Impossible Turn Practice

[azure]

Chris, are you thinking that the increase in kinetic energy comes from the loss of gravitational potential energy in the descent? Then how do you explain that the loss of G.P.E. is exactly the same in a reference frame co-moving with the wind as in a ground-stationary frame, yet the change in kinetic energy is different between the two frames?

Harry's Bob Hoover example is a good one -- even better if Bob adds power and maintains altitude. The power expended by the airplane is the same (I think -- need to check this) in both frames, but again you see that the kinetic energy changes only in the ground frame.

The "extra" kinetic energy in the ground frame has to come from the wind. Where else can it come from?

This is similar to analyses of elastic collisions between two balls where the total energy is conserved in all reference frames, but the energy exchanged between them is definitely NOT a Galilean invariant.

 

[TripleZ]

Please explain what this means in mathematical terms. Are you saying that one choice of frame of reference is more correct than another for the purposes of this problem?

Can you address the questions I raised in my previous posting? You're saying the two maneuvers result in different altitude loss. Do they take a different amount of time to complete?
-harry

In the math model I set up, rate of turn was constant at 60 seconds for 180 degrees, for both setups. The turn was identical for displacement in the X and Y directions (360/180 and 90/270) respectively. Time to completion of the turn was identical (in the model).

Airspeed was simply calculated from the magnitude of Vx and Vy. This leaves out Vz, which when you include the slightly higher altitude loss, it would change the actual airspeed seen by the plane (the magnitude of Vx, Vy, and Vz) slightly. Airspeed was held constant through the turn, with drag being the only negative effect (based on time). Potential energy of the aircraft was held constant, minus the energy loss due to drag.

With the addition of Vz, since you're covering a slightly longer arc distance, I would expect the time of turn to take slightly longer. That gets very complicated for equations though.........



edit:\\
Ok.... Well Vz would not effect "airspeed" since its not air flowing over the wings, rather the vertical speed of the airplane is independent of the "turn" speed of the airplane. Which is why I originally ignored Vz.

The turn 360 to 180, would take the exact identical time to complete. The distance traveled in the X and Y directions should be identical, only the distance in the Z direction and Vz would change. The distance the plane travels is greater, but airspeed (Vx and Vy) remain the same, while total velocity (magnitude of Vx, Vy, Vz) would increase.

 

[tonycondon]

if the rate of turn (degrees/second) and the angle of turn (degrees) were the same between both turns, why does one take slightly longer than the other?

 

[TripleZ]

Oh, crap.... it gets more complicated. Based on the angle of attack of the airplane, Vz must be accounted for as being SOME component of airspeed.......

my head hurts.

 

[tonycondon]

that's because you're thinking too hard

 

[TripleZ]

that's because you're thinking too hard

You know this goes back to "Do engineers make good pilots?" thread.
(after all the over thinking, over analyzing...... Ill be underground, before I make the turn)

 

[flyingcheesehead]

And what are you doing during the "impossible turn" practice?

I immediately push forward and turn -- simultaneously.

Guess what happens to the G-load?

The G-load momentarily decreases with the push and downward acceleration, and as the downward acceleration slows it goes up through 1 G and when the downward acceleration hits 0 (constant rate of descent) then the G-load is exactly the same as it would be in a level turn.

You're not going to be accelerating downward through an entire 180º turn, especially since your turn rate will be lower at a particular bank angle until the downward acceleration stops.

 

[mantakos]

... The turn 360 to 180, would take the exact identical time to complete...

Note that if both aircraft (wind and no-wind) take the same amount of time to do the maneuver, but one loses more altitude, that would require the "wind" aircraft to have a higher rate of descent.

So why would the rate of descent vary between the aircraft? Again, aerodynamically, the situations are identical.
-harry

 

[dmccormack]

The G-load momentarily decreases with the push and downward acceleration, and as the downward acceleration slows it goes up through 1 G and when the downward acceleration hits 0 (constant rate of descent) then the G-load is exactly the same as it would be in a level turn.

You're not going to be accelerating downward through an entire 180º turn, especially since your turn rate will be lower at a particular bank angle until the downward acceleration stops.

