I’ve only got 4 spare NEMA-17s in the garage, I need to step (pun intended) up my stepper motor game.
I hope you post when you’ve got your Maslow set up, I’d like to see how you approach the steppers & counterweights.
how would you automate this measurement? The sled is too heavy (and of
unpredictable weight) to just pull the chain tight
nitpick, it’s only constant if the holes are round (not ovaled 
)
here are advantages to both approaches, we’re beating on them to figure out what
they are.
If the motors are running at full power, each one will put 66 pounds of force on
it’s chain (torque/radius of sprocket), we have seen people literally pull their
sleds to pieces
that shackle will not work with #25 roller chain.
Think bicycle chain, but half the size (1/4" per link instead of 1/2" per link)
attaching the chain in such a way that we have a known pivot point is non-trivial. That’s why I design the arms to be cut out of 3/16 metal, that’s the thickest metal that you can use with a master link, which gives you a known pivot point at the hole.
The stock maslow sticks the entire chain through the hole, then uses a pin through the chain to keep it from coming out. This doesn’t give you a consistant length or pivot point (small errors to be sure, but we are trying for tight accuracy)
it adds enough complications that I would want to avoid it if we can reach ‘good enough’ with one of the existing designs.
one thing is that the Z axis bracket normally anchors to the sled about where you have the bottom right pantograph mount 
another thing is that in the top mount, all the critical dimensions are in a vertical line, so you could cut everything in one place, moving the plywood sideways for each part. (including the critical holes on the sled)
it’s actually far easier, take a piece of 2x4 and turn it on edge, mount the arms to the 2x4 (tight fitting lag bolts with a good shoulder would work well here), and then anchor the 2x4 to the sled
something like this:
I think the 2x4 on edge like this is about the right distance for balance, and it can be as long as we want (so you can have long verticals to reduce the error)
This link onshape top pantograph assembly should let you grab an arm and move it.
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sounterweights can help, but I question how much (the mount of force needed varies wildly)
the bigger problem with steppers is that when they loose power, they move, so you have to do a re-clalibration of the chain lengths every time you loose power. With the worm drive, the chains don’t move when you loose power.
I am not talking about what the motors are capable of. I’m talking about tension forces in the chain as the angle between the chains increases. This happens to be pretty easy to calculate.
To help envision why the force on a chain increases as the angle of the chain increases imagine tying a rope to a bucket of water then lifting the bucket to the height of a door knob. The force (tension) on the rope and the force your muscles need to exert to lift the bucket are each equal to the weight of the bucket.
Now imagine if you pass the same rope through the bucket handle and tie it to a door knob. Stand back holding the other end of the rope with the bucket between you and the doorknob. Keeping your hand at the level of the door knob you could pull the rope and lift the bucket but it would take quite a bit of work to raise a 10 pound bucket even a foot. Try raising the bucket so the handle is at the same height as the doorknob – you can’t do it!! In fact, the force it takes to raise the bucket to that height spikes to infinity. It literally can not be done in our physical world.
The equation to find the force on two cables under load at a given angle is:
F = (m x 0.5) ÷ cos(a x 0.5)
where:
F is the calculated force on each cable
m is the mass
a is the internal angle between the two cables
The dividend:
(m x 0.5) = (20 x .5) = 10
The divisor:
cos(a x 0.5) = cos(145 x .5) = cos(72.5) = .3007058
Divide for quotient:
10 ÷ .3007058 = 33.26lbs (per chain)
The 33.26lbs per chain is based on normal operation and does not account for drag. It is the force on each chain when the sled is in the top center resting position. It also doesn’t account for runaway software that runs a motor until things break or poorly built sleds or crappy wood or anything else.
If you see a problem with my math or can show me how you calculate 66 lbs on each chain I would love to hear it.
Thank you, I know quite well what #25 roller chain is. Honestly I figured someone would probably mention that you can’t fit that shackle through #25 chain, or that even if you could fit it the orientation of the shackle is off by 90˚. I assumed that was clear and that a jumper ring would need to be used. It could be anything really, like a keyring or a piece of wire, heck it could be a stale piece of free range, fair trade, open source, gluten free bread as long as it could withstand the tension!!! Adding a reasonable link between the end of the chain and the shackle (pivot point) will not change the physics.
It may be non-trivial but it doesn’t have to be complicated. A shackle through plywood is quite accurate, especially if people don’t have access to precisely cut and drill 3/16" metal.
What prevents you from doing this with any other linkage design? The bars are all straight, they could all be cut the exact same way. All of the measurements are critical, even the holes on the arms (“horizontal” pieces) right?
I’m not sure if the benefit of having the Top Mounted version out of the way of the router is enough to outweigh the chain attachment point problem (very steep angle at maximum travel) and the compression/tension slop (which results in the router being below the tip of the chain triangle). Maybe it is worth it? Both of those things can be solved. Who knows… I can’t wait until more people with Maslows start building these different designs and testing them!!
I’d try it right now but someone would have to send me a free Maslow first 
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however, the static tension is not an accurate measurement of the forces involved. the max tension the motors can provide can show up when accelerating (especially when reversing direction) or when the router hits a toughter chunk of wood to cut through. your static analysis significantly underestimates the forces involved.
I wasn’t trying to estimate the highest possible force that could hit the system in a perfect storm. I was simply getting a baseline max static tension on the chains (which is why I stated that it didn’t account for drag or friction or anything else) - I believe that to be a little more than 33lbs on each chain if you have a 20lb sled. That’s all
I’m curious how the 66lb number was calculated? I might be wrong and I’d like to know it if I am!
I like your design! Do you have a way to make those parts? I know a guy with a water jet, I don’t know what it would cost though… Of course it could be done by hand too, a little careful marking and drilling would work just fine. I’d really love to see a metal one working!
That’s an important question for any of these approaches.
The lag bolt mounting better be solid, but as long as it is that should work. The potential for side to side wobble still makes me a little nervous, though.
This looks like a good design for those with access to the steel.
Uniformity of the hole distances is critical and the bars with 3 holes MUST be in a straight line, but the measured distance between points, doesn’t affect the geometry. A parallelogram is a parallelogram as long as the opposite sides (determined by our pivot points) are equal. This is why I wanted to model building it and clamping the arms, drilling all at once.
The link just went to a login page for me
I will, right now I’m gathering materials, deciding on frame design, and clearing space in my shop area to set it up. I plan on doing some testing of the motors and sled design options so I will post those results as well.
20 Kg/cm torque with a 1cm radius sprocket
There was mention much earlier of a person who does laser cutting of parts pretty cheaply. I haven’t had a chance to contact the person mentioned. But I would expect that in moderate (50ish was mentioned) quantities these would be fairly cheap
remember that almost all the force is going to be on the chains, pantograph, and mounting block. The amount of force between the mounting block and the sled is much lower.
create a login (or login if you already have an account) and then the link will take you directly to the assembly
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correction, 30Kg/cm torque for the motors
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