Since you are still in the "scribbling" stage of the design,,
determine what the least amount of travel is that you will need,
THEN, modify the length of the movable part that the cylinder connects to.
With the least travel, you obtain the maximum mechanical advantage,,
If the tonnage ends up too high, simply reduce the hydraulic relief pressure.
If the tonnage is too low,, you get to scrap the press,, and start over.
I think I would prefer too much tonnage,,,
Are you considering mounting BOTH cylinders to move the press actuator,,??
Both would give more tonnage, and AGAIN, MORE tonnage is better.
Actually, having two cylinders supplying the tonnage would require more oil flow to move the ram.
That would be a good thing, as far as accuracy, as long as speed is not a controlling factor.
Usually, for one-off forming,, slow is always better.
The harder you fall, the higher you bounce...
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I was going to put a flow control in to limit the movement speed after I see what the movement speed actually is. Worth noting with a mechanical advantage the cylinder also has to move twice as far to produce press movement.
A second cylinder can't travel inside the upper head of the machine unless they we're stacked I suppose.
My actual concern is if I'm actually getting a 2:1. Everyone has said the advantage is determined by the master pivot only, but how can that be that the slave pivot connected by the linkage doesn't reduce the mechanical advantage with its lesser leverage point.
As it sits it has 4" of travel. It hits the tooling a fair bit before that though. In my mind more is unnecessary and less could be a hinderance.
Couple times a week for a handful of bends each time. But, I still want to do it right if im gonna do it.
Last edited by BrooklynBravest; 01-06-2022 at 07:02 PM.
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Zach
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"A second cylinder can't travel inside the upper head of the machine unless they we're stacked I suppose."
Or they could be horizontally opposed, with one cylinder normally open, one normally closed. In this case with a common attachment point, I don't see a problem plumbing them in series.
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If they’re in series you lose power i believe so it’s not really beneficial
If the mechanical advantage actually works, which I’m still not sure how it does with the second slave pivot, then it’s more than I’ll ever need pressure wise
It’s only a 1.5gpm pump as well. Filling 2 cylinders is gonna be somewhat slow.
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Plumbed in parallel but opposite on one cylinder? I can't see how you would lose power if you are exerting the same pressure on twice as much piston. I would think slower, but more force.
The harder you fall, the higher you bounce...
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Zach
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Your adjustable link will be stressed more than your cylinder, but the force at the vertical links will be the same as each other. There is no mechanical advantage difference between the two dog legs if your distance from corner pivot to adjustable link pivot are the same.
Maybe I didn't say that right. Give me a minute.
Last edited by Boostinjdm; 01-06-2022 at 09:57 PM.
My name's not Jim....
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Zach
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D/A is you mechanical multiplier. All the other dimensions need to match other dimensions with the same letter. If A on one side is 6", then A on the other also needs to be 6". Making B a larger value will reduce stress on the adjustment link itself, but will have no effect on the overall mechanical advantage from the cylinder to the punch.
So let's say D/A is 3/1. That would make 10 tons in at the cylinder 30 tons out at the punch for one piece. Adding the second side to balance the punch doesn't change output. It splits it into 15 and 15 for a total of 30. If A=B, then your adjustment link will see 15 tons. If B is twice A, then adjustment link would see 7.5 tons. All that will vary a little at different points in the stroke, but that is a little more complicated to calculate and I'm just not interested in trying to do it.
Last edited by Boostinjdm; 01-06-2022 at 10:38 PM.
My name's not Jim....
Awesome thank you.
The ones that need to match by your description do in fact match.
The actual mechanical advantage here is 2.33
Cylinders are 4” bore at a potential 2950 psi
4” are is 12.57 X 2950psi = 37,081 X 2.33 = 86,399 lbs.
I’m sure there’s loss in the system, but that would be far more power than I need.
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40 tons will bend a lot. You can always cheat by going with a wider lower V die if you need to.
My name's not Jim....
As long as the ratio between master link and slave link are the same, it's all good.
If the master link overall ratio is 2:1, then you'll need a hydraulic system and cylinder able to exert 20 tons, meaning if your cylinder is 3" you will only need 100 bar pressure... not too bad really. Personally I'd reduce the leverage ratio and increase the pressure. Take the strain off the pivot point and put more onto the rest of the system.
Murphy's Golden Rule: Whoever has the gold, makes the rules.
Which pivot point are you referring to?
I can see additional stress at the point around which the lever pivots, but no increase at any other point. 1" 1018 having a shear strength of around 40Ksi is able to handle it with proper pin sizing. Lever web at the pivot point, and between cylinder yoke, and pivot point, can be increased to take the strain I'd think.
Upsizing the force generated by the cylinder places strain on connections at both ends of the cylinder. This brings into question the formula you're using for the force generated by a 3" cylinder. I did a quickie online calculator search, and at 2500psi hydraulic pressure input, a 3" cylinder only puts out 17,000 pounds of force. I believe you'd have to up your pressure, or go to a bigger cylinder. (I'm using tractor hydraulic pump capabilities as a benchmark.......they generally run around 2500psi on the stuff I own)
So......we go to a bigger cylinder I guess?
I'm a bit foggy, been up most of the night, can't sleep. So I might not be getting this right.
I'm not even sure I see any increase in stress at the main lever mid pivot point. The only additional stress on the system is along the distance between the pivot, and the tip of the lever attached to the cylinder..............it'll want to bend. The pivot doesn't care how long the lever is.
I guess, if we just look at it from a torque standpoint.........................the midway pivot will require a set amount of torque, depending on the length of the arm beyond the pivot. It's a constant.
Increasing the length of the lever on the cylinder side has no bearing on the torque load on the mid pivot.
By "midpoint" I mean the pivot at the corner of the right triangle formed by the main lever........not the drag link.
Aw Hell, I'm looking at the wrong pivot. I should be thinking about the drag link pivot, and the length of the arm above it.......it's doing the work.
And the damn calculator was wrong.............another calculator gives 52,000 force. Geez!!!!!!!!!!!!!!!!!! I'm goin' to bed
edit............yeah the first calculator was right............17K worth of force
Mechanical advantage of 2 to 1 gives 34K
Last edited by farmersammm; 01-07-2022 at 05:56 AM.
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I was just thinking(with no sleep) about the lever.
The work is done by the lever on the left side of your press. Yes, it's inefficient, but it's the nature of this kind of fulcrum with the load in the middle, not at the side opposite the fulcrum(although at least half the load is). You're limited by the amount of work it can do when exerting force on the link. That link is the actual place to look at effective force, because it applies force to the right "slave". The force on the left "slave" is reduced by the force bled off by the right "slave". This I guess is where the equalization occurs.
Your bottom pivot on the main arm bears all of the applied force. It's carrying the load of both slaves. As a fulcrum would.
I think I got it right this time. Takes me a while to figure this stuff out.
Last edited by farmersammm; 01-07-2022 at 09:03 AM.