Engineering/physics formula questions about a Supersplit

[LD1]

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.

 

[Craig Clayton]

In can follow with your formula mostly sort of. In my opion what you could be missing is your FINGERS, how do you stop the inertia if something goes wrong with knotty would in the middle of the stroke. I built a hydraulic electric splitter and it is sitting in my barn but disassembled but I can wing the specs. Very quiet and SAFE.
Craig Clayton

 

[LD1]

In can follow with your formula mostly sort of. In my opion what you could be missing is your FINGERS, how do you stop the inertia if something goes wrong with knotty would in the middle of the stroke. I built a hydraulic electric splitter and it is sitting in my barn but disassembled but I can wing the specs. Very quiet and SAFE.
Craig Clayton

Simple. The flywheels are driven by a centrifigual clutch. So if a knotty peice stalls the ram, the flywheels will stop as well and no longer have any stored energy. Simply raise the handle, let the flywheels spool up, and smack it again.

Their is a lot of people who have bad things to say about how dangerus these things are, but in reality, their are far more dangerous things associated with cutting wood.

But right now, I am missinsg something with the calculations. I know it develops more than the 2424lbs of force as calcuated. So I am missing something obvious. It wont stop me from building it, cause it is proven to work, it is just one of those things that will drive me crazy until I figure it out.:laughing:

 

[om21braz]

Been too long since I've had to make any calculations such as you have. Are you planning on driving it with some type of hit-and-miss engine that is in the 300rpm range at 1:1 or a typical gas engine turning far more rpm and going thru a gearbox to get the 300rpm? I just question the speed, as the speed with which the ram pushes the wood to the blade is everything. The pitch of the gear/rack is going to determine how far one revolution of the gear actually moves the ram in linear measurement. Different pitches will change the speed quite a bit.

 

[HenryBrink]

What you are missing is the mechanical advantage of the "screw" The distance of screw thread vs the length of the screw. the screw is 3" in diameter and makes two revolutions per inch. 3*3.14 = 9.42 circumference *2 divided by 1" = mechanical advantage of19 x ft*lbs of force or about 22 tons with some force being lost through friction

Henry Brink

 

[BHD]

I was trying to figure out a similar set of numbers the other day for the energy of a a dropped weight, and first of all it does't not really come out in pounds or kilograms, but in newtons, and it can be converted but I do not know if it really converts to mass or weight, and some of it is how fast it is stopped, or how far it penetrates, the substance it hits,

I called SS and they said the unit works much like a maul and wedge being pounded into the log,

and there videos show wood that splits with one blow, I asked about elm wood, (which can take a hydraulic splitter a lot of work to push through with force, as it slowly splits) they said theres can do elm, but you would need to send the ram in many times to work your way through the wood, much like repeated blows on maul,

my guess is that if you had "X" wood and it would take a 12 to 24 ton hydraulic splitter to do that wood theres would do the same,

the trick is in the speed and the momentum and stopping that momentum,

if you stopped the ram in say 1/10 of an inch, or less, and stopped the fly wheels and all stone cold, it would equate a lot of tonnage,
hydraulic is just that pressure, as the speed is nearly no existent, it is just force, but the SS is weight in motion,

(say your figures are correct, and take that 2400 pound and put that in motion moving (I think they say the cycle is 2 seconds), it is moving at a reasonable rate of speed, and you in a since have a ton of force moving, now stop that weight in like I said a fraction of a inch and you have multiplied that force many times over.

it is like taking your 8 pound splitting maul or sledge and swinging it, how much is your 8 pounds multiplied into, when it hits the wedge, it is enough to split many types of wood with one blow,

note: the wedge on the SS is narrow and not that tall so to amplify the effect of the stroke, of the ram, most hydraulic splitters have tall wide wedges on them that would take more force to drive into and through the log,

 

[LD1]

Been too long since I've had to make any calculations such as you have. Are you planning on driving it with some type of hit-and-miss engine that is in the 300rpm range at 1:1 or a typical gas engine turning far more rpm and going thru a gearbox to get the 300rpm? I just question the speed, as the speed with which the ram pushes the wood to the blade is everything. The pitch of the gear/rack is going to determine how far one revolution of the gear actually moves the ram in linear measurement. Different pitches will change the speed quite a bit.

The flywheels turn much slower than the motor because the flywheel is ALSO the pully. And that is driven via belt and a small pully on the motor. No gearbox required.

What you are missing is the mechanical advantage of the "screw" The distance of screw thread vs the length of the screw. the screw is 3" in diameter and makes two revolutions per inch. 3*3.14 = 9.42 circumference *2 divided by 1" = mechanical advantage of19 x ft*lbs of force or about 22 tons with some force being lost through friction

Henry Brink

You lost me here. Their is no screws. Just straight gears. The pinion is 3" diameter. (which it just dawned on me that the force would be double what I posted earlier because I was using diameter rather than radius.) So a R of 1.5" is 1/8th of a foot. So that 606 ft-lbs times 8 gives me ~ 4848lbs force.

