AA311.Lecture20-1

# AA311.Lecture20-1 - AA 311 Lecture 20 Static Stability...

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Unformatted text preview: AA 311 Lecture 20: Static Stability Transferring moments. Recall that in the previous lecture we began discussing static longitudinal stability. We obtained requirements on the dimensionless pitch moment coefficient as a function of the angle of attack α . Specifically, we found that static longitudinal stability requires C m α < 0 and C m > 0. At this point, it will be helpful to review the basic notion of equivalent representations of forces and moments. z x a F M O z x F (- ) M Fx = x a Figure 1: Equivalent force and moment diagrams. Consider the planar rigid body shown on the left in Figure 1. The body is subject to a force F acting at the point x a and a moment M , which is a pure couple. For this system, we have X F z = F and X M O = M- Fx a . One may easily transfer a set of forces and moments acting at a given point to any other point. For example, one may transfer the force and moment above to the origin O, as shown at the right in Figure 1. L D M V c a a a a L D M V c b b b b ® ® x x Figure 2: Equivalent force and moment diagram for a wing. Now consider a rectangular wing. We assume that lift force, drag force, and aerodynamic moment are known, as functions of angle of attack, at the point x a . (In keeping with aerodynamics convention, the signed distance x is measured positive aft from the wing leading edge.) Suppose we wish to transfer the forces and moment from the point x a to another point x b along the chord. The forces are equal at either point: L b = L a = L and D b = D a = D. It remains to determine the moment M b given M a , L a , and D a . First, compute the moment of the system on the left about a particular point, say the leading edge: M l . e . = M a- L ( x a cos α )- D ( x a sin α ) . Next, compute the moment of the system on the right about the same point: M l . e . = M b- L ( x b cos α )- D ( x b sin α ) . 1 Equating the two expressions for M l . e . and solving for M b gives M b = M a + ( L cos α + D sin α )( x b- x a ) . Dividing through by ( 1 2 ρV 2 ) S ¯ c gives C m b = C m a + ( C L cos α + C D sin α ) x b ¯ c- x a ¯ c . (Note: Since we are considering a rectangular wing, the mean aerodynamic chord ¯ c is simply the constant chord length c .) Define the nondimensional distances h a = x a ¯ c and h b = x b ¯ c . For small angles of attack 1 , we have C m b = C m a + ( C L cos α + C D sin α )( h b- h a ) ≈ C m a + C L 1 + C D C L α ( h b- h a ) . For a well-designed wing operating below stall, C D C L 1 (ignoring, as pathological, the case where C L → 0). Since we have already assumed that α is small, the product C D C L α may be neglected. We therefore have the following approximate equation for transferring an aerodynamic moment between points on a wing: C m b ≈ C m a + C L ( h b- h a ) ....
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AA311.Lecture20-1 - AA 311 Lecture 20 Static Stability...

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