Stall training during PPL

[dmspilot]

for those that need pictures.

Useless analogy. Rolling and turning aren't the same thing. The plane isn't turning in an aileron roll. You end up on the same heading you started. More evidence of your lack of understanding what you're talking about.

 

[Shepherd]

When I started in the mid 1960s SPIN training was required by the guys at the seaplane base where I was training.
I'm glad it was. When I'm in the Cub I usually do a couple of spins in each direction, just to keep in practice.
I have yet to spin the PA-17 because of the placard on the instrument panel.
I have no idea when the FAA abandoned the requirement for the PPL, but I think it was a mistake.
JMHO

 

[Dan Thomas]

Citabrias typically have G-meters. Those that think that the load factor, and therefore the stall speed, decreases in a descending turn should find a Citabria and go for a ride in it and watch that G-meter during various maneuvers, including descending turns.

The only times you'll see a decrease is (1) if the descent rate is constantly increasing, representing an arc of travel rather than a steady descent rate. The drawback here is that extra load factor will be applied to stop that high descent rate, and the stall speed goes up even more. (2) If the descent is steep enough, more of the airplane's weight is taken by drag forces rather than lift, and a small decrease in the G-reading might be noted. Taken to the extreme, a vertical nose-down dive, no forward speed at all, has all the weight off the wings, some carried by drag, and the rest causing Vne overspeed and probable structural failure. But nobody other than aerobats descend like that. The rest of us descend at small angles. A 10:1 glide angle is less than 6 degrees and, if my math is right (I'm no great mathematician) the load factor might decrease by .005 or so.

 

[Dan Thomas]

Not necessary.

In a bank, Load factors increase and stall speed decreases. Period.

Um...

 

[PaulS]

SCIENCE, I HAVE SCIENCE ON MY SIDE!!!!! I'm enjoying this, but I'm keeping my mouth shut, except for below.

This is where I repeat my biweekly post recommending that all new PPL's go get some basic aerobatics training.

It is one thing to talk about the theory of steep turns. It is another to actually roll into an 80 degree bank and pull 3.5G to keep the turn level, or ease off the stick and watch the altitude decrease and airspeed increase.

Immelmans in a low power aircraft like a Citabria or Decathlon are fantastic for developing a feel for stalls. When you roll upright at the top of the half loop, you will often be at or below stall speed. If you are too aggressive trying to hold level flight, the airplane will just fall out of the sky. Gotta relax the stick and feel for the plane to begin flying. (Note this will not be the case if you get your training in an overpowered aircraft like an Extra.)

And of course there is no substitute for doing a dozen spins with recovery within 5 degrees of desired heading to take the mystery out of that event.

This, I did upset training in an Extra, included at least a dozen spins, it was awesome. Highly recommend. We did a few aerobatic maneuvers too, fun stuff. (You can pull the power and fly slow in an Extra if you want, but I'm no expert.)

 

[Ed Haywood]

You can pull the power and fly slow in an Extra if you want.

Yeah, but who wants to do THAT?!

 

[Lindberg]

You can descend at 1,000fpm or 4,000fpm or 8,000fpm and the load factor doesn't decrease just because you're not level.

You can be level at 1g and you can descend at 1g, but you can't get from level flight at 1g to a 4,000fpm descent at 1g while maintaining 1g.

A 45 degree bank in a 1,000fpm decent results in the same load factor as a level turn.

It might or it might not, depending on whether there's any vertical acceleration.

 

[dmspilot]

You can be level at 1g and you can descend at 1g, but you can't get from level flight at 1g to a 4,000fpm descent at 1g while maintaining 1g.

You are correct, however, the point being, after a 4,000fpm descent is established, the load factor is 1g. Despite what unsafe says, descending is not a sufficient condition to reduce load factor below what it would be in constant altitude flight, whether straight or turning. If the flight condition is 45° banked coordinated turn, the load factor is 1.41 whether constant altitude or constant descent.

