Bucket lifting capacity (math)

[apease]

Prompted by the thread "Pallet Forks for BX or B-Series", I wanted to figure out how much my BX24 FEL can lift with forks. I've had to make some assumptions on figures since I'm not near my tractor. If anyone can make this more precise, I'd be grateful.
For a start, to keep the math manageable, I assume that the FEL is at level lifting height. I'm also ignoring any change in power the hydraulic cylinder may have from minimum to maximum extension.
I'm guessing on the distance from the FEL pivot point to the cylinder attachment, from the diagrams on 5-6 of the operators manual, since that's not given. It looks to be the same as distance "F", which is 54.7". The "bucket bottom mid point" also isn't precisely specified - is it the midpoint of the 500mm bucket width, or the mid-point of the actual bucket bottom, which might be 100mm farther from the FEL pivot? I made two guesses at 63.8" and 65.8" from FEL pivot.
I use the basic lever arm equation of

F = W*X/L
or
payload capacity = lifting force * payload to pivot distance / lifting force to pivot distance

(sorry for all the dots in the diagram and table below - the TBN editor seems to remove spaces and tabs so I'm not sure how else to show this)
rough diagram:
..........-X-...........F
.........................|
o-------------------v
.........^
...-L-...|
.........W

................................................................F=WxX/L.....W = FxL/X
lever arm (L).................................54.00in...750.00lbs.....1620.00lbs*
arm at mid-bucket (L) (guess #1).....63.84......529.00.........1350.91*
arm at mid-bucket (L) (guess #2).....65.81......529.00.........1392.56*
arm at bucket edge (L)...................73.69.......518.00.........1526.75*
arm at 4' pallet midpoint (L)............97.69.......383.89*

Numbers with an '*' have been calculated. I get some widely disparate estimates of cylinder upward force (W). I'd love to know why. For the calculation of F I just assumed 1500lbs for W.

All comments welcome.

 

[dennisgw]

Not sure where you got a value of F=1500lbs. Here are some bucket lift capacities for BX and B that I found on a TBN post. You are not going to be any higher than these. You would need to subtract the weight of the forks from the lift capacity and if the center of gravity of the pallet load is out from the bucket edge then you REALLY have to lower what you think you can lift. I was trying to unload my BX42 chipper from a trailer with my Bxpanded Forks. If the lifting strap was a foot out from the bucket edge, I could not life (chipper is somewhere around 400 lbs). I could only lift it if the sling slid down the fork arms to the edge of the bucket. I would love to lift like a fork lift but I think you will not exceed what your current lift capacity is and more than likely less.

BX 2200 LA 211 - 460 LB
B 7400 LA 272 - 495 LB
B 7500 LA 302 - 660 LB
B 2410 LA 352 - 770 LB
B 2710 LA 402 - 882 LB
B 2910 LA 402 - 900 LB
B 21 TL 421 - 926 LB

 

[LD1]

For the nature of trying to keep things simple, your 1500lb estimate/average which will result in a 389lb capacity is fairly close give or take a few lbs.

No need to try to hash out and get exact figures because the pressure will vary a little form tractor to tractor and the MFG's numbers are just calculations as well.

But as far as the varying #'s, it could be a cew things. First, when measuring from the cylinder attachment point on the boom to the bucket pivot, were you measuring in a straight line, or were you following the curve of the boom.

If you were measuring in a straight line, that # is correct.

But what isn't accouted for in the other measurments is the bucket curl. No matter what, the measurment from the pivot pin to the cylinder point CANNOT change. BUT, the bucket mid point and bucket edge CAN change depending on where in the curl cycle the measurments were taken. The bucket edge is going to be closest to the cylinder point when curled all the way back. But as you dump, it will move farther away. Thus changing the length and the figures all together.

Another think you should account for is the bucket curl force. The curl moves in a different arc than the lift. So usually at the bucket edge, your curl cylinders will have more force than the lift cylinders. But with an attachent on like forks, there comes a point somewhere out in front that the curl cylinders will have LESS force than the lift cylinders.

The reason fo this is that 4' in front of the bucket as opposed to the edge is a larger percentage of change from the curl cylinders than it is from the lift cylinder.

