Weight Distribution on Four Wheels 4

[somarp]

I need some help - I am working the CG work for a new machine. I know the estimated weight of the unit and now I am trying to determine based on the CG location weights on each wheel.

I have done the reverse in the past by using weights from each wheel and calculating the CG location, but for some reason going in reverse is causing me problems.

Here is the example I did in the past - even with this one I can not work it in reverse to calculate the weights on each wheel.

Machine weight = 42100

Left Front Tire (LF)= 6980
Right Front Tire (RF)= 14900
Left Rear Tire (LR)= 12620
Right Rear Tire (RR)= 7600

Wheel Base = 72.65
Tire to Tire Dimension = 91.22

To find "X" location of CG I did:
(7600 + 14900) * 72.65 / 42100 = 38.8"

To find "Y" location of CG I did:
(6980 + 14900) * 91.22 / 42100 = 47.4"

Any advice or comments on calculating this backwards would be greatly appreciated.

 

[GregLocock]

borjame, people in glass houses...

quote from first post "I have done the reverse in the past by using weights from each wheel and calculating the CG location,"

SO you have succesfully solved the problem he has already succesfully solved. Thank you for etiquette advice.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[NormPeterson]

Actually, I'm a little suspicious of the measured weights, even though the CG position calculations are unaffected. Since the ratio of the RF to RR measured loads is not the same as the ratio of the LF to LR loads, there is some "weight-jacking" going on that's shifting the corner weights around a bit. About 802 lbs per corner with the RF and LR high and the other two low, according to an automotive corner-weight spreadsheet.

When everything has a high degree of rigidity, it takes very little in the way of deflection or gap to change the results by a considerable amount. In assuming infinite machine rigidity, we're assuming in addition to the infinitely large numerical rate that it's uniform as well. Your machine may not be 'uniformly rigid' with respect to all four corners. Tires have a finite amount of compression under load, even high-durometer solid ones, and can be out of round or have developed a flat spot. The floor matters, too. If one wheel is over a high or low spot (or a soft spot, for that matter), the measured weights will be affected. That would hold true for the load-measuring condition, and for this condition the dimensional uniformity of the measuring equipment itself from pad to pad becomes a potential variable. Set your test rig up elsewhere (even in the same room, or even in the same place but facing a different wall) and I would not expect that you'd duplicate your posted wheel load data. Not that they'd be entirely valid elsewhere anyway.

Suppose the effect of all those conditions was taken as a spring rate of, say, 40,100 lb/in at each corner. Is the flatness of your floor (plus the measuring equipment thickness, tire deformation effects, etc.) within +/-0.020"? Or within +/-0.010" if the effective 'spring rate' is 80,200 lb/in? And so on. I'm inclined to doubt it. Is the stiffness of the floor itself well beyond those numbers at all of the machine support/load measurement points, say, by at least an order of magnitude?

As has been noted previously, this is a statically indeterminate problem, so you have to approach it from the standpoint of deformations. But in this instance, you can't ignore them because they happen to be of small absolute magnitude.

When you roll this thing across a typical floor, the wheel loads are likely to vary wildly. Perhaps you need to determine the greatest load that each wheel could be reasonably expected to carry instead.

Norm