stall practice question

[Dan Thomas]

You're either stalled or not. I haven't been convinced otherwise.

Depending on wing planform and washout, the stall will start at the trailing edge root and progress outward and forward as AoA increases. As lift is lost the nose will drop. With most GA airplanes, the whole wing never gets a chance to stall this way. You can still rock the wings with aileron in a 172 as the nose is falling, for instance.

So it's not "either stalled or not." The only way to get most light airplanes to completely stall the wing is in an aggressively accelerated stall such as the sharp pullup some unfortunate guys do after a low pass, and the airplane flips over and kills them.

Dan

 

[Dan Thomas]

The prop creates prop wash which reduces AoA, and forward Cg assists by initiating corrective actions for the pilot. (Most of our airplanes also contain washout at the wing tips which help keep the outboard potions flying a bit longer that the inboard portion of the wing.) Thrust also assists by encouraging the relative wind, and it is this (relative wind) that is responsible, not the increasing whirring of the fan. May seem a trifle distinction but it's more than semantics.

Sure, over the really short inboard span that the prop blast covers. Thrust otherwise reduces AoA by increasing forward speed relative to the downward flightpath, which changes the relative wind over the whole span even if pitch attitude doesn't change.

Dan

 

[Jaybird180]

That isn't what I was looking for. You wrote, re: 0G, "...I KNOW a wing can never stall there (even beyond critical AoA)..." So, I asked:


because "even beyond critical AoA" there is still some lift and you won't be at 0G. If you think you can have one without the other, I suspect you're mistaken--at least as far as normal flight regimes go. Then the next thing is, you'll be "unloading your wing" from base to final, so as to tighten your turn without any penalty. That's a big no-no, IMO.

dtuuri

Push real hard on the yoke and tell me if you can make 0G

 

[Jaybird180]

Sure, over the really short inboard span that the prop blast covers. Thrust otherwise reduces AoA by increasing forward speed relative to the downward flightpath, which changes the relative wind over the whole span even if pitch attitude doesn't change.

Dan

We're in agreement

Depending on wing planform and washout, the stall will start at the trailing edge root and progress outward and forward as AoA increases. As lift is lost the nose will drop. With most GA airplanes, the whole wing never gets a chance to stall this way. You can still rock the wings with aileron in a 172 as the nose is falling, for instance.

So it's not "either stalled or not." The only way to get most light airplanes to completely stall the wing is in an aggressively accelerated stall such as the sharp pullup some unfortunate guys do after a low pass, and the airplane flips over and kills them.

Dan

The reason I stated the C-172 is a poor example is that the designers intentionally designed the aircraft for training purposes to assist with recovery. It is by design that the aircraft resist stalling at the wingtips. It is by design that the aircraft is nose heavy and thus pitches down.

For this reason, I should have said (and did) that my points were about the behavior of an airfoil and not necessarily a wing (greater than the sum of the parts and all that jazz).

 

[Jaybird180]

That isn't what I was looking for. You wrote, re: 0G, "...I KNOW a wing can never stall there (even beyond critical AoA)..." So, I asked:


because "even beyond critical AoA" there is still some lift and you won't be at 0G. If you think you can have one without the other, I suspect you're mistaken--at least as far as normal flight regimes go. Then the next thing is, you'll be "unloading your wing" from base to final, so as to tighten your turn without any penalty. That's a big no-no, IMO.

dtuuri

If I miss the timing on base-final, I accept simply that I will run wide continue the turn until I am lined up then make a small correction to get on proper heading. At the airports I have operated at, there is no penalty. Doesn't look as cool, but it keeps me from even thinking about kicking in bottom rudder, which really is the no-no, not the steep turn.

