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When gust of wind hit plane form side,does plane allways rotate about center of gravity like weather vane about pivot point or CG as pivot point is just "math areegment" for easier solve equation?

If one force(wind gust) act at body which is not at CG,body will translate and rotate,but center of rotation will not be at CG.

Because only rigid object in space far from other influences will rotate about its center of mass.

If does ,why plane allways rotate about CG? (boats dont do that, but only differnce in air and water is fluid density?)

enter image description here

quiet flyer
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3 Answers3

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It's purely a mathematical trick, but a very good one at that. In reality, pure rotations around the center of gravity are rare.

The key is that it's often convenient to split movement of a body into translations and rotations. So we need some definition to distinguish between a rotation and a translation. For example, consider lifting a pencil from a table on one end. Intuitively, there's definitely some rotation as the pencil is now at an angle, but is there also a translation? The end of the pencil resting on the table has not translated at all, while the end you picked up has translated a lot. So we can describe the same situation by pure rotation around the point resting on the table, or a combination of a translation of the point you're holding and a rotation around that point, or anything in between.

By defining rotations and translations with respect to the center of gravity, you simplify a lot of equations, and you can analyse the system with a free body diagram. If all forces in one direction sum to exactly zero, you will have no translational acceleration, and only rotational acceleration. In reality of course, forces rarely sum to exactly zero, so there's always some translational acceleration of the center of gravity.

