Aileron DesignChapter 12Design of Control SurfacesFrom: Aircraft Design: A Systems Engineering ApproachMohammad Sadraey792 pagesSeptember 2012, HardcoverWiley Publications12.4.1. IntroductionThe primary function of an aileron is the lateral (i.e. roll) control of an aircraft; however,it also affects the directional control. Due to this reason, the aileron and the rudder areusually designed concurrently. Lateral control is governed primarily through a roll rate(P). Aileron is structurally part of the wing, and has two pieces; each located on thetrailing edge of the outer portion of the wing left and right sections. Both ailerons areoften used symmetrically, hence their geometries are identical. Aileron effectiveness is ameasure of how good the deflected aileron is producing the desired rolling moment. Thegenerated rolling moment is a function of aileron size, aileron deflection, and its distancefrom the aircraft fuselage centerline. Unlike rudder and elevator which are displacementcontrol, the aileron is a rate control. Any change in the aileron geometry or deflectionwill change the roll rate; which subsequently varies constantly the roll angle.The deflection of any control surface including the aileron involves a hingemoment. The hinge moments are the aerodynamic moments that must be overcome todeflect the control surfaces. The hinge moment governs the magnitude of augmentedpilot force required to move the corresponding actuator to deflect the control surface. Tominimize the size and thus the cost of the actuation system, the ailerons should bedesigned so that the control forces are as low as possible.In the design process of an aileron, four parameters need to be determined. Theyare: 1. aileron planform area (Sa); 2. aileron chord/span (Ca/ba); 3. maximum up anddown aileron deflection ( ? ?Amax); and 4. location of inner edge of the aileron along thewing span (bai). Figure 12.10 shows the aileron geometry. As a general guidance, thetypical values for these parameters are as follows: Sa/S = 0.05 to 0.1, ba/b = 0.2-0.3, Ca/C= 0.15-0.25, bai/b = 0.6-0.8, and ?Amax = ? 30 degrees. Based on this statistics, about 5 to10 percent of the wing area is devoted to the aileron, the aileron-to-wing-chord ratio isabout 15 to 25 percent, aileron-to-wing-span ratio is about 20-30 percent, and the inboardaileron span is about 60 to 80 percent of the wing span. Table 12.17 illustrates thecharacteristics of aileron of several aircraft.1bAba/2CaSa/2 bai/2 Aa. Top-view of the wing and aileron?Aup?Adownb. Side-view of the wing and aileron (Section AA)Figure 12.1. Geometry of aileronFactors affecting the design of the aileron are: 1. the required hinge moment, 2.the aileron effectiveness, 3. aerodynamic and mass balancing, 4. flap geometry, 5. theaircraft structure, and 6. cost. Aileron effectiveness is a measure of how effective theaileron deflection is in producing the desired rolling moment. Aileron effectiveness is afunction of its size and its distance to aircraft center of gravity. Hinge moments are alsoimportant because they are the aerodynamic moments that must be overcome to rotate theaileron. The hinge moments governs the magnitude of force required of the pilot to movethe aileron. Therefore, great care must be used in designing the aileron so that the controlforces are within acceptable limits for the pilots. Finally, aerodynamic and massbalancing deals with techniques to vary the hinge moments so that the stick force stayswithin an acceptable range. Handling qualities discussed in the previous section governthese factors. In this section, principals of aileron design, design procedure, governingequations, constraints, and design steps as well as a fully solved example are presented.12.4.2. Principles of Aileron DesignA basic item in the list of aircraft performance requirements is the maneuverability.Aircraft maneuverability is a function of engine thrust, aircraft mass moment of inertia,and control power. One of the primary