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

[jenkinsph]

if it were me I would remove the log and hang a plumb bob and string at the same point on the forks. Then measure the distance from plumb bob to the front axle center. The angle of the forklift mast isn't important to solve the problem. To solve you would need the front axle weight, rear axle weight, and wheel base which you did not provide.

 

[xtn]

AxleHub,

If you have a 6' wheelbase, and you hang 2,000 pounds at 3' aft of your rear axle, you will reduce the weight at the front axle by 1,000 pounds. Period.

Sure, you might start with a bunch of weight in the bucket way out in front, and this will make the weight at the front axle a lot higher. But regardless, that counterweight will reduce it by 1,000 pounds.

So it might go from 1000 pounds (subcompact tractor with empty bucket) down to zero. Or it might go from 5000 pounds (bigger tractor with a big bucket full of stuff way out front) down to 4000. Whatever.

Any new weight * distance torque placed on one side of a lever results in an equal but opposite weight * distance torque on the other side of the lever. This is true regardless of how balanced or unbalanced the lever was to begin with.

Does that make sense?

 

[4570Man]

if it were me I would remove the log and hang a plumb bob and string at the same point on the forks. Then measure the distance from plumb bob to the front axle center. The angle of the forklift mast isn't important to solve the problem. To solve you would need the front axle weight, rear axle weight, and wheel base which you did not provide.

I don't have an actual weight of the load so we still can't get an accurate axel weight. Not that it matters. I was just fueling the fire.

 

[jenkinsph]

I don't have an actual weight of the load so we still can't get an accurate axel weight. Not that it matters. I was just fueling the fire.

I understand that but you did provide a number for the load of 1500 lbs, The axle weights would be in the loader specifications hopefully without a load on it.

 

[AxleHub]

AxleHub,

If you have a 6' wheelbase, and you hang 2,000 pounds at 3' aft of your rear axle, you will reduce the weight at the front axle by 1,000 pounds. Period.

Sure, you might start with a bunch of weight in the bucket way out in front, and this will make the weight at the front axle a lot higher. But regardless, that counterweight will reduce it by 1,000 pounds.

So it might go from 1000 pounds (subcompact tractor with empty bucket) down to zero. Or it might go from 5000 pounds (bigger tractor with a big bucket full of stuff way out front) down to 4000. Whatever.

Any new weight * distance torque placed on one side of a lever results in an equal but opposite weight * distance torque on the other side of the lever. This is true regardless of how balanced or unbalanced the lever was to begin with.

Does that make sense?

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

 

[LD1]

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.

Nope.

Adding weight behind the rear unloads the front axle. Not the load on the loader. The wheelbase is the correct calculation.

As he said, if wheelbase is 6' and load is 3' behind rear axle....its a 2:1. For every 2# added 3' behind rear axle takes 1# off front axle regardless of what the load is forward of the rear axle, how far forward it is, and how much it weighs.......provided the rear axle is on the ground that is.

That picture of the JD with teh forks stuck under the pallet......Adding 100# to the rear is only gonna add weight to the front axle. because until the rear tires touch the ground again to re-establish themselves as a fulcrum, there will be no weight transfered.

 

[jenkinsph]

Nope.

Adding weight behind the rear unloads the front axle. Not the load on the loader. The wheelbase is the correct calculation.

As he said, if wheelbase is 6' and load is 3' behind rear axle....its a 2:1. For every 2# added 3' behind rear axle takes 1# off front axle regardless of what the load is forward of the rear axle, how far forward it is, and how much it weighs.......provided the rear axle is on the ground that is.

That picture of the JD with teh forks stuck under the pallet......Adding 100# to the rear is only gonna add weight to the front axle. because until the rear tires touch the ground again to re-establish themselves as a fulcrum, there will be no weight transfered.

I agree^^^, and will add
the forces on the JD in the picture are equal front and rear about the front axle but are in opposite directions. And that is true for the entire arc where the rear wheels clear the ground to the highest point before it tips over.

 

[AxleHub]

Nope.

Adding weight behind the rear unloads the front axle. Not the load on the loader. The wheelbase is the correct calculation.

As he said, if wheelbase is 6' and load is 3' behind rear axle....its a 2:1. For every 2# added 3' behind rear axle takes 1# off front axle regardless of what the load is forward of the rear axle, how far forward it is, and how much it weighs.......provided the rear axle is on the ground that is.

That picture of the JD with teh forks stuck under the pallet......Adding 100# to the rear is only gonna add weight to the front axle. because until the rear tires touch the ground again to re-establish themselves as a fulcrum, there will be no weight transfered.

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

 

[LD1]

Ding Ding....Round 4......

I'll admit, I didnt go through the long thread that S219 posted with the formula. So have no idea what he was figuring.

As far as the axle is concerned....the load IS right over top of it. The axle doesnt "know" that the load is 3' here or 3' there or whatever.

Axle has XXXX pounds on its pivot pin pushing it straight down into the ground.

If wheelbase (that pivot pin) is 6' and you hang a load 3' off the back of the tractor........its 2:1. For every 2# you put on the back, tries to lift the front with a 1# force.

Take a long 2x4, and place OVER the rear axle and UNDER the front axle. Are you trying to tell me that just because there is a loader/load out past the front axle, that automatically changes the simple 2:1 lever into a 3:1 lever that you now have to exert more force (on the same lever) to see a different result.......

Would love to stay and debate this all day. But got firewood to cut and I havent figured out how to get the saws to run themselves yet.

 

[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