I am looking for a project this winter and in need of a woodsplitter so I figured I would search for plans, not going so well. I am looking to build an electric powered that is pretty rugged I would also like a little zip in the boom, I have seen too many that are like watching paint dry. Anyone know of a set of plans or maybe built one themselves?
Logsplitter plans/design
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Problem you run into with plans is by the time you go out and buy new, individual components to meet the plan you will spend as much or more than it cost to buy one. If you can scrounge a bunch of parts, or have some already laying around, that can save some money, but obviously no one has plans which use the exact stuff you have laying around in the basement. Either way, plans come out pretty useless. You could go and look at commercial units, make a couple sketches, run some hydraulic equations and see what you can come up with. But really, they are pretty simple to build - it's just the part about keeping costs low enough so it's not cheaper to go out and buy one.
PS - as far as the cycle time, hydraulics are basically just like gears. For a given power level, you can either go really fast or you can have lots of power. A two stage pump will help a bit, but it can only go so far. Also, generally the faster you go, the less power you have or vice versa. So you'd just have to scale your pump / cylinder / motor to a point where you're happy with the power and the cycle time.
PS - as far as the cycle time, hydraulics are basically just like gears. For a given power level, you can either go really fast or you can have lots of power. A two stage pump will help a bit, but it can only go so far. Also, generally the faster you go, the less power you have or vice versa. So you'd just have to scale your pump / cylinder / motor to a point where you're happy with the power and the cycle time.
You might find a decent splitter with a dead motor and just repower it. A 5hp motor works pretty well with a 16 gpm pump.
SolarAndWood said:
Possibly a 5hp ELECTRIC motor... In the gas world you will need at least an 8hp, preferably closer to a 9 or 9.5... However I agree on the repower idea.
Gooserider
Gooserider said:
That is where I was going with the 5hp. You can get a single phase 7.5hp pretty easily as well. Given my experience with the 5hp/16gpm combo, the 7.5 would probably drive a 22gpm.
I am having trouble figuring out the how to achieve a 6-8 second cycle. I have seen one manufacturer who uses a 2.75x24 cylinder and an 11gpm pump with a 6hp motor with 6second cycle but a 2.75 cylinder seems very small too me although those cheesy $300.00 electric splitters can't have much of a cylinder either. The other end of the spectrum would be a 10 hp motor and 22 gpm pump with a 4" cylinder.
Essentially you are running into the basic problem with hydraulic systems - and it sounds like you are figuring it about right... Essentially you can figure the volume of a given diameter cylinder (area x length) and compare that to the pump gpm to see how long it takes the pump to fill the cylinder. Remember the retract stroke will always be a little faster than the extend stroke (but less powerful) since your cylinder volume is decreased by the volume of the piston rod, but it isn't a huge difference - I would say figure 1.5 - 1.75 x the time on the extend stroke for a complete cycle in and out...
Splitting power to the wedge is a function of the cylinder diameter x the working pressure, which I always assume to be 3,000psi unless I have reason to know differently, as that is a pretty standard value for hydraulics...
Unloaded cycle time is a function of cylinder size x pump volume - as the pump size goes up cycle time goes down, as the cylinder size goes up, so does the cycle time... With the possible exception of some of the machines where marketing has decided to push tonnage over all else, you will see that most machines seem to be set up for about a 15 second cycle
The bigger the pump, the bigger the motor it takes to drive it, with associated increases in operating costs - although this really isn't likely to be a big increase - going up a few HP on a gas engine doesn't really have that much of a fuel consumption penalty. With an electric motor, it gets a bit trickier, as the increased power draw is relatively negligible from a cost / KWh standpoint, but you do start running into limitations on voltage choices, and wiring sizes. I'm not sure how practical it is to run an electric motor big enough to drive a 22gpm pump off of standard household 220 power, and it can be a challenge to get 3-phase to your house... The practical limit might be 16gpm.
I haven't done the math, but I suspect a 4" cylinder w/ a 16 gpm pump would give you ~10 second cycle times, and 20 ton capacity which seems to be the "sweet spot" for splitter sizing. Dropping down to a 3 or 3.5" cylinder would get you a faster machine, but cost you some tonnage, though a 15ton machine is still going to do pretty well...
Gooserider
Splitting power to the wedge is a function of the cylinder diameter x the working pressure, which I always assume to be 3,000psi unless I have reason to know differently, as that is a pretty standard value for hydraulics...
