# Why is it, with less airflow, you'll have less control effectiveness?

I understand the principle of less airflow, less control, but why is that the case?

• A control surface deflects air. So the more air deflected, the more...? – Pondlife Aug 26 '17 at 23:19
• How is this not intuitive? With zero airflow, controls just sit there & flap. With a tiny amount of airflow, the control might accomplish a little bit. With lots of airflow, the control has lots of moving molecules to work with. – Ralph J Aug 26 '17 at 23:21
• @RalphJ In practice, "intuitive" is hard to identify or even define (YMMV). I'd say this is a naive question, but not a bad or unclear one. Although, depending on how much detail the OP needs, it might be better on physics. SE. – Pondlife Aug 26 '17 at 23:39

## Because moments of inertia don't change with speed

Control effectiveness means that the controls effect a change in the balance of moments which results in the desired attitude change. The smaller the control deflection for the same change in attitude, the higher their effectiveness. If $\ddot{\Theta}$ is the pitch acceleration, $∆F_H$ the force change on the horizontal tail due to a control deflection, $x$ the lever arm of that control around the center of gravity and $I_y$ the moment of inertia around the lateral axis, the formula for $\ddot{\Theta}$ is: $$\ddot{\Theta} = \frac{∆F_H\cdot x}{I_y}$$

Both $x$ and $I_y$ are fixed, so only $∆F_H$ has the potential to increase pitch acceleration. $∆F_H$ is proportional to

• Deflection angle $\eta_H$
• Tail size $S_H$ (again fixed)
• dynamic pressure $q = \frac{v^2\cdot \rho}{2}$

A given object will change its attitude more quickly when more force can be created. Therefore, more speed $v$ means more force change and a higher angular acceleration for the same deflection.

When deflected, the control surfaces (ailerons, elevator, rudder) cause an aerodynamic moment about the Aerodynamic Centre. A moment has a moment arm and needs to have a length reference - the aerodynamic moments are defined with reference to wing dimensions: wing span for rolling and yawing moments, and Mean Aerodynamic Chord for pitching moments. If we have a look at the pitching moment P:

$$P = C_{r_{\delta e}} \cdot \delta_e \cdot q \cdot S \cdot MAC$$

With:

• $C_{r_{\delta e}}$ = elevator coefficient (dimensionless)
• $\delta_e$ = elevator deflection
• $q$ = dynamic pressure = $\frac {1}{2} \cdot \rho \cdot V^2$
• $A$ = wing area
• MAC = Mean Aerodynamic Chord

$C_{r_{\delta e}}$, A and MAC are constants. So: pitching moment of the aircraft is proportional to elevator deflection, and to the square of the airspeed. Fly twice as fast, and the pitching moment from a certain elevator deflection will be four times as high.

Basically what keeps your plane suspended above the ground despite gravity pulling it to the surface is the fact that your aircraft constantly pushes (and pulls) air molecules downwards; one of Newton's Laws says that this generates an equal and opposing (i.e. upward) force on your aircraft.

In straight and level flight this force is due to the positive angle of attack that the wings make with the relative wind (NOT THE FLIGHT PATH) which essentially forces air molecules downwards: molecules below the wing are deflected dowards along the bottom of the wing while molecules above the wing are pulled downwards along the top surface of the wing as it moves through them. When you go slower you deflect fewer air molecules downwards per unit time which demands a higher angle of attack in order to keep you suspended; this generally translates to more elevator deflection needed on the pilots part, or in other words: your controls are less effective.

• This explanation of lift is incorrect - lift is generated due to Bernoulli's principle and the creation of a low pressure region on the top of the wing due to wing camber. – ANDY-S Aug 27 '17 at 18:56
• Certainly too: high pressure below wing deflects air molecules below downwards, low pressure above the wing "sucks" molecules above downwards. Bernoulli's principle is oft misunderstood by laymen, requiring careful consideration of the Lagrangian (rather than Eulerian) interpretation of streamlines. See also Https://www.grc.nasa.gov/www/k-12/airplane/wrong1.html – David DeVine Aug 27 '17 at 19:08
• @ANDY-S No. This is a fine intuitive explanation, alluding to conservation of momentum. Bernoulli is just a different way of looking at things, and the results are equally valid. Not so your remark on camber - a symmetric airfoil can easily generate lift, and regardless, lift due to camber can also be explained with Bernoulli. – Sanchises Aug 28 '17 at 7:02
• @ANDY-S you may read this section to understand that many principles are to take into account to explain lift generation – Manu H Aug 30 '17 at 15:31
• I fully agree that a wing produces lift using several different methods - including upwash, downwash, conservation of momentum etc. I was attempting to clarify the common misconception that downwash lift is the primary lift generator on a wing. Fully aware that an explanation of what causes lift would be an entire website in and of itself. Also, @Sanchises, yes, camber is critical to lift generation. A symmetric wing produces 0 lift at 0 AOA. It is the introduction of AOA that creates a pseudo-dissimilar shape of the wing – ANDY-S Aug 31 '17 at 1:24

Control authority comes from the size of moments you can generate, which result from forces acting on the plane (the elevator, the ailerons or rudder), which come from pressure differences, which have a squared relation to velocity. If the airflow speed halves, your control authority gets cut in 4. If the airflow speed doubles, you get 4 times the control authority, etc.

