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This subject keeps coming up in the discussions and questions such as this one, which asks if lift equals weight in level flight. Good answers there, pointing out that upwards force has many sources. But also some that need clarification.

enter image description here

It is also mentioned at several places on this Aviation SE site, in question & comments, that lift always equals weight if the aeroplane is not accelerating upwards or downwards, since only an acceleration requires extra force according to Newton.

With zero wind, lift is always defined as the force perpendicular to the flight path, but gravity does not tilt with the aircraft axes. My question therefore is also about the sum of all vertical forces: in a steady climb, is the total upwards vertical force from all sources (wing, tail, engines, fuselage) larger than, or equal to the weight of the aircraft.

  • If larger: please quantify.
  • If equal: please explain why.

Update

TL;DNR It depends on how the aircraft climbs. Sometimes yes, sometimes no.

Related to this question and the question with the propeller train. The situation is quite complicated, as @PhilFrost answer indicates: when climbing due to nose up attitude the reference axes change, and cosines and Euler angles are introduced etc. The clear situation of aerodynamics, which regards from the viewpoint of the aircraft axes, is now gone. From this viewpoint, the two questions referenced above have resulted in very good answers. Yes for small angles the vertical component of thrust grows quicker than the vertical component of lift decreases - but is the total vertical force greater than or equal to weight?

Lets regard the aircraft then with unchanged angle. Situation 1: level flight in cruise. Situation 2: aircraft attitude unchanged compared to situation 1, but now climbing due to increased power setting and speed. What happens:

  • The aircraft accelerates at first, until $\Delta{drag} = \Delta{thrust}$
  • The increase in speed causes an increase in aerodynamic lift, and the aircraft climbs.
  • The climb causes a change in local angle of attack of the wing, lowering the lift force. This is an auto-stabilising feedback.
  • The aircraft is now in a steady climb - with a vertical speed component.

So there is the thing, there is now a vertical aerodynamic drag component which needs to be overcome by aerodynamic lift. Not by a great amount, but aerodynamic lift in situation 2 is greater than in situation 1. That is why I've accepted xxavier's answer.

Update 2

The propeller train question filters the problem down to a situation where none of the confusing Euler angles need to be accounted for. I've run a real-time simulation of the situation in this answer. Lift is higher than weight in a climb due to increase in airspeed.


Addition

An answer.

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    $\begingroup$ I would contest that list is not always defined as a force perpendicular to the flight path. $\endgroup$ – J Walters Jul 12 '17 at 16:33
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    $\begingroup$ Related: aviation.stackexchange.com/questions/12714/… $\endgroup$ – Peter Kämpf Jul 12 '17 at 17:03
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    $\begingroup$ "there is now a vertical aerodynamic drag component with needs to be overcome by aerodynamic lift" But this is wrong. Climbing requires less lift. The additional upwards force to overcome drag comes from thrust. In fact, any attempt to climb without reducing lift can not result in a statically stable configuration. $\endgroup$ – Phil Frost Jul 13 '17 at 13:57
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    $\begingroup$ You are supposing that additional drag can be overcome by additional lift. But this is not possible under the conventional definition where lift is always orthogonal to lift. If lift is "up" by definition, then drag is "to the rear". You can't offset a force pushing to the rear by pushing up. $\endgroup$ – Phil Frost Jul 13 '17 at 15:11
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    $\begingroup$ This question needs to be improved. The title does not match the body of the question. The sixth sentence is "My question is purely about the sum of all vertical forces: in a steady climb, is the total upwards vertical force from all sources (wing, tail, engines, fuselage) larger than, or equal to the weight of the aircraft." That is not a question about lift, that is a question about net vertical force. The title is misleading. $\endgroup$ – quiet flyer Oct 24 '18 at 1:39
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In an aircraft that is climbing at a constant vertical velocity, the total of the upward-directed vertical forces is the same as the total of the downward-directed vertical forces.

Were it not so, the vertical velocity would not be constant, since any non-zero balance of the vertical forces would result in an acceleration...

