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I have been struggling with the following question for a while, would greatly appreciate some help and explanation why.

Scenario:

  • I'm cruising in my Cessna- cruise attitude 2300 RPM trimmed for 95.

  • I pitch up, say 1 inch of back pressure, and hold it.

Initially the plane starts climbing and slows down, but what happens next?

How will the plane look in say 5 minutes with me still holding that 1 inch pitch up, what attitude, speed, and altitude will it have?

quiet flyer
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leha007
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  • Are you talking about one inch of back pressure as measured by the back pressure to get that displacement on the artificial horizon or the back pressure to initially get one inch of backward movement of the yoke? – Terry Jul 22 '20 at 06:17
  • It may interest you to read the well-known aviation book "Stick and Rudder". Chapter 1 deals with the relationship between attitude, speed, weight, and angle-of-attack. – Jamiec Jul 22 '20 at 11:00
  • You will fly more slowly and climb, but the amount of change depends on the static stability of the plane. Without knowing that, I cannot be more precise. – Peter Kämpf Jul 22 '20 at 11:35
  • Terry- I think he means just what effect does pitch have on the 4 forces. To answer your question I think he means just any elevator deflection relative to the trimmed cruise position, just to pull the control column 1 inch back relative to cruise meaning the slightest deflection of the elevator upwards. – leha007 Jul 22 '20 at 14:49
  • My attitude improves with pitch and speed... – Ron Beyer Jul 22 '20 at 16:39
  • An inch is not a measurement of pressure. Do you mean you move the yoke one inch aft of its trim position under the existing circumstances? If so please modify question to say so. – quiet flyer Jul 23 '20 at 17:25
  • PS why not try it and see? – quiet flyer Jul 23 '20 at 17:46
  • PPS after those 5 minutes pass, are you still holding the same back PRESSURE as you initially applied, as best as you can? Or are you still holding the yoke exactly 1 inch aft of its original trim POSITION? If the latter, how do you know? Did you mark the position with a grease pencil or something, on the stem that protrudes from the instrument panel? Did you wrap (and tape) a paper around the stem so that you could make pencil marks? Did you tighten a sheet-metal-clamp vice grips (w/ padded jaws of course) on the stem at trim to show the place where it protruded from the panel? – quiet flyer Jul 23 '20 at 17:46
  • I've done all those things and they are good ways to explore the effects of change in the POSITION of the control yoke. without regard to PRESSURE. Also, you can simply tighten your fingers tightly around the stem where it protrudes from the panel to mark the spot, and try not to let them slip. – quiet flyer Jul 23 '20 at 17:52
  • Or did you mean, you continually apply the same force or PRESSURE (as best as you can) that INITIALLY moved the yoke one inch aft of its trimmed position, without any regard to its CURRENT position? That would be a whole different scenario. So, the question needs clarification before it can be answered The question could lead to a rather interesting set of answers but it needs to be cleaned up a little first so that it is more clear what the actual question is – quiet flyer Jul 23 '20 at 17:52
  • The point being, as airspeed changes, so does the trimmed position of the control yoke, and so does the force or pressure needed to hold the yoke a given distance away from its current-- or original-- trimmed position. So your question is juggling a lot of variables, without making it clear which are the ones that you intend to hold constant. – quiet flyer Jul 23 '20 at 17:53

3 Answers3

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Well, you knew to be specific about the plane, the RPM, the airspeed, and the attitude.

Pulling the yoke back without changing anything else will create excess lift, and a climb begins. This is the only point in the scenario you have excess lift because: increasing AOA increases drag.

The plane then slows down for 2 reasons. First a higher AOA means more drag. Secondly, as the plane pitches up, the gravity vector increasingly contributes to the drag vector. (A descending aircraft, nose "down", allows gravity to contribute to thrust).

Specifically for a 172, there are 2 outcomes. In this case a steady climb is likely at a lower airspeed. If the yoke is pulled harder, the plane continues to pitch up and stall (with more power and/or airspeed it may loop, but I would not bet on it at 90 knots, 75% power from level flight in a 172).

