I found what I thought was a good deal on a 4" x 24" cylinder for a log splitter. But the rod is 1.5" Have heard of a few problems bending the rod but not sure how likely this is to be a problem. Should I be worried about this and go for 2"? Planned to run this at 2250 PSI from tractor hydraulics.
1.5 vs. 2" rod on log splitter
- Thread starter JimS
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oldnslo
Super Member
Jim,
this will depend on how well the ram is guided and how the cylinder is mounted. If there are very minimal bending moments induced by the design you should be fine.
Note: For NFPA cylinders The "standard rod size is 1 3/4" on 4" bore.
Roy
oldmech
Gold Member
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- Two Gravely Two L One with 10 h.p Dek-Diesel engine one Grillo 107d and one BCS 735 diesel
Hi! 2 in. rod is a lot more stong and more fast for return. But I think 1.5 rod cylinder dont bend if work in line with logsplitter knife. Good luck! Oldmech
wdchyd
Elite Member
Jim,
It will work but less column strenght to resist bending, 2" will not give you any more power to split but will increase your speed on return stroke, the only benefit to 1.5" rod would be if you were pulling under load for more power (which you are not)....
Speed vs Cash are the trade-offs:confused2:
arrabil
Veteran Member
JimS, I've been asking myself this exact same question without getting anywhere either. You'd think with a proper setup there would be no way to bend the rod but all the commercial ones are 2". Even Harbor Freight.
wdchyd, why does the 2" increase the speed of the return stroke over the 1.5"? Does it increase the speed of the push stroke too then?
wdchyd, why does the 2" increase the speed of the return stroke over the 1.5"? Does it increase the speed of the push stroke too then?
the speed is based on how long it takes the fluid to build up behind each side of the piston. for the split stroke, it's the area of the piston, which is pi*r^2, or 3.14 * 2^2 = 12.56 sq in. the the return stroke, you have the same area, but the fluid does not build up where the rod is. for a 2" rod, you have the area of the piston minus the area of the rod, or 12.56 - (3.14*1^2) = 9.42 sq in. since the return side has a smaller overall area, the fluid fills faster, making the time to return faster than the time to split. the larger the rod, the less area on the return side.
wdchyd
Elite Member
Arrabil,
basically the 2" rod takes up more displacement in the front end than the 1.5", the amount of oil on the return stroke side of the cyl is less in the 2" thus retracts faster...
No, it does not change the speed of the push stroke (4" bore is 4" bore)
I'm sure KennyD, Aakkman, JJ and others can share the math with you:drink
hint-hint)
wdchyd
Elite Member
Lostcause,
you're a much faster typer than I, guess I'll have to learn to use more than 1 finger :laughing::laughing: Thanx for the numbers (can't count any more than 21) :laughingwell maybe 20 1/2)
you're a much faster typer than I, guess I'll have to learn to use more than 1 finger :laughing::laughing: Thanx for the numbers (can't count any more than 21) :laughing
well maybe 20 1/2)arrabil
Veteran Member
Ah, makes sense. Thank you both.
I guess that makes the 2" rod worth the extra cost since its stronger AND faster.
I guess that makes the 2" rod worth the extra cost since its stronger AND faster.
yup, faster on the return stroke. just remember that if you are using it in a fashion where you need the cylinder's force on the return stroke, then the larger rod gives you less power. you have to make a decision between power & strength then.
