At what angle-of-attack (sideslip angle) would a symmetric vertical fin plus a deflected rudder have a lift coefficient of exactly zero?

For rudder deflection angles of 5, 10, 20, 30, and 45 degrees, at what (negative) angle-of-attack does a fin-rudder combination (including dorsal fin if present) have a lift coefficient of exactly zero?

For the purpose of this question, define angle-of-attack in relation to the chord line of the fin alone, not including the deflected rudder. If we ignore things like spiral slipstream, deflection of airflow around cabin, etc, it would be the same as the sideslip angle.

Loosely speaking, at this angle-of-attack we could say the force from the deflected rudder is being exactly cancelled by the force from the fin, though in reality each influences the airflow around the other and the fin-rudder combination acts as a single unit.

To allow for a definitive answer, I'll specify to use the fin (including dorsal fin) and rudder from current -production Cessna 172 S, but feel to post an answer addressing the fin and rudder from any other aircraft.

Note that an aircraft could never sustain a sideslip angle that caused the fin plus deflected rudder to have a lift coefficient of zero, unless some other yaw torque due to something like asymmetric thrust or extreme adverse yaw from deflected ailerons was acting to maintain the slip angle.

Approximations are fine-- obviously an exact answer would involve a complicated computer modelliing project, or actual wind tunnel experiments. My hope is that existing information relating to the lift curves of symmetrical airfoils with deployed flaps may be adapted to give an approximation of an answer to this question.

• Despite specification, this is a very broad question. Obviously, for each combination of airfoil, relative rudder chord and other minor factors there will be 'angle of zero lift' for each deflection; it's just a matter of mundane calculations or testing. What information each of these random points can give you? (Say, NACA 0012, 25% rudder, 10° deflection -> -5° $\alpha_0$. So?) Trends may be interesting, such as this angle per deflection vs. relative chord, but that's not what you're asking... – Zeus Aug 5 '19 at 0:45
• Besides, "an aircraft could never sustain a sideslip angle that caused the fin plus deflected rudder to have a lift coefficient of zero" - why, this always happens in a sustained sideslip (even without adverse ailerons, fuselage interference etc.) – Zeus Aug 5 '19 at 0:49
• @Zeus - why do you say that? I would say that in a sideslip in a Cessna 152, where you are pushing hard on the (let's say right) rudder pedal to keep the rudder deflected, it's likely that the NET contribution of the rudder + fin is a yaw torque toward the right, acting against the leftwards yaw torque created by the fuselage. (Basically I'm saying that the center of lateral area or effective center of aerodynamic yaw torque of the fuselage alone is likely well behind the CG, since the rear parts of the fuselage act at such a large moment-arm.) – quiet flyer Aug 5 '19 at 16:37
• Why do you include fuselage? You excluded more relevant things like "spiral slipstream, deflection of airflow around cabin, etc.", why should I assume that fuselage should be calculated? Its influence can be quite complicated. A slender body has its AC quite far forward actually, but when you include interference of other parts (wing, "deflection of airflow around cabin, etc"), there may be no AC as such. With fuselage, the exercise cannot be reduced to "lift curves of symmetrical airfoils with deployed flaps", so you either should exclude it or define exactly. – Zeus Aug 6 '19 at 0:56