But back to your numbers. The circumfence is 9.42" for the spur, or screw as you call it. So what exactally is making 2 revolutions per inch??? The spur is directly coupled to the flywheels. and mated to a linear rack. So 2 reolutions of the "screw" or spur gear is ~19" of ram travel.

if you stopped the ram in say 1/10 of an inch, or less, and stopped the fly wheels and all stone cold, it would equate a lot of tonnage,
hydraulic is just that pressure, as the speed is nearly no existent, it is just force, but the SS is weight in motion,

(say your figures are correct, and take that 2400 pound and put that in motion moving (I think they say the cycle is 2 seconds), it is moving at a reasonable rate of speed, and you in a since have a ton of force moving, now stop that weight in like I said a fraction of a inch and you have multiplied that force many times over.

it is like taking your 8 pound splitting maul or sledge and swinging it, how much is your 8 pounds multiplied into, when it hits the wedge, it is enough to split many types of wood with one blow,

note: the wedge on the SS is narrow and not that tall so to amplify the effect of the stroke, of the ram, most hydraulic splitters have tall wide wedges on them that would take more force to drive into and through the log,

I think your onto it here. The faster the ram comes to a stop, the more instantanious (impact) force their is. So I think I can actually calculate it now. All the formulas I was running across had to do with either acceleration OR rate of speed change. I was looking fo something with just velocity of the ram. But it didn't dawn on me at the time that their actually is a velocity change, or decelaration as you will. The quicker the decel, the more force. So theoretically, something that stops it cold in its tracks, is generating the MAX tonnage the machine can be rated at.

 

[DuckHunterJon]

I could be off in left field (often am), but don't you also need to take into account the motor running? While much of the energy is stored in the flywheels, the motor is also taking on some of the work since it's directly coupled (through the belt) to the flywheel - or is it decoupled through a 1 way clutch or other mechanism?

 

[LD1]

I could be off in left field (often am), but don't you also need to take into account the motor running? While much of the energy is stored in the flywheels, the motor is also taking on some of the work since it's directly coupled (through the belt) to the flywheel - or is it decoupled through a 1 way clutch or other mechanism?

Yes it needs to be counted but it is relatively small. A 5HP motor is only ~7.5 ft-lbs of torque @ 3600RPM. And with 12:1 reduction to the flywheels, this would add ~ 90ft-lbs to the flywheels. With the 3" spur gear, this would gain us ~ 720lbs of linear force to add to the 4848lbs fron the flywheels. So the ~5500lbs of force is still no where near the 12-24tons these things are rated at.

And upon further calculating and trying to figure this out, I am not getting anywhere. With all the formulas for force, impact force, etc. Basically, with an impact, you cannot have a sudden stop. (a value of 0 for either the distance or time). Something has to give. Wether it be the solid object deforming a bit, or the ram. So this makes it difficult to calculate.

And the more I think about it, I really dont think I consider it an impact force like the maul or axe example. While it is fast for a splitter, a 3" pinion @ 300rpm is only 47.1" per second. So ~ 4fps (1.1m/s) is not very fast when trying to figure impact force. And watching all the youtube videos, it doesn't really look like an impact force either. Once the inital contact with the log is made, impact is over. It is just a stout push to finish the job.

But I am sure I am still missing something.

 

[HenryBrink]

Some racks are pitched. For example if the spur makes one revolution, ie travels 9.4 inches does the rack move more or less then 9.4 inches. If it isn't pitched ( perpendicular) and you have 1:1 ratio. If the rack (screw) is pitched, then one revolution of the spur drives the rack 1/2 the distance, then your MA is 2. The other mechanical advantage would come from the the change in drive diameters. IE 18" input/3" output = 6 MA . So your input force would be multiplied by each of the MA generated. The flywheels store the torque, so that the engine doesn't bog down. Looking at the video of one operating it would appear to have very little deceleration. Splitting wood by hand involves the ratio between the time of the velocity changes. IE the acceleration of the ax takes 1 sec/ the deceleration .01 sec or MA of 100.

 

[HenryBrink]

It appears the difference in the way the flywheel splitter and a hydraulic splitter works has to do with power. Power is work/time. If the power to split wood is the same, and the distance the ram travels is the same, then the flywheel needs less force because it does the work in less time. The hydraulic uses more force because it is applied over a greater time.

 

[Transit]

LD1, I'm taking a stab here, from my experience back in the old days, I found it much easer to do all the calculations in one system and not converting back an fourth. If you want the results in pounds, feet, etc. do not work with meters, Kg. There is a possibility that you are missing a conversion factor. Just a thought.

