I am currently developing a flight simulator but there is one force I can't figure out how to simulate using real world physics. When the pilot pushes the stick to the right, the plane will reach a roll rate that it will keep; and when the stick comes back to its neutral position, the plane will stop rolling, more or less quickly.

Is there a formula, or an approximation that explains how the plane will stop its roll? On what does it depend? Apart from wing area, does airspeed and lift impact this? Or does it depends on angular velocity?

I need a formula that I can compute to slow my roll down.

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    $\begingroup$ If you were using "real world physics", things like this would just happen, and you'd be looking at your equations trying to understand why roll (rate) stops. In truth, you seem to be trying to simulate phenomena (roll damping in this case) using pre-existing knowledge about flight mechanics. This is how it's usually done, but this is a big subject in itself. You might now incorporate roll damping using given answers, but you'll invariably miss something else equally important. You need a good book on flight dynamics. $\endgroup$ – Zeus Apr 23 at 0:59
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    $\begingroup$ Do NOT underestimate the roll resistance of wings. I want you to try this experiment to get an intuitive feel of how much wings resist rolls: get yourself a rubber powered propeller (actually, even a battery powered one works but will run a long time). Install the prop onto a stick, wind it up and release it - you will see the stick roll due to the torque from the prop. Now add a piece of card/sheet foam to the stick to make small wings and repeat - you will see the roll rate reduced. Now make the wings longer and repeat. With long enough wings there will be hardly any roll $\endgroup$ – slebetman Apr 23 at 5:18
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    $\begingroup$ Thank you both for your answers. I finally succeed to implement roll daming naturally by taking in account the lift of each wing independently. Now with the AOA of each wing being different during the roll, the roll tends to stop when the stick is released. $\endgroup$ – Anselme Apr 23 at 12:09
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    $\begingroup$ The section about damping of how it flies can help you understand related concepts and mechanisms. $\endgroup$ – Manu H Apr 23 at 18:00

What you want is the roll constant $\text{T}_R$. This is basically one of the characteristics wich determines the equations of motion of an aircraft. It gives the slope of the roll speed increase over time with full aileron deflection and an ideally stiff wing, and equally the rate of decrease once the ailerons are set to neutral during a rolling maneuver.

roll rate over time

With p the dimensionless roll rate, m the aircraft mass, i$_x$ its radius of the momentum of inertia about the roll axis, S its wing surface, b the wing span, v the flight speed, $\rho$ the density of air and $\text{c}_{lp}$ the roll damping coefficient, the formula is

$$T_R = \frac{2\cdot m\cdot\left(\frac{2\cdot i_x}{b}\right)^2}{\rho\cdot v\cdot S\cdot c_{lp}}$$

When the pilot moves the stick, the aircraft will accelerate into the roll but the acceleration will diminish with the square of roll speed until an asymptotic value is reached. This acceleration grows with increasing air density, flight speed, roll damping coefficient and lower wing loading and smaller square of the ratio of roll inertia over wing span. Same in reverse: Stopping will be fastest under the same conditions.

For the roll damping coefficient, use this approximation for wings with an aspect ratio AR larger than 4:

$$c_{lp} = -\frac{1}{4}\cdot\frac{\pi\cdot AR}{\sqrt{\frac{AR^2}{4}+4}+2}$$

Since this is a damping coefficient, it makes sense to be negative.

The asymptotic value is reached when the propelling moment from aileron deflection equals the retarding roll damping:

$$c_{l\xi} \cdot \frac{\xi_l - \xi_r}{2} = -c_{lp} \cdot \frac{\omega_x \cdot b}{2\cdot v_\infty} = -c_{lp} \cdot p$$

For an explanation of this equation and all the terms used please see this answer.

Please note that this is all only valid for a stiff airframe. Increasing dynamic pressure reduces aileron effectiveness because the wing will warp when ailerons are deflected. Assume a linear decrease with dynamic pressure until only a fraction of the ideal roll acceleration remains at top speed and low level flight.

Now you ask about roll damping torque and that cannot be explained in the comments. Look at the last equation – it is already there, albeit dimensionless. To get from there to an actual torque, multiply by wing area, semispan and dynamic pressure:

$$T=\frac{\rho}{2}\cdot v^2_\infty\cdot S\cdot\frac{b}{2}\cdot c_{lp}\cdot p = \frac{\rho}{8}\cdot v_\infty\cdot S\cdot b^2\cdot c_{lp}\cdot\omega_x$$

with $\omega_x$ the actual angular speed in rad/s. Do the unit check - this is actually a torsion moment [Nm]. Please note that I used the reference length for lateral moments used in Germany; the US uses the full span instead of the semispan. So make sure you check which reference length your sources use!

