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For example:

if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.

Can someone explain these situations why exactly are they producing different results in AoA?

Can someone explain these what is going on in terms of forces? enter image description here Please explain in reference with these forces ( in the pic )

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    $\begingroup$ Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! $\endgroup$ Nov 8, 2018 at 15:36
  • $\begingroup$ Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. $\endgroup$ Nov 8, 2018 at 18:57
  • $\begingroup$ This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. $\endgroup$ May 31, 2021 at 22:40

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First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )

Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?

Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.

If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!

From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.

A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.

Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.

In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.

The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.

Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?

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  • $\begingroup$ Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram... $\endgroup$
    – Zeus
    Nov 8, 2018 at 23:42
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    $\begingroup$ An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset. $\endgroup$
    – Zeus
    Nov 8, 2018 at 23:53
  • $\begingroup$ "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. $\endgroup$ Nov 9, 2018 at 19:51
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Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:

Pitch Attitude + Incidence = Angle of Climb + Angle of Attack

The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.

If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.

As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.

What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.

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"Are we changing the angle of attack by changing the pitch of an aircraft"

Yes! Let's start with an aircraft in straight and level steady state flight.

We are taught "pitch controls speed, power controls altitude".

If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.

If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.

Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.

But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.

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The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.

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    $\begingroup$ Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. $\endgroup$ Nov 8, 2018 at 15:49
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    $\begingroup$ Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. $\endgroup$ Nov 8, 2018 at 15:51
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    $\begingroup$ For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. $\endgroup$ Nov 8, 2018 at 15:57
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In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.

The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)

Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.

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    $\begingroup$ Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. $\endgroup$ May 31, 2021 at 22:43
  • $\begingroup$ Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. $\endgroup$ May 31, 2021 at 22:45
  • $\begingroup$ @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. $\endgroup$ Jun 1, 2021 at 1:43
  • $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$
    – Bianfable
    Jun 1, 2021 at 6:58
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    $\begingroup$ @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. $\endgroup$ Jun 1, 2021 at 8:11

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