16.333: Lecture #2

Static Stability

Aircraft Static Stability (longitudinal)

Wing/Tail contributions

0 Fall 2004 16.333 2–1 Static Stability

• Static stability is all about the initial tendency of a body to return to its equilibrium state after being disturbed

• To have a statically stable equilibrium point, the vehicle must develop a restoring force/moment to bring it back to the eq. condition

• Later on we will also deal with dynamic stability, which is concerned with the time history of the motion after the disturbance – Can be SS but not DS, but to be DS, must be SS ⇒ SS is a necessary, but not sufficient condition for DS

• To investigate the static stability of an aircraft, can analyze response to a disturbance in the angle of attack

– At eq. pt., expect moment about c.g. to be zero CMcg = 0 – If then perturb α up, need a restoring moment that pushes nose back down (negative) Fall 2004 16.333 2–2

• Classic analysis:

– Eq at point B – A/C 1 is statically stable

• Conditions for static stability ∂C C = 0; M ≡ C < 0 M ∂α Mα note that this requires C | > 0 M α0

• Since CL = CLα (α − α0) with CLα > 0, then an equivalent condition for SS is that ∂C M < 0 ∂CL Fall 2004 16.333 2–3 Basic Aerodynamics

Lw c.g. M acw iw Z Dw cg

�FRL �w Fuselage Reference Line (FRL) Xac Wing mean chord

Xcg

• Take reference point for the wing to be the aerodynamic center (roughly the 1/4 chord point)1

• Consider wing contribution to the pitching moment about the c.g.

• Assume that wing incidence is iw so that, if αw = αF RL + iw, then

αF RL = αw − iw

– With xk measured from the leading edge, the moment is:

Mcg = (Lw cos αF RL + Dw sin αF RL)(xcg − xac)

+(Lw sin αF RL − Dw cos αF RL)(zcg) + Macw

– Assuming that αF RL � 1,

Mcg ≈ (Lw + DwαAF RL)(xcg − xac)

+(LwαF RL − Dw)(zcg) + Macw – But the second term contributes very little (drop)

1The aerodynamic center (AC) is the point on the wing about which the coefficient of pitching moment is constant. On all airfoils the ac very close to the 25% chord point (+/­ 2%) in subsonic flow. Fall 2004 16.333 2–4

• Non­dimensionalize: L M C = C = L 1 2 M 1 2 2 ρV S 2 ρV Sc¯ • Gives: x x C = (C + C α )( cg − ac ) + C Mcg Lw Dw F RL c¯ c¯ Mac • Define:

– xcg = hc¯ (leading edge to c.g.)

– xac = hn ¯ c¯ (leading edge to AC)

• Then

CMcg = (CLw + CDw αF RL)(h − hn¯) + CMac

≈ (CLw )(h − hn¯) + CMac

= CLαw (αw − αw0 )(h − hn¯) + CMac

• Result is interesting, but the key part is how this helps us analyze the static stability: ∂CMcg = (h − hn¯ ) > 0 ∂CLw since c.g. typically further back that AC – Why most planes have a second lifting surface (front or back) Fall 2004 16.333 2–5 Contribution of the Tail

lt

Lw L c.g. t Mac M w act Zcg Z w cgt D D iw w i t FRL t FRL

�W �FRL V V � V'

• Some lift provided, but moment is the key part • Key items:

1. Angle of attack αt = αF RL + it − �, � is wing downwash

2. Lift L = Lw + Lt with Lw � Lt and S C = C + η t C L Lw S Lt 2 2 η = (1/2ρVt )/(1/2ρVw ) ≈0.8–1.2 depending on location of tail

3. Downwash usually approximated as d� � = � + α 0 dα w

where �0 is the downwash at α0. For a wing with an elliptic distribution 2C d� 2CL � ≈ Lw ⇒ = αw πAR dα πAR Fall 2004 16.333 2–6

� • Pitching moment contribution: Lt and Dt are ⊥, � to V not V

– So they are at angle α¯ = αF RL − � to FRL, so must rotate and then apply moment arms lt and zt.

