# Why does elevator trim hold a given airspeed? [duplicate]

I'm trying to understand a principle described in the book Stick and Rudder by Wolfgang Langewiesche between angle of attack and airspeed in relation to the elevator trim. He states that as angle of attack increases, speed decreases (given a constant power setting), and vice versa. In this section of the book, he's talking about this in relation to the use of the elevator trim for maintaining a specified airspeed. The book is focused on giving pilots practical explanations of how to fly, but is a little lacking and outdated on some scientific explanations as it was written in 1944.

My question is, what is a more scientific and aerodynamic explanation for why the change of AOA of the elevator (trim on elevator) determines the speed of the aircraft in a stabilized condition? What is it about the different camber of the airfoil that affects how fast the aircraft will fly?

I understand the practical use, I just want to understand why it works.

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– Ralph J
Feb 26 at 19:00
• Quite a few links to highly related questions/ answers were removed by the move. Go to the chat (chat.stackexchange.com/rooms/143195/… ) to see those linked related questions/ answers. Feb 26 at 20:08

The math and physics theory behind the phrase "pitch for speed, throttle for rate of climb" is what you're after. First, let's talk about what it means.

In a video game or in your first intuition, it may make sense to 'nose up go up, nose down go down'. Likewise, your intuition and simple video game may behave by 'more throttle is faster, less for slower'. But that isn't how airplanes really fly in equilibrium (steady) states.

"Pitch for speed" means that controlling pitch (usually via trim) is the pilot's primary control for airspeed.

"Throttle for rate of climb" means that changing the throttle is the primary pilot's control for rate of climb.

In level equilibrium (steady, level flight, the state we spend most of our time in), Lift is equal to Weight, and Thrust is equal to Drag.

In a steady climb, we have (where $$\theta$$ is the climb angle and $$RoC$$ is the rate of climb).

$$L=W \cos (\theta)$$

$$RoC = V \sin (\theta) = (T-D)V/W$$

We can derive these equations if you'd like, but we can answer your question with just the steady-level flight version (i.e. $$\theta$$=0) $$L=W$$. The rest is needed to explain 'throttle for rate of climb'. Which is likely to be your next question.

Next, we write the lift equation in coefficient form.

$$C_L = \frac{W \cos (\theta)}{q\,S}$$

Where $$q=0.5 \rho V^2$$, the dynamic pressure. And $$S$$ is the wing reference area.

Let me know if you don't know where the definition of lift coefficient comes from. For now I'm assuming that you already understand that.

We solve that equation to put $$V$$ on the LHS -- just to emphasize it.

$$V = \sqrt{ \frac{2\, W \cos (\theta)}{C_L\, \rho\, S}}$$

Of course, starting with the level flight version...

$$V = \sqrt{ \frac{2\, W}{C_L\, \rho\, S}}$$

This equation was derived based on 1) Summing forces and setting them equal to zero (Newton's laws) and 2) The definition of lift coefficient (how aerodynamic forces change with density, size, and airspeed). That is it. This equation must be true (in steady-level flight) unless you find fault with Newton's laws or the definition of lift coefficient.

How does this answer our question? How does this lead to "pitch for speed"?

You want to change (control) the aircraft's equilibrium airspeed -- $$V$$. To do that, while keeping this equation true, you must change something on the right-hand-side to make it true. What on the RHS does the pilot have control over? Let us consider our options.

$$2$$ - The pilot is unable to change the value of two.

$$W$$ - The pilot has some control over the weight of the aircraft at takeoff. And during flight, fuel burns reducing the weight slowly -- but not adequately to 'control' airspeed. The pilot could throw payload overboard (reducing speed), but there is no practical way to gain weight during flight. Weight is an unsatisfactory way for the pilot to control airspeed.

$$S$$ - Some high lift devices can be considered to change the wing area (though engineers would likely leave the reference area unchanged) -- and a morphing wing might be able to change wing area in flight. So someday, wing area may be a control for airspeed. However, for most conventional aircraft today, the pilot can not change the wing reference area in flight.

$$\rho$$ - The air density changes with altitude and with temperature (hot or cold day conditions), but a pilot that wants to fly at a given altitude (say 10,000 ft.) does not have the ability to change the density of air.

That leaves us with $$C_L$$, the lift coefficient.

$$C_L$$ - The lift coefficient is controlled by the pilot by either the elevator, or the elevator trim. Since we are talking about equilibrium, let's focus on the trim change. Changing the trim changes the lift coefficient (and angle of attack) where the pitching moment on the aircraft is zero. (If you don't understand the details of trim, let us know, we can explain that too.)

By decreasing $$C_L$$ (nose down trim, decrease pitch), we increase speed.

By increasing $$C_L$$ (nose up trim, increase pitch), we decrease speed.

There you have it, "Pitch for Speed".

