"Distance from pin" math

[LD1]

Well then the issue isnt really needing to know what you can and cannot lift with various front attachments. Because the loader only has so much power.

The issue is how to make the tractor SAFE.....while using the loader to its fullest potential. Because IF the front end loader can lift something....no matter with forks or bucket.....you want it to be safe. So just how much weight you are "actually" lifting is irrelevant....if the loader will lift it, its within the capacity of the machine if properly ballasted.

Since I have the same tractor (just an older version).....I have TWO ballasts that both allow full use of the loader. Simply meaning the loader is no where close to strong enough to make the tractor feel un-stable with either of them on. And those two ballasts are
#1.....A 1250# rear blade (your rear blade....if it is insufficient....isnt heavy enough
#2....A 1620# rotary cutter


My rear blade has its weight balanced about 3' behind the pins. And the cutter is about 2' back. (If I take the loader off, I can do one heck of a wheelie with either)
Using my above math.....with the ball ends being 2.67' behind the rear axle....

The 1250# blade gives 7087 lb-ft of counter ballast effect about the rear axle
The 1610# cutter gives 8323 lf-ft

For equal effect.....keeping a ballast box centered 1' behind the pins would require somewhere around 2000#

I also have loaded tires and wheel weights. Though I can take the wheel weights off and feel just as stable with either of the above implements. And while fluid in the tires is ballast....and does help....it doesnt shift any weight off the front axle to give a counter balance effect.

Also worth noting that at one time I had a 1000# 6' cutter with its load center 3' back. The way it felt....I would consider that the bare minimum if you plan on working the loader to its fullest potential. That math says 5670 lb-ft.....or at a minimum 1500# centered 1' behind the pins.

I also have a 1200# 55gal drum filled with concrete ~1000#. It has a 2" square tube and it slips in the trailer mover (so the leading edge of the barrel is ~6" behind the 3ph pins). Which puts its weight center at 1.25' behind the pins. That yields 3920 lb-ft of ballast effect. And can definitively tell you it isnt enough for the MX.

Alot to digest.....but basically......a 55gal barrel of concrete if kept in tight is simply not enough.You either need more weight or weight further back.

Try to get to something in the 7000lb-ft range. Work the math backwards based on how you design a box (how far back it sits) and see how heavy you need to get it.

 

[ning]

[The following is a theorycrafting post which is nearly useless to the OP except to say "you're probably not going to be able to calculate this directly without a lot more information"]

It's tempting to say that "double the distance from the pins, halve the capacity" but this is far from the truth. You can see how this is by realizing that there's a lifting capacity right at the pins (zero distance) and adding an inch (which is an undefined multiple of zero distance) doesn't drop the capacity to zero at all.

Lifting capacity is a combination of multiple things: 1, can the hydraulics lift it; 2, can the axles structurally support the load; 3, can the loader structurally support the load. #2 will be affected by counterweight on the 3ph, which can help by reducing loads on the front axle (at the cost of considerable increase of load to the rear axle, though it's a lot tougher).

From a pure "will the tractor lift it at all ('i don't care it if breaks!')" standpoint, it's all about #1 - hydraulics. In this case, the capacity depends on the distance from the pivot point on the system (like where the fulcrum is on a lever arm), and also the geometry of the hydraulic cylinder attachments - at some points in the lifting arc, it takes a larger amount of cylinder travel to make the loader arm move, while at other points a small amount of cylinder travel will move the arm more.

Since cylinders are basically linear things but they're operating on a boom/arm that's moving in an arc, the force they subject the arm to is going to follow an equation similar to (cylinder force)*sin(angle) where (angle) is the changing angle of the arm vs cylinder - it's strongly dependent on that angle, and it's not a linear relationship (ie, the change in force of the cylinder from inch 1 to inch 2 isn't the same ratio as from inch 2 to inch 3 and isn't the same as from inch N to inch N+1; it's a curve like the edge of a circle or a parenthesis -> ).

We can disregard the angle measurements if we're not looking for a precise "how much" but instead of a relative "how much more/less" vs distance. In this case, the ratio should be a fairly simple ratio of the measures of the distances.

For the loader, you'd probably measure 1, distance from the arm pivot (ie near the steering wheel) to the lifting point where the cylinder attaches to the arm; 2, distance from the center of mass of the entire system (loader + bucket + load) to that same pivot point. Don't forget, the cylinder is lifting the loader, too. The loader is more or less a "third class lever".

