# Example Calculation of Stick Force

Again, the Cherokee 180 will be used as an example, and I emphasize that the numbers are in no way endorsed by the manufacturer. The tail dimensions and angular movements were measured by students during a course on techniques of flight testing for which the Cherokee 180 was used. Being an easy and forgiving airplane to fly, it is a good vehicle for such a course.

The horizontal tail of the Cherokee, shown in Figure 8.16, is of the type pictured in Figure 8.7c. This type of tail is sometimes called a “stabilator.” The gain between the elevator and the horizontal stabilizer, ke, as well as the gearing, G, can be determined from the graphs presented in Figure 8.16. The stick position and angles shown here are arbitrary. They were obtained by moving the control wheel to a given position and measuring the distance from an arbitrary reference point on the wheel to an arbitrary reference point on the instrument panel. At each position, a protractor with a bubble level was placed on the horizontal stabilizer and then on the elevator to obtain the angle of these surfaces relative to the horizontal. Using these graphs, you should be able to determine that

ke = -1.50 G = +0.5 rad/ft.

The following quantities are estimated from the tail geometry assuming the airfoil to be symmetrical with the aerodynamic center at 0.25 c and a section lift curve slope equal to 0.106C//deg.

a, = 0.0642CJdeg (Equation 3.74)

t = 0.55 (Figure 3.32)

 Figure 8.16 Geometry and linkage for the Cherokee 180 tail.

г) = 0.80 (Figure 3.33)

-0.185 (Figure 3.31)

AC;

For the symmetrical airfoil with 5, = 0, CMac = 0, so CM at the pivot line, which is 1.7% of the chord behind the C/4 point, will be

Cm = 0.017a, a,

Thus,

The increment in CLl due to the elevator deflection will be

A CL, = а, тт) 8e = 1.616 5,

The increment in CM about the pivot line produced by 5, will equal the sum of the increment in CMac and the moment produced by the increment in

CL acting ahead of the pivot.

ДСМ = -0.185(1.616) Se + 1.616(0.017)5,

= 1.616 5Д-0.185+ 0.017)

Thus, it follows that

The following numbers are typical of the Cherokee 180 with four pas­sengers, full-fuel, but no baggage.

W = 2255 lb

h = 0.197 (dimensionless location of center of gravity)

For ea of 0.447, h„ = 0.442, so that

Cm„ = CLa(h — hn)

= 4.50(0.197-0.442)

Also,

Cm, = ViVh^i

= (1)(0.384)(3.68)

= 1.41

 = -0.575

The constant, A, in Equation 8.34 becomes

A = 0.0476

The stick force, from Equation 8.35, becomes

Р = С5Л(|)л[і-(0]

= (0.5)(25)(2.5)(14.1)(0.0476)[l –

■2o4‘-(0]

The estimated stick force, P, is presented in Figure 8.17 as a function of airspeed for different trim speeds.

 Figure 8.17 Calculated stick force versus velocity for a Cherokee 180.