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  • $\begingroup$ By total of downward forces, do you mean the weight? $\endgroup$ – Koyovis Jul 12 '17 at 10:13
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    $\begingroup$ @Koyovis Not only the weight. The downward component of drag is a force to be reckoned with too... $\endgroup$ – xxavier Jul 12 '17 at 10:21
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    $\begingroup$ Which means that the total upward force is greater than the weight if the airplane is rising, because, as a bare minimum, there is always a downward component to drag in addition to the weight in that case. $\endgroup$ – Mad Physicist Jul 12 '17 at 19:17
  • $\begingroup$ @MadPhysicist Exactly right. Lift vector tilts back, therefore drag vector tilts back, creating a downwards pointing component of aerodynamic drag. $\endgroup$ – Koyovis Aug 6 '17 at 1:02
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It depends on exactly how you define "lift" and "weight". You might say intuitively that lift is all the forces acting on the aircraft in the upward direction, like this:

enter image description here

In this case, lift must equal weight, otherwise the aircraft would be accelerating. That is, it's rate of climb would be changing.

But it's more usual to define lift this way:

enter image description here

Here, lift and weight are equal in magnitude, but in different directions. Of course lift doesn't need to be equal in magnitude: it can be adjusted by the angle of attack. But let's suppose lift is equal to weight and see what happens.

Let's do all our calculations with Earth as the frame of reference1. It's useful to decompose lift into a sum of vertical and horizontal components so we can analyze the horizontal forces and the vertical forces separately:

enter image description here

Comparing the vertical component of lift with weight, we can see they are not equal:

enter image description here

Considering only the vertical forces drawn here, there is a net downward force on the aircraft. So why then is the rate of climb not decreasing?

A similar transformation happens to thrust. In a climb, thrust provides an additional upwards component. And of course we must also consider drag. Point being in a steady climb, lift (by the conventional definition) is not equal to weight, but the sum of all the vertical components of lift, thrust, and drag do equal weight.

Let's add an arbitrary amount of drag, and enough thrust to balance the vertical forces.

enter image description here

Now the vertical forces are balanced, but the horizontal forces must also be balanced if we want stable flight. Adding all the horizontal forces in my drawing, there's a net force to the left. So this aircraft may be maintaining a steady rate of climb at this instant, but it's losing speed and probably headed for a stall.

enter image description here

Remember, we initially set lift equal in magnitude to weight, and this is what happens. Without changing the direction or magnitude of lift, there's no solution that results in stable flight.

Therefore, a climbing aircraft requires less lift. To maintain this direction and velocity, this pilot must reduce lift by reducing the angle of attack, and increase thrust such that the vectors add to zero and there's no net force on the aircraft. Reducing lift will also reduce drag.

enter image description here


1 Any other frame of reference could work. For example we could use the aircraft as the frame of reference, which would mean lift is always up, but weight would change direction.

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  • $\begingroup$ So we're transforming from an aircraft frame of reference to an earth frame of reference, correct? $\endgroup$ – Koyovis Jul 12 '17 at 13:34
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    $\begingroup$ You could do both. When you use the aircraft frame of reference in a climb, aircraft down is not aligned with gravity. The gravity vector in that case would then point slightly towards the tail (as much as the lift vector is from the earth as a point of reference). Using earth frame of reference might be more common though since gravity vector then is mostly constant. $\endgroup$ – Adwaenyth Jul 12 '17 at 14:55
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    $\begingroup$ I don't think there's any particular reason why the lift in your second image would have to have the same magnitude as the weight. After all, it can be modulated by varying the pitch (and so the angle of attack), just as in horizontal flight. $\endgroup$ – Henning Makholm Jul 12 '17 at 16:35
  • $\begingroup$ @HenningMakholm Right. I just arbitrarily picked something to illustrate. $\endgroup$ – Phil Frost Jul 12 '17 at 17:40
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    $\begingroup$ Perhaps you should edit your answer to reflect Makholm's comment - right now, the answer can be very confusing, leading readers to believe lift should always equal weight. $\endgroup$ – Sanchises Jul 12 '17 at 18:01
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Short answer: No.

Long answer: When the flight path is not horizontal, lift will not be vertical but perpendicular to the direction of motion (in still air). Thrust will also have a vertical component and is different in magnitude from drag, because excess thrust is needed to increase the potential energy of the plane. Note that the vertical component of lift is proportional to the cosine of the flight path angle while the vertical component of thrust is proportional to the sine of the flight path angle, so the thrust part grows more quickly at small flight path angles. Therefore, when climbing, thrust will add some vertical component, so less lift is needed.