For at least 100 degrees of the turnback there is downward acceleration (we're pushing and turning steeply simultaneously), and then gradually decreasing forward pressure as Vg is maintained but bank is reduced once 200 degrees or so are past.

Therefore I'll argue that G loading never exceeds 1 G since in the steepest part of the turn there is forward push. Then bank is being reduced as Vg is attained and maintained.

 

[jesse]

Because my math is not a fictional character. It would be nice if someone observed this. But in reality, even in a strong headwind the difference in altitude is minimal to the entire change. So the casual observer, seeing a 7% difference in altitude (WITH NO MEASURING EQUIPMENT), won't ever "see" the difference.

On top of this, how often do pilots have the opportunity to do two perfectly identical turns with no power. Both with wind and without wind?

If you use the fluid as the point of reference, you are ignoring inertia of the plane. Remember, in all of these examples we have been using pure straight line wind speeds. In reality, the wind gusts and changes speed.

In order to make up the airspeed from a drop in wind speed....... you drop the nose (loose altitude) or add power. You add energy to accelerate the mass. It can come from either altitude or the engine.

The plane does not instantaneously change speed with the wind. In frames of the turn, you accelerate the mass from 50kts to 80kts. In terms of a wind gust, you may only need to accelerate 15kts (small drop in altitude).

You need to think about this harder. You are not right and you are aruging with experienced pilots and some very smart folks. Don't go into this thinking you are right, realize you are wrong and try to figure out why.

An airplane doing circles in an airmass is not going to lose airspeed or altitude by cause of it's movement over the earth. The airplane does not give a **** about the earth - only the airmass. If you maintain a constant 100 knot at constant altitude in a turn your airspeed won't change even if that airmass is moving 100 knots over the ground.

A gust of wind is completely different then a constant airmass.

 

[denverpilot]

I once hovered a Cutlass at 11,000 feet over Puget Sound. I decided to pick an altitude with less of a headwind!

There's a reason its nickname is "Gutless"...

 

[denverpilot]

You know this goes back to "Do engineers make good pilots?" thread.
(after all the over thinking, over analyzing...... Ill be underground, before I make the turn)

My Gallean invariant hurt once, so I went to the doc and he gave me something for it.

 

[gismo]

The plane does not instantaneously change speed with the wind. In frames of the turn, you accelerate the mass from 50kts to 80kts. In terms of a wind gust, you may only need to accelerate 15kts (small drop in altitude).

Chris, the resulting altitude loss in a turn cannot be different in different reference frames such as the air mass and the ground (assuming no vertical air movement). Any calculation that shows otherwise has to be flawed.

Also you keep talking about accelerating the airplane (relative to the ground) when turning from upwind to downwind it totally bogus. The only acceleration affecting the plane is from the horizontal component of the lift vector and that's constant for a given airspeed and bank angle regardless of any (constant) wind. Perhaps if you were to try and visualize this non-existent acceleration using the airmass as your frame of reference. Rather than looking at it as if the air was moving over the ground, consider instead that the ground is moving under the air. Now where is the force generating your acceleration coming from?

 

[flyingcheesehead]

For at least 100 degrees of the turnback there is downward acceleration (we're pushing and turning steeply simultaneously), and then gradually decreasing forward pressure as Vg is maintained but bank is reduced once 200 degrees or so are past.

Therefore I'll argue that G loading never exceeds 1 G since in the steepest part of the turn there is forward push. Then bank is being reduced as Vg is attained and maintained.

No way...

Let's say you're climbing at 600 fpm and your Vg descent rate is 600 fpm, and you do a pushover where you're at 0.8G during the push. It will only take 3.125 seconds to go from +600fpm to -600fpm.

Even at a 45º bank and only 60 knots, rate of turn will be 18.2 degrees per second, so it'll take 10 seconds to make a 180. If we include the push in that, it'll take even longer (less total lift = less horizontal lift component to make the turn).

Let's say we make the push a 0.9G push instead - Your vertical rate will change by 1920 fpm (or, for example, 1000 fpm climb to 920 fpm descent) in 10 seconds.

Sorry Dan, the math just doesn't add up. You're going to be at >1G for a good part of the turn. And I'd happily sit on a scale in your plane to prove it.

 

[COFlyBoy]

A gust of wind is completely different then a constant airmass.

Ding! Ding!

