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Chapter 2: Static aeroelasticity-Unswept wing structural loads and performance Homework problems: Problem 2.5~2.18

Tran Tien Anh@HCMUT

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Chapter 2: Homework problems Problem 2.5 Two wing sections are mounted on shafts and attached to each other and to a wind tunnel wall, as indicated.

Note that the torsion spring stiffnesses are equal, but are offset an amount d. This configuration is similar to, but not identical to, the example 2.8. When the airfoils are placed at an angle of attack a0 , the springs deform. Lifts on the two identical, un-cambered wing sections are

L1 = qSC La (a0 + q1 ) L2 = qSC La (a0 + q2 )

(a) Derive the matrix equations of torsional static equilibrium for this model when it is placed at the angle of attack a0 . Express these equations in the form ì ü ì ü éK ù ïïí q1 ïïý = ïïíQ1 ïïý . Do not solve for the deflection, but identify the structural stiffness êë ij úû ï q ï ïQ ï ïî 2 ïþ ïî 2 ïþ matrix and the aerodynamic stiffness matrix in these equations (“Identify” means put some words together and attach an arrow.) (b) Solve for the divergence dynamic pressure; solve for the value of d that eliminates divergence. The shaft attaching the two segments can be re-positioned, but the outer wing segment always remains in-line with the inner segment. (c) Place an aileron on the outer (right-hand) section. Develop the static equilibrium equations for the system when there is no initial angle of attack, but the control surface is deflected downward an amount d0 .

(d) Solve for the rolling moment generated by the aileron. Solve the reversal speed VR in terms of general parameters such as C Ld and C MAC ,d Problem 2.6 A side view of two idealized wings is shown in the figure. This problem is similar to that discussed in example 2.5, but has one spring removed. These wings are connected by a single spring, with spring constant k [lb./inch]; each wing has planform area S. The spring is pinned to each wing and develops an internal force in response to relative deflection between its ends. This configuration differs from that discussed in example 2.5 because it can rotate without stretching or compressing the spring. This is called rigid body freedom.

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Figure P2.6: Tandem wing geometry with single spring The lift on each wing is given by L1 = qSC La (a0 + q1 ) L2 = qSC La (a0 + q2 )

Where a0 is an initial angle of attack common to both wings; and q1 and q2 are the two twist angles. (a) Solve for the characteristic equation ∆ for static stability (b) If e1 = e2 = e , identify a dynamic pressure parameter q and plot the characteristic equation as a function of this parameter. (c) Find the value(s) of the pressure parameter at which D = 0 . Solve for the mode shape(s). Will the system diverge? Why or why not? Problem 2.7 A wing test article consists of a low density, symmetrical stiffness metallic layer, wrapped around the two layer piezoelectric plates create an electric field E shown in Figure P2.7(b) that causes surface electrode is formed. The wing idealization is shown in Figure P2.7(c); for analysis purposes we have only drawn the chord line. This model has two lift components; the first is due to an initial angle of attack and the wing twist. The second is due to the applied voltage that creates camber.

Figure P2.7(a,b) We will represent the added lift and pitching moment combination created by camber as a single force L2 acting on the wing at a distance e2 after of the shear center (this is the center of pressure). This location of the center of pressure for the additional force and we assume that it does not move with applied voltage. The relationship between lift L2 and the applied voltage V is: L2 = (qSC L,V )V V is voltage, not airspeed. Since the change in camber produces lift and pitching moment, Tran Tien Anh@HCMUT

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¶C L ¶camber tells us how much lift we get per volt ¶camber ¶V across the active material per unit wing area per unit dynamic pressure.

