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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
  • $\begingroup$ This answer overlooks the fact that trim fundamentally controls angle-of-attack which is NOT the same as airspeed. Since lift is less in a climb, for a given a-o-a we'll trim slightly SLOWER in a climb. And same for a dive. Excluding complexities like height of thrust line relative to CG, downthrust angle, etc. $\endgroup$ – quiet flyer Nov 3 at 13:24
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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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In a steady-state climb in an airplane, the basic formula for the magnitude of the Lift vector is Lift = Weight * cosine (climb angle). For much more on this, see the vector diagrams and calculations in this related ASE answer. As long as the Thrust line is aligned with the flight path rather than being tilted up or down, the relationship Lift = cosine (Weight) is "baked into the physics" of a steady-state climb, even when we're flying at a high angle-of-attack that yields a high lift coefficient and a high L/D ratio. As long as Thrust is aligned with the flight path, Lift is less than Weight in a steady-state climb.

If any downthrust or upthrust is present relative to the direction of the flight path, then the situation becomes more complicated. Downthrust makes the Lift vector larger, and upthrust makes the Lift vector smaller. In the limit case where Drag is zero, Lift = Weight * cosine (climb angle) + Thrust * sine (downthrust angle), where the downthrust angle is defined relative to the direction of the flight path. (Treat upthrust as negative downthrust.) When Drag is not zero, then this formula no longer applies. When Drag is not zero, if there is downthrust, the Lift vector will be larger than the value given by the above formula, and if there is upthrust, the Lift vector will be smaller than the value given by the above formula. In such cases, the exact values of the Lift and Drag and Thrust vectors may be found by a vector diagram, if the L/D ratio and climb angle and downthrust or upthrust angle are known.

If enough downthrust is present, Lift will be greater than Weight in a steady-state climb. We can show that in the limit case where Drag is zero, whenever the downthrust angle relative to the direction of the flight path exceeds half the climb angle, Lift will exceed Weight. When Drag is not zero, the downthrust angle that makes Lift equal Weight becomes smaller than half the climb angle.

The cart of the "propeller train" can only push horizontally forward on the wing. So in the case of the "propeller train", as soon as the wing starts rising, the direction of the instantaneous flight path of the wing is no longer parallel to the thrust line, so downthrust exists. In fact, the downthrust angle is always exactly equal to the instantaneous climb angle of the wing. So whenever the wing of the "propeller train" is climbing, Lift must be greater than Weight, even in the steady-state case where acceleration is zero.

Vector diagram of forces in climb-- "Propeller train" vs conventional airplane

The vector diagrams immediately above show the forces in a steady-state climb for the wing of the "propeller train" on the left, and for a conventional airplane (where the downthrust angle at the given angle-of-attack happens to be zero) on the right.

For simplicity, the Lift to Drag ratio is 3:1 in all cases. 10-degree and 30-degree climb angles relative to the horizon are illustrated for both the "propeller train" and the conventional airplane. The diagrams are drawn to the same scale-- the Weight vector is identical in each diagram. The magnitudes of the airspeed vectors are scaled in proportion to the square root of the magnitude of the Lift and Drag vectors.

The right-hand diagram may look a bit odd with the "unattached" Drag vectors, but look closely and you'll see two closed right triangles, each comprised of Weight, Lift, and (Thrust minus Drag), with Lift = Weight * cosine (climb angle) and (Thrust minus Drag) = Weight * sine (climb angle).

(As an aside, the right-hand diagram may be easily modified for any L/D ratio by erasing or extending the right-hand ends of the Drag and Thrust vectors, keeping the value (Thrust-Drag) the same. It's not so simple to modify the left-hand diagram for different L/D ratios--the diagram must be completely re-drawn.)

Here is a table of the forces involved, with L/D fixed at 3:1, Weight arbitrarily set at 100, and airspeed ("A") in arbitrary units. We also have included the Power required (P). Since Power = force * the velocity component along the direction that the force is acting, Power is computed by multiplying Thrust times Airspeed times the cosine of the downthrust angle.

(3/1 L/D)
             "Propeller Train"              "Conventional" Airplane       
Climb angle  L    D    T    A    P          L     D   D-T   T    A    P
      0      100  33   33   100  3300       100  33   0     33   100  3300
     10      115  38   69   107  7260       98   33   17    50   98   4920
     30      145  48   114  120  11900      87   29   50    79   93   7300

It's clear that more Power must be applied to the wing of the "propeller train" when the wing is rising upward than when it is not, just as with the "conventional airplane".

Just for comparison, here are the forces and power requirements resulting from an L/D of 10/1 rather than 3/1-- (the vector diagrams corresponding to these cases are not reproduced here)--

(10/1 L/D)
             "Propeller Train"             "Conventional" Airplane       
Climb angle  L    D    T    A    P          L     D   D-T   T    A    P
      0      100  10   10   100  1000       100  10   0     10   100  1000
     10      104  10   29   102  2890       98   9.8  17    27   99   2700
     30      125  12   75   112  7280       87   8.7  50    59   93   5460

Warning-- we shouldn't assume that the 3/1 L/D table and the 10/1 L/D table could both reflect the same aircraft or wing, at two different angles-of-attack. If we want to use the tables in that way, we must ignore the "Airspeed" and "Power" columns. The reason for this is that the Airspeed and Power columns are arbitrarily set for each table so that a Lift vector of 100 correlates to an Airspeed of 100. In reality, when we put the stick or yoke forward and decrease the wing's angle-of-attack and L/D ratio, we also decrease the lift coefficient, causing a large increase in airspeed that it is not reflected in the tables. The "Airspeed" and "Power" numbers on the tables are only meant for comparison between different climb angles at the same L/D ratio, or for comparison of the "propeller train" and "conventional airplane" cases at the same L/D ratio and climb angle, assuming that the lift coefficient is the same in each case.