But 5 minutes later? As you climb, the engine produces less and less thrust. Assuming static stability (a hallmark of a properly CG balanced 172, checked pre-flight), the rate of climb will decrease until the plane has insufficient thrust to continue climbing. It will then fly level at its trimmed airspeed. (If you have sufficient oxygen and no head wind, your ground speed will be higher).

But this provides an insight into what it is doing when it climbs in the first place. Since the wing is around 4x more efficient than the prop in producing lift, pitching up and "using the prop to climb" does not work very well. It turns out the mechanism of static stability works as well in a climb as anywhere else. As the plane pitches up and slows, enough lift is lost to cause the plane to sink (even though the nose is skyward). Sinking pitches the nose down, which increases speed, which increases lift. this works at any power setting, from gliding to full power..

We can see, "5 minutes" later, the plane no longer can climb, but still will be staticly stable at its trim speed so... cut the power, and enjoy the glide home at that airspeed (if you can make it), or, at least remember CARB HEAT ON.

And also, consult the POH for that flight envelope chart, and remember, excessive pulling the yoke can exceed G limits as well.

Robert DiGiovanni
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  • Thanks guys. So what you are saying is, from our cruise at 95 IAS say 2000 altitude we will enter a climb, our speed will decrease, and we will slowly climb until say at 10,000 feet finally power required curve meets the power available curve, our plane stops climbing (at say the new speed of 80/70), And speed will increase back to the cruise 95? – leha007 Jul 23 '20 at 00:52
  • Sorry Robert, not sure Im following you. I understand it will climb and when it reaches power required=power available the climb will stop. But what would the new cruise IAS be? You are saying it will cruise at its Trimmed airspeed. Wasn’t that trimmed airspeed 95? – leha007 Jul 23 '20 at 00:55
  • No, because you changed trim by pulling the yoke back. – Robert DiGiovanni Jul 23 '20 at 01:02
  • Ohh, I think Im starting to get it. So our new cruise altitude will be higher, speed will be lower: Thank you. – leha007 Jul 23 '20 at 01:07
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    @leha007 well, you're getting comments anyways. Trim speed will not change, rate of climb does. If you go to max altitude, then let go of the yoke, your trim speed returns to 95, and your plane will return to its original altitude, because it has insufficient thrust to do 95 indicated at that altitude. – Robert DiGiovanni Jul 23 '20 at 01:29
  • I would guess that it would take much more than 5 minutes for the plane to stop climbing under the circumstances described in the question, assuming that we are starting the experiment not too high above sea level. – quiet flyer Aug 15 '23 at 01:59
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The general answer goes something like this:

Level flight, at it's most basic involves a balance of 3 factors:

  1. How fast are you going, relative to the wind?
  2. What is the angle of attack (AOA) of your wings, relative to the wind?
  3. How dense is the air that you are flying through, based on altitude, temperature, humidity, etc.?

Speed and AOA are variables that you indirectly control, via throttle (by increasing thrust) and yoke (by using elevator to alter pitch) respectively. The thing is, changing one variable (speed, AOA, or density) requires that one or both of the others change to maintain equilibrium. Want to stay in level cruise at a given density altitude after lowering throttle? AOA must increase. If the AOA stays the same, the aircraft will begin to sink because the current density cannot provide enough lift at that combination of AOA and speed.

That explains some of the "Why". Now let's consider your scenario. You have taken throttle out of the equation. Thrust is fixed and cannot be changed. This means that for any change to AOA, density must provide the equilibrium. Unless you are already at your stall speed, a slight increase in AOA without lowering throttle will begin a climb at slower forward speed because this new combination of AOA and speed are producing excess lift at the current air density. You have traded forward speed for extra lifting force. Because the throttle and new pitch are now fixed, the airplane will simply climb at it's new speed and AOA until altitude and temperature lower the outside air density to the new equilibrium point. At this point, the aircraft will once again be configured for level flight at the new density altitude. Also, because the density will change gradually, so too will the rate of climb decrease gradually. It will probably take a good while to fully level out.

