×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# Weight Distribution on Four Wheels4

## Weight Distribution on Four Wheels

(OP)
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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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

### RE: Weight Distribution on Four Wheels

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

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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

### RE: Weight Distribution on Four Wheels

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?

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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).

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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?

### RE: Weight Distribution on Four Wheels

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!)

### RE: Weight Distribution on Four Wheels

(OP)
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.

### RE: Weight Distribution on Four Wheels

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

### RE: Weight Distribution on Four Wheels

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...

### RE: Weight Distribution on Four Wheels

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

### RE: Weight Distribution on Four Wheels

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...

### RE: Weight Distribution on Four Wheels

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?

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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

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

Remember...

### RE: Weight Distribution on Four Wheels

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...

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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

### RE: Weight Distribution on Four Wheels

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...

### RE: Weight Distribution on Four Wheels

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)

### RE: Weight Distribution on Four Wheels

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...

### RE: Weight Distribution on Four Wheels

ivymike,

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

### RE: Weight Distribution on Four Wheels

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

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

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.

### RE: Weight Distribution on Four Wheels

2
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

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!