control surfaces which cause the aircraft to besteered along its three-dimensional flight path (i.e. maneuver) to its specified destinationis aileron. Ailerons are like plain flaps placed at outboard of the trailing edge of the wing.Right aileron and left aileron are deflected differentially and simultaneously to produce a2rolling moment about x-axis. Therefore, the main role of aileron is the roll control;however it will affect yaw control as well. Roll control is the fundamental basis for thedesign of aileron.Table 12.1. Characteristics of aileron for several aircraftTable 12.12 (lateral directional handling qualities requirements) providessignificant criteria to design the aileron. This table specifies required time to bank anaircraft at a specified bank angle. Since the effectiveness of control surfaces are thelowest in the slower speed, the roll control in a take-off or landing operations is the flightphase at which the aileron is sized. Thus, in designing the aileron one must consider onlylevel 1 and most critical phases of flight that is usually phase B.Based on the Newton’s second law for a rotational motion, the summation of allapplied moments is equal to the time rate of change of angular momentum. If the massand the geometry of the objet (i.e. vehicle) are fixed, the law is reduced to a simplerversion: The summation of all moments is equal to the mass moment of inertia time ofthe object about the axis or rotation multiplied by the rate of change of angular velocity.In the case of a rolling motion, the summation of all rolling moments (including theaircraft aerodynamic moment) is equal to the aircraft mass moment of inertia about x-axismultiplied by the time rate of change (???t) of roll rate (P).Inboard aileron 1Outboard aileron 23No Aircraft Type mTO(kg)b(m)CA/C Span ratio ?Amax (deg)bi/b/2 bo/b/2 up down1 Cessna 182 Light GA 1,406 11 0.2 0.46 0.95 20 142 Cessna CitationIIIBusinessjet9,979 16.31 0.3 0.56 0.89 12.512.53 Air Tractor AT-802Agriculture 7,257 18 0.36 0.4 0.95 17 134 Gulfstream 200 Businessjet16,080 17.7 0.22 0.6 0.86 15 155 Fokker 100A Airliner 44,450 28.08 0.24 0.6 0.94 25 206 Boeing 777-200 Airliner 247,200 60.9 0.22 0.321 0.762 30 107 Airbus 340-600 Airliner 368,000 63.45 0.3 0.64 0.92 25 208 Airbus A340-600Airliner 368,000 63.45 0.25 0.67 0.92 25 25? L cg ? I xx ?P?t (12.7)or?P ? ? LI xxcg (12.8)Generally speaking, there are two forces involved in generating the rolling moment: 1.An incremental change in wing lift due to a change in aileron angle, 2. Aircraft rollingdrag force in the yz plane. Figure 12.11 illustrates the front-view of an aircraft whereincremental change in the lift due to aileron deflection (?L) and incremental drag due tothe rolling speed are shown.The aircraft in Figure 12.11 is planning to have a positive roll, so the right aileronis deflected up and left aileron down (i.e. +?A). The total aerodynamic rolling moment ina rolling motion is:? M cgx ? 2?L ? y A ? ?D ? yD (12.9)The factor 2 has been introduced in the moment due to lift to account for both leftand right ailerons. The factor 2 is not considered for the rolling moment due to rollingdrag calculation, since the average rolling drag will be computed later. The parameter yLis the average distance between each aileron and the x-axis (i.e. aircraft center of gravity).The parameter yD is the average distance between rolling drag center and the x-axis (i.e.aircraft center of gravity). A typical location for this distance is about 40% of the wingsemispan from root chord.+?A?Dright?Lleft?Lrightdyyyoyicg ?DleftzyDyA+?AFront viewFigure 12.2. Incremental change in lift and drag in generating a rolling motion4In an aircraft with short wingspan and large aileron (e.g. fighter such as GeneralDynamics F-16 Fighting Falcon (Figure 3.12)) the drag does not considerably influenceon the rolling speed. However, in an aircraft with a long wingspan and small aileron;such as bomber Boeing B-52 (Figures 8.20 and 9.4); the rolling induced drag force has asignificant effect on the rolling speed. For instance, the B-52 takes about 10 seconds tohave a bank angle of 45 degrees at low speeds, while for the case of a fighter such as F-16; it takes only a fraction of a second for such roll.Owing to the fact that ailerons are located at some distance from the center of gravity ofthe aircraft, incremental lift force generated by ailerons deflected up/down, creates arolling moment.LA ? 