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Thoughts on Chain Sag Calculation


That was the theory, but for supports at same level, this formula is provided:


H = W * L^2 / (8 * S) where S is the sag and L is the span…

So that’s why I said the formula may be based upon a chain without extra tension… just held up by two ends.


make the chain heavier (until L1 is positive) and see if that makes it work.
There may be an assumption here that the line always sags below both endpoints
(i.e., it doesn’t allow a1 to be negative where it really should be)

David Lang


I see one issue here. I haven’t figured out the source, but S1+h should = S2.

Got it. L1 and L2 need to be updated.

Another thing. Based on the way you are using it, “chain length” = L0, not Lc.


looking at the diagram more, I’m sure this is the case.

the formulas assume that the left point is to the left of the center of the
cable, and in our case it’s to the right.

so I think what we are going to need to do is use the diagram, and re-figure the
appropriate formulas.

David Lang


chain length is the “straightChain” as calculated by Maslow. Lc is the chain length after compensated by sag. So, in my example, I just chose a straight line 5 feet for L0 and would expect Lc to be a little longer.



What happens when you update the -102.636 and 109.696 to be -146.377 and 153.4367?


but if you flip this around to solve for S given a specific tension, you will be
able to ramp up the tension and reduce the sag

the resulting Lc that they calculate is with the line under no tension other
than it’s own weight

take a look at this page and see if it helps

found with the search: catenary sag supports at different levels


I had, just forgot to update L1 and L2 in the formula… the results had been updated. Do I still have it messed up?


Yeah, head hurts. I think we need to work on this more… I’ve got some life to manage at the moment. I’m sure the answer is out there.


hmm, looking at this we may need to go back to the differential equasion and go
from there.


we need to get a math geek in here :slight_smile:

I approach that but we need someone much more hardcore than I am

David Lang


it still blows up because L1 gets squared, it assumes that s1 is positive


I think we are approaching it wrong. In the static condition, we have a situation where a chain is stretched across two equal height supports with a point load (i.e., sled) placed exactly where the router bit is located. Something more like this:

scroll down to Cable with central point load… looks like what I’m talking about.

It’s very much like a zipline problem. except those web pages that talk about it don’t seem to care about the length of the cable, just how much sag there is.


I have stared at this for a while, and here are my conclusions: There are two solutions: one when the minimum-point is in-between the end-points, and one when the minimum-point is outside the end-points. We are using the equation for the case when the minimum is in-between the end-points. However, the problem is such that the minimum is not between the end-points.


except that we are adjusting the length of each side independently, so it’s not
a central point load


Yeah, I know… but it’s an example of what we are trying to calculate. Zip line, ski lift, etc. I’m doing lots of googling.


so where can we find an example where the minimum is not between the end-points?

David Lang


Sounds like a good problem for a stackoverflow question.


I just looked at the Wikipedia article on a Catenary. Here: I actually think it would be pretty easy to implement the full Catenary equation. Here are a few key parts:

The Catenary equation

The parameter “a”

s = length of Catenary (chain)
T0 = horizontal component of tension
lambda*g = Weight of chain (Need to make sure units work out)


Going back to @madgrizzle’s original example:

a=T0/lambda*g => a=15/0.1 => a=150

Set your coordinate system s.t. x1 is at the origin => x1=0

x2 = 5/sqrt(2) => x2=3.5355

s=asinh(x2/a)-asinh(x1/a) =>
s=150sinh(3.5355/150)-150sinh(0) = 3.5358