Chapter 6 Lateral static stability and control - 2 Lecture 20 Topics

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1 Staility and control Chapter 6 Lateral static staility and control - Lecture Topics 6.1 Roll control Aileron, diferential aileron and spoiler aileron 6.1. Rolling moment due to aileron Damping moment Rate of roll achieved Aileron power Control force due to aileron Example 6.1 Example 6. Example Roll control A control in roll is needed to give a desired rate of roll (p).it may e recalled from section that p/v =.7 and.9 are desirale for cargo and military airplanes respectively. The desired rate of roll is provided y the ailerons Aileron, differential aileron and spoiler aileron Roll control is achieved in the following ways: (a) The deflections of the left and right ailerons are same in magnitude ut opposite in direction (±δ a ). () Differential aileron: In this case the deflections of the right and the left ailerons are unequal. This avoids adverse yaw. The up going aileron moves through a larger angle than the down going aileron. (c) The spoiler ailerons, on the two wing halves, are shown in Fig When a spoiler is deployed, it disturs the flow on the upper surface of that wing half. Dept. of Aerospace Engg., IIT Madras 1

2 Staility and control This causes loss of lift on that wing half and a rolling moment is produced. The spoilers are generally used at high speeds on large airplanes to counter the loss of effectiveness due to aileron reversal. For further discussion on spoiler aileron see Ref.1.13, chapter 6.The phenomenon of aileron reversal can e riefly explained as follows. Though the airplane is assumed to e rigid in the present discussion, the structure of an actual airplane is elastic. It deflects and twists under loads. When an aileron is deflected down it increases the lift on that wing half ut, it also makes M ac more negative. Consequently, the wing twists and the angle of attack decreases. The twist increases with flight speed. There is a speed, called aileron reversal speed, at which the reduction in the angle of attack due to twist will nullify the increase in the lift due to deflection of aileron. Beyond this speed a downward deployment of aileron would actually decrease the lift. This is called aileron reversal. It may e added that the interaction etween aerodynamic and elastic forces is discussed under the topic Aeroelasticity. Remarks: i) See section for frise aileron. ii) To calculate the rate of roll when the ailerons are deflected, the following two aspects need to e discussed. (a) The rolling moment due to aileron deflection. () The damping in roll i.e. opposite rolling moment caused due to rolling motion. The final rate of roll will e the alance etween these two effects which are estimated in the next two susections Rolling moment due to aileron The deflections of ailerons cause changes in the lift distriutions on the two wing halves which produce a rolling moment. The changes in the lift distriutions can e calculated using wing theory. However, this is a complicated task and a simple method called Strip theory is used for preliminary estimates. In this theory, it is assumed that the change in the lift distriution is confined to the portion of the wing span over which the aileron extends. Further, the change in local lift coefficients (ΔC l ) is given y: Dept. of Aerospace Engg., IIT Madras

3 Staility and control ΔC = ( C / δ )δ l l a a Where, δ a is the aileron deflection and Cl C/ l δ a = ail ; ail= aileron effectiveness α parameter. Fig.6.5 Strip theory Rolling moment due to aileron from a strip of width Δy (Fig.6.5) is: 1 ΔL'= ρv c dy y ΔCl ; c eing the local chord (6.15) ΔC' = l 1 ρ V c dy y ΔC l 1 ρ V S c y Cl = aail δa dy; a = of theairfoil S α ; (6.16) Hence, integrating over the portion of span where aileron extends, gives the rolling moment coefficient due to oth ailerons as: k a ail δa l) aileron= S (6.17) Dept. of Aerospace Engg., IIT Madras 3