 

[gismo]

I suppose that could be true if the accelerating, corkscrewing dive doesn't kill you first. I'd like to see the mathematics with realistic numbers for salvaging a turn gone bad from base to final. How much altitude would you lose, speed would you gain vs radius you would shorten with the bank angles you gave?

dtuuri

I could work that out and the math is pretty simple. But to be clear here, I'm not advocating steep turns with reduced g loading in the pattern. As I mentioned already (and you appear to understand) the result is a constant acceleration towards the ground which has to be arrested with more positive g force in the near future lest an impact with the ground provide that for you.

The one situation where unloading in a too steep turn would be ultimately beneficial is when that's combined with rolling the wings to a more level attitude to avoid a stall. In that case the immediate unloading keeps the wing unstalled allowing the ailerons to do their job.

 

[dtuuri]

Push real hard on the yoke and tell me if you can make 0G

When YOU push on the yoke and achieve 0G, tell ME if you've still got an angle of attack like you said you could have.

dtuuri

 

[dtuuri]

I could work that out and the math is pretty simple.

I'd like to see it. In steady horizontal flight (and a 3° glide path is pretty close to horizontal) a heavy airplane and a light one have the same turn radius at the same speed and bank. If the heavy one salvoes its bomb load and becomes the same weight as the light one--it'll speed up and increase its radius. How will "unloading the wing" be any different than that?

If turn radius and G factor in a vertical dive subscribe to different rules, I won't argue against it. I'd just appreciate a good explanation. Then it would be nice to see exactly how much more similar to the turn radius of the vertical case a 3° glide path becomes than to a level horizontal turn.

Anybody who can debunk the thumb rules for wind correction angles inbound in a holding pattern ought to be able to do this too, don't you think?

dtuuri

 

[gismo]

I'd like to see it. In steady horizontal flight (and a 3° glide path is pretty close to horizontal) a heavy airplane and a light one have the same turn radius at the same speed and bank. If the heavy one salvoes its bomb load and becomes the same weight as the light one--it'll speed up and increase its radius. How will "unloading the wing" be any different than that?

I don't see a difference. Either way it would require a reduction in power to maintain the initial airspeed for long. And with constant weight even before the plane accelerates the turn rate would decrease the instant the load was reduced, just not as much as it would decrease if the bank angle was reduced enough to allow level flight at the same loading.

In simplest terms for the example I gave (45° bank pulling 1.22 g) the vertical force (lift perpendicular to the Earth) would be reduced by the same factor as the horizontal (i.e. 1.22/1.4). As a result the plane's vertical speed would be increasing about (1-1.22/1.4) *32 = 4.1 ft/s^2 or almost 250 FPM every second. That will add up quickly and at some point you would have to pull more g's just to level off. After 5 seconds you'd be descending at 1250 FPM and would have lost a mere 50 ft. But if you rolled wings level at that point and maintained the 1.22 g you'd lose another 30 ft. The actual duration of the turn would depend on how many degrees of heading change were involved, the initial airspeed, and how much if any the airspeed was allowed to increase. I'm not up to calculating that tonight.

Anybody who can debunk the thumb rules for wind correction angles inbound in a holding pattern ought to be able to do this too, don't you think?

This is easy, that took a lot more effort.

 

[Jaybird180]

When YOU push on the yoke and achieve 0G, tell ME if you've still got an angle of attack like you said you could have.

dtuuri

Geez Dave. Now you're purposely being obtuse.

 

[dtuuri]

I don't see a difference. Either way it would require a reduction in power to maintain the initial airspeed for long. And with constant weight even before the plane accelerates the turn rate would decrease the instant the load was reduced, just not as much as it would decrease if the bank angle was reduced enough to allow level flight at the same loading.

I had to metabolize the two glasses of medicinal red wine I had at dinner before attempting to follow this post. Not sure I can understand what a "constant weight" of a bomber dropping its load means, but I am pretty sure just the opposite happens. The "instant" reaction would be that the plane rises and the radius tightens causing a faster turn rate not a decreased one. Unless, that is, the pilot simultaneously adjusts the angle of attack to the same as the lighter airplane being compared to. Next, the acceleration in speed would decrease rate and radius.

