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As answered in this question, aircraft need excess power - not excess lift - to climb. This is plausible when the aircraft's thrust vector has a vertical component (its nose and engine points upwards), but I challenge the requirement of excess power for every case.

Please take a look at the following cart. The thrust gets delivered by a propeller at the rear and the thrust vector is always horizontal. A wing attached to a vertical beam is free to move up and down.

cart front view

cart rear view

When the cart gets accelerated and reaches a certain speed, the lift acting onto the wing gets greater than the wing's weight, leading to a climb of the wing. Please notice that - because thrust is horizontal - the chemical energy burned goes into kinetic energy of the cart and/or heat energy (due to overcoming drag). No power invested by the propeller goes into potential energy of the wing; the climb of the wing is done purely by lift.

Did I miss something?

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    $\begingroup$ You did an excellent job of illustrating your question! I wish others would pose their question with such clarity. $\endgroup$ – Peter Kämpf May 31 '15 at 12:36
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    $\begingroup$ The first thing you missed is drag: Once the wing moves through air, it will create not only lift, but also drag, and that drag will be higher when the wing accelerates up the pole. This drag increase will at least reduce the acceleration the cart receives from the engine. If the wing would not produce lift, the cart would accelerate more quickly and would settle at a higher speed. $\endgroup$ – Peter Kämpf May 31 '15 at 18:00
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    $\begingroup$ I really like these these illustrations! Did you make them yourself? If so, what tools did you use, I need those skills as well ! $\endgroup$ – DeltaLima May 31 '15 at 18:27
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    $\begingroup$ @DeltaLima: I made them with SketchUp. $\endgroup$ – Chris Jun 1 '15 at 10:09
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    $\begingroup$ The drag of the wing changes with the square of the speed of the car, and when the wing moves up or down, it changes in addition with the third power of the angle given by the ratio of vertical to horizontal speed. A square comes from the amount of lift created, and must be multiplied by the angle again to account for its change of direction - therefore the third power. $\endgroup$ – Peter Kämpf Jun 5 '15 at 18:29
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As the answers to your original question already explained, you do need extra lift to accelerate upwards. Once the wing is set into a vertical motion, however, lift again exactly equals weight to keep the wing at a constant vertical speed (if we neglect thrust and drag for a moment). No extra lift is needed to maintain that vertical speed. Only when you want to accelerate further up, extra lift is needed.

The increase in potential energy comes indeed from the propeller, because the lift vector of the climbing wing is tilted backwards, adding a horizontal component that needs to be compensated by extra propeller thrust.

Now let's look at your experiment in detail: I assume the wing has some mass, is rotationally locked and slides up and down that pole without friction. If you accelerate the car, at some point its speed will just be right for the wing to create exactly the lift to cancel out its own weight. At this speed the wing will be stable at any position along the pole. If it slides down a little, its angle of attack $\alpha$ will increase and create more lift, stopping the downward motion. The reverse is true for any upward motion. See below for an illustration of the principle. The cyan vector is the vector sum of the flow due to forward motion (blue) and vertical motion (red), and this is what the wing will "notice".

flow angles at wing in rest and in motion

When the car accelerates further, the lift will increase and now become greater than the weight. The wing will accelerate upwards until its vertical speed will reduce its angle of attack by enough to reduce the vertical aerodynamic forces to exactly equal its weight. Now you have the same situation as before, but not at zero vertical speed, but at a positive vertical speed which will make sure that the wing pops out at the top of the pole unless there is some stop. When the wing hits the stop, the vertical motion ceases, the angle of attack increases and the wing will lift up not only itself, but also part of the car's weight.

Note that I now spoke of the vertical components of the aerodynamic forces, not lift. When drag is added, it will add a vertical component when the wing is in motion. Lift is defined as the sum of aerodynamic forces perpendicular to the flow direction at infinity and drag parallel to it. This cumbersome definition makes sure that local distortions in the flow field do not impact the direction of lift and drag. The direction of lift for the climbing wing will point slightly backwards and the direction of drag slightly downwards. This will add some drag component to the sum of the vertical aerodynamic forces, and lift needs to increase to compensate for this. The horizontal component of lift will now add to the drag and the forces on the pole, so more force from the propeller is needed to push the climbing wing through the air. This extra force is needed to increase the potential energy of the wing on its way up. For a descending wing, the reverse is true: Now drag will add some vertical component and lift will be slightly slower. The forward component of lift will now push against the pole, reducing the force the propeller needs to provide. The reduction in potential energy now reduces the horizontal aerodynamic forces.