It might or it might not, depending on whether there's any vertical acceleration.

I wrote "1,000fpm descent" which should imply no vertical acceleration.

 

[Silvaire]

...Illusion, Below 1000 you start getting illusion of speed, especially when flying down wind. Turns also start to look different...

This is the factor in low level stall/spin accidents that cannot be simulated in practice at higher altitude. When there is wind and you are concentrating on lining up with a runway 500 feet below it is much easier to get uncoordinated than when you are at 3,000 feet because of the illusion created by your track across the ground and you fixation on it. In all of the incidents that resulted in a spin this is the condition that existed. A better training curricula for this are ground reference maneuvers where you learn to gauge the wind and disassociate your ground track from where the nose of the airplane is pointing or how much bank you have.

 

[Ed Haywood]

Citabrias typically have G-meters. Those that think that the load factor, and therefore the stall speed, decreases in a descending turn should find a Citabria and go for a ride in it and watch that G-meter during various maneuvers, including descending turns.

Mr G Meter will do all kinds of momentary things depending on what you do with the G meter control lever, also known as the stick. Throw in a nice steep bank at maneuvering speed and you can oscillate the G meter between 0 and 3G by alternately pulling and relaxing the stick to load and unload the wings. A big part of acro is varying your pull to avoid exceeding limits at various speeds and attitudes. Fun stuff!

 

[dmspilot]

For those that have not read my previous posts on this subject, I will explain it from scratch again.

We all know about the four forces, lift, weight, thrust, and drag. We also know the opposing forces are equal in unaccelerated (steady-state) flight.

We also know from Newton's first and third laws that an unbalanced force will result in an acceleration. F=MA. If the vertical component of lift is greater than weight, acceleration in the upward direction will occur. If the vertical component of lift is less than weight, acceleration in the downward direction will occur.

In a bank, the lift is deflected at an angle, and can be split up into two components, the horizontal component and the vertical component. Tilt the lift 45 degrees, and the vertical component is only 71% of what it was before, a decrease of 29%.

The total lift must be increased in order for the vertical lift to equal the weight of the aircraft again. Increasing the lift by 41% (1/71%) will increase the vertical component to the aircraft's weight again.

The definition of load factor is the ratio of total lift to aircraft weight. In the previous example, the lift has increased to 41% more than the aircraft's weight, so the load factor is 1.41.

What if we do not increase the total lift, allowing it to remain the same as before, and simply allow the vertical component to decrease by 29%? In other words, turn at a 45 bank with a load factor equal to 1?

The aircraft will accelerate downward at a rate equal to 29% of the acceleration due to gravity (being 32 feet per second per second) which equals 9.28 ft/s/s. After one second, the aircraft's descent rate is 9.28 ft/s, or 557 ft/minute. After only four seconds, the descent rate is 2,227fpm, pegging the VSI in most light aircraft.

That is why we have to increase total lift in a turn, even when descending, and why a descending turn will cause an increase in load factor just as much as a constant-altitude turn.

My procedure when I taught people how to perform the "Emergency Descent" task for the private pilot practical test was to initiate the maneuver by rolling into a 45 bank while applying zero elevator pressure. The angle of attack remains relatively constant, allowing the lift to remain relatively constant, maintaining a 1g load factor on the airplane. Within about four seconds we peg the VSI, just like the math says.

 

[Adam Weiss]

I was just thinking about this as I read about another suspect stall/spin during base to final. The nature of stalls that I learned during my PPL training don't seem to coincide with the types of stall accidents that I read about. The power off stall especially doesn't seem relevant at all with the type of stalls seen. It seems that it would make more sense to train accelerated stalls since (correct me if i'm wrong here) those are what kill people in the pattern.

Any opinions on this? I asked my instructor to take me up and practice these but never got a chance to since he moved shortly after I received my PPL.

