Turn dynamics

[[email protected]]

Hi there,

Whoa! It's been a couple of years since I've posted on here. In fact,
looking through some of my previous threads, I was quite the confused
boy back then....and, in a moment of reflection, I've managed to come
up with a topically-similar inquiry. I read through all my old posts
and I couldn't quite get my hands on a definitive answer. So here it
is...

Suppose you're flying trimmed at 100 knots in your Cessna 172, straight
and level, on the front side of the power curve. You enter a 60 deg.
bank that doubles your load factor. Your lift requirement doubles.
Further suppose your goal is to maintain altitude in this turn. To
meet achieve twice the lift, you pull back on the stick to augment the
angle of attack, maintaining airspeed. Everything's good so far? The
up/down forces are balanced. Let's look at drag now.

Banking at 60 deg. and thus pulling 2 g's has shifted every point on
the power required vs. airspeed to the right and up.

Recall that in this scenario, the airspeed has NOT changed. We've only
increased the angle of attack as required, to get twice the lift.

1st question: Is the problem setup flawed? If so, please tell me. I
haven't flown in years :(

2nd question: In the turn, my (L/D)max airspeed has increased by 41%
(sqrt 2) as the (L/D)max airspeed point for the 1g condition has
shifted to a higher airspeed for the 2g condition. Given this
scenario, the airplane in question, and the fact that my airspeed has
NOT changed, am I likely to find myself flying on the FRONT or BACK
side of the power curve in this constant-airspeed turn? What is your
rationale?

3rd question: Given this scenario and the airplane in question, is it
likely that at that same airspeed, the drag and power required at 2g
are HIGHER than at the 1g condition? Why?

4th question: If the answer to (3) is yes, is throttling up the only
way of maintaining altitude in this turn?

5th question: Further to question 4, suppose that I am already at full
throttle, unable to increase thrust, and wish to maintain altitude.
The only remaining variable that I can change is airspeed via yoke
position, by pulling or pushing. Would you agree that I would have to
push on the yoke to maintain altitude if I was on the backside of the
2g power curve and pull on the yoke to maintain altitude if I was on
the front side of 2g power curve? Does it seem counterintuitive to
push on the yoke to maintain altitude in a turn? Something sounds
fishy.

Thanks for any insight.
Alex

[Peter Duniho]

> wrote in message
oups.com...
> [...] Does it seem counterintuitive to
> push on the yoke to maintain altitude in a turn? Something sounds
> fishy.

Sure, of course something sounds fishy. That would be your clue that your
analysis is wrong.

I don't fully even understand what you're trying to propose. For one, you
can't increase your load factor (and thus lift) without either increasing
thrust or decreasing airspeed. The moment you increase the lift, drag
increases and the airplane will start to slow down until thrust equals drag
again.

As far as the parts of your question that I think I comprehend go...

From personal experience, at 100 knots in a C172, you will remain on the
"front side of the power curve" with the proposed maneuver. That is, a
reduction in airspeed will compensate for the increased drag. That said, of
course if you select a power setting that provides less thrust than the
minimum drag at the doubled lift, you cannot expect to enter a 60 degree
bank and remain in level flight. Even at L/Dmax, engine thrust will be less
than drag and the airplane will slow further into the "back side of the
power curve". Only by descending can you get enough thrust to balance drag.

Note that in this latter case, being on the "back side of the power curve"
won't help you. Yes, speeding up would reduce drag, but the reason you get
into that situation in the first place is that there's NO airspeed at which
you can reduce drag enough to be below or equal to the available thrust.
Even if you speed up to exactly L/Dmax airspeed, you still need to maintain
a descent, because there's just not enough engine thrust. All that speeding
up gets you is a shallower descent angle (just as would be the case in a
straight-ahead descent).

In other words, being in a 60 degree bank doesn't change the basic
concepts...only the specific airspeeds at which things happen. This all
assumes, of course, that you *start* on the "front side of the power curve".
If you are in level flight on the "back side", then yes...speeding up allows
you to fly with less thrust, the excess then which you can then use to
compensate for drag induced by a bank.