An airplane is slightly different, because it is free to pitch up or down and the angle of thrust will pitch with it. This will enable the pilot to select the flight path and the amount of lift the wing creates, but again the vertical motion will make sure that any excess lift will translate into increased vertical speed and a lower angle of attack, so the excess lift vanishes. In a climb, thrust needs to be bigger than drag in order to increase the potential energy of the airplane, and now the vertical component of the tilted thrust vector will support some weight, reducing the amount of lift needed to support the weight.

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    $\begingroup$ If the wing doesn't rotate and the airflow is at a constant angle (i.e., in the observer's frame of reference, still air and the cart is moving along horizontal ground), how can the angle of attack change? $\endgroup$ – David Richerby Jun 1 '15 at 6:52
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    $\begingroup$ @DavidRicherby: Due to the wing's motion. I guess I'll update the answer with a sketch - that will be better than a wordy explanation here. $\endgroup$ – Peter Kämpf Jun 1 '15 at 8:43
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    $\begingroup$ Ah, I understand now: the angle of attack is the same whenever the wing is stationary but it alters while the wing is moving up or down. $\endgroup$ – David Richerby Jun 1 '15 at 10:16
  • $\begingroup$ @PeterKämpf: The change of the angle of attack is actually a change in the direction of the relative wind (as seen from the perspective of the wing). Since lift is perpendicular to relative wind, there is an additional lift-induced drag onto the cart when the wing accelerates up. $\endgroup$ – Chris Jun 1 '15 at 10:29
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    $\begingroup$ @PeterKämpf: With vertical drag I mean the component of the drag force acting opposite to the vertical movement of the wing, i.e. the drag force that acts downwards when the wing moves up. $\endgroup$ – Chris Jun 8 '15 at 10:37
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When you say,

No power invested by the propeller goes into potential energy of the wing; the climb of the wing is done purely by lift.

you're missing where the energy of the wing comes from. Lift isn't a magical power that creates potential energy out of nothing: it just turns airspeed (kinetic energy) into height (potential energy). In your example, the power invested by the propeller turns into kinetic energy of the whole cart, including the wing. That's how the energy gets from the propeller (or its fuel) into the potential energy of the wing. You need to use more thrust to drive the cart with the wing attached, than you would if you took the wing away.

There are two ways to look at the forces produced during a climb. Remember that as a wing produces more lift, it also produces more induced drag. That's why you need excess thrust, to generate the excess lift.

For a certain power setting, you can fly level at a certain speed. If you pitch up, the wings will create excess lift, but also more drag. Even though some of your thrust is acting vertically, there isn't any excess thrust, because the drag is greater. You'll slow down, the lift will decrease, and you'll stop climbing.

Instead, you can keep the aircraft level, and add more thrust. This will increase your speed, which will also increase the lift from the wings. This in turn increases the induced drag, which will eventually balance the excess thrust at a new, higher airspeed. Because you've increased the lift by doing this, you'll climb, even though your wings are level. You can only do this because you added power in the first place.

(I feel obliged to point out that you wouldn't usually climb like this: to get a better rate of climb, you'd generally add power and also pitch up, letting your airspeed decrease to the speed where the wings produce the most lift for the least drag.)

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  • $\begingroup$ Induced drag in a steep climb is actually less than in a shallow climb, simply because lift is less (more of the net upwards force generated by thrust). By definition, Lift and Drag are perpendicular and parallel to the flight path (relative wind), not the earth horizontal plane. The increase in lift (and Di) s only momentary, to accelerate to create an upwards velocity as indicated by Peter Kämpfs answer $\endgroup$ – Waked May 31 '15 at 16:38
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    $\begingroup$ Believe it or not, induced drag goes down when speed increases. $\endgroup$ – Peter Kämpf May 31 '15 at 17:54
  • $\begingroup$ @PeterKämpf Because the angle of attack is decreasing, you mean? That's a point. I'd hoped to keep the explanation simpler than that, but maybe I tried to make it too simple. $\endgroup$ – Dan Hulme Jun 1 '15 at 9:53
  • $\begingroup$ Induced drag goes down when speed increases because wingtip vortices decrease at higher speed. $\endgroup$ – Chris Jun 1 '15 at 11:19
  • $\begingroup$ @DanHulme: "You need to use more thrust to drive the cart with the wing attached, than you would if you took the wing away." Of course, the reason is additional drag, which dissipates to heat. I am fully aware that this violates conservation of energy. But remember that energy conservation is a "macro principle" that gets induced by more basic principles, e.g. mechanics. You have to give mechanical reasons to show that energy conservation is in place. $\endgroup$ – Chris Jun 1 '15 at 11:40
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I sort of feel that the rest of the answers are unnecessarily complex, given how simple the fundamentals here are:

Momentum

Question: Is it necessary that L>m.g (or as you put it, an excess of lift) in order to climb?

