Steady Maneuvering

A wing-tail combination is pictured in Figure 8.19 undergoing a steady pull-up maneuver. For clarity, the angle Да, is shown much larger than it

Figure 8.19 Wing-tail combination in a steady pull-up maneuver.

actually is; that is, the ratio of /, to the radius of the flight path, R, is much less than one. The acceleration, an, of the airplane toward the center of curvature requires a lift in excess of the weight given by

L=w(l+^j (8.47)

The lift-to-weight ratio is known as the load factor, n.

The relationships derived thus far that relate CM to CL (or a) hold for the maneuvering case. However, an additional increment must be added to CM as a result of the increment in a, due to the angular velocity, Q.

AM = — |pV2T/,S, a, Да,/,

But

(8.49)

Thus, in coefficient form,

Notice that we can also visualize the airplane as translating with a linear velocity of V while rotating about its center of gravity with an angular velocity of Q. The tail is then moving down with a velocity of l, Q, which leads to Equation 8.49.

Horizontal Stabilizer-Elevator Configuration: Elevator Angle per g
Including the effect of Q, we can now write Equation 8.7 as

См = Cm„ + Cm„& + Cms 5e + CmqQ (8.51)

However, neither Q nor CMq is dimensionless (although their product is). Thus, it is sometimes desirable to write Equation 8.51 as

Cm = CM<) + См„а + Cms Se + CMfl

where q is a dimensionless pitch rate defined by

The choice of cl2 as a reference length is simply according to convention and has no physical significance.

From Equation 8.50, CMq is seen to equal approximately

CM<, = – n, a,VH^ (8.54)

Cm, is referred to as “pitch damping,” since it represents a negative moment proportional to the rotational velocity about the pitch axis.

In a steady pull-up maneuver, there is no angular acceleration, so CM must be zero as it is in trimmed level flight. Assuming Сщ to remain constant, we can therefore equate the increment CM resulting from the steady maneuver to zero. Since CM is assumed linear with a, S, and q, we can write

0 = См„ Да + Cm5 AS, + Cm,<? or

/См Aa + Cm <?

a*-(–c„, ’) <855>

Remember that q now represents the dimensionless pitch rate and not the dynamic pressure. Since

V = QR

and

it follows from Equation 8.53 that

«=^(7) ,8’561

The quantity (ajg) is the acceleration expressed as the ratio to the ac­celeration due to gravity. When, for example, we speak about “pulling 4g’s,” we mean that ajg – 4.0.

Да can be obtained from Equations 8.47 and 8.48.

or

Cl + A CL 1 an CL g

Thus,

ACl=Cl —

= CLa A a + Cls A Se

so that

where CL is calculated for 1 g flight. Combining Equations 8.55, 8.56, and 8.57 results in an expression for the elevator angle per g of acceleration.

A Se Cm CL + gc/2 V2CMqCLa (ajg) CMfLa ~ CMCLs ( )

Substituting Equation 8.13 for CMq, Equation 8.58 becomes

ClCl

CmsClr — CmCl where m is the mass of the airplane. We now define a relative mass parameter, p.

2m

pcS

cl2 is the same reference length used in the definition of q (Equation 8.53). This length multiplied by the wing area gives a reference volume. The parameter p is then the ratio of the airplane mass to the air mass within this volume. Using the definition of p, equation 8.58 becomes

Similar to the neutral point, there is a particular value of h, known as the stick-fixed maneuver point, hm, for which no elevator deflection is required to produce an acceleration. Indeed, if the center of gravity moves aft of this point, the effect of the elevator is reversed. From Equation 8.60,

hm = hn-^CMq (8.61)

Since Cm, is negative, hm lies aft of the neutral point.

In terms of the neutral point,

The quantity (hm – h) is known as the stick-fixed maneuver margin. Again, it is emphasized that CL is calculated for straight and level flight. It is not the total airplane CL required during the steady pull-up maneuver.

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