Yaw stability -- center of gravity must be ahead of center of pressure

[John Mcrain]

If we just prove (https://www.physicsforums.com/threads/does-every-object-rotate-around-its-center-of-gravity.998359/) that object don't rotate about CG ,why then center of gravity must be ahead of center of pressure ,for yaw stability?
Can you explain physics behind this phenomen?

 

[Filip Larsen]

In short, with CP ahead of CM any torque from the total aerodynamic forces will act to increase yaw or pitch, i.e. the aeroplane is not passively stable. If CP is behind CM any torque will instead act opposite, that is, the torque will act to counter any small change in yaw or pitch and thus the airplane is considered passively stable.

I haven't followed the thread you quoted, but in general CM is a nice reference point when describing dynamics of a rigid body in general since it allows separation into linear and rotational dynamics. As a result this means a torque-free body will rotate around its CM independently from any linear motion (i.e. forces acting on the CM, like uniform gravity). In aerodynamics the notion of a torque-free body is of course not very useful, since aerodynamic forces obviously in general will involve a lot of torque (CP != CM in general).

Edit: Pitch stability is nicely explained on https://en.wikipedia.org/wiki/Longitudinal_static_stability but is traditionally described a bit different than yaw stability. If this confuse you then just ignore my comments on pitch.

 

[A.T.]

If we just prove (https://www.physicsforums.com/threads/does-every-object-rotate-around-its-center-of-gravity.998359/) that object don't rotate about CG ,why then center of gravity must be ahead of center of pressure ,for yaw stability?

Because these are two different issues:
1) Conditions to change the rate of rotation
2) Definition of center of rotation
Point 1) by itself says nothing about point 2).

This was just explained (again) to you in the old thread:
https://www.physicsforums.com/threa...und-its-center-of-gravity.998359/post-6442912
Why start a new thread to ask the same question?

 

[A.T.]

... in general CM is a nice reference point when describing dynamics of a rigid body in general since it allows separation into linear and rotational dynamics. ...

Exactly. Using the CM is often practical because the acceleration of the CM is already fully described by the linear laws of motion (F = ma).

 

[John Mcrain]

Why start a new thread to ask the same question?

Because they tell me that is not my orignal question,so I ask new..

 

[A.T.]

Because they tell me that is not my orignal question,so I ask new..

But you keep conflating the two issues, like you did in the old thread. The point was to clarify the difference, not to have the same confusion in two threads.

 

[John Mcrain]

But you keep conflating the two issues, like you did in the old thread. The point was to clarify the difference, not to have the same confusion in two threads.

I already said that my IQ is to low to understand some physics abstraction,plus I struggle with english explanation.
I generaly don't understand "relativity" in members thinking/explantion at this forum..

I say seesaw rotate around pivot point.
Here say, no, it rotate about "Milky Way"!

I say car travel 100km/h.
Here say ,no, it trevel milions km/h about some galaxy!

 

[russ_watters]

I already said that my IQ is to low to understand some physics abstraction,plus I struggle with english explanation.

To be frank, what I see in many of your posts/threads looks more like a stubbornness/lack of desire to learn than lack of ability to learn(among other issues). In order to learn, you first have to accept that what you currently believe is wrong, and you don't seem to want to do that. This particular issue is simple enough that pretty much anyone who can write a complete sentence should be able to understand it:

One of these objects rotates and the other doesn't. Which is which, and why? Note, the question does not contain the words "axis of rotation" or "center of gravity" and your answer shouldn't either.

 

[John Mcrain]

To be frank, what I see in many of your posts/threads looks more like a stubbornness/lack of desire to learn than lack of ability to learn(among other issues). In order to learn, you first have to accept that what you currently believe is wrong, and you don't seem to want to do that. This particular issue is simple enough that pretty much anyone who can write a complete sentence should be able to understand it:

View attachment 276092

One of these objects rotates and the other doesn't. Which is which, and why? Note, the question does not contain the words "axis of rotation" or "center of gravity" and your answer shouldn't either.

Object 2 rotate.
If F1 and F2 is equal magnitude,this is pure rotation about half the distance between two forces,zero translation.

 

[A.T.]

I say seesaw rotate around pivot point.

Yes, even when you sit on one end, so that is the seesaw part at rest in your reference frame, your brain will still tell you that it's rotating around the axis in the middle.

But if you didn't see the ground or anything else except the seesaw bar, you would see it rotating around your seat.

Your brain is not objective.

 

[russ_watters]

Object 2 rotate.

Correct!

If F1 and F2 is equal magnitude,this is pure rotation about half the distance between two forces,zero translation.

It isn't known if F1 and F2 are equal. You didn't explain why and you are bringing in axis of rotation when I explicitly told you not to. Explain why it rotates, and don't discuss axis of rotation.

 

[John Mcrain]

Correct!

It isn't known if F1 and F2 are equal. You didn't explain why and you are bringing in axis of rotation when I explicitly told you not to. Explain why, and don't discuss axis of rotation.

Object 2 rotate because forces are not lie at same line...

 

[Arjan82]

Stability and rotational kinematics are not related. You get stable yaw if a small offset in the yaw results in a restoring moment. This could mean that the center of gravity needs to be in front of the center of pressure.

However, this does not mean that the object 'rotates about' the center of gravity, this is unrelated. And, which point an object rotates about is free to choose, as has been pointed out.