Answer: No, at least not a sustained excess of lift. Newton's Laws state that an object in motion will remain in that state unless a force acts upon it. A force imbalance is required to set the aircraft into a climb, but once this has been achieved the forces can be balanced and the aircraft will continue to climb. As such, an excess of lift is not a condition required for an aircraft to sustain a climb.

Energy

Question: Is it necessary that we add energy to the system (in the form of increasing our power output) in order to climb?

Answer: Yes, if energy is conserved then in order to gain altitude (and by extension gravitational potential energy), we must add energy. We could add no energy, not increase the power output of our engines, and simply pull up, increasing AoA but also drag, and we would climb for a short time as we trade kinetic energy for gravitational potential energy, however we would find that our aircraft quickly slows and we are required to dive to below our original altitude to return to steady level flight.

Hence a power excess is necessary for climb, but a sustained lift excess is not.

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  • $\begingroup$ I disagree with your statement that "an excess of lift is not a condition required for an aircraft to sustain a climb." Do you have any authorities on that which you can provide. Conventionally, an excess of lift results in a climb, and a short fall of lift results in a descent, compared to balanced lift and weight. Accordingly, I also look for an authority on your concluding statement that sustained left does not require a "power excess." $\endgroup$ – mongo Aug 5 '17 at 18:55
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    $\begingroup$ In a stabilized climb (constant airspeed, constant direction of flight path through space), lift is LESS than weight. See my answer. $\endgroup$ – quiet flyer Oct 15 '18 at 8:58
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The simple answer is easy to demonstrate. Start with an aircraft TRIMMED for straight and level flight. For example 1000 feet, 100 mph, 1500 rpm fixed pitch prop.

Lift = aircraft weight and thrust = aircraft drag.

Now increase engine rpm by 150 rpm (10% more thrust), which increases thrust. The aircraft will for a moment accelerate, the increased airflow over the wing and stabilizer increases lift and the aircraft will gain altitude. In a few seconds the system will balance once again, the airspeed will return to the trimmed 100 mph, and the excess thrust will show up as climb rate. The aircraft will now be slightly pitched up, but the angle of attack remains constant since it is controlled by the stabilizer trim setting, which we did not touch.

Next roll the elevator trim forward, which will lower the nose a bit. The airspeed will increase slightly and the climb rate will reduce. When trimmed once again to straight and level flight the aircraft rate of climb will be 0, the airspeed will be above 100 mph. Now the extra thrust shows up as increased speed.

To continue the example, reduce the rpm back to the original 1500 rpm. leave the trim alone. The aircraft should now show a decent rate, at the new slightly higher airspeed.

All this was done without input from the control stick.

Anytime the pilot maneuvers the primary flight flight controls, there is a nearly instant trade between angle of attack, speed, lift, drag, inertia, climb rate or decent. Jerry S.

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  • $\begingroup$ Lift = aircraft weight is valid only in a specific scenario (and the requirement for trimmed flight does not cover it): pitch angle null and engine mounting pitch null; alternatively, thrust vector "pitch" null. In any other case, including trimmed conditions, lift != aircraft weight $\endgroup$ – Federico May 31 '15 at 18:37
  • $\begingroup$ @Federico, in a systemic sense, lift of the caused by engine pitch is lift. Just as body lift, tail lift (or negative lift) all sum to the body or system lift. If the aggregate lift goes up, the airplane can climb. If it becomes less than the weight of the aircraft, the aircraft descends. $\endgroup$ – mongo Aug 5 '17 at 19:01
  • $\begingroup$ In a stabilized climb (constant airspeed, constant direction of flight path through space), lift is LESS than weight. See my answer. $\endgroup$ – quiet flyer Oct 15 '18 at 8:57
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The above answers beautifully explain the theoretical solution to your problem, but since you haven't accepted any one of them as of now, i'd be illustrating the solution numerically.

