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Thus, to pull higher load factors during the maneuver, more up elevator deflection
is required compared to that in level fiight at the same altitude and forward speed.
This incremental elevator deflection.can be expressed as
                                    A8, = _C (xc8 - Nm)An
                          -        ' (3.204)
            Cm8
   Stick-free maneuver point.   Suppose that the e     or is left free to float
during a pull-up maneuver. Then, the horizontal ta,'~_;'e.   at'a'k is given by
                  ctt - (CtW -/LU+/t -6+A :.,) (1-     -,,.,)          (3.205)
so that
     dar
   - =[.1.(1-~)+-,,,1(,-,-....) (3.206)
               dCL =
Then,
 G~ )..=-ti,+(;g ),--.. (i-:)+2v-,](i-,cC,,.,)
           (3.207)
                                                                      ~
       N;" =  ac*w - GC/  )f + -V  7  [(1 - i) + 2p~,] (1 -,CC,,.,)
                             .  (3.208)
STATIC STABILITY AND CONTROL                           253
We observe that the stick-free maneuver stability is higher than the stick-free
stability in level flight and the stick-free maneuver point is aft of the stick-free
neutral point. This increase in stick-free stability during the maneuver, as said
 before,is caused by the angular velocity S2 and the associated increase in tail angle
of attack Acti.r. The stick-free maneuver margin is given by
For a stable aircraft, H:n > 0.
HL = NL - xcg
(3.209)
    Stick force gradient.    As we know, the stick force is directly proportional to
A8e = 8e.R - 8V:/, the difference between the required elevator deflection and the
floating angle of the elevator. Recall that these two quantities are also related to
the stick-fixed and the stick-free stability levels of the airplane, respectively.
    From Eq. (3.101), we have
8e.R = - CC.,  &~i ),                                  (3.210)
The floating angle of the elevator is given by
so that
= - C;~B as                                                             (3.211)
 Cha
= - C,,.,  [C..  (1 - :  ) + s-2j ]                     (3.212)
                  A8e = 8e.R - 8e.f                                                                (3.213)
        CL
      =-cC.,GC~),+~C [C (i-J)+'2J-] (3.214)
Differentiating with respect to CL and rearranging, we obtain
     dA8e
　
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