2?L ? yA (12.10)However, the aerodynamic rolling moment is generally modeled as a function of wingarea (S), wing span (b), dynamic pressure (q) as:LA ? qSClbwhere Cl is the rolling moment coefficient and the dynamic pressure is:(12.11)q ? 12 ?VT2 (12.12)where ? is the air density and VT is the aircraft true airspeed. The parameter Cl is afunction of aircraft configuration, sideslip angle, rudder deflection and aileron deflection.In a symmetric aircraft with no sideslip and no rudder deflection, this coefficient islinearly modeled as:Cl ? Cl?A ? A (12.13)The parameter Cl? A is referred to as the aircraft rolling moment-coefficient-due-to-aileron-deflection derivative and is also called the aileron roll control power. The aircraftrolling drag induced by the rolling speed may be modeled as:DR ? ?Dleft ? ?Dright ? 12 ?VR2 StotCDR (12.14)where aircraft average CDR is the aircraft drag coefficient in rolling motion. Thiscoefficient is about 0.7 – 1.2 which includes the drag contribution of the fuselage. Theparameter Stot is the summation of wing planform area, horizontal tail planform area, andvertical tail planform area.Stot ? Sw ? Sht ? Svt5(12.15)2CL?w ?? A yo?y CydyThe parameter VR is the rolling linear speed in a rolling motion and is equal to roll rate(P) multiplied by average distance between rolling drag center (See Figure 12.11) alongy-axis and the aircraft center of gravity:VR ? P ? yD (12.16)Since all three lifting surfaces (wing, horizontal tail, and vertical tail) are contributing tothe rolling drag, the yD is in fact, the average of three average distances. The non-dimensional control derivative Cl? A is a measure of the roll control power of the aileron; itrepresents the change in rolling moment per unit change of aileron deflection. The largerthe Cl? A , the more effective the aileron is at creating a rolling moment. This controlderivative may be calculated using method introduced in [19]. However, an estimate ofthe roll control power for an aileron is presented in this Section based on a simple stripintegration method. The aerodynamic rolling moment due to the lift distribution may bewritten in coefficient form as:?Cl ? ?LAqSb ? qCLA Ca y AdyqSb ? CLA Ca y AdySb (12.17)The section lift coefficient CLA on the sections containing the aileron may be written asCLA ? CL? ? ? CL? d?d?A? A ? CL? ? a ? ? A (12.18)where ?a is the aileron effectiveness parameter and is obtained from Figure 12.12, giventhe ratio between aileron-chord and wing-chord. Figure 12.12 is a general representativeof the control surface effectiveness; it may be applied to aileron (?a), elevator (?e), andrudder (?r). Thus, in Figure 12.12, the subscript of parameter ? is dropped to indicate thegenerality.Integrating over the region containing the aileron yieldsCl ? Sbi(12.19)where CL?w has been corrected for three-dimensional flow and the factor 2 is added toaccount for the two ailerons. For the calculation in this technique, the wing sectional liftcurve slope is assumed to be constant over the wing span. Therefore, the aileron sectionallift curve slope is equaled to the wing sectional lift curve slope. The parameter yirepresents the inboard position of aileron with respect to the fuselage centerline, and yothe outboard position of aileron with respect to the fuselage centerline (See Figure 12.11).6? 2?1 ? 2? b ? y? ydy? y?C ? Cr ?1 ? 2??y Cydy2 ? ? ?1? 