In simplest terms for the example I gave (45° bank pulling 1.22 g) the vertical force (lift perpendicular to the Earth) would be reduced by the same factor as the horizontal (i.e. 1.22/1.4). As a result the plane's vertical speed would be increasing about (1-1.22/1.4) *32 = 4.1 ft/s^2 or almost 250 FPM every second.

Trying to follow you here... 32 feet/sec² is the acceleration of a falling body if I remember my physics. You're saying it will fall, like a rock? But a plane with more lift (i.e. 1.4/1.4) would fall at 32 ft/sec² and a plane with less lift (i.e. 1.22/1.4) falls less fast?? I'm afraid I don't get it. Probably just me.

That will add up quickly and at some point you would have to pull more g's just to level off. After 5 seconds you'd be descending at 1250 FPM and would have lost a mere 50 ft. But if you rolled wings level at that point and maintained the 1.22 g you'd lose another 30 ft. The actual duration of the turn would depend on how many degrees of heading change were involved, the initial airspeed, and how much if any the airspeed was allowed to increase. I'm not up to calculating that tonight.

I can't even attempt to follow this paragraph until I can understand the first two. I wonder if I'm alone.

dtuuri

 

[dtuuri]

Geez Dave. Now you're purposely being obtuse.

Now, now. I'm only trying to make sure you understand that zero Gs mean zero angle of attack (or come up with a creative scenario where it isn't). You hedged on the point, the way I read what you wrote, when you said,

"Personally, I am comforted by knowing that I can unload the wing to break any stall, provided I can reduce load factor to the necessary value approaching 0G, because I KNOW a wing can never stall there (even beyond critical AoA)..."​

If, I say "if", you believe you can have both at once then I see trouble ahead for you someday. Even if you don't, conjuring up an off-beat scenario like, say, during a whip stall where you could, is a worthwhile mental exercise.

Btw, have you tufted your plane yet?

dtuuri

 

[redtail]

Stalled wings do in fact create quite a bit of lift. So if you go up in your 172, pull power off, and hold the yoke fully aft, you think this is the same thing as a rock falling to the ground? The same thing as the airplane suddenly not having wings at all? The airplane is still flying - just not very efficiently.

Critical AOA in most airplanes is typically between 15-20 degrees. The bottom surface of the wing will still push quite a bit of air downward when flying at this AOA. The top surface may be turbulent, which reduces the total downwash of air, or "lift", but quite a bit of lift is still generated when the wing is stalled

Stalled wings still produce quite a bit of lift - it's just less than unstalled wings, and typically not enough to maintain level flight

My old Sporty's Private Pilot course (VHS) had an inflight demonstration of this (using a Cherokee) that made it very clear to me during my primary training. They used tufts of yarn all over the wing for flow visualization. Very effective demonstration.

 

[Jaybird180]

Now, now. I'm only trying to make sure you understand that zero Gs mean zero angle of attack (or come up with a creative scenario where it isn't). You hedged on the point, the way I read what you wrote, when you said,

"Personally, I am comforted by knowing that I can unload the wing to break any stall, provided I can reduce load factor to the necessary value approaching 0G, because I KNOW a wing can never stall there (even beyond critical AoA)..."​

If, I say "if", you believe you can have both at once then I see trouble ahead for you someday. Even if you don't, conjuring up an off-beat scenario like, say, during a whip stall where you could, is a worthwhile mental exercise.

Btw, have you tufted your plane yet?

dtuuri

If the plane we are flying is at 18* AoA, it is stalled. If we give a swift shove of the yoke, we are reducing the G load on everything, wings, cabin and its occupants. We can push swiftly enough to reduce G to an amount that will break the stall (this need not be 0G but an amount less that what we previously had and approaching 0 will do based on certain factors discussed below), allowing the boundary layer to reattach. Now we have some lift.