A hundred points for Jesse.

Simple Newtonian mechanics only applies in intertial (ie. non-accelerating) reference frames.

Add in an accelerating reference frame (using gusting wind as your reference) and you've got some complicated math to do.

 

[dmccormack]

No way...

Let's say you're climbing at 600 fpm and your Vg descent rate is 600 fpm, and you do a pushover where you're at 0.8G during the push. It will only take 3.125 seconds to go from +600fpm to -600fpm.

Even at a 45º bank and only 60 knots, rate of turn will be 18.2 degrees per second, so it'll take 10 seconds to make a 180. If we include the push in that, it'll take even longer (less total lift = less horizontal lift component to make the turn).

Let's say we make the push a 0.9G push instead - Your vertical rate will change by 1920 fpm (or, for example, 1000 fpm climb to 920 fpm descent) in 10 seconds.

Sorry Dan, the math just doesn't add up. You're going to be at >1G for a good part of the turn. And I'd happily sit on a scale in your plane to prove it.

So, how much? +1.001 G?



I've tried it -- several times. If I maintain 60 MPH in the Chief throughout the steep turn (60 degree bank initially while pushing), I'm golden.



Sadly, we'll need to wait until December for it to be cold enough to lift you and me and a scale.

 

[flyingcheesehead]

So, how much? +1.001 G?

Nope - As previously stated, whatever it'd be in a level turn. Going with the 45º bank example, 1.414 G. And you can't neutralize all of that in the turn without increasing your vertical acceleration quite a bit.

I've tried it -- several times. If I maintain 60 MPH in the Chief throughout the steep turn (60 degree bank initially while pushing), I'm golden.

Hmmm. In a 60-degree turn, are you pushing, or ruddering?

Sadly, we'll need to wait until December for it to be cold enough to lift you and me and a scale.

Hey, I'm all for winter flying.

 

[TripleZ]

You need to think about this harder. You are not right and you are aruging with experienced pilots and some very smart folks. Don't go into this thinking you are right, realize you are wrong and try to figure out why.

An airplane doing circles in an airmass is not going to lose airspeed or altitude by cause of it's movement over the earth. The airplane does not give a **** about the earth - only the airmass. If you maintain a constant 100 knot at constant altitude in a turn your airspeed won't change even if that airmass is moving 100 knots over the ground.

A gust of wind is completely different then a constant airmass.

I don't think I'm right..... but I don't see where Im wrong yet. If that makes sense. I keep working at the problem until I find the solution, and I have to start at my end..... because I don't know why your end is correct yet.

It's not to be belligerent to the other pilots, just my way of working through it. Every response brings a new piece to the puzzle.

Chris, the resulting altitude loss in a turn cannot be different in different reference frames such as the air mass and the ground (assuming no vertical air movement). Any calculation that shows otherwise has to be flawed.

Also you keep talking about accelerating the airplane (relative to the ground) when turning from upwind to downwind it totally bogus. The only acceleration affecting the plane is from the horizontal component of the lift vector and that's constant for a given airspeed and bank angle regardless of any (constant) wind. Perhaps if you were to try and visualize this non-existent acceleration using the airmass as your frame of reference. Rather than looking at it as if the air was moving over the ground, consider instead that the ground is moving under the air. Now where is the force generating your acceleration coming from?

For the frame of reference of the air, a turn in dead calm or wind has nothing different. Identical. I completely agree. My math model even shows that. The forces acting on the wings does not care about ground speed, only airspeed.

But the forces acting on the BODY of the airplane, inertia, are referenced to the earth.

Hmm.... lets see. Another way of looking at it.

If an airplane is flying at 65knots in dead calm. (65knot groundspeed). And a headwind increases to 65knots, and you maintain airspeed. The plane has not accelerated with reference to the wind... correct?
Your ground speed has decelerated with respect to the ground... to 0kts... correct?

You would feel that deceleration, the ball hanging from the mirror per say, would swing forward. Right?

 

[azure]

But the forces acting on the BODY of the airplane, inertia, are referenced to the earth.

No! Forces are not referenced to a frame. Because acceleration is a time rate of change of velocity, accelerations (and therefore forces) are the same in all inertial frames.

(This is Galilean relativity btw, NOT Einsteinian where the force is a three vector and not invariant.)

