Calculating FEL Lift Capacity

[tcartwri]

As mentioned in my post #1 above, I was trying to figure out the factory relief valve setting on the loader. After doing all the drawing and using 1500 psi as a starting point, I found the relief setting of 1638 psi (IIRC) for the loader in the B9200 tractor manual.
Incidentally, I did up the relief to 2000 psi by putting a .012" shim under the relief valve spring. I know, this may not be wise as far as loader and tractor longevity, but the bucket is so small that it doesn't require that much psi. It's not going to take any more psi to carry a full bucket now than before changing the relief. Trying to lift too much weight with a chain, however, is a different story and requires some judgement. Guess who gets to fix it if I break it.
What it does is give me more "breakout digging power" when loading the bucket and also sends this higher psi on to the rear remotes via power beyond where it can be utilized.

Have you measured the real increase in lifting power after the mod? I bumped mine by trial and error. Pushed into a load of black earth that I figured it should lift, and adjusted the bypass until it did. Hot Rodder mentality.....

 

[Rock knocker]

How did you come up with the psi and load? Just curious.

I'm just a stupid geologist, so I may have made fundamental assumptions in my math using triangles, resultants and the like. I scaled off the drawing, rounded some numbers because this isn't a classroom test and figured the mechanical FEL assembly had no weight.

If I'm wrong, I accept that

 

[npalen]

Have you measured the real increase in lifting power after the mod? I bumped mine by trial and error. Pushed into a load of black earth that I figured it should lift, and adjusted the bypass until it did. Hot Rodder mentality.....

The increase in lifting power should be at least 33% since the FEL relief was set at 1500 psi now 2000 psi. Might be even more than that since a good portion of the 1500 psi was used to overcome the loader weight and friction. Would that be correct?

 

[oldnslo]

The increase in lifting power should be at least 33% since the FEL relief was set at 1500 psi now 2000 psi. Might be even more than that since a good portion of the 1500 psi was used to overcome the loader weight and friction. Would that be correct?

No, the loader weight should be a constant regardless of system operating pressure.

 

[gladehound]

The increase in lifting power should be at least 33% since the FEL relief was set at 1500 psi now 2000 psi. Might be even more than that since a good portion of the 1500 psi was used to overcome the loader weight and friction. Would that be correct?

I believe you are correct that the increase lifting power has to be more than 33%.

For example. If it took 500 psi to get the loader off the ground, you previously only had 1000 PSI going into lifting additional weight (1500 minus 500). Now, you have 1,500 for additional weight (2,000 psi minus 500 psi). So in this hypothetical example you would have 50% more lift (1500/1000).

Wow, that's quite a big bump of from the design spec! Operate with caution :thumbsup:

 

[npalen]

Wow, that's quite a big bump of from the design spec! Operate with caution

Your point is well taken. It's amazing that a shim of only .012" bumps the pressure that much. There is a screw adjustment in the end of the relief cartridge that was staked at the factory. I assume it was used to set the factory pressure rather than using shims and then staked as a precautionary/liability measure.

 

[davedj1]

While we are on this subject. I am in the process of building a front end loader for a Deutz Allis 1920. I have the mounts done and the arms are more or less built (need to clean up the welds and weld cross bar in). I am planning on using 1.5" bore with 1" rod cylinders from SS, they have some specific for loader arms. I don't need to lift a ton but I would like to lift a minimum of 500 lbs. The lift cylinders will be around 25-30 deg. I guess my question is should I use the 1.5" cylinders or bump up to 2" cylinders? The calculation I posted yesterday above here has the 1.5 cyl at about 1700 lbs each and a 2" cylinder at about 2200 lbs each. I know there is more to the math and geometry than I am posting so if you need any more info just shout. I'm suing a pressure of 1500 psi.
Thanks in advance.

edit: Ok so going off of rock knocker's drawing is it safe for me to assume I can use the same formula? Overall distance from pivot to pin over distance of pivot to cyl mount?
I'll get some measurements and report back.

 

[Rock knocker]

edit: Ok so going off of rock knocker's drawing is it safe for me to assume I can use the same formula? Overall distance from pivot to pin over distance of pivot to cyl mount?
I'll get some measurements and report back.

No engineers have reported back with pros or cons of my math

 

[IceStationZebra]

I used to perform these tests for a previous employer. It was almost 8 yrs ago but as I recall breakout force is the maximum vertical force you can generate at the cutting edge using the bucket or lift cylinders. This force may exceed the tipping limit of the machine, which we countered by tying the rear end down. This is done to account for the dynamic nature of people trying to pry something.

The capacity, at least for wheel loaders and skid steers, is done with bucket curled to carry the most load. Then pull down though the load CG with the CG at the most forward position. This is normally done with 10% fuel and no operator. (wheel loaders are positioned at full turn) You have to assume the lightest configuration of the machine and heaviest configuration of bucket. (with or without cutting edge, etc.) Sometimes we did families of buckets that where close to each other in mass and just tested the heaviest one. As the front tires squat and the rear tires unload the machine's CG moves up and forward a little, so you lose a little bit of leverage there too.

ISZ

 

[gladehound]

While we are on this subject. I am in the process of building a front end loader for a Deutz Allis 1920. I have the mounts done and the arms are more or less built (need to clean up the welds and weld cross bar in). I am planning on using 1.5" bore with 1" rod cylinders from SS, they have some specific for loader arms. I don't need to lift a ton but I would like to lift a minimum of 500 lbs. The lift cylinders will be around 25-30 deg. I guess my question is should I use the 1.5" cylinders or bump up to 2" cylinders? The calculation I posted yesterday above here has the 1.5 cyl at about 1700 lbs each and a 2" cylinder at about 2200 lbs each. I know there is more to the math and geometry than I am posting so if you need any more info just shout. I'm suing a pressure of 1500 psi.
Thanks in advance.

edit: Ok so going off of rock knocker's drawing is it safe for me to assume I can use the same formula? Overall distance from pivot to pin over distance of pivot to cyl mount?
I'll get some measurements and report back.

Not sure I follow what Rock Knocker is doing. I might just be dense but I think I need a more step by step description to follow. I'm not even sure what he's trying to solve for?

[Edited for 1500 psi] You want to lift 500 pounds.

To what height do you want to lift it and what is the distance from loader pivot to the cylinder center line at the point where a perpendicular line drawn from the cylinder center line would intersect the loader pivot? Call this Dma

At the height you want to lift, what is the horizontal distance from the loader arm pivot to the center of the load? Call this Dl

What is the weight of your loader arms + attachment when lifted from the center of the load that you want to lift in question above? Call his W

Then you have...

lift = (Cylinder force x Dma / Dl) - W

For 1.5 inch cylinders at 1500 psi cylinder force = 5299 pounds

lift = (5299 pounds x Dma / DI) - W

For the purpose of the example, assume Dma is 12 inches, Di is 60 inches, and the weigh of the bucket + loader arms when lifted from the bucket pin is 500 pounds.

Then you have....

lift = (5299 pounds x 12/60) - 500

lift = 560 pounds at center of the load (note - this is just an example, need Dma, Di & W from you)

These calculations do not include friction so you'll have to account for some margin. You'll have to plug in your own numbers at the loader position that is important to you and decide what size cylinders to go with. Or just send Dma, Di, and W to me and I'll plug it into the Excel sheet that I just set up while writing this response.

On the flip side, if you just get the 2" cylinders, unless you have some very unusual loader geometry, they will lift 500 pounds in any reasonable scenario. Once you factor in the weight of the loader / attachment, they'll lift more than twice what the 1.5" cylinders will lift.