Sanchises
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  • When side gust hit,plane is not rotate around CG.If we know that plane lateral center of pressure is somewhere behind CG,then we know that pivot point is somewhere in front of CG.Do you agree? –  Jan 11 '21 at 22:00
  • @EBV821, if you try to describe the motion as pure rotation, yes. But it basically never makes sense to describe the motion that way. – Jan Hudec Jan 11 '21 at 22:49
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    @EBV821 Indeed, there's no one unambiguous way to split rotations and translations. CG is nice for a mathematical description. The point ahead of CG you mention is nice for attaching your plane in a windtunnel because that's where lateral displacement due to a gust is initially zero. – Sanchises Jan 12 '21 at 07:32
  • @Sanchises I post new question about testing yaw stability at gusts in wind tunnel –  Jan 12 '21 at 09:32
  • @JanHudec What if I must test yaw stability from wind gusts in wind tunnel?https://aviation.stackexchange.com/questions/83468/where-placed-attached-point-at-aircraft-when-simulate-wind-gust-yaw-stability –  Jan 12 '21 at 09:33
  • @Sanchises So your answer is ,plane is not rotate about CG? –  Jan 12 '21 at 09:57
  • @EBV821 My answer is that "it depends". For a co-moving inertial observer: yes, the plane will appear to momentarily have a pure rotation about a point ahead of the CG. But a physicist sees a superposition of translation and rotation around the CG. But perhaps for your purposes, that's a bit too much "spherical cow in a vacuum". – Sanchises Jan 12 '21 at 10:07
  • @RobertDiGiovanni I'm sorry but what on Earth are you talking about? The question is about weathervaning in a gust (i.e., sideslip). AC and CP are about pitch which is a different axis altogether. Please use the comment section only for asking clarification or suggesting improvements for this particular answer. – Sanchises Jan 12 '21 at 11:36
  • @Sanchises a fuselage in side slip acts as an airfoil. It will have an aerodynamic center too. Have a look at an aerobatic plane flying on its side. – Robert DiGiovanni Jan 12 '21 at 13:34
  • @RobertDiGiovanni That was totally unclear from your comment. You mention camber, and I am yet to see a fuselage that is cambered in the sideslip direction (perhaps after a crash). Anyway, was your comment meant as a request for clarification or suggesting an improvement to my answer? – Sanchises Jan 12 '21 at 13:53
  • @RobertDiGiovanni PeterKampf wrote in comment that accepted answer is not correct.I agree with accepted answer for now.. – user707264 Jan 05 '23 at 19:51
  • When I took analytical mechanics, I was taught that any set of forces acting on a body in free space, (i.e. unconstrained by being attached to something, like a weathervane is, or a span of a bridge), that any set of forces, no matter how many or how complex, can be converted into an single equivilent force acting though the center of mass, and a single equivilent torque or moment acting about the Center of Mass. And I was taught how to perform this conversion. So bodies in free space always rotate about their center of mass. – Charles Bretana Jan 07 '23 at 00:53
  • @CharlesBretana Yes, but they maybe didn't teach you that can do this trick around any point in space, not just the CoM. A common example is the aerodynamic center. – Sanchises Jan 07 '23 at 04:44
  • Sorry, I assumed you would get it. If any set of forces can be treated equivalently to a single force and torque acting through the CoM, it follows that free objects rotate about the CoM. Even a single force, acting on a line that does not pass through the CoM can be converted to and treated, equivalently, to a single force and a torque, acting on the CoM. – Charles Bretana Jan 07 '23 at 12:45
  • And you can't do this trick with any other point . If you do it with a point other than the CoM, then the single force you posit that is acting through a point other than the CoM, will itself cause a torque about the CoM, causirotation. translation and a roration. – Charles Bretana Jan 07 '23 at 12:50
  • But perhaps I was not clear in my first post about this (the limit on the size of these posts!). What I was taught, is that. "any set of forces, no matter how many or how complex, can be converted into an single force acting though the center of mass, *that will produce only an equivilent translation*, and a single torque or moment acting about the Center of Mass, that will produce an equivilent rotation". – Charles Bretana Jan 07 '23 at 12:56
  • @CharlesBretana You are of course completely right that the CoM is the only point where a pure force (real or "equivalent") will not induce a rotation (although the location at which torque acts is completely irrelevant), which is exactly what I mean with a mathematical convenience. I disagree however that from this it follows that objects rotate around their CoM, although for most problems that is of course the most logical description. You can however derive all equations of motion around any point you dare choose ... Ctd – Sanchises Jan 07 '23 at 19:39
  • Indeed, if any set of forces and torques can be reduced to a single force acting on the CoM and a single torque, it logically follows that this force and torque can be split into an arbitrarily complex set of forces and torques acting at an arbitrary set of locations, if one so desires. Often not useful but occasionally insightful. – Sanchises Jan 07 '23 at 19:41
  • Of course this is all quite academic, but it's a bit of a pet peeve of mine when people claim absolute truths based on a convenient description. Consider a Newtonian apple: does it fall because of a force called gravity, due to gravitational acceleration without an opposing force, or even potential energy being converted to kinetic energy? They're all the same equation rearranged to your liking. – Sanchises Jan 07 '23 at 19:47
  • @Sanchises, of course you are right that mathematically, you can represent any set of forces and torques with an infinite number of sets of different forces and torques that are equivalent, Also, the set of forces and torque sthat will generate the same translation and rotation is completely dependent on the choice of the frame of reference that you choose to do the analysis in, and what it's inertial linear and angular velocity and acceleration of that frame of reference. Indeed, that's what the transform tensor does. (cont) – Charles Bretana Jan 08 '23 at 01:52
  • But the OP's question (although he doesn't come right out and say this), is about the assumption that free bodies (those that are unconstrained to be fixed or immobile at some part of their structure), are intrinsically, inertially, bound to move, in response to any set of external forces, in such a way that the CoM can be treated as the point where all the mass is located for the purposes of translation, and that any total resultant torque on the body, can be created as acting about the CoM. This is true, and all the special cases and semantics that implies otherwise only confuses the issue. – Charles Bretana Jan 08 '23 at 01:56
  • @Sanchises, and as to Newton's apple, as it turns out, it is not falling. It will appear to be traveling in an absolutely straight line in any inertial (unaccelerated) frame of reference. See the diagram on page 33 of http://xdel.ru/downloads/lgbooks/Misner%20C.W.%2C%20Thorne%20K.S.%2C%20Wheeler%20J.A.%20Gravitation%20%28Freeman%2C%201973%29%28K%29%28T%29%281304s%29_PGr_.pdf – Charles Bretana Jan 08 '23 at 02:02
  • I think we're splitting hairs here, but I do maintain that the CoM description is a mathematical convenience. In the end all we're doing is coordinate substitution, and while some coordinate systems are more convenient than others, one is free to choose their coordinate system. Sure, officially gravity is a fictitious force but must we then abolish all books that say lift=weight? Can we never choose the landing gear as a pivot? Or describe the aircraft dynamics from the IMU sensor instead of the CoM? Any equation of motion defined around the aerodynamic center is illegal? – Sanchises Jan 08 '23 at 08:15
  • Anyway, feel free to write your own answer :) and I will probably upvote it because I do believe your viewpoint is equally valid and valuable. – Sanchises Jan 08 '23 at 08:29
  • @Sanchises Plane always rotate around c.g. ,when plane goes in any curve he still rotate around c.g. Rember that moon rotate about itself when revolve around earth.. – user707264 Oct 07 '23 at 13:42
  • @user628075 Yes, it always rotates around the cog. It will also rotate around any other point in the universe. Mathematics are funny like that. – Sanchises Oct 07 '23 at 18:44
  • @Sanchises But does c.g. of plane move is straight line or not ,during plane rotate? – user707264 Oct 07 '23 at 19:38
  • @user628075 Probably not in a straight line. In a coordinated turn especially, there's a well defined point that has zero translation and only rotation (compare with a pylon turn) – Sanchises Oct 08 '23 at 05:22
  • @Sanchises But when crosswind hit plane, than net force is non zero, does that mean c.g. must move in curve path which mean plane is not rotate around c.g.? – user707264 Oct 08 '23 at 15:27
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In any inertial frame of reference, no motion (linear or rotational), can occur without a real force to cause it.