3 ?The aileron roll control derivative can be obtained by taking the derivative with respect to?A:Cl? A ?2CL?w ? yoSbi(12.20)?0.80.60.40.20.1 0.2 0.3 0.4 0.5 0.6 0.7Control-surface-to-lifting-surface-chord ratioFigure 12.3. Control surface angle of attack effectiveness parameterThe wing chord (C) as a function of y (along span) for a tapered wing can be expressedby the following relationship:? ? ? ?1? ?? ? b ? ?(12.21)where Cr denotes the wing root chord, and ? is the wing taper ratio. Substituting thisrelationship back into the expression for Cl? A (Equ. 12.20) yields:Cl? A ?2CL?w ?Sbyo? Cyir? ? ? ?1? ?? ? ? ?(12.22)orCl? A ?2CL?w ?Cr ? y 2?Sb ? ? ? y ?3? b ? ? yiyo(12.23)This equation can be employed to estimate roll control derivative Cl? A using the ailerongeometry and estimating ? from Figure 12.12. Getting back to equation 12.12, there aretwo pieces of ailerons; each at one left and right sections of the wing. These two piecesmay have a similar magnitude of deflections or slightly different deflections, due to theadverse yaw. At any rate, only one value will enter to the calculation of rolling moment.Thus, an average value of aileron deflection will be calculated as follows:7? Aleft ? ? Aright? A ? ? ???P ? yD ? ?Sw ? Sht ? Svt ?CDR ? yD? ? ? ? dP1 (12.24)2The sign of this ?A will later be determined based on the convention introduced earlier; apositive ?A will generate a positive rolling moment. Substituting equation 12.9 intoequation 12.7 yields:?LA ? ?D ? yD ? I xx P?As the name implies, P is the time rate of change of roll rate:(12.25)?P ? ddt P (12.26)On the other hand, the angular velocity about x-axis (P) is defined as the time rate ofchange of bank angle:P ? ddt ? (12.27)Combining equations 12.26 and 12.27 and removing dt from both sides, results in:?P d? ? PdP (12.28)Assuming that the aircraft is initially at a level cruising flight (i.e. Po = 0, ?o = 0), bothsides may be integrated as:? ?? P d? ?0Pss? PdP0(12.29)Thus, the bank angle due to a rolling motion is obtained as:PP?where P is obtained from equation 12.25. Thus:(12.30)Pss? ? ?0I xx PLA ? ?D ? yD dP (12.31)Both aerodynamic rolling moment and aircraft drag due to rolling motion are functions ofroll rate. Plugging these two moments into equation 12.31 yields:?1 ?Pss?0 qSCl b ?12I xx P2dP (12.32)The aircraft rate of roll rate response to the aileron deflection has two distinct states: 1. Atransient state, 2. A steady state (See Figure 12.13). The integral limit for the roll rate (P)in equation 12.32 is from an initial trim point of no roll rate (i.e. wing level and Po = 0) toa steady-state value of roll rate (Pss). Since the aileron is featured as a rate control, thedeflection of aileron will eventually result in a steady-state roll rate (Figure 12.13). Thus,unless the ailerons are returned to the initial zero deflection, the aircraft will not stop at aspecific bank angle. Table 12.12 defines the roll rate requirements in terms of the desired8bank angle (?2) for the duration of t seconds. The equation 12.32 has a closed-formsolution and can be solved to determine the bank angle (??) when the roll rate reaches itssteady-state value.Roll rate(deg/sec)Psstss t2 Time (sec)Figure 12.4. Aircraft roll rate response to an aileron deflectionBankangle(deg)?2?1t1 t2 Time (sec)Figure 12.5. Aircraft bank angle response to an aileron deflectionWhen the aircraft has a steady-state (Pss) roll rate, the new bank angle (Figure 12.14) after?t seconds (i.e. t2-tss) is readily obtained by the following linear relationship:?2 ? Pss ? ?t2 ? tss ? ? ?1 (12.33)Due to the fact that the aircraft drag due to roll rate is not constant and isincreased with an increase to the roll rate; the rolling motion is not linear. This implies9y ? k ? 