Here’s a comparison of relative magnitudes of control surface hinge moment and “self-centering” bellcrank moment. First the control surface hinge moment of an Impact with flaps at 30 degrees flying a 5.25-second lap on 70-foot lines:

H

_{f}=(ρV

^{2}/2)*S

_{f}*c ̅

_{f}*Ch

_{f} =(ρV

^{2}/2)*S

_{f}*c ̅

_{f}*(∂Ch

_{f}/∂δ

_{f}*δ

_{f} +∂Ch

_{α}/∂α*α),

where

H

_{f} is hinge moment, the torque reqired to deflect the flap

ρ is air density, currently .002246 slugs/ cubic ft. at my house

V is true airspeed = 70 ft. * 2π / 5.25 sec. = 83.8 ft./sec. for a 5.25-second lap on 70-ft. lines, neglecting the variation in airspeed from the inside tip to the outside tip

S

_{f} is flap area, call it 144 sq. inches = 1 sq. ft.

c ̅

_{f} is flap mean aerodynamic chord, call it 2.4 inches = 0.2 ft.

Ch

_{f} is flap hinge moment coefficient, which is nondimensionalized hinge moment, expanded as shown and probably not as scientific as it looks.

∂Ch

_{f}/∂δ

_{f} is rate of change of hinge moment with flap deflection at constant angle of attack: = -.011/degree*

δ

_{f} is flap deflection, 30 degrees for this example

∂Ch

_{α}/∂α is rate of change of hinge moment with angle of attack at constant flap deflection: = -.005/degree*

α is angle of attack, degrees. Assume 6 degrees for this example.

H

_{f}=((.002246 slugs/ cubic ft.) *83.8 ft./sec.

^{2})/2*1 sq.ft.*0.2 ft.*(-.011/degree*30°-.005/degree*6°)

=-1.14 ft.lb.

I figure an Impact with conventional control mechanism should have -8.8 degrees of flap per inch of differential line pull (ciphering upon request). That gives 2.1 lb. of differential line tension, I think, at -30 degrees of flap. If total bellcrank travel is restricted to 60 degrees, leverage is -14.7 degrees of flap per inch of differential line pull, which gives 3.5. lb. of differential line tension at -30 degrees of flap. Hope I came close on all those signs.

*Gary Letsinger said, “Hinge moment calculations are easy to do, but hard to believe.” These ∂Ch

_{f}/∂δ

_{f} and ∂Ch

_{α}/∂α data come from an NACA report shown in two classic textbooks as an example not to be used in general. Bill Netzeband in his Control Line Aerodynamics Made Painless series,

https://miniatureaeronautics.com/Dave_Day/netze/control.htm , used it in general, though, and if it’s good enough for Wild Bill, it’s good enough for me.

Now the differential line tension from a “self-centering” bellcrank. Force on bellcrank from “centrifugal” force only = mV

^{2}/r,

where

m= airplane mass. We usually refer to airplane weight: 73 oz. for my massive Impact. To convert to mass in slugs, one converts ounces to pounds and divides by the acceleration of gravity, about 32.2 ft./sec.

^{2}. 1 lb. = 1 slug ft./sec.

^{2}m=73 oz.*(slug ft./sec.

^{2})/(16 oz.*32.2ft./sec.

^{2})=.142 slug

V is inertial speed = true airspeed if there’s no wind = 83.8 ft./sec. from above calculation

r is circle radius = 70 ft.

mV

^{2}/r=(.142 slug*83.8 ft./sec.

^{2})/70 ft.=14.2 lb.

Plugging this number into the second graph of the first post, one gets pounds of differential line tension caused by bellcrank hole arrangement. The red and orange lines on the plot show the contribution of hinge moment to differential line tension. I’ll assume that flap hinge moment varies linearly with flap deflection from -30 degrees to +30 degrees.

Differential line tension is the sum of the contributions of bellcrank bend angle and hinge moment. I recommend lots of bellcrank movement (yellow line) so’s to shy away from the Netzeband wall. That gets one into the perverted part of the bent-bellcrank contribution. If bellcrank movement is limited to 60 degrees total (red line), bent-bellcrank contribution is better behaved.