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.

 

[JStephen]

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.

Visualize what happens when you jack up a wheel on a car. You start applying force to that one wheel. So the load at that point changes, and load on other tires changes. But the CG doesn't change appreciably.

In other words, I think you're missing some information. If you can't assume that both front wheels are the same, or that both rear wheels are the same, I think you'd need to know spring stiffness, etc., to calculate loadings at each wheel.

 

[ivymike]

part of the problem would seem to be that the weight supported at each wheel will depend on suspension deflection at that wheel. While you should still be able to find the cg location in x and y based on the weight measured at each wheel, you won't necessarily be able to determine the weight at each wheel by working in the opposite direction. You can find (LF+RF) and (LR+RR) and (LF+LR) and (RF+RR) from the cg location and mass, but finding LF by itself requires more information or an assumption of some kind.

Note that the measured weights you indicated above show that the RF+LR wheels are carrying 65% of the vehicle weight. If the vehicle is "tiptoeing" on the scales, then maybe that makes sense...

If your vehicle was a rectangular lump of rock supported on four rigid points, then the cg location and mass you listed above should result in the following weight distribution:
front-right 11694
front-left 10186
rear-right 10806
rear-left 9414

 

[IRstuff]

I thing that given the data you have, you can't get there from here. The reason is that given the CG and weight, you only have 3 knowns, but 4 unknowns.



TTFN

 

[MintJulep]

You have four unknowns and only two equations so far.

Third equation is W = LF+RF+LR+RR

I can't figure a fourth independant equation. I believe the problem to be statically indeterminate for more than three wheels.

 

[Timelord]

The weight on each wheel is proportional to its distance from the CG. This gives you the other equations. I don't think the suspension deflection makes much difference until it significantly alters the distances to the CG.

Timelord

 

[ivymike]

timelord, check the measured figures against your theory.

Alternatively, picture a customized car with a set of hydraulic actuators that allow it to pick up one wheel from the pavement and hold it in the air ("three wheel motion" - extreme case shown here

http://onebadpup.150m.com/images1/high4.jpg

). Have the distances to the CG (projected to the ground plane) been significantly altered? Would you expect the airborne wheel to carry a load in any way proportional (inverse or otherwise) to its distance from the CG?

 

[ivymike]

sorry, that link didn't work. See the photo here:

http://www.onebadpup.150m.com/page12.html

 

[cowski]

I'm going to make a couple simplifying assumptions:
1) there is no suspension (none was mentioned in the original post)
2) the 4 wheels rest perfectly on the same plane (ie if the cg was in the exact middle of the wheels, all 4 wheels would be loaded equally)

We have 4 unknowns (as has already been stated). To get our 4 equations why not sum moments about 4 different lines?
To make life easier, I would suggest summing moments about the "front axle" (line between front wheels), the "rear axle" (line between the rear wheels), a line from the left front to right rear, and the line from the right front to the left rear.

I haven't done the math, so I don't know if it works out; but that is where I would start.

 

[ivymike]

I think what you'll end up with is the weight distribution I listed above.

I should note that my description above was incorrect - if it was a rigid body on four rigid supports, then one of the supports could be removed without affecting the balance (the one in the quadrant opposite the cg).

 

[MintJulep]

It is not sufficient to have four equations. You need four INDEPENDANT equations.

Summing moments around different planes (such as a diagonal) does not provide independant equations.

As noted by several, it is possible for any one of the four legs to carry no load at all. Thus there are an infinate number of permutations of load distribution.

 

[cowski]

If they are not independent equations you can at least sum moments about 1 line to get another independent equation to use with the others listed above.

 

[borjame]

You're forgetting moment balanace equations for the two car planes. Suspension doesn't matter.


(LR+RR)*X1 = (LF+RF)*X2
X1+X2 = forward to rear wheel dim
X1 = dist from rear to CG
X2 = dist from front to CG
Two eqns, two unknowns: pretty easy

Then do again for side to side.

 

[JStephen]

Borjame, what you find when you do that is total load for the front, or total load for each side...but not load PER WHEEL.

If you assume no suspension, then you still have rubber tires and still have the same problem. (Some trucks, truck trailers, front end loaders, etc., are built this way).

In real life, for a car, you figure the car is pretty much symmetrical, and then it's easy enough to find the loads, but the example loads given above are very unsymmetrical.

You might also consider what happens if you bend the frame out of plane...load distribution changes, but CG doesn't, distance to wheels doesn't.

 

[ivymike]

I don't think anyone is forgetting those. They happen to be insufficient to solve the posted problem, unless I'm missing something. I'm surprised that you didn't post a solution, if it is in fact as easy as you suggest... (I see that you did suggest a method for calculating (LF+RF),(LR+RR), (LF+LR), and (RF+RR), now how about finishing off with a solution that splits out each wheel separately?)

Going forward in the thread, would it be reasonable to request that posters test their suggestions prior to posting?

 

[Dennyd]

After reading ivymike's last post I rolled up this thread and notice that somarp has only started this post and not commented since. Though I appreciate the generally good content on these forums I'd like us to maintain a level of professionalism in our posts - both the initial ones and the replies. To me that means thinking through your request making sure it is: a) appropriate, b) clearly written, and c) sufficiently supported with data/problem definition (how many times are requests made for more problem definition?). Similarly, that replies are well thought out and directed to the point of the post or recent reply, rather than to a tangent (start a new post if necessary).

A star for you, ivymike!

- - -Dennyd
(Only related to Dennis Miller by this short rant!)

 

[somarp]

Thanks for everyone's replies - I appreciate it very much.

Before my initial post I had derived the same results as ivymike, but when my numbers did not match my starting numbers I made the assumtion that I was making an error somewhere in my calculations.

Just to give everyone a little more detail about my design -these weights are from a rubber tracked unit that we manufacture, so there is no suspension that must be accounted for. The way we generated those numbers was by lifting the unit and then placing digital load cells under the idler centerline and also the drive sprocket centerline.

For now I am planning on staying with my numbers as what ivymike has suggested. I do appreciate everyone's comments and thoughts.

 

[ivymike]

note that I later raised a question as to what "real" conditions would actually result in the numbers I calculated (my description was incorrect).

 

[meintsi]

You don't have enough information to solve this problem.

You cannot assume rigid supports on a single plane because the value's given (if correct) are physically impossible. (Simple 3D CAD model with gravity applied will provr this!)



In order for the values stated above to be correct, the cg must be located somewhere very close to the line between the front right tire and left rear. (slightly forward and to the right of the solid body center) The kicker is that either the left-front, or right-rear tire is offset low.

The other variables that need to be acounted for in this case are tire pressure (or at least some sort of material rigidity at the support point), and the difference in elevation between the LF and RR.

Four point support in a single plane is not possible with the values stated above.



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

 

[ivymike]

Four point support in a single plane is not possible with the values stated above
why not?