Engineering/physics formula questions about a Supersplit

[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.