quote:

Originally posted by William_C._Foiles:
Good to see you on the board Hudson. I don't guess you've been to the Fudrukers in Dharan recently.

If you are still with the double cross company, you are a member; write and I shall give you a contact within the xx company if that is where you are.
--------
I am not a member, at least yet, as I have been reminded.

*No, I haven't been to Dhahran lately, but am thinking about going to Qatar soon. Maybe they have a Fuddrukers?

*Yes, I am still with the xx company, who maintains membership with ROMAC. I've contacted our member about the software.

Thanks,

Hudson

David,

Sorry, I didn't mean to ramble on... My point was never use only a computer program when you can do it graphically too.

Now that Hudson has the power, this should not be too difficult. Let's try a balance question.

Suppose we have the following balance data.


r0 = initial vibration
[ 7.1414284 @ 98.049467 end 1
7.1414284 @ 81.950533 end 2]

Add a trial weight to end 1
tw1 on end 1 = 7.1 @ 270 degrees

The resulting vibration r1=
[ 0.8584877 181.93131 at end 1
0.8584877 358.06869 at end 2]

Nice shot, but not perfect.

Remove tw1 and add tw2 to end 2
tw2 = [7.1 @ 180 degrees to end 2] - After all we had the 7.1 weight from the other end

r2=
1.1423664 181.45126 end 1
1.1423664 358.54874 end 2]



Questions:

1. What would be a good balance? What would you do to balance this?

2. What if we need to keep the balance weight less than 10? Can you do better than the first trial weight?

This message has been edited. Last edited by: William_C._Foiles,

Part #2 - how about 10 < 270.6 on end 1 and 2.95 < 92.2 on end 2. This gives a magnitude around 0.74 on both ends unless I calculated wrong.

TwoPlaneWork.xls (112 Kb, 31 downloads)

ElPete,

That's pretty good for a limit of 10 (try the angles of 270 and 90 - this is not a couple).

How about a limit on the weights of less than 4?

How much worse is the vibration?

A rough sketch of the original vib, trial runs, and solution limited to 10 is shown here.

We have to overshoot with weight 1 a little bit to get the benefit of weight 2.

For weight limit 4, we don't overshoot, the best we can do is 4 on end 1 at 270 and 0 on end 2. Gives 3.4. A lot worse.

I cleaned up the graph a little and posted here

Graph.ppt (25 Kb, 26 downloads)

Pete,

There is a much better solution with weight limit around 3.5, giving a residual of around 1 - not much worse.

You're right. I was trying to solve it in my mind using the graph plot but came to the wrong conclusion. I should have seen looking at the graphical plot that roughly equal weights at 270 would do a better job.

The computer tells me if my weight limit is 4, I should put 4 at end 1 and 3 at end 2 (both at 270 degrees) and I would get 0.98 on both.

By the way, for anyone interested, the excel solver tool/add-in provides all the tools needed to solve these, including constrains on the weights. But you have to give it a good initial guess or else it will give an answer which is a "local minimum" but maybe not the best answer within the global allowable bounds.

Also Bill published an interesting article on an approach using scilab programming

This was an interesting example with a lot of different twists... three different types of solutions depending on the constraints.

This message has been edited. Last edited by: electricpete,

Most of the time you can't put but so much weight, because of space, excessive force generated, or means of attachment.

If you go all the way up to 4 that's probably about optimal.

Interesting balance

Bwt =
! 3.5355339 270 !
! 3.5355339 180. !

Residual vibration

[1 @180
1 @ 0]

-----------------------
With the way you approach this, this might be interesting.

I saw this type of result on gas turbines and steam turbines occasionally. If you have a least squares program notice what happens when you balance this.

The vibration is measured on 4 bearings (say vertical). Only 1 balance plane is used.

r0=
[ 2 @ 0
2 @ 0
2 @ 0
2 @ 0]

tw= 2 @ 0 degrees

The response with the trial weight

r1=
[ 2.0880613 @16.699244 degrees
2.0880613 @16.699244
2.0880613 @-16.699244 or 343.30076 degrees
2.0880613 @16.699244]

Ok - bad shot, it happens to the best of us.

What does your balance program predict for the corrective balance weight and the residual vibration with the corrective balance? Did the vibration get better or worse?

My spreadsheet tells me to put 3.33 at angle 90 which gives vibration of 1,1,3,1. The worst vibration went from 2 to 3. I don't think the plant people would like that!