But the reason I bring this up is, pallet forks can cause a great deal of pressure and possible rupture hoses (especially on smaller tractors) in the curl curcuit. The curl cylinders may only be capable of "curling" 150lbs at the tip of the forks. BUT if the lift cylinders are ABLE to lift 300lbs, with that kind of load on the tips, you are in a sense pressurizing the curl curcit to double the PSI the tractor pump is putting out. And there is NO pressure relief for this cause the valve is blocking the flow. So if you have a heavy load in the tips of the forks, and a little bouncing, it is very easy to damage the curl cylinders or hoses. (Ask me how I know)

That is why pallet forks that actuall replace the bucket are better IMO if you are doing a lot of pallet moving. Getting rid of the bucket keeps the load much closer to the pivots. Which results in greater capacity and less chance for a blown hose.

 

[LD1]

Not sure where you got a value of F=1500lbs. Here are some bucket lift capacities for BX and B that I found on a TBN post. You are not going to be any higher than these. You would need to subtract the weight of the forks from the lift capacity and if the center of gravity of the pallet load is out from the bucket edge then you REALLY have to lower what you think you can lift. I was trying to unload my BX42 chipper from a trailer with my Bxpanded Forks. If the lifting strap was a foot out from the bucket edge, I could not life (chipper is somewhere around 400 lbs). I could only lift it if the sling slid down the fork arms to the edge of the bucket. I would love to lift like a fork lift but I think you will not exceed what your current lift capacity is and more than likely less.

BX 2200 LA 211 - 460 LB
B 7400 LA 272 - 495 LB
B 7500 LA 302 - 660 LB
B 2410 LA 352 - 770 LB
B 2710 LA 402 - 882 LB
B 2910 LA 402 - 900 LB
B 21 TL 421 - 926 LB

I think he was reverse figuring how much vertical force is applied at the cylinder point by using lift specs that kubota provided.

That is how he came up with a much greater # that you listed. Because it takes more force because the lift point is so far behind the actual load.

 

[dennisgw]

LD1, Yep. Thanks for reminding me. Wasn't aware of the curl issue either.

 

[apease]

dennisgw - LD1 is right - in my message I gave a calculated figure of 1500lbs for W - which is the force applied by the hydraulic cylinder. Since its closer to the pivot than the load, it has to be a much greater force than the weight of the load.

LD1 - good point about the bucket being a potential point of failure. We have two "links in the chain" and the actual lifting force the tractor can generate is the minimum of what the two joints can handle. So, I tried to calculate the force on the bucket for the same pallet fork load.

breakout force at pivot......................1415.00lbs
breakout force at lip...........................992.00

423.00lbs less per 19.69in
21.49 lbs less per inch

breakout force at 4' pallet midpoint.......476.28lbs > 383.89 for loader

It looks to me that in this case the loader pivot is still the limiting factor and we have a margin of safety on the allowable forces on the bucket hydraulics.

 

[Transit]

The lifting force of most all FEL is given at the bucket pivot at the end of the lift arm, so many thousand pounds. This is the lifting force not including the weight of the bucket. The net lifting force is lifting force at the pivot (? the weight of the bucket or any other attachment.

This than becomes a problem in High School Physics, The Sum of the moments around a fixed point, or Torque.

Were the torque = the distance from the reference pivot point X the weight.

What does this all mean?

Pick the reference pivot point. Lets say, we use the vertical post at the cab. Now this point does not have to be a POINT, it may be a vertical line through the reference point, but always use the same reference [now called the reference plain]. The way you measure this distance is from the ref plain is at a right angle to the plain and the distance to the load.

The first calculation:

With the bucket on the ground, measure the distance from the ref plain to the bucket pivot.

Lets say that distance is 56 inches and the lift force is 2,500 pounds. There fore the torque is 56 X 2,500 inch-pounds.

Calculation 2:

With the pallet arms in place, measure the distance from the ref plain to the center of the pallet arms, say 72 inches

Now divide the torque in calculation 1 by 72 to obtain the new lift weight.

56 x 2,500 / 72, this is the lifting force not counting the weight of the bucket and pallet arms.

 

[LD1]

dennisgw - LD1 is right - in my message I gave a calculated figure of 1500lbs for W - which is the force applied by the hydraulic cylinder. Since its closer to the pivot than the load, it has to be a much greater force than the weight of the load.