All of the above is posted in good faith. You are correct when you say you
were confused back when you posted previously, but I have to say that
today's post exhibits some of the same "overcomplicating" behavior we saw
back then. You don't get something for nothing, and no matter how you
manage to state the question, the airplane always flies the same. I hope
this question doesn't devolve into a similar quagmire as we saw "back then".
I know I don't have the patience for it.

Pete

[[email protected]]

Hey Pete,

I knew I could count on you to give me the first cuttingly informative
reply. I was going to post some follow-up questions but nah...it's not
worth the patronizing.

Thanks for your time, and the warm welcome.


Peter Duniho wrote:
> > wrote in message
> oups.com...
> > [...] Does it seem counterintuitive to
> > push on the yoke to maintain altitude in a turn? Something sounds
> > fishy.
>
> Sure, of course something sounds fishy. That would be your clue that your
> analysis is wrong.
>
> I don't fully even understand what you're trying to propose. For one, you
> can't increase your load factor (and thus lift) without either increasing
> thrust or decreasing airspeed. The moment you increase the lift, drag
> increases and the airplane will start to slow down until thrust equals drag
> again.
>
> As far as the parts of your question that I think I comprehend go...
>
> From personal experience, at 100 knots in a C172, you will remain on the
> "front side of the power curve" with the proposed maneuver. That is, a
> reduction in airspeed will compensate for the increased drag. That said, of
> course if you select a power setting that provides less thrust than the
> minimum drag at the doubled lift, you cannot expect to enter a 60 degree
> bank and remain in level flight. Even at L/Dmax, engine thrust will be less
> than drag and the airplane will slow further into the "back side of the
> power curve". Only by descending can you get enough thrust to balance drag.
>
> Note that in this latter case, being on the "back side of the power curve"
> won't help you. Yes, speeding up would reduce drag, but the reason you get
> into that situation in the first place is that there's NO airspeed at which
> you can reduce drag enough to be below or equal to the available thrust.
> Even if you speed up to exactly L/Dmax airspeed, you still need to maintain
> a descent, because there's just not enough engine thrust. All that speeding
> up gets you is a shallower descent angle (just as would be the case in a
> straight-ahead descent).
>
> In other words, being in a 60 degree bank doesn't change the basic
> concepts...only the specific airspeeds at which things happen. This all
> assumes, of course, that you *start* on the "front side of the power curve".
> If you are in level flight on the "back side", then yes...speeding up allows
> you to fly with less thrust, the excess then which you can then use to
> compensate for drag induced by a bank.
>
> All of the above is posted in good faith. You are correct when you say you
> were confused back when you posted previously, but I have to say that
> today's post exhibits some of the same "overcomplicating" behavior we saw
> back then. You don't get something for nothing, and no matter how you
> manage to state the question, the airplane always flies the same. I hope
> this question doesn't devolve into a similar quagmire as we saw "back then".
> I know I don't have the patience for it.
>
> Pete

[alexy]

wrote:

>Hey Pete,
>
>I knew I could count on you to give me the first cuttingly informative
>reply. I was going to post some follow-up questions but nah...it's not
>worth the patronizing.
>
>Thanks for your time, and the warm welcome.

I neither know nor care to take the time to find out what is behind
that comment. But the reply that you appear to be complaining about
looked pretty informative. It mentioned a prior history, but didn't
get into personalities or petty stuff. So what's the problem?
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

[Peter Duniho]

"T o d d P a t t i s t" > wrote in message
...
> Just because it does not seem to make sense, does not
> necessarily mean that it's wrong.

My point isn't that something not making sense means it MUST be wrong, just
that it's a suggestion that it probably is.

> [...]
>>For one, you
>>can't increase your load factor (and thus lift) without either increasing
>>thrust or decreasing airspeed.
>
> I don't follow you. He can increase load factor only by
> increasing bank angle unless you want to leave the steady
> state constant speed turn regime he's using here.

My statement isn't about HOW to increase load factor. It's about the
consequences OF increasing load factor.

> He's not talking about changing lift or bank angle.

Sure he is.

> He
> wants a steady lift equal to twice the aircraft weight at
> sixty degrees of bank.

