I also read a post that warned about not “reversing” the tension on the sprocket. That avoids any backlash in the gearbox.
what are the units of tension in this spreadsheet? KG.cm? Ft.lb?
Tension units are a force, so they are lbf. Essentially they are the geometric component of the force in B7 (and B25 in the diff version).
Note that if you want a metric version of this spreadsheet, just make a copy and replace all the cells that say “in” with “mm” and all the cells that say “lb” (or should) with “kg” and then adjust the numbers in B3:B7 and B21:B25 to the metric equivalents (or leave the numbers where they are and build a really small Maslow with a surprisingly heavy sled).
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1kg is 2.2 lbs and
the radius of the 10T gear sprocket is about 1cm.
motor manufacture says these motors have a torque of 30kg.cm
so that means they can pull a 66 lb weight? at the end of the 25# roller chain? seems hard to believe.
With two motors working together that means 132lbs of lift potential?
Yeah, that doesn’t square with our experience. In fact, that seems almost an order of magnitude high.
The two motors have literally ripped sleds apart for people when things have
gone wrong enough, 132 pounds of force at stall is not unreasonable.
they are small motors, but almost 300:1 on the gearbox gives them a LOT of
torque.
David Lang
I think you need to do more than that, does this include anything in it that
mixes weight and distance?
David Lang
the tension is intended to just keep the slack in check. But when you use
elastic pulling on the chain, it’s applying the most force when you need it the
least, and the least force when you need it the most.
switching to counterweights won’t let you get by with smaller motors, but it
will avoid the tension on the slack side being higher than the tension on the
sled side, which will cause inaccuracy as you pass the point where the tensions
are equal.
you want to have counterweights, and you want the counterweight to be
effectively less than the minimum tension (which means if you have the chain
doubled, and a min tension of 3.2 pounds, you can have up to about 6 pounds of
counterweight)
David Lang
can you make two versions of that graphic, one with the sled in the top center, one with it in the bottom corner to show the two most critical locations (it may be worth showing a top corner, just to show that it’s not as bad as the other two positions, even though it is the shallowest angle for the far chain)
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low corner suffer from low tension on the long chain. As the angle on the short chain gets closer to vertical the tension on the other chain gets smaller
Nope, it uses the distances to figure out proportions (nondimensional), then multiplies that by sled weight to come up with the tensions.
The way I like to think of the difficult places is that there are three forces acting on the sled in the x,y plane. 1) the left chain, 2) the right chain, and 3) gravity. If any two of those three get lined up so that they are getting close to parallel, you run into problems. At the top center the two chains are almost parallel. At the bottom corners, one chain and gravity are almost parallel.
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cough
i think i’ve got all the formulas updated, but i don’t know what conditional flags on the metric forces should be, so they’re not there right now…
Edited to add ability to switch between the stock machines on the fly, and enable use without copying it to your own page. (green fields are editable by anyone)
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I find this a very interesting topic.
The only thing is that I don’t totally understand what the outcome says. Despite the comments I read in this post. So please correct me if I’m wrong.
I want the tension to be as high as possible and angles as close as possible to 45 degrees?
If yes, than it should be possible to calculate the (theoratical) optimum of the set-up with something like Excel-solver (within the boundaries of my garage). Right?
In theory there should be a optimum ratio. From my opinion the ratio of a ply sheet, 4x8 = 1:2. However the Maslow design has some real life factors that will show deviation from the theoretical optimum.
We are more or less at 15° angle, have sled friction, chain sag, are limited by the amps the motor shield can handle and the power supply, a moving CG on Sled y-axis with triangular depending on what angle the chains come in and perhaps a few more I can’t come up with.
My search for the ‘holy grale’ from back in the days: Is it true that there is an optimum ratio between width and hight of the frame?
No, if the tension gets too high (top center, typically), the motors will have trouble moving the sled. Conversely, if the tension gets too low (lower corners, typically) then the motor isn’t having much influence on the sled (potential loss of control).
Most projects I have seen cut don’t even use up the entire 4x8’ sheet of plywood, so the easiest solution is to just use the central area and make a series of holders/clamps to hold wood where it is cut best.
or cut half the sheet 4x4 area in the center and then move it over and then cut the other 4x4 half.
if you really must cut the entire 4x8 sheet at once then move the motors up from 18" to about 30". That will make the machine about 90" tall and many ceilings are only 96" tall, so not much room left.
and have the distance apart about 144 inches or more. some garages don’t have much more space than that.
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that is a good explination (and also explains why adding a simple third motor or
bungee to the bottom center doesn’t work)
David Lang
how do you define “where it cuts best”? 
if you have a good area that you can get really good accuracy in, then let’s
figure out what the min-max force are in that area, build a machine that gets
those same min-max forces across the entire area, and see if we get equally good
accuracy everywhere.
If we do, we have a solution to the problem about what size the maslow ‘should’
be.
If we still suffer accuracy problems, we know to look at other causes.
David Lang