Hanging cables


A hanging cable takes the shape of a catenary, a suspension bridge cable is a parabola. What’s the connection between these?

Inline math works: $$F_M(N, N_R) = {N-1 \choose N_R-1}{M-N+1 \choose N_R}$$, and so does display math:

$$F_M(N, N_R) = {N-1 \choose N_R-1}{M-N+1 \choose N_R}$$


*cable weight negligible (weight is proportional to horizontal distance): cable forms parabola *cable weight significant (weight is proportional to arc length): cable forms catenary *flexible, weight significant (weight is proportional to arc length): parabola again?

i remember a derivation where one simple change in an integrand (for the weight distribution - constant per horizontal distance vs constant per arc length) makes the difference between parabola and catenary. this looks sort of like it:

$$y’ = \frac \mu T x$$ - parabola $$y’ = \frac \mu T {\int_0^x \sqrt{y’^2+1}dt$$ - catenary

*y’ = \frac \mu T {\smashmargin2{\int\nolimits_{\tiny 0}}^{x}} \sqrt{y’^{\tiny 2}+1}dt - catenary

this is probably the integral i remember which requires recognizing as being equivalent to the form of the arclength integral for cosh() - derivative being equal to arclength. equivalently, second derivative = one plus square of first derivative

μ is linear density (?) and T is magnitude of tension at minimum point of curve (?) - but how to get to these diffeqs?


Tanzania 5: Post-mortem

A webgl shader experiment