How to calculate the catenary curve — step by step (with a worked example)
A flexible chain hanging from just its two ends settles under its own weight into the catenary y = a·cosh(x/a). The parameter a controls the shape, and cosh is the hyperbolic cosine. Worked example: a = 2, x = 3 → y ≈ 4.70. Hyperbolic functions belong to upper-secondary calculus (Grade 11).
Quick answer
A freely hanging chain follows the catenary y = a·cosh(x/a). The parameter a sets the shape and equals the height of the lowest point. Example a = 2, x = 3: y = 2·cosh(1.5) ≈ 4.70.
At a glance
| Formula | y = a·cosh(x/a) |
|---|---|
| Method | Evaluate the hyperbolic cosine |
| Steps | Ratio x/a, cosh, times a |
| Example | a=2, x=3 → ≈ 4.70 |
| Lowest point | x = 0, where y = a |
| Grade level | Grade 11 (upper-secondary calculus) |
Worked example: y = 2·cosh(3/2)
We want the height of the chain at x = 3 when the shape parameter is a = 2.
The steps to find the catenary height
These steps give the height y at any position x for a given parameter a.
Step 1 · Start
y = a · cosh(x ⁄ a)The general catenary formula.Step 2 · Substitute
2 · cosh(3 ⁄ 2)Insert a = 2 and x = 3.Step 3 · Ratio
2 · cosh(1.5)Evaluate x/a = 3/2 = 1.5.Step 4 · Result
≈ 4.70cosh(1.5) ≈ 2.3524, times 2 gives ≈ 4.70.Step 5 · Check
Rise: 4.70 − 2 = 2.70Height above the lowest point (y − a).
Why y = a·cosh(x/a) describes the chain
At every point of the chain, gravity and the cable tension must balance. Solving that force balance leads exactly to the hyperbolic cosine cosh(t) = (eᵗ + e⁻ᵗ)/2. The parameter a sets the scale: at x = 0, cosh(0) = 1, so y = a — that is the lowest point. As |x| grows, y rises ever more steeply. The catenary resembles a parabola but falls away more gently; the two agree only near the lowest point.