We can do the same thought experiment with the rate of climb equation, but it is already in the form we need...

$$RoC = (T-D)V/W$$

At a given moment in flight, if the pilot wants to change rate of climb, what can they do to the RHS of the equation to make it happen?

The only satisfactory control variable is Thrust -- which is controlled by throttle.

"Throttle for Rate of Climb".

One fun thing about this is to imagine that instead of a pilot operating a given aircraft, you are an aircraft designer. The equations are the same, but your perspective on them changes...

$$V = \sqrt{ \frac{2\, W}{C_L\, \rho\, S}}$$

Now perhaps you want the aircraft to fly efficiently at a given flight condition ($$V$$, $$\rho$$).

Perhaps by 'efficiently', you mean that you want to fly at the best lift-to-drag ratio. This occurs at a specific value of lift coefficient, we'll call it $${C_L}^*$$.

That leaves the aircraft designer with control over the wing loading $$W/S$$ -- one of the most important aircraft design parameters. It is the scaled version of "how big is the wing".

Similarly, the other main aircraft design parameter is "how big is the engine". The scaled version of that is the thrust to weight ratio "T/W". If you look back at the rate of climb equation and imagine yourself an aircraft designer, I think you'll see the importance of $$T/W$$.

Let us know if you need further explanation of any of the steps I skipped.

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– Ralph J
Feb 26 at 18:59

My question is, what is a more scientific and aerodynamic explanation for why the change of AOA of the elevator (trim on elevator) determines the speed of the aircraft in a stabilized condition? What is it about the different camber of the airfoil that affects how fast the aircraft will fly?

This answer will tackle the problem from the view point of stick-fixed stability, i.e. assuming that the elevator trim wheel position completely determines the actual position of the elevator (or the all-moving horizontal tail). That's admittedly a bit of an oversimplification.

For the aircraft to be in equilibrium, net pitch torque must be zero. For a given angle-of-incidence of an all-moving horizontal tail, or a given elevator position of a "conventional" horizontal tail, the downlift or uplift created by the tail, and the lift created by the wing, can only create equal and opposite pitching moments around the C.G. when the wing is at one particular angle-of-attack. At any other angle-of-attack, the pitch torques from the wing and tail won't be in balance, and the aircraft will pitch up or down. If the aircraft has been designed to have good stick-fixed stability, this pitching effect will bring the wing to the angle-of-attack where the pitch torque from the wing exactly balances the pitch torque from the tail.1 So the position of the elevator trim determines the position of the elevator, and the position of the elevator determines the angle-of-attack of the wing.

By setting the wing's angle-of-attack, we are also setting the airspeed. The reason for this is that for any given wing, lift is determined by angle-of-attack and by the square of airspeed, and total lift must equal weight, or else the flight path will curve up or down. Any upward or downward curvature of the flight path will make the weight vector become partly parallel to the flight path, so that it acts like a thrust vector or a drag vector, driving a change in airspeed.2

Actually, the statement that "total lift must equal weight" is only true for wings-level, horizontal flight. For a given angle-of-attack, banking actually causes the airspeed associated with any given angle-of-attack of the wing (and therefore with any given elevator position) to increase3, because lift must be greater than weight. And in a wings-level climb or descent, there's actually a very slight decrease4 in the airspeed associated with any given angle-of-attack of the wing (and therefore with any given elevator position), because the total lift vector must equal weight times the cosine of the climb angle or descent angle.5

For more detail, see various answers to the related ASE question Does “trimmed for current speed” mean that the plane will do whatever to maintain the current airspeed? , especially this one.

Footnotes:

1. An excellent resource for better understanding the "balancing act" between the tail and the wing, and why a given elevator position will only "balance" the wing at one particular angle-of-attack of the wing, is John S. Denker's excellent "See How It Flies" website.

2. When the airspeed, and therefore the lift vector, are "too high" or "too low" for the angle-of-attack of the wing, the complete dynamics by which the airspeed changes and eventually settles down to the "correct" value for the angle-of-attack are beyond the scope of this answer. Typically--in the absence of some "help" from the pilot or autopilot-- a series of slowly damping pitch "phugoid" oscillations are involved, with the flight path alternately curving upward and downward until the aircraft settles into a new flight path, with the climb or descent angle "correct" for the power setting, and the airspeed "correct" for the angle-of-attack. To a first approximation, angle-of-attack can be viewed as remaining constant throughout the pitch phugoid, but the airspeed (and therefore the lift vector) is constantly "out synch" with the value that would be required to maintain linear flight at any given instant. Also the direction of the weight vector relative to the direction of the flight path is constantly changing, so the thrust-like or drag-like effect from the weight vector is constantly changing.

