Much ado about chain sag


What is chain sag and how does it influence the accuracy of the Maslow?

To clarify, I know what chain sag is in a closed drive system - that is, a loop of chain between a drive and a driven sheave. What is chain sag in a system like the Maslow? - a mass suspended on a chain between two drive sheaves…

Thanks :slight_smile:

it’s pretty much the same thing. the chain between the sled and the sprocket
isn’t a stright line, it’s an arc

As a result, the actual distance between the sled and the sprocket is a little
less than the amount of chain that has been let out.

This means that the sled isn’t quite where the computer thinks it is.
(currently, we think this means that it’s off by ~1mm when in the bottom

The amount of sag depends on the horizontal distance the chain is spanning, the
weight of the chain in that span (which is higher if the sled is in the bottom
corner than the top corner as there is more chain needed to cover the diagonal),
and the amount of tension on the chain.

The amount of tension on the chain depends on the weight of the sled, and the
angle of the two chains to the sled.


Can’t answer that with the precision and maths that is discussed in this chain-sag-inaccuracy 108 long topic, but will give a novice opinion.
From what I understand, if you cut in the low left corner, the chain from the right motor will be the longest and introduce a ‘pull up right’ of the sled due to the gravity of the chain.

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Thanks for the link. I read almost all of it. Much was above my pointy head, but I think I have the gist of it.

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You nailed it.

This effect is compounded (in your example scenario) when the sled is moving away from the right motor. Basically, the cutting force required to move in that direction is taking tension off of that chain, allowing it to sag more.

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