By pushing, we have reduced weight momentarily, which means the wings do not need to support as much weight as they did before but aerodynamics being what it is, we get the resulting lift force based upon the amount of relative wind we have. And even if the boundary layer hesitates in reattaching, at 0G we are not dependent upon lifting force from the wings (FOR THAT MOMENT - we can't push indefinitely, right)

When we cease the push, obviously we are no longer at 18* AoA, we are at a value less than that.

I think this video covers the subject clearer
http://apstraining.com/blog/2009/10/13/the-v-g-diagram-or-v-n-diagram/


Sorry, I haven't done the yarn yet. I thought about it last week, and I don't recall the post where that was discussed.

 

[dtuuri]

I think this video covers the subject clearer
http://apstraining.com/blog/2009/10/13/the-v-g-diagram-or-v-n-diagram/

The point of using the seat of your pants as a poor-man's AoA indicator is valid and is what I'm trying to make sure you understand, but you keep going 'round and 'round the rosemary bush with talk of boundary layers, etc. Although the instructor has a couple things just backwards, i.e., he motions the wrong way when discussing how weight causes the stall curve to shift (EDIT: I stand corrected, I thought he said something different while motioning) and he states that load factors have a direct "impact" on your angle of attack when it's just the opposite--the AoA is the direct cause of any load factor, I very much like his even style of delivery. Can't say I much like the statement, "The V-G Diagram is a relatively unfamiliar diagram to most non-military pilots," at the top of the webpage. Such arrogance.

dtuuri

 

[gismo]

I had to metabolize the two glasses of medicinal red wine I had at dinner before attempting to follow this post. Not sure I can understand what a "constant weight" of a bomber dropping its load means, but I am pretty sure just the opposite happens. The "instant" reaction would be that the plane rises and the radius tightens causing a faster turn rate not a decreased one. Unless, that is, the pilot simultaneously adjusts the angle of attack to the same as the lighter airplane being compared to. Next, the acceleration in speed would decrease rate and radius.

The "constant weight" remark was an attempt to refer only to the non-bomber "unloading the wing" slightly. As to the decreased turn rate
(also WRT the non-bomber) I meant the immediate result of reducing the g load. Due to inertia it will take time for the speed to increase but the turn rate is a direct function of the lateral component of the wing's lift (force) so the rate will diminish exactly as soon and by the same amount as the g load is reduced.

Trying to follow you here... 32 feet/sec² is the acceleration of a falling body if I remember my physics. You're saying it will fall, like a rock? But a plane with more lift (i.e. 1.4/1.4) would fall at 32 ft/sec² and a plane with less lift (i.e. 1.22/1.4) falls less fast?? I'm afraid I don't get it. Probably just me.

I could probably be more clear but I'm too lazy. The example was an airplane starting in a 45° bank level coordinated turn. The plane would "feel" 1.4 g in that turn and if the g-force was reduced to 1.22 by "unloading" the turning force and the vertical lift would decrease by a factor of 1.22/1.4 (=0.87). With that reduction in vertical lift (from 1g) the airplane will accelerate towards the ground at 1-0.87 or 0.13 g. That's where the (1-1.22/1.4) * 32 ft/s^2 comes from. Due to the reduction in the vertical lift the plane's vertical speed will increase 4.1 ft/s^2 (250 FPM per second).

Depending on the airplane it may be necessary to reduce power as the plane accelerates towards the ground (vertical speed increasing) to prevent the airspeed from rising but if the power is adjusted so as to keep the airspeed constant the result is easier to calculate since the turn rate and radius aren't affected by airspeed changes.

I can't even attempt to follow this paragraph until I can understand the first two. I wonder if I'm alone.

dtuuri

Hope that helps (and hope I didn't make any mistakes).

 

[dtuuri]

The "constant weight" remark was an attempt to refer only to the non-bomber "unloading the wing" slightly. As to the decreased turn rate (also WRT the non-bomber) I meant the immediate result of reducing the g load. Due to inertia it will take time for the speed to increase but the turn rate is a direct function of the lateral component of the wing's lift (force) so the rate will diminish exactly as soon and by the same amount as the g load is reduced.

