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



 
 
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  #1  
Old November 27th 03, 04:28 AM
Koopas Ly
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Default Angle of climb at Vx and glide angle when "overweight": five questions

Happy Thanksgiving to ya'll,

This is related to my previous "overweight" post. I want to know if
you agree with the following reasoning:

Context: you take off over your maximum gross weight, and wonder how
your climb speeds and performance are affected.

Vy: The AOA corresponding to the best rate of climb remains the same.
However, the airspeed at which the best rate of climb speed increases
by the square root of the ratio of the current weight and maximum
gross weight. The pilot then pitches to obtain that new, higher Vy
airspeed.

Even though you're flying at the Vy corresponding to your new weight,

Q1: Is it correct to say that that your climb rate is now LOWER than
your max. gross weight Vy, due to the increase in power required at
the higher load factor? If so, you'd agree that you're flying at a
"higher speed" but yet climbing slower? Is that "higher speed" the
speed along the flight path line or the horizontal velocity that's
parallel to the ground? Would clarifying this last question open
another can of worms? The rate of climb is the vertical velocity with
respect to the ground, that I am sure. I think I am confusing
aircraft airspeed with velocity along the flight path vector or
horizontal velocity with respect to the the inertial frame.

Q2: Further, is it correct to write that the thrust required to
counter the drag is higher at that moment? As such, is the angle of
climb at Vy is also lessened?


Vx: Again, the AOA corresponding to the best angle of climb for
obstacle clearance remains the same. However, the airspeed to now
achieve the best angle of climb at the current weight increases by the
same square root of the ratio of the current weight and maximum gross
weight. The pilot then pitches to obtain that new, higher Vx speed.

Q3: Is it correct to infer that your thrust required to counter the
drag is higher at that moment? Hence, is the maximum climb angle less
that what it would be at max. gross weight?

Q4: Further, is it correct to write that your power required at that
moment is higher than at max. gross weight and as such, the rate of
climb at the maximum climb angle is reduced?


Best Glide Speed (best range): From a simple FBD with no thrust
vector, one can find that the best angle of glide is only dependent
upon the inverse tangent of the reciprocal of the lift to drag ratio
or:

Tan(glide angle) = 1 / (L/D), assuming small angles, glide angle ~
(L/D)^(-1)

L/D is purely angle of attack driven. Therefore, the glide angle does
not change with respect to weight.

Here's question 5: I would make the risky proposition of stating that
drag and power required ALWAYS increase with increasing weight. To
me, when you have neither thrust nor power available during gliding
flight, increasing weight/load factor STILL increases your drag and
therefore results in a higher descent / glide angle. Likewise, the
increase in weight/load factor under no power STILL triggers an
increase in power required and therefore higher sink rate.

Right now, I am seeing a contradiction between the 1/(L/D) equation
and what I've described above. I can't seem to figure out why.

Referring back to the best angle of climb question above, it appears
that increasing your weight decreases your max. climb angle. I just
don't see a difference between climb and descent angle insofar as both
being excess thrust or drag phenomena. Why would increasing your
weight decrease your max. climb angle but not affect your glide angle?
In both cases, I see an increase in drag with higher load factor
resulting in a decrease in max. climb angle and an increase in glide
angle. Max. climb angle is very much defined by best L/D (i.e. min.
drag) and excess thrust while min. glide angle (best glide angle) is
defined by best L/D alone. That best L/D is still analogous to its
"powered counterpart", as it still represents the point of minimum
drag. However, that drag has increased with the higher weight,
lessening the L/D ratio. As a result, without an engine, I expect
that the glide angle to increase (reducing range). Perhaps the fact
that there's no thrust vector has something to do with this fiasco.

Ok, I'll stop rambling. I hope I conveyed my thoughts clearly enough.

Happy Thanksgiving,
Alex
  #2  
Old November 27th 03, 08:26 AM
Koopas Ly
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I did more brainstorming at work before going home early with a
headache...this is regarding the last point in my post.

Just to reiterate, I wanted to know why your best glide angle was
independent of weight whereas your maximum angle of climb was.

Here's what I came up with with a FBD:

For an aircraft climbing, unaccelerated, where theta is the flight
path angle.

T = D + W*sin(theta)
L = W*cos(theta)

For small theta, cos(theta) = 1 and sin(theta) = theta

so we now have

T = D + W*theta
L = W

flight path angle is given as (T - D)/W [Eq. 1]

or T/W - D/W or T/W - D/L or

theta = T/W - 1/(CL/CD) [Eq. 2]

First, you'll notice that the first term of Eq. 2 is the thrust to
weight ratio and the second is the lift to drag ratio. To increase
theta during a climb or minimize theta during a glide, one must
minimize the second term by flying at best L/D AOA. For a given
thrust, decreasing weight will increase theta.