For bucket tilt, you need to calculate 1, distance from where the cylinder attaches to the linkage behind the bucket to the fulcrum (pins); 2, distance from the pins to the center of mass of everything hanging off of the pins (ie the bucket/forks/grapple plus your cargo). The tilt is more of a "first class lever" that we're familiar with (see Archimedes moving the earth) and the distance question is easier to answer because we don't have the mass & distance of the loader system to consider; in this case, tilt is much closer to a "move it twice as far out and get half the capacity", though you still have to consider the mass of the bucket (and ssqa system, minus some small amount of counterbalance weight of the linkage between the hydraulic attachment point and the pins).

 

[ning]

When using pallet forks or other attachments where one end of the attachment is far from the other, what are the equations to calculate effective load? (Whether it's loader or 3ph)

For example, the rear of my rotary cutter is 8-9 feet from the 3PH, and the front edge of my new bucket is over 4 feet from the loader pin. Is there equation I can use (with simplifying generalizations where possible) to calculate how much load I can manage?

I have two primary cases where I'd like to know:

1) When using my pallet forks and/or new (very deep) bucket. Taking the simple case of a pallet, how many pounds can my LA1065 loader (~2300 lbs at pin) lift with a load uniformly distributed across the pallet?
2) I'd like to make a rear ballast that has at least the same ballast value as my 700 lb but very long rotary cutter but in a much more compact form so I can maneuver. Actually, I'd like to add quite a bit more ballast value than my rear attachments provide, which is why I'd make one - but for starters understanding some simplified math with the rotary-cutter-ballast-replacement scenarios is probably informative.

Given:

• (U) Lift Capacity (Bucket pivot pin, max height)
• Standard 2361 lbs (1071 kg)
• (V) Lift Capacity (500 mm forward, max height
• Standard 1784 lbs. (809 kg)

Here you can see the ratio for having your point mass (or center of mass if uniform) being half a metre in front of the pins. If you have a one-metre pallet with a load spread uniformly on it, you can roughly consider the max load to be 809kg minus forks weight... assuming your pallet starts at the pivoit center. Of course, it doesn't and instead starts six inches or more in front, so your capacity is reduced from that. Plus, your pallet is more than a metre long, which puts your uniform mass farther forwards, so it's further reduced, but we don't know how much without knowing the weight and weight distribution of the loader.

2) is a lot easier to do. #1 weigh the cutter. #2 find the balance point fore/aft. #3 carve out a smidge (#1's weight) of neutronium (effectively a point mass, a teaspoon of it would weigh 4 billion tons) from a passing neutron star and mount it at (#2's distance). You now have a much more compact weight which will act identically as a counterbalance without hitting as many things.

More seriously, though, if we're looking for an effective counterweight for the loader, this becomes a simple question, because the pivot point is easy to find - it's the center of the rear axle that we want the lever arm (of the entire tractor + counterbalance) to pivot around.
A: Find the distance from the center of the axle (center of rear wheels) to the center of mass of your cutter, and multiply that distance by the weight of the cutter.
B: Now measure distance from the center of the axle to, say, a foot behind the 3ph arms, which will be the center of your ballast box. Divide this distance into the number you got in part A, and you have the mass you need at that distance to be equivalent to the cutter. Add more mass there, and you have more counterweight.

 

[DieselBound]

One can LIFT stuff, but moving it is an entirely different matter! IF one is near max then one really doesn't want to be running around with that weight.

Not that it's a great practice, but one can always put weight on the cutter deck for added ballast. Likely OK in a pinch.

If you're pushing the limits and you need to do it often then perhaps more power is the solution. Trying to "just manage" is a recipe for an eventual "oh crap!" moment.

 

[Danno1]

2) ... carve out a smidge ... of neutronium (effectively a point mass, a teaspoon of it would weigh 4 billion tons) from a passing neutron star and mount it …

This is what I did to my tractor!!!



.

 

[Gee Ray]

Purchase pallet forks that fit on the 3 point hitch. 3 Point Forks | Northstar Attachments for example.

Purchase an IBC tote. A 275 gallon size will work. Add water until it works for you.

If the idea of a liquid moving in the tank is a problem fill the container with sand.

Easy on, easy off. Cheap. Plus you have forks for those very heavy pallets.

 

[Bullwinkle123]

Purchase pallet forks that fit on the 3 point hitch. 3 Point Forks | Northstar Attachments for example.

Purchase an IBC tote. A 275 gallon size will work. Add water until it works for you.

If the idea of a liquid moving in the tank is a problem fill the container with sand.

Easy on, easy off. Cheap. Plus you have forks for those very heavy pallets.

Interesting idea! Of course once i have the forks it might be easier just to make a box for rocks.

 

[Captain Dirty]

. . . Of course once i have the forks it might be easier just to make a box for rocks.