In 700 hours of flying, I had never had a "real" stall.
However, this summer (and after almost 300 hours in my plane) I had two instances where I actually stalled on approach.
It was exactly the same both times.
It was a 1900ft runway, with no go-around option.
I was flat and slow on the approach to make sure I didn't float or go long.
However, my sink rate was too high in the flare, and instead of adding a touch of power, I yanked back on the stick and stalled instantly.
Fortunately, both times I was only a foot or two off the runway, but it was a wake-up call.
Had I been another foot or two off the runway, it could have been ugly.
I had gotten really lazy and never would have thought I could actually stall the plane.
I've since revisited my numbers and done some more slow flight to get reacquainted with that end of the envelope.
Bottom line: power off stalls are a real thing. Stay vigilant and don't get lazy like me.

 

[Ed Haywood]

In 700 hours of flying, I had never had a "real" stall.
However, this summer (and after almost 300 hours in my plane) I had two instances where I actually stalled on approach.
It was exactly the same both times.
It was a 1900ft runway, with no go-around option.
I was flat and slow on the approach to make sure I didn't float or go long.
However, my sink rate was too high in the flare, and instead of adding a touch of power, I yanked back on the stick and stalled instantly.
Fortunately, both times I was only a foot or two off the runway, but it was a wake-up call.
Had I been another foot or two off the runway, it could have been ugly.
I had gotten really lazy and never would have thought I could actually stall the plane.
I've since revisited my numbers and done some more slow flight to get reacquainted with that end of the envelope.
Bottom line: power off stalls are a real thing. Stay vigilant and don't get lazy like me.

Well shucks, that just sounds like a well executed short field landing.

 

[Adam Weiss]

Well shucks, that just sounds like a well executed short field landing.

Maybe if I had anticipated the stall and not stunk up the cockpit.

 

[Martin Pauly]

I wish that diagram that shows load factor and stall speed as a function of bank angle had never been created. It is valid only under specific circumstances, and has caused a lot of grief not just in online forums but also in oral exams/checkrides.

A wing stalls if, and only if, the angle of attack exceeds the critical angle of attack. Plain and simple.

We have also all heard (I hope) that the airplane can stall at any airspeed, in any attitude. But sometimes I question if beyond rote memorization of this sentence pilots really understand what it means.

Pulling the stick back gets an airplane closer to a stall. Pushing the stick forward buys us margin from a stall. Forget stall speed - if you feel a buffet, push the stick forward. Plain and simple.

If you want to determine stall speed, then load factor is the primary variable to look at. (Aircraft gross weight, configuration, and air density are also factors, but they don't change as much or as quickly as we can change load factor.)

Load factor is lift over weight. Thus, whenever we are asking the wings to produce more lift than the weight of the aircraft, our load factor is greater than 1. That can happen in a turn, that can happen entering a loop, that can happen while transitioning from level flight to a limb, that can happen in the landing flare or during a go-around. When it happens, stall speed is increased from straight-and-level stall speed.

Consequently, we have a load factor of less than 1 whenever we ask for less lift than the weight of the aircraft. That can happen when pushing the nose down to start a descend, or to increase the rate of descent we already had going. That can happen at the top of a loop when gravity adds to the lift produced by the wings, helping us pull the aircraft towards the center of the loop. An aircraft parked on the ramp has a load factor of 0 - no lift is needed.

We can combine things that increase the load factor with things that decrease the load factor, so that the changes cancel each other out. A coordinated turn while simultaneously increasing the rate of descent can be done at a load factor of 1 (and thus no change in stall speed from straight-and-level flight). Note that descending alone isn't sufficient - it's an increase in the rate of descent (or a reduction of the rate of climb) that creates the less-than-1 load factor. So we can only do this for a short time before the nose of the airplane is pointed straight down.

In a steady descent (constant rate of descent, wings level), the load factor is actually a teeny tiny bit smaller than 1. The reason is that there is now a vertical component of drag which helps offset the weight of the aircraft. Similarly, in a steady climb the load factor is also a little bit smaller than 1, because thrust now has a vertical component. But unless we fly something like an F-15 which can climb almost vertically, the difference is so small that it's completely negligible for stall speed. I have some diagrams at the end of a video introduced in this thread which show the math behind it.

