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Thoughts on Chain Sag Calculation


I was wondering how accurate the chain sag calculations were and stumbled across a webpage that referenced the “Reference Data for Engineers” book. In that book is a set of formulas that can provide a catenary length based upon differing endpoint heights, weight of chain/unit length, and tension on the chain. I’m still trying to sort through the math, but iirc, we have a means to calculate tension in the chain (that spreadsheet of @dlang’s) and we can pretty much figure out the weight of chain/unit length. So, I was wondering if this method might not be an improvement. In the calibration model, we can adjust both weight of sled and weight of chain to try to arrive at a fit. Here’s a link to google that has a preview of the book… I was able to find the page by 49-20 by searching for sag.

Accuracy Issues and Calibration Overview

That sounds VERY promising. Right now we don’t take into account the change in tension on the chain which could make a big difference


And of course, after expanding all the stuff out, I might end up with the exact same equations… I notice some possible similarities. But I don’t see tension in the current model so there’s something… It might be ‘bound’ together with chain weight in the chainSagTolerance.


That one seems pretty important. The chain tension varies by at least a factor of 5, depending on where the sled is located on the surface. I didn’t know this was ignored.


Well, something seems off. If the weight per unit length is zero, you’d expect no sag. Instead, the formula blows up with a divide by zero. sigh.


What does it do with a trivially small value instead of zero?


Blows up. The terms that get computed don’t even make sense with the diagram.


Is this for the current Maslow kinematics calculation, or the calculation in the book?


In the book


The offending formula is:

W is the weight/foot. So as the weight decreases, the latter term becomes larger and larger and eventually greater than L/2… which then suggests that L1/2 is negative (i.e., a negative length). I can understand L1 becoming 0, but not negative.


Looks like the next formula divides by L_1, meaning as L_1=>infinity, the second term in the following equation=>0.


Using 0.1 lbs/foot, 15 lbs of tension, 6.56 feet of chain at a 45 degree angle, L1 = -101.426 feet… something is off with the formula our how I’m interpreting it


I think the problem is H is related to W… and I think that’s the issue with using this formula. I was treating H as the tension caused by the weight of the sled but I think in this formula, H is the tension caused by the chain links pulling on each other… as W drops, so should H


I think it is possible to have a negative number here. Imagine if there was enough tension in the chain, and the angle from horizontal was high enough, that the chain never sags below the lowest end-point. This is common for the Maslow. The chain is mostly straight, and the bottom point shown in the picture is not between the two ends. It is at some hypothetical location outside of the end-point bounds.

Further, as the chain becomes lighter, it becomes straighter, and that end-point location gets further and further away.


we need to use these formulas, but since the weight is really an unknown (we
don’t have the real weight, and there’s the question of if the framt tilt
adjusts the effective weight), treat it as a variable to solve for

#25 roller chain is 0.09lb/ft (easily found if you search for roller chain

David Lang


In theory it wasn’t ignored. (lthough the current forumula is wrong)

The formula has the weight and several other constants in it. The idea was to
fold these all together into one value (k) and then solve for it. It doesn’t
matter which of the values is what, what matters is the combined result.

David Lang


Here’s my data… W=0.1 lb/ft, T=15 lbs, a = 45 degrees, chain length = 5 feet.

H, the horizontal component of tension, equals T * cos(a) = 10.6
L, the horizontal span, equals chain length * cos(a) = 3.53 feet
h, the vertical span, equals chain length * sin(a) = 3.53 feet (edit: had used L and not Lo)


L1 / 2 = 3.53 / 2 - (3.53 * 10.6 * 0.707107 / (0.1 * 3.53)) = -73.1883 feet
L2 / 2 = 3.53 / 2 + (3.53 * 10.6 * 0.707107 / (0.1 * 3.53)) = 76.71834 feet

L1 = 2 * L1/2 = 2 * -73.1883 = -146.377
L2 = 2 * L2/2 = 2 * 76.7194 = 153.4367

S1 = 0.1 * (-146.377)^2 / (8 * 10.6) = 25.2667
S2 = 0.1 * (153.467)^2 / (8 * 10.6) = 27.76275

Lc = 3.53 + 4/3 * ( 25.2667^2 / -146.377 + 27.76275^2 / 153.4367) = 4.41264

Chain length originally equaled 5 feet, now it equals 4.41264 feet… so it shrank?

Somewhere either the formula is wrong, an assumption is wrong, or I can’t do the math correctly.


L and h should be the same. sin(a) and cos(a) should be equivalent at 45deg. I get 3.53 for both L and h.


but as I think about it more, the tension doesn’t just depend on the angle of
the chain we are checking sag on, it depends on the angle of the other chain as
well (if you are in the bottom left, the right chain has far less tension on it
than if the right chain is at the same angle, but much shorter)

David Lang


No, H is related to T and the angle