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

[LD1]

Still not enough info. What's the wheelbase? And what's the tilt of the mast? Basically need to know how far forward of the front axle the load is centered.

11.8k lift with a 1500# load brings the total to 13.3k.

Assuming a 60" wheelbase, and the load centered (due to tilted mast) at 24".....

That's a 2:5 ratio. So 1500# on the forks at 24" should take 600# off the rear.

So, add the 600# transfer to the 150$ load and that is how much more weight the front sees over the unloaded figure. And the rear...600# less.

 

[xtn]

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.

Sure. But we're talking about the one relating to how much load can be removed from the front axle by hanging ballast on the 3-point hitch. In this case, the rear axle is the pivot point. If you hang 2000 pound at a distance of 3' behind that pivot point, you can measure a reduction of 1000 pounds at a distance of 6' forward of that pivot point. Plain and simple. Having a full bucket 9' forward of that pivot point, or whatever, does not change that simple math.

 

[xtn]

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.

Weight and balance calculations are simple... IF you can separate the total mass of the object into known smaller masses.

You pick a "station" base point to use consistently for all inputs. Let's say you pick the very rear of the fork lift. The rear axle has 4500 pounds on it, at a known distance forward of your station base point. The front axle has 7300 pounds on it at a known distance forward of your station base point. That covers the unloaded fork lift.

All we need is how much the load weighs, and how far forward from your station base point its center of gravity is.

 

[Xfaxman]

Do I have too much weight behind the rear axle?

 

[RaydaKub]

Sure. But we're talking about the one relating to how much load can be removed from the front axle by hanging ballast on the 3-point hitch. In this case, the rear axle is the pivot point. If you hang 2000 pound at a distance of 3' behind that pivot point, you can measure a reduction of 1000 pounds at a distance of 6' forward of that pivot point. Plain and simple. Having a full bucket 9' forward of that pivot point, or whatever, does not change that simple math.

Exactly my point, again, I'm attempting to help explain in yet another way that might help others understand. You are only addressing the second lever while others were trying to add the first lever to the calculation. That's why I pointed out that there were 2 different levers AND described where they were and the effects each were responsible for.
I'm on your side. Honest! ☺

 

[ovrszd]

Do I have too much weight behind the rear axle?

Since a staged pic, no.

 

[SPYDERLK]

Weight and balance calculations are simple... IF you can separate the total mass of the object into known smaller masses.

You pick a "station" base point to use consistently for all inputs. Let's say you pick the very rear of the fork lift. The rear axle has 4500 pounds on it, at a known distance forward of your station base point. The front axle has 7300 pounds on it at a known distance forward of your station base point. That covers the unloaded fork lift.

All we need is how much the load weighs, and how far forward from your station base point its center of gravity is.

we dont know the wheelbase

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.

unloaded axel weight. Front 7300, rear 4500

 

[LD1]

we dont know the wheelbase

Or the angle of the mast to determine how far forward the load actually is. And how far forward the mast is mounted from the front axle when verticle. So the 30" out on the forks is a meaningless number as well

 

[jenkinsph]

Or the angle of the mast to determine how far forward the load actually is. And how far forward the mast is mounted from the front axle when verticle. So the 30" out on the forks is a meaningless number as well

That is why I suggested using a plumb bob and measure to the center of the axle. Knowing the angle without the length(height) or the distance from the front of the mast I figured it was easy to use the plumb blob and an tape measure. Of course none of this is supplied or given. But this would be an easy way to do it in the field.

 

[TomSeller]

Oh god, kill me now. Oh wait, I can unsubscribe.