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5 Answers

Center Of Gravity – Torque

Asked by: 6414 views , ,
Aerodynamics, Student Pilot

I was wondering the other day about the actual role of the CoG...

On the ground, the pivot points of the aircraft are it's wheels and, unless the CoG is exactly over the wheels, it will create a torque (either a nose-heavy situation or a tail-heavy situation). So the gravitational force basically creates a torque. But, as the aircraft gets airborne, CoG becomes the pivot point, right? So gravity will no longer produce a torque on the aircraft...

We all know that CoG has to be within certain limits in order for the pilot to be able to safely control the aircraft in terms of stability and maneuverability. We also know that if CoG is beyond the fore limit of the aircraft it will be nose-heavy but also have a lot of stability (because the distance from the horizontal stabilizer and the vertical stabilizer is relatively long). Also when the CoG is beyond the aft limit, the aircraft will be tail heavy and have a decreased stability.

But why these situations happen? What I want to say is that it is not the weight of the aircraft which produces this nose-heavy and tail-heavy tendency, right? It is the lift, because, if the CoG is beyond the fore limit, the lift creates a torque to raise the tail. In the opposite, if the CoG is too aft, the lift will create a torque to raise the nose. What I am also thinking is that the rolling tendency of an aircraft (which has it's CoG beyond the left-right limits) works in the same way. If the CoG is well left, the lift created from the right wing creates greater torque than the lift from the left wing and that is the reason why the aircraft banks to the left.

 

I don't really know if I'm right or completely wrong, but I'm really looking forward to your answers.

Thank you in advance.

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5 Answers



  1. Wes Beard on May 07, 2013

    On the ground, the airplane will pivot around the main wheels. If the CG is too far aft, it is quite possible for the airplane to rotate around the main wheels and the tail to hit the floor. The same is true if the CG is too far to one lateral side…. though the CG location would need to be on the wing somewhere. If the fuselage interior was not finished and the left wing tank was full while the right wing tank was empty, it might cause the airplane to tip over.

    In flight, the airplane will rotate around the CG. Draw a side view of an airplane… Place the wings, tail and horizontal stabilizer on the airplane. Draw a vertical arrow down from the first third of the wing and mark that as the CG. Draw a vertical arrow up just aft of the previous line and mark that as center of pressure (CP). Draw a vertical arrow down from the horizontal stabilizer and mark it as the tail down force (T-Down).

    To answer how the airplane responds to changing CG we need to remember the weight and balance formula: WEIGHT * ARM = MOMENT. In this case, we need to calculate the ARM between the CG and CP as well as the ARM between the CG and T-Down force. The weights we are using is the vertical lift vector from the CP and the vertical T-Down force.

    As the CG moves forward, the ARM between the CG and CP increases as well as the ARM between the CG and T-Down force. The T-Down force has to increase to make the equation equal which in turn makes the airplane more stable. With more T-Down force, the angle of attack is going to be higher and a slower true airspeed in cruise should be expected as well as a higher fuel burn.

    As the CG moves closer to the CP, the T-Down force will substantially decrease. This will cause lower lift from the wings, longitudinal instability and a faster cruising TAS.

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  2. Radu Poenaru on May 08, 2013

    Just wanted to drop a small correction:

    Even while sitting on the ground, the gravitational force does not produce torque. It is the surface Normal force that applies at the wheels point of contact that create the torque you mentioned about, not gravity.

    Also, ok, a second correction: it is true that on the ground the airplane will seem to rotate about the main wheels contact point, but that is, well, ~ factually incorrect: if you raise the nose wheel (eg. while taking off) what happens is a rotation of the body around the CoG and a translation up of the CoG that produce a combined effect of seeming to rotate around the main wheels. The body will always rotate arround the CoG (it has no concept of on-the-ground vs. in-the-air), and always as a result of other forces (gravity creates no torque)

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  3. Brian on May 08, 2013

    Patamis,

    Your analysis appears correct though some of it has little relevance to a pilot. Lateral CG, for instance, can be ignored in most cases. So long as you keep the fuel tanks or baggage compartments (in a twin) loaded per the POH specs you won’t find yourself in any trouble.

    As far as fore and aft CG, when operating within their limits there is no appreciable effect on lateral or directional stability. Your concern here is with pitch stability and control. An aft loading is nearly always superior to fore loading so long as you remain within the aircraft’s limits. You’ll have better control and faster cruising speeds.

    —

    Radu,

    The pivot point is over the mains when you’re on the ground. Just like the pivot point of a seesaw is where the bolt attaches to the center, not where the CG is located. In each example the object may desire to rotate about the CG, but it is forced to rotate about the point that is firmly fixed to the ground.

    Patamis, your analysis here was spot on. A CG aft of the mains causes an unstable yawing condition on the ground. This is the reason tail draggers easily ground loop without pilot input and tricycle aircraft don’t.

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  4. patamis on May 08, 2013

    So we reach the following conclusions:

    1) Weight -as long as the aircraft is airborne- does not produce any torque (because the CoG is also the pivot point and therefore the arm is equal to zero). Weight produces torque only then the CoG differs from the pivot point (when the aircraft is on the ground).

    2) Any torque that is produced is a result of a force different than weight (also speaking when the aircraft is airborne). Therefore it is the position of the CoG and not the weight that affects the torques as well as the “effort” made by the stabilizers.

    But, if the weight does not produce any torque (again when the aircraft is in the air), why the CoG has to be within limits concerning it’s distance from the datum line AND concerning the actuall weight of the aircraft. I mean, ok, the maximum weight of the aircraft in every state of it’s flight is known and will not be exceeded. But, if the CoG is, for example, 42 inches aft of datum, it doesn’t matter if the aircraft has a mass of 1.500 kg or 2.000 kg as the torque of weight will always be 0. Is it because greater weight requires greater lift and by using that MOMENT = ARM * FORCE greater lift means dangerously “strong” moment (and a lot more “effort” from the stabilizer to “deal” with this moment)?

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  5. Brian on May 12, 2013

    “It doesn’t matter if the aircraft has a mass of 1.500 kg or 2.000 kg as the torque of the weight will always be zero. Is it because greater weight requires greater lift.”

    True the weight is not producing a torque. However, lift is producing a torque. That torque is proportial to the arm between cgcp and the quantity of the lifting force. Total weight means more lift, thus more torque from the production of lift. Likewise, a shift in cg means a change in cgcp arm and a subsequent change in the torque produced.

    I encourage you to reread Wes’ post with regard to the cg/cp relationship. That seems to be the missing piece for you from what I can see. It is this relationship that determines pitch stability.

    You might also consider visiting http://www.faa.gov/regulations_policies/handbooks_manuals/aviation/media/00-80T-80.pdf

    Read chapter 4 starting on page 243 and read the definitions and longitudinal stability sections of this chapter. It ends in page 283.

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