Pondering the center of gravity (CG) of my tractor

[ctgoldwing]

Now that I have installed 3" wheel spacers I feel more confident on some of the grades I have to cross. Not being a complete idiot (despite what my friends say) I really want to know how much of an improvement of the 'tip angle' it actually made.
My trig may be a little rusty but I still understand the sine of an angle My problem is I do not know how high the CG is. I have read threads that say 15" for a machine like mine - an L3800. In my googling I even found one that said 10" (a number I think is impossible). I'm even leery of 15". 20" may be closer to the mark. The assumption is its in the middle of the tractor (side to side). Otherwise your tip angle would be different for each side.

Using 15 or 20 inches as a guess I come up with about a 2-1/2 deg improvement on the tip angle. Of course variables like raising the FEL or even a rear implement can change everything.

My math may be way off but as a starter I really would like to know how high the CG is. Has anyone ever found this?

 

[ericm979]

With most modern CUTS the crankshaft is at the same height as the rear axle. Therefore the CG has to be at least as high as the axle. With the weight of the block and head, loader frame and arms (even with the bucket down), operators's station and operator, it should be significantly higher than the axle.

 

[buck12]

Now that I have installed 3" wheel spacers I feel more confident on some of the grades I have to cross. Not being a complete idiot (despite what my friends say) I really want to know how much of an improvement of the 'tip angle' it actually made.
My trig may be a little rusty but I still understand the sine of an angle My problem is I do not know how high the CG is. I have read threads that say 15" for a machine like mine - an L3800. In my googling I even found one that said 10" (a number I think is impossible). I'm even leery of 15". 20" may be closer to the mark. The assumption is its in the middle of the tractor (side to side). Otherwise your tip angle would be different for each side.

Using 15 or 20 inches as a guess I come up with about a 2-1/2 deg improvement on the tip angle. Of course variables like raising the FEL or even a rear implement can change everything.

My math may be way off but as a starter I really would like to know how high the CG is. Has anyone ever found this?

How much more stable do the spacers make it feel?

 

[Cat385B]

Are your rear tires loaded with liquid ballast?

 

[CobyRupert]

I’ve always guessed that a bare tractor’s CG is probably 6” above the axle, but that’s just a guess.

FEL’s, cabs, attachments and loaded tires can alter CG.

I’ve wondered how CG could actually be determined. I’ve never worked the math out, but there’s got to be a way of calculating it using 4 load cells, one under each tire and seeing how much the weight shifts as the “level-ness” changes.

 

[RSR]

Check out: https://journals.tubitak.gov.tr/agriculture/issues/tar-13-37-2/tar-37-2-10-1103-19.pdf

The take home message is that there is a lot more that goes into tractor stability than just the CG. Traction and velocity, as well as bumps, can make a big difference. As you mentioned, loader height and load will influence CG as well.

Nevertheless, you can get a feel for how widening the stance influences the static tipover angle. Even if you don't know the exact CG, you can make an educated guess (see comments above) and then play around with numbers. The article above quotes specific for a particular Mitsubishi tractor. Using their values, the static angle of tipover is 43.5 degrees. Widening the rear tires by 10% increases the angle to 46.3 degrees.

 

[neverfailyou]

I’ve always guessed that a bare tractor’s CG is probably 6” above the axle, but that’s just a guess.

FEL’s, cabs, attachments and loaded tires can alter CG.

I’ve wondered how CG could actually be determined. I’ve never worked the math out, but there’s got to be a way of calculating it using 4 load cells, one under each tire and seeing how much the weight shifts as the “level-ness” changes.

Remembering back to my engineering classes.... the theoretical vs practical CG is a complex calculation with point mass free body diagramming and assumptions clearing stated for a classroom solution to the exam question. This thread highlights many of the factors (and assumptions) effecting the practical calculation.
Yes -your guess is as good as mine-

 

[RSR]

You could always just measure the CG too. (Something I've always wanted to do with my tractor). Let the tractor have an outside-to-outside wheel width of W, center of mass/gravity height of H_tractor, and tractor mass M.

Park on a level surface next to a solid wall/tree. Chain your tractor laterally to said structure. Then, jack up the side of the tractor next to the structure until the tractor is "balanced" and right at the tipping point. Then, measure the angle, theta, at which it is leaning right at the tipping point.

The angle will be related to to the center of mass as:

tan(theta) = (W/2H_tractor)

Or, H_tractor = W/(2tan(theta)).

If you know the mass of the tractor, you can then directly calculate how the center of mass will be changed by adding weight at different heights to the tractor (e.g., operator, bucket load, filled tires) as

H_total = M_tractor(H_tractor/(H_tractor + H_addedmass)) + M_addedmass(H_addedmass/(H_tractor+H_addedmass)).


This does have some inherent assumptions (e.g., the tractor will tip about a fixed rigid point).

Also, keep in mind that movement DRASTICALLY change the safe angle (You really want to know the moment of inertia to determine that influence). So, don't assume that because the static tipping angle is 40 degrees, that you can drive on a 30 degree slope no problem.

 

[chim]

You could always just measure the CG too. (Something I've always wanted to do with my tractor). Let the tractor have an outside-to-outside wheel width of W, center of mass/gravity height of H_tractor, and tractor mass M.

Park on a level surface next to a solid wall/tree. Chain your tractor laterally to said structure. Then, jack up the side of the tractor next to the structure until the tractor is "balanced" and right at the tipping point. Then, measure the angle, theta, at which it is leaning right at the tipping point.

The angle will be related to to the center of mass as:

tan(theta) = (W/2H_tractor)

Or, H_tractor = W/(2tan(theta)).

If you know the mass of the tractor, you can then directly calculate how the center of mass will be changed by adding weight at different heights to the tractor (e.g., operator, bucket load, filled tires) as

H_total = M_tractor(H_tractor/(H_tractor + H_addedmass)) + M_addedmass(H_addedmass/(H_tractor+H_addedmass)).


This does have some inherent assumptions (e.g., the tractor will tip about a fixed rigid point).

Also, keep in mind that movement DRASTICALLY change the safe angle (You really want to know the moment of inertia to determine that influence). So, don't assume that because the static tipping angle is 40 degrees, that you can drive on a 30 degree slope no problem.

This is one of the most sensible methods I have seen. Saying that adding spacers makes a tractor feel more stable is a pretty unquantifiable piece of misinformation produced by someone standing behind his tractor and being impressed with the wider stance.

 

[WranglerX]

One might try to duplicate this test......

BX25D Tipover Challenge: Stock vs Wheel Weights vs Spacers - YouTube

It sort of sold me on spacers when I analyzed the numbers (just didn't do wheel weights though)....

Dale