Engineering/physics formula questions about a Supersplit

[Jerry/MT]

Long story short, I am wanting to try to build a flywheel/inertia logsplitter.
About the only thing I will have to buy is the rack/spur gear and an engine.

But I was just trying to calculate how they come up with the splitting force and seem to be at a loss here.

The moment of Inerta (I) for a solid disk is 1/2 x M x R(squared)

Mass in Kg's and R in meters.

The Supersplit splitter has 72lb and 18" flywheels.

So, 1/2 x 32kg x .2286 meters squared gives me a I of .8361

Formula for stored energy is 1/2 x I x Angular velocity squared. Angular velocity in radians/second.

300RPM is 31.41 rad/s

1/2 x .8361 x 31.41 squared is 412 Joules of energy per flywheel. Times two flywheels is 824 Joules of flywheel energy.

Convert that to ft-lbs and come up with 606 ft-lbs of stored energy.

Since the pinion/spur gear is 3" diameter (or 1/4 of a foot) 606 x 4 is 2424lbs of force along the circumfrence of the spur being transmitted to the rack gear. I know that is not nearly enough to split wood. And SS rates the splitters @ 12-24 tons. So...What am I missing here??? I know it is probabally something simple that I am just over looking and I am sure you guys will come through for me once again.

Hint:
Find out how much power you are developing and then use the relationship between Power, Torque, and rpm to figure out the available force

 

[SPYDERLK]

Hint:
Find out how much power you are developing and then use the relationship between Power, Torque, and rpm to figure out the available force

Sorry. That isnt going to do it. This is adding complexity and without extreme care the issue will be hopelessly muddied. The energy in the flywheel is the work that is available to be applied to split the wood. That E will be released as a Force times a Distance. Stick with that. That FxD=606ftlb. As distance decreases to near zero, force goes to near infinity ... so does the instant delivered power. A million HP for a micro second is the same amt of E as 1 HP for a second. Splitting a log takes varying amounts according to resistance, from fractional up to several HP seconds of energy. Just stick with the known E. From there you can figure out the force applied in any specific instance by finding out how far the wedge moved. ... however, the "available" force is limited only by lack of rigidity in the mechanism or failure of its weak link.
larry

 

[BHD]

YouTube - WORLDS FASTEST LOGSPLITTER here is another fly wheel splitter, most likely the idea for the one posted above a few posts,

 

[jdjllee]

I think your fatal flaw is in the last step where you take the stored energy and try to convert it to force using your pinion radius. I would also propose that the problem is more complex than it seems as there is a prime mover adding energy to the fly wheels at all times and driving the ram regardless of the fly wheels. The engine is turning several thousand RPM yet the flywheels are low speed at 300 RPM so there must be a pretty good reduction gearing and some power applied to the ram ignoring the flywheel stored energy. But to take a look at the effect of the flywheels alone let's just ignore the engine as you suggest. The formula I would suggest you play with is T=Ia where I is the moment of inertia of the wheels and "a" is the angular acceleration ( de-acceleration really) of the wheels as the ram strikes the wood and plows in til it cracks the wood or stalls the wheels. Let's just assume a complete stall and a constant de-acceleration during the stall. This happens very quickly and the quicker it happens the greater the impulse applied. The harder the wood the faster the stall. If you try to split a steel block the stall time will be so short, the impulse so large, the machine will likely be destroyed. I say play with the formula because the nature of the problem is such that the time to stall and therefore the acceleration is unknown. But you can play with it and vary the stall time to see the trend. The inertia I is known from the weight and geometry of your wheels. I would suggest you use a formula for I that will give you units lbinsec*2. I = wR*2/772.8 for a solid disk flywheel. From this I calculate I=10.48 lbinsec*2 for w=100lb wheel and R=9 inch radius. So I =21 for two solid wheels. (I can be made much larger with proper design wheels) Now moving on to "a", "a" is the angular acceleration of the wheels. If the wheels are rotating 300 RPM they are rotating 31.4 rad/sec. To get an idea what happens as you strike the wood and stall the wheels you can assume this stall happens in 1 second (an unreasonably long time for this machine perhaps) so assuming a constant acceleration thru the stall you have; a =31.4 rad/sec /1 sec = 31.4 rad/sec*2. Now you can plug this in to the formula T=Ia and get a torque required to stall the wheels in 1 second. Using the pinion radius you can then calculate a force on the ram from this torque. Now you can do it all again but stalling the wheels in .1 seconds, then .01 seconds and you can see the force gets huge as the time to stall the wheels decreases. I got 438.6lbs for 1 sec stall. (Note below that the machine will run out about full stroke in that time under no load), 6580lbs for .1 sec stall, and 65,814lbs for .01 sec stall. And you can see you have probably destroyed the machine if you stall it as quickly as .01 sec. You get the idea. When I calculate free running ram speed based on 3 inch pinion pitch diameter and 300 rpm wheels I get about 47 in/sec. So the machine runs out full stroke rather quickly, on the order of a second under no load. In operation , the impulse times when the wood is struck and cracked are going to be short but the tonnage in that instant will be large as above. I have over simplified it for sure. My lack of mathematical rigor leaves much to be desired and some of my Professors (deceased) might be turning over in their graves

