Calculating and comparing splitting force of flywheel and hydraulic splitters.

[KiwiBro]

We've all seen the BS claims about how the 3" cylinder with pump of questionable pressure splitter is capable of 40t of ram force. It's fairly easy to calculate the force of a hydraulic ram, but how about a ram on a kinetic/flywheel splitter?

From here:
http://www.engineeringtoolbox.com/flywheel-energy-d_945.html

we can work out the kinetic energy a flywheel has, but I'm not sure how to compare that to a hydraulic ram force or tonnage rating.

Let's take a DR pro XL model and assume:

weight of each flywheel=74 lbs
max rpm=400
inertial constant=.75 (I'm guess with this)
diameter=18 1/4"

So, moment of inertia = .75 x 74 x .76 x .76 = 32
and kinetic energy = .5 x 32 x (400/9.55) x (400/9.55) = about 28000 ft lb per flywheel

Is that correct or where have I gone wrong with the calcs please?

Then, how to compare that kinetic energy to a traditional hydraulic ram tonnage rating?

 

[lefturnfreek]

There is no direct comparison cause your not comparing apples to apples. Totally different styles of splitting, different force styles involved.

Kinetics are good at knot free easy wood to split but if you through this bad boy on there.... that's a 12in knife for scale












...it would just stall or throw the block, cause one of them broke my splitter. Got a knotty piece or a bunch of Y's....ya ain't doing them either....

 

[nathon918]

We've all seen the BS claims about how the 3" cylinder with pump of questionable pressure splitter is capable of 40t of ram force. It's fairly easy to calculate the force of a hydraulic ram, but how about a ram on a kinetic/flywheel splitter?

From here:
http://www.engineeringtoolbox.com/flywheel-energy-d_945.html

we can work out the kinetic energy a flywheel has, but I'm not sure how to compare that to a hydraulic ram force or tonnage rating.

Let's take a DR pro XL model and assume:

weight of each flywheel=74 lbs
max rpm=400
inertial constant=.75 (I'm guess with this)
diameter=18 1/4"

So, moment of inertia = .75 x 74 x .76 x .76 = 32
and kinetic energy = .5 x 32 x (400/9.55) x (400/9.55) = about 28000 ft lb per flywheel

Is that correct or where have I gone wrong with the calcs please?

Then, how to compare that kinetic energy to a traditional hydraulic ram tonnage rating?

you need to do a lot more calculating...your not splitting with the fly wheel...its only driving a "transmission", you would have to calculate the gear reduction into it, (size of pinion gear relative to size of flywheel etc.)
then if you figured that out, it would only be peak force...its not constant force, just peak force for a split second then it drops to near nothing. its basically just a big hammer
they both have their place, but as lefturnfreek said they're not comparable

 

[Marshy]

Its not a direct comparrison for sure and a lot of assumptons would have to be made. I have a lot of opinions about where to start but unless your ready to write your eningeering thesis on "an integrated approach to determining the similitude of hydraulic vs kinetic mechanisms for separation of duramen" I wouldnt bother.

 

[Motorsen]

You should calculate the work/energy in relation to time exerted. Which can be hard as in the case of the hydraulic splitter that it is dependent on how much pressure/energy is needed to split the wood piece. Maybe calculate the potential max energy in case of the hydraulic splitter which is related to the maximum pressure delivered from the pump/pressure release valve setting. The kinetic splitter is easier to calculate as it is kinetic energy released and is the same every stroke.
Hope this will help you.

Motorsen

 

[Marshy]

You should calculate the work/energy in relation to time exerted. Which can be hard as in the case of the hydraulic splitter that it is dependent on how much pressure/energy is needed to split the wood piece. Maybe calculate the potential max energy in case of the hydraulic splitter which is related to the maximum pressure delivered from the pump/pressure release valve setting. The kinetic splitter is easier to calculate as it is kinetic energy released and is the same every stroke.
Hope this will help you.

Motorsen

I agree. Every chunk needs a different amount of energy so thats why you should focus on max work energy delivered. It would be easy to calcuate for the hydraulic splitter and fairly easy to calculate for the kinetic splitter for instantanious but to get the real answer you need to account for the time that the force is being applied to the split before the rack disengages.

 

[Chris-PA]

You would be comparing the continuously applied force of the hydraulic ram, which should not be too hard to calculate, with the force derived from the stored kinetic energy of the flywheel. One is a high speed impact and the other a slow continuous force, and I'd bet they will have different effectiveness in ripping apart wood fibers. There are likely to be all sorts of complicated variables (including wood type, etc.) for which you'll have no real data.

Even in calculating the force of the ram, you are only calculating the force the ram applies along its axis, not the force the wedge imparts to break the wood fibers or how effective that is.

 

[Marshy]

You dont have to get into the forces regarding the wedge and splitting fibers. The hydraulic splitters are rated based on the force the rams are capable of at their pressure relief setting. So you take the same analogy with the kinetic splitter. The rack engages exerts energy and then disengages. You figure out how much energy is transfered in that portion of the cycle and you have the bounds of you problem.

