Question

You are analyzing and designing a simple cambered wing in flight. Figure Q5 shows the forces and moments act on this wing.

a. Check (supporting by analysis) the location of the aerodynamic center with respect to the center of gravity if the design criterion to achieve is the static equilibrium.

b. Check (supporting by analysis) the location of the aerodynamic center with respect to the center of gravity if the design criterion to achieve is the static stability.

c. Conclude your results obtained in Parts Q5(a) and Q5(b).

d. Select a design that satisfies the combined criterion for stable trim in case of wing alone design.

e. Evaluate the selected design in terms of aerodynamic efficiency and handling qualities

L 12 ma V ►D W Figure Q5.

Answer 1

SOLUTION: The location of the aerodynamic centre can be determined from the knowledge of how the normal force coefficient and moment coefficient about any point on the chord line very with angle of attack. For most of the airfoils, the position of aerodynamic centre is very nearly constant at quarter chord. For x = lw. We have AWON CmLE Cmo --Cy = Cmac Cmac = Cmc + Ics From defination of aerodynamic centre, ac mac да = 0 mac TUT Ww ac да 1) OCN = 0 + да c да OC ly da OC da с NOTE: Location of aerodynamic centre does not depend on magnitude of the aerodynamic coefficient. It depends on the derivative of the aerodynamic coeffiecient with repsect to nl of attack. lu Cm . da 2x C da In contrast, location of centre of pressure depends on the magnitude of aerodynmic coefficient. I cp . Cmi с Су Now, for static stability a small increase in angle of atttack must produce a negative pitching moment about th centre of gravity, to decrease the angle of attack back towards trim. Conversaely, a samll decrese in angle of attack must produce a positive pitching moment to increase the angle of attack to restore the trim. Thus, "The pitchiing moment about CG must vary with angle of attack such that any change in angle of attack prodces a change of opposite sign in the ptching moment about CG" дС 1 LOM да <0 да 3pv?se Thus, for static stability, am <0 да This is also called pitch stiffness Cm = Cmac,ne - WCL =0 At trim/equilibrium am amaclw #CL да да <0 с да Static stabilityrequirement From defination of aerodynamic centre, ac mac да = 0 1 дС <0 с до ac да > 0 c This is for a less than a stall Knowing the airfoil terminology, we will now explore the requirement for tream and then examine the pitch stability of the equilibrium state.

Assumptions: • Wing is symmetric in the span wise direction, • Motion of the wing throgh the air is in a direction normal to the span. . This results in no ide force, no rolling moment and no yawing moment. . CG, a.c. are allign with the thrust vector which is alligned with direction of flight For this symmetric flight condition, the aerodynamic forces acting on the wing can be resoved into a lift force L, a drag force D and a pitching moment about the aerodynamic centre of the wing Mac. For wing to be trimmed (i.e. equilibrium) the summation of forces in both the horizontal and vertical directions must be zero. This requires, TED LEW Mcg = 0 Mcgr Mac-LW™L At trim, M = Mac-LWL 3pv2 sacim 2 S&C.. = 3pv?sëlma, dan Cz) = 0 Cm Cmac,w-C=0 NOTE: For a given weight and airspeed, the lift coefficient is fixed by the trimrequirement (L=W). The moment coefficient about about aerodynamic centre is fixed by the wing geometry Cmac. Thus, for a given geometry, weight and airspeed, Iwisgiven by, Cmac "cfortrimm=0 CL Cmec c CL CL is always positive Cmac is negative for cambered airfoil Thus, <0 For equilibrium (trim), the aerodynamic centre of cambered wing must be forward of the centre of gravity > 0 implies, for static stability, the aerodynamic centre must be aft of the centre of From Abovec gravity For trim, ac must be ahead CG(cambered) For stability, ac must be behind / aft of CG. The above cases are opposite. Thus a simple cambered wing is not staticallystable in free flight For trim, Cmoc ho C CL For stability, La с да <0 substitute lw as in above equation Cmac BCL С да <0 дС > 0 For trim Da for a below stall as coefficient must be positive to support the weight Thus to get stable trim Cmac must be positive.

If stable trim is to be maintained, a single wing with no tail must always produce a positive pitching moment coefficient about aerodynamic centre. we know that, symmetric airfoil produces Cmac=0 cambered aerfoil produces Cmac<0 To produce Cmac, airfoil selction must have negative camber over at least some section of the chord. Note: An airfoil with negative camber through the chord is inefficient in producing positive lift and has low maximum lift coeffiecient. A better choice is an airfoil that has negative camber over only some portion of the chord near th trailing edge that is reflex airfoil. It is possible to design an aircraft consisting of only a single flying wing with no tail, so that stable trim flight can be achieved. However, such designs are not preffered as this is prone to poor handling qualities (damping is less). A better option is usually to combine a wing with a conventioanal tail. Positive art measured fromag WAL Voo (aite Mae w Dia - Reflex ed airfoil

Please give some positive rating because my CF is less...

Thank you so much...