LD1 - good point about the bucket being a potential point of failure. We have two "links in the chain" and the actual lifting force the tractor can generate is the minimum of what the two joints can handle. So, I tried to calculate the force on the bucket for the same pallet fork load.

breakout force at pivot......................1415.00lbs
breakout force at lip...........................992.00

423.00lbs less per 19.69in
21.49 lbs less per inch

breakout force at 4' pallet midpoint.......476.28lbs > 383.89 for loader

It looks to me that in this case the loader pivot is still the limiting factor and we have a margin of safety on the allowable forces on the bucket hydraulics.

What would your breakout force be at the tip of those 4' forks as opposed to the loader lift??

There may be just that one time when you have a load right on the tip tht I am concerned with.

Like when you need to push that pallet just a little farther, and you back up and lift with just the tip, or when you cant quite reach what you are trying to lift and use just the tip, etc.

 

[apease]

Transit - aren't we saying the same thing? Torque of the load and of the cylinder must be equal. I've calculated an approximate force from the cylinder, and so knowing the distance of that force from the pivot and the force from gravity on the load at a known distance from the pivot I solve for the load allowed.

 

[J_J]

You all can pencil whip those figures all day, but the trial and error method will get it real close.

Why not just use 55 gal drums of water. Each 55 gal of water weighs about 440 lbs. Two full drums of water would weigh about 880 lbs. Add another drum if you can lift the two, and add gal of water until you can not lift the load. Each gal of water weighs about 8 lbs.

Or cement blocks.

 

[LD1]

The lifting force of most all FEL is given at the bucket pivot at the end of the lift arm, so many thousand pounds. This is the lifting force not including the weight of the bucket. The net lifting force is lifting force at the pivot (? the weight of the bucket or any other attachment.

This than becomes a problem in High School Physics, The Sum of the moments around a fixed point, or Torque.

Were the torque = the distance from the reference pivot point X the weight.

What does this all mean?

Pick the reference pivot point. Lets say, we use the vertical post at the cab. Now this point does not have to be a POINT, it may be a vertical line through the reference point, but always use the same reference [now called the reference plain]. The way you measure this distance is from the ref plain is at a right angle to the plain and the distance to the load.

The first calculation:

With the bucket on the ground, measure the distance from the ref plain to the bucket pivot.

Lets say that distance is 56 inches and the lift force is 2,500 pounds. There fore the torque is 56 X 2,500 inch-pounds.

Calculation 2:

With the pallet arms in place, measure the distance from the ref plain to the center of the pallet arms, say 72 inches

Now divide the torque in calculation 1 by 72 to obtain the new lift weight.

56 x 2,500 / 72, this is the lifting force not counting the weight of the bucket and pallet arms.

This is incorrect. You cannot pick ANY point. You have to pick the common point at which both points are rotating round, which is the rear loader pivot.

With your fomula, all you woudl have to do would be to choose a point farther away and come up with a larger number Which cant be.

In your own example, you are giving distances of 56 and 72. What you are saying is that the lift force is 56/72 of the origional capacity (77.7%)

The difference in distance between the pivot @ 56 and the forks @ 72 is 16". This distance does not change no matter what point you choose.

So now lets choose a "point" farther back. Say we are now 90" from the pivot. This would make the forks @ 106" away because they are 16" appart and that cannot change. So 90/106 is 84.9 %.

Nothing else changed except the refrence point.

So you have to use the point at which both are pivoting aroung and NOT ANY point.

Unless I am misunderstanding you on something.

 

[LD1]

You all can pencil whip those figures all day, but the trial and error method will get it real close.

Why not just use 55 gal drums of water. Each 55 gal of water weighs about 440 lbs. Two full drums of water would weigh about 880 lbs. Add another drum if you can lift the two, and add gal of water until you can not lift the load. Each gal of water weighs about 8 lbs.

Or cement blocks.

Now where's the fun in that

 

[apease]

LD1 - good point to emphasize safety, and easy to calculate now that I have everything set up - loader should be able to generate 308lbs at the tip of 4' pallet forks (4' from the edge of the bucket that is). But I think there was a flaw in my previous bucket force calculation. Using the same equations as for the loader, and assuming 12" from cylinder to pivot I get roughly the same force for the bucket cylinder of 1600lbs (when using the BX24 manual figure of 992lbs breakout force at bucket edge). That give us a slightly different figure of 447lbs breakout force for a load centered on a 4' pallet, and 288lbs for a load at the tip of the forks.

So, the bucket does become the weak link in the chain at the tip of the forks, although only by 20lbs.