That means that unless an external force is acting on the body through some point other than the center of mass, rotations can only occur about the center of mass.

Bianfable
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Charles Bretana
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  • okay, picky, picky picky! Unless the aggregate vectorial total of all forces is acting on the body through a point other than the center of mass... Which means that it is hardly ever the case in aviation, since by definition, if an aircraft is in steady state flight, the aggregate vectorial total of all forces *Must* be zero. – Charles Bretana Jan 05 '23 at 22:30
  • We're talking about bodies free to move without constraint. A Body that is held or constrained in any way obviously does not have the same six degrees of freedom that an aircraft has. The Weathervane example is irrelevant. – Charles Bretana Jan 06 '23 at 03:31
  • Again, (unless constrained so that it cannot move at some point), a body in free space must rotate about its center of mass. The center if Mass does not have to be part of the body or inside the body (consider a sickle-shaped object, or a dumbell-shape with an extremely curved shaft connecting two massive weights ), but for any object, if any force is applied that does not act through the center of mass it will generate a torque or moment about the Center of mass. – Charles Bretana Jan 07 '23 at 00:47
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There's more to this question than first meets the eye at first-- it opens to the door to a question about what do we mean that we say an object is "rotating about its CG".

Do we mean that if we have a movie camera following (or travelling above) an aircraft in linear flight, such that the CG of the aircraft is in the middle of each movie frame, and then "something happens"-- a gust of wind strikes the airplane, or the pilot makes a control input-- the CG of the aircraft will stay in the middle each new movie frame as the camera continues travelling along linearly?

If that's what we mean, the answer is "no"-- that wouldn't happen when a gust of wind struck an airplane, or when a gust of wind struck a weathervane that was somehow levitating in the air. It also wouldn't happen when the pilot pulled back on the control stick or stepped on one rudder pedal.

In many maneuvers we can identify some other point, often far beyond the physical edges of the aircraft, as the center of rotation of the maneuver. Obvious examples include perfectly round turns, or perfectly circular loops. If we are only interested in describing the actual motion of the body, and not the forces and torques involved, then we can arguably describe these maneuvers as simple rotations around the point in the center of the maneuver, with zero translation involved.

Similarly, this related answer considers the simple case of describing the motion of a pencil being lifted from a table while keeping one end on the table-- we can say the pencil is rotating around its end with zero translation, or we can say the pencil is rotating around it CG while also translating diagonally.

However all these alternative descriptions that take some point other than the CG as the center of rotation fail to accurately describe the linear acceleration of the body in terms of the net force acting on the body.

The Center of Gravity is unique in the sense that it is the only point on the body whose instantaneous linear acceleration is directly proportional to the actual net force acting on the body, while the instantaneous linear acceleration of all other points on the body can be viewed as the sum of the instantaneous linear acceleration of the CG plus a rotational acceleration about the CG.

If we define rotational acceleration as occurring around some point other than the CG of a body, then we'll find we have to introduce a pseudoforce representing the gain or loss of "apparent weight" due to G-loading, acting at the CG, which is equal in magnitude and opposite in direction to the real net force acting on the body. Now the forces we're considering to be acting on the body (including this pseudoforce) no longer add up to be equal to the actual net acceleration of the body (times mass).