The equation above would apply for both climbs and descents. What's
interesting to note is that in a glide when T = 0, Eq. 2 simplifies to
theta = - 1/(CL/CD), independent of airspeed or weight. The
dependence on load factor mathematically vanishes when the engine goes
out. How strange!

However, when flying with engine thrust and at the best (L/D) AOA, an
increase in load factor will IMPACT theta, lessening the climb angle
or making it more "negative".

Don't you find it coincidental that the absence of thrust makes the
dependence of load factor upon theta vanish? Can it only be explained
from the math?

So somehow, the magic equations seem to explain my confusion...but I
am not quite at peace. Au contraire. I've only talked about L/D,
weight, and thrust. What about drag? I am still claiming that you'll
experience an increase in drag with a weight increase REGARDLESS of
whether or not you have thrust. Thus, due to the increase in drag and
REGARDLESS of thrust or not, your angle THETA will be affected (i.e.
theta will get larger, more negative, airplane will pitch DOWN). In
the case of gliding flight, everybody but me agrees that your glide
angle won't be affected. Obviously, I am wrong but I don't know why.
To reiterate, that's why I've been writing all this stuff, 'cause I
don't get why THETA happens to change only when you have thrust but
doesn't when you don't have any and are gliding.

Alright, so the increase in drag with respect to load factor is shown
as:

D = V^2*CD where CD = (1 + CL^2)

By virtue of increasing your load factor (either via CL and/or V),
your drag goes up...[right?]

Now, try to tie that in to Eq. 2 in a desperate attempt to explain why
an increase in DRAG won't cause a increased glide angle (less range)
but will cause a logical decrease in climb angle.

From D = V^2 * (1 + CL^2) and theta = T/W - 1/(CL/CD)

Well, I'll stop here because I seemed to have taken the wrong path, as
I am pretty much stuck. ARRGHH! I AM GOING TO KILL A TURKEY!
  #3  
Old November 27th 03, 02:02 PM
Andrew Rowley
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I didn't go through all your math, but consider this:

Your engine makes the same power regardless of weight, so the higher
the weight the slower the climb.

However, as you get heavier you have stored more energy by climbing
(that's one reason why it took you longer with the same engine power).
With more weight, you have more stored energy per 100 feet. At a
higher airspeed you are also descending faster.

As you lower a weight you are releasing energy, the rate at which this
is released is effectively power (this is what maintains your airspeed
without the engine.) So your heavier weight and higher rate of descent
gives you more "power" in your glide.
  #4  
Old November 27th 03, 04:34 PM
cddb
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Read a book. I think you'll find the two main factors in climb are power
to weight ratio and span loading. Guess what, more weight hurts both.

In article ,
(Koopas Ly) wrote:
I did more brainstorming at work before going home early with a
headache...this is regarding the last point in my post.

Just to reiterate, I wanted to know why your best glide angle was
independent of weight whereas your maximum angle of climb was.

Here's what I came up with with a FBD:

For an aircraft climbing, unaccelerated, where theta is the flight
path angle.

T = D + W*sin(theta)
L = W*cos(theta)

For small theta, cos(theta) = 1 and sin(theta) = theta

so we now have

T = D + W*theta
L = W

flight path angle is given as (T - D)/W [Eq. 1]

or T/W - D/W or T/W - D/L or

theta = T/W - 1/(CL/CD) [Eq. 2]

First, you'll notice that the first term of Eq. 2 is the thrust to
weight ratio and the second is the lift to drag ratio. To increase
theta during a climb or minimize theta during a glide, one must
minimize the second term by flying at best L/D AOA. For a given
thrust, decreasing weight will increase theta.

The equation above would apply for both climbs and descents. What's
interesting to note is that in a glide when T = 0, Eq. 2 simplifies to
theta = - 1/(CL/CD), independent of airspeed or weight. The
dependence on load factor mathematically vanishes when the engine goes
out. How strange!

However, when flying with engine thrust and at the best (L/D) AOA, an
increase in load factor will IMPACT theta, lessening the climb angle
or making it more "negative".

Don't you find it coincidental that the absence of thrust makes the
dependence of load factor upon theta vanish? Can it only be explained
from the math?

