Pitching Moment of an Airplane

In summary, an airplane's wing always has a natural pitching moment, either positive or negative, due to the lift force acting at the center of pressure. This pitching moment is countered by the horizontal stabilizer in the back and the force of gravity. The position of the center of gravity and the center of pressure is important for stability, and most aircraft have the center of gravity placed ahead of the center of lift for positive pitch stability. However, some aerobatic aircraft have neutral pitch stability to allow for more precise control during maneuvers. The pitching moment is mostly affected by the curvature of a wing, and symmetrical airfoils do not have a pitching moment. The aerodynamic center is the point at which the moment due to the lift force is
  • #1
fog37
1,568
108
Hello Everyone,

From what I read, an airplane's wing always has a forward-pitching moment which is a torque that tends to rotate the plane nose down. Is this moment nonzero for any angle of attack (zero, positive, negative)?

A torque needs a force and a distance (lever arm). I think the force in the pitching moment is the lift force ##L## and the lever arm ##d## is the distance from the lift to some specified point, for example he center of gravity ##CG##. The force ##L##, applied at the center of pressure CP, varies in strength and position with changes of the angle of attack. For stability, in case of wind gusts, the center of gravity ##CG## of the plane is placed ahead of the CP. Thanks to the horizontal stabilizer in the back, an aerodynamic downward force is generated that keeps the plane stable. By itself, the wing would be highly unstable. So there are two torques: the CCW torque due to ##L## about the ##G## and the CW torque due to the stabilizer force about the ##CG##. The force of gravity produces no torque about its own position.

My question: why is there this natural forward pitching moment in the airfoil? Why aren't planes designed so the ##CP## and the ##CG## are in the same spot? If ##CP## and ##CG## were at the same point, we would not need the rear horizontal stabilizer...

Thanks!
 
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  • #2
If I understand correctly, you want to rely on CG and CP acting in opposite direction at the same point. This is known as an unstable condition.
 
  • #3
fog37 said:
My question: why is there this natural forward pitching moment in the airfoil? Why aren't planes designed so the CPCPCP and the CGCGCG are in the same spot? If CPCPCP and CGCGCG were at the same point, we would not need the rear horizontal stabilizer...

You mentioned stability. It sounds to me like this too is a stability problem, that might lead to flutter or pitch oscillations.

If the torques are exactly balanced, then deviation in either direction causes a restoring torque which would reverse direction when passing through zero. That sounds like a recipe for flutter. Said another way, it would be good design to make sure that none of the forces on the structure ever reverse direction during normal flight with perturbations.

The pivot point of a boat's rudder is also moved to nearly cancel torques. But it always stops shy of total cancellation, because we need negative feedback at all times despite perturbations.
 
  • #4
fog37 said:
Hello Everyone,

From what I read, an airplane's wing always has a forward-pitching moment which is a torque that tends to rotate the plane nose down. Is this moment nonzero for any angle of attack (zero, positive, negative)?

It depends on the wing section. Different wing sections have different pitching moments. Those designed for non swept tailless aircraft have a positive (pitch up) moment rather than a negative pitch down moment.

A torque needs a force and a distance (lever arm). I think the force in the pitching moment is the lift force ##L## and the lever arm ##d## is the distance from the lift to some specified point, for example he center of gravity ##CG##. The force ##L##, applied at the center of pressure CP, varies in strength and position with changes of the angle of attack. For stability, in case of wind gusts, the center of gravity ##CG## of the plane is placed ahead of the CP. Thanks to the horizontal stabilizer in the back, an aerodynamic downward force is generated that keeps the plane stable. By itself, the wing would be highly unstable.

Thats true of conventional aircraft but not generally as non-swept tailless aircraft wouldn't be possible.

You might be interested in how tailless aircraft get their stability...

https://www.mh-aerotools.de/airfoils/flywing1.htmPS I think Martin's website has been around for perhaps 20 years.
 
  • #5
Thanks everyone!

So, in essence, from what I understand, a wing by itself always has a natural pitching moment, either positive (CW) or negative (CCW) and will always be unstable. However, once the wing(s) are mounted on the fuselage, the plane can become stable thanks to the correcting moment due to the rear horizontal stabilizer and the moment due to the force of gravity. Planes without the tail, tail-less airplanes are also called wing airplanes since they are essentially one big wing.

