Angle of climb at Vx and glide angle when "overweight": five questions

"Koopas Ly" wrote in message
om...
Gerry,

I agree with everything you said.

Would you agree that augmenting an airplane's weight undoubtedly
increases its drag? Doesn't matter if the airplane is powered or not.

Nope. Weight has no direct effect on drag. Drag is composed of parasitic
drag and induced drag. Parasitic drag = Cd*1/2*rho*V^2*S where rho is
density of the fluid and S is the wing area. Cd and V should be obvious.
Drag increases in denser air, with greater speed, and greater wing area
(fowler flaps). Cd can be changed with flaps or speedbrakes. Induced drag
is proportional to lift and AOA. Think of it this way; at high AOA part of
that lift vector (which is perpendicular to the wing) is lifting towards the
rear, which is drag.

I think your confusion comes from the idea that the heavier plane has to fly
faster at the same AOA to generate lift equal to weight. At that higher
speed, the total drag will be higher. That is why the power required curve
moves up (more power to overcome more drag) and right (higher AS to generate
the needed lift.) Weight moves the curve because it drives the needed lift,
it does not alter the relationship between lift and drag.

Now, apply that to an airplane in a steady climb at full power and
best L/D speed and another one that's gliding at best L/D. Both
airplanes are identical so ideally, they would be flying at the same
speed since I don't think best L/D speed changes with power settings.

OK.

I hope we can agree that both airplanes are operating at their minimum
drag points, points of least thrust required.

The minimum drag point is the point of least thrust required. But that's
not best L/D. For a glider, it's the speed for minimum sink rate. You'll be
sinking at a steeper angle than best L/D, but your speed is lower so the
vertical speed is at a minimum.

This means that in the
case of the airplane climbing, its climb angle is maximized.

Nope. While Vx and Vy have no specific correlation with speed for min drag
or best L/D. They are driven by excess thrust and power only. In my plane,
Vy is 79 kts, best L/D is 72 kts.

In the
case of one gliding, its descent angle is minimized.

That's correct.

I hope you're still in agreement.

Next, double both airplanes' weights, and fly them faster by 41%
(sqrt(2)) so that they are still operating at their best L/D angle of
attack.

Would you agree that in both cases, both the drag and power required
increased?

Yes. Which is why rate of climb suffers for the climbing a/c. For the
glider, the angle stays the same because L/D is the same. The extra power
required comes from the higher rate at which you trade altitude (potential
energy) for speed (kinetics energy.)

To me, I immediately relate the increase in drag as a decrease in
climb angle (for the powered airplane) and an increase in descent
angle (for the gliding airplane). Likewise, the increase in power
required means a decrease in climb rate (for the powered airplane) and
an increase in sink rate (for the gliding airplane).

Correct for the climbing case. For the glider, the higher sink rate
generates more power in the same proportion as the drag increase. So speed
increases and the descent angle stays the same.

Final conditions: For the powered plane that's ascending, as you've
said, the airspeed will be higher, maximum climb angle SHALLOWER,
climb rate lessened, ground speed increased.

If the speed is higher and the climb rate is less, the angle will be
shallower. Plot it out.

For the gliding airplane, the airspeed will be higher, glide angle
UNCHANGED, sink rate higher, ground speed higher.

The best glide angle is driven by Cl/Cd. At higher weights, you have to go
faster to maintain best L/D (or Cl/Cd). But since the ratio doesn't change,
the glide angle doesn't either.

I'd recommend a copy of *Aerodynamics for Naval Aviators* It's a good start
at explaining it all without getting too involved in the math.

Gerry

"Gerry Caron" wrote in message
om...
I think your confusion comes from the idea that the heavier plane has to
fly
faster at the same AOA to generate lift equal to weight. At that higher
speed, the total drag will be higher.

The drag would be higher even for equal airspeeds. As you already stated,
induced drag is proportional to angle of attack and *lift*. At higher
weights, lift is necessarily higher as well, increasing induced drag.

Now, you may consider that a secondary effect rather than a "direct effect"
(though I don't). But you ought to at least mention it. Your post seems to
be saying that weight does not in and of itself increase drag, but weight
does do exactly that.

The minimum drag point is the point of least thrust required. But that's
not best L/D.

Why not? Lift is constant for a given weight, so the best ratio of lift to
drag will occur where drag is at its minimum. Seems to me that the minimum
drag point IS exactly the same as the best L/D point.

Pete

Gerry,

My comments below.


"Gerry Caron" wrote in message . com...
"Koopas Ly" wrote in message
om...
Gerry,

I agree with everything you said.

Would you agree that augmenting an airplane's weight undoubtedly
increases its drag? Doesn't matter if the airplane is powered or not.

