My understanding of trim: When you say you have trimmed an aircraft for a constant speed, say 100 mph, you are actually trimming the horizontal stabilizer so that there are no forces on stick (or tail down force is balanced with nose weight) at that speed. To do that you either control the horizontal stabilizer/trim tabs until no force is felt on the stick.

My question is, (if my understanding is right), can I say that we are trimming for constant angle of attack instead of speed (and the constant angle of attack is represented in form of speed). By trimming the aircraft, aren't we making the aircraft fly at constant angle of attack?

If so, can I make these following statements about trim: (ignoring the things like propeller blast etc..,)

  1. When you are cruising in a trimmed condition and decrease( or increase) the engine power, the aircraft will pitch down (or up) until it finds the old angle of attack where all the forces are balanced and the trim is reestablished.

  2. When you are cruising and closed (or opened) the throttle completely, the aircraft will descent (or climb) at the same trimmed angle of attack.

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  • $\begingroup$ For a 1g situation I would generally agree that you are holding a constant angle of attack which is evidenced by a constant indicated airspeed. However, for maneuvers with a grater than 1g this doesn't hold true. We know the airplane stalls from exceeding the critical angle of attack and the airplane can stall at any pitch attitude and airspeed. $\endgroup$ – wbeard52 May 5 '16 at 2:29

Short answer: Yes

Generally you are right. You are trimming the aircraft for a point on its polar, and that point is reached at a specific angle of attack.

You say yourself we should neglect secondary effects like the propeller blast, and under this condition you are right. Trimming means to set the distribution of lift between wing and tail surface, and this directly sets the angle of attack.

Now the more elaborate part of the answer

You trim the aircraft for a specific dynamic pressure, that is the product of air density and flight speed. If you slightly change your question from "trimming for constant speed" to "trimming for constant indicated speed", you gain precision and the answer is still yes. With a change in altitude you will also affect the Reynolds number the aircraft flies at, and the same polar point is reached at a slightly higher angle of attack the lower the Reynolds number gets when the aircraft climbs into colder air.

In turns the lift demand is increased. Not changing trim during a turn will let the aircraft speed up to increase lift at an unchanged angle of attack. That is why you pull more when turning - you want to increase the angle of attack, so speed can stay constant. When speeding up, you again change the Reynolds number and indirectly also the angle of attack, but this effect is very small.

When you change the power setting, you add energy to the aircraft at a different rate. Thrust becomes bigger and is counteracted by the sum of drag and the component of weight which acts in parallel to the thrust vector. Now the vertical component of weight is reduced, and so is your need for lift - but ever so slightly. This effect by itself means you will fly at a slightly smaller speed which will also very slightly affect the angle of attack. But since you did not change trim, the distribution of lift between wing and tail is unchanged, so the angle of attack is the same, save for a minuscule Reynolds number effect.

Increasing thrust will also increase the prop blast over the tail surfaces, making them more effective and slightly changing the balance of the aircraft. Depending on the direction of tail lift, the result is an increase or a decrease in angle of attack, but this is precisely the effect you wanted us to disregard.

Another secondary effect is a constant loss of weight as the engine burns fuel. This makes the aircraft lighter, so it requires less lift and a smaller angle of attack over time to remain at the same speed. Since you trimmed a specific polar point, the aircraft would need to slow down in order to stay at this polar point at its lower weight. But the lower weight also creates less drag, so instead of slowing down the aircraft will climb. Now the reduced density will require more angle of attack and will reduce engine output so in the end the same angle of attack is maintained at a higher altitude but a slightly lower indicated airspeed.

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  • $\begingroup$ And the weight of the aircraft changes in time, so instantaneously the proposed equality holds, but in some time the aoa will decrease (due to consumed fuel). ;) $\endgroup$ – Gürkan Çetin May 4 '16 at 20:28
  • $\begingroup$ @GürkanÇetin: No, weight loss will reduce trim speed. AoA stays the same (again, save for small Reynolds number effects). $\endgroup$ – Peter Kämpf May 4 '16 at 20:55
  • $\begingroup$ What I mean is if you keep constant airspeed (suppose no climb) aoa would decrease. (Would be noticeable for a slow flying aircraft). The question was trimming for speed, so I took this side. If u keep aoa, then speed would decrease, you're right. $\endgroup$ – Gürkan Çetin May 4 '16 at 20:59
  • $\begingroup$ @GürkanÇetin In order to keep constant speed with reduced weight you need to re-trim the aircraft. This must be commanded by the pilot. If the pilot sets trim once and then flies horizontally for a while, speed goes down with the square root of the weight loss. It's the inverse case to the speed increase in a turn. $\endgroup$ – Peter Kämpf May 4 '16 at 21:09
  • $\begingroup$ Gotcha! I see what the source of my confusion was. Descent doesn't mean that net lift and weight aren't still equal when there is a 0 net acceleration. Answer removed. Thanks for clarification. $\endgroup$ – Ryan Mortensen May 4 '16 at 23:42

This is an interesting question. I believe that both of your statements are wrong though.

"Trimming for constant speed" is technically not possible, as the trim itself has nothing to do with the speed. As you probably know, you have to adjust both throttle and trim to have the plane fly level at a constant speed. Trim only controls your elevator (neglecting the other two trims that larger aircraft usually have as well). When you "trim for constant speed", you usually adjust your power setting while keeping the plane flying at a certain altitude. Then (or even while you adjust your speed) you trim the plane to take work of your arms ;)

Once you're flying at a constant speed, you can say that the angle of attack remains the same when you're not descending/ascending. No comes the tricky part though:

  1. You're saying that the aircraft will find a balanced condition again - which is right. Yet you have decreased the power, thus you've changed one force. Because of that, the other forces must change as well to balance each other out. The angle of attack will thus NOT be the same when the throttle is different.
  2. Again, when closing the throttle, the aircraft will find a new balanced condition, and again, the angle of attack will change.

Another interesting question is: When you are flying at a trimmed condition and decrease the throttle, then after a while increase it back to the original setting, will the aircraft return to level flight? - The answer is yes, but no ;) Yes, because now all forces are back to the original condition, thus the airplane must have the same attitude. No, because by descending, the air density will change, thus resistance of the air becomes bigger which enforces a bigger force or resistance on the plane messing up the balance again.

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  • $\begingroup$ I believe "you have to adjust both throttle and trim to have the plane fly level at a constant speed" should really be "at a given speed". You can do a levelled fly at any reasonable power setting, and the speed will be constant (in time). I just think this can be a source of confusion. $\endgroup$ – yo' May 4 '16 at 20:03
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    $\begingroup$ One important point is that you have decided to constrain your answer to level flight, which changes a lot (especially since the OP even refers to pitch and altitude changes in his question). Because of this, your answer doesn't really accurately answer the question. $\endgroup$ – Lnafziger May 4 '16 at 21:44

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