1
$\begingroup$

At 0g, stall is theoretically impossible.

In practice, can a maneuver be performed that, by bringing the aircraft to, or very close to 0g, also brings the aircraft to a temporary standstill, or near standstill, in terms of true air speed?

Can such a maneuver be performed without exceeding 30° pitch?

If so, how?

|improve this question
$\endgroup$
  • 2
    $\begingroup$ At 0g, stall is theoretically impossible. Are you in orbit or free falling? In the second case stall is happening already $\endgroup$ – jean Jul 21 '17 at 16:07
  • $\begingroup$ @Jean "In the second case [free falling (assumed within the atmosphere where planed are capable of flight)] stall is happening already" Are you sure about that? Take a look at a Coefficient of lift vs AoA diagram and explain why what is in quotes is incorrect. A free-falling skydiver experiences more than 0g because of air pressure. A stalled wing does not = 0 lift. $\endgroup$ – Ryan Mortensen Jul 21 '17 at 17:45
  • $\begingroup$ 0kts is different than 0g. I can have 0 knots ground speed flying into a very strong head wind and not stall at all. I would recommend the title to make this more clear. $\endgroup$ – 0tyranny 0poverty Jul 24 '17 at 2:48
  • $\begingroup$ Also, you could go from straight and level flight and nose over at just the right rate and create 0gs, and dive nose down, and maintain 0gs for a few seconds with proper airflow over the wings (i.e. not stalled). $\endgroup$ – Devil07 Jul 24 '17 at 13:48
4
$\begingroup$

No, stalls can occur at any airspeed, attitude, or load factor. All that is needed is to exceed the critical angle of attack and you have stalled the airplane.

Some places this can occur and does happen if mishandled by the pilot is at the top of inside loops, where the airplane can come very close to a zero airspeed if it's done wrong. Another place this does occur is a hammerhead stall where the stall occurs in the vertical, just as the aircraft is at or comes very close to zero airspeed.

|improve this answer
$\endgroup$
  • 6
    $\begingroup$ There is a massive difference between "No stalls can occur..." and "No, stalls can occur....". That comma literally makes the sentence mean the exact opposite. $\endgroup$ – abelenky Jul 21 '17 at 13:27
  • 3
    $\begingroup$ To wit: Eats shoots and leaves $\endgroup$ – msanford Jul 21 '17 at 16:00
3
$\begingroup$

You're asking 3 questions.

How can you fly at 0kts without stalling?

Have a thrust:weight ratio greater than one. John Klatt's Screaming Sasquatch is one such airplane.

In practice, can a maneuver be performed that, by bringing the aircraft to, or very close to 0g, also brings the aircraft to a temporary standstill, or near standstill, in terms of true air speed?

Sure. Pull to vertical (or near vertical) and wait. You'll hit zero airspeed right before gravity reasserts itself. As Carlo Felicione mentioned in his answer, a hammerhead maneuver will do this. There are other maneuvers as well. The main idea is to convert your forward momentum into a vertical climb and stop depending on the wings for lift.

When I perform a loop (I fly an aerobatic plane -- DON'T try this in a Cessna!) the aircraft is essentially ballistic across the top of the loop. It will indicate 0 knots airspeed regularly (that's the trick to making round loops)

Can such a maneuver be performed without exceeding 30° pitch?

Probably not, and the fact that you're asking is kind of scary.

Probably not because you won't have enough upward momentum to offset gravity between the time your wings stop producing lift and when you hit zero airspeed.

Scary because these are aerobatic maneuvers, and attempting them in the wrong aircraft without training is stupid dangerous.

|improve this answer
$\endgroup$
  • $\begingroup$ Thank you. I was just asking out of curiosity. The Screamin Sasquatch's vertical climb and the other aerobatic maneuvers are exactly the kind of thing I was interested in hearing about. I'm confused about your last sentence though: Students regularly practice stalls and spins in non-aerobatic aircraft; is there anything more dangerous than that at <30° pitch and <60° bank? $\endgroup$ – Zaz Jul 23 '17 at 1:07
  • $\begingroup$ "Is there anything more dangerous..." would make a good separate question. The short answer is that it becomes really easy to over-stress the air frame, tumble the gyros, void your insurance, etc. $\endgroup$ – Dan Pichelman Jul 23 '17 at 13:33
  • $\begingroup$ Being stalled is a property of the wing and the wing only, so thrust/weight ration has nothing to do with it. What thrust/weight ration greater than 1 does is allow maintaining altitude even if stalled—but the plane would still be stalled, it would just not lose altitude or control because of it. $\endgroup$ – Jan Hudec Jul 23 '17 at 18:47
  • $\begingroup$ @JanHudec You're completely correct. I was thinking more in terms of "what will the pilot experience" rather than the formal definition of a stall. How should I best improve this answer? I read the question as "the OP wants to hit 0 IAS without falling out of the sky" $\endgroup$ – Dan Pichelman Jul 24 '17 at 1:17
1
$\begingroup$

Can you fly at 0 knots airspeed? Sure you can. Helicopters do it all the time. It's called hovering. Technically speaking, it is flying, i.e. being held aloft by lift generated from airfoils. True that the airfoils are not stationary, but the rest of the aircraft is.

If you want to get picky and say airplane with wings, I present the V22 Osprey.

A pure aircraft can achieve 0 knots in the air, but technically speaking, it's not flying, because it's not being held aloft by an airfoil. It would be kept aloft either by pure engine power in the case of very high performance aircraft or non tilt rotor VTOL such as the Harrier or F35B, or as part of a ballistic arc. Neither mode of operation uses an airfoil for the duration of the zero kts airspeed, so neither would be 'flying' by the traditional definition.

Zero g, on the other hand, can only be achieved briefly at the top of an arc where the wings are unloaded, or in a dive. 45 degrees nose down, 360 KIAS, and you get near or at zero g. Popularly known as the vomit comet.

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

I would offer that a tailslide would be a negative or zero-g maneuver. The aircraft is brought into a vertical pitch of 90° (exceeding the 30° in the question) and held through the stall with rudder alone. Once the tailslide begins, the direction of travel with reverse airflow is controlled by rudder alone. At this point the airspeed is negative with regards to the normal direction of flight but positive with relation to the aircraft as a whole (although in the opposite "wrong" direction). Recovery occurs when the rudder is offset allowing the tail to "slide out" from underneath the nose and snap the nose to the new direction of flight (down in this case).

Not sure if it answered your question but thought I'd chime in since I've seen this performed a few times.

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

0 knots of airspeed in a heavier-than-air fixed-wing aircraft and not stall? No.

0 knots of groundspeed in a heavier-than-air fixed-wing aircraft and not stall? yes.

0g and stall? No a stall is not a 0g maneuver. Even if you were in a vacuum, there is no relative wind because there would be no air, hence no angle of attack, hence no angle of attack to exceed to enter a stall. As soon as you leave said vacuum then the molecules of air are now applying some force to the airframe meaning it is no longer experiencing 0g, although this small force would be imperceptible to human somatosensory systems if they were onboard.

The reduction in g-forces are associated with stall recovery, which is done by pilot control input reducing the angle of attack causing the stalled condition to cease, and therefore, also reducing the lift while the aircraft regains kinetic energy by trading off its potential energy. This reduction in g-forces is not technically a result of a stalled flight condition.

Stalls do not equal zero lift; keep that in mind.

|improve this answer
$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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