CFI task: Flight at minimum controllable airspeed (slow flight)

[flyingcheesehead]

Is load factor the only delta between straight and level and turning flight?

What about increased induced drag on the high wing?

Isn't there form drag imposed by the rudder required to counter adverse yaw?

Okay, good points.

What does low wing/high wing have to do with induced drag? I'm guessing you didn't mean to say "high".

Let's see what the increase in induced drag would be. The load factor in a level 10-degree banked turn is 1/cos(10º) = 1.015 g. To find the ratio between the existing induced drag and the final induced drag, we'll look at the induced drag equation

and note that the only thing that changes is L, so everything else will cancel and we find that the ratio of induced drag in the turn to induced drag in level flight is the square of the load factor or about 3% more than it was in level flight. Assuming we were at, say, 15" MAP in level flight, that's going to require a change of only slightly less than a half inch of MAP.

As for form drag, since we're very slow it's relatively low. Each airplane will be different here, since some have more adverse yaw than others, but those same airplanes require some rudder deflections even in level flight due to the various air burbles you run into in a non-ideal air mass - So, this shouldn't be a significant additional source of drag.

Here's another thought that brings up: If you are truly in ideal, theoretical, MCA in a turn, when you attempt to level out, you will spin.

 

[flyingcheesehead]

Why this is still continuing I'm not sure.

Because I'm an engineering student (again) and I have a mental disease that makes me enjoy going through this learning process. Plus, I need a break from mutual inductance and damping ratios.

The point of the maneuver (at the stage where we make turns in Slow Flight) IMHO is to teach the loss of lift due to the Horizontal Component taking away from the Vertical Component of Lift.

I thought we were talking about CFI-level stuff here, not private.

Also, the horizontal component does not directly "take away from" the vertical component - That implies subtraction.

With a load factor of 1g, at a 10-degree bank the horizontal component is 0.174g but our vertical component is still .984g. The sum of the two is NOT equal to 1.

I'll say this again, if truly at the edge of a stall (varying portions of the wing stalling independently aside) any loss of the Horizontal Component (which is keeping you at your altitude) will result in a loss of altitude.

The Vertical component is what is keeping us at our altitude - But I'm sure you knew that.

In addition, by completely tossing out the non-ideal factors as you have in your quote, you make this all a moot point. Yes, if you toss out the non-ideal factors and make this a perfectly ideal slow-flight-at-the-critical-angle-of-attack exercise, you're absolutely right. But we're talking about real airplanes in the real sky here.

 

[flyingcheesehead]

Simple, since Kent's calculation of load factor at 10 degrees increases stall speed by .5 knots he thinks no power adjustment is necessary.

... no power adjustment is necessary due to load factor.

What he fails to consider is the loss of vertical lift to horizontal is in addition to load factor.

No, the loss of vertical lift to horizontal is *what causes the need for* the additional load factor. They are the same effect, and the stall speed increase is part of that same effect:

Bank -> loss of a small part of vertical lift component -> increased load factor to maintain altitude -> increased stall speed

What *is* in addition, as you pointed out, is the 3% increase in induced drag due to the increased load factor.

I still maintain that in the real world where the stall is non-ideal that an application of power isn't the way to go, for a couple of reasons: First, I know that I can't move the throttle accurately enough to gain only the 1/2" MP, thus I would theoretically be speeding up and getting away from true MCA flight, and second, real-world stalls are non-ideal as discussed in a previous post and it's not possible to fly so close to the edge of the stall in level flight that a 10-degree or less bank will cause an immediate stall.

More likely, while leaving the throttle alone (assuming just the right setting to begin with), the extra 3% induced drag will *eventually* cause enough of a slowdown that the wing fully stalls. However, the time it takes for this to happen means that on a checkride, the maneuver is probably already over.

 

[dmccormack]

What does low wing/high wing have to do with induced drag? I'm guessing you didn't mean to say "high".

"High wing" = the one that's up in a bank (in a left bank, the right wing would be up or "high" compared to the left).

As for form drag, since we're very slow it's relatively low. Each airplane will be different here, since some have more adverse yaw than others, but those same airplanes require some rudder deflections even in level flight due to the various air burbles you run into in a non-ideal air mass - So, this shouldn't be a significant additional source of drag.

Excellent discussion on load factor.

Yet t in a High L/D scenario (such as flight at MCA), induced drag is a greater proportion of the total equation (see AFNA).

While the drag imposed by rudder deflection is minimal, it's still drag.

Here's another thought that brings up: If you are truly in ideal, theoretical, MCA in a turn, when you attempt to level out, you will spin.

You may drop a wing -- not sure why you'd spin.

 

[dmccormack]

Also, the horizontal component does not directly "take away from" the vertical component - That implies subtraction.