No. He starts with straight and level flight, and then considers what
happens when one enters a 60-degree bank. That's the initial condition for
the problem.

Pete

[Jose[_1_]]

> you pull back on the stick to augment the
> angle of attack, maintaining airspeed. Everything's good so far?

No, not really. When you pull back on the stick, you will lose
airspeed. In order to maintain airspeed, you will need power. If you
also apply power at the same time as you pull back on the stick (at the
same time you are in the 60 degree banked turn) then you can maintain
airspeed and altitude.

Once you get your head around the need for power at this point, the
answers to the rest of your questions should become clearer.

Jose
--
"Never trust anything that can think for itself, if you can't see where
it keeps its brain." (chapter 10 of book 3 - Harry Potter).
for Email, make the obvious change in the address.

[Dan Luke]

> wrote:

> Hey Pete,
>
> I knew I could count on you to give me the first cuttingly informative
> reply. I was going to post some follow-up questions but nah...it's not
> worth the patronizing.
>
> Thanks for your time, and the warm welcome.

If your feelings get hurt that easily, usenet is not the place for you to be
asking questions.

--
Dan
C-172RG at BFM

[[email protected]]

There's plenty of people on usenet who answer questions without being
condescending.

Dan Luke wrote:
> > wrote:
>
> > Hey Pete,
> >
> > I knew I could count on you to give me the first cuttingly informative
> > reply. I was going to post some follow-up questions but nah...it's not
> > worth the patronizing.
> >
> > Thanks for your time, and the warm welcome.
>
> If your feelings get hurt that easily, usenet is not the place for you to be
> asking questions.
>
> --
> Dan
> C-172RG at BFM

[[email protected]]

Hi Jose,

When I formulated the scenario, I broke it down into two phases.

The first phase was related to the forces perpendicular to the relative
wind, or the action of pulling back on the stick and increasing your
angle of attack enough to meet the increased lift requirement, with no
change in airspeed. This essentially satisfied lift = 2*weight. I
then looked at the changing forces parallel to the relative wind
(thrust/drag) in the second phase.

The increase in drag due to the higher load factor would slow you, and
you'd soon find yourself pitched down and descending, followed by a
restauration of the initial airspeed flown before and during the turn
(with no change in power, that is). As I understand, one could apply
additional power to maintain altitude and airspeed in the turn though.

My questions pertained to the second phase, and regarded the increase
in drag/power required and their effect on the maintenance of altitude.

Perhaps we're saying the same thing? (and perhaps not...in which case
I'd appreciate another round of clarification)

Thanks for your input.
Alex


Jose wrote:
> > you pull back on the stick to augment the
> > angle of attack, maintaining airspeed. Everything's good so far?
>
> No, not really. When you pull back on the stick, you will lose
> airspeed. In order to maintain airspeed, you will need power. If you
> also apply power at the same time as you pull back on the stick (at the
> same time you are in the 60 degree banked turn) then you can maintain
> airspeed and altitude.
>
> Once you get your head around the need for power at this point, the
> answers to the rest of your questions should become clearer.
>
> Jose
> --
> "Never trust anything that can think for itself, if you can't see where
> it keeps its brain." (chapter 10 of book 3 - Harry Potter).
> for Email, make the obvious change in the address.

[Jose[_1_]]

> When I formulated the scenario, I broke it down into two phases.

If you mean "what happens first" and "what happens second" then I follow
you. Otherwise,

> pulling back on the stick and increasing your
> angle of attack enough to meet the increased lift requirement, with no
> change in airspeed.

.... doesn't work.

That is, the airspeed won't change instantaneiously, but it will change,
because the steady state is unsustainable without more power. Maybe
that's what you mean by the second phase: (The increase in drag due to
the higher load factor would slow you...)

If that's what you meant, then you're ok.

> and
> you'd soon find yourself pitched down and descending, followed by a
> restauration of the initial airspeed flown before

What are you doing with the stick? (same position? same forces? same
airspeed?)

It's slightly different, but you might be able to get your head around
it better, if you think of Uncle Bob parachuting into your airplane.
There you are, fat, dumb, and happy, flying straight and level. (I'm
going to make all the numbers up here out of whole cloth). Uncle Bob
parachutes right into your airplane, and amazingly lets go of the canopy
and fixes the airframe faster than you can say "337". But he's now in
the plane.

