KEVOS® MAKEOS · Mechanics
Catenary Cable Sag, Tension and Length
What tension and length does a cable or chain of this weight have when it sags this much over the span?
Exact catenary for a uniform cable or chain hanging between two supports at the same level: catenary parameter, horizontal and maximum tension, cable length and support angle, compared with the parabolic approximation.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Span between supports | m | Horizontal distance between the supports (at the same level). |
| Sag at mid-span | m | Vertical sag at mid-span. |
| Cable weight per metre | N/m | Weight per unit length (mass per metre × 9.81, plus any uniform ice or attached load). |
Method and formulas
- y = a cosh(x ÷ a): the parameter a solves d = a [cosh(S ÷ 2a) − 1].
- Horizontal tension H = w a; maximum tension at the supports = w (a + d).
- Length = 2a sinh(S ÷ 2a); parabolic estimate H ≈ w S² ÷ 8 d.
- Example: 10 N/m over 100 m with 5 m sag carries 2,558.3 N at the supports.
Limitations
- Uniform weight along the cable, supports at equal level, inextensible cable.
- Concentrated loads and elastic stretch change the shape.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Check the tension in a hanging cable, hose or chain
- Work out the cable length to order for a span and sag
- Set sag for festoon and catenary supply systems
Who it is for
- Structural engineers
- Electrical engineers
- Rigging supervisors
What you get
- Horizontal and maximum tension
- Cable length
- Support angle
Interpreting the result
- Halving the sag roughly doubles the tension; tight cables carry very high loads.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 2999. Topic reference; no source tables or recommended settings are reproduced.
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