ADDENDUM:

The "lift" topic always sparks arguments, because technically, lift can be generated in any direction, depending on the definition of lift you are considering. An aircraft moving through the air on its side, may still create some small amount of force laterally, which opposes gravity, even though the wings may be producing no force at all. A helicopter in hover is utilizing all of its accelerated airflow for to counteract gravity, but a change in rotor disc tilt divides that total force between lift (opposing gravity) and thrust (opposing drag). An aerobatic airplane hanging on its prop is utilizing accelerated airflow in much the same way. An aircraft flying upside down is creating lift that is simultaneously against the pull of gravity AND toward the bottom of the aircraft.

Additionally, Bernoulli's principles work equally well in any orientation. An asymmetrically cambered airfoil mounted vertically will still create an area of lower pressure on one side than on the other, and is still capable of altering angle of attack to adjust deflection. Is this lift? The simple physics definition would say "No" because it is not opposing gravity. An aerodynamic analysis of the airfoil would say "Yes", the airfoil is creating lift according to its orientation, which in this case, affects only yaw.

It would be nice if the entire world could agree that Lift is the sum of all forces that act in opposition to gravity, and Thrust is the sum of all forces that counteract drag - regardless of whether these forces are provided by airfoil or prop (which is still an array of airfoils!). Unfortunately, for now, clarification is often required when discussing this problem in polite company.

In closing, I would like you to consider this: in your scenario, the reason that your airplane slows down is because, much like the helicopter, your prop disc is being tilted, dividing its power between lifting and thrusting. The drag on the airplane remains, so the added burden of lift is what leeches away the speed. The wings, on the other hand, no longer bear the full burden of opposing gravity, so the lift that they are producing DOES become less, but only because they are now sharing that load with the prop.