Lets assume that your cart is moving with a constant velocity of 'v'

Then, K.E. = 1/2 (mv^2)
D = 1/2((density)(v^2)S(Cd))
and total energy E = K.E. + D*distance (Assuming frictionless interaction of surfaces everywhere)

now, Cd = Cd0 + K(Cl)^2
distance = v*t
so T.E. = 1/2(v^2)(m + (density)SVt(Cd0 + K(Cl)^2))

Here it can be seen that total energy is being used for

  1. The kinetic energy part of the Cart
  2. The coefficient of lift part of the Cart's wing

The coefficient of lift part is hence responsible for the energy use3d up in lifting the wing upwards and hence the whole system obeys conservation of energy

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Updated simulation without vertical drag force

In this situation, for the wing only, lift in a climb is greater than the weight. The vertical force stabilises to equal the weight, but since the lift vector is tilted backwards slightly due to the upward velocity, aerodynamic lift increases.

Peter Kämpfs answer describes what happens to the wing in this situation, but what we did not have was a quantification. I've run a real time simulation of the forces on the wing in the drawing of the OP, as a function of airspeed $V_{air}$ and vertical wing velocity $\dot{z}$. The forces on the wing are drawn below, I've taken a NACA 0012 profile with an $ \alpha_0$ of 2 degrees:

enter image description here

$$L = lift = C_L \cdot \frac{1}{2} \cdot \rho \cdot {V}^2 \cdot A \tag{1}$$

$$D = C_D \cdot \frac{1}{2} \cdot \rho \cdot {V}^2 \cdot \tag{2}A$$

For NACA 0012, $C_L$ is proportional to $\alpha$: $C_L$ = 1 at $\alpha$ = 10 degrees, hence $$C_L = k_L \cdot \alpha \tag{3} $$

When the wing goes up, the angle of attack changes: $$ \Delta \alpha = arctan(\frac{\dot{z}}{V_{air}}) \tag{4}$$

We now lump all the constants together: $K_L = k_L \cdot \frac{1}{2} \cdot \rho \cdot A$, $K_D = 0.01 \cdot \frac{1}{2} \cdot \rho \cdot A$ ($C_D$ for standard roughness at Re = 6 x $10^6$ = 0.01 for angles up to 4 deg)

The lift of this angle of attack is found by combining (1), (3) and (4):

$$ L = K_L \cdot (\alpha_0 - \Delta \alpha) \cdot V^2 \tag{5}$$

resulting force $F$ is divided by mass to result in wing acceleration, which is then integrated with a digital Euler integrator to yield $\dot{z}$

L and D are aligned with the free stream vector V, while the weight is always aligned with the vertical. We take the cosine of the L vector minus the sine of the D vector

$$ F_{up} = L \cdot cos(\Delta \alpha) - D \cdot sin(\Delta \alpha) \tag{6}$$ Now for:

  • m = 1 kg
  • A = 1 $m^2$
  • $\alpha_0 = 2 deg$
  • $k_L$ = 0.1

We get L = 9.81 N at $V_{air}$ = 8.949 m/s. If we then increase $V_{air}$ from 8.949 to 10.5 m/s in 1.5 second, the wing gets an initial acceleration upwards. After 2.4 seconds acceleration is zero, the wing goes up with constant velocity $\dot{z}$ = 0.1 m/s. The angle of attack has then reduced from 2 deg to 1.45 deg

Values printed for beginning of test through to 3 seconds:

enter image description here

There are some 2nd order effects in the response which may be from digital instability due to the large time step of the Euler integrator. Time to check this is not available at the moment.

So in the end situation, L is 9.82 N which is larger than weight in a climb due to increase in airspeed. Not by much - the lift vector is tilted backwards at a small angle, determined by the ratio of $\dot{z}$ and V which is 0.01. The total vertical force is $ L \cdot cos\alpha - D \cdot sin\alpha - W$