- Martin

 

[dmspilot]

I wish that diagram that shows load factor and stall speed as a function of bank angle had never been created. It is valid only under specific

I am glad it was created, but I wish people didn't say it was "only" for "level flight", because that is bull.

Plus, it is safer to assume the relationship is true when it's not, than to believe the relationship is false when it actually is true, like some are here. The latter is outright dangerous.

 

[Martin Pauly]

I think we are on the same page, @dmspilot
My point is that thanks to this diagram, too many pilots are under the impression that bank angle is all they need to think about when it comes to stalls and stall speed.

- Martin

 

[Martin Pauly]

imho - Not worth adding that which may be grasped on and misinterpreted badly and loose the entire larger point you were making. Interesting certainly. But for a subset of readers, it may leave them with the entirely wrong message.

Yes. Thus my disclaimer that "the difference is so small that it's completely negligible for stall speed."

- Martin

 

[francisco collazos]

Excellent write up Martin...

 

[Lindberg]

Despite what unsafe says, descending is not a sufficient condition to reduce load factor below what it would be in constant altitude flight, whether straight or turning.

Agreed. And banking is not a sufficient condition to increase it.

 

[MauleSkinner]

I think we are on the same page, @dmspilot
My point is that thanks to this diagram, too many pilots are under the impression that bank angle is all they need to think about when it comes to stalls and stall speed.

- Martin

For the most part it is. When that ceases to be the case, those who are limited to that basic understanding will learn the full process from their aerobatic instructor.

 

[Clip4]

Any CFI can and should teach accelerated stalls. Just because it’s not part of the ACS doesn’t mean it can’t be covered. As a CFI the ACS is just the baseline of what I teach, I usually go way beyond the ACS into real world scenarios.

An accelerated stall is a stall that occurs at an airspeed higher than normal due to a higher load factor (g loading). A 20°bank results in ~5% increase in stall speed.


From the ACS for both power on and off stalls, Maintain a specified heading ±10° if in straight flight; maintain a specified angle of bank not to exceed 20°, ±10° if in turning flight, while inducing the stall.

The DPEs have a tool to test for the base to final turn coordination and stall recovery.

 

[Dom Uliano]

I'm only 17 hrs into my ppl but my instructor has had me practice both power off and power on stalls several times in separate lessons. We also went over, and he demonstrated, emergency procedures but I haven't done that yet. We've done a lot of slow flight, did some forward slips ( a lot of those), and some steep turns so far.

 

[MauleSkinner]

However, my sink rate was too high in the flare, and instead of adding a touch of power, I yanked back on the stick and stalled instantly.

Technically the accelerated stall that the OP was asking about.

 

[unsafervguy]

For those that have not read my previous posts on this subject, I will explain it from scratch again.

We all know about the four forces, lift, weight, thrust, and drag. We also know the opposing forces are equal in unaccelerated (steady-state) flight.

We also know from Newton's first and third laws that an unbalanced force will result in an acceleration. F=MA. If the vertical component of lift is greater than weight, acceleration in the upward direction will occur. If the vertical component of lift is less than weight, acceleration in the downward direction will occur.

In a bank, the lift is deflected at an angle, and can be split up into two components, the horizontal component and the vertical component. Tilt the lift 45 degrees, and the vertical component is only 71% of what it was before, a decrease of 29%.

The total lift must be increased in order for the vertical lift to equal the weight of the aircraft again. Increasing the lift by 41% (1/71%) will increase the vertical component to the aircraft's weight again.

The definition of load factor is the ratio of total lift to aircraft weight. In the previous example, the lift has increased to 41% more than the aircraft's weight, so the load factor is 1.41.

What if we do not increase the total lift, allowing it to remain the same as before, and simply allow the vertical component to decrease by 29%? In other words, turn at a 45 bank with a load factor equal to 1?

