Four Motor Maslow.. sort of

For 12 feet of 1/16-inch rope, which is a little more than you need for a 12-foot top beam, a 4-inch spool end (see below) seems to work well. My spool cylinder is a little over 1.5 inches in diameter so that I can use these for mounting it to the motors (MDF isn’t strong enough to directly mount to the motor’s shaft… I tried, didn’t work well):

If you go to 1/8-inch rope, you need ~5-inch spool. I use this website for the calculations:

http://www.giangrandi.ch/soft/spiral/spiral.shtml

For the bottom motors, I don’t think you have to go with 12-feet wide spacing since all its doing is orienting the sled vertical and assisting the pull into corners… For that, I’m planning on a 9-10 foot “bottom beam” and made some 3.75-inch spool.

I designed the spool cylinder such that the wrap around the cylinder stays “neat” in hopes of maybe one day being able to calculate how much is spooled out using only the motor’s encoders… but as I’m progressing through my testing, I’m finding it very challenging to maintain a perfectly neat wrap around the spool and think it’s unlikely to be reliable enough to depend on.

image

The center piece (spool cylinder) is 1/16-inch Delrin and the ends are 1/4-inch MDF. I started with 1/8-inch ends but they were too flexible and the force from the rope pushed them apart to allow the rope to slide past the previous layer resulting in some binding… Going to a 1/4-inch really helped… hopefully that will be all that’s needed.

1 Like

if the only purpose of the bottom chains is to keep the sled vertical, then have
the top lines join at a single bolt on the top and the bottom lines join at a
single point on the bottom. That significantly simplifies your math.

David Lang

2 Likes

The math is really simple as it is.

*aChainLength = sqrt( pow( w/-2 - (xTarget-dxy), 2.0) + pow( h/2 - (yTarget+dxy), 2.0)) - clip; //topLeft
*bChainLength = sqrt( pow( w/2 - (xTarget+dxy), 2.0) + pow( h/2 - (yTarget+dxy), 2.0)) - clip; //topRight
*cChainLength = sqrt( pow( w/-2 - (xTarget-dxy), 2.0) + pow( h/-2 - (yTarget-dxy), 2.0)) - clip; //bottomLeft
*dChainLength = sqrt( pow( w/2 - (xTarget+dxy), 2.0) + pow( h/-2 - (yTarget-dxy), 2.0)) - clip; //bottomRight

(dxy is the horizontal/vertical distance from the center of the sled to where the cable exits the spool. these equations assume the spools are configured in a square arrangement… but that could easily be changed if needed). It’s simple because it assumes there’s no tilt in the sled

But the advantage to having the motors spaced to the corners is that it moves everything away from the center of the sled, giving room for the router and a clear downward path for the dust collector hose.

1 Like

Thanks, those numbers are helping me visualize it a lot. Did you make it so that the rope stacks on top of itself into only one layer? ie each turn lands exactly on top of the previous one, not like this:

image

Edit: More like a roll of tape where each bit of tape is directly on top of the lower layer…sorry this I’m sure there is some vocabulary to express exactly what I’m trying to ask

As I understand it, he’s planning for something where each wrap is beside the
prior wrap, not on top of. With a relatively large spool and short length of
line, this should not require a very long spool. This should result in a
constant diameter.

David Lang

I know exactly what you are asking. I’m trying to use the “roll of tape” method. As much as I’d like to do a constant diameter spool, I think the complexities of keeping the winding neat is too great (though I have some ideas, but not sure if they can be realistically implemented). To do something like above, I think you’ll need a level winder to achieve such a neat wrap and that makes a build much more difficult. As I alluded to in post above, the hope for the “roll of tape” method was to be able to calculate how much rope was spool out at any given time based upon the formulas used in that website I linked to. I just think at this point that’s too optimistic and am just focusing on using external encoders and wrapping cable up in the spool as neat as possible without overlaps/pinches/etc.

2 Likes

cut a grove in the spindle for the rope to follow and keep tension on it and you
should be able to have it pretty well controlled.

David Lang

search on youtube for ‘cable driven robots’ and you will find videos of a lot of
heavy duty robots, but you will also get some videos of some small ones that
show the cable on a spool with a single layer of winding

David Lang

1 Like

hmm… lost my post… I’ll abbreviate what I was typing.

That’s something I’ve looked at quite a bit over these many months, but I have yet to find a grooved spool that I can use. I’ve spent hours/days looking and the closest to a COTS spool is one that’s used for garage door openers. If anyone has suggestions, I’m all ears (my 3D printer is a bit challenged at the moment and not sure I could print one… long story). I think as long as you don’t constrain the rope once it comes off the spool (i.e, pass it through an eyelet or rollers), it should wrap neatly.