Aaron Holmes
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  • Thanks guys. So what you are saying is, from our cruise at 95 IAS say 2000 altitude we will enter a climb, our speed will decrease, and we will slowly climb until say at 10,000 feet finally power required curve meets the power available curve, our plane stops climbing (at say the new speed of 80/70), And speed will increase back to the cruise 95? – leha007 Jul 23 '20 at 00:52
  • Yes, you will climb. Yes, your speed will decrease. 10,000 feet is speculative - it depends on factors that affect the air density and how much trim you have already applied. I doubt that you will climb 8,000 feet from one inch of aft yoke (especially before running out of gas). Yes, when the curves are in equilibrium you will once agin be in level flight. – Aaron Holmes Jul 23 '20 at 01:04
  • No, you will not increase speed once you get back to level flight. This will be the new density altitude that supports level flight for this new AOA/Speed combination - the air is thinner, so less thrust, so less excess power to make you climb, right up until you are level and can maintain the new, slower speed (80 or 70 kts, whatever it is). – Aaron Holmes Jul 23 '20 at 01:04
  • Got it! Thank you. – leha007 Jul 23 '20 at 01:10
  • And another one to see if I understand, say from cruise i apply back pressure, but this time don’t lock control column and release it. Will the plane climb up, then down then up then down in a series of decreasing oscillations until it is back at the same altitude at 95? – leha007 Jul 23 '20 at 01:12
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    @Aaron Holmes thanks, your explanation re-inforces a concept that is in the process of being improved on this site: lift is greater than weight in a climb. A great many people here thought it was less. At any power/trim setting, a staticly stable plane will pitch up (if too fast) or pitch down (if too slow). The throttle determines angle to the horizon and vertical lift determines rise or sink. As a 172 pitches up, the vertical thrust component is insufficient to replace the vertical lift (1:4 ratio), and the plane climb angle equilibrates based on *thrust*. – Robert DiGiovanni Jul 23 '20 at 07:26
  • @RobertDiGiovanni -- Re " lift is greater than weight in a climb. "-- nope. Not in a stabilized steady-state climb. In such a case, lift is less than weight. See for example https://aviation.stackexchange.com/a/56476/34686. – quiet flyer Jul 23 '20 at 17:44
  • @quiet flyer: See the top rated answer on the link you posted. In this instance, we are defining lift in the way that a physicist would, as "the sum total of all forces acting against gravity", instead of "the sum total of all forces acting perpendicularly toward the top of the aircraft". – Aaron Holmes Jul 23 '20 at 18:40
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    This one will die hard, but it must. First vertical lift is cosine lift, there for the lift must be greater than weight if the plane is pitched up or down. Remember, using the wing is always more efficient than thrust or drag! Second, any vertical thrust contribution is only 1/4 (for a 172) the vertical lift lost from pitch. In a "stabilized steady state" all upwards forces = weight, essentially rendering the aircraft "weightless" and thrust acts against drag. So happy with Aaron, because Rho got in there too. Rho is also part of the Lift equation, as is increased AOA. – Robert DiGiovanni Jul 23 '20 at 18:55
  • @Robert, how did you arrive at your 1/4 figure? If you are talking about "vertical thrust contribution", then it is not directly comparable to lift and highly depends on pitch. Say, we were level (pitch = 0), 0% thrust component, 100% lift. Pitch up 5°: now the lift loss $1-\cos(5°) \approx 0.4$%, thrust gain $\sin(5°) \approx 8.7$%. If we account for L/D ≈ 10, then there is still 0.87% gain in absolute numbers vs 0.4% loss. – Zeus Jul 24 '20 at 00:54
  • @Zeus at 95 knots 1/4 thrust to lift ratio would actually be a bit generous, and (for the Clarke Y) L/D is much greater than 10 (at optimal AOA). But I very much like your effort to plug in numbers rather than to argue passionately. I work backwards from reality, both Vy and Vbg are faster than V min sink because...? Vy performance is much better than Vx because...? In the end "the vertical thrust component" is moot for best performance (that you need at maximum altitude). Thrust only opposes drag (in the direction the plane flies). The wing does the lifting. – Robert DiGiovanni Jul 24 '20 at 04:35
  • So, the best way to climb is to set AOA at optimal and speed up slightly (Vy). Increasing altitude brought this home so well, the higher you go, the less you can pitch up (because you can no longer go faster). The thrust vs airspeed curve for the 172s fixed pitch prop makes it a bit more confusing (and the much slower Vx even possible). – Robert DiGiovanni Jul 24 '20 at 04:57
  • @Robert, it's fairly obvious that wing is the most efficient aerodynamic device. But still, I can't see how you tie it to the prop efficiency. The literal interpretation of your "any vertical thrust contribution is only 1/4 the vertical lift lost from pitch" is the calculation I've done, only with a different result. C172's best L/D is known to be just under 11, so at 95 kt it's probably around 8, and the lower it is, the greater the thrust must be, as well as its vertical contribution. – Zeus Jul 24 '20 at 07:49
  • Well, happy to continue working on it. A 2400 lb Cessna flies fine with 600 lbs of thrust, but cannot climb on the prop (it falls short 1800 lbs!). The best way is to use the wing. In this case, increasing AOA at the same thrust setting causes a climb. An airspeed reduction from 95 to 70-80 (see drag curve) gives the plane (less drag aerodynamicly and more thrust) the ability to climb, and it does until Rho factor no longer allows it. Now, if you were on the other end of the power curve, like Langewische, pulling the elevator would cause a higher drag "mushy glide". – Robert DiGiovanni Jul 24 '20 at 12:02
  • But the "vertical lift component" may exist for an F-16, but not for a 172. Thrust must be used to move the plane, letting the more efficient wing lift it. – Robert DiGiovanni Jul 24 '20 at 12:05
  • As the airspeed decays and then stabilizes at a lower value, lift will end up less in the climb than it was in horizontal flight. See https://aviation.stackexchange.com/q/40921/34686 – quiet flyer Aug 15 '23 at 00:42
  • Re " in your scenario, the reason that your airplane slows down is because, much like the helicopter, your prop disc is being tilted, dividing its power between lifting and thrusting." -- that's definitely not the primary reason that the airplane slows down when the yoke is moved aft to increase the wing angle-of-attack. Higher angle-of-attack equals higher drag coefficient from wing which causes the airspeed to decrease. – quiet flyer Aug 15 '23 at 02:28
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Part one: generally speaking--

