converting g-mm unbalance to mils visa-versa

[micjk]

Is there an equation for converting g per mm to mils pk-pk unbalance. We have a shop balance machine that we have had to have special tooling fabricated for to balance spindles and rolls for our winders and godet cabinets. The problem is that I don't think the setup that we are using is reliable as far as results comparing the original design of the balance machine versus utilizing our special tooling on same. The geometry of the machine has changed considerably. We have stiffened one padastal and actually added @ 80 lbs of mass to the other. I am, in the mean time contacting the manufacturer (CEMB) to get there input. For now what I am doing is using my field balancing equipment ( manufactured by CSI) to perform the task. My analyzer gives results in mils pk-pk and I would like to convert this to g-mm. Any assistance would be deeply appreciated. Bear in mind that this is a 2 plane configuration. I have an equation for single plane but it tells me due to cross effect of balance in each plane, it may or may not be reliable for 2 plane. I can solve for each plane individually but would like to combine the 2 results into a single "unbalance result" for the entire roll, spindle,etc.
Hope this makes sense.
Regards,
MICJK

 

[Ralph2]

Not that I have ever seen. You are trying to compare apples with oranges. The gm/mm is the measurement of a force. The mills pk to pk is the result of that force. But is influenced by so many variables that no one conversion would hold true.
Can you not test your balance machine for to check that it gives the correct response. A balance machine is every bit like a electronic bathroom scale. It can be checked the same way. On your bathroom scale if you add a precise weight and the scale does not reflect that it is in error. Same with your balance machine. Start with a "semi" balanced rotor. Add a known weight at a known radius. If the resulting unbalance shown is not what you have added within 5% your balance machine needs to be re calibrated. (correction 1 plus or minus correction 2 = your test piece)
To do a more comprehensive test one does this process every 30 degrees, plots out the results. Somewhat complicated and is spelled out in most ot the API specs on balancing. API 611 gives some graphs and templates in the appendix (third edition 1988)
Good luck
Ralph

 

[electricpete]

There have been a number of discussions on a similar topic at reliability-magazine.com. One is here:

http://www.reliability-magazine.com/ubb2000/ubb/Forum2/HTML/000801.html

I'm certainly not familiar with all of the conventions involved in unbalance world. Here is my way of fitting it into a physics frame of reference. There is a theoretical bounding conversion which can be obtained from physics.
(1) M = rotor mass
(2) e = eccentricity
(3) m = unbalance mass
(4) r = distance of unbalance
(5) r*m = unbalance
(6) w = radian speed = 2*pi*f
(7) M*e=m*r
(8) e = m*r/M is the radius that the mass will move at
(9) The associated centrifugal force has a familiar form F=M*V^2/R where R=e and V=2*Pi*r=2*Pi*e
(10) F = M(2*Pi*f*e)^2/e =M (2*Pi*f)^2*e
(11) acceleration a = F/M = (2*Pi*f)^2*e
(12) velocity v = a/(2*Pi*f) = (2*Pi*f)*e
(13) This is the conversion from eccentricity to [maximum theoretical] velocity.
G2.5 limit is 0.008 millimeter at 50hz. V=2*Pi*50(sec^-1)*0.008 = 2.5 mm/sec

This would be the conversion between imbalance and velocity for a rotor spinning about it's center of mass with no spring action. (picture is spinning in space). Since no spring it is by definition above resonance. Actual vibration level in real-world scenario (with spring) for the same unbalance would be lower.

You mentioned g per mm. Doesn't mean anything to me. If you are interested maybe you can explain that part to me. (no guarantees that I have any answers you are looking for).

 

[GregLocock]

<electricpete> That is NOT an upper bound for the vibration seen, since it ignores resonances of the rotor/pedestal assembly. For instance, our driveshaft weighs around 8 kg, yet the 'effective' mass is about 2 kg, for balancing purposes. It could be an upper bound for the bearing motion, ie relative between shaft and pedestal, but I wouldn't know for sure (and don't think it is in practice). Of course if the shaft itself resonates then all bets are off.

<micjk> I think you probably meant to type g mm, not g per mm. You need to interrogate your machine to find the influence coefficents, of which you will have 3 (well 4, but the two crosscoupling ic s should be identical). This will give you a figure in terms of vibration per unit imbalance. Converting from vibration back to mm pk-pk can be done in the usual way. You do have a problem, if you want to get back to a single figure, since you can't really reduce a 2 plane machine to a single number. Cheers

Greg Locock

 

[electricpete]

You're right greg it is not bounding in that sense. It does has some significance in defining the system characteristics when far above resonance. I'll have to think how to put that better.