Note that in reality, unless the wing of the "propeller train" is free to pivot in pitch, the angle-of-attack of the wing of the "propeller train" cannot actually stay constant regardless of climb rate-- the higher the climb angle, the more the "relative wind" comes down from above, relative to the horizon. This decreases the airfoil's angle-of-attack. Therefore the L/D ratio cannot actually stay constant as the climb angle increases. Unless the wing is fixed at a very nose-high pitch attitude so that it is essentially stalled when it is not rising, higher climb angles will be correlated with lower lift coefficients, and therefore with higher airspeeds. And as long as we are on the "front side" of the Drag curve, the lower angles-of-attack caused by higher climb angles will be correlated with lower L/D ratios. With the "propeller train", in the steady-state condition, a decrease in L/D ratio is reflected not as a decrease in Lift, but rather as an increase in Drag, Thrust, and Lift. This is best visualized by looking at the left-hand vector diagram and seeing what else changes when we increase the length of the Drag vector, while holding the climb angle constant.

In essence, the vector diagrams and tables above give some insights into how the downthrust angle affects the forces acting on the wing of the "propeller train", and they also clearly show that the wing of the "propeller train" demands more Thrust and Power when it is rising than when it is not, but they pertain most directly to a wing free to pivot to maintain a constant angle-of-attack and L/D ratio regardless of the fuselage pitch attitude, along the lines of the "freewing" concept (link to PDF) (better link needed).

Another interesting variation of the "propeller train" concept would be , or a spinning cylindrical wing using the "Magnus" effect (Wikipedia link) in which case the magnitude of the Lift and Drag vectors would be insensitive to the direction of the relative wind.

What insights does the "propeller train" give us about flight in a conventional airplane? It tells us that it is possible to enter a climb without pitching up at all, so that the aircraft's pitch attitude stays the same as it was in horizontal flight, but in most cases it is extremely inefficient. If we are initially flying very slowly, on the "back side of the Drag curve", at a very high angle-of-attack and a nose-high pitch attitude, adding power while holding the pitch attitude constant will indeed cause us to gain airspeed and start climbing. In this part of the flight envelope, this strategy works just fine. However if we are in anything like normal cruising flight, we'll find that adding power while holding the pitch attitude constant can only produce a very modest climb rate and climb angle. As we enter the climb, to prevent the aircraft from pitching up to maintain the same angle-of-attack it had in horizontal flight, we'll be holding forward pressure on the control stick or yoke to decrease the angle-of-attack, or we'll re-trim the aircraft to accomplish the same thing. Unless we're on the "back side" of the Drag curve, the decrease in angle-of-attack associated with climbing will degrade the L/D ratio, and unless our aircraft had substantial upthrust relative to the flight path when it was flying horizontally, any positive climb angle will introduce downthrust relative to the direction of the flight path. If we're starting from normal cruising flight on the "front side" of the Drag curve, all these relationships work against us, and we'll find that it takes a very large increase in airspeed, Thrust, and Power to achieve a very modest climb rate and climb angle.

The "propeller train" thought experiment makes it clear why we normally enter a climb by allowing the aircraft to pitch up so that the angle-of-attack is not decreased by virtue of the aircraft's rising trajectory, and the thrust line is not pointed downward relative to the flight path, regardless of whether or not we actually choose to increase the angle-of-attack as we enter the climb.

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?

Yes-- with the "propeller train", when the wing is rising, thrust certainly is doing work along the direction of the flight path through the airmass, which is never purely vertical. Here's an analogous situation-- imagine a kite flying on a windy day, with the string at a 45-degree angle. Imagine that we have a lightweight ball that can dangle from a hook that we can clip over the kite string. The purely horizontal force from the wind will make the ball run up the kite string. The wind is doing work that increases the potential energy of the ball, and the propeller of the "propeller train" is doing work that increases the potential energy of the wing.

Related links:

Does lift equal weight in a climb? -- includes vector diagrams

Are there any situations where having high lift but low lift to drag ratio would be beneficial? -- includes a focus on lift and drag coefficients as well as the actual force vectors, in climbing flight

Is excess lift or excess power needed for a climb? -- examines the present question in exhaustive detail, not limited to the steady-state condition of constant climb rate, and includes a modified version of problem where the wing of the "propeller train" is free to pivot to maintain constant angle-of-attack

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  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – quiet flyer Nov 5 at 12:56
  • $\begingroup$ I decided to delete the other answer so as not to distract from the present one, but it is still there those w/ +10K to see-- still contains a few significant insights not included in the present answer, and is entirely accurate as far as I can see-- aviation.stackexchange.com/questions/15243/… $\endgroup$ – quiet flyer Nov 7 at 15:09
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    $\begingroup$ You have been asked before to stop making excessive edits to posts. I am locking this to prevent the same abuse of the system you have previously been notified of. $\endgroup$ – Jamiec Nov 9 at 7:56
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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. $\endgroup$ – user 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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