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

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