I am clear on what the CG and the CP are: the CG is the center of gravity where the weight force is applied. CP is the center of pressure where the lift and drag forces are applied. What about the aerodynamic center? It is the point about which the aerodynamic moment is constant for various angles of attack. What does that mean?

I guess the aerodynamic moment is due to the lift force, always applied at the CP. The moment of this force varies depending on which point we calculate it about. But if we pick the aerodynamic center, the moment due to the lift force, is constant as the angle of attack changes. How is that possible? As the angle of attack change,s the position of CP also changes (moves forward).
Thanks
 
  • #6
The pitching moment is mostly related to the net curvature of a cambered wing. As a curved wing moves through the air, it tries to pitch in the direction of curvature. Symmetrical airfoils don't have a pitching moment. From wiki article: "In the case of a symmetric airfoil, the lift force acts through one point for all angles of attack, and the center of pressure does not move as it does in a cambered airfoil. Consequently, the pitching moment coefficient for a symmetric airfoil is zero."

https://en.wikipedia.org/wiki/Pitching_moment

As for aircraft stability, most aircraft have center of gravity a bit in front of center of lift for some positive pitch stability, but aerobatic aircraft usually have neutral pitch stability, so that the pilot doesn't have to compensate to keep the aircraft moving in a precise straight line during a dive or a climb.
 
  • #7
Thanks rcgldr. I understand your comments completely. One only clarification if possible:

Wikipedia says that "the pitching moment on an airfoil is the moment (or torque) produced by the aerodynamic force on the airfoil if that aerodynamic force is considered to be applied, not at the center of pressure, but at the aerodynamic center of the airfoil. "

"...One of the remarkable properties of a cambered airfoil is that, even though the center of pressure moves forward and aft, if the lift is imagined to act at a point called the aerodynamic center the moment of the lift force changes in proportion to the square of the airspeed..."


Isn't the aerodynamics force represented by the lift force? The aerodynamic center is a real point on the chord. But it seems strange and artificial to have to have to imagine the light force, which truly acts always at the center of pressure CP, at that other position, the aerodynamic center...

thanks
 
  • #8
fog37 said:
Isn't the aerodynamics force represented by the lift force?

In reality there are forces acting all over the surface of a wing. Typically we sum all these forces then resolve them into two forces we call lift and drag plus a pitching moment.

When you resolve a vector into components you have some freedom to choose which direction they act. In this case where you put the lift vector effects the pitching moment needed to make the sums add up.

The aerodynamic centre is a special point. If we put the lift component there then you get a pitching moment that doesn't change with angle of attack. That turns it into a constant making things easier to analyse.
 
  • #9
Thanks CWatters.

I see how the pressure field distribution gives rise to a distributed force, the aerodynamic force which can be decomposed, as you mention, into lift and drag. Lift is perpendicular to the wing and drag is horizontal. What do you mean when you say "lift and drag plus a pitching moment"? What force or couple of forces cause the pitching moment?

Also, the lift force is applied to the center of pressure CP. How can I just change its application point and apply it to the aerodynamic centre AC? That seems not physical. Is it just a mathematical trick?
 
  • #10
fog37 said:
Thanks everyone!

So, in essence, from what I understand, a wing by itself always has a natural pitching moment, either positive (CW) or negative (CCW) and will always be unstable.

You can make a wing that is stable on its own but it tends to produce less lift and be less efficient. You are better off making a wing that does pitch nose down and fix the problem by adding a tail.

So yes you are correct in that on most conventional aircraft there is a down load on the tail to counter the nose down pitching moment of the wing. This down force on the tail behaves like extra weight so the wing has to produce a bit more lift than it would otherwise.

On a canard the tail is at the front and contributes to the overall lift as well as countering the pitching moment. Some say this is why the Wright Brothers put the tail at the front but I don't know if they appreciated the advantage it gives over a rear tail.

The centre of gravity is also important. The more rearward it is the less stable the aircraft is but a rearwards center of gravity also counters the nose down pitching moment. So moving the centre of gravity back means the tail doesn't have to provide so much down force nor does the wing have to provide extra lift to compensate. This means the drag of the tail and wing are reduced. So a rearward centre of gravity can save fuel at the expense of stability.
 
  • #11
Thank you.