2yD?Sw ? Sht ? Svt ?CDa 2 ? (12.41)?y ?Sw ? Sht ? Svt ?CDRthat the variation of the roll rate is not linear; and there is an angular rotation about x-axis. However, until the resisting moment against the rolling motion is equal to theaileron generated aerodynamic rolling moment; the aircraft will experience an angularacceleration about x-axis. Soon after the two rolling moments are equal, the aircraft willcontinue to roll with a constant roll rate (Pss). The steady-state value for roll rate (Pss) isobtained by considering that the fact that when the aircraft is rolling with a constant rollrate, the aileron generated aerodynamic rolling moment is equal to the moment of aircraftdrag in the rolling motion.LA ? ?DR ? yD (12.34)Combining equations 12.14, 12.15, and 12.16, the aircraft drag due to the rolling motionis obtained as:DR ? 12 ??P ? yD ?2 ?Sw ? Sht ? Svt ?CDR (12.35)Inserting the equation 12.35 into equation 12.34 yields:LA ? 12 ??P ? yD ?2 ?Sw ? Sht ? Svt ?CDR ? yD (12.36)Solving for the steady-state roll rate (Pss) results in:Pss ? 2 ? LA??Sw ? Sht ? Svt ?CDR ? yD3(12.37)On the other hand, the equation 12.32 is simply a definite mathematical integration. Thisintegration may be modeled as the following general integration problem:xdxx ? a 2According to [20], there is a closed form solution to such integration as follows:(12.38)y ? k 12 ln?x 2 ? a 2 ? (12.39)The parameters k and a are obtained by comparing equation 12.38 with equation 12.32.k ? 3D2I xx (12.40)V 2 SCl b3RHence, the solution to the integration in equation 12.32 is determined as:112?1 ? ?o ? 1 ? 2???3 ???ln? P 2 ??y ?Sw ? Sht ? Svt ?CDR??1 ? ???I xx?yD3 ?Sw ? Sht ? Svt ?CDR??PssV 2 SCl b?Sw ? Sht ? Svt ?CDR yD ??? 0(12.42)Applying the limits (from 0 to Pss) to the solution results in:?1 ? 3DI xx ln?Pss ? (12.43)Recall that we are looking to determine aileron roll control power. In another word, it isdesired to obtain how long it takes (t2) to bank to a desired bank angle when ailerons aredeflected. This duration tends to have two parts: 1. The duration (tss) that takes an aircraftto reach the steady-state roll rate (Pss), 2. The time (?tR) to roll linearly from ?ss to ?2(See Figure 12.14).t2 ? tss ? ?tRwhere(12.44)?tR ? ? 2 ???1Pss(12.45)Comparing Figures 12.13 and 12.14 indicates that t1 = tss. The time (tss) that takes anaircraft to achieve a steady-state roll rate due to an aileron deflection is a function of?angular acceleration ( P ). Based on classical dynamics, this accelerated roll can beexpressed as:P tss (12.46)2It is assumed that the aircraft is initially at a wing-level flight condition (i.e. ?o = 0).Hence:t1 ? tss ? 2?1? (12.47)Pwhere ?1 is determined from equation 12.43. In addition, in an accelerated rollingmotion, the relationship between final roll rate (P1) and initial roll rate (Po) is a function?of the rate of roll rate ( P ) and the final bank angle (?1). Based on classical dynamics, anaccelerated rolling motion can be expressed as:?P12 ? Po2 ? 2 P ?1 (12.48)It is assumed that the aircraft is initially at a wing-level flight condition (i.e. Po = 0) andthe new roll rate is the steady-state roll rate (i.e. P1 = Pss). Thus:?P ? Pss22?1(12.49)where Pss is determined from equation 12.45.11Bank angle(deg)?reqtreq Time (sec)Figure 12.6. Bank angle versus timeFor a GA and transport aircraft, the time to reach the steady state rolling motion(t1) is long (more than ten seconds). Thus, the application of equations 12.48 and 12.49 isoften not needed for aileron design, since the roll requirement is within a few seconds.However, for a fighter aircraft and missile, the rolling motion (See Figure 12.15) is veryfast (the time t1 is within a few seconds), so the application of equations 12.48 and 12.49is usually needed for aileron design. For this reason, when the bank angle (?1)corresponding to steady state roll rate (Pss) is beyond 90 degrees, the equation 12.46serves as the relationship between the required time to reach a desired bank angle.Therefore, the duration required (treq) to achieve a desired bank angle (?des) will bedetermined as follows:t2 ? 2? des? (12.50)PThe equations and relationships introduced and developed in this section providethe necessary tools to design the aileron to satisfy the roll control requirements. Table12.12 addresses the military aircraft roll control requirements; for a civil aircraft, it issuggeste