the aerodynamic coefficient C L,V =

Figure P2.7(c): Active airfoil with applied loads Questions: (a) When the airfoil initial angle of attack is zero (a0 = 0) solve for the lift on the wing as a function of the applied voltage V. (b) Find the wing divergence dynamic pressure q D as a function of the aerodynamic and structural parameters. (Voltage should not appear). (c) Let’s create a feedback system where we measure the twist θ and feed this back to the voltage, so that V = qk (k is an arbitrary number that we can choose). In this case, the voltage term should move to the left side of your equation. If the wing is placed at an initial angle of attack a0 , solve for the wing twist and solve for the divergence dynamic pressure. Hints: The total airfoil lift is written as L = L1 + L2 = qSC La + qSC L,VV where a = a0 + q In this problem, a0 = 0 . When V = qk , the equation of torsional static equilibrium is e1 éêëqSC La (a0 +q)ùúû - e2 éêqSC L,V qk ùú - KT q = 0 ë û

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Problem 2.8: The 1 DOF idealized wing whose cross-section is shown in Figure P2.8(a) has leading edge and trailing edge control surfaces. There is no initial angle of attack when the two control surfaces are un-deflected. Three lift components act on the idealized wing. Each force has its own location with respect to the shear center pin shown in the figure. Twisting is resisted by the torsional spring with torsional stiffness KT .

Figure P2.8(a) The entire assembly is mounted at the end of a long bar a distance r from the center of rotation. This assembly will move upward at a terminal speed v = pr when the surfaces are deflected. The leading edge surface and the trailing edge flap are geared together so that the three lift components are as follows: (where v = pr ) L1 = qSC Laq + qSC La - v V L2 = qSC Ld d0

(

)

L3 = qSC Ld (0.25d0 )

The rolling moment is M roll = rL = r (L1 + L2 + L3 )

Figure P2.8(b): Top view of wing assembly showing rotational rate, p. Question: a) Solve for the twist q in terms of the aerodynamic derivatives and

v . V

b) Solve for the equation for the steady state roll rate p . c) Solve for the reversal dynamic pressure Hints: Assume that we are rolling at a constant roll rate so that the rolling moment due to the aileron is balanced by a damping in roll moment (see the notes) due to an upward velocity. In this case, the total lift on the surface is L = L1 + L2 + L3

æ ö v 1 or L = qS ççC Laq - C La + C Ld d0 + C Ld d0 ÷÷÷ çè V 4 ø÷ Tran Tien Anh@HCMUT

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The FBD is shown below.

Figure P2.8(c): Free-body diagram for θ computation e qSC La Sum moments using +q as the positive direction (clockwise). Let q1 = 1 . We get KT

é æ öæ ö ù æ ö ê-q çç v ÷÷ + q çç 1 e3 - e2 ÷÷ ççC Ld ÷÷ d ú ÷÷ çç ÷ 1ç ê 1 çèV ÷÷ø C La ø÷ 0 úú èç 4 e1 e1 øè ê ë û q= (1 - q1 ) The rolling moment is é æ öù v 1 M roll = rL = r êêqS ççC Laq - C La + C Ld d0 + C Ld d0 ÷÷÷úú ÷øú V 4 êë çè û ìï v ü é ù ïï- + êq æç -5e1 - 4e2 + e1 ö÷÷ + 5 ú æçC Ld ö÷÷ d ïïï ÷÷ ÷ 0ï ïï V ê 1 ççç úç 4e1 ø 4 ûú èççC La ø÷ ïï ê è ï ë M roll = rL = rqC LaS í ý=0 ïï ïï 1 - q1 ) ( ïï ïï ïï ïï î þ é æ -5e - 4e + e ö 5 ù æC ö v pr 1 2 1÷ ÷÷ + ú çç Ld ÷÷÷ d0 = = êêq1 ççç with v = pr ÷ø 4 úú ççèC La ÷ø V V 4e1 êë çè û æ -5e1 - 4e2 + e1 ö÷ 5 pr ÷÷ + For reversal = 0 = q1 ççç çè V 4e1 ø÷ 4 The answer is qreversal =

5KT

SC La (5e1 + 4e2 - e3 )

The aileron reversal problem has two meanings and two approaches. In the first case we restrict the upward velocity, v , to be zero and solve for the lift (or rolling moment) generated by an aileron deflection. Reversal is defined as the airspeed (or dynamic pressure) at which the lift is zero. By definition, at reversal in this case, both lift and upward speed are zero, but only because the upward speed, v, is restrained from the beginning and we solve for the value of q to make lift (or moment) zero. In the second case, the upward airspeed (or steady-state roll moment) is constant, unrestrained and non-zero. To have a constant upward speed we must have the lift (or rolling moment) equal to zero. This is a constraint that allows us to solve for v . At reversal, both v and lift (or moment) are zero. We use the “ v equation” to find the value of q at reversal. Because, in the end, no matter which of the two approaches we use for reversal, the reversal dynamic pressure must be the same since both conditions result in a zero upward velocity and zero lift due to aileron deflection. Only the path to reversal is different.

