hapter 5 Directional static stability and control - Lecture 17 Topics 5.5 ontribution of power to nβ 5.6 ontribution of ertical tail to nβ 5.6.1 Influence of wing-body combination on contribution of ertical tail ( nβ ) 5.6. Expression for ( nβ ) 5.7 Directional static stability 5.7.1 Pedal-fixed static directional stability 5.7. Weather cock effect 5.7.3 Pedal-free static directional stability 5.7.4 Desirable leel of nβ Example 5. 5.8 Directional control 5.8.1 Aderse yaw and its control 5.8. ontrol in cross wind take-off and landing 5.5 ontribution of power to nβ Figure 5.5 shows a tractor propeller in sideslip. It produces a side force Y p and yawing moment Y p l p. Since, the moment depends on angle of sideslip, there is a direct contribution to nβ. It is denoted by nβp. It is seen that the contribution is negatie and hence, destabilizing. If the airplane has a pusher propeller, then the contribution is positie and stabilizing. A jet engine in sideslip will also produce nβp whose alue will depend on the engine location. A propeller also has an indirect contribution to nβ. The slipstream of a propeller in sideslip would be asymmetric (Fig.5.6). It is obsered that for a positie alue of β, the left wing will hae a larger region influenced by the Dept. of Aerospace Engg., IIT Madras 1
propeller slipstream than the right wing. Since, the dynamic pressure in the slipstream is higher than the free stream dynamic pressure, the left wing, with larger region influenced by slip stream, will hae higher drag than the right wing. This would result a slight destabilizing contribution to nβ. Fig.5.5 Propeller and ertical tail in sideslip Dept. of Aerospace Engg., IIT Madras
Fig.5.6 Slip stream of a propeller in sideslip Remark: An accurate estimate of nβp is difficult due to the influence of arious factors.it is generally small and ignored during initial estimate of nβ. 5.6 ontribution of ertical tail In subsection.4.4. it is shown that the horizontal tail at an angle of attack produces lift L t and a pitching moment M cgt. Similarly, a ertical tail at an angle of attack (α ) would produce a side force (Y ) and a yawing moment (N ) (See Fig.5.5).The side force Y is perpendicular to the elocity V t as shown in Fig.5.5. Howeer, the angle α is small and Y is taken perpendicular to FRL. Now, Y = - 1 ρ V S Lα t (5.13) Note that as per conention Y is positie in the direction of y-axis. Hence, positie β gies negatie Y. The yawing moment due to ertical tail is gien as: 1 N = ρ V S Lα t l (5.14) Dept. of Aerospace Engg., IIT Madras 3
and = n 1 ρ V S 1 ρ V S b Lα t l 5.6.1 Influence of wing-body combination on contribution of ertical tail The wing body combination has the following influences. (5.14a) (a) The angle of attack (α ) at ertical tail is different from and (b) The dynamic pressure (½ ρv t) experienced by it (ertical tail) is different from (½ ρv ) (Figs.5.5, 5.7). The angle of attack is modified as: α = β + σ ; (5.15) where, σ is called side wash. The dynamic pressure experienced by tail is expressed as: ½ ρv t = η (½ ρv ) (5.16) Fig.5.7 Side wash on ertical tail Dept. of Aerospace Engg., IIT Madras 4
5.6. Expression for nβv Taking into account the interference effects Eq.(5.14a) becomes: 1 (β+σ) ρ V S = n Lα t 1 ρ V Sb =V η Lα (β+σ) (5.17) where, 1 ρv S l t V = and η = ; S b 1 ρv Differentiating Eq.(5.17) by β gies : nβ = V η Lα (1+ ) dβ (5.18) (5.19) As mentioned earlier, the wing and fuselage influence σ and η (Fig.5.7). Based on Ref.., the following empirical formula gies the influence of wing-body combination. S zw η (1+ ) = 0.74 + 3.06 S + 0.4 + 0.009 Aw dβ 1+cos Λc/4w d (5.0) where z w is the distance, parallel to z-axis, between wing root quarter chord point and the FRL ; d is the maximum depth of the fuselage; and cos c /4w is sweep of wing quarter chord line. Remark: The slope of lift cure of the ertical tail ( Lα ) can be calculated by knowing its effectie aspect ratio (A eff ) and using methods similar to those for estimation of the slope of lift cure of the wing. The aspect ratio of the ertical tail (A ) is b / S, where b is generally the height of ertical tail aboe the centre line of the portion of the fuselage where the ertical tail is located and S is the area of the ertical tail aboe the aforesaid centre line. The effectie aspect ratio Dept. of Aerospace Engg., IIT Madras 5
