What do you call a fall that is due to too little engine power, not due to cessation of aerodynamic lift?

Question

If you raise your plane's pitch but the engines can no longer counter the gravity and you begin to fall (because the engines are too weak, not because of too little lift), what do you call that? This could also happen on planets without atmospheres (to rocket-propelled craft). Is it called a "stall" too, a different kind of stall? Or doesn't it have any special designation?

Answer 1

A "stall" may be terminology going back to the very origins of flight, when early aviators did not fully understand why there was a massive increase in drag (and accompanying loss of airspeed) when AoA exceeded a certain limit.

The AoA definition lives on today, but it is interesting to consider airspeed loss due to insufficient thrust or excessive vertical pitch.

The result of airspeed loss under more benign conditions is simply the mechanism of static stability. A properly designed aircraft will pitch down before it stalls to maintain airspeed. This is a function of CG placement and tail volume. As the aircraft loses airspeed, it's flight path should begin to curve downwards, gaining airspeed.

In more extreme cases, such as a failed loop, the aircraft can lose all airspeed and even fall backwards. But because it will then be highly directionally unstable, manuevers such as the "hammerhead" will allow safe recovery.

The inability to climb on all worlds is "insufficient thrust", and is also defined as "service ceiling".

Answer 2

A stall is a stall. When the angle of attack exceeds critical, turbulent drag rises sharply and lift no longer rises with it. It doesn't necessarily imply that you fall; a kite (like a common child's toy) normally flies in a stalled condition.

A stall can't happen without atmosphere, because the term implies that aerodynamic lift and drag are changing and that angle of attack exists (which it doesn't without air to flow over the wing).

Answer 3

In this case it's simply called: fall.

In aviation, stall is associated with lift-driven machines, and it seems from your question that you are not talking about this type of machinery.

You can find the definition of stall in this pdf document: Glossary of Terms from Flight Research: Problems Encountered and What They Should Teach Us, by Milton Thompson with J. D. Hunley, NASA:

Stall - loss of lift due to an acute wing angle, often caused by insufficient speed
stall - A condition in which an aircraft or airfoil experiences an interruption of airflow resulting in loss of lift and a tendency to drop.

Notice that both definitions involve lift loss.

You can find many scientific papers that use the term "fall".

Answer 4

"If you raise your plane's pitch but the engines can no longer counter the gravity and you begin to fall (because the engines are too weak, not because of too little lift), what do you call that?"

At this moment you've described, you DO have "too little lift"- otherwise you wouldn't be falling. If the wing is stalled (not flying) due to insufficient relative wind going over the airfoil (for whatever reason) you are in a stalled condition.

An airplane at altitude doesn't need an engine or thrust to fly, it only needs a pilot to keep the nose (pitch attitude) low enough that the wing's critical angle of attack is not being exceeded, thereby ensuring that the airflow remains attached to the upper surface of the wing creating lift. Same as a glider.

A falling airplane is considered to be in the "stalled" condition, regardless of how or why the pilot mismanaged the airspeed or pitch attitude to put the aircraft in that condition. If you put the nose down it will become un-stalled once you reach the wing's minimum flying speed (exceed stall speed).

Answer 5

If you raise your plane's pitch but the engines can no longer counter the gravity and you begin to fall (because the engines are too weak, not because of too little lift), what do you call that

Well, for starters, a steady-state descent happens when thrust is less than drag, but equal to drag minus the component of the weight vector that is acting parallel to the flight path. A steady-state climb happens when thrust is greater than drag, but equal to drag plus the component of the weight vector that acts parallel to the flight path. (Also, to maintain the steady-state condition, the lift vector must be appropriate as well-- the appropriate value is equal to weight * cosine (climb angle or glide angle). Otherwise, the flight path will curve up or down-- which will typically change the airspeed as well, by upsetting the balance between thrust, drag, and the component of the weight vector acting parallel to the flight path.

Is a steady-state descent a type of falling? Is a steady-state descent which happens to be straight downward, with the wings at the zero-lift angle-of-attack and the airspeed at terminal velocity, a type of falling?

Or does "falling" only refer to a downward curve of the flight path? Or is a better definition any situation where the aircraft's vertical speed is growing more negative or less positive? Or is it simply any situation where we find ourselves to be travelling rapidly towards mother earth?

Your question doesn't appear to be referencing a steady-state condition. Maybe what you mean by "fall" is the situation where you gain some extra airspeed and then you arc the flight path up into a steep climb and try to keep the airplane climbing along that line, but you don't have enough thrust to maintain airspeed-- since to keep the airspeed constant, the thrust must oppose not only the drag force, but also the component of gravity which is acting parallel to the flight path. Otherwise, something has to "give". Now the situation could unfold in several different ways, depending on the pilot's control inputs.

