Formula for a catenary: x(t) = t y(t) = 1/2 a [e^(t/a) + e^(-t/a)] = a cosh (t/a) t = 0 corresponds to the vertex
y(t) = 1/2 a [e^(t/a) + e^(-t/a)] = a cosh (t/a)
t = 0 corresponds to the vertex
(s^2 + a^2)k = -a s(t) = a sinh (t/a) k(t) = -1/a sech^2 (t/a) o(t) = -2 tan^-1 [tanh (t/2a)]
k(t) = -1/a sech^2 (t/a)
o(t) = -2 tan^-1 [tanh (t/2a)]