(A eff ) of the ertical tail is much higher than A and depends on the interference effect due to fuselage and horizontal tail. For details see appendix, and Ref.1.8b, Ref.1.1, chapter 3 and Ref. 1.7, chapter 8. 5.7 Directional static stability Haing obtained the contributions of arious components to nβ, arious aspects of directional static stability are discussed below. 5.7.1 Pedal-fixed static stability As regards the action of pilot to effect the moement of control surfaces, the following may be pointed out. (a) The eleator is moed by the forward or backward moement of the control stick. (b) The ailerons are operated by the sideward moement of the control stick. (c) The rudder is moed by pushing the pedals. In longitudinal static stability analysis the stick-fixed and stick-free cases were considered. These deal with the eleator-fixed and eleator-free cases respectiely. Similarly, in directional static stability, rudder-fixed and rudder-free stability is considered. These cases are also referred to as pedal-fixed and pedalfree stability analyses. In the analysis of directional static stability carried out so far the contributions of wing, fuselage, nacelle, power and ertical tail to nβ hae been considered. Noting that for the pedal-fixed stability, the rudder deflection is constant, ( nβ ) pedal-fixed is gien by adding these indiidual contributions i.e. nβ = ( nβ) w + ( nβ) f,n,p + V V η Lα (1+ ) (5.1) dβ 5.7. Weathercock effect Wheneer an airplane, originally flying with zero sideslip, deelops a sideslip (β), the ertical tail tends to bring it back to the original position of zero sideslip. This effect is similar to that of the ane attached to the weathercock which is used to indicate the direction of wind and is located on top of buildings in meteorological departments and near airports (Fig.5.8). When the ane is at an angle of attack, it produces lift on itself and consequently a moment about its hinge. This moment becomes zero only when the ane is aligned with the wind Dept. of Aerospace Engg., IIT Madras 6
direction. Hence, the ane is always directed in a way that the arrow points in the direction opposite to that of the wind. The action of ertical tail on the airplane is also similar to that of the ane and helps in aligning the airplane axis with wind direction. Hence, the directional stability is also called weathercock stability. Fig.5.8 Weathercock or weather ane (Source:www.pro.corbis.com ) 5.7.3 Pedal-free static directional stability: As noted in section 3.1 the hinge moment on the eleator is made zero by suitable deflection of the eleator tab. Similarly, the hinge moment of the rudder is also brought to zero by suitable deflection of the rudder tab. The analysis of static stability when rudder is left free to moe is called rudder-free or pedal-free stability. The equation for hinge moment about rudder hinge ( hr ) can be expressed as: hr = hα + hδrδ r + hδrtab δ rt (5.) Where, δ r is the rudder deflection and δ rt is the deflection of the rudder tab. The floating angle of rudder, δ rfree is obtained when hr is zero i.e. Dept. of Aerospace Engg., IIT Madras 7