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  • $\begingroup$ What you gave me is PERFECT. I knew it existed equations I can fit into my simulation and see what I get. And there is one thing I dont understand, the speed of the plane is involved here but I don't see any angle of attack. Is the AOA considered of 0 for the equation ? $\endgroup$ – Anselme Apr 22 at 19:48
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    $\begingroup$ But actually what I am searching for is the equation of the roll damping torque. Something I could apply constantly and that would change with speed, AOA, etc. Because the roll damping coefficient and the roll constant does not describe what happens regarding those parameters. For example, what happens when the airplane is rolling and suddenly stalls ? With 0 airpseed, how will the roll rate vary ? I guess a plane will roll longer until it stops if put on a rotating axis on the ground, rather than flying 200km/h through the air. I don't know if you see what I mean ? $\endgroup$ – Anselme Apr 22 at 20:05
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    $\begingroup$ @Anselme: Roll damping is independent of AoA. Almost. If the aircraft stalls, things get nonlinear and very individual for different designs. Roll damping can even become positive, so the aircraft accelerates into a roll without aileron command. That's quite scary when it happens at low speed and can easily ruin your day. $\endgroup$ – Peter Kämpf Apr 22 at 20:47
  • $\begingroup$ @PeterKämpf That's a "snap roll", "wing drop" or asymmetric stall, isn't it? If the aircraft stalls while in a sideslip, then the downwind wing may stall before, or more severely than, the upwind wing, producing a sudden roll away from the relative wind. (You know how all this works, of course; I'm asking if my understanding is correct.) $\endgroup$ – Terran Swett Apr 23 at 20:46
  • $\begingroup$ @TerranSwett: Yes, your understanding is correct. $\endgroup$ – Peter Kämpf Apr 23 at 21:03

Your question is an essence a question about aerodynamic damping in the roll axis. As the aircraft rolls, the rolling motion increases the angle-of-attack of the descending wing and decreases the angle-of-attack of the rising wing, eventually causing the lift created by each wing to be equal despite the deflected ailerons. At this point roll torque is zero and the roll rate can no longer increase. If the stick is brought back to center, the roll torque from aerodynamic damping will soon cause the rate will soon drop to zero, or to near zero. (Yes, it is possible for an airplane to tend to roll toward a steeper or shallower bank angle with the stick centered, but the roll rate is much lower than we see with the stick highly deflected to one side.)

Read more about in this section of John S. Denker's excellent "See How It Flies" website.

For the same wing area, roll damping will be greater with a large wingspan than with a small one. This is why airplanes with smaller wingspans generally achieve higher roll rates than aircraft with larger wingspans.

For the same external physical shape, a high moment of inertia in the roll axis (due for example to having the wings full of fuel or loaded with bombs) will cause the roll damping effect to take more time to bring the roll rate to zero (or near zero) after centering the stick.

To get actual formulae, you are probably going to have take a deep dive into some textbooks on flight dynamics or aircraft design.

Rolling also creates an adverse yaw torque, even with ailerons centered, which creates sideslip which interacts with any dihedral or sweep that is present to slow the roll rate. See this section of the "See How It Flies" website. But that's probably a finer-scale effect than you are looking to take into account. (The complex subject of the roll torque generated by the interaction between sideslip and the aircraft's 3-dimensional geometry has been touched on in several different questions on the ASE site.)

You'll find plenty of other content in the "See How It Flies" website that should help you with your project as well.

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    $\begingroup$ Thank you for your help. So if I understand well, the downgoing wing is producing more lift that the upgoing one, causing a torque opposing the rolling motion of the plane ? Is this what is responsible of roll damping ? If thats right then I now know how to simulate roll damping as I already have functions to calculate wing lift with AOA. Can you confirm I understood well ? Thank you $\endgroup$ – Anselme Apr 22 at 16:55
  • $\begingroup$ Yes--sounds like you understand. $\endgroup$ – quiet flyer Apr 22 at 16:56
  • $\begingroup$ Ok, thank you very much for your help, I appreciated it $\endgroup$ – Anselme Apr 22 at 16:58
  • $\begingroup$ @quiet flyer I assume the second "increases" on the second line should be "decrease"? $\endgroup$ – John K Apr 22 at 18:25
  • $\begingroup$ @JohnK -- good catch $\endgroup$ – quiet flyer Apr 22 at 20:46

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