw lt
tail
G. Leng, Flight Dynamics, Stability & Control
Moral
Aircraft stability depends critically on cg location
Note : All aircraft have forward and aft cg limits
Question : What happens if either forward or aft limits are violated ?
G. Leng, Flight Dynamics, Stability & Control
2.0 Quantifying static longitudinal stability
W
c.g.
wing
Lw Lt
Xw Xt
tail
Xcg
V
em0
G. Leng, Flight Dynamics, Stability & Control
The trim conditions require that
Lw + Lt = W
m = 0
Where m = m0 + Lw (Xw – Xcg) + Lt (Xt – Xcg)
= m0 + Lw lw + Lt lt
G. Leng, Flight Dynamics, Stability & Control
2.1 : Lift curve slope
Lw = ½ V2 Sw CLw
The key is to express aircraft lift and pitching moment in terms ofcontributions from the wing and the tail.
= ½ V2 SW aw
G. Leng, Flight Dynamics, Stability & Control
2.2 : Effect of downwash on the tail lift
Lt = ½ Vt2 St CLt
= ½ (V2) St at [ (1 - /) + e ]
Recall the tail surface is affected by downwash
G. Leng, Flight Dynamics, Stability & Control
2.3 : Aircraft lift coefficient
Define the aircraft lift as L = Lw + Lt . In non-dimensional form
½ V2 Sw CL = ½ V2 Sw CLw + ½ Vt2 St CLt
CL = CLw + (Vt /V)2 (St/Sw) CLt
= CLw + (St/Sw) CLt
G. Leng, Flight Dynamics, Stability & Control
The aircraft lift curve slope a = (CL)/ is given by :
a = aw + (St/Sw) at ( 1- )
G. Leng, Flight Dynamics, Stability & Control
2.4 : Aircraft pitching moment coefficient
Define the aircraft pitching moment as :
m = m0 + Lw lw + Lt lt
Non-dimensionalise i.e. divide by ½ V2 Sw c
Cm = Cm0 + (lw/c) CLw + (St lt) /(Swc) CLt
G. Leng, Flight Dynamics, Stability & Control
In terms of aoa and tail deflection
Cm = Cm0 + (lw/c) aw + VH at [(1-) + e]
Cm = Cm0 + [ (lw/c) aw + VH at (1-) ]
+ [ VH at ] e
How would you interpret this ?
Collecting terms…
G. Leng, Flight Dynamics, Stability & Control
2.5 : Neutral point & static margin
Differentiate Cm with respect to
Cm = (lw/c) aw + VH at (1- )
Express in terms of cg location…
Cm = [(Xw - Xcg )/c] aw + (St/Sw) [(Xt - Xcg) /c] at (1- )
= [ (Xw/c) aw + (St/Sw) (Xt/c) at (1- ) ]
- [ aw + (St/Sw) at (1- ) ] (Xcg/c)
G. Leng, Flight Dynamics, Stability & Control
The neutral point is the cg location where Cm = 0, i.e.
Xnp/c = [(Xw - Xt)/c] (aw/a ) + (Xt/c)
Cm = [(Xw - Xt)/c] aw + a (Xt/c) - a (Xcg/c)
Rewrite the pitching moment curve slope in terms of Xnp/c …
Cm = a (Xnp/c) - a (Xcg/c)
= - a [ ( Xcg – Xnp) /c ]
