# How can I calculate the rolling moment of an aileron for a given plane based on its performance?

Let's say I have an already-built airplane with known basic characteristics like weight, wing span and wing surface, and I can measure the time of all possible manoeuvres at different speeds.

How can I calculate the roll moment of an aileron in its maximal deflection? The precision of hundreds of [kg*m] is sufficient.

If you know the rolling speed at a given flight speed, you can calculate the aileron effectiveness and use that to calculate the forces. The final rolling speed is reached when roll damping and the aileron-induced rolling moment reach an equilibrium: $$c_{l\xi} \cdot \frac{\xi_l - \xi_r}{2} = -c_{lp} \cdot \frac{\omega_x \cdot b}{2\cdot v_\infty} = -c_{lp} \cdot p$$ Thus, your aileron effectiveness is $$c_{l\xi} = -c_{lp}\cdot\frac{\omega_x \cdot b}{v_\infty\cdot(\xi_l - \xi_r)}$$ The roll damping term is for unswept wings $$c_{lp} = -\frac{1}{4} \cdot \frac{\pi \cdot AR}{\sqrt{\frac{AR^2}{4}+4}+2}$$ and the moment per aileron now is $$M = c_{l\xi} \cdot \xi \cdot S_{ref} \cdot b \cdot q_\infty$$ Calculate the moment for each aileron separately; normally the left and right deflection angles are not exact opposites, which helps to reduce stick forces.

If you only need an approximation, maybe do it like this:

You first need to have all dimensions and the deflection angles. I expect you don't have lift polars of the wing section, so you need to approximate the lift increase due to aileron deflection with general formulas. This is $$c_{l\xi} = c_{l\alpha} \cdot \sqrt{\lambda} \cdot \frac{S_{aileron}}{S_{ref}} \cdot \frac{y_{aileron}}{b}$$ and the moment per aileron now is $$M = c_{l\xi} \cdot \xi \cdot S_{ref} \cdot b \cdot q_\infty = c_{l\alpha} \cdot \sqrt{\lambda} \cdot \xi \cdot S_{aileron} \cdot y_{aileron} \cdot q_\infty$$ Nomenclature:
$$p \:\:\:\:\:\:\:\:$$ dimensionless rolling speed (= $$\omega_x\cdot\frac{b}{2\cdot v_\infty}$$). $$\omega_x$$ is the roll rate in radians per second.
$$b \:\:\:\:\:\:\:\:$$ wing span
$$c_{l\xi} \:\:\:\:\:\:$$ aileron lift increase with deflection angles $$\xi$$
$$\xi_{l,r} \:\:\:\:\:$$ left and right aileron deflection angles (in radians)
$$c_{lp} \:\:\:\:\:\;$$ roll damping
$$c_{l\alpha} \:\:\:\:\:\;$$ the wing's lift coefficient gradient over angle of attack. See this answer on how to calculate it.
$$\pi \:\:\:\:\:\:\:\:$$ 3.14159$$\dots$$
$$AR \:\:\:\:\:$$ aspect ratio of the wing
$$\lambda \:\:\:\:\:\:\:\:$$ relative aileron chord
$$S_{aileron} \:$$ Surface area of the aileron-equipped part of the wing
$$S_{ref} \:\:\:\:\:$$ Reference area (normally the wings's area)
$$y_{aileron} \:$$ spanwise center of the aileron-equipped part of the wing
$$v_\infty \:\:\:\:\:$$ true flight speed
$$q_\infty \:\:\:\:\:$$ dynamic pressure

Depending on the relative chord length of the aileron, this formula is good for maximum deflections of 20° of a 20% chord aileron or 15° deflection of a 30% chord aileron. Remember: This is a rough estimate for straight wings.

• What are the units in the first equation? The right hand side appears to have units of [seconds], assuming p is dimensionless, while the left-hand-side appears to be dimensionless. – supergra May 14 at 19:11
• @supergra: No, it's dimensionless. I just realised that I confused p with $\omega_x$ in the first two equations. Thank you for finding this mistake! – Peter Kämpf May 14 at 21:25