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:
$$F_M(N, N_R) = {N-1 \choose N_R-1}{M-N+1 \choose N_R}$$
http://whistleralley.com/hanging/hanging.htm
*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
*
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?