Here's further explanation if anything isn't quite clear.

For control authority, you need to be able to apply your desired moment to the aircraft. Moments are forces acting at some distance from your rotation center. In an aircraft, say you want to roll the aircraft. The ailerons deflecting create a pressure difference between the right and left wings. This ends up as different forces acting basically at the ailerons, creating that roll moment. That's just the basics of roll. Now, for the airflow part.

First, I mentioned that for roll, it's those pressure differences caused by airflow over the wing and the aileron. The forces (the ones we're concerned with here) are created by the pressures on a surface. Remember, pressures are forces over areas. Now, lets look at the pressures. The equation for dynamic pressure is $\frac{\rho V^2}{2}$, that's the density times velocity squared over 2. We will assume our density isn't changing here, so in order to change the pressure, we change the velocity of the flow. BUT, its squared. Without airflow, it's obvious that no roll moment is created because the velocity is zero. A plane on the ground with no airflow over the wing doesn't try to roll.

In general, for roll, pitch and yaw authority(that's all of them), you can consider the feeling when you put your hand out of the window in a moving car. If you deflect air downward, your hand gets pushed up. In reality, it's the difference in pressure between the top and bottom, due to flow speeds. The faster you go, the more airflow, the greater the pressure differences you can generate, because of the squared relation. The slower you go, any flow speed differences might become negligible, meaning no pressure difference, thus no force acting.

With some numbers, let's say that at a high speed, the elevator gets deflected. Let's say that the flow over the top is going 100 (arbitrary velocity units), and the flow under is going 110. The pressure on top will be $\frac{\rho}{2}*100^2 = \frac{\rho}{2}*10000$ lets ignore the $\frac{\rho}{2}$ term, and just be aware that it linearly converts our number into a pressure. so we have 10000 pressure somethings on top, and we have 12100 pressure somethings on bottom (using the same formula). That means we have a net of 2100 pressure somethings pushing up on the tail now. Great, the tail has enough control authority to push the nose down as commanded.

Now, lets slow the speeds down by a factor of ten. The top air is going 10, and the bottom now goes 11. Lets see the pressure change compared to before. The pressure on top will be 100 pressure somethings, and on bottom it will be 121. The resulting net pressure acting on the tail is then 21 pressure units, 100 times less than before, even though the speeds only changed by a factor of ten. now, you have 100 times less force acting on the tail (resulting in equivalently less moment), and you might not be able to control the pitch as much as you want to.

Control surfaces are used to change the effective camber of the airfoil they are controlling. For example, a downward deflected aileron would increase the effective camber of a wing along the aileron's span. An increase in camber will increase the lift generated at a certain airspeed over that area of the wing, causing the desired rolling moment. It is PARTIALLY due to this change in developed lift that generates adverse yaw, requiring rudder to coordinate turns.

At higher airspeeds, the wing is producing more total lift, and therefore more responsive to changes in camber.

Additionally, control surfaces also respond according to Newton's 3rd law - the ailerons deflect the passing airflow in a direction other than parallel to the wing skin, resulting in a reactive force causing roll. As with the camber change, this phenomenon becomes more pronounced at increased airspeeds, and conversely less pronounced with a reduction in airflow.

A simplified explanation can be found at FAA Pilot's Handbook

• I think OP wants to know the WHY behind the theoretical equations rather than simply THAT "Hey look when V increases in this equation it must be true that L also increases, c.p." – David DeVine Aug 27 '17 at 19:31

This can be explained by Newton's second law, $F = m\times a$ and third law, every force has an equal force to the opposite direction.

$m$ here is the mass of airflow, $a$ is the acceleration caused to airflow (seen as changed direction of the airflow). A force equal to $a\times m$ is exerted to the control surface. More airflow, more mass, more force.

The very same reason why an airplane stays in the air in the first place.

• You should enhance your answer by expanding some sentences. One important feature is the dynamic pressure. You may include it in your answer. You may also explicit your last sentence. – Manu H Aug 30 '17 at 15:36