If an airplane is flying at 65knots in dead calm. (65knot groundspeed). And a headwind increases to 65knots, and you maintain airspeed. The plane has not accelerated with reference to the wind... correct?

Yes.

Your ground speed has decelerated with respect to the ground... to 0kts... correct?

Yes.

You would feel that deceleration, the ball hanging from the mirror per say, would swing forward. Right?

Yes... but note that the wind is no longer an inertial reference frame. That makes the analysis harder.

In your previous example, both the ground frame and the wind frame are inertial. And the acceleration of the plane (as a vector) with respect to either is exactly the same. That's true even though the speed changes in the ground frame and not in the wind frame, because in the ground frame that acceleration isn't purely centripetal.

 

[dmccormack]

Nope - As previously stated, whatever it'd be in a level turn. Going with the 45º bank example, 1.414 G. And you can't neutralize all of that in the turn without increasing your vertical acceleration quite a bit.

Hmmm. In a 60-degree turn, are you pushing, or ruddering?

You're intentionally confusing the issue.

A 60 degree bank is achieved early in the maneuver -- push, turn aggressively -- as always -- a coordinated turn (I'm not sure why you would expect "ruddering" just because it's 60 degree bank?)

The push achieves several goals immediately-- it unloads the wing, allows rapid roll, reduces the Gs felt in the bank, and helps maintain airspeed (since we were climbing when we lost power).

The nose is down significantly to maintain 60 MPH in that steep bank.

The earth comes around pretty quickly and then I reduce the bank gradually as the push pressure is reduced. Now we're in energy conservation mode -- don't maintain that nose down attitude -- we need every foot.

I've practiced this only power to idle and at altitude -- other Aeronca pros have done this with engines out but I'm too chicken (so far).

My real-life turnback had enough residual power that I was able to make it back to the airport and then had to deal with too high, too close. Thank God for slips and draggy airframes -- Fairmont has a 30' hill on the southern edge of the field and there was no way I could climb over it if I wasn't down and stopped (see 4G7).

 

[bbchien]

Chris, Read Azure's post (#99) carefully. She is a Ph.D. physicist. VERY carefully.

 

[TripleZ]

No! Forces are not referenced to a frame. Because acceleration is a time rate of change of velocity, accelerations (and therefore forces) are the same in all inertial frames.

(This is Galilean relativity btw, NOT Einsteinian where the force is a three vector and not invariant.)


Yes.

Yes.

Yes... but note that the wind is no longer an inertial reference frame. That makes the analysis harder.

In your previous example, both the ground frame and the wind frame are inertial. And the acceleration of the plane (as a vector) with respect to either is exactly the same. That's true even though the speed changes in the ground frame and not in the wind frame, because in the ground frame that acceleration isn't purely centripetal.

Got it. Understand.

Simple way in my mind.
Mid-point of the turn for a 15kt head wind is NOT 0kt ground speed, its -15kt ground speed. A deceleration of 50kt to -15kt for a total of 65kt. The plane then accelerates from -15kt to 80kt for a total of 65kt.

Which is identical to the calm wind condition, except there, the midpoint of the turn ground speed is 0kts. The same energy is spent accelerating the mass of the aircraft during the turn.

The shapes of the turns from calm to windy should have clued me in earlier.



Thanks for your patience guys.

 

[flyingcheesehead]

You're intentionally confusing the issue.

Well... Not really. I was unintentionally confusing it.

A 60 degree bank is achieved early in the maneuver -- push, turn aggressively -- as always -- a coordinated turn (I'm not sure why you would expect "ruddering" just because it's 60 degree bank?)

Meh... Forget that. Rudder would get your nose down more than elevator once you were in a 60º*bank... But that's irrelevant. So never mind.

The push achieves several goals immediately-- it unloads the wing, allows rapid roll, reduces the Gs felt in the bank, and helps maintain airspeed (since we were climbing when we lost power).

No doubt on any of those. Where we differ is in how long the push is actually reducing the G's. Especially at a 60º*bank!

I think there was an E-R prof who studied the issue and found that the optimum bank angle was 45º as well. 60º gets you around the turn faster, but significantly increases stall speed and drag...