I pitch up, say 1 inch of back pressure, and hold it.

I'm going to assume that you mean that you move the control yoke aft one inch of its original trim position and keep it there, regardless of how the force needed to keep it there changes during any subsequent airspeed changes. (If you don't re-trim, you'll need to maintain some amount of back pressure on the yoke, but "inches" aren't really a unit of "pressure", and the exact force needed to hold the new control yoke position will change somewhat as the airspeed changes.)

Initially the plane starts climbing and slows down...

Correct

...but what happens next?

How will the plane look in say 5 minutes with me still holding that 1 inch pitch up, what attitude, speed, and altitude will it have?

Well, it depends on whether you were initially flying on the "front side" (fast side) or "back side" (slow side) of the drag curve.

We'll assume for simplicity that thrust remains constant throughout, though that's not exactly true.

If you were flying on the "front side" (fast side) of the drag curve, then slowing down reduces drag, and the airplane will end up in a climb.

If you were flying on the "back side" (slow side) of the drag curve, then slowing down increases drag, and the airplane will end up in a descent. Be careful-- you are in the "slow flight" part of the flight envelope, and perhaps may be close to the stall speed.

In either case the airspeed will be reduced.

In the first case (the "front side" case), the pitch attitude will certainly end up more nose-high than before you moved the control yoke. In the latter case (the "back side" case), that may or may not be true.1,2

In either case the angle-of-attack will have increased. Think of the control stick or yoke as being most fundamentally an angle-of-attack control, not a pitch attitude control. Pitch attitude is simply the result of angle-of-attack plus climb angle (or minus descent angle.)3

Part two: since you specified that you were initially flying at 95 mph or knots--

You were initially well on the "front side" (fast side) of the drag curve, so you will end up in a climb when the airspeed stabilizes at the new, decreased value. The pitch attitude will be shifted in the nose-up direction. (One inch of change of control yoke position isn't enough to "pull through" to the "back side" of the drag curve.) As for specific numerical values-- why not try it yourself and report back? But it's safe to say that your airspeed will still be well above Vy, not to mention Vx.

Footnotes:

  1. Note therefore that the title of the question is somewhat misleading-- in the long run, you are not necessarily "increasing pitch" (attitude) with your control input.

  2. There is a third case where the airplane starts on the "front side" of the drag curve and ends up on the "back side". In this case the flight path could end up being exactly horizontal after the aircraft settles into stabilized flight at the new angle-of-attack and airspeed. In such a case, the pitch attitude would have shifted in the nose-up direction. And there is yet another case, starting at a slightly lower initial airspeed, where the aircraft ends up in a descent, but at the same pitch attitude as it started with.

  3. To be complete, we should add "minus wing incidence" to this sentence-- if we assume for simplicity that wing incidence is zero, then the sentence is exactly true.

quiet flyer
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  • PS if your meaning was that you apply whatever aft force is needed to keep the yoke one inch aft of its current instantaneous trim position, the answer would remain essentially unchanged, especially "Part two". Because you aren't going to be causing a big enough airspeed change to see much change in yoke position as the airspeed decays a little, while you hold a constant x ounces or pounds or kilos or Newtons of aft force on the yoke. But the yoke will end up a little further aft of the position that you initially pulled it to, if you continue to hold a constant force as the aspd decays. – quiet flyer Aug 15 '23 at 02:52