Velocity, acceleration and force in a system 11

[elogesh]

Hai,

I am having a moving table in a system.The table undergoes following motion,

1) Motion 1: Accleration of the table from zero velocity to max. velocity

2) Motion 2: Movement of the table at constant velocity (Zero acceleration)

3) Motion 3; Deceleration of the table from max.velocity to zero velocity.

We are going to study the vibration behaviour of the system, once the table movment is stopped.We planned to use stationary dynamic model.

We planned to apply the velocity and acceleration characteristics in terms of force vs time characteristics.

Following is the approach used for force calculation,

1)Motion 1- Force, F1 = M a (M-mass of the moving table and a - acceleration of the table).

2)Motion 2- moving at constant velocity,the force same as what was there for motion 1 (F2=F1)

3) Motion 3-Deceleration of the table from max. velocity to zero velocity. Therefore F3= Ma

Regarding this calculation, customer has an issue. He says that force for motion 2 should be zero, because the system is moving at constant velocity and hence zero acceleration (Zero force). But I perceive that there has to be a force to keep it moving at constant velocity.

I am not quite sure that,whether Am I right?

Can someone help me in clarifying this issue/

Looking forward for your help.

Regards,
E.Logesh

 

[TheTick]

F2<>F1. Continue to apply F1 force, and table will continue to accelerate.

F2 = friction

 

[mog69]

There must be a force input to overcome system inefficiencies such as bearing drag / friction / atmospheric drag (don't know what speed we're talking about) etc, unless you've got a frictionless system operating in a vacuum!

 

[elogesh]

Hi,

Thanks to Tick and Mog69.

I completely agree with you. I find it difficult to convince the customer. He says force2=0 in motion 2.

I tried to take example of, acceleration, motion at constant velocity and deceleration of car. But couldn't convince.

How can I go to simple terms in explaining this to my customer?

Regards,
Logesh.E

 

[mog69]

Ask your Customer what would happen if he was maintaining a steady speed in his car on a flat road with no wind, and put the gearbox in neutral gear - would it carry on at steady speed or slow down? (If it carried on at steady speed I think we'd all be very happy indeed.....!)

 

[GBor]

Or ask him if he has ever seen a perpetual motion machine? If he has, we'd all like to know where!

 

[rb1957]

am i missing something, or didn't a guy named Newton have something to say on this ?.

F1 is the force required to overcome the inertia of the tbale, right; to accelerate it from a stationary table into a moving table.

if F2 is the froce required to keep the table moving, at a constant velocity, then it's much less than F1 as it only has to overcome the losses in the system (friction, drag, ...). think about the work done in the two cases, much more work is being done in the latter.

 

[MintJulep]

Motion 1: F1 = ma1 + friction + drag

Motion 2: F2 = friction + drag

Motion 3: F3 = -ma2 - friction - drag

 

[MintJulep]

by the way, fiction and drag are both functions of velocity.

 

[Ashereng]

Or, is your client saying motion 2 force is small enough to be negligible?

If the duration is short, and very little friction, it may be small enough to be negligible.

Real numbers may shed more light on your client's arguement.

"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?

 

[sreid]

Your customer may have a point. When moving at constant velocity no matter what the friction is, there is no NET force that would affect the system.

 

[IRstuff]

You need to separate the causes and effects.

In order to keep the slip table moving against the drag of the oil film, the table needs to be forced.

However, since the control system is set to maintain constant velocity on the UUT, the UUT experiences little, if any, residual acceleration.

Therefore, your control accelerometer mounted at the base of the UUT or the table itself will show no net acceleration.

