-2
$\begingroup$

There has been some discussion in recent questions regarding climbing technique and optimization of the Lift vector (and the resultant vertical Lift vector).

Some writers claim the lift vector must be smaller in a climb.

This does not seem to make sense, as the wing is more efficient than the propeller at generating force, by a factor of 4, for an average GA aircraft.

Although there is more drag at higher speeds, is it not better to climb at a higher speed with a larger Lift vector than it is to pitch up and use the engine thrust to climb?

Climb and distance Vy vs Vx, and glide distance Vbg vs Vmin sink seem to support this concept.

Can I maintain optimal Lift/Drag AOA at a higher airspeed by reducing elevator trim?

$\endgroup$
  • $\begingroup$ "Although there is more drag at higher speeds, is it not better to climb at a higher speed with a larger Lift vector"-- why assume that more airspeed corresponds to a larger Lift vector? This is only true if the angle-of-attack is constrained to be constant. Such a constraint is not consistent with the basic flight dynamics of free flight. I think the question can be improved. It's hard to be sure exactly what you are asking, other than questions that have already been asked like "Is Lift larger than Weight in a climb?" $\endgroup$ – quiet flyer Apr 11 at 16:43
  • $\begingroup$ We need a question about "what does elevator position most directly control: angle-of-attack, or pitch attitude?" There may already be one. $\endgroup$ – quiet flyer Apr 11 at 16:48
  • $\begingroup$ Anyway the answer to the last sentence is "no, starting at a given weight and a given airspeed, and a given thrust setting, if we change the elevator position to improve the L/D ratio, the airspeed will change as well, as will the climb angle." Does that help? Waiting till question is stabilized to post as an actual answer. $\endgroup$ – quiet flyer Apr 11 at 16:50
  • 1
    $\begingroup$ Sorry to aggravate you; just you know it wasn't my dv. I've sworn off dv'ing on ASE. $\endgroup$ – quiet flyer Apr 11 at 17:48
  • 1
    $\begingroup$ I never DV posts either. This isn't PSE, which is a political cesspool dominated by a faction that bullies dissenters off the channel. Unless a post is obviously a troll or someone looking for others to do their homework for them, any question is good with me and if I don't like a question I just ignore it. $\endgroup$ – John K Apr 11 at 19:12
2
$\begingroup$

Here is one of those writers who claim that lift is smaller in a climb than in straight flight.

With good reason.

The sum of all vertical forces will be zero in a stationary climb or the aircraft would accelerate. It is a fallacy to assume that lift is higher at higher speed – lift is just as large as is needed to equalize all other forces. Now it is important to note that the aerodynamic coordinate system (in which lift is defined as the aerodynamic force parallel to the Z axis) is tilted, pointing slightly upwards with its X axis parallel to the flight path. Drag is slightly pointing down and lift is slightly pointing back when seen from an earth-fixed observer.

To equalise those forces, we have only thrust (at least in the absence of wind – or this would be the special case of a glider climbing in a thermal). This thrust must now compensate both the drag force and the backward component of lift. It is, therefore, higher by that backward component of lift compared to straight and level flight. And it is also pointing slightly up due to the climb attitude of the aircraft. Therefore, this upward-pointing thrust component (which is larger than what is needed to compensate the downward pointing drag component) works against gravity and helps the lift in compensating weight. Lift can, therefore, be a bit smaller than in straight and level flight.

Forces and their angles acting on a climbing aircraft in side view

Forces and their angles acting on a climbing aircraft in side view. The light, translucent arrows are copies of the solid arrows shifted into a vector chain to show that they indeed equalize to zero net force.

The extreme case is a vertical climb: Now all the force counteracting weight is supplied by thrust and lift is zero, or the aircraft would accelerate sideways. Climb is simply a linear combination of level flight and a bit of vertical climb.

Now you could argue that the B-52 climbs with its nose pointing down, so thrust must also point down a bit. Maybe, but in level flight that thrust would point even more down and more lift would be required to compensate for that downward thrust component. No matter how you look at it, as soon as the flight path angle pitches up, so does the thrust, thereby reducing the required lift.

Now you argue that climbing at a faster speed is better. But that would create more drag, leaving less excess power for climbing. Also, when flying faster, the aircraft must reduce its angle of attack in order to produce the right amount of lift. Climbing is not about lift maximisation, but about excess energy maximisation.

| improve this answer | |
$\endgroup$
  • $\begingroup$ What Peter is saying is climb is a function of thrust that is in excess of the minimum required for level flight and the maximum thrust surplus occurs at best rate, near maximum L/D. Any faster is less efficient in terms of climbing from x to y altitude with the minimum additional energy. $\endgroup$ – John K Apr 12 at 4:02
  • $\begingroup$ Actually, sustained wind and updraft don't change the geometrical orientation or magnitude of the forces either as long as the airmass is locally uniform on a large enough scale for the glider to come to equilibrium. The only affect the trajectory as seen from the ground. Content in first set of parentheses could simply be omitted. Really have to get that diagram added to my answer to related question. Ditto to Robert's question about want to understand the vectors for Vbg. $\endgroup$ – quiet flyer Apr 12 at 4:32
1
$\begingroup$

No. There is a direct correlation between airspeed and AoA. In level, balanced unaccelerated flight you cannot change airspeed without affecting the angle of attack.

L/D max AoA will occur at a certain airspeed for a given weight. It is not possibly to fly a different airspeed at the same AoA unless you are in accelerated flight. (I.e. pulling Gs in a turn)

| improve this answer | |
$\endgroup$
0
$\begingroup$

Can I maintain optimal Lift/Drag AOA at a higher airspeed by reducing elevator trim?

No, starting at a given weight and a given airspeed, and a given thrust setting, if we change the elevator position to change the angle-of-attack to change the lift coefficient and drag coefficient in a way that improves the L/D ratio, the airspeed will change as well, as will the climb angle or descent angle. There is no other way to end up with closed vector polygon of Lift, Weight, Drag, and Thrust. For a given thrust or power setting and a given weight, if you use the elevator to change angle-of-attack, you will end up with a change in airspeed.

For shallow climb or descent angles, in sustained linear flight, it's a good approximation to say that there's a one-to-one correlation between angle-of-attack and airspeed, or between angle-of-attack and L/D ratio (excluding the "back side" of the polar curve, or between airspeed and L/D ratio (excluding the "back side" of the polar curve). Changing any one of these things will automatically change the others.

For very steep climb or descent angles it gets a little more complex because the amount weight supported by the wing is less than in level flight, so the airspeed associated with any given angle-of-attack and L/D ratio will be less than it was in horizontal flight, just as if we had changed the weight of the aircraft. But even in a very steep climb or dive, for a given power or thrust setting, we can't choose between different L/D ratios (and different resulting climb or dive angles) for the same airspeed, unless we are changing the shape of the airfoil or aircraft by deploying flaps, spoilers, etc.

Nor can we have our pick of different airspeeds for the same angle-of-attack and thrust or power setting.

| improve this answer | |
$\endgroup$

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