IF YOU DO NOTHING, the plane will descend, because lift hasn't changed,
but weight has. To compensate, you pull back on the stick, increasing
lift (at the expense of drag). Now with the nose pointing in the air,
you maintain altitude, but you're going slower because the extra drag,
and also partly because the thurst vector is now pointing more up, and
less in the direction of flight.

To go faster without gaining altitude, you must lower the nose (again)
=and= add power. You will need a higher airspeed than originally to
provide sufficient lift (you need to hold Uncle Bob up too!) if you
attempt to hold the same angle of attack as you originally had, but
that's not what you want. You want your orignal airspeed, AOA be
damned. So, at that original airspeed (which of course requires more
power), you'll need a higher AOA.

You can generate the same lift at a range of AOA and power. Doing this
with more power requires you to go faster. Doing it with a higher AOA
requires you to go slower. You pick the one combination that gives you
the desired (original) airspeed. It will be a higher-than-original AOA,
with higher-than-original power.

Most of this thinking would apply to the turning flight too, but it
might be easier to think of it in this context first. (btw, this is the
context of airdrops in reverse - you drop your load, you need to reduce
power and pitch down if you want to maintain altitude and airspeed).

Jose
--
"Never trust anything that can think for itself, if you can't see where
it keeps its brain." (chapter 10 of book 3 - Harry Potter).
for Email, make the obvious change in the address.

[[email protected]]

> > When I formulated the scenario, I broke it down into two phases.
>
> If you mean "what happens first" and "what happens second" then I follow
> you.

yeah, that's what I meant. What I referred to as the first phase is
"what happens first" and the second phase is "what happens thereafter".


The second phase follows the first.

Are we in agreement?

Alex

[Jose[_1_]]

> The second phase follows the first.
>
> Are we in agreement?

I think so.

My answers to your questions are:
1: No
2: From my experience in a 172, I don't think you'd move to the back
side, however I have not looked at the performance charts or done the math.
3: Yes, for the reasons I explained upthread.
4: Outside of magic, yes.

Jose
--
"Never trust anything that can think for itself, if you can't see where
it keeps its brain." (chapter 10 of book 3 - Harry Potter).
for Email, make the obvious change in the address.

[Julian Scarfe]

> wrote in message
oups.com...

> 5th question: Further to question 4, suppose that I am already at full
> throttle, unable to increase thrust, and wish to maintain altitude.
> The only remaining variable that I can change is airspeed via yoke
> position, by pulling or pushing. Would you agree that I would have to
> push on the yoke to maintain altitude if I was on the backside of the
> 2g power curve and pull on the yoke to maintain altitude if I was on
> the front side of 2g power curve? Does it seem counterintuitive to
> push on the yoke to maintain altitude in a turn?

It's pretty much always counterintuitive to push on the yoke to maintain
altitude. The power curve is the power curve, and while the numbers might
change when you change the load factor, the nature of it doesn't.

For any power curve, if your power available exceeds the minimum power
required, you have the opportunity to maintain altitude or climb. If your
power available does not exceed the minimum power required, you're going
down, whether you push or pull.

Normally, we approach the minimum power required from the higher airspeed
side. You have to do something slightly unusual to get to steady state on
the backside of the power curve, e.g. take power off and then put it back on
again, or zoom up a little to allow the airspeed to fall.

It's the same in a turn. As you pull back to maintain your altuitude in the
turn, your speed bleeds off. If you just allow the speed to bleed off while
maintaining altitude, you'll either reach a point on the normal side of the
power curve where you can maintain speed and altitude, or you'll find you
reach minimum power required and you still can't maintain altitude, in which
case you'll be going down, whatever you do. To get into equilibrium on the
backside of the power curve, you'd have to do something "unusual", more than
just pulling back to maintain altitude. I suppose it's slightly more likely
that you might do that in a steep turn, when you've exceeded the performance
of the aircraft at your chosen bank angle and power setting, the airspeed
has dropped below minimum power required, and you think "Doh, better put
some power on now".