It's not uncommon in aviation to find diagrams that take exactly this approach. See footnote 1 for a specific example. Such diagrams are not necessarily "wrong", as long as their limits are understood-- they do not correctly depict the actual net force acting on the object, and thus they do not correctly depict the instantaneous linear acceleration of the object.

If the linear acceleration of a body is zero, and the rotational acceleration of that body is also zero, then need to consider a "pseudoforce" acting at the CG vanishes, and any point will serve equally well as a "pivot point" for torque calculations.

One way to think about the "pendulum effect" conundrum (see this related ASE question) is to recognize that if we select some point (such as the wing's effective "center of lift") other than the CG as our "pivot point" for torque calculations, then the "apparent weight" vector acting at the CG can contribute a roll torque about this point. The dynamics of turning flight and sideslips are such that this roll torque contribution from the "apparent weight" vector is generally stabilizing (tends to roll the aircraft toward wings level) if the CG is located below our chosen "pivot point". Naturally, the same choice of "pivot point" may also eliminate other roll torque contributions (e.g. aerodynamic sideforce generated by the wing in a sideslip) that would also be stabilizing-- since the choice of "pivot point" for our torque calculations can not ultimately not affect the answer that we get.

"Shifting gears" a bit, if we are simply trying to describe the motion of an object without regard to the forces and torques at play, there is one sense in which there is one particular point other than the CG may sometimes be the most descriptive choice for some purposes. Consider a steady-state turn with constant yaw and pitch rotation rates. A pivot point can be chosen for each axis that indicates that the point where the undisturbed relative wind is (or would be) parallel to the longitudinal axis of the aircraft. This point may lie outside the physical boundaries of the aircraft itself. See for example this section entitled "Long-Tailed Pitch Effect" from the John Denker's "See How it Flies" website. Since the aircraft is rotating in the nose-up, tail-down sense in the pitch axis, this induces a curvature in the undisturbed relative wind that would be sensed at any point on the aircraft. (For an extreme case, think of how yaw strings or telltales would deflect if mounted on the wingtips of an aircraft in a flat spin, or on the tips of a propeller or rotor blade-- all the strings or telltales would not be parallel to each other.) Since the longitudinal axis of the aircraft is straight, not curved-- the fuselage doesn't "bend like a banana" to conform to the curvature in the relative wind induced by the pitch rotation-- the relative wind can't be parallel to the longitudinal axis along the full length of the aircraft. In this particular diagram, the aircraft has been drawn in an orientation where the relative wind is nearly parallel to the aircraft's longitudinal axis at the nose, but not at the tail. In a sense this diagram depicts a situation where the aircraft is rotating "around" a point near the nose, so the resulting aerodynamic damping effect is not greatly affecting the angle-of-attack of the wing, but is creating a strong positive angle-of-attack at the tail. However, if we simply increased the angle-of-attack of the whole aircraft a few degrees, the point where the longitudinal axis is parallel to the curving relative wind would shift forward to some point ahead of the nose, which would arguably be a more realistic depiction of actual flight with a positively lifting wing. So we may find it useful for some purposes to describe aircraft's motion as a pitch rotation around that point ahead of the nose, plus a translation parallel to the direction of the curving relative wind at that point. The same perspective can be applied to yaw rotation in turning flight. In this section entitled "Long-Tail Slip" of the "See How it Flies" website, we are looking down on an aircraft in a turn, and again we can see how the curving relative wind cannot be parallel to the aircraft's longitudinal axis along the entire length of the fuselage. In the upper figure (8-9), the relative wind is depicted as parallel to the fuselage near the tail, and coming from the inside of the turn at the nose. In the lower figure (8-10), the relative wind is depicted as parallel to the fuselage near the nose, and coming from the outside of the turn at the tail. In a sense, by simply adjusting the aircraft's yaw attitude in relation to the relative wind, we've moved the location of the axis "around" which the aircraft is rotating from the tail up to the nose. Along the same lines, consider what happens if we yaw the aircraft 5 or 10 degrees further toward the "outside" of the turn so that the airflow is now striking the "inside" (left) side of the fuselage along its entire length, so that we are have a slipping turn. One way to describe the aircraft's orientation and motion is to view the yaw rotation as occurring around some point well behind the aircraft, while the aircraft is also translating forward parallel to the direction of the curving relative wind at that point. Conversely, in a skidding turn, where the entire "outside" (right) side of the fuselage is exposed to the relative wind, the point where the point where the curving relative wind would actually be parallel to the (extension of) the aircraft's longitudinal axis has moved to some point far in front of the nose of the aircraft. Now the aircraft can be viewed as rotating (in yaw) around that point, while also translating forward parallel to the instantaneous direction of the relative wind at that point. But these are only descriptions of motions, not of the forces at play-- the net centripetal force actually produced by the aircraft at any instant corresponds to the force needed to move the CG through it's arcing path through the sky, not some other point far ahead of the aircraft or far behind the aircraft.