The nose is down significantly to maintain 60 MPH in that steep bank.

... and that's a good indicator of the same.

 

[Palmpilot]

No! Forces are not referenced to a frame. Because acceleration is a time rate of change of velocity, accelerations (and therefore forces) are the same in all inertial frames.

(This is Galilean relativity btw, NOT Einsteinian where the force is a three vector and not invariant.)


Yes.

Yes.

Yes... but note that the wind is no longer an inertial reference frame. That makes the analysis harder.

In your previous example, both the ground frame and the wind frame are inertial. And the acceleration of the plane (as a vector) with respect to either is exactly the same. That's true even though the speed changes in the ground frame and not in the wind frame, because in the ground frame that acceleration isn't purely centripetal.

I was hoping a physicist would jump in here and rescue us. Thanks!

As an engineer, my tendency is to just pick whichever reference frame makes the analysis easy. When someone insists on using a reference frame that makes the analysis complicated, that's where I start floundering!

 

[dmccormack]

I think there was an E-R prof who studied the issue and found that the optimum bank angle was 45º as well. 60º gets you around the turn faster, but significantly increases stall speed and drag...

Bingo!!

The stall speed is not significantly increased since the steepest bank is during the push.

The relatively miniscule increase in drag doesn't matter since I'm trying to turn around in the least possible distance -- I'll trade 5' of altitude for 45 degrees more turn.

 

[Everskyward]

I was hoping a physicist would jump in here and rescue us. Thanks!

As an engineer, my tendency is to just pick whichever reference frame makes the analysis easy. When someone insists on using a reference frame that makes the analysis complicated, that's where I start floundering!

I kept thinking about the "Do engineers make better pilots?" thread. To me, a non-engineer, the concept seems so simple and doesn't need to be described by equations. The boat in a current analogy is good enough for me. Yes I know part of the boat is above the water and subject to the forces of the air but I ignore that part.

 

[jsstevens]

My Gallean invariant hurt once, so I went to the doc and he gave me something for it.

Did you put that on your next medical form?

 

[gismo]

Bingo!!

The stall speed is not significantly increased since the steepest bank is during the push.

The relatively miniscule increase in drag doesn't matter since I'm trying to turn around in the least possible distance -- I'll trade 5' of altitude for 45 degrees more turn.

First of all, your turn isn't "coordinated" if you aren't pulling 2 g's in a 60 degree bank unless your vertical speed is increasing, this is simple physics. It is possible to pull less than 2g in a 60° bank without increasing VS but that requires that the fuselage generates some lift and that means you aren't coordinated. You might not notice this during a brief 60 degree bank but it's still true.

Second there's no point in making the bank steeper if you don't pull the appropriate g force. When you "unload the wing" you are increasing the turn radius and reducing the turn rate.

 

[mantakos]

... When you "unload the wing" you are increasing the turn radius and reducing the turn rate.

That's what I never understood about this "unload the wing" thing. It's the horizontal component of lift that hauls you around in a circle, if you unload the wing you're killing the force that's trying to make you turn.

I think there's sometimes confusion between the aspect of the turn that is changing the direction of travel of the plane's mass and the aspect of the turn that is the change in the plane's orientation. The second is really driven by the first.
-harry

 

[dmccormack]

First of all, your turn isn't "coordinated" if you aren't pulling 2 g's in a 60 degree bank unless your vertical speed is increasing, this is simple physics. It is possible to pull less than 2g in a 60° bank without increasing VS but that requires that the fuselage generates some lift and that means you aren't coordinated. You might not notice this during a brief 60 degree bank but it's still true.

Second there's no point in making the bank steeper if you don't pull the appropriate g force. When you "unload the wing" you are increasing the turn radius and reducing the turn rate.

Vertical speed is increasing -- early in the maneuver there's a significant push to attain and maintain Vg resulting in negative G's early, then increasing Gs as the Vs is arrested once Vg is reached -- by then bank is decreasing for the remainder of the maneuver.

I'm not sure how you conclude that turn isn't coordinated? Coordination implies no skid or slip that would introduce drag unnecessarily.

 

[gismo]

Vertical speed is increasing -- early in the maneuver there's a significant push to attain and maintain Vg resulting in negative G's early, then increasing Gs as the Vs is arrested once Vg is reached -- by then bank is decreasing for the remainder of the maneuver.