Julian

[koopas]

Hi Todd,

Thanks for replying. I thought about what you wrote, specifically the
achievement of a decreased sink rate with a higher weight...and tried
to make sense of the the counterintuitiveness. The present discussion
would not relate to my original post, since we'll be talking about a
non-turning descending glider with added mass (ballast) and not a
powered 172 maintaining AOA and altitude in a 60 deg. banked turn.

You wrote: More weight always means more
> lift required and more lift required always means more drag
> and more power required. OTOH, the power produced by descent
> is the sink rate times the weight. By doubling weight (to
> get load factor n=2), I get twice the power for any descent
> rate, but the power needed to produce the additional lift is
> less than twice, so there's a net decrease in sink rate even
> though there's an increase in sink rate times weight.

So I looked up the equations for Rate of climb/descent and flight path
angle.

Rate of climb/descent = 33,000 (Pa-Pr) / W = 33,000*Pr / W (for glider)
sin (theta) = theta (for small angles) = (T - D) / W = - D / W (for
glider)

So the key lies in that both equations incorporate Weight in the
denominator.

In the case of sink rate, I am concluding that, as you said above, the
new higher Weight is greater than the increased Pr caused by the higher
weight, resulting in a decreased rate of descent, even though Pr has
increased due to higher drag due to the higher weight (I hope you're
following :) In the case of the flight path angle, if the same AOA is
maintained and airspeed is increased to a meet the higher weight
requirement, the flight path angle will remain the same.

Is my line of reasoning correct?

A couple more questions linger though:

1. You tried to highlight the nuance between adding weight (physically
more mass) and inducing a higher "effective weight" in a turn. I
believe your discussion (and mine above) pertained to a glider with
added mass, via ballast. Would the same reasoning hold during a turn
when "W" is constant...and only the lift has been increased to pull the
higher g's? Indeed, if W and power are constant in a constant AOA
turn, then one would expect an increase in drag, power required,
airspeed, sink rate, and the same angle of descent. On the other hand,
in the case of simply adding ballast when flying in a non-turning
glider, I agree with you that you could experience an actual decrease
in sink rate, accompanied by an increase in airspeed at the same AOA
and angle of descent.

2. This is related to question 1. In the case of the non-turning
glider with ballast that maintains its best L/D AOA at a higher
airspeed at the same angle of descent, I can now see how you could
experience a decreased sink rate, as explained above. However, it
seems contradictory to me that you can have an increase in airspeed
coupled with a decrease in sink rate. How do you reconcile these two
seemingly opposite physical effects?