Again, in a descent less lift is needed. Now thrust is smaller than drag, and drag, pointing slightly upwards, contributes a vertical component, counteracting weight. So in both cases lift is smaller than weight.

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  • $\begingroup$ Likewise, when descending, drag will add some vertical component to oppose gravity. The lift component for either climb or descent will be equal to weight x cosine(angle from horizontal). However in the second part of the original question, if the aircraft is not accelerating, then the net force on the aircraft is zero. Gravity would be opposed by some combination of lift and/or thrust, and/or drag. $\endgroup$ – rcgldr Jul 12 '17 at 14:38
  • $\begingroup$ @HenningMakholm: I linked to an answer which has a diagram quite like you wish. I was told not to repeat myself and to better link to existing answers. $\endgroup$ – Peter Kämpf Jul 12 '17 at 17:03
  • $\begingroup$ Is vertical component of thrust the sine of the flight path angle or of aircraft attitude relative to earth? $\endgroup$ – Koyovis Jul 13 '17 at 23:40
  • $\begingroup$ @Koyovis: Only if thrust is aligned with the aircraft's longitudinal axis. Roughly it is, but there can be a few degree of difference. $\endgroup$ – Peter Kämpf Jul 14 '17 at 8:59
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If we define lift as the component of the total aerodynamic forces on the aircraft that is perpendicular to its direction of motion, then lift will be slightly smaller in a stable climb.

It is probably easiest to analyze the situation in a coordinate system that is tilted such that one of the axes is parallel to the direction of motion. Then all of the forces -- lift, drag, thrust -- work just like in an ordinary coordinate system in horizontal flight. The only difference is that the weight force now has a different direction -- but still the same magnitude.

This means that the component of weight that is perpendicular to the motion is now slightly smaller, and lift must be correspondingly smaller too. The plane's angle of attack will be slightly smaller than in level flight at the same (calibrated) airspeed.

On the other hand, the weight vector now gains a significant component parallel to the direction of motion, and this has to be counteracted by more thrust, lest the aircraft would slow down. (This will much dominate over the small decrease in induced drag that results from the slightly smaller lift).

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  • $\begingroup$ Yes seen from the aircraft reference frame, weight now decomposes into lift and drag. $\endgroup$ – Koyovis Aug 7 '17 at 10:25
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It depends on the relative axis orientation.

  • Gravity is always aligned with earth axes.
  • For fixed wing aircraft, lift and drag are aligned with the airflow axes (aligned with the airflow in a steady state starting position). Note that thrust is only aligned with drag at AoA zero.

The thing is that for fixed wing aircraft, a steady state climb is mostly and automatically associated with increasing AoA, which tilts the aircraft axes upwards, resulting in an upward tilt of the airflow axes. But fixed wing aircraft can also climb by increasing speed, which results in a steady state climb with a reduced AoA.

Below is an analysis of the two cases for fixed wing climb, and for helicopters where the airflow axes are rotating with the blades - providing lift, not thrust.

TL;DNR

  • Fixed wing climb by increasing AoA: modulus of the lift vector < gravity vector
  • Fixed wing climb by increasing velocity: mod lift > mod g
  • Helicopter in steady climb: mod lift >> mod g

enter image description here

  • Lift L at angle $\alpha$
  • Drag D at angle $\alpha$
  • Thrust T at angle $\phi$
  • Weight W at the vertical

Force equilibrium in unaccelerated flight:

$$ T\cdot cos(\phi) = L\cdot sin(\alpha) + D\cdot cos(\alpha) \tag{H} $$ $$ L\cdot cos(\alpha) + T\cdot sin(\phi) = D \cdot sin(\alpha) + W \tag{V} $$

Equation (V) states that the total upwards vertical force is equal to weight plus a component of aerodynamic drag - of the whole aircraft, wing + fuselage + tail etc. So the total upwards force will always be greater than weight, unless $\alpha$ = 0

Let's have a look at a couple of cases.