Nope. Weight has no direct effect on drag. Drag is composed of parasitic
drag and induced drag. Parasitic drag = Cd*1/2*rho*V^2*S where rho is
density of the fluid and S is the wing area. Cd and V should be obvious.
Drag increases in denser air, with greater speed, and greater wing area
(fowler flaps). Cd can be changed with flaps or speedbrakes. Induced drag
is proportional to lift and AOA. Think of it this way; at high AOA part of
that lift vector (which is perpendicular to the wing) is lifting towards the
rear, which is drag.

I think your confusion comes from the idea that the heavier plane has to fly
faster at the same AOA to generate lift equal to weight. At that higher
speed, the total drag will be higher. That is why the power required curve
moves up (more power to overcome more drag) and right (higher AS to generate
the needed lift.) Weight moves the curve because it drives the needed lift,
it does not alter the relationship between lift and drag.


I think we're both saying the same thing: increased drag is an effect
of higher weight.

Drag is a function of Cd and V. Cd is a function of Cl. When you
increase weight, you have to increase Cl and/or V so your drag goes
up. Please jump in if I am wrong.



Now, apply that to an airplane in a steady climb at full power and
best L/D speed and another one that's gliding at best L/D. Both
airplanes are identical so ideally, they would be flying at the same
speed since I don't think best L/D speed changes with power settings.

OK.

I hope we can agree that both airplanes are operating at their minimum
drag points, points of least thrust required.

The minimum drag point is the point of least thrust required. But that's
not best L/D. For a glider, it's the speed for minimum sink rate. You'll be
sinking at a steeper angle than best L/D, but your speed is lower so the
vertical speed is at a minimum.


I will agree with Peter Duniho. I believe the speed corresponding to
least drag for a given weight is indeed best L/D speed.



This means that in the
case of the airplane climbing, its climb angle is maximized.

Nope. While Vx and Vy have no specific correlation with speed for min drag
or best L/D. They are driven by excess thrust and power only. In my plane,
Vy is 79 kts, best L/D is 72 kts.


Ok, I understand what you mean. Allow me to clarify. Thrust
available from a for a reciprocating engine-propeller combination is
fairly constant with respect to velocity. If thrust available is
constant, the max. L/D speed (speed of least drag) will give you the
greatest excess thrust and the maximum climb angle. In other words,
the max L/D speed is Vx at maximum thrust available.

For Vy, the best L/D speed is not the point of least power required so
I can't apply the above reasoning. Besides, power available is not
constant so one would have to graph power reqd. vs. power available to
determine Vy.



In the
case of one gliding, its descent angle is minimized.

That's correct.

I hope you're still in agreement.

Next, double both airplanes' weights, and fly them faster by 41%
(sqrt(2)) so that they are still operating at their best L/D angle of
attack.

Would you agree that in both cases, both the drag and power required
increased?

Yes. Which is why rate of climb suffers for the climbing a/c. For the
glider, the angle stays the same because L/D is the same. The extra power
required comes from the higher rate at which you trade altitude (potential
energy) for speed (kinetics energy.)


We've agreed that in both cases, drag increases. When you write
"Which is why rate of climb suffers for the climbing a/c", I assume
you mean rate of climb at Vy? Well, it doesn't matter...the rate of
climb of the climbing a/c will suffer both at Vx and Vy speeds. The
climb angle at both Vx and Vy will also suffer. Still in agreement?

When you dwell into the case of the glider, you've effectively lost
me. I understand that the glide angle stays the same because the
heavier glider is still flying at best L/D speed...I can see that from
the math.

My question remains: the gliding a/c has encountered both an increase
in drag and power required. You've agreed to that. However, its
glide angle does not change. Its sink rate increases though. Why
doesn't the glide angle change due to the increased drag? That's the
part I don't get. Could you please explain it in another way?



To me, I immediately relate the increase in drag as a decrease in
climb angle (for the powered airplane) and an increase in descent
angle (for the gliding airplane). Likewise, the increase in power
required means a decrease in climb rate (for the powered airplane) and
an increase in sink rate (for the gliding airplane).

Correct for the climbing case. For the glider, the higher sink rate
generates more power in the same proportion as the drag increase. So speed
increases and the descent angle stays the same.


Again, I understand the climbing case. I don't understand the glider
case. See above.



Final conditions: For the powered plane that's ascending, as you've
said, the airspeed will be higher, maximum climb angle SHALLOWER,
climb rate lessened, ground speed increased.

If the speed is higher and the climb rate is less, the angle will be
shallower. Plot it out.

I agree. I think my reasoning for determining that the climb angle
will be shallower is different than yours. If you write the equations
of motion of the airplane at the new weight and the new airspeeed (aoa
the same), I find that even though the lift = weight*cos(climb angle)
or using small angle approximations lift ~ weight, the increase in
drag now requires a shallower climb angle so that the fore-aft
equation of motion T - D - W*sin(theta) = 0 is satisfied. As "D"
increases, theta must be decreased for the equality to remain true.

