How this calculator works
The ordinary catenary models a flexible uniform chain under its own weight in ideal equilibrium. The scaled model multiplies its height by b/a, giving y=b cosh(x/a); this is a mathematical vertically scaled curve, not a solver for an arbitrary load distribution. Both curves are symmetric about x=0. Inverse x gives both symmetric positions. Recovering ordinary a at a nonzero x can give two values, found by bracketing either side of the minimum; it is not generally unique. Tables contain at most 101 samples, and no chart is drawn.
Formula
Ordinary y=a cosh(x/a); scaled y=b cosh(x/a). Inverse x=plus or minus a acosh(y/b). Scaled b=y/cosh(x/a); scaled a=|x|/acosh(y/b). For ordinary parameter recovery, minimise cosh(u)/u where u=|x|/a and u tanh(u)=1.
Worked example
For a=2 and x=0, the ordinary ordinate is 2 and its slope is zero. With horizontal scale 2 and vertical scale 3, the scaled ordinate at x=0 is 3. Ordinary a=1 and x=ln(2) gives y=(2+1/2)/2=1.25.
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Frequently asked questions
Is a catenary a parabola?
No. They are close near a shallow minimum, but hyperbolic cosine grows differently away from the centre.
Why are two x values returned?
The curve is even in x. Every ordinate above the minimum appears once on either side.
Can a be zero?
No. It divides x and must be strictly positive in these upward-opening models.
Does the table include the exact end?
Only if the selected step lands on it. The table starts at the first boundary and does not extend beyond the interval.