Ok, you are validating my observation that "unloading the wing" from, say, base to final, results in less turn (not what the pilot wants) as well as more stall protection, so it is a self-defeating strategy to correct an overshoot. I'm always alarmed when an engineer seems to disagree with me. Whew!

I could probably be more clear but I'm too lazy. The example was an airplane starting in a 45° bank level coordinated turn. The plane would "feel" 1.4 g in that turn and if the g-force was reduced to 1.22 by "unloading" the turning force and the vertical lift would decrease by a factor of 1.22/1.4 (=0.87). With that reduction in vertical lift (from 1g) the airplane will accelerate towards the ground at 1-0.87 or 0.13 g. That's where the (1.22/1.4) * 32 ft/s^2 comes from. Due to the reduction in the vertical lift the plane's vertical speed will increase 4.1 ft/s^2 (250 FPM per second).

Depending on the airplane it may be necessary to reduce power as the plane accelerates towards the ground (vertical speed increasing) to prevent the airspeed from rising but if the power is adjusted so as to keep the airspeed constant the result is easier to calculate since the turn rate and radius aren't affected by airspeed changes.


Hope that helps (and hope I didn't make any mistakes).

The only mistake is: "That's where the (1.22/1.4) * 32 ft/s^2 comes from."
Don't you mean: That's where the (1-1.22/1.4) * 32 ft/s^2 comes from?

Otherwise, I'm envious of your ability to speak with numbers. B)

dtuuri

 

[gismo]

Ok, you are validating my observation that "unloading the wing" from, say, base to final, results in less turn (not what the pilot wants) as well as more stall protection, so it is a self-defeating strategy to correct an overshoot. I'm always alarmed when an engineer seems to disagree with me. Whew!

My calculation shows that the turn rate is reduced but not by as much as it would be if you reduced the bank angle so that a level turn was possible as the same (reduced) g load. IOW reducing the g load instead of the bank angle does work better from a turn rate perspective albeit at the expense of vertical acceleration.

The only mistake is: "That's where the (1.22/1.4) * 32 ft/s^2 comes from."
Don't you mean: That's where the (1-1.22/1.4) * 32 ft/s^2 comes from?

Yep, I inadvertently abbreviated the reference (corrected now)

Otherwise, I'm envious of your ability to speak with numbers. B)

No envy required, it's part of a curse. :wink2:

 

[dtuuri]

IOW reducing the g load instead of the bank angle does work better from a turn rate perspective albeit at the expense of vertical acceleration.

But... folks are not reducing g load in order to reduce bank angle, they're reducing g load in order to increase bank angle--to prevent an overshoot. Some folks think they can have their cake and eat it too, i.e., steepen the bank without penalty as long as they also reduce g load.

dtuuri

 

[Jaybird180]

But... folks are not reducing g load in order to reduce bank angle, they're reducing g load in order to increase bank angle--to prevent an overshoot. Some folks think they can have their cake and eat it too, i.e., steepen the bank without penalty as long as they also reduce g load.

dtuuri

Hmmmmm......I recall a discussion about base-final where that was the advice. I will try to find it

 

[Clark1961]

But... folks are not reducing g load in order to reduce bank angle, they're reducing g load in order to increase bank angle--to prevent an overshoot. Some folks think they can have their cake and eat it too, i.e., steepen the bank without penalty as long as they also reduce g load.

Not without penalty, they have to surrender altitude...keep your nose down and you likely won't depart...

I know that you know this, I'm just pointing it out to other folks with perhaps a bit less training and experience.

 

[gismo]

But... folks are not reducing g load in order to reduce bank angle, they're reducing g load in order to increase bank angle--to prevent an overshoot. Some folks think they can have their cake and eat it too, i.e., steepen the bank without penalty as long as they also reduce g load.

Pretty much the same thing. If you increase the bank from 35° to 45° without increasing the g loading, the turn rate will increase but not as much as it would if you banked to 45° and increased the g loading so as to maintain a constant altitude or constant rate of descent.

