You need balast or you will trash your front axle!!!! really?

[LD1]

Read through S219's post again.

You have a mis-understanding of what he was saying I think.

His formula is to find the amount of rear weight needed to offset FEL load.

So yea.....if FEL is 3' out, wheel base 6', and 3PH implement hanging 3' back.............

The 3:1 you imply IS NOT axle load. Rather it takes 3# on the 3PH to offset every 1# in the FEL as to NOT add any weight to the front axle.

In other words..........

Using those same dimensions........ you have a 1000# fel load. That would add 1500# to the front axle. It would take 3000# off the rear to take 1500# off the front axle.

So...2:1 ratio for Front axle unloading
3:1 ratio for offsetting loader load

 

[xtn]

Weight will be reduced on the front axle. Yes
But "wheelbase is NOT" the measurement to calculate the front side of the lever length. The front side of the lever length is rear axle point to front of load.

If rear axle to 3 pt arm end (how weight is attached) is 3 feet and wheelbase is 6 feet. You are saying it is 1 to 2 ratio or 2000 # effects 1000 #. But I'm saying the added distance to the front load (3 ft. From front axle to load) makes it a 1 to 3 ratio which means only 666 # of reduction on the front axle.

Essentially wheelbase is needed to calculate the ground pressure impact but rear axle to front load is needed to calculate what the impact ratio is. The fulcrum is the rear axle. I'm calculating straight line horizontal impact. Then if the front load gets elevated from a pure horizontal . . . Then you alter the impact in a more complex calculation . . Which we are not discussing in this example.

JMHO

The wheelbase IS the measurement to use if what you want to calculate is how much the effect is at the specific point where the front wheel is. Yes, the TOTAL lever length includes the bucket's position, but we aren't calculating how much easier we can tip the tractor back by grabbing it at the bucket. We're calculating how much easier we can tip the tractor back if we grab it by the front axle.

So as a PERCENTAGE, it wouldn't change as much if you've got a big load out front. Like in my example, on a subcompact you could actually reach more than 100% change and tip the tractor back...or with a larger, loaded tractor it may only represent a smaller percentage change.

But as an absolute number, regardless of the overall weight and even regardless of the distribution of that weight...if you hang 2000-lbs * 3-ft on one side of a lever...you're going to have an equal an opposite reaction of 2000-lbs * 3-ft on the other side of the lever. And that 2000-lbs * 3-ft equals 1000-lbs *6-ft, or 666.6-lbs * 9-ft, or whatever. That's how torque measurements work. Since we want to know the effect "AT THE FRONT WHEELS," the effect measured there (assuming the 6-ft wheelbase) is the 1000-lbs * 6-ft amount.

 

[gladehound]

Greetings LD1,

I never claimed that rear weight addition unloads the loader. It unloads the front axle. But what you are saying is that not only mysrlf . . but s219's calculation formula is incorrect also . . and I'm guessing s219 is pretty good at physics and geometry

I'll leave others to present their viewpoints.

Here is mine: glade and you both present a concept that works if the load was directly over the front axle . . and if that were the case s219's formula and my viewpoint would also agree.

However the load is 3 feet further ahead of the front axle . . . And both s219's formula and my viewpoint indicate that leverage is enhanced on the front side because of it and needs to be factored into the efficiency of the counterbalance. I'm not here to argue. In fact apparentlybto entertain because Glade has said numerous times he laughs often from this thread. Debate of viewpoints is just that . . debate. My understanding is that physics has to resolve inconsistencies to be valuable and accurate. How do you logically resolve the impact of leverage of 3 feet forward of front axle?

We agree if load is on the front axle directly . . I resolve the leverage impact of the load 3 feet further forward in the same manner that the 3 feet behind rear axle is resolved . . A factor of leverage position. That is the same resolution that s219's formula handles it.

We all agree that ground pressure is reduced on the front axle. You and glade say its 1000 lbs of change baded on using wheelbase length . . Myself and s219's formula show it woukd be 666 lbs of change on that front axle based on load length/positioning of the lever.