If I rewrite the objective function as a sum of fourth powers of the magnitudes, it tells me to put 1.21 at 90 which gives 1.64,1.64,2.36,1.64. Further increase in power of the vibration magnitudes for the objective function would presumably give smaller and smaller weights at angle 90 which gives closer magnitudes to 2.0. The only way to get a real improvement in all points would presumably be to find some more balancing planes.

SRSS_4brg.xls (82 Kb, 17 downloads)

Some of the least squares programs allows weighting of the measurements for a balance calculation. The original formulation allowed this. Weighting would be a method to deal with this type of problem.

If your program allows weighting, apply a heavier weighting to the one that is going bad. If the program doesn't allow weighting, you can fake it by always multiplying the vibration by a constant for each measurement plane.

Back to the Question by David_G ...

David,

If you are field balancing, stop and check the alignment. I have had misaligned couplings and misaligned hammermill bearing housings cause situations like you are describing.

Something about the misalignment throws off the instrument. I'm sure others could offer a more technical explanation of this.

Take Care,

Bill,

So, why do you think balance programs result in such a hugh correction weight, whereas graphical solution ( at least in my case) or the prediction feature does not? After all, the calculation approach is just a relatively simple and accurate math representation of the graphs. Is this phenomenon common for mostly couple unbalance?

Chris,

I agree, presence of misalignment will do it. The balancing program assumes that all vibration comes from unbalance and therefore will never balance out misalignment. In my case in order to verify misalignment I could not really check for phase relationship across the coupling (since vibration on the motor was very little). Checking phase data on the driven machine rarely makes any sense. And I could not perform alignment physically. So I just went ahead and did a graphical solution, and, unlike balance program, it worked fine.

David

This message has been edited. Last edited by: David_G,

Why do [some] balance programs come up with larege balance weights.

Not trivial problem. It has to do with the numerical properties of the balance problem. This in turn has to do with the physics and the available balance planes. Noise in an unbalancable direction will hurt. This has happened in practice since the first of the multiplane programs.

One example, try to balance a rigid rotor (theoretically rigid or practically rigid) in 3 planes. Well, it only needs 2 planes. That extra linear degree of freedom can create problems. I have a theoretical example on the rigid rotor (if I can find it). Do enough balances and these type of problems will occur.

The first time I had a least squares program calculate a balance that increased the vibration, similar to one of the example problems, blew my mind, but it is not too tough to figure out why it can happen, just matrix algebra.

Balancing requires 'engineering judgement' - no it doesn't always take an engineer, but some common sense will go a long way. I have seen people set machinery on the ground (i.e. knock it off the pedestals by just following the balance program - or at least the results of following the program) -- use sound judgement.

The two-plane solution is based on the following assumptions:
1 - vibration and weight magnitudes and phases/angles are measured accurately.
2 - the system itself is not changing over time (vibrations are constant)
3 - the system is linear
4- the system is far below any flexible-rotor resonance.
5. - did I leave any out?

One would assume if the 2-plane solution doesn't work, then one of those assumptions is bad. To keep the open mind, it could be any of them. But I'd like to focus on #1.

I have a theory that if you have a perfectly symmetric rotor, the 2-plane solution can be badly fooled by small errors in measurement (#1) while the static/couple is much more robust and not as easily fooLed by small errors in measurement. (I think this falls within the category of numerical properties mentioned by Bill)

I'll give a brief explanatinon. I hope to come back later with more detailed math explanation or an example.

The 2-plane model explaining how unbalance U causes vibration V is:
V1 = IC11*U1 + IC12*U2
V2 = IC21*U1 + IC22*U2

Where ICij is influence coefficient representing effect on vibration i of unbalance at point j.

The full two plane solution assumes the cross terms IC12 = IC21 are non-zero. i.e. the vibration and unbalance in plane 1 (V1,U1) are coupled to those in plane 2.

Now do a linear transformation to the static/couple problem
VS = (V1+V2)/2, VC = (V1-V2)/2
US = (U1+U2)/2, UC = (U1-U2)/2

The most general solution (full coupled solution) of the transformed system would be based on the following model:
VS = ICSS*US + ICSC*UC
VC = ICCS*US + ICCC*UC

But the simple (uncoupled) system we assume instead is based on:
VS = ICSS*US
VC = ICCC*UC
i.e. we assume ICCS=ICCS=0 meaning the static and couple unbalance/vibration problems are assumed to be decoupled from each other.

Mathematically it can be shown that the assumption of decoupling of static and couple unbalance are valid when the rotor is perfectly symmetrical, i.e. IC11 = IC22.