Now a related but different questions if I may: an airplane is traveling horizontally at a speed ##v_1## and at an altitude ##h_1## where the density is ##\rho_{1}##. The pilot wants to speed the plane up, say to ##v_2= 2 v_1##. What happens? I guess

a) the propeller starts spinning faster to increase the thrust
b) the lift automatically increases, even without changing the angle of attack because of the larger thrust. The airplane's weight remains fixed so the plane automatically increases altitude, since the lift > weight, and continues to go up until lift and weight are equal to each other again
c) As higher altitudes are reached, the air density decreases and the lift decreases to match the weight. At that point, at that new altitude, the plane is moving at ##v_2## only horizontally not climbing anymore. Is that correct?

I don't think it is physically possible to increase the airplane's speed while remaining at the same altitude and at the same angle of attack.

Next scenario: the plane moves at speed ##v_1## at angle of attack ##\alpha_1##, the thrust is kept the same. What happens if the angle of attack is changed to ##\alpha_2 > \alpha_1##? The lift force increases. What happens to the speed? The speed remains $$v_1$$ (unless the pilot increases the thrust). So the airplane gains altitude while moving at speed ##v_1##. But lift will change, due to the changing air density. Lift will eventually return to be equal to weight. Now the plane travels at speed ##v_1## in an horizontal path.

Am I on the right track?
 
  • #12
fog37 said:
Thank you.

Now a related but different questions if I may: an airplane is traveling horizontally at a speed ##v_1## and at an altitude ##h_1## where the density is ##\rho_{1}##. The pilot wants to speed the plane up, say to ##v_2= 2 v_1##. What happens? I guess

a) the propeller starts spinning faster to increase the thrust
b) the lift automatically increases, even without changing the angle of attack because of the larger thrust. The airplane's weight remains fixed so the plane automatically increases altitude, since the lift > weight, and continues to go up until lift and weight are equal to each other again
c) As higher altitudes are reached, the air density decreases and the lift decreases to match the weight. At that point, at that new altitude, the plane is moving at ##v_2## only horizontally not climbing anymore. Is that correct?

I don't think it is physically possible to increase the airplane's speed while remaining at the same altitude and at the same angle of attack.

If the pilot wants to stay at a constant altitude all he has to do is reduce the angle of attack to keep lift constant. He can do this with the trim wheel.

Next scenario: the plane moves at speed ##v_1## at angle of attack ##\alpha_1##, the thrust is kept the same. What happens if the angle of attack is changed to ##\alpha_2 > \alpha_1##? The lift force increases. What happens to the speed? The speed remains $$v_1$$ (unless the pilot increases the thrust). So the airplane gains altitude while moving at speed ##v_1##. But lift will change, due to the changing air density. Lift will eventually return to be equal to weight. Now the plane travels at speed ##v_1## in an horizontal path.

Am I on the right track?

If the pilot increase angle of attack the lift will increase but so will drag so the speed will reduce and that reduces lift. Exactly what happens depends on where on the lift drag curve the wing moves from and to.

I would forget the effect that changing altitude has on performance for the moment. It's not something you normally consider when investigating small changes to speed or angle of attack.
 
  • #13
Nice! Thanks a lot CWatters.

Let me paraphrase to make sure I get it:

a) Keeping the same flight altitude and the same total weight, the flight speed can be increased by increasing thrust and simultaneously reducing the AOA via the trim wheel (dont know what that is at this moment). Reducing the AOA reduces lift. To maintain equilibrium, i.e. lift = weight, the plane must travel faster to still generate a sufficient lift, which got reduced due to the smaller AOA. Everything works out without having to change altitude. Changing angle of attack on plane is achieved via the lift control surfaces (flaps) that change the camber and drag. The overall wing does not change angle if it is fixed, just these surfaces move up and down.

b) For the same thrust and moving at particular speed, an increase in AOA always increases lift for an aircraft. The increased lift causes the airplane to start climbing (changing altitude) since lift>weight. However, increasing AOA increases drag so the speed is reduced compared to the original speed unless thrust is increased to make up for the speed decrease. I will learn about the lift/drag curve. How should I read it? It clearly tells us how drag increases with lift.

If we kept increasing AOA, the airplane speed would continue to decrease until the smallest value, called stall speed. If the AOA increased further, stall occurs and the plane starts losing lift. So the stall speed is the slowest speed an airplane can fly before it runs into stall. After stall, lift is not lost completely but just decreased and the plane starts losing altitude because it is moving too slow. However, it can gain speed again as it descends and return to equilibrium.

