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.

 

[meintsi]

Step back and look at the distribution pattern in your own calculations and reason it out.

If there is no suspension, or other unknown factors at work, how can the highest two weights be at opposite corners? A physical impossibility for a single plane with gravity being the only force.

Remember...
"If you don't use your head,
your going to have to use your feet."

 

[ivymike]

how about if you balanced the object perfectly on two opposite corners, and then brought third and fourth coplanar points of contact gently into place without letting them carry a significant load?

how about if the object had a cg that was slightly off of the line between the two opposite corners (as in this case), so that it couldn't be supported by only two simple supports, and you supported it on three points (at three corners)? Would you not expect that the "opposite" corners would carry most of the load, while the additional corner carried a small fraction of the total? Could you not then add a fourth support that barely changed anything?

If I really did such a thing... would the universe end abruptly?

 

[zekeman]

The problem is fairly straight forward.
Take moments about each of the 4 lines of centers between the tires I get:
FL+FR=k1M
RR+RL=(1-k1)M
FL+RL=k2M
FR+RR=(1-k2)M
where
FL, RL front left and rear left
etc
M=weight of unit
k1,k2=distr factor ratio of distances to overall
Writing in det form I get
0 0 1 1 FL (1-k1)M
1 1 0 0 x FR = k1M
1 0 1 0 RL k2M
0 1 0 1 RR (1-k2)M
Since the main determinent is not zero , a solution can be obtained.

 

[meintsi]

right - and then would all four points lie on the same plane?

Zekeman - go ahead and solve it then....

Remember...
[navy]"If you don't use your head,[/navy]
[navy]your going to have to use your feet."[/navy]

 

[meintsi]

and is 16% really a value for a 4th support that barely changes anything?

remember, you're dealing with a very simple problem using gravity....

SOMARP - how did you arive at these values per tire using two loadcells located on two centerlines????

centerlines of what??? and where on the centerlines ???

Remember...
[navy]"If you don't use your head,[/navy]
[navy]your going to have to use your feet."[/navy]

 

[ivymike]

right - and then would all four points lie on the same plane?
There is no fundamental reason that they couldn't. The simple fact is that the system is overconstrained with four coplanar points for support, and any combination of forces that satisfy the force- and moment-balance equations represent a physically valid solution.

 

[ivymike]

um, overconstrained is not the right word... I'm feeling a bit inarticulate.

 

[meintsi]

Your statement is correct but,
how does any combination of forces that satisfy the force-moment-balance equations = gravity in this case?

As the problem is stated, there is NO planar solution.

Remember...
[navy]"If you don't use your head,[/navy]
[navy]your going to have to use your feet."[/navy]

 

[rzrbk]

I agree ,as ivymike and others have suggested, that you should be able to calculate the weight distribution given a center of gravity, wheelbase dimensions, and wheight at all four wheels. But this wouuld be an ideal weight distribution. With the data given, this weight distribution is apparently being thrown off.

Whether by one tire being lower, an unlevel floor, flat tire, or other; at least one tire is being caused to take more or less than its share of wheight. In this case a suspension, or a less rigid design would probably cause the measured weight distribution to be closer to the ideal.

Although I'm not sure how to prove it, I suspect that the center of gravity calculated using the measured weights will still be correct due to the opposite corner also being affected creating a sort of self correction. (as long as cart is somewhat level)

 

[meintsi]

Great idea about the internal self-correction. I totally overlooked the possibility.

In order for these values to be correct on a single plane, there MUST be another, unaccounted for, force.



Remember...
[navy]"If you don't use your head,[/navy]
[navy]your going to have to use your feet."[/navy]

 

[MintJulep]

ivymike,

The correct term, as I noted in my first post in this thread, is "statically indeterminate".

 

[ivymike]

ah yes.. obviously... that's what I get for going online whilst fighting the flying monkey flu...

 

[MintJulep]

I can't recall if the cure for flying monkey flu is dousing yourself with a bucket of water or dropping a house on yourself.

 

[ivymike]

any combination of forces that satisfy the force-moment-balance equations = gravity in this case
um, draw a free body diagram and figure it out yourself. start by drawing one big down-arrow at the cg, and four little up arrows at the corners... there are literally an infinite number of combinations of forces applied at the corners that can react the force of gravity at the cg. All valid solutions will give static equilibrium - there will be no net moment acting about any point on the body, and the down-forces will sum to the same quantity as the up-forces.

 

[JStephen]

Y'all go back to my first response and reconsider what it means:

"Suppose the thing weighs 1,000 lbs, with center of gravity exactly at the center. You could have right front tire @ 500 lbs, left rear tire @ 500 lbs, and other two at zero. Or, all 4 at 250. Or some combinatin in between."

This is two different load cases. They give exactly the same CG. They sum to the same moments about the front axle, rear axle, right wheel line, or left wheel line. So given the weight and the CG, you don't have a clue which load case this might be. Furthermore, you can take any combination of these two that totals the same weights and get additional load cases that do the same thing:
Right front: 350 lbs
Left front: 150 lbs
Right rear: 150 lbs
Left rear: 350 lbs.

This is a simplified version because it just so happens that the CG is at the geometrical center. But the same thing happens if it is located off-center; you just don't have enough information to figure the wheel reactions.

 

[zekeman]

Correction to my post
I incorrectly stated that the determinant was not zero when, in fact it is, which means that not all the equations are independent. This is made obvious by adding the first 2 equations to get the same result as adding the last 2; that result is that the summation of all the forces is equal to the load. As stated by others, the physical reason is that only 3 point forces are necessary to sustain the load which is borne out mathematically in that only 3 independent equations are possible.

 

[unclesyd]

Food for thought.

http://www.auto-ware.com/software/ccw/ccw.htm

 

[aviat]

This is a link to one way race car teams find CG. The machine must be leveled first, and weighted, and then raised at one end and reweighted.

http://www.racerpartswholesale.com/longtech6.htm

 

[borjame]

OK, it seems a large number of people forgot what SOMARP was looking for: the CG. Suspension et al is irrelevant to the problem b/c he measured each wheel independently. The supension forces would come into play if you had the total weight and CG and wanted to find the load at the tires. Again, he already has the load would like the CG.

USING MY ORIGINAL EQNS:
Going from left side to right:
19600X1 - 22500X2 = 0
X1 = 1.148X2
X1+X2 = 91.22
X2=42.468
X1=48.75

Going from front to back:
21880Y1 - 20220Y2 = 0
Y1 = .924 Y2
Y1+Y2=72.65
Y2=37.75
Y1=34.89

The CG is located 3.142 units to the right and 1.425 forward of the geometric center. Please double check my math. I don't normally include the actual math b/c that's your job as an engineer, not mine, I chime in just to help point someone in the right direction. In my humble opinion, those that disparage other people's input without adding anything aren't helping. That would be an example of not using your head.

 

[ivymike]

As I recall, he has the CG location and was looking for the weights at each wheel:
I am trying to determine based on the CG location weights on each wheel.