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run for landing. Like ailerons, the spoilers have to be deflected differentially for
roll control. Br9cause they are mounted on the top surface of the wings, only up-
ward deflection is possible. Therefore, if the aircraft wants to roll to the right, only
the right spoiler has to be defiected upward. The left spoiler remains in its neutral
position. This causes the right wing to lose lift, whereas the lift on the left wing
remains unchanged.ll~is imbalance in lift creates a rolling moment to the right or
in the direction of the de:flected spoiler.
The aileron effectrveness is measured by the rolling-moment coefficient per
unit defiection (differential) of the ailerons and is denoted by Cl8a and can be
approximate,ly estimated using the strip theory as follows.
;l
':; I
~
STATIC STABILITY AND CONTROL
Fig. 3.105 Strip theory estimation of aileron effectiveness.
309
on the wing planform. As discussed in Chapter 1, for rectangular wings of high
aspect ratio, high taper ratio, and low sweep, the ailerons retain their effectiveness
all the way up to the wing stall because the stall on such wings originates at the
root sections and progresses towards the wing tip.
Howe'ver, for highly swept-back wings, the stall originates at the wingtip and
spreads towards the wing root. As a result, the ailerons lose their effectiveness
long before the entire wing stalls. In this respect, the swept-forward wing has a
distinct advantage because the stall progresses from root to tip more or less like a
rectangular wing.
Example 3.11
For the uulless vehicle configuration ofExamples 3.2 and 3.8, obtain the lateral
stability derivative Ctp for M -- 0.7 and M -. 2.0.
Solutron. We have forsubst ;peeds
(Cip,W,B, = :/s['(-::)/xcp KMAK f + (-,,)A]
+F [CFp KM,+ Aj~p ] +(ACW)z,
We have A = 2.8396, Ac/2 - 25.3070, and A, - 0.170. Using the data given in
Fig. 3.96 for A:0 and A -.0.5 andinterpolating for A = 0.170, we get (CtplCL) Xty2
= -0.00214. For A]cos Ac/2 = 3.1311, from Fig. 3.97, we get (approximately)
K MA = 1.0. For l}/b = (lf - 0.5cr)/b = 22.4792/17.3228 - 1.2976. From Fig.
3.98, we obtain Kf =1.0, and, from Fig. 3.99, (Cip/CL)A = - 0.0010. From
310 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
Fig, 3.100, we obtain (approximately) C,p/ F = - 0.00015 using interpolation for
the given value of}.. KMr as given by Fig. 3.101 depends on the Mach number. For
M - 0.7, we obtain KMr = 1.07. With d - 3.048 m, A - 2.8396, zw :1.27 m,
and b - 17.3228 m from Eqs. (3.369) and (3.370) wc obtain
ACip = -0.0000261]deg2
~=-
(ACtp)z. = 0.0009105
Substituting, we get
(Ctp)W(B) = CL(-0.00214 - 0.0010) + F(-0.00015 * 1.07 - 0.0000261)
+0.0009105 - -0.00314CL - 0.000186r + 0.0009105
With F : 3.5 deg, we get
' (CLp)WtB) = -0.00314CL+0.000258
The vertical tail contribution is given by
(Ctp)v = -ka, (1+ ?ap) ,,, (SS ) zu cosa ~IU sina
b
From the solution of Example 3.8, we have k -. 0.76,a - 0.0491]deg, and
(i+ Lau/aPDov - 1.2012. With zv - 3.8290 m,lu : 7.7561 m, b -. 17.3228 m,
and Sy/S = 0.1909 and assuming a to be small, we get
(Cip)v - -0.76*0.0491*1.20120*0.1909 3.8290-7.75
17.322~56 )
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