The aircraft will accelerate downward at a rate equal to 29% of the acceleration due to gravity (being 32 feet per second per second) which equals 9.28 ft/s/s. After one second, the aircraft's descent rate is 9.28 ft/s, or 557 ft/minute. After only four seconds, the descent rate is 2,227fpm, pegging the VSI in most light aircraft.

That is why we have to increase total lift in a turn, even when descending, and why a descending turn will cause an increase in load factor just as much as a constant-altitude turn.

My procedure when I taught people how to perform the "Emergency Descent" task for the private pilot practical test was to initiate the maneuver by rolling into a 45 bank while applying zero elevator pressure. The angle of attack remains relatively constant, allowing the lift to remain relatively constant, maintaining a 1g load factor on the airplane. Within about four seconds we peg the VSI, just like the math says.

here in lies the problem. you admit that the bank angle does not cause the load factor to increase. but you are using practical application to try and disprove a physical law.

What if we do not increase the total lift, allowing it to remain the same as before, and simply allow the vertical component to decrease by 29%? In other words, turn at a 45 bank with a load factor equal to 1?
The aircraft will accelerate downward at a rate equal to 29% of the acceleration due to gravity (being 32 feet per second per second) which equals 9.28 ft/s/s. After one second, the aircraft's descent rate is 9.28 ft/s, or 557 ft/minute. After only four seconds, the descent rate is 2,227fpm, pegging the VSI in most light aircraft.

That is why we have to increase total lift in a turn, even when descending, and why a descending turn will cause an increase in load factor just as much as a constant-altitude turn.

in theory, you could continue the 1 g decent, but as dan and you both pointed out, and i agreed with,the exponential increase will cause a very large ever increase in vertical speed (until the really nasty math of drag and other fluid dynamics kick in) so in the practical world, lift must be increased, and hence increasing load factor to arrest that rate. but it does not change the fact that the increase in bank does not change the load factor, its the need to arrest the ever increasing decent rate that causes that. the statement that bank angle increases load factor is just wrong according to physics.

hopefully, all this argument brings a better understanding if what is really happening in a turn will be the end result.

 

[Gilbert Buettner]

I agree with the OP.
Not just stalls...but stalls are a great example. We are really taught to demonstrate getting into a stall...in a very calm 1 G sort of way. Often it's just demonstrating almost stalling.
and then we repeat these again and again...with every rental checkout and every flight review, etc.... year after year

wrote procedure:

1. Climb and clear. Climb to an altitude not less than 1,500 feet agl, and perform clearing turns in each direction. ...
2. Slow to rotation speed. ...
3. Add power. ...
4. Induce stall. ...
5. Keep that ball centered with rudder. ...
6. Break and recover.

Just want to point out something - you need to be able to complete the maneuver more than 1,500 AGL. So, climb to an altitude at least several hundred feet higher just to be able to comply with that requirement.

 

[dmspilot]

Agreed. And banking is not a sufficient condition to increase it.

No, but the thread is about stalling during turns in the traffic pattern. It's not about aileron rolls or the vomit comet. So let's try to not make misleading claims like saying the relationship between bank angle and stall speed is only true for level flight and not for descents. It's wrong and dangerous.

 

[dmspilot]

here in lies the problem. you admit that the bank angle does not cause the load factor to increase. but you are using practical application to try and disprove a physical law.

What physical law am I trying to disprove?

You stated the bank angle load factor graph only applicable to level flight. That is false.

 

[Pinecone]

Not necessary. I believe your post to be false and misleading.

There may be elements of truth in your elaboration however your initial statement is simply not correct.

In a bank, Load factors increase and stall speed INCREASES. Period. If you want to add additional elements to modify that then go ahead but that does not change the reality of the true statement regarding a bank, load factors and stall speed.

I am not getting into a personal dispute with you and hopefully that does not appear so. I just think publicly posting factually incorrect information on a public forum bears a responsibility to be corrected.