Remove "load factor" from the equation.

What makes the airplane turn?

Everything else equal: Is there a reduction in the vertical component of lift in a turn?

 

[gismo]

Everything else equal: Is there a reduction in the vertical component of lift in a turn?

I'm having a hard time isolating "everything else" since so many parameters are interlinked but if you maintain the same average airspeed (each section of each wing sees a different local airspeed) and keep the same AoA the total lift will be constant and the lift vector opposing gravity will be less than it was in wings level flight. As a result the airplane will accelerate towards the ground (i.e. the VS will increase at a constant rate) if the above mentioned parameters are held constant. I believe the rate of acceleration will decrease and eventually diminish due to drag (just as a falling human will do as they approach terminal velocity) but I suspect that the terminal velocity of an airplane falling in an otherwise normal slightly banked attitude would be in the several thousand FPM range.

But the tricky part is maintaining a constant AoA during such a descending turn as the VS itself causes an increase in the AoA unless the pitch attitude is altered.

 

[dmccormack]

I'm having a hard time isolating "everything else" since so many parameters are interlinked but if you maintain the same average airspeed (each section of each wing sees a different local airspeed) and keep the same AoA the total lift will be constant and the lift vector opposing gravity will be less than it was in wings level flight.

It was a somewhat rhetorical question -- the answer is "no, there is a loss of vertical component equal to that exchanged for the horizontal component of lift."

Which is why we have to add power, increase AoA or both to maintain altitude in a turn. Given that we should be at a very high AoA straight and level during MCA, any turn will reduce total vertical lift. We can't increase AoA to maintain altitude, so we have to add power.

 

[Commander Dave]

The Vertical component is what is keeping us at our altitude - But I'm sure you knew that.

You are correct - I meant to say Vertical

It has been edited in my post.

 

[flyingcheesehead]

"High wing" = the one that's up in a bank (in a left bank, the right wing would be up or "high" compared to the left).

Ah, OK, I get it.

Yet t in a High L/D scenario (such as flight at MCA), induced drag is a greater proportion of the total equation (see AFNA).

While the drag imposed by rudder deflection is minimal, it's still drag.

True - All I'm saying is, that drag is insignificant in the real world and would not require a power increase.

You may drop a wing -- not sure why you'd spin.

I want to change my answer. Without increasing your speed from theoretical MCA first, you'd never make it into the turn to begin with. Any application of aileron would drop the outside wing before the turn even started. If you did get into the turn, attempting to level out would drop the inside wing. In either case, the sudden drop of one wing and the loss of half your lift would drop the fuselage, thus the other wing would drop slightly as well. Now you have an increased angle of attack on both wings, with the dropped wing having more of an increase than the upper wing. Both wings stalled, one more than the other: Spin.

But again, in the real world this doesn't happen because theoretical MCA flight is not possible in an actual airplane.

This is a fun discussion.

 

[flyingcheesehead]

Remove "load factor" from the equation.

What makes the airplane turn?

Everything else equal: Is there a reduction in the vertical component of lift in a turn?

Yes, there is a slight reduction of the vertical component of lift in a turn. My entire assertion throughout this discussion has been that in a real air mass with a real airplane, you're not going to be flying right on the edge of the stall because it's not possible to maintain that ideal situation in a real airplane, and that because of that the tiny loss of the vertical component is not enough to cause an immediate stall.

This is mainly a discussion of theoretical issues vs. the real ones.

 

[flyingcheesehead]

It was a somewhat rhetorical question -- the answer is "no, there is a loss of vertical component equal to that exchanged for the horizontal component of lift."

Incorrect!

The relationship between the horizontal and vertical components is NOT linear, you are NOT exchanging an *equal* component.

In a 10-degree bank with a 1g load factor, the vertical component is 0.985g and the horizontal component is 0.174g. 0.985+0.174 = 1.159. Therefore, we have not exchanged an equal component of vertical lift for horizontal lift - We have given up only 0.015g of our vertical lift, and gained over 10 times that amount, 0.174g, in horizontal lift.

A more extreme example would be a 1g load factor, 30-degree bank. In this case, we would still have 0.866g of vertical lift, but we would have 0.5g of horizontal lift! Add the two, and you would have 1.366, but we still have only a 1g load factor!

Make friends with thy sines and cosines, and they will help you turn.

 

[azure]

Incorrect!

The relationship between the horizontal and vertical components is NOT linear, you are NOT exchanging an *equal* component.

While this is correct, there is another way of thinking about doing arithmetic with the lift components in which it is perfectly correct to speak of adding and subtracting or exchanging one for another: vector addition.

