Turn dynamics

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

The second phase follows the first.

Are we in agreement?

I think so.

My answers to your questions a
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
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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

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)