5
$\begingroup$

from https://aneedforairspeed.wordpress.com/2009/06/27/climbing-descending/

Related to questions on forces in a climb: if any aircraft is climbing with a constant velocity, is the total upwards vertical force (lift included) greater than the weight?

$\endgroup$
16
  • 2
    $\begingroup$ We already have a question that specifically asks "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." See Does lift equal weight in a climb? $\endgroup$ Jun 11 at 4:57
  • $\begingroup$ The body of the question asks about the total upwards force, which might be construed to mean either the net vertical force excluding weight, or the sum of the upward force components. Since at least one aerodynamic force has a downward component, these are not the same thing. The title asks about the total vertical force, presumably meaning the net vertical (aerodynamic?) force, i.e. the sum of all vertical forces other than weight. The title and body could use some editing to be better harmonized and be more clear about which of these things you are asking. $\endgroup$ Jun 11 at 5:43
  • 1
    $\begingroup$ (So consider changing title to read "Is the sum of the upwards forces higher than weight in a steady climb?", or "Is the net vertical aerodynamic force greater than weight in a steady climb?", depending on which you really want to know.) $\endgroup$ Jun 11 at 5:45
  • $\begingroup$ Is Lift Greater Than Weight In A Climb was asked 4 years ago, editing it now may make many of the answers a mismatch. It was active again when this answer was added, initially stating that it was. It received quite some downvotes. It is important to realise in the debate that upwards vertical force is higher than weight in a climb, but lift isn't necessarily, depending on the tilt of the aircraft velocity vector. $\endgroup$
    – Koyovis
    Jun 11 at 14:40
  • $\begingroup$ It's interesting how vague simple words can end up being-- (I guess that's why a force vector diagram is worth a thousand words) -- I'm still not completely clear whether you are asking about the net vertical aerodynamic force, or the net vertical force, or you are interested in adding up all the upward aerodynamic force components (while ignoring the downward aerodynamic force components) and comparing that value to the weight-- my answer aviation.stackexchange.com/a/56476/34686 to the other related question is intended to address all those different cases-- $\endgroup$ Jun 11 at 14:45
5
$\begingroup$

This question is purely a definition issue, and the answer is 'yes' or 'no' based only on which definitions you use. In Newtonian physics, a lot of complex interactions are modelled as single, lumped vectors which we call "forces". These forces share nice properties with vectors: notably, that we can decompose vectors into multiple vectors, or sum multiple vectors into a single vector. One of the main reasons to do so is to decompose a vector into components parallel to some (typically orthogonal) coordinate system.

An important observation is that there is no 'true' way of representing the forces acting on the airframe. While some decompositions are more popular than other, all are equally valid (if done correctly). I will take two examples, one of which arrives at your conclusion 'yes', the other 'no'.

Example 1. Decompose the aerodynamic forces on the airplane, parallel and orthogonal to the flight path. Call one 'lift', call the other one 'drag'. Let's assume 'thrust' is also parallel to the flightpath. Weight is represented as a single vector, orthogonal to Earth, and is not decomposed along the flight path. Now take all the forces that we decomposed along the flightpath, and again decompose them but now orthogonal to Earth. Now, only look at the forces pointing 'up', which in a climb (but not in a descent) removes the vertical component of the 'drag' vector, and compare it to the weight vector. With this elaborate procedure, we can conclude the answer is 'yes'.

Example 2. Combine all aerodynamic forces on the plane in a single vector instead of decomposing them into lift and drag, called the 'net aerodynamic force'. Leave the thrust and weight vector unchanged. Again, we decompose all vectors along the Earth reference frame. Now, we find the sum of all the upward components is exactly the same as weight. We can conclude the answer is 'no'.

enter image description here Note: the net aerodynamic force is shown in the left diagram for illustration only to show that it is the sum of lift and drag, and is not actually part of the force balance for example 1.

$\endgroup$
10
  • $\begingroup$ Would this case be example 2 as well? $\endgroup$
    – Koyovis
    Sep 7 at 19:15
  • $\begingroup$ @Koyovis In that case there is no lift, only drag, so there's no difference in the net aerodynamic force and the decomposed lift and drag vectors. If your climb is steep enough, example 2 changes from 'no' to 'yes'. (1/2) $\endgroup$
    – Sanchises
    Sep 7 at 19:18
  • $\begingroup$ Either way, this highlights why I wonder what the practical value is of your question at all, because the answer changes according to definitions and flight conditions. What exactly inspired this question? (2/2) $\endgroup$
    – Sanchises
    Sep 7 at 19:26
  • $\begingroup$ The downwards pointing drag is there as well in shallow climbs, if small enough like with GA planes it can be neglected. When working with aerodynamics of helicopters & military jets, it is very apparent that this force needs to be considered in questions about lift & climb. $\endgroup$
    – Koyovis
    Sep 8 at 0:36
  • 1
    $\begingroup$ @ymb1 No idea either. I lumped parasitic drag with the rest of aerodynamic forces, because the point here is to show that the definitions you use to separate the forces acting on the airframe determine the final answer, not to elaborate a complete description of all forces acting on the aircraft. $\endgroup$
    – Sanchises
    Sep 8 at 6:26
4
$\begingroup$

Yes.