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Problem 2.9 An un-cambered wing with span, b, is attached to both walls of a wind tunnel and placed at an angle of attack a0 . The wing is idealized as a torsionally flexible, uniform property element like that discussed in Section 2.12. Use Figure 2.9e1 as your reference, but with an additional support on the right.

Figure P2.9: Effect of the support stiffness conditions on divergence The differential equation of torsional equilibrium used to model wing twist deformation is: eqca 0a0 d 2q æçeqca 0 ÷ö ÷÷ q = + çç 2 ÷ GJ dx èç GJ ø (a) Find the divergence dynamic pressure of this wing.  in.lb.  (b) Torsion springs with stiffness, k  is placed at the left and right ends of the  radian  test article (at x=0). Solve for the expression characteristic equation for divergence in kb terms of a parameter b = . Plot the divergence dynamic pressure as a function of GJ b to generate a figure similar to Figure P2.9(b).

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Problem 2.10 Classical wing divergence ignores the fact that the wing is attached to the fuselage with pitch and plunge freedom. Consider the idealized, un-cambered wing attached to a torsion spring which is, in turn, attached to an idealized fuselage, as shown in Figure P2.10a. The torsional spring resists torsion by developing a restoring moment M S equal to KT q , where θ is the relative rotation of the wing with respect to the fuselage. The fuselage is a freely flying article, but for our model it is attached to a frictionless pin to simulate the ability of the aircraft to rotate about its own center of gravity (c.g.). The wing has two degrees of rotational freedom a and θ that generate lift. An all-movable tail on the fuselage provides a pitching moment about the aircraft c.g. by generating lift, Ltail , given by the expression

Ltail = qStail C La

tail

(a + b )

(1)

In Eq. 1, b is the tail rotation with respect to the fuselage while a represents the rotation of the fuselage about the airplane c.g. or model pin.

Figure P2.10a: Wing fuselage configuration geometry Question: (a) Solve for expressions for α and θ in response to the tail angle, β using the following definitions eqSC La q = KT æ f öæ S ö÷æçC L ö÷ ÷ç tail ÷ç atail ÷÷ > 1 R =çç ÷ç (2) çèd ÷ø÷çèç S ÷÷øççç C La ÷÷ è ø As noted on Figure P2.10a, the pin representing the c.g. is usually ahead of the wing aerodynamic center. If this is so, then the dimension d will be negative. When the parameter R defined in Eq. 2 is multiplied by -1 the result is similar to the definition of the tail volume used in stability and control studies. If R > 1 (or − R < −1) then the configuration is stable in pitch in forward flight. (b) The purpose of the tail control surface is to create a change in pitch angle. Define control reversal as the condition that occurs when we deflect the tail surface, but there is no change in aircraft attitude angle. Set up an analysis to determine whether or not tail control reversal q occurs. Solve for the tail control reversal dynamic pressure.

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eqSC La as a function of R. KT Plot this divergence pressure parameter as a function of R. What value of R is required to make the divergence pressure parameter equal to 1?

(c) Solve for divergence dynamic pressure parameter q =

Hints: The FBDs are set up in Figure P2.10b. They must be useful. The equilibrium equations are also shown below. So, too, is the characteristic equation.