It would be beneficial to use a constant diameter spool as you could eliminate external encoders and would have “simple” math for kinematics. You’d just have to accommodate specials such as where the rope comes off the spool, like we currently do for chainAroundSprocket, and for improved accuracy (and I don’t know it really will amount to much) consider the deltaZ component between the frame mount point and where the rope is on the spool.

I agree with the idea, but I think that in practice an encoder on the line is the way to go. Ensuring that the cable never coils incorrectly seems like a challenge, and the extra physical space taken up by that type of roller becomes significant when we try to fit everything on the sled.

If you are talking about 1/16" rope, than you need a 16 pitch grouve on the
spool, that’s trivially cut with any metal-working lathe.

If I am doing the math right, a spool will hold pi * D * length / rope diameter

of pi * 2 * 1 / 1/16 = 100" of rope (need ~108" so one wrap over 1" of spool
length)

or, 1.5" fine thread bolts are 12 pitch, so

pi * 1.5 * 2 / 1/12 = 113" so 1.5" diameter bolt using 2" of thread will work
(the line’s distance from the workpiece varying up to 2", but since you can make
the max variation be at the longest distance, you should be able to keep the
angle down)

or, you can get 8 pitch bolts (8UN) up to 4" in diameter

pi * 4 * 1 / 1/8 = 100

one problem with trying to stack the layers is that the rope will compress,
depending on how much tension is on the line as it wraps the next layer up, so I
do’t think you will get a consistant length/angle function.

David Lang

1 Like

Well, I’m liking the idea more and more. Last night I spent a lot of time trying to get the end stops incorporated into the pillars to handle the “slack” condition and it’s getting really, really complicated. Sometimes you have to take a step back and re-evaluate.

My main goal is to avoid the need for custom fabricated parts where you have to pay a company to mass produce a quantity to get it affordable. Now, I’m thinking that as with laser cutting services, you can readily get parts 3D printed and even if they are expensive to produce as one-offs (or four-offs), 3D printed spools can probably eliminate $125-$150 in parts needed for using external encoders (encoders, laser-cut mounts, bolts, etc.) and it might just turn out cheaper…

I’m thinking of something like this:

image

I’ve got 1/4-inch D-Shafts (can’t seem to find 8mm shafts for whatever reason), couplers, and bearings so I have all the parts I need to make it once I get spools designed and printed.

The spools will be under an enormous amount of pressure from the rope… my monoprice mini delta can print PLA and ABS. Do you think that will be strong enough?

2 Likes

http://www.xportcam.com/

https://www.youtube.com/channel/UCPwgdpccKsyzC_gTntAlKcw

3 Likes

I was able to print a spool, but the scale is off a bit and the dshaft won’t fit … I’ve struggled with getting the scaling correct on this printer to accommodate PLA shrinkage. It seems challenging to get a specific outside diameter and a specific inside diameter to come out correctly. I’ll try some calibration cubes to see if I can get it to work better.

The design I was a 1.5 inch spool body with 0.085 in. diameter grooves to accommodate the 0.0625 in. rope (1/16th inch). I wrapped the rope by hand and it seems to wrap neatly, but I think @bar might be right about the risk of something going wrong with the wrap for some unforeseen reason. So, I think I’m going to try to make the grooves deeper (same diameter, just deeper into the spool body) such that the entire rope is “embedded” in the spool… roughly like this.

image

.

1 Like

Is there a chance a ribbon shaped cable could work? Pivoting on the end if need be. Like a seatbelt or cargo strap material… But much smaller.

1 Like

That could actually solve LOTS of problems if we can find a source for low/no-stretch ribbon. 1/4-inch to 1/2 inch width would probably be all that’s needed.

2 Likes

Best I could find was this: 3/8-inch UHMWPE webbing (I thinky dyneema is UHMWPE). Bit pricy ($60 for 50 feet)

2 Likes

http://www.machiningnews.com/2014/04/twincat-3-unites-pc-control-and-science-for-simplified-cable-robot-operation/

http://www.cablecam.com/News.aspx?id=114

5 Likes

I like the ribbon idea, very interesting. It might be harder to find a low stretch ribbon, that geometry just feels less stable to me, but it would be a big help.

The webbing I linked to claims to be 1/32-inch thick, which seems really thin. If true, great as it’s rated to be pretty strong. I think I’m going to order some and see how it works if I can’t find a different, lower cost source. A 3/8-inch webbing would work really well in my current design and solve a number of problems.

2 Likes