Why do you want to achieve and maintain Vg in the turn? That's way faster than optimum for a turnback. Ideally one should hold a speed just above stall during most of the turn. I suppose that if you make the bank steep enough the 1g best glide speed might actually be just above the accelerated stall speed for the bank, is that what you're shooting for? Even then, I suspect that negative g forces aren't needed but I can see the advantage of reducing the wing load and/or preventing it from building as you roll into the bank until you've got enough speed to prevent a stall. But keep in mind that it will require extra g force (beyond 2g in a 60° bank) to stop the increasing vertical speed. The amount of extra force required will be a function of the rate at which you return to a constant rate of descent.

I'm not sure how you conclude that turn isn't coordinated? Coordination implies no skid or slip that would introduce drag unnecessarily.

Yes uncoordination will increase drag and waste precious energy but that won't stop it from happening if you hold a constant rate of descent with a 60° bank and generate less than 2 g. Sounds like my mistake was assuming you weren't letting the VS increase while the wing was "unloaded", I thought that's what you were saying to Kent. It's definitely possible to remain coordinated while pulling less than 2g in a 60° bank but when doing so you must be accelerating towards the ground (increasing VS) at a rate determined by the actual g force generated.

 

[TMetzinger]

I kept thinking about the "Do engineers make better pilots?" thread. To me, a non-engineer, the concept seems so simple and doesn't need to be described by equations. The boat in a current analogy is good enough for me. Yes I know part of the boat is above the water and subject to the forces of the air but I ignore that part.

Ummm... think about the airplane-on-a-treadmill. I'm astounded at how many folks (including many who should know better) get that wrong.

 

[mantakos]

... It's definitely possible to remain coordinated while pulling less than 2g in a 60° bank ...

Maybe I'm not clear on what the textbook definition of "coordinated" is, but my understanding of the term suggests that if you're coordinated then there's a one-to-one mapping between bank angle and "g's".

If you're in a 60 degree bank and pulling less than 2g, then per my understanding of the term "coordinated", you ain't coordinated. I think the little ball in the turn coordinator will agree with me and fall toward the down wing.
-harry

 

[Palmpilot]

Why do you want to achieve and maintain Vg in the turn? That's way faster than optimum for a turnback. Ideally one should hold a speed just above stall during most of the turn.

One reason would be to have a safe margin above stall speed. In a 172N in a 45 degree bank, the stall speed can be as high as 56 knots indicated. The wings-level best glide speed is 65 knots, and a nine knot margin seems plenty close enough to me, given that I'm probably going to be a little shook up and therefore not maintaining airspeed and bank angle as precisely as I would like.

I suppose that if you make the bank steep enough the 1g best glide speed might actually be just above the accelerated stall speed for the bank, is that what you're shooting for? Even then, I suspect that negative g forces aren't needed but I can see the advantage of reducing the wing load and/or preventing it from building as you roll into the bank until you've got enough speed to prevent a stall. But keep in mind that it will require extra g force (beyond 2g in a 60° bank) to stop the increasing vertical speed. The amount of extra force required will be a function of the rate at which you return to a constant rate of descent.

Doesn't 2g in a 60 degree bank result in the vertical component of lift equalling the weight, and therefore a constant vertical speed? I thought that was how the 2g was derived (since the cosine of 60 degrees is 1/2).

Yes uncoordination will increase drag and waste precious energy but that won't stop it from happening if you hold a constant rate of descent with a 60° bank and generate less than 2 g. Sounds like my mistake was assuming you weren't letting the VS increase while the wing was "unloaded", I thought that's what you were saying to Kent. It's definitely possible to remain coordinated while pulling less than 2g in a 60° bank but when doing so you must be accelerating towards the ground (increasing VS) at a rate determined by the actual g force generated.

FYI, it's been theoretically shown that 45 degrees is the bank angle that gives the least altitude loss in the turn. See "The Possible `Impossible' Turn," and "The Feasibility of Turnback from a Low Altitude Engine Failure During the Takeoff Climb-out Phase" on this Web site:

http://www.nar-associates.com/technical-flying/technical_flying.html

 

[Jaybird180]

In my profession, it is common for people to use "high-falootin" terms so they can converse with those that truly understand the substance behind them. Careful listening can discern the people who have a document that says they know such and such from those that have actual experience in it and understand how to communicate concepts before they were given a name.