1. Climb due to increase in speed, fixed wing

A case pointed out some time ago by Chris, who defined totally uncoupled thrust and lift forces by putting a wing on a pole mounted on a train car. If thrust increases, speed will increase and the wing will climb upwards with a constant velocity $V_z$. This will change angle of attack, and will tilt the lift vector backwards. The wing climbs with constant velocity once the total upwards vertical force is identical to weight, plus the vertical component of drag which points downwards.

enter image description here

Note that thrust is nowhere to be seen in this picture, only aerodynamic forces. Thrust is set at angle $\phi$ = 0 and will be equal to L * sin($\alpha$) + D * cos ($\alpha$). Lift L is tilted backwards by angle $\alpha$, and is greater than the vertical upward force by factor $1/cos (\alpha) $.

So in this case (climb by increase in speed):

  • Total upward force is greater than weight by an amount of D * sin $\alpha$.
  • Lift is the only contributor to upward force, is tilted back, and is greater than total upward vertical force.

2. Climb due to aircraft pitch up, fixed wing

Now let's have a closer look at the case of a fixed wing aircraft, climbing because of an increase in pitch angle. All the above forces and both equation (H) and (V) are to be considered. Angle of attack $\alpha$ is defined by pitch angle $\phi$, airspeed V, and climb speed $\dot{z}$.

So in this case:

  • Total upward force is again greater than weight by an amount of D * sin($\alpha$)
  • Both thrust T and lift L are contributors to total upward force. How much each contributes depends on pitch angle $\phi$ and climb speed $\dot{z}$.

3. Helicopter in vertical climb

Now for the helicopter in climb. At first glance, this is a case of only thrust being responsible for the climb action, because the rotor disk delivers vertical thrust downwards. But here is the thing: that is from a fuselage perspective, but now lift is defined relative to the rotating blade airspeed.

Our reference frame is once again earth axes. The vertically climbing helicopter has the same downwards aerodynamic force as the hovering helicopter, except for minor increases due to the vertical drag of the fuselage. The pilot transitioned the helicopter from hover to climb by pulling on the collective, increasing blade pitch and tilting the lift vector backwards (earth axes).

enter image description here

The vertical component of lift is equal to weight plus the downward vertical component of (blade drag + vertical fuselage drag). Lift is greater than its vertical component by a factor of 1/cos $\phi$.

So in this case (climb by increase in pitch):

  • Total upward vertical force is greater than weight by an amount of (D * sin($\alpha$) + vertical fuselage drag).
  • Lift is the only contributor to upward vertical force and is tilted back, so lift is greater than total vertical force by factor 1/cos($\alpha$).

Conclusion

Case 2. is considered multiple times on this site. Aerodynamic lift may be less than weight, depending on relevant angles and speeds. Thrust must always be higher than in steady horizontal flight by an amount of L * sin($\alpha$).

All cases have a higher upward vertical force than weight: a vertical aerodynamic drag component must be compensated for.

@xxaviers answer is accepted. Many other answers are also correct for a steady state fixed wing climb due to aircraft axes tilt relative to gravity.

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  • $\begingroup$ Good analysis! Indeed, lift is bigger than weight if thrust is not tilted (case 1) or is coupled with lift (case 3). In case 2, lift is smaller than weight if the vertical component of thrust is greater than the vertical component of drag. This is easily accomplished in almost all but the most extreme cases of downthrust because in a climb thrust is much bigger than drag in order to add potential energy to the plane. In a descent the reverse is true, $\endgroup$ – Peter Kämpf Aug 7 '17 at 10:34
  • $\begingroup$ and again lift becomes smaller than weight (because now the small thrust points down and the bigger drag points up). $\endgroup$ – Peter Kämpf Aug 7 '17 at 10:34
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The title of the question is misleading.

In the body of the question we read-

"My question is purely about the sum of all vertical forces: in a steady climb, is the total upwards vertical force from all sources (wing, tail, engines, fuselage) larger than, or equal to the weight of the aircraft."

Obviously for acceleration to be zero, net force must be zero, so net vertical force must be equal to weight. This is not a very interesting question.

The title asks a completely different question: "Does lift equal weight in a climb?" This is a much more interesting question.

In the context of fixed-winged flight, Lift is defined to act perpendicular to the flight path through the airmass, and Drag is defined to act parallel to the flight path through the airmass. For the purposes of the answer, we'll assume that Thrust acts parallel to the flight path through the airmass, although this clearly not always exactly true. This simplifying assumption leads to the following vector diagram:

Powered climb at climb angles of 45 and 90 degrees:

Powered climb at climb angles of 45 and 90 degrees

In the vector diagrams above, "angle c" is the climb angle-- it is 45 degrees in the left-hand figure, and 90 degrees in the right-hand figure.