So somehow, the magic equations seem to explain my confusion...but I
am not quite at peace. Au contraire. I've only talked about L/D,
weight, and thrust. What about drag? I am still claiming that you'll
experience an increase in drag with a weight increase REGARDLESS of
whether or not you have thrust. Thus, due to the increase in drag and
REGARDLESS of thrust or not, your angle THETA will be affected (i.e.
theta will get larger, more negative, airplane will pitch DOWN). In
the case of gliding flight, everybody but me agrees that your glide
angle won't be affected. Obviously, I am wrong but I don't know why.
To reiterate, that's why I've been writing all this stuff, 'cause I
don't get why THETA happens to change only when you have thrust but
doesn't when you don't have any and are gliding.

Alright, so the increase in drag with respect to load factor is shown
as:

D = V^2*CD where CD = (1 + CL^2)

By virtue of increasing your load factor (either via CL and/or V),
your drag goes up...[right?]

Now, try to tie that in to Eq. 2 in a desperate attempt to explain why
an increase in DRAG won't cause a increased glide angle (less range)
but will cause a logical decrease in climb angle.

From D = V^2 * (1 + CL^2) and theta = T/W - 1/(CL/CD)

Well, I'll stop here because I seemed to have taken the wrong path, as
I am pretty much stuck. ARRGHH! I AM GOING TO KILL A TURKEY!

  #5  
Old November 27th 03, 04:40 PM
Gerry Caron
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"Koopas Ly" wrote in message
om...
Happy Thanksgiving to ya'll,

This is related to my previous "overweight" post. I want to know if
you agree with the following reasoning:

Context: you take off over your maximum gross weight, and wonder how
your climb speeds and performance are affected.

Vy: The AOA corresponding to the best rate of climb remains the same.
However, the airspeed at which the best rate of climb speed increases
by the square root of the ratio of the current weight and maximum
gross weight. The pilot then pitches to obtain that new, higher Vy
airspeed.

OK so far. As you add weight, the "power required curve" (Power Req'd vs.
AS) for the a/c moves up and to the right -- more power required and it
occurs at a higher AS. Power available doesn't change much -- the engine is
the same, prop efficiency does impart some variation with AS, but it's small
over the variations we're discussing here.

Vy occurs at the point where there is the greatest excess power. (The power
req'd curve addresses level flight, excess power is what causes the climb.)
Since the power required moved up and right and the power avail is approx
the same, Vy increases and the amount of excess power decreases so rate of
climb decreases.

Net affect is higher AS (speed along flight path), lower climb angle, higher
ground speed (cotangent climb angle) and lower rate of climb (tangent climb
angle).


Vx: Again, the AOA corresponding to the best angle of climb for
obstacle clearance remains the same. However, the airspeed to now
achieve the best angle of climb at the current weight increases by the
same square root of the ratio of the current weight and maximum gross
weight. The pilot then pitches to obtain that new, higher Vx speed.

Same as above, except that we replace power with thrust. Curves look the
same, but the key points occur at slightly lower airspeeds.

With added weight; Vx goes up, climb angle goes down, ground speed goes up,
and rate of climb goes down.


Best Glide Speed (best range): From a simple FBD with no thrust
vector, one can find that the best angle of glide is only dependent
upon the inverse tangent of the reciprocal of the lift to drag ratio
or:

Tan(glide angle) = 1 / (L/D), assuming small angles, glide angle ~
(L/D)^(-1)

L/D is purely angle of attack driven. Therefore, the glide angle does
not change with respect to weight.


You are correct. L/D is purely aerodynamic and doesn't change with weight.
But when weight goes up you need more lift to get the same glide angle.
Since you're gliding, the only way to generate more lift is to fly faster.
So your AS for best glide increases as weight goes up. Since you're going
faster, the sink rate also increases; but since the angle is the same, you
still glide the same distance -- you just get there sooner.

That last reason is why some gliders have water ballast tanks. Since they
almost always fly at best L/D when trying to cover ground, the added weight
lets them fly faster. Which is important when racing.

Gerry



  #6  
Old November 27th 03, 11:23 PM
Koopas Ly
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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.

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.
I hope we can agree that both airplanes are operating at their minimum
drag points, points of least thrust required. This means that in the
case of the airplane climbing, its climb angle is maximized. In the
case of one gliding, its descent angle is minimized.

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?

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).

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.

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

I've capitalized the effects I don't understand. In both cases, I
contend there was an increase in drag that should have affected both
airplanes' flight path angle to maintain equilibrium of forces.

Best regards,
Alex
  #7  
Old November 28th 03, 04:56 AM
Gerry Caron
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"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


  #8  
Old November 28th 03, 06:24 AM
Peter Duniho
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"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


  #9  
Old November 28th 03, 10:17 AM
Koopas Ly
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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

  #10  
Old November 28th 03, 05:59 PM
Greg Esres
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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.
 




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