Safety question - forward tipping limits

[loulblades]

To get the 2 front tires on my GC2300 filled with foam, it was $130. I am not sure what it would be for the rear tires. A portion of the charge was setup.

If I recall correctly, if you use the weight box approach, you have the price of the weight box + about $1 a pound if you buy the weights that go with it. Of course once you have a box you could find stuff lying around to throw in there. Including, but not limited to, barbell weights.

 

[Bob_Skurka]

$130 for small fronts sounds about right. I think I recall someone saying they paid about $400 to fill their rear tires on a 30hp CUT. As for the weight box, most I see run about $175 to $225 and typically weigh about 100#. Sand, concrete and discarded scrap metal are common filling. I have 100# of steel sitting on top of concrete & broken bricks and mine is filled to about 3" below the rim. The 100# of steel are in the form of 2 wheel weights that I recessed partially into the concrete and I can remove from the weight box and attach to my front hubs. Just like RonJHall did with his weight box, I filled my BELOW the surface so it can also be used to tote tools and other things to work sites so it becomes a tool carrier & weight box.

I think we all use what works best for our own situations, there probably is no perfect answer.

But for Dave's question, he is using a CUT without any ballast at all, and that is pretty darn dangerous.

 

[jeepcj7]

On any model CUT 4wd tractor, with the loader installed and the bucket empty, 60% of the total weight should be on the rear axle and 40% on the front axle. On a 2wd tractor the ratio is 20% front;80% rear.

 

[scott_vt]

Bob,
I use a 35 gal drum filled with rocks and concrete when doing any real loader work. its just in the way if you need the PTO.Its about 700lbs.

scotty

 

[scott_vt]

I will try that download again.

 

[scott_vt]

Oh well, just go to page 3 on the Massey photo section and you can see what I did. Sorry about the multiple post.

scotty

 

[davesisk]

</font><font color="blue" class="small">( On any model CUT 4wd tractor, with the loader installed and the bucket empty, 60% of the total weight should be on the rear axle and 40% on the front axle. On a 2wd tractor the ratio is 20% front;80% rear. )</font>

So, can I calculate the max I'd need like this (it's 4wd)?

Rated full-height lift weight in FEL: ~600lbs
Rear ballast: 900 lbs!

Ignoring the weight of the tractor + FEL (which is presumably 40/60 weight distribution already and can theoretically be ignored), then I'd need 900 lbs on the rear to keep a 40/60 weight distribution with a fully loaded bucket. This would of course make the rear get about 80% of the weight with an empty bucket.

That seems like a lot of weight just to establish stability! /forums/images/graemlins/shocked.gif Should I be using the 1000 lb breakout force rather than the full-height lift weight? /forums/images/graemlins/confused.gif That would mean I'd need 1500 lbs on the rear, but the 3ph is only rated to lift 1200. Assuming this is a valid way to estimate ballast, which might be a big assumption. /forums/images/graemlins/tongue.gif

Thx!
Dave

 

[davesisk]

OK, I calculate this differently, but still come up with the same number as an estimate:

Tractor + FEL: total 2200, front 880, rear 1320, f/r ratio 40/60

+600 FEL: total 2800, front 1480, rear 1320, f/r ratio 53/47

+600 FEL + 900 ballast: total 3700, front 1480, rear 2220, f/r ratio 40/60

So, it still looks like, to meet the recommended weight distribution with a load in the FEL, 900 lbs would still be necessary. Ouch!

Of course, I guess this is just an estimate and it would really depend on where in the rear the weight is put. On the tires, more weight would be needed. On the 3ph (further away from the center), less weight would be needed.

I guess there's little or no concern about having more weight on the rear when the bucket's empty, right?

If anyone's interested, here's the equation I used to solve for the ballast weight x.

0.6(2800 + x) = 1320 + x

This means you want 60% of the total weight (known total weight plus the ballast weight you don't know) to be equal to the weight of the rear (known weight of the rear plus the unknown ballast weight).

Dave

 

[Bob_Skurka]

Dave, don't want to get into an arguement over any of this, however I look at the numbers posted by jeepcj7 as simply theoretical and I seriously doubt that all the tractors are made that way, although it might be the ideal.

Further, it doesn't explain the center of effort/center of gravity issues with balance (think of a sailboat, the COE is the single point on the sail where if all the winds force is applied, it would move forward as designed).

EXAMPLE 1 - On a tractor, you may very well have 60% of the weight on the rear axle, 40% on the front, but that is without the load and that doesn't take into account the actual location of the COE relative to the ground. I could have a 40/60 ratio and have the COE at 2" above axle height on BRAND X tractor and I could have the COE at 22" above axle height on BRAND Z but still have the proper theoretical weight ratio on each axle. In real world circumstances, the tractor with the COE lower to the ground will require less ballast by a large margin.

EXAMPLE 2 - Take a tractor with a fairly long overall length and a short wheelbase. Much of the front weight of that tractor is going to be in front of the front axle, even if it has the theoretical 40/60 weight ratio. Now when all is static, that tractor is in perfect balance. But when the bucket has a load and the load comes up off the ground, the weight ratio is going to dramatically tip forward on that machine compared to a machine of equal LOA but with a longer wheelbase.

EXAMPLE 3 - Now take yet another type of example. Two identical tractors, exact same model, same year, same everything . . . except the FEL on one unit is the high capacity FEL and the FEL on the other unit is the lower capacity unit (real life example, Kubota L3130 with a 723 loader versus a the 513 loader). The tractor is identical, so it has the identical COE and center of gravity and the same static weight distribution ratio . . . however the 723 loader is MUCH stronger than the 513 loader so using a "static" model, both would require the SAME ballast. Using a real world model, the tractor with 723 loader would require MORE ballast to be safe . . . and in my mind that is correct answer.


What I can tell you is that until you put some weight on the rear of that machine, I think you should not use the FEL except on FLAT ground and only with LIGHT loads and only if you don't carry the load very far off the ground. (and wear your seat belt too)

My smaller tractor has a 750# capacity (measured at bucket center) and requires a weight box PLUS wheel weights or fluid. If your tractor has a 600# capacity (measured at bucket center) then I would suggest you need something less, but still fairly close to what I need in terms of ballast. If you would like me to look up my manual when I get home, I can tell you exactly what is recommended for mine, you can downrate it by the % of capacity you have versus mine, and calculate something that would be safe. Just let me know if you want the info.

 

[jeepcj7]

Do not use the weight in the bucket in your equation. The weight distribution should be 60% rear, 40% front with the bucket empty (no weight in bucket period). You will never be able to add enough ballast to the tractor if you try to go 60:40 with a loaded bucket. Accetable weight is wheel weights, fluid or foam in the tire, something on the 3 point or any combination of the 3.