 

[hedge hog]

I have never seen the claim of a 3" bore splitter produce 40 ton of force
the pump would have to produce 11,320 psi to make it work
but I know what your saying

are you after the cycle times of kinetic splitter?
there are a lot of ways to speed up a hydraulic splitter and keep it safer

 

[Chris-PA]

You dont have to get into the forces regarding the wedge and splitting fibers. The hydraulic splitters are rated based on the force the rams are capable of at their pressure relief setting. So you take the same analogy with the kinetic splitter. The rack engages exerts energy and then disengages. You figure out how much energy is transfered in that portion of the cycle and you have the bounds of you problem.

You can work out the force the ram puts on the round pretty easily, and you can calculate the kinetic energy the other splitter has when it hits, but you cannot directly compare force to energy.

In general force it the rate of energy transfer, so if the KE that is delivered to the round is dissipated quickly then it creates a high peak force. Basically, F=ma, so if the mass of the KE splitter wedge stops fast over a small distance, then the acceleration is high and F will be high. If the wood is soft and the wedge moves a longer distance into the round, then acceleration is smaller and F will be smaller.

The hydraulic splitter does not care about that at all, as the force is continuously applied even as the wedge moves.

 

[nathon918]

If the wood is soft and the wedge moves a longer distance into the round, then acceleration is smaller and F will be smaller

but that doesn't really mean anything...there is still more energy stored in the flywheels if needed, also the ram speed wouldn't add much of anything to the force, its not moving fast enough... acceleration of the ram also isn't much of a factor at all, its 0 to full speed in the time it takes to engage it, under no load it takes nearly no energy out of the flywheels to get moving...not the same as swinging a hammer or maul, where you would have to build momentum (swing) to have it work correctly...the flywheels are "stored" momentum
you could measure peak force on the supersplit, using a hydraulic "scale" basically just a cylinder with a gauge at one end, find some way to secure the scale to the beam and hit it with the ram...then its just area x pressure

 

[Chris-PA]

but that doesn't really mean anything...there is still more energy stored in the flywheels if needed, also the ram speed wouldn't add much of anything to the force, its not moving fast enough... acceleration of the ram also isn't much of a factor at all, its 0 to full speed in the time it takes to engage it, under no load it takes nearly no energy out of the flywheels to get moving...not the same as swinging a hammer or maul, where you would have to build momentum (swing) to have it work correctly...the flywheels are "stored" momentum
you could measure peak force on the supersplit, using a hydraulic "scale" basically just a cylinder with a gauge at one end, find some way to secure the scale to the beam and hit it with the ram...then its just area x pressure

The ram is a continuously applied force - the peak force is determined by the size of the engine, pump, cylinder, etc., but the amount of energy available is limited only by the amount of fuel in the tank.

The KE splitter is a discontinuous process (like swinging a maul), and on any given hit the energy is limited by the mass and rotational velocity of the machine. The only way to get more energy to the round is to hit it again.

In the first case the available energy is (practically) unlimited, but the peak force is fixed. In the second case the available energy is limited but the peak force is really not (at least not in the same way - it depends on how fast the thing decelerates when it hits).

 

[zogger]

Someone who owns one, just call up supersplit and ask them how they measure the power at their shop.

 

[the GOAT]

Find a piece of wood a supersplit won't split. (Shouldn't be too hard)

Split that piece of wood while watching the pressure on a hydraulic splitter and calculate the force applied. Do it with enough pieces and you'll have a pretty good idea of the max splitting force of a supersplit in comparison to a hydraulic splitter.

 

[nathon918]

Find a piece of wood of wood

what kind of wood of wood is that?

 

[the GOAT]

am I troll?

Yes

 

[nathon918]

Yes

NO

 

[CUCV]

The hardest part about calculating the force applied by a supersplit is figuring out that time it takes to consume the stored energy. The force could be practically infinite if its consumed almost instantly.
Back when I added the 3rd flywheel on my SS I back figured the SS claimed tonnage solved for time then was able to recalculate with the extra flywheel. I can't remember all the numbers now but I want to say the j model was rated at 16 ton.

The amount of constant push thru power the motor applies is easy to calculate and minor in comparison to the energy stored in the flywheels applied in a fraction of a second.

 

[Chris-PA]

The hardest part about calculating the force applied by a supersplit is figuring out that time it takes to consume the stored energy. The force could be practically infinite if its consumed almost instantly.
Back when I added the 3rd flywheel on my SS I back figured the SS claimed tonnage solved for time then was able to recalculate with the extra flywheel. I can't remember all the numbers now but I want to say the j model was rated at 16 ton.

The amount of constant push thru power the motor applies is easy to calculate and minor in comparison to the energy stored in the flywheels applied in a fraction of a second.

+1

Someone who gets it!

 

[tla100]

Wedge design is huge. SS have a narrow thin blade. Where most hydraulic are more blunt