electricpete
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Registered:1443797040 Posts: 651
Posted 1515428038
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#1
Here's a coupling I'm looking at: Bendix type diaphram coupling 7600 rpm Total Coupling weight (hubs plus spacer) - 23 pounds Imbalance tolerance in each plane is 4*W/N = 4*23/7600 ~ 0.01 inch-ounce ~ 0.3 inch-gram The coupling includes 12 nuts and 12 bolts in each balance plane, at a radius of approx 3" from rotational axis. The vendor states each nut is controlled to within 0.05 grams and each bolt is controlled to within 0.05 grams. The vendor instructions state that the position of the nuts and bolts does not need to be controlled. The vendor will will assemble the coupling at their facility and perform balance, then they will disassemble and ship to us with the nuts and bolts not labeled for which position they belong to. We don't plan to (don't want to) perform any balance check on site. That is their plan. Their plan does not seem reasonable to me since the fastener tolerance might easily exceed the balance tolerance. I haven't had a chance to talk to them yet, but I'd like to be prepared when I get my opportunity to talk to them. The 0.3 inch gram balance tolerance when considered at a 3" radius translates to 0.1g. If they happen to balance with a +0.05g nut and +0.05g bolt at 12:00 position and -0.05g bolt and -0.05g nut at 6:00 position, and then we reassemble with these four pieces swapped, we have change in balance of 2*0.05 at each of 4 positions at a 3" radius or 1.2 g-inch... blows away our balance tolerance of 0.3g-inch. I realize this exact scenario is very unlikely, but it's unsettling that by just changing position of 4 out of 24 fasteners potentially exceeds our imbalance tolerance. I'd like to be able to evaluate or understand it better. MY QUESTIONS ARE AS FOLLOWS: Q1 - Does above approach proposed by vendor seem reasonable or unreasonable? Q2 - If you were going to attempt to establish a balance tolerance for the nuts and bolts in this situation, how would you do it? Here's my initial rough attempt at item Q2 If we assumed N pieces with normal weight distribution and a standard deviation of sigma = S and , then the standard deviation of the scalar sum of the weights of the N pieces by central limit theorem is N*S/sqrt(N) = sqrt(N)*S. In our case N=24 and the deviation of the scalar sum of weights is [sqrt(24)*S] ~ 5*S. So far we are dealing with scalar sum and I don't know how to translate it to vector sum, but it seems reasonable (conservative?) to estimate the standard deviation of the vector sum (of unbalances) as the [standard deviation of the scalar sum of weights] times the radius. [sqrt(N) * S] * R <= Balance Tolerance [sqrt(24) * S] * 3*inch <= 0.3 gram*inch S <= 0.3/(3*sqrt24) ~ 0.1/5 = 0.02 gram Accordingly I will propose that the vendor either control the tolerance of each nut and each bolt to 0.02gram or else label the position of each nut and bolt. Does this make sense? Is there a better way to do statistical vector sum to estimate the effects of individual piece variability on overall unbalance?

Danny Harvey
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Registered:1442582546 Posts: 1,403
Posted 1515437792
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Pete, I can't comment on your math other than to say that if YOU did it I trust that it is correct. That said, I don't think that marking and tagging the bolts is an unreasonable request. They could probably do it in less time than it took you to prove that they need to. ;-)

John from PA
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Registered:1458487711 Posts: 963
Posted 1515439528
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From API 671 Special-purpose couplings for refinery service, 4th edition 2007, reaffirmed 2010 . There is also an ISO version ISO 10441:2007 Special Purpose Couplings for Petroleum, Chemical and Gas Industry Services.

Noknroll
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Posted 1515454898
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+1 on what Danny said regarding math (way over my head) and tagging the individual nut and bolt weights to allow you to distribute evenly around the coupling on assembly. Or is that what you said mathematically?

Curran919
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Registered:1474360900 Posts: 599
Posted 1515489834
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Pete, The math seems to click with my hindbrain from uni. The only assumption I would contest is the comparison of a controlled threshold to the standard deviation. Remember that only 68% of samples are within the bounds of one standard deviation. Therefore, you cannot equate this to a 0.02 gram control threshold. If anything, 2.5S, or 0.05g would be about 99%, and even a 1% failure of a controlled threshold deviation seems pretty high. Of course, this only makes sense for a normal distribution. If they have a fixed manufacturing process where they accomplish the 0.05g by default, this may hold. However, if they are test-weighing their bolts, rejecting outliers or removing material post-test, then all the math will be bunk because you won't have a normal distribution, as the kurtosis will be much lower. For your peace of mind, I would get a balance scale (we keep one with our balance demo rig to weigh out balance masses). Cheap pocket scales can measure 1/100g accurately enough. Then you can get a much better idea of what the actual population standard deviation is.

Curran919
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Posted 1515490834
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Quote:

Originally Posted by electricpete So far we are dealing with scalar sum and I don't know how to translate it to vector sum, but it seems reasonable (conservative?) to estimate the standard deviation of the vector sum (of unbalances) as the [standard deviation of the scalar sum of weights] times the radius. [sqrt(N) * S] * R <= Balance Tolerance [sqrt(24) * S] * 3*inch <= 0.3 gram*inch S <= 0.3/(3*sqrt24) ~ 0.1/5 = 0.02 gram

Okay, I gave it some more thought, I think you underestimated how conservative using the scalar sum instead of the vector sum is. There is probably a highly technical way of proofing the vector sum CLT, but my better approximation, assuming a 'worst case' scenario of #2-6 being max and #8-12 at min, would be: 24 => 4 + 8*cos(30) + 8*cos(60) ~ 15 S <= 0.3/(3*sqrt15) ~ 0.026 gram just... a bit different.

Curran919
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Posted 1515495140
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Originally Posted by John from PA From API 671 Special-purpose couplings for refinery service, 4th edition 2007, reaffirmed 2010 . There is also an ISO version ISO 10441:2007 Special Purpose Couplings for Petroleum, Chemical and Gas Industry Services.

So Annex E actually has the exact bloody formula you want... for bolts: 0.1g * (6" / pi) * sqrt(12/2) = 0.47 g-in to add in the nuts, you just take the RSS, which would be 0.66 g-in total. Therefore still twice your balance tolerance, so I think you need them to label the bloody bolts.

JuddJones
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Registered:1442594106 Posts: 384
Posted 1515511884
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I love this forum! You guys sure know how to humble a dumb millwright who stumbled into this job. Sorry for the disruption, please continue the genius. I will continue to observe like sitting mid court at a tennis match.

Noknroll
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Posted 1515533168
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Big J, I'm impressed too, just dont know what foriegn language they speak, probably from the planet mathematicious in the formulaen galaxy