 

[John Mcrain]

Yes, even when you sit on one end, so that is the seesaw part at rest in your reference frame, your brain will still tell you that it's rotating around the axis in the middle.

But if you didn't see the ground or anything else except the seesaw bar, you would see it rotating around your seat.

Your brain is not objective.

Yes but for design purpose you mast know that seesaw rotate about pivot point jont,otherwise you can weld joint and seesaw would not work ..

 

[russ_watters]

Object 2 rotate because forces are not lie at same line...

Close enough. The way I would word the answer it is that the object rotates because there is a net torque. Next question:

About which point(s) does the object rotate? Use the previous answer in your response to this one.
(Now you can assume F1 and F2 are equal/opposite.)

 

[John Mcrain]

Close enough. The way I would word the answer it is that the object rotates because there is a net torque. Next question:

View attachment 276097

About which point(s) does the object rotate? Use the previous answer in your response to this one.

I say about point C if F1 and F2 is equal in magnitude...

But you say about any point...

 

[russ_watters]

I say about point C if F1 and F2 is equal in magnitude...

But you say about any point...

The correct answer is any point because of the prior answer: it rotates because there is a net torque. So, let's prove it.

The distance between points is 1. What is the net torque about point C? How about point A?

 

[russ_watters]

Yes but for design purpose you mast know that seesaw rotate about pivot point jont,otherwise you can weld joint and seesaw would not work.

Of course it rotates about the pivot point. What you are suggesting we are telling you here is exactly the opposite of what we are actually telling you.

The fact that you can say it rotates about the kid sitting on one end doesn't mean you can't say it rotates about the pivot point. It's both, not either/or. You are the one insisting it has to be either/or, not us.

 

[John Mcrain]

The correct answer is any point because of the prior answer: it rotates because there is a net torque. So, let's prove it.

The distance between points is 1. What is the net torque about point C? How about point A?

F1=F2=30

point C
MC= F1x2 + F2x2 = 30x2 + 30x2= 120

point A

MA=-F1x3 + F2x7 = -30x3 + 30x7 = 120

 

[russ_watters]

F1=F2=30

point C
MC= F1x2 + F2x2 = 30x2 + 30x2= 120

point A

MA=-F1x3 + F2x7 = -30x3 + 30x7 = 120

You didn't need to assign a value to F and you used the wrong distance, but anyway, you got the same answer for each. There's a net torque about each point and it's the same torque.

One of the things you are doing wrong in many of your examples is you are adding in extra forces, but not accounting for them. With the see-saw, for example, the central pivot point isn't just an arbitrary point; it applies a force.

 

[John Mcrain]

You didn't need to assign a value to F and you used the wrong distance, .

What wrong distance I made?

 

[A.T.]

Yes but for design purpose you mast know that seesaw rotate about pivot point jont,otherwise you can weld joint and seesaw would not work ..

A mechanical joint doesn't imply a unique rotation center. Look at segment between J and F the video below. Around which of its 3 joints (J, H, F) is it rotating?

 

[russ_watters]

What wrong distance I made?

Each interval is 1 unit of distance, so the torque about C is -1F + (-1F) = -2F (clockwise, by defining up and right as positive)

 

[John Mcrain]

Each interval is 1 unit of distance, so the torque about C is -1F + (-1F) = -2F (clockwise, by defining up and right as positive)

I count each square 1 unit..

 

[russ_watters]

I count each square 1 unit..

I told you what the distances were...

 

[John Mcrain]

The correct answer is any point because of the prior answer: it rotates because there is a net torque.

I know if put stick instead line A-E,and two engines with propellers instead forces F1 and F2 , the stick will be rotate in clokwise direction about poin C at this paper, without any translation...

I don't understand if I am really so stupid or I just don't know how to explain what I want to say/ask...

 

[A.T.]

I don't understand if I am really so stupid or I just don't know how to explain what I want to say/ask...

If you want to talk about an unique "center of rotation", you should define precisely (mathematically) what you mean by that. See this post for example:
https://www.physicsforums.com/threa...und-its-center-of-gravity.998359/post-6442666

Saying stuff like "But I can see it" or "There is a joint there" is not sufficient for the reasons explained to you.

 

[Lnewqban]

If we just prove (https://www.physicsforums.com/threads/does-every-object-rotate-around-its-center-of-gravity.998359/) that object don't rotate about CG ,why then center of gravity must be ahead of center of pressure ,for yaw stability?
Can you explain physics behind this phenomen?

Are you talking about yaw stability of an airplane?

 

[John Mcrain]

Are you talking about yaw stability of an airplane?

Yes yaw stability for airplane,rocket..

 

[Lnewqban]

Yes yaw stability for airplane,rocket..

For any reason, the airplane can yaw around its CG.
Because of that, the fuselage and the tail induce some additional lateral aerodynamic drag.
The faster the airplane goes, the stronger is that induced force.

The shape of the airplane should be able to locate the center of that induced lateral drag at a point located some distance aft the location of the CG.
You call that point center of pressure, but I like to call it center of drag.

That distance-drag force combination creates a torque about the CG that opposes the original yaw and tends to return the proper alignment of fuselage with airstream.
If that distance is too big, the airplane tends to crab or weathervane in the presence of strong cross-winds.
The proper distance makes the airplane self-correcting at yaw deviations.