Fascinating stuff!

thanks.
 
  • #14
fog37 said:
Nice! Thanks a lot CWatters.

Let me paraphrase to make sure I get it:

a) Keeping the same flight altitude and the same total weight, the flight speed can be increased by increasing thrust and simultaneously reducing the AOA via the trim wheel (dont know what that is at this moment). Reducing the AOA reduces lift. To maintain equilibrium, i.e. lift = weight, the plane must travel faster to still generate a sufficient lift, which got reduced due to the smaller AOA. Everything works out without having to change altitude. Changing angle of attack on plane is achieved via the lift control surfaces (flaps) that change the camber and drag. The overall wing does not change angle if it is fixed, just these surfaces move up and down.

Flaps do change camber and angle of attack but normally you would use the elevator. Eg push forward on the control column slightly. To avoid physically having to push all the time the trim wheel moves a small part of the elevator called a trim tab.

b) For the same thrust and moving at particular speed, an increase in AOA always increases lift for an aircraft. The increased lift causes the airplane to start climbing (changing altitude) since lift>weight. However, increasing AOA increases drag so the speed is reduced compared to the original speed unless thrust is increased to make up for the speed decrease. I will learn about the lift/drag curve. How should I read it? It clearly tells us how drag increases with lift.

If we kept increasing AOA, the airplane speed would continue to decrease until the smallest value, called stall speed. If the AOA increased further, stall occurs and the plane starts losing lift. So the stall speed is the slowest speed an airplane can fly before it runs into stall. After stall, lift is not lost completely but just decreased and the plane starts losing altitude because it is moving too slow. However, it can gain speed again as it descends and return to equilibrium.

Fascinating stuff!

thanks.

That last part is correct.

Not so long ago some professional pilots didn't realize they had stalled the airliner they were flying. They lost 30,000 feet before crashing into the sea killing all on board.
 
  • #15
fog37 said:
I will learn about the lift/drag curve. How should I read it? It clearly tells us how drag increases with lift.

"Polars" include plots showing how the lift and drag coefficients change with angle of attack. Note they are the coefficients not the lift or drag forces themselves. The forces also depend on speed.

https://www.mh-aerotools.de/airfoils/hdipolar.htm
 
  • #16
Thank you, again.

Let's say the plane, on its own, has a weight ##W_{plane}##. An extra maximum load ##W_{load}## could be added to the plane. That means the total weight would become ##W_{total} = W_{plane} + W_{load}##. For the plane to fly, the motor must spin a propeller fast enough to allow the plane to achieve a sufficient speed to generate a lift Force ##F >= W_{total}##.

Once we know the total weight ##W_{total}##, we can determine the wings size and call their joint area ##A_{total}## (both wings). Thee wing loading (lb/ft^2) is ##W_{total} / A_{total}##. Once the wing loading is calculated, how do we choose the suitable motor power ##P##, i.e. the suitable ##Watts/lb##?

The power of a motor is P = thrust *pitch speed. The pitch speed determines how quickly the propeller spins hence the maximum airplane speed. What about the thrust? The propeller has a pitch and diameter that must be selected.
Thanks.
 
  • #17
The power needed in level flight is

Power = drag * velocity.

In the climb the aircraft is gaining potential energy so I think you have to add that as well so it would be something like

Power = (drag * velocity) + (weight * climb rate)

However I think more/max power is required during take off to minimise the length of runway required.
 
  • #18
YouTube video on trim. From 2 min 40 seconds he demonstrates setting the elevator trim so he doesn't have to push forward all the time...

 
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  • #19
Thanks. Pretty cool.

I am planning to design a RC model plane. I understand there are a lot of different, sometimes competing requirements. For example, I wonder what would make the plane take off and land over a short distance. I think a small wing loading and a small stall speed would achieve a short take off distance. Now I have to backtrack, given a certain weight and wing loading, what type of motor (power) I need...
 
  • #20
Have you flown an RC plane before? If not then I recommend joining a club, it's not as easy as it looks. You can spend months turning a pile of wood into an aircraft and seconds turning it back again! Some clubs have training aircraft you can have a go with.

Yes aircraft with a low wing loading and lots of power will take off quicker than one with a high loading and less power but that's not usually something we worry about very often. Light weight aircraft can break more easily and don't fly as well in windy/gusty conditions.
 