That is the case for many rotors (many rotors are very nearly symmetrical). Let's say it is the case for your machine. Then your machine is happy with the static/couple solution.

Let's also say there is a small error in phase and magnitude of one of the initial measurements.
Now let's say we try to solve with the a FULL (coupled) static/couple (different than what you do which is a simplified uncoupled static/couple solution). You would use a model:
VS = ICSS*US + ICSC*UC
VC = ICSC*US + ICCC*UC
Due to the measurement error, your program would very likely get some small but non-zero ICCS=ICSC, even we know the true value ICCS=ICSC=0 (static and couple are uncoupled due to symmetry).

Now do the 2-plane balance in the static couple coordinates with those incorrect coefficients. The solution may try to use that very small coefficient ICSC within the solution, but it would calculate a very big weight to cause any vibration effect with such a small coefficient.

The simple (uncoupled) static/couple solution that assumes ICCS=ICSC=0 would be better since it is not fooloed by the error of the non-zero ICCS=ICSC.

I think the 2-plane solution in our normal 1 / 2 coordinates will behave identically to the full (coupled) 2-plane solution in the static/couple coordinates, since it is simply a linear transformation. So the simple (uncoupled) static/couple solution outperforms this one as well.

There is an underlying principle that I am struggling to find the words for... something like: small additive errors on a quantity which is already very small or 0 (like KSC) tend to have more effect than small additive errors which are small in comparison to the quantity that they are added to (like KSS, KCC). I will try to see if I can get more precise math explanation of the whole thing or example numbers when I get the time.

At least that's what I came up with sitting here thinking about the math. Then again if it's true, I'm surprised we never would have heard before that the simple static/couple method can be more robust in the presence of measurement errors than the full 2-plane method if the rotor is almost symmetric. Maybe I am missing something (open to comments).

One thing that makes no sense to me. Lots of people have proven through field experience that a simple (uncoupled) static/couple solution works well for overhung rotors (with certain extra effort to remove static first and not remove couple until the phase looks like couple). According to my logic, overhung rotor would be the least suited to simple (uncoupled) static/couple solution. I can't find a way within my math framework to explain why that approach works for overhung rotors.

This message has been edited. Last edited by: electricpete,

ELPete,

Many of these discussions are above where I need to go, and many of the arguments I don't begin to follow.
But what you say about the static couple of overhungs, I have found to be true on some.
Most of the time, a single plane solution will correct any unbalance on both bearings way below an acceptable residual, and the job is done.
When a single plane does not lower both bearings, a normal two plane will bring them down, most of the time. When neither of those two work, a static-couple solution knocks it in the head. As with most contractors, I don't think I ever took the time to see just exactly why a static couple worked, when the other two didn't. I just make notes for when/if I return to that particular machine, and if I return and phase angles are similiar, I just use what worked previously, charge them a day and go home

Refering to the original opening of the thread, I have found that most generally when the handheld calls for a lot of weight, I have often used too much trial to start with, and shook the whole machine hard for the trial run, hence changing its dynamics. I go ahead and let it tell me a solution, pay attention to the angle, and smile at the weight, back up and use a smaller trial, placing it where the previous solution called for. By lessening my weight, and getting much closer to the angle, the soultion MOST OF THE TIME works out very well. I find this to occur most often on those machines right at a resonance or rotor critical state.
It may take a while, may not, but the next time I have to resort to static-couple, I will have much more data to put in this discussion for perusal, and possibly answer your (and my) dilemma, Why does it work?
Dave

Dear Dave,

IN CLOSING, I SAY TO YOU ALL CONTRIBUTORS THAT NO TRUER AND WISE WORDS WERE SAID...

I AM AN ARDENT ADVOCATE OF THE K.I.S.S. principle.

Cheers,

MarkoLeo

quote:

The two-plane solution is based on the following assumptions:
1 - vibration and weight magnitudes and phases/angles are measured accurately.
2 - the system itself is not changing over time (vibrations are constant)
3 - the system is linear
4- the system is far below any flexible-rotor resonance.
5. - did I leave any out?

You have an extra one. Resnance is not precluded.

1. weight is measured consistently. For instance, one can use inches (ok cm) of flat bar for the measurement if the flat bar is uniform thickness and width.

Static couple shots have a better chance of de-coupling the equations (going to single plane solution like ElPete showed) with a flexible system and some symmetry. In this case, the static and couple shots can act as modal balance weights.