enter image description here

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  • $\begingroup$ You start from wrong assumptions. Drag is always perpendicular to lift - it's defined this way. There cannot be a drag component aligned with lift. Next, when the wing climbs, the lift force vector tilts backward and the drag vector downward due to the change in angle of attack. Now some of the drag is in the direction of weight but still perpendicular to lift. What is missing is the horizontal force of the beam which counteracts drag. $\endgroup$ – Peter Kämpf Jul 19 '17 at 18:59
  • $\begingroup$ @Peter I've thought long and hard about your comment. It makes sense to always define lift and drag in the direction of the airstream, *regardless of the direction of the airstream itself", is basically what you are saying. I agree with that, and will rework the simulation model. However, there is a lingering issue that multiple persons here have been trying to express: we accept that in a climb, power is required to overcome the increase in potential energy from the gravitational field. However, where is the power accounted for that needs to overcome aerodynamic resistance in vertical... $\endgroup$ – Koyovis Jul 21 '17 at 4:51
  • $\begingroup$ ...direction. A helicopter taking off vertically needs increased lift to offset the vertical fuselage drag. A rocket taking off needs to provide thrust equal to (weight + drag). Why is this for fixed wing only valid in thrust direction, and not in lift direction? $\endgroup$ – Koyovis Jul 21 '17 at 4:55
  • $\begingroup$ Thrust is opposite to drag (roughly) but larger in a climb. So all vertical drag is compensated by thrust, and some vertical thrust remains to reduce demand for lift. Please see the drawing here and look how the force vectors compare. In a descent, thrust is smaller, leaving some vertical drag uncompensated which now reduces the required lift. Yes, in a descent drag helps to reduce lift. Thrust is in the direction of drag while lift is orthogonal to both. So thrust compensates $\endgroup$ – Peter Kämpf Jul 21 '17 at 21:00
  • $\begingroup$ drag, not lift. In vertical ascent (rocket or helicopter) it is thrust, not lift, which compensates for drag (and needs to be a bit larger than weight). Both produce thrust to counteract weight, and only when the helicopter adds some forward speed, lift will be created in addition to thrust. $\endgroup$ – Peter Kämpf Jul 21 '17 at 21:08
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There is a lot more to this question than initially meets the eye-- it is quite an interesting question.

Normally, in the context of fixed-wing flight, the thrust vector acts roughly parallel to the flight path through the airmass. When the thrust vector is exactly parallel to the flight path through the airmass, and the flight path is linear rather than curving up or down or to either side, then the vector diagram of forces in climb looks like this (left-hand case-- climb angle of 45 degrees-- right-hand case-- climb angle of 90 degrees):

Powered climb at climb angles of 45 and 90 degrees:

Powered climb at climb angles of 45 and 90 degrees

We can see that Lift = Weight * cosine (climb angle). In the left-hand diagram, the climb angle is 45 degrees and Lift = .707 * Weight. In the right-hand diagram, the climb angle is 90 degrees and Lift is zero.

But, these diagrams assume that the Thrust vector acts parallel to the flight path through the airmass. Obviously, if this isn't true, thrust equation lift = weight * cosine (climb angle) is also no longer true. To take an extreme case, note that when the exhaust nozzles of a Harrier "jump jet" are pointed straight down, the wing is "unloaded"-- the plane can hover at zero airspeed with zero lift, supported entirely by thrust. Conversely, during a glider winch launch, the towline pulls steeply downward on the glider. This too can be viewed as a form of "vectored thrust"-- but now the load on the wing is increased, rather than decreased, so the wings must generate a lift force that is much greater than the aircraft's weight.

In the case presented in this question, Thrust does NOT act along the flight path of the "aircraft", if we consider the wing to be the "aircraft". When the wing is rising up the pole, the vertical motion causes a change in the direction of the wing's trajectory through the airmass and also a change in the direction of the "relative wind", but there is no corresponding change in the direction of the Thrust vector. Thus the thought experiment presented in the question is NOT representative of the typical situation in fixed-wing flight. The thrust vector is NOT fixed in direction relative to the chord line of the wing, and is NOT acting roughly parallel to the direction of the "relative wind" experienced by the wing, and the direction of the wing's flight path through the airmass.

Furthermore, the basic mechanism that governs the airspeed of a fixed-wing aircraft is absent. Normally, as an aircraft climbs, if Lift exceeds Weight, the flight path will curve upward, causing the Weight vector to have a greater component acting parallel to the direction of the aircraft's flight path through the airmass, causing a decrease in speed. But in this experiment, since the wing is "locked" into position on the cart in the fore-and-aft sense, if the wing's trajectory curves upward, it appears that the cart will provide however much thrust is needed to hold the horizontal component of the wing's velocity vector component exactly constant. Assuming, that is, that the wing's drag is trivial compared to drag from other sources such wheel drag and wheel bearing drag from the cart, so that variations in drag from the wing have essentially no effect on the airspeed and groundspeed of the cart.

So the forces acting on the wing in this thought experiment will be very different from the forces typically acting on a fixed-wing aircraft in actual flight. It should not come as surprise to discover that in the case of this thought experiment, lift actually must be greater than weight in order for the wing to climb up the pole.

We really could end this answer right here. But it's rather interesting to look a little more deeply at the forces acting on the wing in the thought experiment.

What are some of the notable features of the thought experiment?