Figure P2.10b: Wing/ fuselage free body diagram é ö ö æC æ f ö÷çæ S ö÷æçC La,tail ö÷ùú æ öæ ê ÷÷ a + q q =q çç f ÷÷çç Stail ÷÷÷ çç La,tail ÷÷÷ b q ê1 - çç ÷÷çç tail ÷÷çç ú ÷ ç èçd ø÷èç S ø÷èç C La ø÷÷ú èçd ø÷ èçç S ø÷ èç C La ø÷÷ êë û and q a - éëê1 -q ùûú q = 0 So that D =-q éê(1 - R )(1 - q ) +q ùú = -q (1 - R + Rq ) ë û Problem 2.11: Three identical un-cambered wing segments are connected to each other by torsional springs and to the wind tunnel walls by additional torsion springs, as indicated in the figure. The four torsion springs have the same spring constant KT . The wing segments are mounted on bearings on a spindle attached to the tunnel walls. V

e Area=S 1 K T

e

2 S Area=S K T

S 3 Area=S K T

K T

Figure P 2.11 (a) Develop the equations of static equilibrium at neutral stability. The three degrees of freedom are q1, q2 and q3 . (b) Write the expression for the strain energy stored in this configuration as a function of the torsional displacements. Use the energy method to derive the system stiffness matrix in terms of the three torsional deflections. Compare this result to that found in part (a). They should be identical. (c) Solve for the three dynamic pressures that create neutral stability. Tran Tien Anh@HCMUT

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(d) Find the mode shapes at all neutral stability points. Describe the mode shapes (how do the surfaces deflect relative to each other?). (e) Which of the three dynamic pressures in part (c) is the divergence dynamic pressure? Problem 2.12 Two wing sections are mounted on shafts attached to each other and to a wind tunnel wall, as indicated. Note that the two torsion spring stiffnesses are equal, but are offset an amount d. This configuration is similar to, but not identical to, the problem in example 2.8. When the sections are placed in the wind tunnel at an angle of attack a0 , the two springs deform, as indicated in the diagram. Lift on the two, identical, un-cambered wing sections is L1 = qSC La (a0 + q1 ) L2 = qSC La (a0 + q2 )

V line of aerodynamic centers

e A

d e

1

2

b/2

b/2

A torsional spring stiffness K (in-lb/rad)

torsional spring stiffness K (in-lb/rad)

2+ 1+ view A-A

Figure P2.12 (a) Place an aileron on the outer (right-hand) section. This aileron has aerodynamic coefficients

C Ld and C MAC ,d . Develop the static equilibrium equations for the system when there is no

initial angle of attack, but the control surface is deflected downward an amount d0 . (b) Specialize the result in part (a) by making the aileron flap-to-chord ratio E = 0.15 with

d e = 1 and = 0.10 . Solve for the rolling moment generated by the aileron as a function e c of dynamic pressure, q. (c) Solve for the reversal speed.

Hints: (a) The static equilibrium equations for the system when there is no initial angle of attack, but the control surface is deflected downward an amount d0 .

é 2K ê ê-K êë

é ê 1 -K ùú ïìïq1 ïüï ê í ý - eqSC La ê ê0 K úú ïïq2 ïï ûî þ ê êë

Tran Tien Anh@HCMUT

öïü ïìïæ ö æ öæ öæ d úù çç1 + d ÷÷ççC Ld ÷÷ + ççc ÷÷ ççC MAC ,d ÷÷÷ïï ï ÷ ï ì ü ÷ èç C La ÷ø÷ïïï e ø÷÷çèçC La ø÷ èçe ø÷ç e úú ïïq1 ïï = eqSC d ïïèç ý æ öí ý La 0 í ïï ïï æd ö÷çæC ö÷ çç1 + d ÷÷ú ïïîq2 ïïþ L d çç ÷ç ú ÷ ÷ ï ï ÷ ÷ çè ÷ e øúû çè e ÷øèççC La ø÷ ïîïï ïþïï 10

The aileron coefficients are computed to be:

C C Ld = 0.4805 and MAC ,d = -0.0966 . With C La C La

d e = 1 and = 10 , we get: e c é 2 -1ù ìïq ïü é ù ïì üï ìï-0.0051üï ï ê ú ïí 1 ïý - q ê1 -1ú ïíq1 ïý = q d ï ý o í ê-1 1 ú ïq ï ê0 2 ú ïq ï ï ï 0.4805 êë úû ïî 2 ïþ êë úû ïî 2 þï ï î þï ì ü ì ü q do ïï(0.4754 - 0.4703q )ïï ïïq1 ïï = í ý í ý 2 ïïq2 ïï ïï (0.956 - 0.4805q ) ï 2 q 4 q + 1 î þ ïï îï þ The lift on each section is: ìï 0.4754 - 0.4703q üï ì ü ìü 2.081q ïïL1 ïï = qSC d ïï1ïï + qSC d ï í ý í ýï Ld 0 í ý Ld 0 2 ïïL2 ïï ïï1ïï ï 2q - 4q + 1 ïî(0.956 - 0.4805q )ïïþ î þ îþ é ù ê 1.021q 2 - 3.011q + 1 ú L1 = qSC Ld d0 ê ú ê ú 2q 2 - 4q + 1 ë û é ù 2 ê 1.00q - 2.011q + 1 ú L2 = qSC Ld d0 ê ú ê ú 2q 2 - 4q + 1 ë û ì ü ì ü 1 ïì b 3b ïüï ïL1 ïï = b 1 3 ï ïL1 ï ï ïýí M Roll = ïí ý í ý = (L1b + 3L2b ) ïîï 4 4 ïï ï ï ïL2 þ ïï 4 þïïîL2 þï 4 î

(

)

(

)

(

)

(

)

{

M Roll =

( (2q

}

)

qSC Ld d0 1.005q 2 - 2.261q + 1 2

)

- 4q + 1

Problem 2.13 The wing idealization is shown in the figure will be tested in the wind tunnel at several different airspeeds and in several different configurations. All testing is to be done at sea level conditions and at such low speed that the flow field is incompressible. Changes in the position of the wing box change the wing shear center position and the offset distance between the shear center and the aerodynamic center. For instance, when the shear center is at the mid-chord, when the shear center is at the wing mid-chord the offset distance e is equal to c/4. Preliminary wind tunnel testing shows that the wing divergence dynamic pressure is 100lb / ft 2 .

lift wing box

shear center range

Figure P2.13: Wind tunnel wing segment model cross-section The team aerodynamicist predicts that the ratio C MAC / C La is -0.075 and the lift curve slope is 4.0; the reference planform area is S = 10 ft 2 . The section angle of attack a0 for all tests will be pre-set to 3o.

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Question: (a) Develop the expression for the lift force when the shear center is located at the mid-chord. Find the twist angle q as a function of wind tunnel airspeed. (b) Consider the same wing with different wing shear center locations. The offset distance e0 = c / 4 when the shear center is located at the mid-chord (part (a)) is to be used as the reference and ei is the offset when the shear center is moved. Develop the equations for wing lift L and twist q as functions of the ratio ei / e0 . Write these expressions in non-dimensional form, but use the divergence q for the reference wing as the reference in all expressions, because the divergence speed changes with ei / e0 . (c) Plot wing lift and twist angle vs. airspeed for four ei / e0 values of 0, 0.25, 0.50 and

0.75. Use an airspeed range between zero and 150 ft./s. Problem 2.14: Nonlinear torsional spring stiffness Consider the single degree of freedom un-cambered wing with the cross-section shown in Figure P2.14. A concentrated load P is applied aft of the shear center to create a moment Pd about the shear center to create twist θ. The wing has no initial incidence angle. This single degree of freedom wing model has a linear aerodynamic applied load given as L = qSC Laq but it resists applied torque by developing a nonlinear structural restoring moment given by the relationship: M S = KT q + KT 1q 3