I'm certainly open to being incorrect . . I just don't see how s219 is incorrect either. Show or explain to us how load 3 feet ahead of front axle has no impact change to a tractor than having it directly over the front axle.

I'm always open to learn.

Thanks

JMHO

S219's formulas look right and are entirely consistent with what LD-1, myself and others have been saying throughout this thread. It looks like you might have misapplied some information from S219's post.

 

[RaydaKub]

The wheelbase IS the measurement to use if what you want to calculate is how much the effect is at the specific point where the front wheel is. Yes, the TOTAL lever length includes the bucket's position, but we aren't calculating how much easier we can tip the tractor back by grabbing it at the bucket. We're calculating how much easier we can tip the tractor back if we grab it by the front axle.

So as a PERCENTAGE, it wouldn't change as much if you've got a big load out front. Like in my example, on a subcompact you could actually reach more than 100% change and tip the tractor back...or with a larger, loaded tractor it may only represent a smaller percentage change.

But as an absolute number, regardless of the overall weight and even regardless of the distribution of that weight...if you hang 2000-lbs * 3-ft on one side of a lever...you're going to have an equal an opposite reaction of 2000-lbs * 3-ft on the other side of the lever. And that 2000-lbs * 3-ft equals 1000-lbs *6-ft, or 666.6-lbs * 9-ft, or whatever. That's how torque measurements work. Since we want to know the effect "AT THE FRONT WHEELS," the effect measured there (assuming the 6-ft wheelbase) is the 1000-lbs * 6-ft amount.

You actually have 2 levers in play here. The first is the FEL to the rear axle, with the fulcrum being the front axle. More load equals more lift on the rear axle, regardless of any counterweight. The ground under the rear tires doesn't know where the pressure comes from, it just knows there is some weight and it gets to be less weight when there is a bigger load in the bucket.
The second lever is from counterweight to front axle with the rear axle as fulcrum. Again, the ground under the front tires doesn't know where the weight came from. It just knows that with more counterweight, the pressure on the front tires got to be less.
The 2 levers operate somewhat independently of each other but are constrained by total weight of the tractor. If both levers "win" you just broke your tractor in half.
Its kind of like 2 oxen in a yoke. You need both of them sharing the load to operate efficiently.

 

[SPYDERLK]

The tractor is small enough that it is unlikely to lift that load even enough to drag it off. Rather the forks would just slide out and leave the lumber on the trailer.

But a bundled and banded load of wood, and significant curl back to scoot the load off the trailer without the forks slipping out, it could have started with the load tilted back pretty far.

Clamp on forks dknt have much curl power, but ssqa or pkn on forks, and loader up high, curl would have no trouble curling a load far too heavy to lift with the loader.

And/or in a panic, when the tractor started to nose dive, grabbing the joystick to "hang on" and pulling towards you on the way down do end result in that situation pictures as well.

But most of the time....you're right, the forks would dump and crash the load or pallet

Yeah, but the easiest and most credible way to be like that is to drive up straight and square to a heavy pallet sitting on the ground - insert forks and lift. Pretty safe too if the pallet is waaaay too much for the tractor.

 

[SPYDERLK]

Bear with my ignorance.

Then the physics equation changes as the height increases because the horizontal distance is changed?

If so, then the vertical height does in fact change everything??

The equation would include the angle function; therefore the calculation value would change. - Not the equation.

 

[SPYDERLK]

Anyone care to calculate the load on this front axle? Don't forget to account for the mast being tilted behind the axel. The forklift weighs 11,800 pounds. Let's assume the load weighs 1500 pounds and is held 30 inches from the back of the fork. View attachment 459686

Not enuf info to solve.

 

[kevin37b]

You just need a few pallets on the 3 pt . Everything should be good afterwards .

 

[4570Man]

Not enuf info to solve.

unloaded axel weight. Front 7300, rear 4500

 

[SPYDERLK]

unloaded axel weight. Front 7300, rear 4500

One would have to derive an equation to calculate the front borne weight as a function of the wheelbase "X". Possible, but Im too rusty on the math.