TTFN

 

[desertfox]

Hi elogesh

A body moving a constant velocity has no resultant force acting on it.
mog69 if you do what you suggest with a car then there will be a resultant force acting on it, because as you put the car in neutral gear the resultant force acting on the car is friction at the wheels and drag on the body from displacing air.
When the car is travelling at constant speed the force driving the car balances the forces due to friction and drag.Newtons law states:-

If a body is at rest, or moving with constant velocity,then there is no resultant force acting on it and any forces that do act must balance exactly.


regards
desertfox

 

[mog69]

Hi desertfox
Exactly my point - by definition if a body is at constant speed or rest (steady state) there is no net resultant force acting on it, but those forces that are acting are in equilibrium. I think the we simply need to be more precise in the use of the term "force" , ie do we mean net resultant or do we mean input.
Going back to elogeshs' original enquiry, I think the force we're discussing is the input force required to maintain the table at constant velocity, in equilibrium with the resistive forces in the system (oil film shear, friction etc). In that sense elogesh is correct - there must be an input force unless the system is 100% efficient.
best regards
mog69

 

[solider]

I'm a bit confused with how successful this table movement is going to be.

If you want to control velocity and acceleration then you control it with a position vs time relationship, not a force vs time. In fact I dare say, it is virtually impossible to follow a planned velocity and acceleration profile by controlling force alone.

If you try to control velocity and acceleration with force then you need to know EXACTLY what the various friction and drag factors are. Friction itself is very difficult to model. For example, at start up, before the table starts moving you have to overcome the static friction before it starts moving. Straight away you are delayed and your not following the original velocity plan. When the system comes to a halt, you will likely suffer from stick slip at the low velocities.

To me it doesnt make sense to use. USE A POSITION VS TIME CONTROLLER.

On another note, if you do get round to using a position controller, then the motion profile you propose will excite vibration needlessly in the table movement due to the step changes in acceleration. I get the impression you want them just so that you can measure them as residual vibration.

Great if you want them, but if you don't want them, then change your motion profile to something smoother.

 

[desertfox]

Hi mog69

Your right we do need to be clearer in terms of whether were talking about a resultant force or not.
However the original poster clearly states the force at position 2 is equal to the force at position 1.
Position 1 is a force to accelerate the table.
Ah okay maybe force at position 1 is the force required to
accelerate the table to a constant velocity such that at this constant velocity the original force at position 1 is now cancelled out by friction forces etc.

regards

desertfox

 

[svanels]

Motion 2: Movement of the table at constant velocity (Zero acceleration)

The Sum of forces is almost zero (depending on the friction, air resistance, etc..)

But where does the vibration comes in? Sudden impact at a spring? Free vibration? For motion 1 and 2 use aceleration as initial conditions, for motion 2 use velocity

 

[solider]

I might be missing the point of your experiment here, let me know!

How about this.

It is IMPOSSIBLE to control FORCE AND MOTION at the same time. One is a consequence of the other.

If you control force, you get a MOTION which entirely depends on the RESULTANT FORCE on the table, which is dependent on the motor force, and viscous and coulomb friction (both of which are difficult to model), drag, etc, etc. IE if you want to get a velocity and acceleration with force control, then GOOD LUCK, you have some serious modelling to do.

If you motion control, then your force or torque on the motor are a consequence of the motion,friction etc. THe motion control loop just deals with these disturbances. Here you will get a precise velocity (with good control) and hence acceleration.

 

[elogesh]

Hi all,

Thanks for all your responses.

I was on travel and couldn't reply immmediately.

It is great to have comments/suggestons from experts like you.


Mog 69

"Going back to elogeshs' original enquiry, I think the force we're discussing is the input force required to maintain the table at constant velocity, in equilibrium with the resistive forces in the system (oil film shear, friction etc)".

Yes, you are right.I am talking about the input force. I amn't interested in the resultant force.

I should have been more specific in mentioning "input force' instead of just "force".

we talked to the customer and resolved the issue.

Solidier:
"If you want to control velocity and acceleration then you control it with a position vs time relationship, not a force vs time. In fact I dare say, it is virtually impossible to follow a planned velocity and acceleration profile by controlling force alone".

We are doing this calculation as a part of giving inputs for FE analysis.He provided us the velocity and accleration profiles, based on that we arrived out the input force with some assumptions. But we proposed to the customer that, it is possible to determine the forces during the motion using multi-body motion simulation packages like "Adams, Nastran-4D and also module from LMS".Then this forces can be used for the FE calculation. But unfortunetly, neither we nor customer have time and money to perceive it further.

Customer is happy that the results trend is as per his expectation.

Once again thanks for your support.

Regards,
Logesh.E