Calculating Rigid body motion from accelerometer data 1

[machineryguy]

Hello all,
I was hoping to get some feedback/advice from the group.
I have a frame for which I would like to get a much better handle on what is going on with it.

I have 3-axis acceleration data at 5 locations on a frame with respect to time (about 60 seconds worth). What I am trying to calculate is how to represent the frames motion over time given the acceleration data. For now, I am going to keep it simpler, by assuming planar motion (setting aside the z accelerations for now). (let X be parallel to ground, and Y vertical, with Z perpendicular to long axis of frame)
I pulled out my old dynamics text, and I THINK that I am going to need at least 3 of the accelerometers data to be able to obtain the general,planar motion of my "rigid" body.
I'm looking at a chapter labeled "Relative-Motion Analysis:Acceleration"
To state the acceleration of a point on a rigid body, you relate it to another points acceleration and the relative acceleration of the two. I will need to know alpha (angular acceleration of the rigid body), and omega (angular velocity of the rigid body). Obviously, I know the relative position vectors, as I know where the accelerometers were put.

A1 = ACG +A1/ACG = ACG + (alpha X r1/CG) - (omega^2*r1/CG)
A5 = ACG +A5/ACG = ACG + (alpha X r5/CG) - (omega^2*r5/CG)
A3 = ACG +A5/ACG = ACG + (alpha X r3/CG) - (omega^2*r3/CG)
here,
A1, A3, A5 = accelerations at point 1, 3, and 5.
ACG = acceleration at the frames center of gravity.
alpha = frames angular acceleration
omega = frames angular velocity
r1/CG, etc = position vector from point 1 to center gravity

Does this look correct to everyone? I appreciate your thoughts, Thanks!

 

[GregLocock]

The right hand result looks about right, but I haven't checked thoroughly. I don't like the middle column one little bit.

For 2d motion you need to measure X and Y vibrations at two points only, and if you choose the directions sensibly, one of those is redundant.

Again for 3 d motion you'll need 3 locations, but you won't need x y and z at every one, there is some redundancy there.



Cheers

Greg Locock

 

[EnglishMuffin]

I've just noticed this post, so I suppose it's either resolved or a dead issue. Provided one interprets the notation correctly (ie reading A1/ACG as meaning the vector acceleration of point 1 relative to the CG and not having anything to do with division), then in my opinion the equations are correct. One of my textbooks also uses this strange "slash" notation. The expression "alpha X r1/CG" presumably refers to the vector cross product of the angular acceleration and the displacemnt vector between the CG and point 1, which is a vector at right angles to r1/CG in the plane of the lamina, as it should be. So if this is what Mr. Locock means by the "middle column", I don't quite follow him. There are three unknowns : ACG, alpha, and omega, and consequently in general one needs three independent equations to find them. Of course, if one also knew the instantaneous angular velocity of the lamina explicitly, one would only need two accelerations. If instead of a two dimensional lamina one had a one dimensional object (ie a line), one would need accelerations at only two points. Textbooks on dynamics state that there are in general four unknowns for a plane accelerating body - two translational accelerations, an angular velocity, and an angular acceleration, but in the above equations the two translations are combined in one vector, giving just three unknowns.
So on this occasion, I disagree with Mr. Locock.

 

[EnglishMuffin]

Having thought about it a bit more - I was too hasty. You only need two locations to determine the motion, either for a line object or a plane lamina. If you resolve in two orthogonal directions, that gives four equations, for A1V, A1H, A2V and A2H, which you can then solve for the four unknowns ACGV, ACGH, alpha and omega. So basically I do agree with GregLocock. Nobody seems to be following this anyway !

 

[machineryguy]

Thanks for your help guys. Ahem....I am following this thread!
I have set up the data in a MathCad worksheet and using two sets of accelerations, I appear to be getting reasonable accelerations as functions of time. I am using these values to approximate wheel loading on my frame.
Coincidentally, I did an FFT on the data, and I am getting peaks at frequencies which correlate to those I calculated given a "rigid" frame, and tires having a given stiffness. (Had to go way back to advanced vibs days for that calculation). There is one more peak that is showing up in the frequency domain, but my suspicion is that it is the frame's first natural frequency. As soon as I have a chance, I will come up with an approximate number on this.
There is also seeming to be a consistent broadband response at the low frequency end (zero to 5'ish Hz). This I would presume is the response from repeated holes, bumps, edges, etc?
So, the data, from a big picture sense seems to be following what I would "expect", making me believe I am on the right track here. Thanks again!

 

[EnglishMuffin]

Many FFT analyzers produce meaningless responses up to about 5Hz - I'd be surprised if that data means anything. And reading very low accelerations with the average accelerometer can be a problem.
So its got "tires" huh ?
By the way - I didn't mean to imply you weren't following it - just that there seems to be a general lack of response - not even a follow up from Mr. Locock about his "middle column". Now that the subject of FFT's has come up he'll probably respond.

 

[GregLocock]

Oh, I was following it, just waiting for info. I understand the / means 'with respect to' in which case the middle column makes sense.

I'd say the FFT analyser is usually OK down to DC typically (after all if you switch high pass filtering off they make perfectly good if rather expensive voltmeters), the problem is that the rest of the instrumentation is not capable of reliably responding at low frequency - most piezo accelerometers chop out at about 3 Hz, and charge amps the same. It is possible to buy DC accelerometers - based on strain gauges. They are fragile, but a delight to calibrate, since you just turn them upside down to get a 2g offset.

These low frequency cutoffs are not especially well defined, although I think B&K gear still gives you good linearity between channels at 3 Hz, even using different accelerometrs and charge amps.

Typical road surfaces have a 1/f spectrum which probably explains the LF stuff.

The best way of getting RBMs from a wheeled vehicle using operational data is to run at high speed on a smooth road, and fix an out of balance weight to one wheel. Then do a coast down and synchronous sampling. Clean data!





Cheers

Greg Locock

 

[EnglishMuffin]

I had always assumed, apparently wrongly, that the meaningless data at up to, say, 3Hz was a limitation of the analyzers, having never used anything but piezoelectric accelerometers. So it disappears if you can set up a strain gage type? I presume the DC offset can be handled with a capacitor as usual. What produces the cut-off in the accelerometers? Is it related to built-in electronics or is it an intrinsic limitation of the piezoelectric design ? If you had a large enough acceleration amplitude at say 2 Hz, would you still have a problem ?

 

[GregLocock]

Switch your AC/DC coupling in the FFT analyser off, and get a torch battery. Play. Reason is that an A/D converter is essentially a DC device, with a half scale offset.

With a strain gauge you can see this down to zero - it is how we calibrate them.

I don't know why piezos roll off at low frequency, it could be the charge amps. I can't see an intrinsic reason.

The roll off still happens - we get a lot of signal at 1-2 hz as that is primary ride in a car, and the piezos are great at filtering it out!










Cheers

Greg Locock

 

[MikeyP]

Greg, the piezo roll-off an example of "the observer disturbing the observerd" and it is, as you say, to do with charge amps. Take the DC case. Squeeze the crystal (put a brick on it or something) and it produces a charge. In order to measure that charge we need to draw some of it to create a current. A short while later we take a second sample and find that there is less charge because we took some of it away with the first measurement ... repeat until no charge left ... Voila! a high pass filter. Charge amps have a very high input impedance but it is not infinite.

M