Catenary Curve Calculator
A cable hanging freely between two points forms a catenary curve: y = a·cosh(x/a). Enter the span (distance between points) plus the sag or the cable length — the calculator solves parameter a numerically (Newton), then computes cable length, sag, relative horizontal tension T₀ = w·a, a point table, and a curve plot.
Compute the curve
| x | y (from lowest point) |
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Formulas used
y = a·cosh(x/a) · sag d = a·(cosh(S/2a) − 1) · length L = 2a·sinh(S/2a)T₀ = w·a (horizontal tension, constant along the cable) · Tmaks = w·a·cosh(S/2a) = T₀ + w·d (at the suspension point)
The equation for a is transcendental — there is no closed-form algebraic solution — so it is solved with Newton's method starting from the parabolic approximation a₀ = S²/8d. The solver usually converges in 5–8 iterations.
Worked examples (verified)
Span 60 m, sag 3 m.
a = 150.4974 m · L = 60.3982 m · T₀ = 150.4974·w · Tmax = 153.4974·w
Span 250 m, sag 12 m.
a = 653.0319 m · L = 251.5295 m · T₀ = 653.0319·w · Tmax = 665.0319·w
Span 40 m, panjang kabel 41,5 m.
a = 42.3988 m · sag = 4.8052 m · T₀ = 42.3988·w · Tmax = 47.2041·w
Span 12 m, sag 0,6 m.
a = 30.0995 m · L = 12.0796 m · T₀ = 30.0995·w · Tmax = 30.6995·w
All figures were recomputed with 60-digit precision arithmetic (independent oracle) and match this page's engine.
Frequently asked questions
What is a catenary curve?
A catenary curve is the shape taken by an ideal cable (perfectly flexible, uniform weight, no bending stiffness) hanging between two points under gravity. Its equation is y = a·cosh(x/a), with the parameter a determining the "flatness" of the curve: a large a means the cable is nearly straight, a small a means it hangs very loose.
How is parameter a computed from span and sag?
From the sag relation d = a·(cosh(S/(2a)) − 1), where S is the span. This equation is transcendental, so it is solved with Newton’s method starting from the parabolic guess a₀ = S²/(8d). Example: a 60 m span and 3 m sag give a = 150.4974 m (parabolic guess 150 m, a correction of only 0.33%).
Can I compute from the cable length instead of the sag?
Yes. From the length L = 2a·sinh(S/(2a)), substituting u = S/(2a) gives sinh(u)/u = L/S, which Newton solves for u, then a = S/(2u). Example: a 40 m span with 41.5 m of cable gives a = 42.3988 m and sag = 4.8052 m.
What does T₀ = w·a mean?
T₀ is the horizontal component of the cable tension, constant along the cable and equal to the weight per unit length (w) times the parameter a. The maximum tension occurs at the suspension point: Tmax = w·a·cosh(S/(2a)), which simply equals T₀ + w·d. Example: a 60 m span with 3 m sag gives T₀ = 150.4974·w and Tmax = 153.4974·w.
What is the difference between a catenary and a parabola?
A parabola is the small-sag approximation of a catenary (the cosh series truncated at the quadratic term). For a 60 m span with 3 m sag, the parabolic approximation gives a₀ = 150 m versus the exact 150.4974 m — a 0.33% difference. The looser the cable, the larger the difference; this calculator uses the exact catenary, not the approximation.
What happens if the cable length equals the span?
The cable becomes a perfectly straight line: sag = 0 and parameter a tends to infinity. This is a degenerate case — physically it means the cable is pulled with infinite tension, which is impossible. The calculator shows it as sag 0 with infinite a.
Why must the solver be numeric — is there no direct formula?
Because a appears both inside and outside the cosh function in the equation d = a·(cosh(S/(2a)) − 1) — a transcendental equation with no closed-form algebraic solution. Newton’s method with a parabolic starting guess converges very fast, usually 5–8 iterations to machine precision.
Does the cable’s weight affect the curve’s shape?
No — the curve’s shape (parameter a, length, sag) depends only on the span and sag/length geometry, not on the weight. The weight w only scales the tension: T₀ = w·a. So a steel cable and a nylon rope with the same span and sag have identical shapes, but their tensions differ in proportion to their weights.