 

[apease]

J_J - theory and practice work together. It's very important to test the theory in practice, because the calculations or assumptions might be wrong. But I think it's also a good idea to know the theory for cases where you need to extrapolate to a case you can't test, and to allow for explaining things to others. It's always good to know *why* something is the way it is.

 

[LD1]

LD1 - good point to emphasize safety, and easy to calculate now that I have everything set up - loader should be able to generate 308lbs at the tip of 4' pallet forks (4' from the edge of the bucket that is). But I think there was a flaw in my previous bucket force calculation. Using the same equations as for the loader, and assuming 12" from cylinder to pivot I get roughly the same force for the bucket cylinder of 1600lbs (when using the BX24 manual figure of 992lbs breakout force at bucket edge). That give us a slightly different figure of 447lbs breakout force for a load centered on a 4' pallet, and 288lbs for a load at the tip of the forks.

So, the bucket does become the weak link in the chain at the tip of the forks, although only by 20lbs.

I knew it would, I just did not know how far it would on your tractor.

A real world example, we set the trusses on my polebarn with my little L3400 and they were 24' to the peak.

We strapped a 16' section of 3-leg TV antenna tower to the bucket. The trusses were only ~150-200lbs each and the bucket would NOT curl them. But it would lift them with pleanty of power to spare.

So by lifting them, I was exerting more than 2500psi on the curl cylinders. Since my tractor makes 2500PSI and that wasnt enough to curl. Fortunatally I did not blown any hoses or damage anything, but this was the point I was getting at.

 

[Grandad4]

Interesting and informative discussion... thanks!

Once you guys get the math sorted out for a load that's perfectly centered on the forks, you might have a go at figuring out how far off center that load can be (i.e. unbalanced side to side) before the tractor lifts a wheel. The calculations start getting more complex, but "seat of the pants" math tells me there's very little lateral tolerance with a load hanging way out in front like that.

 

[apease]

Grandad4 - If you treat the wheels as fixed, and assume a static situation, I don't see how you'd get sideways tipping. Since the bucket is only as wide as the tractor, the load can't be beyond the fulcrum (the wheel). In reality the situation would be a little more complex I think since the wheel is going to compress under force so maybe there's something going on that I don't know how to model.

However, as another poster mentioned from experience, you can certainly get tipping over the length of the tractor. So, I'd tried to figure that out. I'll start with the worst case of no backhoe and nothing on the three-point. Tractor alone (BX24) weighs 1520lbs. The wheelbase is 55.1" so we can assume a point mass (W) of 1520lbs at the midpoint 27.5" (X) from the fulcrum. Reach of the bucket is 54.7" and we assume a point mass in the middle of the 48" forks so 78.7" (L). F=W*X/L which gives us 531.13lbs which is greater than the 383.89lbs the loader can lift. The Kubota engineers are smart - they gave us a margin of safety so that it's not possible to tip the tractor even in a worst case scenario. However, a dynamic load (i.e. dropping and stopping the load, dropping the front wheels into a depression while driving) is another matter, which could easily raise the force to the tipping point.

Anyway, that's my amateur analysis.

 

[apease]

oops, two error - forgot to include the weight of the loader, and you probably also meant a load the extends beyond the bucket, like moving a long tree laid the long way on the bucket. Analysis coming right up...

 

[dennisgw]

Granddad/Appease:
I can appreciate Granddad's query on stability and things being off center. I had my BXpanded forks on and was carrying a load of branches (probably overhanging 6-7' each side of the bucket forks) across a part of my property that had a "slight incline". I did have the bucket up a bit so the hanging branches would clear the ground and dodge some flowers coming up. At a certain point Grandad's concerns occurred, I could feel the tractor lift slowly lifting on the back...........not a good feeling! I leaned over to the lifting side and dropped the bucket, the tractor immediately sat flat again. I kept the bucket very low for the rest of my journey. So center of gravity of the load in combination with angle is a powerful combination. Lesson learned.

 

[apease]

ok, revised lengthwise tipping calcs - assume the center of mass of the loader+bucket is X=14" forward of the wheel. LA240 loader and bucket together weigh W=507lbs. So, the effect on our forklift load is L=78" is again F=W*X/L giving us 92lbs, or a tipping point weight on the forks of only 438lbs vs lifting capacity of the loader at that point of 383lbs. That's cutting it a bit close, so theory matches practice and says to have the backhoe or something on the three point when using forks.

Sideways tipping calcs next...