Calculate chain sag from span, chain weight, and tension, or find the tension needed to keep sag within a set limit in feet, inches, or meters.
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Chain Sag Formula
The calculator uses the common parabolic sag approximation for a chain hanging between two supports at the same height. It is most accurate when the sag is small compared with the span.
- s = chain sag at the midpoint
- w = chain weight per unit length
- L = span between supports
- T = horizontal tension in the chain
For the tension mode, the same formula is rearranged to solve for the tension needed to keep the sag at or below a chosen value:
- T = tension needed
- w = chain weight per unit length
- L = span between supports
- s = maximum allowed midpoint sag
The find sag function takes the span, chain weight, and tension, then calculates the expected midpoint droop. The find tension needed function takes the span, chain weight, and maximum sag, then calculates the pulling force needed to stay within that sag limit. The calculator converts the selected units internally so the formula uses consistent length, weight, and force units.
Typical Chain Weights and Sag Interpretation
Use these values only as rough references. Actual chain weight varies by grade, link shape, coating, and manufacturer.
| Chain type | Typical weight | Common use |
|---|---|---|
| #10 jack chain | 0.10 lb/ft | Light hanging loads |
| 1/8 inch proof coil chain | 0.21 lb/ft | Small utility chain |
| 1/4 inch proof coil chain | 0.85 lb/ft | General-purpose chain |
| 3/8 inch proof coil chain | 2.10 lb/ft | Heavier utility chain |
| 1/2 inch proof coil chain | 3.45 lb/ft | Heavy hanging or barrier chain |
| Sag as percent of span | How it usually looks |
|---|---|
| Less than 2% | Very taut, almost straight |
| 2% to 5% | Light visible droop |
| 5% to 10% | Noticeable curved shape |
| More than 10% | Deep sag, usually needs more tension or a shorter span |
Chain Sag Examples
Example 1: Find the sag
A chain spans 20 ft between supports. The chain weighs 1.5 lb/ft, and the applied tension is 50 lb.
Convert to calculator formula units:
- L = 20 ft = 240 in
- w = 1.5 lb/ft = 0.125 lb/in
- T = 50 lb
The midpoint sag is 18 in, or 1.5 ft.
Example 2: Find the tension needed
A 1/4 inch proof coil chain weighs about 0.85 lb/ft. It spans 12 ft, and you want no more than 4 in of sag.
- L = 12 ft = 144 in
- w = 0.85 lb/ft = 0.07083 lb/in
- s = 4 in
You need about 45.9 lb of tension to keep the sag at 4 in or less.
Chain Sag FAQ
Is this the exact catenary formula?
No. This calculator uses the parabolic approximation, which is simpler and works well for many light-to-moderate sag cases. A real hanging chain forms a catenary curve. If the sag is large compared with the span, the exact catenary result can differ from this estimate.
Why does sag increase so quickly when the span gets longer?
Sag depends on the square of the span. If you double the span and keep the same chain weight and tension, the sag becomes about four times larger. Shortening the span is often one of the most effective ways to reduce sag.
Does the tension result include a safety factor?
No. The tension result is the estimated force needed for the sag limit only. It does not include a safety factor, shock loading, wind, movement, support strength, anchor strength, or the working load limit of the chain. For load-bearing or safety-critical use, check the rated hardware and apply the proper safety factor.