1. The pilot keeps pulling the control stick back to try to maintain the linear climb, and eventually the wing is placed at the stall angle-of-attack and the wing stalls.

2. The pilot keeps pulling the control stick back to try to maintain the linear climb, but as he approaches the max L/D ratio he realizes this isn't going to work so he refrains from pulling the stick back any further, so the wing's angle-of-attack remains constant after this point. As the airspeed (and therefore the lift vector) decreases further, the flight path eventually starts arcing downward. What happens next could range from a gentle transition back to steady-state flight along a more sustainable flight path, be it climbing, descending, or level, or a rather abrupt pitch-over from a steep climb to a steep dive, depending on how extreme the initial climb was. There certainly could be a "falling" sensation involved, even if the wing never reaches the stall angle-of-attack (and in some cases even if the aircraft never actually enters a descent). Think of a rocket being blasted straight up on a vertical trajectory, without enough energy to escape the atmosphere-- what happens at the top of that trajectory might be quite exciting, even if it has nothing to do with a wing reaching the stall angle-of-attack.

3. At some point after establishing the linear climbing trajectory, the pilot starts moving the control the stick forward as needed to hold airspeed constant, rather than allowing it to further decrease. The flight path will arc downward into a more sustainable trajectory, be it climbing, descending, or level. The wing will certainly not reach the stall angle-of-attack, but the arcing transition could in a sense be described as "falling". An interesting exercise would be to precisely describe how the wings' lift vector changes throughout this maneuver-- it cannot stay constant, since the angle-of-attack is being decreased but the airspeed is not increasing. The G-meter mounted on the panel would show a decrease during the push-over maneuver. Depending on the airspeed at the time the pilot begins moving the control stick forward, it may or may not be possible to get the plane stabilized on a steady-state trajectory without ever having the airspeed increase, but in all cases the airspeed could be held constant for a while as the aircraft floats "over the top" of the change in trajectory.

4. We could undoubtedly conceive of several more permutations, again depending on exactly what control inputs the pilot makes.

So, the answer to your question is "it depends on what the pilot does". But examples like rockets, bullets, thrown rocks, etc show that an object certainly can "fall" without having wings that are meeting the air at or above the stall angle-of-attack!

On a more practical (and less verbose) level, consider the case where a pilot "unloads" the wing to the zero-lift angle-of-attack to avoid a stall. This maneuver can be carried out regardless of whether the flight path was climbing, descending, or level at the time the aircraft approached the stall, and regardless of whether the aircraft's airspeed was decreasing or increasing as the aircraft approached the stall. (Consider steeply banked, tightly turning cases as well.) What happens next? A "fall".

Note however that in all cases, even including wingless objects, the only way that the flight path can arc downward into a different trajectory (be it climbing, descending, or level, and regardless of how abrupt the downward arc is) is if the actual lift vector has become smaller than the lift vector that is required to sustain that trajectory. The required lift vector is equal to weight * cosine (climb angle or glide angle) -- as noted in this related ASE answer. (We're now assuming wings-level (non-turning, non-banked flight for simplicity.) In a very steep climb (or a very steep dive), this value will be much less than weight, but if the wings aren't making that much lift (or if there are no wings!), the flight path will still arc (curve) downward. If the wings are making that much lift, the flight path can't arc (curve) downward. So your thesis about "begin to fall (because the engines are too weak, not because of too little lift)" doesn't make sense. Regardless of how weak or strong the engines are-- regardless of whether the thrust-to-weight ratio is 10-to-1, 1-to-1, 1/10-to-1, or 0 (as in a glider), the only way the flight path can arc downward ("fall") is if the lift vector is smaller than the lift vector that is required to sustain the trajectory that exists at that instant, and that required lift vector is equal to weight * cosine (climb angle or glide angle.)

The bottom line is that extreme climbing maneuvers conducted with weak engines tend to cause place an aircraft in a position where the lift vector is less than weight * cosine (climb angle), and the flight path therefore curves downwards. This happens because lift depends on airspeed as well as angle-of-attack, and if you are steeply climbing with a weak engine, you can't maintain airspeed. So you lose lift, regardless of whether you place the wing anywhere near the stall angle-of-attack or not.

Standard caveats apply-- for simply, we're assuming that the thrust line is parallel to the instantaneous flight path, and except where stated otherwise, all actual formulae are intended for the simple case of non-turning (i.e. non-banked) flight.