We can see that in a powered climb, Lift = Weight * cosine (climb angle), where the climb angle is measured relative to the airmass (an important distinction in the case of gliding flight-- an unpowered climb in a thermal updraft is still a descent in relation to the airmass!)

Clearly, Lift is less than Weight in a powered climb. For example, if the climb angle is 45 degrees, Lift = .707 * Weight. If the climb angle is 90 degrees, Lift must be zero.

The same is also true in a descent-- Lift = Weight * cosine (descent angle), so Lift is less than Weight. This is explored in more detail in some of the links given at the end of this answer.

Note that we've taken the approach of combining the Thrust and Drag vectors into a single (Thrust-Drag) vector, and then we've arranged this vector into a closed vector triangle with Lift and Weight. Whenever vectors can be arranged nose-to-tail into into a closed polygon-- a triangle in this case-- this shows that net force must be zero, meaning that acceleration is zero and velocity i constant. For clarity we've also drawn the individual Thrust and Drag vectors outside the vector triangle. These are redundant with the (Thrust-Drag) vector.

Varying the climb angle and/or the L/D ratio:

Varying the climb angle or the L/D ratio

Note that for a given aircraft in a given configuration, any given angle-of-attack is associated with specific values for the lift coefficient, drag coefficient, and ratio of lift coefficient / drag coefficient. Lift is proportional to lift coefficient * airspeed squared, and drag is proportional to drag coefficient * airspeed squared, so the ratio of the lift coefficient / drag coefficient is also the ratio of Lift / Drag. So for a given aircraft in a given configuration, any given angle-of-attack is associated with specific ratio of Lift to Drag.

If the left-hand diagram above and the middle diagram above both represent the same aircraft in the same configuration, then the aircraft must be flying slightly slower in the middle diagram. That's the only way the L and D values can both be slightly smaller, for the same L/D ratio. Adding power to increase the climb angle, while holding the angle-of-attack constant, makes the airspeed decrease slightly. However in the case illustrated here, the change in airspeed would be too tiny to ever notice in practice-- it would be equal to the square root of the change in the value of the magnitude of the lift vector or drag vector.

If all the diagrams represent the same aircraft in the same configuration of flaps etc, then the right-hand diagram (5:1 L/D ratio) would represent a lower angle-of-attack than the left-hand or middle diagrams (10:1 L/D ratio). (We'll ignore the other possibility that the 5:1 case represents mushing flight very near stall, where drag is very high.) A lower angle-of-attack means a lower lift coefficient, yet the size of the lift vector is the same, so the airspeed must be higher in the case illustrated in the right-hand diagram. Therefore the climb rate is also higher. In short, when we increase thrust to increase our climb rate, we also must reduce angle-of-attack, if for some reason we wish to keep our climb angle constant rather than allowing it to increase.

Powered climb at 45-degree climb angle at 8 different ratios of Lift to Drag:

Powered climb at a 45-degree climb angle at 8 different ratios of Lift to Drag

Note that as we decrease our L/D ratio, it takes more and more thrust to maintain the same 45-degree climb angle. In the case where the L/D ratio is 2/1, thrust must actually be greater than weight! This is a bit counterintuitive, as we could obviously climb straight up with some small but non-zero airspeed if thrust were only slightly greater than weight. However, that vertical climb would be conducted at a very low airspeed. In the diagram above, if all cases represent the same aircraft in the same configuration, by constraining the climb angle to be constant, so that L also must remain constant, we're constraining the airspeed to get progressively higher and higher as we reduce the angle-of-attack, lift coefficient, and L/D ratio. Hence the huge increase in drag, and thrust-required, as we reduce the angle-of-attack, lift coefficient, and L/D ratio.