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  • #21
CWatters said:
Yes aircraft with a low wing loading and lots of power will take off quicker...
I agree ...
 
  • #22
OCR said:
I agree ...

Helps if you have a bit of a headwind as well :-)



The thing to remember is that after landing like this you must remember to retract the flaps before you climb out.
 
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  • #23
Hello again CWatters,

I am thinking about take-off phase and the coefficient of light ##C_L## of the airfoil. The angle of attack and the flaps determine the magnitude of this coefficient for a specific airfoil geometry and wing size.

On the runway, once the plane reaches a certain speed ##v##, the lift force ##L= \frac {1}{2} A C_{L} v^2## will become larger than the weight ##W=mg## and the plane will take off. The cambered wing of the plane is attached to the fuselage at a nonzero angle of attack. This means that there is a nonzero lift coefficient ##C_L##.
Cambered airfoils have a nonzero lift coefficient even at zero angle of attack. The flaps are used during take off to create a large lift coefficient ##C_L## while the plane is still moving on the runway.

Once the plane is just off the ground and ##L> W## the pilot will further increase the lift coefficient ##C_L## by changing the angle of attack by trimming the elevators. The plane will climb because ##L> W##. The flap are not used anymore once in the air because they create drag.

To determine the minimum take off speed, it seems we would need to know the maximum lift coefficient ##C_Lmax## of the airfoil. But ##C_Lmax## occurs when the plane is close to stall and at a large angle of attack, larger than when the plane is moving horizontally along the runway. When the plane is lifting itself off the ground its angle of attack is not the stall angle of attack. So How are the stall speed and the minimum take off speed related? How does the knowledge of ##C_Lmax## for the airfoil help us determine the minimum take off speed?

Thanks!
 
  • #24
fog37 said:
Hello again CWatters,

I am thinking about take-off phase and the coefficient of light ##C_L## of the airfoil. The angle of attack and the flaps determine the magnitude of this coefficient for a specific airfoil geometry and wing size.

On the runway, once the plane reaches a certain speed ##v##, the lift force ##L= \frac {1}{2} A C_{L} v^2## will become larger than the weight ##W=mg## and the plane will take off. The cambered wing of the plane is attached to the fuselage at a nonzero angle of attack. This means that there is a nonzero lift coefficient ##C_L##.
Cambered airfoils have a nonzero lift coefficient even at zero angle of attack. The flaps are used during take off to create a large lift coefficient ##C_L## while the plane is still moving on the runway.

Once the plane is just off the ground and ##L> W## the pilot will further increase the lift coefficient ##C_L## by changing the angle of attack by trimming the elevators. The plane will climb because ##L> W##. The flap are not used anymore once in the air because they create drag.

The benefits of flaps is that they increase lift at low speeds. They also reduce the stalling speed. The pilot will keep using them until the aircraft is flying fast enough not to need them.

To determine the minimum take off speed, it seems we would need to know the maximum lift coefficient ##C_Lmax## of the airfoil. But ##C_Lmax## occurs when the plane is close to stall and at a large angle of attack, larger than when the plane is moving horizontally along the runway.

It's too dangerous to fly close to the stall so most aircraft are designed to take off at a higher speed. In that case the required angle of attack on the ground is lower.

During WW2 a number of aircraft were designed for short field use and these typically had tall undercarriage to present the wing at a greater angle of attack on the ground so they can take off at close to max lift/ stall. Example...

http://storch-aviation.com/assets/images/storch-sld-01.jpg

When the plane is lifting itself off the ground its angle of attack is not the stall angle of attack. So How are the stall speed and the minimum take off speed related? How does the knowledge of ##C_Lmax## for the airfoil help us determine the minimum take off speed?

Thanks!

They are related by the safety margin. I am not really sure how that's calculated, but it will be related to things like how the aircraft performs during an engine failure, wind gusts etc. I guess they may even allow for the pilot setting the wrong flap position.
 