 

[LD1]

Some racks are pitched. For example if the spur makes one revolution, ie travels 9.4 inches does the rack move more or less then 9.4 inches. If it isn't pitched ( perpendicular) and you have 1:1 ratio. If the rack (screw) is pitched, then one revolution of the spur drives the rack 1/2 the distance, then your MA is 2. The other mechanical advantage would come from the the change in drive diameters. IE 18" input/3" output = 6 MA . So your input force would be multiplied by each of the MA generated. The flywheels store the torque, so that the engine doesn't bog down. Looking at the video of one operating it would appear to have very little deceleration. Splitting wood by hand involves the ratio between the time of the velocity changes. IE the acceleration of the ax takes 1 sec/ the deceleration .01 sec or MA of 100.

I am not sure what you mean by pitched racks. The teeth either have a 14.5* or 20* pressure angle. It is a straight 1:1. The rack will move 9.4 inches per revolution. No MA their.

The flyhweels are "storing" 606 ft-lbs of torque as mentioned earlier. A spur with a radius of 1.5" is 1/8th of a foot. So 606 x 8 = 4848lbs of force @ that 1.5" away from the center pivot.

4848lbs is still about 6-12x's less than what the SS model claims. I am obviously missing something and I am at a loss for now.

 

[om21braz]

I am not sure what you mean by pitched racks. The teeth either have a 14.5* or 20* pressure angle. It is a straight 1:1. The rack will move 9.4 inches per revolution. No MA their.

Your particular gear moves the rack 9.4" per revolution. A gear with a different diametral pitch would move it a greater or lesser distance.

 

[dynasim]

Stored energy can only be converted to force if given a distance(or time if a velocity is known). The ft-lbs in stored energy is only indirectly related to device diameter.

If 606 ftlbs of energy were dissipated(by hitting a log) in 1 inch of travel, the average force would be 7200 lbs(606 ft lbs/(1/12 ft)). As is observed by hitting something with a sledge hammer, the impulse force as the hammer stops(by hitting the wedge) can be nearly infinite.

That is why spongy wood is much more difficult to split by hand.

Chris

 

[LD1]

Your particular gear moves the rack 9.4" per revolution. A gear with a different diametral pitch would move it a greater or lesser distance.

The spur and rack have to have the same diametral pitch. So it dont matter weather they are 6 pitch, or 32 pitch. A 3" spur gear is going to move its mating rack that same 9.4 inches.

Now if we were talking a 18tooth spur of a different pitch, that would be a different story. But the diameter would NOT be 3" and therefore the circumfrence would NOT be 9.4.

 

[LD1]

As is observed by hitting something with a sledge hammer, the impulse force as the hammer stops(by hitting the wedge) can be nearly infinite.



Chris

This is kinda what I was thinking as well. But it would still be nice to know exactally what they use to come up with their "rated" tonnage.

Maybe I'll send them an e-mail.

 

[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 probabally 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

 

[dfkrug]

Basically, with an impact, you cannot have a sudden stop. (a value of 0 for either the distance or time). Something has to give.

You are correct that you do not have instantaneous transfer of energy
from the splitter to the wood. If you graphed the force vs. time
experienced at the point of impact, you would see an initial spike, then
a rapid fall off to zero as the wood fractured. That would apply to
some pieces of wood anyway. Others that are more fibrous, or with
knots would experience an initial drop in force, but then the F v. T
line would decline much slower until the wood fully split. The latter
piece would have high "toughness", the energy needed for deformation.

 

[LD1]

they are saying that their splitter will pop a ~18" log that takes 12-24T to get to crack.

I suspect this is all they have done when rating their splitters.

Either way, I think I am still going to try to build one. Just have to come up with suitable flywheels and go from their. I pretty much have everything else figured out how I want to do it.

Allthough I am not sure weather they are just using a steel rack, or wether theirs is hardned at all?? I can pick up just a steel rack and pinion gears that are 6 pitch from Mcmastercarr for ~$150. But I dont know what the brand is. Some other companies like motion and grainger have 6p gear racks that are over $300, so I am not sure what the difference is yet. I dont want to spend $300-400 for a rack and spur If I dont have to. But I also dont want to waste $150 on a set that will wear out within a few cords of wood.

Another question for you fellows if you know the answer. They make two pitch types. 14.5* and 20*. Everything I have read says the 20* are better for higher loads. So what exactally is that a measurment of? Angles of the gear teeth? If so, wouldn't a 20* rack have a higher force trying to un-couple the rack and spur?? Which I have read has been and issue on some other homemade SS knock-offs. Bending the cam-lock assembly that holds the rack into the spur.