But you're right, this can also be deduced by simple geometry knowing
that your climb rate has decreased due to the increase in power
required and the decrease in excess power. The decrease in climb
rate, regardless of airspeed, invariably translates into a shallower
climb angle.



For the gliding airplane, the airspeed will be higher, glide angle
UNCHANGED, sink rate higher, ground speed higher.

The best glide angle is driven by Cl/Cd. At higher weights, you have to go
faster to maintain best L/D (or Cl/Cd). But since the ratio doesn't change,
the glide angle doesn't either.


I agree with everything you've said.



I'd recommend a copy of *Aerodynamics for Naval Aviators* It's a good start
at explaining it all without getting too involved in the math.

Thanks for the tip. I seem to collect every book except the right
one!


Gerry

GC: I'd recommend a copy of *Aerodynamics for Naval Aviators* It's a
good start at explaining it all without getting too involved in the
math.

KL: Thanks for the tip. I seem to collect every book except the right
one!
------------------------------------------------------

You don't seem to have any trouble with the math, so I'm not sure what
value that AFNA would have for you. You're looking for more intuitive
explanations than what the math provides, and that's really hard to
come by. Most aerodynamics books don't have a lot of interest in
providing what you want.

The minimum drag point is the point of least thrust required. But
that's not best L/D. For a glider, it's the speed for minimum sink
rate.

I vote with Peter. Minimum sink is least POWER required, not least
THRUST.

On Fri, 28 Nov 2003 at 16:59:43 in message
, Greg Esres
wrote:

I vote with Peter. Minimum sink is least POWER required, not least
THRUST.

But surely minimum sink rate is only relevant when there is no power or
thrust? It requires finding a minimum value of V*sin(theta) where theta
is the angle of climb (negative when descending).

As I recall for a gliding aircraft minimum sink comes roughly at the AoA
where (Cl^(3/2)/Cd is a maximum. This is normally at a higher AoA than
maximum Cl/Cd and is some cases is quite close to the stall.

Power is drag (or thrust) times velocity.
--
David CL Francis

But surely minimum sink rate is only relevant when there is no power
or thrust?

I suppose so. If power were constant with airspeed, then minimum sink
would also be Vy.

It requires finding a minimum value of V*sin(theta) where theta is
the angle of climb (negative when descending).

Sounds good.

As I recall for a gliding aircraft minimum sink comes roughly at the
AoA where (Cl^(3/2)/Cd is a maximum.

I'd have to look it up to be sure, but it looks right.

This is normally at a higher AoA than maximum Cl/Cd and is some
cases is quite close to the stall.

Agreed, except, according to the books, the velocity of minimum power
is ALWAYS less than least drag.

But surely minimum sink rate is only relevant when there is no power
or thrust?

I suppose so. If power were constant with airspeed, then minimum sink
would also be Vy.

Yes, if power available was constant with airspeed, Vy would also be
your min. sink rate speed. The books indicate that power for a
piston-engine propeller combination increases with velocity. The
increase is not linear though. The derivative dP/dV looks to be of
the form y = -mx + b. Kinda like an upside down smiley face with only
the left side showing.

Thrust available looks fairly constant w.r.t. velocity.



It requires finding a minimum value of V*sin(theta) where theta is
the angle of climb (negative when descending).

Sounds good.

As I recall for a gliding aircraft minimum sink comes roughly at the
AoA where (Cl^(3/2)/Cd is a maximum.

I'd have to look it up to be sure, but it looks right.


That's correct. Min sink rate would occur at the AOA where
1/[(Cl^3/2)/cd] is a minimum as it is the point of least power
required. Min. thrust required is proportional to 1/(Cl/Cd).



This is normally at a higher AoA than maximum Cl/Cd and is some
cases is quite close to the stall.

Agreed, except, according to the books, the velocity of minimum power
is ALWAYS less than least drag.


What book are you reading Greg? My references indicate the same for a
Cessna Skylane. If you assume no power available, then, your min.
sink (best endurance) speed is less than your best glide speed.

My Cessna 172SP POH lists a best glide speed of 68 kts. Best
endurance speed isn't exactly mentioned but I've heard it's close to
stall. I just recall someone saying "If you want to stay up for as
long as you can, fly close to stall". Do you know why the POH doesn't
mention that speed? Would that be giving away too much?

Oh, by the way, could you please give a shot at answering my questions
that are contained in my Nov. 27 reply to Gerry Caron? (in this same
overall post)

Thanks,
Alex

On Sat, 29 Nov 2003 at 01:19:53 in message
, Greg Esres
wrote:
This is normally at a higher AoA than maximum Cl/Cd and is some
cases is quite close to the stall.

Agreed, except, according to the books, the velocity of minimum power
is ALWAYS less than least drag.

Sounds reasonable and I don't think it is inconsistent.
--
David CL Francis