As we've already noted, the wing is locked in place relative to the cart in the fore-and-aft direction. The wing cannot speed up or slow down in the fore-and-aft direction, relative to the cart. Furthermore, IF the drag from the wing is trival compared to the other sources of drag acting on the cart, so that the drag from the wing has essentially no effect on the cart's airspeed, this means that the wing cannot speed up or slow down in the fore-and-aft direction relative to the airmass (or relative to the ground). The cart will transmit to the wing however much thrust is needed to hold constant the fore-and-aft component of the wing's airspeed vector. This is very different from the typical situation in fixed-wing flight.

Also. as the thought experiment was originally worded, the wing is locked into a constant pitch attitude.

Here is a vector diagram illustrating the situation when the cart is moving at some constant airspeed that is NOT high enough to allow the wing to lift off the cart:

enter image description here

The forces illustrated include Lift (L), Drag (D), Weight (W), Thrust (T), and the upward force (C) exerted by the cart on the wing as the wing rests on the cart. Net force is zero. The L/D ratio is 10:1.

Now assume that we hold the wing down with a catch as we increase power and accelerate to some higher airspeed, and then allow everything to stabilize. Then we unlatch the catch. The diagram below shows the situation at the instant we unlatch the catch--

enter image description here

The wing has not yet begun to rise upward, so there is no change in the direction of the wing's trajectory through the airmass, or the direction of the lift and drag vectors. The wing's angle-of-attack has not changed, so the lift and drag coefficients have not changed, so the L/D ratio is still 10/1. The dashed line represents the net force vector, which is simply the vector sum of all the other force vectors. Acceleration = force / mass, so we can also label the net force vector as "Acceleration * mass".

What happens as the wing starts to rise (accelerate) up the pole? The wing's upward velocity causes a change in the "relative wind" experienced by the wing. The wing's angle-of-attack immediately decreases or becomes negative, so the lift coefficient decreases, and the L/D ratio decreases. (The drag coefficient might decrease too, but not as much as the lift coefficient.) If the wing has a cambered, non-symmetrical airfoil, it will still produce lift at some small negative angle-of-attack, but not very much-- the lift coefficient will be low. When the wing reaches some given upward vertical velocity, Lift will have decreased to the point such that the vertical component of the net aerodynamic force acting on the wing will no longer be larger than Weight, the net force on the wing will drop to zero, and the wing will no longer be able to accelerate, but rather will move up the pole with a constant vertical velocity. The figure below illustrates this situation:

enter image description here

The direction of the wing's path through the airmass is parallel to (and opposite to) the direction of the Drag vector. The climb angle-- labelled "C" in the diagram-- is the acute angle formed between the Thrust and Drag vectors, and also between the Lift and Weight vectors. This is also the angle between the Drag vector and the horizon, and also the angle between the Lift vector and the vertical direction. The vectors can be all arranged head-to-tail in a closed figure, so the net force is zero. Lift is slightly larger than Weight, and Thrust is quite a bit greater than Drag. If the wing is mounted on the cart with zero incidence, then the wing's angle-of-attack must be slightly negative-- in fact it must be equal to negative "C" degrees. We've drawn the L/D ratio as 2/1, to represent the decrease in the wing's lift coefficient caused by the change in angle-of-attack. The wing is moving up the pole at a constant velocity.

Interestingly, this situation is virtually identical to the situation experienced by the rising wing as an aircraft rolls to a steeper bank angle, especially if the roll is driven by a spoiler deployed on the descending wing with no modification to the shape of the rising wing. The change in angle-of-attack caused by the wing's rising motion through the airmass limits the vertical speed that the rising wing can attain-- this is called "roll damping". The lift and drag vectors are "twisted aft" or "twisted backwards" from the direction they pointed before the rolling motion started-- you can also see this "twist" illustrated in this diagram https://www.av8n.com/how/img48/adverse-yaw-steady.png from this section https://www.av8n.com/how/htm/yaw.html#sec-adverse-yaw of the excellent "See How It Flies" website https://www.av8n.com/how/ .

The situation is also exactly like the situation we'd have if we were towing a glider under the following conditions: 1) We have a very long tow rope-- so long that the angle of the tow rope relative to the horizon is not influenced at all by the glider's climb rate relative to the tow plane. 2) The towplane is flying in such direction such that the glider's end of the rope is pulling exactly horizontally on the glider 3) The glider's drag is trivial compared to the towplane's thrust and drag, so the glider's aerodynamic situation has no influence on the towplane's airspeed. 4) The glider pilot is giving pitch control inputs in such a way as to force the glider's pitch attitude to stay constant relative to the horizon, regardless of climb rate.