Beware of those that study glossaries at the back of a textbook.

Just my $.02

 

[Jaybird180]

...and BTW, yes, I read this entire thread before commenting.

 

[flyingcheesehead]

Maybe I'm not clear on what the textbook definition of "coordinated" is, but my understanding of the term suggests that if you're coordinated then there's a one-to-one mapping between bank angle and "g's".

If you're in a 60 degree bank and pulling less than 2g, then per my understanding of the term "coordinated", you ain't coordinated. I think the little ball in the turn coordinator will agree with me and fall toward the down wing.

You *can* be coordinated in a 60-degree bank and not pulling 2 G's, but you'll be accelerating downward. In any sustained 60-degree banked coordinated turn, you'll eventually have to pull 2 G's if you want to stay coordinated and alive.

 

[dmccormack]

Maybe I'm not clear on what the textbook definition of "coordinated" is, but my understanding of the term suggests that if you're coordinated then there's a one-to-one mapping between bank angle and "g's".

If you're in a 60 degree bank and pulling less than 2g, then per my understanding of the term "coordinated", you ain't coordinated. I think the little ball in the turn coordinator will agree with me and fall toward the down wing.
-harry

"Coordinated" doesn't always mean "ball centered."

 

[gismo]

Maybe I'm not clear on what the textbook definition of "coordinated" is, but my understanding of the term suggests that if you're coordinated then there's a one-to-one mapping between bank angle and "g's".

If you're in a 60 degree bank and pulling less than 2g, then per my understanding of the term "coordinated", you ain't coordinated. I think the little ball in the turn coordinator will agree with me and fall toward the down wing.
-harry

Would you think your flight was uncoordinated if the wings were level while you pitched down from a climb to a descent? The g-force will be less than one g but unless the plane is allowed to yaw (point in a different lateral direction than it's going) it will be coordinated. The same thing applies in a banked turn, you can adjust the pitch attitude for more or less g force than required for a constant vertical speed and still remain coordinated. If you ignore asymmetric thrust and drag, coordinated flight exists when the lift is perpendicular to the average wing surface (i.e. the ball is in the center).

 

[gismo]

One reason would be to have a safe margin above stall speed. In a 172N in a 45 degree bank, the stall speed can be as high as 56 knots indicated. The wings-level best glide speed is 65 knots, and a nine knot margin seems plenty close enough to me, given that I'm probably going to be a little shook up and therefore not maintaining airspeed and bank angle as precisely as I would like.

OK, that's what I began to suspect. What you're really targeting is some margin above the accelerated stall speed and Vbg happens to be close.

Doesn't 2g in a 60 degree bank result in the vertical component of lift equalling the weight, and therefore a constant vertical speed? I thought that was how the 2g was derived (since the cosine of 60 degrees is 1/2).

Yes. What I was referring to is that you will need a larger vertical lift component than 1g at some point to decrease your vertical speed. Looking at what I wrote I see that this wasn't very clear.

FYI, it's been theoretically shown that 45 degrees is the bank angle that gives the least altitude loss in the turn. See "The Possible `Impossible' Turn," and "The Feasibility of Turnback from a Low Altitude Engine Failure During the Takeoff Climb-out Phase" on this Web site:

http://www.nar-associates.com/technical-flying/technical_flying.html

I'm acquainted with Dr Dave and many of his papers including the rather detailed one on turnback. He does prove that the optimal bank for minimal altitude loss in a gliding turn is 45° but he also admits this is a simplified model. Among other things he's ignoring the fact that a steeper turn during the first 180 degrees of turn will put you closer to the runway so for that portion of the turn, a 45° bank is less than optimal and similarly beyond the 180° point it would be optimal to achieve the maximum progress towards the runway for each unit of altitude loss which may require more than 1.05 times the stall speed in the turn. That said, I suspect the difference between 45° and what's truly optimal isn't great and since the wind will affect all of that plus the fact that most of us aren't going to hold a target bank angle to within a degree 45 degrees seems like a good target.

So why were you using 60 degrees?