enter image description here

Vertical forces are forces in the earth axes reference frame. So remaining in this reference frame, assuming $\gamma$$\alpha$ in a steady climb, the vertical forces are:

$$W + D \cdot \text{sin }\gamma - L \cdot \text{cos } \gamma - T \cdot \text{sin } \gamma = 0$$

  • Pointing downwards = + = $W + D \cdot \text{sin }\gamma$
  • Pointing upwards = - = ($L \cdot \text{cos } \gamma + T \cdot \text{sin } \gamma$)

So in order to continue climbing, the total upwards vertical force - consisting of a combination of Thrust and Lift - must be larger than weight by a factor of D * sin $\gamma$

Notes:

  1. Upwards aerodynamic force is often called lift. But lift is defined in the airframe aerodynamic axis, and tilts with the direction of the airspeed. So for a fixed wing aeroplane in a steady climb, total vertical force is higher than weight, but lift is smaller than weight.
  2. If the upwards force changes, the climb speed changes accordingly. There is an acceleration, causing a change in aerodynamic drag which stops when the forces are in equilibrium again.
$\endgroup$
14
  • 1
    $\begingroup$ Only in a constant density atmosphere. In reality the airplane loses climb speed with increasing altitude, so there is a vertical deceleration term which reduces the vertical force ever so slightly. $\endgroup$ Jun 11 at 4:31
  • $\begingroup$ Lift is NOT total Upwards aerodynamic force! It is the component of all aerodynamic forces perpendicular to the plane formed by the wings and the flight path of the aircraft. If in a climb, then the flight path is inclined upwards and not vertical, so lift (if it to have any same meaning) is also not vertical. You CAN define it any way you want I suppose, but defining it as the vertical component when the aircraft is in a steep climb or descent, (or for that matter a steep bank), serves only one purpose, to confuse. $\endgroup$ Jun 12 at 0:11
  • $\begingroup$ @PeterKämpf The plane is also burning fuel, reducing the weight over time as well. The plane in question is climbing at a steady rate. $\endgroup$
    – Koyovis
    Jun 12 at 0:15
  • $\begingroup$ @CharlesBretana Thank you for pointing out the value of asking this question twice, once about lift and once about net upwards force. Mentioned in note 1. in the question. $\endgroup$
    – Koyovis
    Jun 12 at 0:17
  • 1
    $\begingroup$ And what is net upwards force in an F-16 in a pure vertical climb? Do we count the force of the dynamic air pressure on the front surfaces of the aircraft as negative upwards force? what about the force of air pressure on the rear surfaces of the airframe? They are of course pushing the aircraft upwards. Are they "Lift"? Breaking up aerodynamic forces into defined components is done to create understanding and simplify and enable calculations. Applying these arbitrary definitions in scenarios where they do not help to do that only confuses. $\endgroup$ Jun 12 at 0:21
1
$\begingroup$

Yes, always, unless aerodynamic drag does not exist.

For winged aircraft, the above is impossible, therefor the answer is yes, always.

It is important to realize excess thrust is required to climb. Excess thrust closes the weight/lift triangle but does not account for aerodynamic drag, which equals the amount of additional thrust required to maintain airspeed.

Add this "handle" (in the direction of flight) to the closed vector diagram for vertical lift and voila! There is your vertical aerodynamic drag component (decomposed from the aerodynamic drag vector).

Always there, in any climb, for any aircraft.

An aircraft that has a thrust to weight ratio of less than one simply must use a ramp while maintaining airspeed against drag.

The combination of excess thrust and lift support the weight, enabling steady state flight with 0 acceleration (from Gravity), while the remaining thrust at a given velocity opposes aerodynamic drag (part of which is being used to create Lift).

enter image description here

Like this.

In level flight only around 300 lbs of thrust is needed, as the far more efficient wing can now bear all the weight.

$\endgroup$
1
$\begingroup$

If the net forces are zero, the movement will be steady, as per Newton's second law.

If upwards vertical forces equals weight, we will have net zero vertical forces and a no vertical movement (hover).

If the total upward forces are greater than weight we will have a vertical acceleration until drag brings velocity to a steady state.

If all vertical forces = weight the aircraft may be rising, hovering, or descending with 0 acceleration.

$\endgroup$
4
  • $\begingroup$ Thanks Robert for the edits! Could you check again to see if there is a possible conflict between paragraph two and four. If so, I think it is pragraph four that should be kept. $\endgroup$
    – Mats Lind
    Sep 9 at 12:00
  • $\begingroup$ In paragraph 2 "downward" force from drag = 0, so 2 and 4 do not conflict. Please excuse my passion (in editing). Paragraph 4 covers all 0 vertical acceleration cases, indeed, vertical forces essentially make the plane "weightless" leaving the drag from velocity. (Now we can think about airships too). $\endgroup$ Sep 9 at 12:04
  • $\begingroup$ Ah, I see, drag has a downward component in steady climb. And we presume it is a climb. And we exclude downward compents from upwards vertical forces. Then its the slowest climb = hower. Couldn't it still give the reader the expression that zero force always have to give a standstill and not as in Newtons second a steady movement? $\endgroup$
    – Mats Lind
    Sep 9 at 12:16
  • $\begingroup$ An object in an environment with drag and 0 net force will slow to 0 velocity. (Unless drag force is included!). Since we are comparing all upward force to weight, if there is upward velocity, there is a downward drag component, therefor... $\endgroup$ Sep 9 at 13:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.