Figure P2.14: Single degree of freedom airfoil with linear aerodynamic load and nonlinear structural spring KT 1 can be either positive or negative. When KT 1 is positive we have a so-called "hardening spring". When KT 1 is negative the spring is called a "softening spring". When KT 1 is zero, we have the usual linear spring relationship. K (a) When T 1 = 0.2 derive the nonlinear equations of static equilibrium when an initial KT moment M s = eqSC Laa0 is applied at a flight dynamic pressure q. Plot four different values of dynamic pressure parameter q =

qD =

M0

KT

vs. θ for

q = 0.1, 0.5, 1.0 and 1.1 with qD

KT for -0.5 < q < 0.5 eSC La Answer

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M0 = (1 - q ) θ + 0.2θ 3 KT

(b) The effective torsional stiffness is defined as the change in M 0 with respect to θ, ¶M 0 effective torsional stiffness is defined . The normalized ¶q K 1 ¶M 0 as Ke = eff = so that K e = (1 - q ) + 0.6q 2 . Using the same values of KT KT ¶q dynamic pressure parameter as in part (a) plot normalized effective stiffness as a function of q for -0.5 < q < 0.5 . (c) Remove the external load and verify that the static equilibrium equation becomes. Keff =

q3 +

KT KT 1 - q ) q = 0 . This equation has three solutions q = 0 or q = ( (1 - q ) . KT 1 KT 1

When q < q D then there is only one equilibrium point qe = 0 because the term under the

KT 1 £ 1 (note that this is the KT inverse of the term in the twist solution above and is a measure of the spring nonlinearity

radical is negative. Plot these solutions as a function of 0 £

Problem 2.15 Consider an airfoil of unit span and chord c = 0.5m hinged at 0.35m from the leading edge in a wind tunnel. A torsional spring of constant KT is attached to the airfoil at the hinge point, and a translational spring of constant K h = 100 éêëN / m ùúû is attached to the airfoil at its leading edge. At zero airspeed, the airfoil is at 5 degrees angle of attack and at 10m/s airspeed, the airfoil is at 10 degrees tota1 angle of attack. Assume C La = 4.5 / rad and r = 1.225 kg / m 3 . a) Calculate the divergence dynamic pressure, the torsional stiffness KT b) Calculate the percentage increase in the aerodynamic moment about the hinge point due to the aeroelastic effect. Problem 2.16 Consider a wind-tunnel model of a wing with 2m span and 50cm chord which has an ai1eron that spans from b/3 to b/2. The wing is attached rigidly at the root to the wind- tunnel æ hö wall. The torsional rigidity of the wing varies along the span and is given by 5 * 105 çç1 - ÷÷÷ . çè 4 ø÷ Assume C La = 4.5;C Ld = 2.5;C M ,d = -1.5 and e=0.1. Assume a single mode shape of f (h ) = h for the twist distribution.

a) Calculate the reversal pressure. b) Calculate divergence dynamic pressure Problem 2.17 An airfoil of unit span and 0.5m chord is mounted in a wind tunnel at its quarter chord point via a torsiona1 spring of constant 300Nm/rad. The airfoil is free to oscillate in pitch direction only. The airfoil mass is 5kg. Its moment of inertia about the center of gravity is 0.1 kgm 2 , which the center of gravity is located at the mid chord. Assume standard air density 1.225 kg / m 3 . Assume quasi-steady aerodynamics and write the complete and simplified equation of motion of the airfoil based on the given parameters. (a) Write the characteristics equation of the system, (b) Based on that discuss the stability of the system and possibility of flutter and divergence. Tran Tien Anh@HCMUT

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Problem 2.18: Flexible airfoil lift effectiveness and divergence A flexible, symmetrical  CMAC  0  airfoil is supported by two springs, K1 and K 2 as shown in Figure 1. At zero airspeed, with no forces applied, the airfoil has a nose-up pitch angle, αo, with respect to the horizontal. Small angles arealways assumed. The lift on the rigid airfoil is Lrigid  qSCL 0 When lift is applied at the aerodynamic center of the airfoil (the 1/4 chord point on the airfoil) the airfoil moves upward an amount h as it twists an amount θ. The springs in Figure P2.18 are undeformed when θ = 0 and h = 0. The flexible airfoil lift L flex is calculated from the equation

L flex  qSCL 0   

Figure P2.18: Configuration geometry Question: a) Derive the algebraic expression for the ratio Lflex / Lrigid b) Derive the expression for divergence dynamic pressure c) Find solution by using matrix approach d) Find position of center of twist and shear center location c c 5c e) Calculate with given values a = ; b = ; a + b = d = 8 2 8

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