And if I am wrong, by all means I stand ready to be corrected. By a different poster.

EDIT: I am wrong. Typo. lol. Correcting. Thanks for pointing it out.

Yes, you are wrong.

 

[Pinecone]

So much lack of understanding of aerodynamics in this thread.

Go back to you basic bank any diagram from primary training.

If you are at 1 G in level flight and bank to 60 degrees, you incline the lift vector by 60 degrees. If you do nothing else, your load will still be 1 G and the stall airspeed will by the same as 1 G straight and level flight.

BUT, to maintain level flight, you need to pull back and increase the load factor to 2G to maintain level flight. That increases the stall speed.

Attitude and load factor are not linked without the addition of flight path.

You can stall a plane going straight up. And straight down. And you can positive G stall an plane while inverted.

As for load factor in a descent, think about straight down. What is the load factor?

 

[Lindberg]

here in lies the problem. you admit that the bank angle does not cause the load factor to increase. but you are using practical application to try and disprove a physical law.

What if we do not increase the total lift, allowing it to remain the same as before, and simply allow the vertical component to decrease by 29%? In other words, turn at a 45 bank with a load factor equal to 1?
The aircraft will accelerate downward at a rate equal to 29% of the acceleration due to gravity (being 32 feet per second per second) which equals 9.28 ft/s/s. After one second, the aircraft's descent rate is 9.28 ft/s, or 557 ft/minute. After only four seconds, the descent rate is 2,227fpm, pegging the VSI in most light aircraft.

That is why we have to increase total lift in a turn, even when descending, and why a descending turn will cause an increase in load factor just as much as a constant-altitude turn.

in theory, you could continue the 1 g decent, but as dan and you both pointed out, and i agreed with,the exponential increase will cause a very large ever increase in vertical speed (until the really nasty math of drag and other fluid dynamics kick in) so in the practical world, lift must be increased, and hence increasing load factor to arrest that rate. but it does not change the fact that the increase in bank does not change the load factor, its the need to arrest the ever increasing decent rate that causes that. the statement that bank angle increases load factor is just wrong according to physics.

hopefully, all this argument brings a better understanding if what is really happening in a turn will be the end result.

In 4 seconds, you can turn 90*. And descend 75'. Even if it takes 6 seconds, you'll descend 168'. You'll still have plenty of altitude to lose between pattern and the runway. This Aircraft Turn Information Calculator says that at 80 knots and 45* of bank, it will take 13.2 seconds to make the 180* turn from downwind to final, which, if my math is correct, means I'll lose about 800' and roll out on final at 200' AGL. That seems perfect. Maybe we'd have fewer pilots with a tendency to haul back on the yoke in pattern turns if we didn't tell them to fly stupid-big stabilized patterns and 3* descents.

 

[brcase]

The aircraft will accelerate downward at a rate equal to 29% of the acceleration due to gravity (being 32 feet per second per second) which equals 9.28 ft/s/s. After one second, the aircraft's descent rate is 9.28 ft/s, or 557 ft/minute. After only four seconds, the descent rate is 2,227fpm, pegging the VSI in most light aircraft.

An excellent run through the numbers.
But say we pull 1.2G with only a 10% (5kts?) increase in stall speed. After all I think most pilots would have to work pretty hard to not make the turn without pulling back a little in a turn like we were taught basically since lesson #1. At a 1.2G 45 degree bank turn the downward acceleration is only 16% or about 5ft/s^2.