Unloaded cycle time is a function of cylinder size x pump volume - as the pump size goes up cycle time goes down, as the cylinder size goes up, so does the cycle time... With the possible exception of some of the machines where marketing has decided to push tonnage over all else, you will see that most machines seem to be set up for about a 15 second cycle
The bigger the pump, the bigger the motor it takes to drive it, with associated increases in operating costs - although this really isn't likely to be a big increase - going up a few HP on a gas engine doesn't really have that much of a fuel consumption penalty. With an electric motor, it gets a bit trickier, as the increased power draw is relatively negligible from a cost / KWh standpoint, but you do start running into limitations on voltage choices, and wiring sizes. I'm not sure how practical it is to run an electric motor big enough to drive a 22gpm pump off of standard household 220 power, and it can be a challenge to get 3-phase to your house... The practical limit might be 16gpm.
I haven't done the math, but I suspect a 4" cylinder w/ a 16 gpm pump would give you ~10 second cycle times, and 20 ton capacity which seems to be the "sweet spot" for splitter sizing. Dropping down to a 3 or 3.5" cylinder would get you a faster machine, but cost you some tonnage, though a 15ton machine is still going to do pretty well...
Gooserider
I am thinking a 3" cylinder and the 16gpm pump will work out for my needs. That coupled with a 5hp 3600 RPM electric should be very feasible. I do wish I would stumble across someone who may have some trial and error experience
You can get a 10 h.p. electric motor for single phase 220 current. Not common or cheap but available and worth it, in my book. Next step, two stage pump, probably 22 gallon. That and a four inch cylinder and you should be blazingLEE fast and still have plenty of power. My setup is 5h.p. single phases 220, one stage pump, with a 6" piston. Way excess power, I don't need to split logs, I can shear them if I want to, but a little slow for my taste.For this years pile of rounds, I will get past the speed issue by adding a four way wedge, for a three fold increase in productivity, but by next fall, the plan is pretty much what I said in my opening sentence (I already have the 10 h.p. single phase electric motor), a 22 gallon two stage pump (Northern Tools) and a dedicated splitter valve, with auto retract.
As for plans, just go stare at someone elses for a bit. It you can't remember everything, take a picture. By far your best bet is to buy a beater and fix it up.
As for plans, just go stare at someone elses for a bit. It you can't remember everything, take a picture. By far your best bet is to buy a beater and fix it up.
A good source for single phase motors is abandoned barns, with permission of course.
You'll find 10 HP on most silo unloaders.
3-5 HP on barn cleaners.
I just had a 10HP Baldor totally rebuilt and it emptied my wallet to the tune of $750. Don't bother checking new motor prices unless your pace maker has a new battery in it.
You'll find 10 HP on most silo unloaders.
3-5 HP on barn cleaners.
I just had a 10HP Baldor totally rebuilt and it emptied my wallet to the tune of $750. Don't bother checking new motor prices unless your pace maker has a new battery in it.
I figured I chime in for selection of an AC induction motor for a DIY logsplitter. To select an induction motor one must take into account of how much breakdown torque the motor can provide when the wedge will encounter some very tough logs. Once this is established a percentage of 75% of the breakdown torque will be what the motor will be required to provide for the max set point of the relief valve. The reason the use of 100% of the breakdown torque is not used is because voltage sagging greatly effects induction motor torque. In fact torque varies as the square of the voltage for induction motors. If we back calculate by taking the square root of 75% we get about 87%. So if the nominal voltage the motor expects is 240 volts AC then at 87% of this value will be 208 volts. So if the voltage sags this low from voltage drops in the wires, the motor will still be able to provide the torque for the really tough wood without stalling.
Typical torque curve of induction motors:
(broken image removed)
Here is an example for someone who may want to build a REAL 16 ton log splitter. The majority of logs require about 4 to 12 tons, but one may encounter some logs that require tonnage beyond these values. In these cases the breakdown torque of the induction motor will come in handy when a log may require 16 tons. For induction motors this requirement can be met with 3 full load horsepower (i.e. the horespower on the motor nameplate). For gasoline engines the value must be 5-10 horespower depending upon the relief valve setting since they do not have breakdown torque.