Footnotes:

  1. For example in the book "Basics of R/C Model Aircraft Design: Practical Techniques for Building Better Models: Practical Techniques for Building Better Models" by Andy Lennon, we find a diagram showing the balance of torques in steady-state turning flight. The center of lift of the wing, not the CG, is taken as the "pivot point" for the analysis of rotational acceleration. The diagram correctly shows the net torque on the aircraft to be zero, but only because a force labelled "centrifugal force" is introduced, pulling downward at the CG, in addition to the actual weight vector. This is the G-loading pseudoforce mentioned above. The torque acting on the aircraft can indeed be correctly analyzed this way--the diagram correctly shows that the net torque is zero-- but the net force (if we include the pseudoforce) is now zero, and no longer reflects the actual net force that must be acting on the aircraft in turning flight. This issue would be avoided if we took the CG as our pivot point for our analysis of rotational acceleration. Lennon then incorrectly goes on to state that aircraft stress analysis often suffers from the "error" of failing to consider the "centrifugal force" imposed by maneuvering flight, considering instead "only" the actual lift force created in the turn or loop

  2. This is especially relevant when we are analyzing the pitch and roll torques created by a pilot's body on a hang glider. Both of these two methods are equally valid: 1) Assuming that at any given instant the pilot is holding himself rigidly fixed in some particular desired position, note the resulting CG of the glider-pilot system and analyze the resulting pitch and roll torques created by the aerodynamic forces acting around this pivot point. 2) Take the CG of the glider alone as the pivot point, and in addition to the aerodynamic forces generated by the glider around this pivot point, also consider the pitch and roll torques created by force exerted by the pilot's arm muscles on the control bar, and by the pilot's "apparent weight vector" (not purely vertical, yet also not necessarily perfectly "square" to the wingspan) acting at the point where the flexible "hang strap" connects to the rigid structure of the glider.

quiet flyer
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  • When I took analytical mechanics, I was taught that any set of forces acting on a body in free space, (i.e. unconstrained by being attached to something, like a weathervane is, or a span of a bridge), that any set of forces, no matter how many or how complex, can be converted into an single equivilent force acting though the center of mass, and a single equivilent torque or moment acting about the Center of Mass. And I was taught how to perform this conversion. – Charles Bretana Jan 07 '23 at 00:37
  • @CharlesBretana -- can you help me understand the purpose of this comment? Are you suggesting that some specific part of this answer contradicts that idea? – quiet flyer Jan 07 '23 at 13:45
  • Its hard to nail down exactly... but the ops question, "does it always rotate about the center of mass?" Is easily answered, YES, and it seemed that your answer dances around that. Your comment ".. what do we mean when we say that an object is rotating about its CoM..." seems to imply that this is not a clearly defined concept. It is. What we mean is, (as outlined in Newtons laws), that the CoM always follows a straight line in inertial space, unless acted on by an external force. – Charles Bretana Jan 07 '23 at 18:59
  • .... Or on a parabolic ballistic curve in an accelerated, non- inertial frame ( like the earth) which is consistent with the acceleration of that frame of reference. Any change in orientation or rate of rotation caused by those forces will always be based on the vectorial sum of those individual moments, calculated based on the dot product of the force vector and the distance vector between the point of force application and the CoM. I.e, the normal distance between the extension of the force vector and the CoM.0 – Charles Bretana Jan 07 '23 at 19:03
  • Your description of how yaw strings mounted in different places will point in different directions (true), seemed confusing as it suggests that somehiw this implies something about mutiple ewually vailid choices fie a center of rotation, that would be equivilent to the CoM. Not so. If analyzed, I would posit that yaw strings only indicate the relative orientation of the free stream flow at that point, and do not indicate anything about the position of the center of rotation. – Charles Bretana Jan 07 '23 at 19:12