4 Staility and control Note that the aileron extends from (k 1 /) to (k /). To apply correction for the effect of finite aspect ratio of the wing, the slope of the lift curve of the aerofoil (a ) in Eq.(6.17) is replaced y the slope of the lift curve of wing a (Ref.1.7, chapter 9). Note: a = C Lαw Hence, k a ail δa l) aileron= S (6.18) Damping moment As the airplane rolls in flight, it produces an opposite rolling moment or a damping moment. The explanation for this is as follows. Consider an airplane rolled to right i.e. positive rolling motion with the right wing going down and the left wing going up. Let, the angular velocity e p. Now, a wing section at a distance y from the c.g. experiences, on the right wing, a downward velocity of magnitude py or a relative wind of py in the upward direction (Fig 5.9). Similarly, a section at a distance y on the up going left wing experiences a downward relative wind of py (Fig.5.9). Thus, the section on down going wing experiences an increase in angle of attack of Δα = py / V. The section on the up going wing experiences a decrease in angle of attack Δα = -py / V. These changes in angles of attack would produce changes in the lift on the two wing halves and hence a rolling moment of positive sign is produced. However, this positive moment in present only when there is a rolling velocity p. Hence, it is called damping moment. It should e noted that the change in angle of attack, though dependent on py, takes place over the entire wing span. Again, using strip theory, the rolling moment due to a strip of length Δy is: (ΔL ) damp = ½ ρv c dy (ΔC l ) damp y Or (ΔC' ) = l damp 1 ρ V c dy (ΔC l ) damp y 1 ρ V S py (ΔC ) = a Δα = a, V Noting l damp (6.19) Dept. of Aerospace Engg., IIT Madras 4

5 Staility and control c(δc l ) damp y dy a p (ΔC' l) damp= S VS (6.) Integrating over oth the wing halves and noting that the rolling moment on oth the two wing halves reinforce each other, yields: a p / l) damp= V S (6.1) To apply correction for the effect of finite aspect ratio of the wing, the slope of the lift curve of the aerofoil (a ) is replaced again y the slope of the lift curve of wing (a) i.e / ap l ) damp= V S (6.) Rate of roll achieved The airplane would attain a steady rate of roll when the moment due to the aileron deflection equals the moment due to damping i.e. k a aail δ ap = S V S Simplifying, k 1 / (6.3) p = k ail V δa / (6.4) p δ Hence, = V k ail a / Some times the deflections of aileron on the up going wing (δ aup ) and on the down going wing (δ adown ) may not e equal. In this case δa is taken as : (6.5) Dept. of Aerospace Engg., IIT Madras 5

6 Staility and control δ + δ δ δ a= = aup adown atotal, in this case, p (δ ) = V 4 k ail atotal / (6.6) Aileron power Taking derivative of Eq.(6.18) with respect to δ a gives the aileron power as: k C' l CLαw =C' ail lδa = δ S a (6.7) Control force due to aileron Ailerons are operated y sideward movement of the control stick. An analysis of the control force required can e done in a manner similar to that for rudder and elevator. However, in practice it is more complex as the ailerons on the two wing halves move in opposite direction. See Ref. 1.7 Chapter 9 for some information. Example 6.1 The lift curve of a light airplane wing of rectangular planform is almost straight etween angle of zero lift (-3 ) and the incidence of 1 at which C L =1.66. The wing chord is.14 m, the aspect ratio is 8.3 and the dihedral angle is 5. Assuming that the level flight speed is m/s, calculate rolling moment set up y a sudden yaw of 5 (Adapted from Ref.1.4, chapter 14 with permission of author). Solution: The data supplied are as follows. α L = -3, C L =1.66 at α = 1. Hence, (dc L /dα) wing = C Lαw = (1.66/13) =.8 deg -1. Dept. of Aerospace Engg., IIT Madras 6

7 Staility and control c = c =.14 m, A = 8.3, Γ= 5, V = m/s, The rolling moment due to wing dihedral (L w ) Γ is to e calculated when β = 5. From Eq.(6.3) dc 1 dα L (L' w ) Γ= -βγ ρv S y y = S / In this case c is constant, hence / c y c 1 c y = = = S S 4 S 4 c =.14 m, S = x c, A = / S = / x c Hence, = 8.3 x.14 =17.76 m ; / = m S = x c = x.14 = 38.1 m Hence,.14 (17.76) y = = m And L W = (41.15) = 5 o,β = 5 o Remarks: = Nm i) From the availale data we can otain (C lβ ) Γ. From Eq.(6.4a) dc y dα L -1 lβ) Γ= - Γ = = -.1 rad Hence, -1 =-.179 deg lβ Γ ).179 = - = Γ 5 Dept. of Aerospace Engg., IIT Madras 7