  • #25
Hi again CWatters, I would like to ask you about the aerodynamic center AC and the pitching moments if possible. This is what I know:

a) The total weight W is applied to the CG (center of gravity) of the airplane.

b) The Lift force ##L_{wings}## is applied to the center of pressure ##CP_{wing}## of the wing. The CP is a point along the chord that changes with angle of attack ##\alpha##. Variations of ##\alpha## produce changes in the position of CP and in the magnitude of ##L_{wings}##.

c) The horizontal stabilizer is an inverted wing also producing a lift force ##L_{tail}## pointing downward at its center of pressure ##CP_{tail}##.

d) The aerodynamic center ##AC## is a point that does not vary with changes in ##\alpha##. The moment of ##L_{wings}## about this point AC is constant for variation of ##\alpha##. However, I have noticed that we can change the point of application of the force ##L_{wings}## from the point CP (where it physically exists) to AC. If we do so, we need to add a couple moment. What is the magnitude of this couple moment? Is it constant? Is it CCW (nose down)? Are the forces in the couple both equal to the the lift force ##L_{wings}##?That said, I would like to clearly understand how many pitching moments are applied to the entire airplane. First of all, we need to choose an arbitrary pole ##P## about which to calculate the moments. We could choose the CG of the plane or the leading edge of the wing. These are the moments:
  • The moment due to the weight ##W##.
  • The moment due to ##L_{wings}##.
  • The moment due to ##L_{tail}##.
  • The intrinsic moment due to the wing. What is exactly this moment and how large is it? Is it the moment I described in d) above?

Thanks!
 
  • #26
I think that, choosing ##CG## as the reference pole, these are the five moments that need to add to zero:
  1. The force ##W## produces no moment (zero lever arm): ##\tau_W = 0 ##
  2. The lift force ##L_{wings}##, applied at ##AC_{wing}##, produces a moment given by ##L_{wings} ## times the distance between ##CG## and ##AC_{wing}##. This moment produces CCW rotation.
  3. The ##L_{tail}##, applied at ##AC_{tail}##, produces a torque given by ##L_{wings}## times the distance between ##CG## and ##AC_{tail}##. This moment produces CW rotation.
  4. The intrinsic aerodynamic moment ##M_a## solely due to the wing. This is a CW moment.
  5. The intrinsic aerodynamic moment ##M_{tail}## solely due to the horizontal stabilizer CW. This is a CW moment.
 
  • #27
Perhaps have a look at the Wikipedia definition of AC...

https://en.m.wikipedia.org/wiki/Aerodynamic_center

It's the point about which the pitching moment doesn't vary with small changes in AOA. I am not an expert but there are two aspects of stability that need to be got right...

Static stability. This is what you are talking about in your post #26. Eg The moments sum to zero.

Dynamic stability: This determines what happens when the aircraft is disturbed, eg what happens if turbulence lifts the nose? Does the aircraft stay nose high? Return to level flight? Diverge (eg pitch up more and more)? Oscillate? Normally you want it to return to level flight without overshooting/oscillating. I've never looked at the maths but my understanding is that the maths is made simpler if you take moments about the AC rather than the GoG or some other point, because the pitching moment doesn't change with AOA.

I'm afraid you are getting to the limits of my knowledge at this point.

PS regarding #26...

The down force on the tail produces a nose up torque. If that's CW then the pitching moment of the wing (nose down) must be CCW. So 4. appears to be wrong.
 
  • #28
Ok I'm a bit wrong with my description of static and dynamic stability. See this YouTube video...

 
  • #29
Thanks CWatters! I truly appreciate your help.

My lengthy discussion is due to the fact that I am trying to build a model airplane and I am trying to figure out he wing placement (hence the lift which is applied to the aerodynamic center AC located at 25% of the chord) must be placed related to the total ##CG## and to the rear horizontal stabilizer. For most airplanes, stability is achieved when the CG is in front of the AC which is in front of the AC of the rear stabilizer wing...

Have you designed a RC plane before? If so, how did you determine those distances?

thanks.
 
  • #30
I cheated. Mk 1 eye ball used to get the basic dimensions, the basis that if it looks right it probably is!

CoG set somewhere between 25% and 33% initially and small adjustments made later - perhaps by moving the battery pack backwards or forwards. Typically forward for improved stability on scale models and training aircraft, back for aerobatic and speed models.

I flew fast electric powered competition gliders some years ago and for those we would set them up with a slightly rearward CoG as that allowed us to reduce the down load on the tail. That in turn meant the wing had to produce less lift and hence less drag so they went faster. I understand commercial jets use a similar slightly rearward cog trick to reduce fuel consumption. If you move it back too far they get unstable and tricky to fly.