Have a good weekend.
Alex


T o d d P a t t i s t wrote:
> "Peter Duniho" > wrote:
>
> >> [...] Does it seem counterintuitive to
> >> push on the yoke to maintain altitude in a turn? Something sounds
> >> fishy.
> >
> >Sure, of course something sounds fishy. That would be your clue that your
> >analysis is wrong.
>
> Just because it does not seem to make sense, does not
> necessarily mean that it's wrong. For example, If I fill
> the wing tanks on my glider with water, my sink rate
> decreases at 100 knots as compared to flying with them
> empty. Does it make sense that carrying more weight reduces
> the sink rate? Not to most pilots. Nevertheless, that's
> what happens. It works in airplanes too. Descent rate
> decreases for most higher airspeeds at higher weights.
>
> Let's work through his questions to see what he's getting
> at:
>
> >>trimmed at 100 knots...straight and level, on the front side of the power curve.
> >>60 deg. bank that doubles your load factor.... goal is to maintain altitude
>
> >>Is the problem setup flawed?
>
> No
>
> >>In the turn, my (L/D)max airspeed has increased by 41%
> >> am I likely to find myself flying on the FRONT or BACK
> >>side of the power curve in this constant-airspeed turn? What is your
> >>rationale?
>
> You'll be on the front side if the min power required speed
> for level flight was less than 100 knots/sqrt(2). (Which it
> is for a 172) You are correct in thinking you could move
> from front side of the power curve to the back side, just
> like you could move all the way to stall.
>
> >>Given this scenario and the airplane in question, is it
> >>likely that at that same airspeed, the drag and power required at 2g
> >>are HIGHER than at the 1g condition? Why?
>
> Yes. It's not only likely, it's definite that more power is
> required and drag is higher. It's this question that made
> me start with my glider water ballast comment above. The
> sink rate of my glider will reduce by adding water ballast
> at 100 knots (as compared to no ballast). However, that
> does not mean that drag decreased. Nor does it mean that
> power required decreased. More weight always means more
> lift required and more lift required always means more drag
> and more power required. OTOH, the power produced by descent
> is the sink rate times the weight. By doubling weight (to
> get load factor n=2), I get twice the power for any descent
> rate, but the power needed to produce the additional lift is
> less than twice, so there's a net decrease in sink rate even
> though there's an increase in sink rate times weight.
>
> In your case (load factor increased by turning flight, not
> by mass increase), you need additional power for the
> additional lift and drag, but you have not added mass, and
> you are not getting additional power by the descent of that
> mass in the gravitational field. You must get it from the
> engine. It will always take more power to produce more lift
> at the same speed.
>
> >> If the answer to (3) is yes, is throttling up the only
> >>way of maintaining altitude in this turn?
>
> Yes - it takes more power to run at n=2 and 100 knots than
> n=1 and 100 knots. The only place to get that power is from
> the engine. If my glider is fully ballasted and I'm on the
> back side of the power required curve (close to stall), I
> can reduce descent rate and reduce power required by going
> faster. It seems analogous to the situation you are asking
> about, but it's not. In the turn, you are not getting power
> from descent in a gravitational field and you are not
> carrying more mass. If you plot comparative sink rate
> curves, they cross each other. If you plot power required
> curves, they do not cross (although both do have a back
> side/front side.)
>
> >>Further to question 4, suppose that I am already at full
> >>throttle, unable to increase thrust, and wish to maintain altitude.
> >>The only remaining variable that I can change is airspeed via yoke
> >>position, by pulling or pushing. Would you agree that I would have to
> >>push on the yoke to maintain altitude if I was on the backside of the
> >>2g power curve and pull on the yoke to maintain altitude if I was on
> >>the front side of 2g power curve?
>
> Yes. It takes less power to fly at the min power required
> airspeed for the n=2 condition. If you are on the back side
> of the power required curve, then going faster (yoke
> forward) will decrease sink rate if you hold the 60 degree
> bank.
>
> >>Does it seem counterintuitive to push on the yoke to maintain altitude in a turn?
>
> Yes it does, but then the region of reversed control is an
> unusual flight regime.
>
> >>Something sounds fishy.
> I would have said "smells fishy," :-)
>
>
> >I don't fully even understand what you're trying to propose.
>
> I think I understand where he's at. I've been through this
> analysis a lot of times with glider ballast computations and
> comparisons to load factor in turn charts. I *think* he's
> hung up over the difference between load factor in a turn
> versus load factor by weight increase. There are some
> subtle but interesting differences.
>
> >For one, you
> >can't increase your load factor (and thus lift) without either increasing
> >thrust or decreasing airspeed.
>
> I don't follow you. He can increase load factor only by
> increasing bank angle unless you want to leave the steady
> state constant speed turn regime he's using here.
>
> >The moment you increase the lift, drag
> >increases and the airplane will start to slow down until thrust equals drag
> >again.
>
> He's not talking about changing lift or bank angle. He
> wants a steady lift equal to twice the aircraft weight at
> sixty degrees of bank. He's talking about changing the
> speed and turn diameter. If he's above the min power speed
> for load factor 2 he can slow and increase AOA to reduce
> power required.
>
> >As far as the parts of your question that I think I comprehend go...
> >
> >From personal experience, at 100 knots in a C172, you will remain on the
> >"front side of the power curve" with the proposed maneuver.
>
> I agree.
>
> (I had some major dental work yesterday and less than 4
> hours sleep last night - I hope I didn't make too many
> stupid errors in the above. :-)
>
>
> --
> Do not spin this aircraft. If the aircraft does enter a spin it will return to earth without further attention on the part of the aeronaut.
>
> (first handbook issued with the Curtis-Wright flyer)