 

[SPYDERLK]

I think your fatal flaw is in the last step where you take the stored energy and try to convert it to force using your pinion radius. I would also propose that the problem is more complex than it seems as there is a prime mover adding energy to the fly wheels at all times and driving the ram regardless of the fly wheels. [snip]

Good! ... [ I agree about the prime mover input as well.] That had occurred to me too. I think theres some sort of slip clutch tho, so prime mover mass is decoupled greatly in proportion to the magnitude of the rapid force event. So steady state drive torque makes almost no difference in a ram strike.
larry

 

[DuckHunterJon]

Here is a neat 'flywheel' log splitter - no idea how it works - no ram seems to be involved

YouTube - Flywheel Wood Splitter

-Nick

Forget the splitter, I just want wood that splits that easily. I'd love to see a video of some nice hickory or twisted up elm being split.

 

[LD1]

Forget the splitter, I just want wood that splits that easily. I'd love to see a video of some nice hickory or twisted up elm being split.

From everything I have read from owners that have the SS, this splitter holds its own.

There is a lot of reading on AS about these splitters. AND guys that have made their own. They say rarley do they have to whack a peice twice. Adn even if you had to hit a tough one 3 or 4 times, it is still faster than a hydraulic unit.

 

[SPYDERLK]

Long story short, I am wanting to try to build a flywheel/inertia logsplitter.
About the only thing I will have to buy is the rack/spur gear and an engine.

But I was just trying to calculate how they come up with the splitting force and seem to be at a loss here.

===================================================
Convert that to ft-lbs and come up with 606 ft-lbs of stored energy.
===================================================
Since the pinion/spur gear is 3" diameter (or 1/4 of a foot) 606 x 4 is 2424lbs of force along the circumfrence of the spur being transmitted to the rack gear. I know that is not nearly enough to split wood. And SS rates the splitters @ 12-24 tons. So...What am I missing here??? I know it is probabably something simple that I am just over looking and I am sure you guys will come through for me once again.

Thanks for going thru all that. Your equations cover the right stuf, so Im trusting your answer. The part I have set off is all you need to play appropriate head games. The E tells how much work you can do. The gearing has nothing to do with it and will mire you in confusion til that insight hits. You have figured out that your F x D product is 606. So 48000 x D'=606. D=606/48000 = .013feet [less mechanism friction force in that distance]... or at 12T, twice that ... or at 48T, half that. ---- Probably they are saying that their splitter will pop a ~18" log that takes 12-24T to get to crack. In many cases in that range the extreme drop in force at pop will allow the ram to follow thru for a full split.
larry

Hi LD1. I just saw this thread again and went thru it. It seems to me that the alert you sensed was actually a mix/match application of Energy as Torque. Yes? - - Did you ever make one?

 

[LD1]

Hi LD1. I just saw this thread again and went thru it. It seems to me that the alert you sensed was actually a mix/match application of Energy as Torque. Yes? - - Did you ever make one?

No I never made one. I made a conventional hydraulic one....only with a 4"cylinder paired with a 22gpm pump to get cycles about twice as fast as typical store bought ones

 

[Renze]

Anyways. What you are missing in the calculation, is that you calculate the max CONSTANT force the flywheel can excert over the full stroke. It can excert far greater peak forces when it is stopped by a knot, theoretically infinite unless you take deformation of the crank and rod into account.

When you put a pressure gauge on a hydraulic splitter you will see that it hardly ever reaches full pressure.