(hα α + hδrtab δ rt ) δ rfree = - hδr dδ dα = - = - (1+ ) dβ dβ dβ rfree hα hα Hence, Remarks: hδr i) onention for rudder deflection hδr (5.3) (5.4) A rudder deflection to left is taken as positie. It is a general conention that a positie control deflection produces negatie moment. This is consistent with the conention that rotation is taken positie clockwise when looking along the axis about which the rotation takes place. ii) hα and hδr and leel of static direction stability with rudder free Noting the conention for β, and with the help of pressure distribution shown in Fig.3.3, it is obsered that a positie alue of β would produce a negatie force on the rudder and hence a positie hinge moment. onsequently, hα is positie. In a similar manner it is obsered that a positie rudder deflection would produce a positie side force and hence, negatie hinge moment. Thus, hδr is negatie. Hence, from Eq.(5.4), (dδ r ) free / dβ is negatie. Thus, when a disturbance produces positie β, the rudder will take such a position that the rudder deflection is negatie. The result is a negatie change in yawing moment or reduced static stability. Thus, the leel of static directional stability will be reduced when the pedal is free. 5.7.4 Desirable leel of nβ In longitudinal static stability, the shift of c.g. has a profound effect on the leel of stability ( ) as the contribution of wing to m m depends directly on (x cg - x ac ). Thus, the shift in the position of c.g, during flight, almost decides the area of the horizontal tail. Howeer, a shift of c.g. does not cause a significant change in nβ because such a change may only hae a secondary effect by way of slightly affecting l. Hence, to arrie at the area of the ertical tail, a criterion to prescribe a desirable alue of nβ is needed. Reference 1.7, hapter 8 gies: W 1 ( nβ) desirable = 0.005 ( ) deg b -1 (5.5) Dept. of Aerospace Engg., IIT Madras 8
where, W is the weight of the airplane in lbs and b is the wing span in feet. Reference 1.1, Appendix B gies characteristics of seen airplanes. It is obsered that for the subsonic airplanes nβ lies between 0.0013 to 0.006 deg -1. Howeer, the final alue of nβ is decided after the dynamic stability analysis. From lateral dynamic stability analysis (chapter 9) it will be obsered that a large alue of nβ leads to some unacceptable response of the airplane to the disturbance. Example 5. A model of an airplane is tested in a wind tunnel without the ertical tail. ontributions of arious components gie nβ = -0.001 deg -1. If the ertical tail is to be positioned at a point on the aft end of the fuselage giing a tail length of 4.8 m, How much ertical tail area is required to gie an oerall ηβ = 0.001 deg -1? Assume that the ertical tail would hae an effectie aspect ratio of, the wing area is 18 m, wing span is 10.6 m and the wing is set at the middle of the fuselage. Solution: nβ = ( nβ ) w + ( nβ ) f,n,p + ( nβ ) t Gien: ( nβ ) w + ( nβ ) f,n,p = -0.001 ( nβ ) required = 0.001 Hence, ( ηβ ) t = 0.001 - (- 0.001) = 0.004 From Eq.(5.19), ( nβ ) = V η Lα (1+ ) dβ I) Estimation of Lα : A eff =. The expression for Lα is (Ref.1.8b): πa Lα = in rad A βm (1+tan Λ c/) + +4 K β where, β M = (1-M ) 1/, K = lift cure slope of airfoil / π. Λ c/ = sweep of mid chord line -1 Dept. of Aerospace Engg., IIT Madras 9
As Mach number is low subsonic β M 1. Let K =1, onsequently, = L + A +4 Veff πa Lα + +4 Λ c = 0 w 4 π For A of.0, = =.60 rad = 0.0454 deg 10.6 A w = = 6.4 18-1 -1 Remark: Reference 1.7, Fig.8.8 gies Lα =0.044 deg -1 for A eff of.0. II) Estimation of η (1+ ) dβ The expression for η (1+ ) dβ as gien by Eq.(5.0) depends on S / S but S / S is not known at this stage. Hence, as a first approximation it is assumed that S / S = 0.1. The quantity z w /d can be taken as zero for the mid wing configuration. 0.1 Hence, η (1+ ) = 0.74 + 3.06 ( ) + 0 + 0.009 6.4 = 0.964 dβ 1+1 onsequently, the first estimation of V is: 0.004 = V 1 0.0454 0.964 or V = 0.05484 Noting that, V =, gies S = 0.05484 18 =.18 m S l 10.6 S b 4.8 To improe the estimation of S,its alue in the preious step is substituted in the expression for η (1+ ) dβ i.e..18 / 18 η (1+ ) = 0.74 + 3.06 ( )+0+0.009 6.4 = 0.9655 dβ 1+1 The second estimation of V is: Dept. of Aerospace Engg., IIT Madras 10