8 Staility and control ii)the procedure to estimate lβ) Γgiven in Ref.1.8, which is ased on the one given in Ref.., gives for this case with Λ =, λ = 1 and A = 8.3: lβ) Γ = -.7 Γ. Example 6. A light airplane has a wing of rectangular planform 1.8 m span,.14 m chord and C Lmax of 1.5. The wing loading is 85 N/m. The airplane is rolled through 45 in one second when flying at three times its stalling speed. Estimate the rolling moment created y the ailerons assuming steady motion (Adapted from Ref.1.4, chapter 14 with permission of author). Solution: The prescried data are as follow. = 1.8 m, c =.14 m, C Lmax =1.5, W/S = 85 N m - The rate of roll (p) = 45 s -1 =.785 rad -1. Flight velocity = V = 3 V stall W 85 V stall= = = 3.41m/s ρ S C Lmax V = 3 V stall = 91.3 m/s. To determine L due to aileron we assume that the rolling moment due to aileron equals the damping moment. From Eq.(6.) / ' ap l damp VS (C ) = S = 1.8 x.14 = 7.39 m, A = /S = (1.8) / = Using πa C = which is approximately valid for unswept wing at low + A +4 Lα Mach numer (see example 5.), C = Lα π = 4.55 rad -1 / 3 3 y3 / c.14 (1.8) Further, = c [ ] = = = 187 m Dept. of Aerospace Engg., IIT Madras 8

9 Staility and control l) damp = 187 = L' damp= ρv S l) damp 1 = = Nm Example 6.3 An airplane has a straight tapered wing with taper ratio ( λ ) of.4 and aspect ratio of 8. It has. c ailerons extending from.55 semi span to.9 semi span. If aileron defects up 18 and down 1 at full deflection, estimate p/v for the airplane. If the wing span is m, otain the rate of roll in degrees per second at sea level for air speeds etween 15 to 5 kmph. Solution: The data supplied are as follows. λ =.5, A = 8, = m Hence, S = / A = (13.64) / 8 = 3.6 m. Ailerons of.c extends from.55 / to.9 /. (δ a ) up = 18, (δ a ) down = 1 hence (δ a ) total = 3. From Eq.(6.6), p (δ ) = V 4 k ail atotal / To evaluate the integrals an expression is needed for c as function of y. The root chord (c r ) and the tip chord (c t ) are otained as : S = (c r + c t ) = (c r +.4 c r ) 3.6 Or c r = =.436 m c t =.974 m ; (/) = 6.8 m For a straight tapered wing : Dept. of Aerospace Engg., IIT Madras 9

10 Staility and control y c = cr - (cr - c t ) / y Hence, c = ( ) = y 6.8 Consequently, cy =.436 y y The aileron extends for.55 / to.9 / or from y = to m. Hence, y y (.436 y y ) dy = = m / / = ( y) y dy 3 y = y = =141.6 m The quantity ail can e roughly estimated using Fig..3. For c a /c =., τ =.4 Hence, p = =.867 V The quantity p 1 V δ a is a measure of aileron effectiveness. In the present case: p 1 =.867/15 =.5378 deg V δ a The variation of p with V is given in the tale elow. p V p = =.867 V =.1183 V V V (kmph) V (m/s) p (rad/sec) Dept. of Aerospace Engg., IIT Madras 1

11 Staility and control Remark: The quantity p/v will remain constant upto certain speed, then decrease due to reduction in aileron effectiveness owing to flexiility of the structure. It (p/v) would e zero at aileron reversal speed. Dept. of Aerospace Engg., IIT Madras 11