Mt = −lt [Lt cos α¯ + Dt sin α¯]

−zt [Dt cos α¯ − Lt sin α¯] + Mact

• First term largest by far. Assume that α¯ � 1, so that Mt ≈ −ltLt 1 L = ρV 2S C t 2 t t Lt

1 2 M l ρV StCL C =t = − t 2 t t Mt 1 2 1 2 2 ρV Sc¯ c¯ 2 ρV S l S = − t t ηC Sc¯ Lt

ltSt • Define the horizontal tail volume ratio VH = Sc¯ , so that

CMt = −VHηCLt

• Note: angle of attack of the tail αt = αw − iw + it − �, so that C = C α = C (α − i + i − �) Lt Lαt t Lαt w w t

where � = �0 + �ααw

• So that C = −V ηC (α − i + i − (� + � α )) Mt H Lαt w w t 0 α w = V ηC (� + i − i − α (1 − � )) H Lαt 0 w t w α Fall 2004 16.333 2–7

• More compact form: C = C + C α Mt M0t Mαt w

⇒ C = V ηC (� + i − i ) M0t H Lαt 0 w t

⇒ C = −V ηC (1 − � ) Mαt H Lαt α

where we can chose VH by selecting lt, St and it

• Write wing form as C = C + C α Mw M0w Mαw w ⇒ C = C − C α (h − h ) M0w Mac Lαw w0 n¯

⇒ CMαw = CLαw (h − hn¯)

• And total is:

CMcg = CM0 + CMα αw

⇒ C = C − C α (h − h ) + V ηC (� + i − i ) M0 Mac Lαw w0 n¯ H Lαt 0 w t

⇒ C = C (h − h ) − V ηC (1 − � ) Mα Lαw n¯ H Lαt α Fall 2004 16.333 2–8

• For static stability need CM = 0 and CMα < 0

– To ensure that CM = 0 for reasonable value of α, need CM0 > 0 ⇒ use it to trim the aircraft

• For CMα < 0, consider setup for case that makes CMα = 0 – Note that this is a discussion of the aircraft cg location (“find h”)

– But tail location currently given relative to c.g. (lt behind it), which is buried in VH ⇒ need to define it differently

• Define lt = c¯(ht − h); ht measured from the wing leading edge, then l S S V = t t = t (h − h) H cS¯ S t which gives

St CM = CL (h − hn¯) − (ht − h)ηCL (1 − �α) α αw S αt St St = h(CL + η CL (1 − �α)) − CL hn¯ − htη CL (1 − �α) αw S αt αw S αt

• A bit messy, but note that if LT = Lw + Lt, then S C = C + η t C LT Lw S Lt so that St CL = CL (αw − αw ) + η CL (−�0 − iw + it + αw(1 − �α)) T αw 0 S αt = CL + CL αw 0T αT

St with CL = CL + η CL (1 − �α) αT αw S αt Fall 2004 16.333 2–9

• Now have that

St CM = hCL − CL hn¯ − htη CL (1 − �α) α αT αw S αt

• Solve for the case with CMα = 0, which gives � � C CLαw St Lαt h = hn¯ + η ht (1 − �α) CL S CL αT αT

CL αT • Note that with γ = C ≈ 1, then Lαw h + (γ − 1)h h = n¯ t ≡ h γ N P which is called the stick fixed neutral point

• Can rewrite as � � � � CL S CL C = C h − h αw − η t h αt (1 − � ) Mα LαT n¯ t α CL S CL αT αT = CL (h − hNP ) αT which gives the pitching moment about the c.g. as a function of the location of the c.g. with respect to the stick fixed neutral point

– For static stability, c.g. must be in front of NP (hN P > h) Fall 2004 16.333 2–10

• Summary plot: (a = CL ) and (hn = hNP ) αT

Cm h > hn CG aft C = C + a(h - h )� m m0 n C m0 h = hn

0 �

h < hn CG forward

• Observations:

– If c.g. at hN P , then CMα = 0

– If c.g. aft of hN P , then CMα > 0 (statically unstable) – What is the problem with the c.g. being too far forward? Fall 2004 16.333 2–11 Control Effects

• Can use elevators to provide incremental lift and moments

– Use this to trim aircraft at different flight settings, i.e. CM = 0

Horizontal tail

�e

• Deflecting elevator gives:

dCL ΔCL = δe = CL δe, with CL > 0 dδe δe δe ⇒ C = C (α − α ) + C δ L Lα 0 Lδe e

Cm �e = 0 Original trim �

0

�Cm Final trim � �e > 0

dCm ΔCm = δe = Cm δe, with Cm < 0 dδe δe δe ⇒ C = C + C α + C δ m m0 mα mδe e

CL

�e > 0

�e = 0

Final Original trim pt. trim pt.

�CL

0 � Fall 2004 16.333 2–12

• For trim, need Cm = 0 and CLTRIM � � � � � � Cm Cm α −C α δe T RIM = m0 C C (δ ) C + C α Lα Lδe e TRIM LTRIM Lα 0 So elevator angle needed to trim: C C + C (C + C α ) (δ ) = Lα m0 mα LTRIM Lα 0 e TRIM C C − C C mα Lδe Lα mδe • Note that typically elevator down is taken as being positive 2 • Also, for level flight, CLTRIM = W/(1/2ρV S), so expect (δe)TRIM to change with speed.