0.004 V = = 0.05475 0.9655 0.0454 Or S =.176 m Since, the two estimates are close to each other, the iteration is terminated and S =.176 m is taken as the answer. 5.8 Directional ontrol ontrol of rotation of the airplane about the z-axis is proided by the rudder. The critical conditions for design of rudder are: (a) Aderse yaw, (b) ross wind take-off and landing, (c) Asymmetric power for multi- engined airplanes and (d) Spin 5.8.1 Aderse yaw and its control When an airplane is rolled to the right, the rate of roll produces a yawing moment tending to turn the airplane to the left. Similarly, a roll to left produces yaw to right. Hence, the yawing moment produced as a result of the rate of roll is called aderse yaw. To explain the production of aderse yaw, consider an airplane rolled to right, i.e. right wing down. Let, the rate of roll be p. The rate of roll produces the following two effects. a) A roll to right implies less lift on the right wing and more lift on the left wing. This is brought about by aileron deflection in the present case an up aileron on the right wing and a down aileron on the left wing. Since, L on the right wing is less than L on the left wing, the induced drag coefficient ( Di ) on the right wing is less than Di on left wing. This results in a yawing moment causing the airplane to yaw to left. b) Due to the rolling elocity (p) a section on the down going wing at a distance y from the FRL experiences a relatie upward wind of magnitude py. At the same time a section on the up going wing at a distance y from FRL experiences a relatie downward elocity of magnitude py. This results in the change of Dept. of Aerospace Engg., IIT Madras 11
direction of the resultant elocity on the two wing hales (Fig.5.9). Now, the lift ector, being perpendicular to the resultant elocity, is bent forward on the down going wing and bent backwards on the up going wing.onsequently, the horizontal components of the lift on the two wing hales produce a moment tending to yaw the airplane to left. An approximate estimate of the effect of aderse yaw is (Ref.1.1, chapter 3): L ( n) aderseyaw - 8 V pb (5.7) where, p = rate of roll in radians per second; b = wing span and V = flight elocity. Fig.5.9 Effect of rate of roll An airplane is generally designed for a specific alue of (pb/v). For example, Ref.1.7 hapter 9 prescribes that up to 80% of V max the airplane should hae: pb / V = 0.07 for cargo/ bomber pb / V = 0.09 for fighter Hence, one of the criteria for rudder design is that it must be powerful enough to counter the aderse yaw at prescribed rate of roll. 5.8. ontrol in cross wind take-off and landing An Airplane sometimes encounters side winds during take-off and landing. As regards control during this eentuality, the following three points may be noted. (1) When an airplane flying at a elocity V, encounters a side wind of elocity, the resultant elocity ector makes an angle Δβ to the plane of symmetry; Δβ = /V. Dept. of Aerospace Engg., IIT Madras 1
()The tendency of an airplane possessing directional static stability, is to align itself with the wind direction (weather cock effect). (3) During take-off and landing the pilot has to keep the airplane along the runway. Hence, when a cross wind is present the airplane is side-slipping with angle Δβ. Thus, another criterion for the design of the rudder is required. It must be able to counteract the yawing moment due to sideslip produced by the cross wind ( nβ x Δβ). This criterion becomes more critical at lower speeds because (a) the effectieness of the rudder, being proportional to V, is less at lower flight speeds and (b) Δβ being proportional to 1/V, is high at low flight speeds. According to Ref.1.1, chapter the rudder must be able to oercome = 51 ft / s or 15 m/s at the minimum speed for the airplane. It may be pointed out that on a rainy day, with heay cross winds, the landing on the airport may be refused if the cross wind is more than that permitted for the airplane. Dept. of Aerospace Engg., IIT Madras 13