Now, what if we modify the experiment by allowing the wing's pitch attitude to vary, while holding the wing's angle-of-attack constant relative to the wing's trajectory through the airmass-- perhaps by adding a stabilizing vane or tail to the back of the wing?

Now what happens when the wing starts to rise?

The figure below illustrates a situation where lift is exactly equal to weight. The drag vector is horizontal, so the wing cannot be rising or falling through the air-- it must be staying in a fixed position on the pole.

enter image description here

Note that we've chosen to illustrate a 5:1 L/D ratio for this version of the thought experiment.

Now what if we give the wing the tiniest upward push, so that it starts to rise? As soon as it starts to rise, its velocity through the airmass is augmented by its vertical motion. And now the wing is free to pivot in such a way that its angle-of-attack can stay constant, so we don't have the "damping" effect that we had in the earlier version of the experiment. The resulting increase in airspeed and lift is much like we see when a glider rises on a winch tow, except that in the case of the wing on the imaginary cart, the thrust vector stays horizontal, rather than pointing partly downwards. The wing's climb angle through the airmass will get steeper and steeper as its vertical speed increases. This causes the airspeed to increase, which causes the Lift vector to get larger.

The figure below illustrates the situation that we see when the wing's climb angle through the airmass reaches 60 degrees. Again, the dashed line represents the net force vector, which is simply the vector sum of all the other force vectors. Acceleration = force / mass, so we can also label the net force vector as "Acceleration * mass".

enter image description here

In this particular case, we've sized the lift and weight vectors to represent the situation where the horizontal component of the wing's speed through the airmass is exactly the same as it was in the previous diagram above, where lift was exactly equal to weight in the case where the wing's trajectory was horizontal. Simply by rising upward, the wing has experienced a doubling of airspeed and a four-fold increase in the magnitude of the lift vector. The sum of the vertical components of the lift and drag vectors is now 1.3 times the weight of the wing. Of course, we could modify the diagram to represent a case where the wing was experiencing a net upward force even before it started accelerating upward, simply by decreasing the size of the weight vector relative to the other vectors.

If the cart's velocity stays constant, and cart is able to transfer however much thrust is needed to the wing to keep it locked in place in the fore-and-aft direction relative to the cart, will the wing keep accelerating faster and faster up the pole?

It turns out it will not. Even if the wing has no weight at all, it will stop accelerating upwards once its climb angle equals the inverse tangent of the L/D ratio. For the 5/1 case illustrated here, that climb angle is 78.7 degrees. If the wing does have weight, the maximum achievable climb angle will be less. In the particular case illustrated above, where the Weight vector is exactly equal to the Lift vector that existed when the wing had zero upward velocity, the maximum achievable climb angle is somewhere between 70 and 75 degrees. Above this maximum achievable climb angle, the vertical components of Lift and Drag no longer add up to a value that is greater than Weight. So even when the wing is free to pivot to maintain a constant angle-of-attack, and the cart has infinite thrust available to allow it to maintain a constant airspeed in spite of changes in the wing's drag force, there is a limit to the climb angle that the wing can achieve.

Here's an interesting table

ca- cos- sin- aspd- L- D- vcL- vcD- net aero vc- net vert

0 1.00 0.00 1.00 1.00 .200 1.00 0.00 1.00 0.00

30 .866 .500 1.15 1.33 .267 1.15 .133 1.02 .021

45 .707 .707 1.41 2.00 .400 1.41 .283 1.13 0.13

60 .500 .866 2.00 4.00 .800 2.00 .693 1.31 0.31

70 .342 .940 2.92 8.55 1.71 2.92 1.61 1.32 0.32

75 .259 .966 3.86 14.9 2.99 3.86 2.88 .980 -0.02

80 .174 .985 5.76 33.2 6.63 5.76 6.53 -.773 -1.773

Assumptions--

  • Constant horizontal component of airspeed in all cases

  • 5/1 L/D ratio.