We need 16 tons and have decided to use a hydraulic cylinder with an internal piston diameter of 4" and this cylinder will have an 18" stroke when fully extended. So we take 16 tons*2000LBS per ton to get 32,000 pounds of force. Now we must determine how much PSI is required to meet this amount of force. So we find the area of the 4" diameter piston with Radius^2*PI which is 12.57 square inches. We take 32,000# divided by 12.57 to get a PSI value of 2,545 PSI. It is best to go a little higher to account for expansion so we will pick 2,700 PSI. Final step we apply is to see how much torque the prime mover must provide. So the formula is DP/(6.28*12*Eff) = Torque, where D is displacement of the pump in in^3/rev, P is the pressure in PSI, and Eff is the pump efficiency which is typically 85%. A typical hydraulic pump with 11 GPM with 650 PSI low pressure and 2.9 GPM for high pressure mode with 2,700 PSI will have corresponding displacements of .706 in^3/rev and .186 in^3/rev. With all data provided the torque the prime mover will need to provide is 7.16 pound-feet for low pressure of 650 PSI and for 2,700 PSI for the high pressure is 7.84 pound-feet.
A 3450 RPM, 3HP induction motor #120341 from: http://www.electricmotorwarehouse.com/single_phase_TEFC.htm. Has breakdown torque of 13.5 pound-feet according to the Leeson website. Using 75% of this value is 10.12 pound-feet which meets our requirements for this motor to provide the torque for 16 tons.
Cycle time can be found by taking the stroke length and multiplying this by the area of the internal piston then take 231*GPM which converts this to cubic inches per minute and multiply this to the volume found to get a decimal form of minutes that can be converted to seconds with multiplication of 60. The retract time will be less since the rod will take up some of the volume.
Typical torque curve of induction motors:
(broken image removed)
Here is an example for someone who may want to build a REAL 16 ton log splitter. The majority of logs require about 4 to 12 tons, but one may encounter some logs that require tonnage beyond these values. In these cases the breakdown torque of the induction motor will come in handy when a log may require 16 tons. For induction motors this requirement can be met with 3 full load horsepower (i.e. the horespower on the motor nameplate). For gasoline engines the value must be 5-10 horespower depending upon the relief valve setting since they do not have breakdown torque.
We need 16 tons and have decided to use a hydraulic cylinder with an internal piston diameter of 4" and this cylinder will have an 18" stroke when fully extended. So we take 16 tons*2000LBS per ton to get 32,000 pounds of force. Now we must determine how much PSI is required to meet this amount of force. So we find the area of the 4" diameter piston with Radius^2*PI which is 12.57 square inches. We take 32,000# divided by 12.57 to get a PSI value of 2,545 PSI. It is best to go a little higher to account for expansion so we will pick 2,700 PSI. Final step we apply is to see how much torque the prime mover must provide. So the formula is DP/(6.28*12*Eff) = Torque, where D is displacement of the pump in in^3/rev, P is the pressure in PSI, and Eff is the pump efficiency which is typically 85%. A typical hydraulic pump with 11 GPM with 650 PSI low pressure and 2.9 GPM for high pressure mode with 2,700 PSI will have corresponding displacements of .706 in^3/rev and .186 in^3/rev. With all data provided the torque the prime mover will need to provide is 7.16 pound-feet for low pressure of 650 PSI and for 2,700 PSI for the high pressure is 7.84 pound-feet.
A 3450 RPM, 3HP induction motor #120341 from: http://www.electricmotorwarehouse.com/single_phase_TEFC.htm. Has breakdown torque of 13.5 pound-feet according to the Leeson website. Using 75% of this value is 10.12 pound-feet which meets our requirements for this motor to provide the torque for 16 tons.
Cycle time can be found by taking the stroke length and multiplying this by the area of the internal piston then take 231*GPM which converts this to cubic inches per minute and multiply this to the volume found to get a decimal form of minutes that can be converted to seconds with multiplication of 60. The retract time will be less since the rod will take up some of the volume.
Nice explanation...
As a partial shortcut on some of the math, Burdens' Surplus Sales (Which I've linked to several times in the past as a good on-line source for semi-low cost hydraulic gear) has a pretty decent set of technical info pages, with calculators in the hydraulic section that can automate a good bit of it...
Gooserider
As a partial shortcut on some of the math, Burdens' Surplus Sales (Which I've linked to several times in the past as a good on-line source for semi-low cost hydraulic gear) has a pretty decent set of technical info pages, with calculators in the hydraulic section that can automate a good bit of it...
Gooserider
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