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Exact parabolic cable length from span and sag
Arc length of a parabolic cable of horizontal span L and midspan sag f, with k = 4f/L.
- Ref
- CIV-42
- Rev
- v1
- Field
- Civil
- Result
- S [m]
The arc length of a parabolic cable of span L and sag f is S = (L/2)[√(1 + k²) + arsinh(k)/k] with k = 4f/L. It is exact for the parabolic shape and a little more than the chord L. For small sag the series S ≈ L(1 + 8f²/(3L²)) is a good approximation. A cable under its own weight hangs as a catenary and has a slightly different length.
Arc length of a parabolic cable of horizontal span L and midspan sag f, with k = 4f/L. This integrates the parabolic profile’s length exactly. A cable that hangs under its own weight per unit length forms a catenary, not a parabola, and so differs slightly. At f = 0 the page returns L.
Calculator
- Extra length over the chord, S − L0.7818 m
- Small-sag approximation, L(1 + 8f²/(3L²))30.8 m
- Sag to span ratio, f/L0.1
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The cable is 30.782 m long, 0.7818 m (2.606 %) longer than the straight span of 30 m.
- The small-sag approximation gives 30.8 m, very close at this sag ratio.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| L Horizontal span | 5 to 500 m | 8.176 | 500 | 491.9 | +10% | +9.52 % |
| f Midspan sag | 0 to 30 m | 30 | 69.7 | 39.7 | +10% | +0.519 % |
Sweep of Midspan sag
| Midspan sag (m) | Cable length (m) |
|---|---|
| 0 | 30 |
| 3.75 | 31.21 |
| 7.5 | 34.43 |
| 11.25 | 38.99 |
| 15 | 44.37 |
| 18.75 | 50.27 |
| 22.5 | 56.53 |
| 26.25 | 63.03 |
| 30 | 69.7 |
Sweep one variable
Variables and how each one contributes
Cable length
What the equation gives you. Everything else feeds it.
Horizontal span
The horizontal span, in m. Raises the length in direct proportion for a fixed sag ratio. Raising it by 10 % (to 33) moves the result by +9.52 %, from 30.78 to 33.71 m.
Midspan sag
The midspan sag, in m (zero gives a straight cable of length L). A deeper sag lengthens the cable, at first with the square of f. Raising it by 10 % (to 3.3) moves the result by +0.519 %, from 30.78 to 30.94 m.
Worked example with the values above
- Slope parameter: k = 4f/L = 4 × 3 / 30 = 0.4.
- Bracket: √(1 + k²) + arsinh(k)/k = 1.07703 + 0.975088.
- S = (L/2) × bracket = 30.7818 m.
Assumptions and limits
- Supports at the same level and vertical loads only, for arches and for cables.
- A flexible cable with negligible bending stiffness and a load uniform along the horizontal span (parabola).
Common mistakes
- Using the cable formulas for a cable under its own weight, which hangs as a catenary.
- Using sag measured from the wrong reference line.
- Forgetting the vertical component of tension at the supports.
Statistics and uncertainty
Cable tension is inversely proportional to sag, so the formulas are very sensitive to the measured sag: a 5 % error in sag gives a 5 % error in tension. Real cables also stretch and have some bending stiffness, which shifts results by a few per cent.
Where it is used
- Sizing cables, anchors, abutments and ties.
- Finding thrust and moments in three-hinged arches.
Maths you need
- Funicular shape: the shape that carries load by axial force only
- Three-hinge condition: the moment at the crown hinge is zero
- Horizontal tension: the constant horizontal component along a loaded cable