Angle of attack during descents

Mig29 · Post by **Mig29** » Tue May 16, 2006 9:18 pm

ok...I have a hard time convincing students about why does inner wing drops first during descending/turning stalls, then when climbing/turning, where outside wings drops obviously and requires less convincing

anyone has any solid theory on 'why' or simple method on explaining?

many thanks!

Degrassi · Post by **Degrassi** » Wed May 17, 2006 10:25 am

Inside wing in a descending turn stalls first, due to the simple fact that the relative airflow is coming more from below then infront, causing it to be at a higher angle of attack...

Lommer · Post by **Lommer** » Wed May 17, 2006 1:41 pm

The inside wing is on the inside of the "circle" and is therefore travelling slower through the air than the outside wing (the path it travels is shorter). Thus, it stalls first.

cyrus · Post by **cyrus** » Wed May 17, 2006 8:39 pm

I have 2 pictures that will explain it to you 100%. PM me for the images.

C-GPFG · Post by **C-GPFG** » Mon Jun 05, 2006 6:31 am

Here's the diagram I used...

laticsdave · Post by **laticsdave** » Mon Jun 05, 2006 8:44 am

Think of a spiral staircase, and the difference between the inner and outer handrails...............

C-KEEP · Post by **C-KEEP** » Mon Jun 05, 2006 10:59 am

Just to add to the confusion, you can also try using a piece of paper to represent the relative airflow. Place it (relatively) horizontal, slightly angled up to represent the airflow coming from below. Take a model airplane and place it in a bank with the nose down.

Now, this is the part where you need a third hand...take two pencils and put them under the wingtips pointing out (to amplify the angle of attacks). Now, compare the angle that the pencil makes with the paper on the inside wing, to the angle that the pencil makes with the paper on the outside wing. These would be the angle of attacks. You'll see that the inside wing has a greater angle than the outside wing.

That's what my instructor used to make it clear to me...

mellow_pilot · Post by **mellow_pilot** » Mon Jun 05, 2006 10:07 pm

As far as turning goes, open the door. Hinges move slower than the handle. Elloquent proof that the outside wing is moving faster.

Now just use the standard lift equation to compare the lift on both wings. All other things being equal (which they're not, but let's keep it simple) the 'V' squared makes the inside wing have less lift than the outside and stalls first.

I think the other explanations for the relative airflow component are bang on. I really don't know how to simplify that, short of a feild trip to a wind tunnel.

Blakey · Post by **Blakey** » Tue Jun 06, 2006 1:03 am

Too easy if you're instructing on a low-wing. Duct tape a yard-long piece of yarn to each wing tip and compare the angles. Even I couldn't miss that! Works great for yaw on landing too; tie a knot in the end of a piece of yarn and put the knotted end in the oil filler door. If it's laying straight back, you're co-ordinated!

quicksilver · Post by **quicksilver** » Tue Jun 06, 2006 9:53 pm

Is that TC approved yarn??

quick

x-wind · Post by **x-wind** » Thu Jun 08, 2006 11:02 am

The inside wing is on the inside of the "circle" and is therefore travelling slower through the air than the outside wing (the path it travels is shorter). Thus, it stalls first.

That's probably the easiest way to understand it. Just make sure they know its dependent on A of A and not speed.

polythene_pam · Post by **polythene_pam** » Tue Jun 13, 2006 12:44 am

The inside wing is ...travelling slower through the air than the outside wing ....Thus, it stalls first.

Here is where your explanation lacks logic... The leap from 'wing travelling slower' to 'wing stalling'... we're teaching that stalls result from high angles of attack, NOT from lack of airspeed.

also,

Now just use the standard lift equation to compare the lift on both wings. All other things being equal (which they're not, but let's keep it simple) the 'V' squared makes the inside wing have less lift than the outside and stalls first.

soooo... You're saying that
lack of airspeed = lack of lift
lack of lift = stall
Sorry, that's just wrong. please stop teaching that.

In your initial stall lesson, you should be teaching:

lack of airspeed = increased AOA to maintain vertical speed(lift stays the same)
excessive AOA = stall

So get the lift equation right out of there. It's got nothing to do with the amount of lift, other than the fact that AOA must increase TO KEEP LIFT THE SAME.

NB: In a balanced turn, if the attidude and bank are constant, whether climbing or descending, lift is virtually EQAL between the two wings. Otherwise, your bank would be changing. Now, when you go to explain the need for inside and outside aileron in climbing and descending turns, you'll have to explain that you lied in a previous lesson. WAY TO CONFUSE YOUR POOR STUDENT!!

I must disagree with the "inside wing moves slower therefore stalls first" explanation in favour of the spiral staircase one.

I haven't done the math on the airspeed difference between the two wings in a turn, but I think its effect is negligible compared to the effect of the AOA difference.

Please, correct me if I'm wrong.

I see the "airspeed difference" explanation as being actually detrimental, as it is an attempt to simplify things but results in confusion. Any critically thinking student will naturally question your explanation.

Their thoughts will develop:In a one-minute turn with a radius of several hundred meters, at speeds of 45-80 knots, how much effect will a few meters difference between the wings be? Maybe a half knot or so from wingtip to wingtip? Can that really be the cause of what can be a fairly drastic wing drop? Or is it infact more, like 15 knots or so? Besides, in a very steep bank (90 degrees), the difference in turn radius from one wing to the other becomes almost nil, yet the wing drop is still pronounced, hmm, how do we explain this?? Besides, didn't the instructor just tell me that stalling is a function of AOA, NOT airspeed?

