Chain sag inaccuracy

@dlang
To store the cable, I was thinking of unhooking the cables from the sled and just hooking them on to something on the opposite end of the frame. Having to take things apart could certainly add more inaccuracy into the mix though.

remember, we are aiming for ~0.4mm of accuracy, and currently, chain sag is
thought to cause ~1mm of inaccuracy, so any ‘set’ or stretch in the line, or in
the connection needs to be substantially less than that. I don’t think it’s
likely to be a significant improvement over the current ± 1mm problem of sag

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Gotcha. The only other thing I could think of along these lines would be some type of worm drive. I did some digging on McMaster and found this: https://www.mcmaster.com/#6542k64/=1aft47n

Its basically a worm drive hose clamp except the slotted material comes in a ~100ft roll for $60 and you cut it to the length you want. The only tensile information given says that it will fail at 550 lbs.

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that stuff is designed for tighten once and leave, it wont last any consistent use.
why not just use a regular lead screw rod? sure its not flexible and thus would more room around/above/below your work area, but could be a way to prove/eliminate the source of the inarticulacy.

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is that likely to be as flexible as the chain? I would expect it to start
holding it’s shape, which will be far more error than the chain sag.

remember, once we can figure out how to measure the amount of chain sag, we can
allow for it in the software. The big problem is figuring out how to measure the
amount of sag that we have.

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With a 20 pound sled at the top center of the work area each chain (or cable) will have about 33.26 pounds of tension. (reference)

David Lang and I have a bit of a long-running disagreement on the forces produced on the sled. The 66 pounds that he is referring to is the stall force for the motors - it’s what the motors are capable of producing; it is not the force on the chains in a static system at the top center. The actual tension in the chains (33.26 pounds) is easy to calculate and is explained quite clearly in this post.

In order for a 20 pound sled to put 66 pounds of force on the chains the internal chain angle would have to be about 162.55˚, that puts the sled just over 217mm (8.54 inches) above the top of the work piece. So it’s easy to see that the system should not see 66 pounds of force under normal operation. I do agree that there will be brief, momentary, spikes in the forces due to variables such as knots, sled friction, path direction change, dull bits, etc. but this will not be constant and therefor any error will also be momentary. (this math is assuming a 20 pound sled with motors that are 440mm above the workpiece and 2900mm apart)

I have not only done the math but I’ve also done real world tests with inline spring scales and come to the same conclusions (some of which is documented here).

Again, if anyone can show me how they are calculating 66 pounds per chain I would love to see the math. I could be wrong and I would be happy to concede if that’s the case. My math, real-world tests, and logic shows me ~33 pounds per chain, and I’ve shown all my work and explained the reasoning so hopefully someone can show me where I’m wrong if I’m wrong.

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@pyrosrock
I can’t speak to the long term reliability of that strap but I agree that the one I linked would probably have issues since it uses ridges rather than an actual slot that is cut through the material. I thought about using a lead screw but I figured that the extra space requirements would keep it from being a viable option for many people who have their Maslow setup in a smaller space. It certainly makes a lot of sense from a precision standpoint.

@dlang
Yes, the strap would be less flexible than a chain and it would retain the shape that it is stored in. But it wouldn’t have to be wound up into a coil at any point after shipping. It could be straightened out before being used. The advantage over a lead screw is that the flexibility of the strap would allow it to deflect if pushed into a wall when the sled was in the upper left or right hand corner.

Cool! I didn’t know the chain sag was something that could be factored into the software. Before I ask any obvious questions on that front, is there a post or an article discussing that topic? Thanks!

@pillageTHENburn
Its hard to argue with theory that’s reinforced by empirical data. Using the wire rope stretch calculator, 33 pounds on a 1/8" diameter equates to 0.0155% stretch. If there’s 33 pounds on a 6 foot length of cable, that’s about 0.28mm of stretch. Additionally, the load on the motors could be used as a way to quantify just how much tension there is in the cables and use that to compensate for whatever stretch there is, right?

Cool! I didn’t know the chain sag was something that could be factored into
the software. Before I ask any obvious questions on that front, is there a
post or an article discussing that topic? Thanks!

there are posts talking about it (somewhere), but no articles

@pillageTHENburn

Its hard to argue with theory that’s reinforced by empirical data. Using the
wire rope stretch calculator, 33 pounds on a 1/8" diameter equates to 0.0155%
stretch. If there’s 33 pounds on a 6 foot length of cable, that’s about 0.28mm
of stretch. Additionally, the load on the motors could be used as a way to
quantify just how much tension there is in the cables and use that to
compensate for whatever stretch there is, right?