3. To a first approximation, the increase in trim airspeed associated with banking is in proportion to the square root of the increase in wing loading--so for shallow bank angles, it will be minor. Actually, the increase in trim airspeed is somewhat greater than given by this formula, because the curving nature of the relative wind in turning flight "bends" the relative wind upward at the tail, so that the tail's angle-of-attack relative to the local relative wind is more positive or less negative than it would be in linear flight, which shifts the pitch torque generated by the tail in the nose-down direction and causes the wing to trim to a slightly lower angle-of-attack. (This can also be understood as a "pitch damping" phenomenon, i.e. an aerodynamic resistance to pitch rotation.) This trim change associated with turning flight is covered to some extent in the resource linked in footnote 1. That's why glider pilots often hold a lot of aft stick input when circling in a thermal updraft-- even if the glider is trimmed to fly at the min. sink angle-of-attack when wings-level.

4. Re "very slight" -- the effect is "very slight" for typical light general aviation airplanes, because typical climb or glide angles are so shallow, and may well be completely over-shadowed by other pitch trim effects relating to the thrust line. In practice, reducing power to enter a descent may well cause a slight increase in airspeed. However, in a very-high-performance airplane with a thrust-to-weight-ratio exceeding one-to-one, we certainly wouldn't expect the same relationship between control stick position and airspeed in horizontal flight at full power, as in a steady-state sustained vertical climb at full power! See footnote 5 for more.

5. For more see Does lift equal weight in a climb? and other answers to related ASE question "Does lift equal weight in a climb?" Also, it may seem extremely counterintuitive to imagine that angling the flight path down into a dive would ever decrease the airspeed--after all, the weight vector is gaining a (thrust-like) forward component, relative to the flight path. However, keep in mind that in the context of this question, which assumes a given position of the elevator trim, i.e. a given position of the elevator, and therefore a given angle-of-attack of the wing, we are causing the flight path to descend by reducing power, or by lowering the landing gear, or by deploying some other sort of speed brake (or drogue chute?) that doesn't affect the wing's lift coefficient. We're not talking about simply moving the control stick or yoke forward to reduce the angle-of-attack of the wing.

• Thank you so much for this answer, I honestly found it (and all of the related articles you shared) to be the most useful in bridging the gap between my previous understanding to the more complex and mathematical explanation in the answer I chose. I had a hard time choosing between your answer and his but eventually settled on his as it explains it to an even greater detail. But I really appreciate the help you provided in giving me the steps to understanding the topic. Much appreciated and have a great day! Feb 25 at 23:18

In my other answer, I assumed you understood how trim works. Perhaps that is incorrect, so let's talk about pitch trim.

The airplane lift coefficient is modeled as:

$$C_L=C_{L,0}+C_{L,\alpha}\,\alpha+C_{L,\delta}\,\delta$$

Where $$\delta$$ is the control deflection. And $$C_{L,\alpha}$$ is the derivative of the lift coefficient with respect to $$\alpha$$, and the other control derivatives are similar. $$C_{L,0}$$ is the lift coefficient at zero alpha and zero elevator deflection.

And the pitching moment coefficient is similarly modeled as:

$$C_M=C_{M,0}+C_{M,\alpha}\,\alpha+C_{M,\delta}\,\delta$$

The trim problem is solving the above two equations and two unknowns simultaneously. We want to find $$C_L$$ equal to some desired target value -- $${C_L}^*$$, and CM equal to zero. We can do this manually, but we usually set up as a matrix problem...

$$C_L={C_L}^*$$

$$C_M=0$$

$$\begin{bmatrix}C_{L,\alpha}&C_{L,\delta}\\C_{M,\alpha}&C_{M,\delta}\end{bmatrix}\begin{bmatrix}\alpha\\\delta\end{bmatrix}=\begin{bmatrix}{C_L}^*-C_{L,0}\\-C_{M,0}\end{bmatrix}$$

This is a static trim question -- we're looking at the steady equilibrium value of angle of attack and control deflection required to achieve any desired lift coefficient at zero moment. It ignores the oscillations and time dynamics that occur when you change from point to point.

When the pilot changes the control deflection (or the trim tab position), they move the airplane to a new trim solution -- i.e. a different angle of attack where moment is zero. There is a new $$C_L$$ that corresponds to that new trim position.

In this set of equations, the only thing the pilot had direct control of is the elevator angle $$\delta$$. So, by controlling $$\delta$$, they control $$C_L$$, and per my other answer, by controlling $$C_L$$, they control speed.

• After reading all three of the answers and provided sources and links, I feel like I finally have a deep understanding of the greater picture of the relationship of AoA, trim, and airspeed. Thank you for the mathematical explanation and time spent on spelling it out for me. It is much appreciated! I also apologize for my delayed response, been a busy day. Thanks again and have a great day! Feb 25 at 23:24
• @RattMuscle -- the concept of ASE is that it's fine to take time craft a good answer, and similarly, a rapid response in terms of upvoting etc is not necessarily expected. Comments are considered entirely peripheral and a "thank you" is not expected, much less a rapid one, so no need for apologies of any kind. But I'm sure all appreciated your kind words all the same-- Feb 26 at 15:13