If the total lift vector is the same magnitude as before bank, then the remaining vertical lift vector is just what is left of the total lift vector when you (vectorially) subtract off the horizontal component of lift.

But yep, to get the magnitudes of those horizontal and vertical components, it helps to make friends with your sines and cosines.

 

[Cap'n Jack]

Incorrect!

0.985+0.174 = 1.159.

You may wish to check this again- that's not how you add vectors. It should be (0.985**2+0.174**2)**0.5=1, which is your 1g load factor.

A more extreme example would be a 1g load factor, 30-degree bank. In this case, we would still have 0.866g of vertical lift, but we would have 0.5g of horizontal lift! Add the two, and you would have 1.366, but we still have only a 1g load factor!

Again, incorrect. (0.866**2 + 0.5**2)**0.5 = 1 again.

The conclusion is correct but the way you got there isn't.

 

[dmccormack]

Incorrect!

The relationship between the horizontal and vertical components is NOT linear, you are NOT exchanging an *equal* component.

My statement was oversimplified, but reaches the same conclusion -- there is an exchange -- vertical or horizontal or some combination.

And yes -- even in "real world" flying, when you go from straight and level to turning, there will be power adjustment if you want to maintain altitude.

At 38 MPH a standard rate turn is about 9 degrees, and I'll tell you -- it looks like 10 degrees to me.

 

[flyingcheesehead]

While this is correct, there is another way of thinking about doing arithmetic with the lift components in which it is perfectly correct to speak of adding and subtracting or exchanging one for another: vector addition.

If the total lift vector is the same magnitude as before bank, then the remaining vertical lift vector is just what is left of the total lift vector when you (vectorially) subtract off the horizontal component of lift.

But yep, to get the magnitudes of those horizontal and vertical components, it helps to make friends with your sines and cosines.

I'm missing something here. Whether you're thinking of graphical vector addition or sines and cosines, the magnitude of the horizontal component plus the magnitude of the vertical component does not equal 1 when the total lift vector equals 1 unless one of them equals 0 (level or knife-edge).

Right?

 

[flyingcheesehead]

You may wish to check this again- that's not how you add vectors. It should be (0.985**2+0.174**2)**0.5=1, which is your 1g load factor.



Again, incorrect. (0.866**2 + 0.5**2)**0.5 = 1 again.

The conclusion is correct but the way you got there isn't.

All you're doing is multiplying by 2 and dividing by 2 - That doesn't change anything, and doesn't equal 1. ?!?

 

[flyingcheesehead]

And yes -- even in "real world" flying, when you go from straight and level to turning, there will be power adjustment if you want to maintain altitude.

And I still disagree. I have never added power to make a turn in slow flight, and I've never stalled, and I've passed all my checkrides...

Plus, as I stated before, the amount of power necessary is so small it'd be nearly impossible to adjust properly.

I think we should go up together and test it out.

 

[azure]

I'm missing something here. Whether you're thinking of graphical vector addition or sines and cosines, the magnitude of the horizontal component plus the magnitude of the vertical component does not equal 1 when the total lift vector equals 1 unless one of them equals 0 (level or knife-edge).

Right?

Right, and I never said it did. I was only speaking to the usual rough language where one talks about losing some vertical lift to the horizontal component. It's correct to talk that way only in the context of vector addition. It doesn't change the fact that the magnitudes of the components don't add arithmetically -- they add "in quadrature".

 

[flyingcheesehead]

Right, and I never said it did. I was only speaking to the usual rough language where one talks about losing some vertical lift to the horizontal component. It's correct to talk that way only in the context of vector addition. It doesn't change the fact that the magnitudes of the components don't add arithmetically -- they add "in quadrature".

Aha! OK, that's what I thought but I misunderstood what you were trying to say.

 

[Clark1961]

All you're doing is multiplying by 2 and dividing by 2 - That doesn't change anything, and doesn't equal 1. ?!?

Naw, He's using FORTRAN nomenclature. ** is the same as ^ so the formula is squaring the value. He coulda just said something about the square of the hypotenuse is equal to sum of the squares of the legs of a right triangle...

(if'n yer gonna be an electrical engineer then ya better have that one down cold...or I suppose you could just work DC)

 

[flyingcheesehead]

Naw, He's using FORTRAN nomenclature. ** is the same as ^ so the formula is squaring the value. He coulda just said something about the square of the hypotenuse is equal to sum of the squares of the legs of a right triangle...

(if'n yer gonna be an electrical engineer then ya better have that one down cold...or I suppose you could just work DC)

Holy crap... It's been, um, *cough*enteen years since I've looked at any FORTRAN code! Yikes.

But yes, in that case it does in fact equal one.

 

[dmccormack]

And I still disagree. I have never added power to make a turn in slow flight, and I've never stalled, and I've passed all my checkrides...