A 75kt 90 degree 45degree bank turn at 1.4Gs takes about 5 seconds so I assume the turn rate is reduced about the same at a 1.2G, or 6 seconds for a 90 degree turn. While this still results in a 4800ft/min descent max it only results in about an 215ft loss of altitude over the 6seconds Plus maybe another 215ft to decelerate back to a 1G condition. BTW this is vs 116ft lost in a 20degree bank 90 degree turn at a constant 500ft/min decent at 1.1G', and the turn radius goes from 1037ft to 654 ft (BTW I think 2,227fpm you stated is actually 4,320ft/min, since you still accelerating through 3 remaining seconds, correct me if I am wrong, and I figure the 1G 4second 45degree bank turn results in a loss of 135ft)

And this assumes we hold the 45 degree bank for the full 90 degrees of turn instead of maybe just one or two seconds to prevent the overshoot to begin with.
These are exponential numbers so doing for even a little bit shorter time results in a lot less altitude loss, a 4second turn on results in a 75ft loss with a 75 foot recovery.

And then back to one of my original statements. I find getting an aerodynamic indication of a stall out of a steep turn to be rather challenging, much harder than from a 20 degree bank turn.

Brian

 

[Crashnburn]

Did he gradually increase the angle of attack during the turn until it broke or was it a quick yank and pop?

Have seen accelerated stalls taught each way.

My first two instructors wanted me to do a quick yank and stall. My previous CFI, and my current CFI want me to hold the nose up just high enough to let the airspeed bleed off until the plane stalls.

 

[dmspilot]

This Aircraft Turn Information Calculator says that at 80 knots and 45* of bank, it will take 13.2 seconds to make the 180* turn from downwind to final, which, if my math is correct, means I'll lose about 800' and roll out on final at 200' AGL. That seems perfect.

Not so fast. The calculator is for balanced, coordinated, level turns. As @dtuuri pointed out, you won't turn at the same rate if you do not increase the load factor. The horizontal component of lift will be less.

You are arguing that a 1g steep turn is practical way to fly a traffic pattern and using a calculator that assumes a normal turn to support your argument. It doesn't work that way.

Also, at the end of your hypothetical 13.2 second maneuver, although the aircraft has lost 800 feet, it is now descending at over 7,000fpm. How do you arrest such a descent? By increasing the load factor. And likely stalling or over-stressing the airframe. You have solved absolutely nothing.

 

[MauleSkinner]

Also, at the end of your hypothetical 13.2 second maneuver, although the aircraft has lost 800 feet, it is now descending at over 7,000fpm. How do you arrest such a descent? By increasing the load factor. And likely stalling or over-stressing the airframe. You have solved absolutely nothing.

And how much speed have you gained in the process?

 

[chemgeek]

The power off stall especially doesn't seem relevant at all with the type of stalls seen.

The only low-time pilot fatal we have had out of our airport was a mismanaged power-off emergency descent following a likely carb ice induced loss of power. The pilot apparently got too slow while maneuvering to make an off-airport landing in a nearby field, resulting in a low altitude stall/spin. Learning to fly an aircraft in the slow flight regime while recognizing incipient stalls is a critical pilot skill, not only for this kind of emergency, but also for controlling aircraft in the landing phase.

 

[TommyG]

An accelerated stall is a stall that occurs at an airspeed higher than normal due to a higher load factor (g loading). A 20°bank results in ~5% increase in stall speed.


From the ACS for both power on and off stalls, Maintain a specified heading ±10° if in straight flight; maintain a specified angle of bank not to exceed 20°, ±10° if in turning flight, while inducing the stall.

The DPEs have a tool to test for the base to final turn coordination and stall recovery.

But CFIs need to go beyond the ACS, throw that plane into a 30 degree bank, pull up and stall that plane. I would say a large majority do not. They teach the straight and level nice and safe stall.

 

[Ed Haywood]

But CFIs need to go beyond the ACS, throw that plane into a 30 degree bank, pull up and stall that plane.

A lot of CFI's would be terrified of dropping into a spin while doing that.

 

[Greg Bockelman]

Not me. And a 30 degree bank stall doesn’t even raise my pulse rate.

 

[PaulS]

I think you have pilots training in million dollar aircraft where recovery from a spin is to pull the chute. In the skipper, in which spins are approved in the utility category, we did stalls like they were going out of style. I think instructors are more relaxed in a $35,000 trainer, but that's just me.