Determining tractor center of gravity.

[PapaPerk]

Hello,

I'm starting my 3 point log skidder attachment project.

Included on the attachment will be a bracket for suitcase weights for loader counterbalance. In an effort to determine max counter weight needed, I need to determine the tractors current center of gravity.

To determine center of gravity I need to measure tractor's front and rear axle weight. To do this cheaply I'm thinking about using a bath scale coupled with a long moment arm. I will have to weigh each wheel... or at least one front and rear wheel. Anyone ever done such a thing? :anyone:

I know the bath scale, moment arm technique works well for measuring trailer tongue weight.

 

[zzvyb6]

Just remember to keep the wheels on the axle being measured at exactly the same height (especially the rears), otherwise load transfer will affect the results (although you just need F/A CG height for this only..... Does your manual contain axle weights (but loaded tires affect this greatly).

 

[melanelly]

What th... Bathroom scale? what size tractor you got. One of those JD peddle jobs?:laughing:

 

[PapaPerk]

What th... Bathroom scale? what size tractor you got. One of those JD peddle jobs?:laughing:

6000 lbs tractor. By using a lever arm u reduce load applied to the scale to around 250. Simple mechanics.

 

[Chain Bender]

Do like I did today. I was pushing some tops and logs trying to clean up behind a bunch of slob loggers that clear cut my neighbors 52 acres. I was looking at my next push instead of where I wsa backing and ran squar up on a 20" diameter stump that was about 1/2 taller than my tractor (B3200). Even the FEL wouldn't lift enough to push the back timres to the ground. wound up taking a chaing saw and cutting a 5-6" diameter pole about5' long. urned the bucked about 3/4 way down and captured the lople and lifted hard and rolled the bucked just enough to slide me off the stump. Thank goodness, cause it was gonna be a long walk home to get a chain and a com-a-long.

Chain Bender

 

[hizoot]

Could someone explain how to set up the moment arm for this? I understand about fulcrums/leverage...but cant envision how to do this without mounting the bathroom scales on the ceiling


thanks!

 

[284 International]

Set a piece of pipe on a hard, flat surface some distance away from the scale. Put a mark on a heavy beam where it rests upon the pipe. Run the beam to the scale, where another short piece of pipe is resting on the scale. Mark that distance and measure the distance between them, and zero the scale.

Now you have a repeatable, known distance with fairly discrete reference marks. Then, jack up the tractor and set up the apparatus underneath, using the same reference marks, and then set the tractor onto the beam, much closer to the pipe on the ground than the scale (Say, by a factor of 10 or 20 times). It would be more accurate to run a 2x4 or something perpendicular to the beam, to keep the tire up, and give a more precise location for the weight of the tractor.

Read the weight on the scale, and multiply by whatever the multiple of distance is between the tractor and the pipe on the ground and the tractor to the scale. So, if the tractor is 1 foot away from the pipe on the ground, and 10 feet away from the scale, and the scale reads 100 lbs, you multiply 100 lbs by 10, and learn that the tractor weighs 1000 lbs on that tire.

I don't understand why the OP needs to know the CG, but it sounds like something interesting to look at his posted photographs when he does it!

 

[Gordon Gould]

Why not just put a block in the center of the scale. Place a sturdy beam that will hold your tractor like a 6X6 on the block with the other end on the floor so you have a ramp. Make a dummy for the other wheel so you go up level. Drive up the ramp until you get a good weight at about 3/4 capacity of the scale. Measure lever arms ( beam lenghts, floor end to scale and floor end to tractor) Wt = (beam lenth/tractor length)X(scale reading)

 

[PapaPerk]

Set a piece of pipe on a hard, flat surface some distance away from the scale. Put a mark on a heavy beam where it rests upon the pipe. Run the beam to the scale, where another short piece of pipe is resting on the scale. Mark that distance and measure the distance between them, and zero the scale.

Now you have a repeatable, known distance with fairly discrete reference marks. Then, jack up the tractor and set up the apparatus underneath, using the same reference marks, and then set the tractor onto the beam, much closer to the pipe on the ground than the scale (Say, by a factor of 10 or 20 times). It would be more accurate to run a 2x4 or something perpendicular to the beam, to keep the tire up, and give a more precise location for the weight of the tractor.

Read the weight on the scale, and multiply by whatever the multiple of distance is between the tractor and the pipe on the ground and the tractor to the scale. So, if the tractor is 1 foot away from the pipe on the ground, and 10 feet away from the scale, and the scale reads 100 lbs, you multiply 100 lbs by 10, and learn that the tractor weighs 1000 lbs on that tire.

I don't understand why the OP needs to know the CG, but it sounds like something interesting to look at his posted photographs when he does it!

Exactly! That's how u do it. I just have to make sure to keep the tractor level on all four corners while doing it! Lol!

The reason I want the CG is to calculate how much ballast I need on the back of the tractor. I've made an excel spreadsheet that calculates tire axle loading based on loader loads and rear ballast load. I'm trying to reduce stress on front axle when picking up heavy logs.

I'll be sure to take photos and share when I do the weight measurements.

 

[PapaPerk]

Why not just put a block in the center of the scale. Place a sturdy beam that will hold your tractor like a 6X6 on the block with the other end on the floor so you have a ramp. Make a dummy for the other wheel so you go up level. Drive up the ramp until you get a good weight at about 3/4 capacity of the scale. Measure lever arms ( beam lenghts, floor end to scale and floor end to tractor) Wt = (beam lenth/tractor length)X(scale reading)

This is a good idea with some minor changes. The lever arm has to be level and the rear of the tractor also has to be level. But I think this is how I'll do it. Thanks for the input!

 

[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.