We would also make trial flights adjusting the elevator trim until it flew as required. Then we would change the rigging angle of the tail plane until no elevator trim was required, again to reduce drag. Our tail planes used thin virtually symmetrical sections.

Another issue for us was adverse yaw. We did away with the rudder to save weight and used differential aileron movement (more up aileron than down) to control yaw in turns. If you didn't set this up right you lost speed in turns.

I flew in several international competitions but never placed higher than about 13th out of 50-70 competitors. This is back in the 1980/90s. My main problem was making them light enough. I always seemed to need more resin than necessary to wet out the glass/Kevlar. The rules set a minimum wing loading so a heavy aircraft was also bigger and slower.
 
  • #31
Thanks CWatters. That is awesome and impressive. I am trying to make a RC model high-lifting airplane (that can carry a lot of weight, given a certain take off speed, etc.). I see how I can determine the minimum take off speed given a certain wing areas, and motor power. I wonder how quickly the plane will reach that speed. That will determine the minimum takeoff distance (runway). Do you have any insight in that?

Just to make sure, you mention that the CG of the entire airplane should be placed at distance, from the leading edge of the wing, that is 25%-33% the mean chord line (starting from the trailing edge of the wing 25-33$ of the chord backward).

Thanks and happy Easter!
 
  • #32
CWatters said:
I flew fast electric powered competition gliders some years ago

Wow, that sounds like fun. But you don't say if the gliders were remote control models, or full size with you on board as pilot.
 
  • #33
They were F5B class RC models. I also flew full size gliders for awhile but not long enough to go cross country, all local flying.

There are quite a few videos of F5B models on YouTube. An example...

 
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  • #34
fog37 said:
Thanks CWatters. That is awesome and impressive. I am trying to make a RC model high-lifting airplane (that can carry a lot of weight, given a certain take off speed, etc.). I see how I can determine the minimum take off speed given a certain wing areas, and motor power. I wonder how quickly the plane will reach that speed. That will determine the minimum takeoff distance (runway). Do you have any insight in that?

For that sort of aircraft you want a thick wing section with quite a bit of camber and possibly flaps. It's tricky working out how fast it will accelerate to that speed. More power the better. A fine pitch prop will also accelerate faster but don't go too fine or the top speed will be too slow.

Just to make sure, you mention that the CG of the entire airplane should be placed at distance, from the leading edge of the wing, that is 25%-33% the mean chord line (starting from the trailing edge of the wing 25-33$ of the chord backward).

Thanks and happy Easter!

Yes 25-33% of the chord back from the leading edge.
 
  • #35
fog37 said:
I guess the aerodynamic moment is due to the lift force, always applied at the CP.
This is wrong and I don't know if it is clearly corrected later. There is a torque on the wing that has nothing to do with the location of the wing versus the CP. In creating lift, airflow is diverted down. That produces a twisting torque on the wing which is different from the leverage arm that the lift is applied to. Both must be accounted for.
 

Related to Pitching Moment of an Airplane

What is the pitching moment of an airplane?

The pitching moment of an airplane is the force that causes the aircraft to rotate around its lateral axis, also known as the pitch axis. It is a measure of the aircraft's stability and can be affected by various factors such as airspeed, angle of attack, and center of gravity.

How is the pitching moment calculated?

The pitching moment is calculated by multiplying the force acting on the aircraft by the distance between the force and the aircraft's center of gravity. This distance is known as the moment arm and is typically measured in feet or meters.

What factors affect the pitching moment of an airplane?

Airplane design, weight and balance, angle of attack, airspeed, and control inputs are all factors that can affect the pitching moment of an airplane. Changes in any of these factors can result in a change in the aircraft's pitching moment, which can impact its stability and handling.

How does the pitching moment impact an airplane's flight?

The pitching moment has a significant impact on an airplane's flight characteristics. A positive pitching moment will cause the aircraft to nose up, while a negative pitching moment will cause it to nose down. This can affect the aircraft's climb, descent, and level flight, as well as its ability to maintain a stable attitude.

How do pilots control the pitching moment of an airplane?

Pilots control the pitching moment of an airplane through various means, including the use of control surfaces such as elevators, trim tabs, and flaps. By adjusting these surfaces, pilots can change the amount of lift and drag on the aircraft, which in turn affects the pitching moment and allows them to control the aircraft's attitude and stability.

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