  • The last column (but only the last column) assumes that the value of Weight is selected in such a way that Weight is exactly equal to the value of the Lift vector in the particular case where the climb angle is zero, meaning that Weight and Lift are exactly in balance in this case. The value of Weight has no effect on any of the other columns.

ca= climb angle in degrees

cos= cosine of climb angle

sin= sine of climb angle

airspd= speed of wing through air in arbitrary units

L= Lift in arbitrary units

D= Drag in the same units as L

We've chosen an L/D ratio of 5/1

vcL= vertical component of lift (acts upwards) = L * cosine (climb angle)

vcD= vertical component of drag (acts downwards) = D * sine (climb angle)

net aero vc = vertical component of net aerodynamic force= (vcL-vcD) -- a positive sign means that the net aerodynamic force acts upwards, while a negative sign means that the net aerodynamic force acts downwards.

net vert = net vertical force = (net aero vc - weight), assuming that Weight is selected in such a way that Weight is exactly equal to the value of the Lift vector in the case where the climb angle is zero.

If the last column (net vert) is negative, this means that in the case where Weight is set to the particular value described above, the climb rate must decelerate (and the climb angle must decrease).

If the second-to-last column is negative, the climb rate must decelerate (and the climb angle must decrease) even if Weight is zero.

This version of the thought experiment-- where the pitch attitude of the wing is free to vary to maintain a constant angle-of-attack-- is somewhat like what happens at the start of glider winch tow, especially near the beginning of the tow when the towline is very long and the tow force stays almost horizontal for a while, even if the glider starts to climb rapidly.

Finally-- the original question contained the following line: " Please notice that - because thrust is horizontal - the chemical energy burned goes into kinetic energy of the cart and/or heat energy (due to overcoming drag). No power invested by the propeller goes into potential energy of the wing; the climb of the wing is done purely by lift." Thrust certainly does do work along the direction of the flight path through the airmass, which is never purely vertical. The situation appears to be analogous to a lightweight cube (say made of balsa wood) being blown up a slippery ramp by a wind that is blowing horizontally. Is the wind increasing the potential energy of the cube?

For more on the more "conventional" climbing situation-- a fixed-wing aircraft with thrust acting parallel to the direction of the flight path-- see these related answers to related questions:

:Does lift equal weight in a climb?"

"What produces Thrust along the line of flight in a glider?"

"'Gravitational' power vs. engine power"

"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?"

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


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The other answer (and the basic premise) are misleading. Aviation is a careful balance of just about everything, and you rarely get more of one without losing some of the other.

Consider a rocket. No wings, and thus no lift. Sea level, 0m/s to orbit on 8 minutes. Everything done with a rather ridiculous amount of power.

Now consider an airship. Also no wings, but excess lift. Goes up all by itself. The engines are purely for maneuvering, if you remove them we usually call the thing a balloon.

One of the most ridiculous things I ever heard was a flight instructor who claimed that the throttle controls altitude and the elevator controls speed. I called complete BS on the statement and asked if he found himself flying toward some cumulo-granitus would he prefer to add power or yank back rather sharply on the controls?

What he was trying to get across was that in the extremely narrow regime of straight and level flight*, adjusting power will affect speed which affects lift, and the plane will eventually stabilize at a different altitude after a power adjustment. Adjusting pitch will change your speed almost immediately, but your altitude will change as well. You could arrange a demonstration of adjusting power and pitch simultaneously and having nothing else change, but that basically proves my point. If you want to slow down (for example, landing) do you leave the throttle at full and pull the stick all the way back? Of course not. It's a very delicate balance, and it's something pilots spend a lot of time learning. Or if you work for certain asian airlines you just have some very expensive computers handle it for you.

  • Yes, we spend a lot of time there, but consider how precise you need to be. A fraction of a degree, uncorrected, in any axis will result in you crashing before you run out of fuel.
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    $\begingroup$ "The other answer" which one? There are now three other answers and there may well be more by the time you read this. $\endgroup$ – David Richerby Jun 1 '15 at 6:55
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    $\begingroup$ Also, -1 for the racist reference to "certain asian airlines". $\endgroup$ – David Richerby Jun 1 '15 at 6:57
  • $\begingroup$ Not only that, but there were at least two answers at this time this was posted. $\endgroup$ – a CVn Jun 1 '15 at 8:34
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    $\begingroup$ @paul OK. But, as you say, they just happen to be located in Asia. Saying they're Asian doesn't identify them and their ethnic origin isn't relevant to their preference for autoland. Using the word "Asian" as the only description makes it look like you're talking about Asian airlines in general so it's best avoided. $\endgroup$ – David Richerby Jun 1 '15 at 14:55
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    $\begingroup$ @DavidRicherby That's why I said "certain asian airlines" and not "asian airlines". I wonder what your reaction would have been if I said "certain european airlines". $\endgroup$ – paul Jun 1 '15 at 23:04

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