Determining tractor center of gravity.

[tcartwri]

I think if you read your manual you will see the manufacturer has already done this for you. There should be a suggested weight and offset for your balast box to counteract the max lifting force of the fel. From that you should be able to calculate the CofG.

 

[284 International]

Like tcartwri said, why not just put 85% or so of the rated capacity of your lift on the hitch? It doesn't matter where the CG is, you're just interested in minimizing weight on the front axle. Hang as much on the back end as your capacity will allow, and that gives you the minimum weight on the front axle possible.

The Nebraska tractor test data usually (at least in the ones for my machines) has CG information, including height from the ground, as well as fore/aft position from datum.

 

[PapaPerk]

I think if you read your manual you will see the manufacturer has already done this for you. There should be a suggested weight and offset for your balast box to counteract the max lifting force of the fel. From that you should be able to calculate the CofG.

Hmmm I'll have to look at the manual again.

I may still do my experiment... it's just for the fun of it! :thumbsup:

 

[PapaPerk]

Like tcartwri said, why not just put 85% or so of the rated capacity of your lift on the hitch? It doesn't matter where the CG is, you're just interested in minimizing weight on the front axle. Hang as much on the back end as your capacity will allow, and that gives you the minimum weight on the front axle possible.

The Nebraska tractor test data usually (at least in the ones for my machines) has CG information, including height from the ground, as well as fore/aft position from datum.

Yeah I could do that. But there's not point in overloading the rear end either. I plan to purchase suitcase weights at equipment auctions... and I need some idea of how much weight I need to buy.

I could also just load the tires with Rimguard and be done with it. But... when have I ever taken the easy route?! :laughing:

 

[284 International]

Loading the rear tires won't reduce the apparent load on the front axle. It will help in traction, which may keep the fronts from doing all the work, but the front axle will still see the same load.

I don't think you'll be able to overload the rear axle hanging a weight the 3 point can lift, especially with a loader up front.

What kind of tractor is it?

 

[Snaker]

Couple of problems with plan A.

Assuming your tractor has a pivot on the front axle, weighing one front wheel won't be accurate. You would have to include both I believe. That would be total weight for the front end, needing to be compared to total weight of the rear.

CG is a indication of a 3 dimensional position (2 if you skip side to side). You would be looking for a point along the length of the tractor and also the height. You can't find that by using the weight at the wheels.

What your trying to do would be called weight bais I think. Once you have accurate weights you would have to calculate to find the balance point.

Aother idea would be to lay two planks on a pipe, drive on the planks until the tractor is near balanced, and plumbob from the pipe to the tractor.

For adjusting for ballast though I guess you would actually be looking for a ratio of front end weight to total weight, thinking of the rear axle being the pivot point.

 

[PapaPerk]

Loading the rear tires won't reduce the apparent load on the front axle. It will help in traction, which may keep the fronts from doing all the work, but the front axle will still see the same load.

I don't think you'll be able to overload the rear axle hanging a weight the 3 point can lift, especially with a loader up front.

What kind of tractor is it?

Yes you are right. Loading the rear tires will not reduce weight on front axle.

When I said overloading... I meant there's no point in me putting more weight on the tractor than needed. :confused2:

The tractor is a Kubota L3830 with LA723 loader... quick attach bracket... MDS forks and Kubota bucket. I also have wheel weights. All of this hardware is why I can't just reference a manual... I want true weight. :thumbsup:

 

[PapaPerk]

Couple of problems with plan A.

Assuming your tractor has a pivot on the front axle, weighing one front wheel won't be accurate. You would have to include both I believe. That would be total weight for the front end, needing to be compared to total weight of the rear.

CG is a indication of a 3 dimensional position (2 if you skip side to side). You would be looking for a point along the length of the tractor and also the height. You can't find that by using the weight at the wheels.

What your trying to do would be called weight bais I think. Once you have accurate weights you would have to calculate to find the balance point.

Aother idea would be to lay two planks on a pipe, drive on the planks until the tractor is near balanced, and plumbob from the pipe to the tractor.

For adjusting for ballast though I guess you would actually be looking for a ratio of front end weight to total weight, thinking of the rear axle being the pivot point.

I'm only measuring CG in the L (x) axis. It's safe to assume tractor is balanced evenly from side to side. And I'm not concern with CG height, H (z) axis.

As long as tractor is level when being weighed it doesn't matter if tractor has a front pivoting axle.

Yes if I had heavy enough planks I could balance the tractor to find CG. They do this at antique tractor shows for fun. However I still would not know weight on each axle.

 

[JoeL4330]

I am a little slow on these things; you are trying to determine the fore-aft CG to estimate how much weight you have to add TO THE REAR to counter-balance the LOADER to make your 3PT HITCH SKIDDER work efficiently??

Incidentally, if you had turfs (or maybe industrial tires...not R-1s) you could just measure the area of the contact patch in sq. in. and multiply by the psi in that tire to get a pretty good approximation ...old LEO trick for estimating overweights ..which brings to mind the fact that they have portable scales per wheel that would do what you want ...maybe you can prevail on a friendly trooper (or a trucker) to borrow a set...too spendy to buy

 

[dugman]

One can get a ballpark idea of where the center of mass (CM) is located between the rear and front axles using the CUTS calculator (for John Deere tractors). This does NOT yield any vertical or lateral CM info. Plugging my 4720 cabbed tractor data with 400X loader, MX6 cutter and loaded R4 rear tires I get a total tractor weight of 7917 lbs with 37% on the front axle and 63% on the rear. This means that 2929 lbs on the front axle and 4988 lbs on the rear. Looking up (or measuring) the wheelbase I get 128.2". The following formula can be used to calculate the CM in front of the rear axle:

Xr = location of CM in inches in front of the rear axle
L = wheelbase (128.2 in my case) CORRECTION 71.5 in
Wt = Total Weight of tractor (7917 lbs in my case)
Wf = Weight on front axle (2929 lbs in my case)

Xr = (L * Wf) / Wt

so in my case:

Xr = (128.2 * 2929) / 7917 CORRECTION: Xr = (71.5 * 2929)/7917

Xr = 47.4 inches in front of rear axle. CORRECTION: Xr = 26.5 in front of rear axle (thanks for pointing out my lookup error jenkinsph!)

It is reasonable to assume that the lateral CM is approx. the center of the tractor (unless the lunch bag is REALLY big beside the operator!). Keep in mind the center of mass DOES NOT CHANGE with the tractor's orientation or inertial frame of reference. Again this doesn't tell us anything about the vertical location of the CM.