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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 . Specifically, we found that static longitudinal stability requires C m < 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 )- D ( x a sin ) . Next, compute the moment of the system on the right about the same point: M l . e . = M b- L ( x b cos )- D ( x b sin ) . 1 Equating the two expressions for M l . e . and solving for M b gives M b = M a + ( L cos + D sin )( x b- x a ) . Dividing through by ( 1 2 V 2 ) S c gives C m b = C m a + ( C L cos + C D sin ) 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 + C D sin )( h b- h a ) C m a + C L 1 + C D C L ( 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 is small, the product C D C L 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 ) ....
View Full Document

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online