Plus, as I stated before, the amount of power necessary is so small it'd be nearly impossible to adjust properly.

I think we should go up together and test it out.

Passing checkrides <> practing MCA.

:wink2:

The checkride exercise equals getting the stall horn on and flying around that way. I took my Comm in an A36 with VGs - I probably had 12 knots from stall with the horn on. Made it sorta easy.



No horn, lights, or other newfangled doodads in the Chief -- just an eerie quiet and then you stop flying -- but even then it doesn't drop. With any power over idle it basically mushes (36' Clark Y airfoil and 1200 lbs means it converts from airplane to parachute).

Adverse yaw really is adverse on those long wings, even if the ailerons are frise type. The big rudder is there for a reasons.

When I do slow flight I am constantly sawing at the throttle -- may be habit, but the lightness in the seat tells me I'm losing lift. Maybe I'm just losing brain cells?

Anyway, once I get it all re-assembled I will go out and test.



If you want to come up with me we may need a day below 50, no gas, no shoes, and be sure you eat a very light breakfast.

 

[nosehair]

First, I know that I can't move the throttle accurately enough to gain only the 1/2" MP,

Ah, but you can! You can. That's one of the primary controls that you learn to use in coordination with the elevator to stay on top of it.

You can move the throttle an infinite amount just like the elevator. You can't 'see' this movement on the guage, but you don't need to. You don't have an angle of attack indicator for the elevator, and even if you did, you wouldn't use it for elevator control during the landing.

The training purpose of slow flight is to learn to feel the throttle and elevator, as well as rudder/aileron coordination during the almost stalled condition during the landing.

I can 'feel' my way in and out of the stall with ever-so-slight throttle and elevator feather-like motions. But I have to be very in-touch with the throttle along with the rest of the airplane flight controls.

 

[flyingcheesehead]

The checkride exercise equals getting the stall horn on and flying around that way. I took my Comm in an A36 with VGs - I probably had 12 knots from stall with the horn on. Made it sorta easy.

Cheater.

When I do slow flight I am constantly sawing at the throttle -- may be habit, but the lightness in the seat tells me I'm losing lift. Maybe I'm just losing brain cells?

But that would cause your weight to go down and thus you wouldn't need to add power.

If you want to come up with me we may need a day below 50, no gas, no shoes, and be sure you eat a very light breakfast.

Yup. Or we could go up in the 182, in which case we might want to throw your airplane in the back seat to help us stall. I've held it in a full-power MCA situation and tried to stall it - After it climbed 4,000 feet and I couldn't get it to break, the CHT's got to the point where I had to quit trying. (Solo on a cold day.)

FWIW, the Diamond doesn't really "break" in a stall either, it'll just kind of mush and dance around. Does a great falling leaf though - I guess Diamond has pointed out to some people on demos that the descent rate of a DA40 held in the stall is significantly less than the descent rate of a Cirrus under chute.

 

[dmccormack]

Yup. Or we could go up in the 182, in which case we might want to throw your airplane in the back seat to help us stall. I've held it in a full-power MCA situation and tried to stall it - After it climbed 4,000 feet and I couldn't get it to break, the CHT's got to the point where I had to quit trying. (Solo on a cold day.)

Same in the C205 -- will simply mush.

 

[nosehair]

the descent rate of a DA40 held in the stall is significantly less than the descent rate of a Cirrus under chute.

Now that's worth repeating.

 

[flyingcheesehead]

Now that's worth repeating.

Ummm... It's true.

 

[Cap'n Jack]

All you're doing is multiplying by 2 and dividing by 2 - That doesn't change anything, and doesn't equal 1. ?!?

Naw, He's using FORTRAN nomenclature. ** is the same as ^ so the formula is squaring the value. He coulda just said something about the square of the hypotenuse is equal to sum of the squares of the legs of a right triangle...

(if'n yer gonna be an electrical engineer then ya better have that one down cold...or I suppose you could just work DC)

FORTRAN among others. I think some versions of BASIC also used ** instead of ^. I get them all mixed up now.

Holy crap... It's been, um, *cough*enteen years since I've looked at any FORTRAN code! Yikes.

But yes, in that case it does in fact equal one.

It's cleaner than calling what are essentially subroutines in C/C++ (where ^ is an XOR) or in PASCAL.

 

[gismo]

Holy crap... It's been, um, *cough*enteen years since I've looked at any FORTRAN code! Yikes.

But yes, in that case it does in fact equal one.

Shoot, the last time I looked at any FORTRAN was over 25 years ago but I still recognized the exponentiation operator. Then again I can't recall the name of the pilot I just met yesterday. Funny how some things stick better than others.