Chain Sag and cable stretch both depend on the same two factors

  1. the length of the chain/cable
  2. the tension on the chain/cable

In both cases, if you know these two factors, then you can calculate the
sag/stretch and compensate for it in software.

The problem is exactly how to determine the tension in the chain/cable since it
depends on the weight of the sled, which we don’t have a way to measure to any
accuracy. We can measure it indirectly by measuring how much the chain sags and
calculating how much weight there would need to be to have that much sag.

But I haven’t been able to think of any other way to measure this. We are trying
to detect very small differences in length (on the order of one part in 10,000)

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The problem is exactly how to determine the tension in the chain/cable since it
depends on the weight of the sled, which we don’t have a way to measure to any
accuracy

Just jumping in here (with all intentions to be very respectful), but how accurate do you need to be on the weight of the sled? I’d think that if you had calculations that could account for sag based upon the weight of the sled (along with length of chain which you should know at any given time) that even coming close to the real weight would be an improvement rather than a detriment. Can’t you just tell people to put their sled on a scale and measure to the nearest lb/kg? If you measure it as, idk, 20 lbs. and the real weight is 20.34432 lbs, I can’t see that the delta would throw off the very small sag adjustment so that it makes Maslow less accurate than before.

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I wouldn’t trust home scale to be accurate within ± 5 lbs. :wink:

That said, I’ll have to do some calculations to see how much it matters. It may
be that an error of 25% is ‘close enough’

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Might be able to get by with a ‘rule of thumb’ value based upon a sled with a R22002 router and two bricks.

Question though, is weight of the sled the only thing important? Just thinking about it (and not sure how much of a factor it would be), the chain that’s “pulling” the sled toward it would seem to me to have less sag than the chain that is just supporting the sled’s weight. And if it’s cutting something, there’s more resistance and therefore less sag? Again, not sure how much of a factor it would be in comparison to the weight.

I’m guessing the orientation of the chain would also play a role in the amount of sag. At the same length and tension, there should be more sag the closer it is to horizontal, right?

What I was trying to say earlier is that the current required by the motor will need to vary as the tension varies. In that case, we should be able to estimate the tension in the chain/cable at any point in time based on how much current the motor needs. I’m certainly not an electrical engineer though so I could be way off on this one.

I’m guessing the orientation of the chain would also play a role in the amount
of sag. At the same length and tension, there should be more sag the closer it
is to horizontal, right?

actually I’m not sure how much that hurts (although it may be that what matters
is the horizontal distance of the chain, not the total distance)

the bigger factor is that the closer to horizontal, the less force on it (or
more precisely, the closer to vertical the other chain is, the less force is
left to act on the long, close to horzontal chain)

it may be that both need to be taken into account.

What I was trying to say earlier is that the current required by the motor
will need to vary as the tension varies. In that case, we should be able to
estimate the tension in the chain/cable at any point in time based on how much
current the motor needs. I’m certainly not an electrical engineer though so I
could be way off on this one.

in theory yes, but since the worm gear prevents back-driving the motor, 0
current will tell you nothing about the tension on the chain.

you could possibly correlate the power to the motor and how much it’s moving,
but we’ve found that these motors are not that consistant, and the power to
generate a specific amount of force varies drastically based on how fast you are
moving (that’s the reason the PID loops were introduced, prior to that it tried
to generate tables to calibrate electrical power to motor movement)

The idea is a sound one, but we don’t ‘currently’ have a way to read the value. :smile:

Can you relate power to pwm frequency and percent, and compare it to actual speed from the encoder?

An interesting idea. That info is readily available.

This thread has gotten TLDR for me, so ignore this if it has been covered. Could you just add a wait measurement to the calibration prosses? Most people have a bathroom scale. Step on the scale with the sled and step on the scale without. Now you have the sled weight (within a pound or 2) which should be close enough to cover stretch conditions. Or am I wrong?

That would work. I use a fishing scale (never caught a fish as big as a Maslow sled, though :smile:). I confirmed the accuracy on the bathroom scale as you described.

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the tension on the cables/chains is not just the weight of the sled, but also effected by the angle of your frame/workpeice the ammount of friction between the sled and the workpiece and the change of speed of the motors.
just saying…

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very true, but we know theangle, and speed (we don’t know the friction)

we assume that friction has no (or only momentary) effect on the position.
between the vibration from the motor and the effect of gravity swinging the
sled. We work to keep the chain angles such that there is sufficient force to
move the sled.

It’s seemed to me that the big unknown was the weight

this thread is making me reconsider how big an unknown that is, but until I get
some time to play around with a chain sag calculator, I don’t know.