And, wait a minute, if there is less lift coming from the inside wing, shouldn't the wing drop be due to lack of lift, and NOT due to a stall? which is it?

Suddenly, your student is pondering this when they should be thinking about AOA.

Reminds me of the old "air going over the wing travels further in the same time, so therefore it is faster" explanation of lift... simple, but WRONG.

Or, my favorite, the old "coreolis causes your toilet to drain clockwise" explanation... um, NO!!

Instead of "oversimplifying", why not just explain it right the first time and avoid confusion. If you don't want to explain it properly, just tell them which wing will drop sooner. That's really all they need to know anyway. As for the "why", say: "I'll get into that in more detail later" and move on, instead of making up some neat little incorrect explanation.

Just a question for you "slower wing: less lift: stall" people....

Put the following planes in order from most to least lift:
(all planes are equal, zero bank, travelling in a straight line)

A: an aircraft in level flight, 100 mph
B: an aircraft in 500fpm descent, 100 mph
C: an aircraft in 500 fpm climb., 100mph
D: an aircraft in level flight, 2 knots above stall speed

square · Post by **square** » Tue Jun 13, 2006 9:37 pm

climbing plane, about to stall plane, level plane, descending plane
i say about to stall has more than level plane because the resultant vector (ie the force the wing creates) is a lot bigger than the one the level plane makes. but if you're talking about Lift(tm) the two level planes should have an equal amount.

anyway, someone mind explaining the non-fairy tale reason that the two asymmetric stalls happen?

ill make a stupid-student guess and say outboard stalls in climbing turn cause of slipstream? the slipstream goes straight back, the high wing is above that slipstream whereas the low wing is below it, SO: high wing has slipstream hitting its bottom surface increasing its aoa and low wing has slipstream hitting top, decreasing aoa. i dont get the descending asymmetric stall though.. unless it really is because of a smaller amount of air passing under the inboard wing than the outboard.

quicksilver · Post by **quicksilver** » Tue Jun 13, 2006 11:42 pm

D, A, C, B...maybe. I think Pam must be too smart for me.

quick

JW · Post by JW » Wed Jun 14, 2006 8:54 pm

In option D, what if the stall speed was 98kts?

JW

polythene_pam · Post by **polythene_pam** » Wed Jun 14, 2006 9:15 pm

square: I should have been more concise: I'm asking about TOTAL lift. I implied that by saying "which plane" has more lift, as opposed to which wing. my apologies.

see, I always just like to say that the total lift of all those planes is the same!

If the aircraft is not accelerating vertically, then w = L. It's that simple.

the only time that W =/= L is when the flight path is curved, ie. beginning a climb/decsent, or levelling off... or any time the load factor =/= 1.

I only asked the question to demonstrate the fact that many people, even instructors lack a total understanding of lift and vector geometry.

It relates to banking as well. The only time one wing has more total lift than the other is when the aircraft is banking. Once at a set bank angle, the two wings have virtually equal lift. I say virtually because other factors such as wing dihedral are also involved, but we won't get into that now.

SO, I have a problem with the theory that one wing is going slower, therefore has less lift, yet the wing is not dropping due to lack of lift as in the lift equation... sure, it'll drop due to lack of lift from a stall, but that's a completely different phenomenon which has nothing to do with the lift equation.

Having said that... I think that diagram above (C-GPEG's) is excellent. It demonstrates well the difference in angle of attack while avoiding any mention of the speed difference between the wingtips (which, in this exaggerated diagram, would appear to be quite severe! One could infer that the inside wing is half the speed as the outside one, but thankfully the diagram makes NO mention of airspeed)

I'd like if gpeg had a matching one for climbing!

Strega · Post by **Strega** » Wed Jun 14, 2006 9:18 pm

Lommer · Post by **Lommer** » Wed Jun 14, 2006 10:04 pm

polythene pam, i'd be interested to hear your explanation of why the outside wing stalls in a climbing turn. I think the airspeed explanation makes sense so long as one recognizes the relationship between airspeed and angle of attack. Still, I'm always interested to hear other ways of explaining things (I'm not an instructor and have no aspirations of becoming one, but these things really help in loud debates at the bar).

boozy · Post by **boozy** » Sun Jun 18, 2006 9:09 pm

I think "A" should make more lift than "D" (assuming the stall speed is ya know something around 60 something). I think this is because the effective weight of the plane is higher when level at 100 mph....because, the tail is producing a higher downward force. So the summ of all the downward vectors (known as weight) would be more and the lift would have to be greater to compensate for all this. So I think the total lift is not the same.

quicksilver · Post by **quicksilver** » Sun Jun 18, 2006 11:46 pm

Pam:

see, I always just like to say that the total lift of all those planes is the same

I don't think thats exactly true Pam. After thinking about this for a bit and spending about 1 minute looking in a text book (because I'm tired and already have my licence, not because it was easy to find) I think that each plane is for sure producing a different amount of total lift, close to the same but not. The only one I'm really not sure about would be the one in slow flight, because it has so much drag its resultant of lift to drag will be a lot greater than the one cruising at 100mph. So slow flight aircraft I think might be producing the most lift, but again that is the only one I'm not quite sure about. Other than that aircraft A then C, and last but not least would be B. A,C, and B are all explained pretty well in Kersh*** but I didn't look hard enough to find slow flight aircraft. Anyone else care to explain what I can't?

quick

Angle of attack during descents

Angle of attack during descents

Ok

hmmm