As we explore climb angles closer and closer to 90 degrees, the L/D ratio has less and less influence on the thrust required. A figure similar to the one above, but for a climb angle of 60 or 70 degrees, would show less increase n thrust-required as we decrease the angle-of-attack, lift coefficient, and L/D ratio than we see at a climb angle of 45 degrees. This also implies that we're forcing less of an increase in airspeed as we decrease the angle-of-attack, lift coefficient, and L/D ratio in such a case. That makes sense-- as thrust carries more and more of the aircraft weight, the dynamics of the wing have less and less influence upon airspeed. In the case of a truly vertical climb, the wing must be at the zero-lift angle-of-attack and the L/D ratio must be zero. In such a case, of course, the drag force still varies with airspeed, and so the faster we want to fly straight up, the more thrust we need.

For the sake of clarity, this answer has focussed on some rather steep climb angles. It's important to also keep in mind that for shallow climb angles (or descent) angles that are typical of general aviation light aircraft, the cosine of the climb angle is not much smaller than 1, and so Lift is nearly equal to Weight (specifically, Lift is only slightly less than Weight.) Since Weight doesn't vary with climb or dive angle, we can conclude that for shallow climb or dive angles-- with no other accelerations going on (specifically, 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 nearly constant, regardless of whether the aircraft is climbing, descending, or neither. This means if the climb or descent angle is shallow and the net G-load is one, the airspeed indicator can also be interpreted as an angle-of-attack gauge. Why should this be so? To keep lift nearly constant, it must be approximately true that the lift coefficient is varying in inverse proportion to the square of the airspeed. This establishes a nearly fixed relationship between airspeed and angle-of-attack, for shallow climb or descent angles and net G-loadings near one. If the airspeed is low, the lift coefficient and angle-of-attack must be high, and if the airspeed is high, the lift coefficient and angle-of-attack must be low, regardless of whether the aircraft is climbing at a shallow angle, descending at a shallow angle, or flying horizontally. So the airspeed indicator is in essence an angle-of-attack gauge at shallow climb or descent angles. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- a given angle-of-attack will be associated with a lower airspeed than in horizontal flight, and a given airspeed will be associated with a lower angle-of-attack than in horizontal flight. In the most extreme case where the aircraft is climbing straight up, lift must be zero, so the lift coefficient must be zero, and the 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. Clearly the airspeed indicator cannot serve "double duty" as a guide to angle-of-attack in such a situation.

We've also assumed throughout this answer that the Thrust vector acts parallel to the flight path through the airmass. Obviously, if this isn't true, then the 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. At any rate, it's best to thoroughly understand the simple case where the thrust vector acts parallel to the flight path, before going on to consider more exotic cases.

To see a vector diagram of the forces in climbing flight from an outside reference source, see the diagram below. This diagram shows the same relationships as the other diagrams included in this answer, but the forces have not been arranged into a closed vector polygon, so it is less obvious that the net force is zero.

forces in climb

Above is a vector diagram showing the forces in a stabilized, linear, constant-airspeed climb -- from https://systemdesign.ch/wiki/L%C3%B6sung_zu_Steigflug

FS = thrust

FW = drag

FGp is the component of weight that acts parallel to the flight path, and is ALSO exactly equal in magnitude and opposite in direction to (thrust - drag)

FGs is the component of weight that acts perpendicular to the flight path, and is ALSO exactly equal in magnitude and opposite in direction to lift.

FA = lift

FG = weight

Angle beta is the climb angle-- the angle between the flight path and the horizon.

See these related answers to related questions:

"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?"

"Is excess lift or excess power needed for a climb?"

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  • $\begingroup$ Recent changes in this answer were motivated by my decision to delete another answer from another related question-- I noticed it had some content that fit well here. $\endgroup$ – quiet flyer Nov 13 '18 at 5:16
  • $\begingroup$ The first of the 5 links given at the end of this answer has a typo-- the url should be aviation.stackexchange.com/questions/56352/… $\endgroup$ – quiet flyer Nov 13 '18 at 15:00
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No the lift will not equal the weight for an aircraft climbing (at constant velocity).

I cannot draw where I am at, so bear with me.

For an aircraft travelling at constant velocity, not undergoing acceleration, either vertically or horizontally, the lift generated by the wing will be less than the aircraft weight. You can see that the lift component will be less than the weight vector as you increase the angle of attack. For example at a climb angle of 45 degrees, the lift component will equal square-root(2)/2 of the weight (or roughly 71% of the weight).

So how can the aircraft continue in a straight path upward? The engines provide thrust that applies a force equal to the difference in lift and weight. You can see this if you draw a force-balance diagram (which I will try to do later).

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