 

[dtuuri]

Pretty much the same thing. If you increase the bank from 35° to 45° without increasing the g loading, the turn rate will increase but not as much as it would if you banked to 45° and increased the g loading so as to maintain a constant altitude or constant rate of descent.

That's what I've been saying--if you unload the wing, you defeat the purpose of banking more. No free lunch.

dtuuri

 

[gismo]

...the turn rate will increase but not as much as it would if you banked to 45° and increased the g loading so as to maintain a constant altitude or constant rate of descent.

That's what I've been saying--if you unload the wing, you defeat the purpose of banking more. No free lunch.

I don't think you're saying the same thing. I'm saying that banking steeper without adding g load does increase the turn rate immediately. The resulting increase in vertical speed is the payment (so I do agree about the cost of lunch except that it's deferred).

 

[dtuuri]

I don't think you're saying the same thing. I'm saying that banking steeper without adding g load does increase the turn rate immediately. The resulting increase in vertical speed is the payment (so I do agree about the cost of lunch except that it's deferred).

So, you think if you just keep the same bank, but push the stick forward it'll tighten the turn too, just not as much as if you steepen the bank also?

Edit: Well, on second thought, I'm sure you don't think that. I don't know why you think we differ then. It might be that you see some trivial trigonometric truth that I don't have any use for. I'm not disputing what you're saying, I just don't see any application for it. If the problem is lining up on final when your bank is currently insufficient for the job, I'd say "Steepen 'er up." To make use of your point, on the other hand, it seems I'd have to say, "Bank more than needed then release back pressure to unload the wing." In order to make that work, you'd need to keep one eye on the G-meter and practice it a lot. Uh, do you have a G-meter?

dtuuri

 

[marcoseddi]

Right Rudder


Sent from my iPhone using Tapatalk

 

[MAKG1]

I don't think you're saying the same thing. I'm saying that banking steeper without adding g load does increase the turn rate immediately. The resulting increase in vertical speed is the payment (so I do agree about the cost of lunch except that it's deferred).

But increasing the turn rate is an increase in acceleration. If you're going to avoid increasing "G load" (which is really just the total acceleration), that has to go somewhere. Where does it go? Acceleration due to constant speed is zero. You could do this with a downward loop or appropriately modified inverted barrel roll, I suppose...

If you just let go of the elevator, you'll get less turn rate, not more and not the same, as you split the straight and level lift vector between turning and holding the plane up.

 

[gismo]

So, you think if you just keep the same bank, but push the stick forward it'll tighten the turn too, just not as much as if you steepen the bank also?

Edit: Well, on second thought, I'm sure you don't think that. I don't know why you think we differ then. It might be that you see some trivial trigonometric truth that I don't have any use for. I'm not disputing what you're saying, I just don't see any application for it. If the problem is lining up on final when your bank is currently insufficient for the job, I'd say "Steepen 'er up." To make use of your point, on the other hand, it seems I'd have to say, "Bank more than needed then release back pressure to unload the wing." In order to make that work, you'd need to keep one eye on the G-meter and practice it a lot. Uh, do you have a G-meter?

dtuuri

Actually I do have a g-meter but the AoA on the glareshield would work better for this. That said I want to make it clear that I'm not advocating "extra-banking" (i.e. increasing bank without increasing g-load) as a means to correct a potential overshoot (or anything else for that matter). I'm just trying to explain that "extra-banking" does increase the turn rate even thought I think it's a bad idea.

 

[Ianovic]

Necro-post, but I made a diagram way back when to describe this whole increase bank to keep from overshooting final, without increasing g-load deal. From above, the flight path of the "overbanked" u turn looks more like an ellipse. The normal turn is more rounded. From a horizontal viewpoint, it's obvious that the "overbanked" plane flew farther, but used the same amount spacing from downwind to final, it just used more altitude because the lift vector was biased more in the horizontal than the vertical. No one cares now but...