Model a Hanging Chain

Find the position with minimal potential energy of a chain or cable of length suspended between two points.

Set parameter values for the length of the chain , the left-end height , and the right-end height .

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L = 4; a = 1; b = 3;

Let be the height of the chain as a function of horizontal position, with .

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xf = 1; nh = 201; h := xf/nh;

Set up variables for the height of the chain .

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varsy = Array[y, nh + 1, {0, nh}];

Denote the slope at position by and set up variables for it.

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varsm = Array[m, nh + 1, {0, nh}];

Denote the partial potential energy from to by .

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varsv = Array[v, nh + 1, {0, nh}];

Denote the length of the chain at position by and set up variables for it.

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varss = Array[s, nh + 1, {0, nh}];

Join all variables.

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vars = Join[varsm, varsy, varsv, varss];

The objective is to minimize the total potential energy .

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objfn = v[nh];

Here are the boundary value constraints from the geometry.

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bndcons = {y[0] == a, y[nh] == b, v[0] == 0, s[0] == 0, s[nh] == L};

Discretize the ODEs: , , .

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odecons = {Table[ y[j + 1] == y[j] + 0.5*h*(m[j] + m[j + 1]), {j, 0, nh - 1}], Table[v[j + 1] == v[j] + 0.5* h*(y[j]*Sqrt[1 + m[j]^2] + y[j + 1]*Sqrt[1 + m[j + 1]^2]), {j, 0, nh - 1}], Table[s[j + 1] == s[j] + 0.5*h*(Sqrt[1 + m[j]^2] + Sqrt[1 + m[j + 1]^2]), {j, 0, nh - 1}]};

Choose initial points for the variables.

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tmin = If[b > a, 0.25 , 0.75]; init = Join[Table[4*Abs[b - a]*((k/nh) - tmin), {k, 0, nh}], Table[4*Abs[b - a]*(k/nh)*(0.5*(k/nh) - tmin) + a, {k, 0, nh}], Table[(4*Abs[b - a]*(k/nh)*(0.5*(k/nh) - tmin) + a)*4* Abs[b - a]*((k/nh) - tmin), {k, 0, nh}], Table[4*Abs[b - a]*((k/nh) - tmin), {k, 0, nh}]];

Minimize the total potential energy, subject to the constraints.

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sol = FindMinimum[{objfn, Join[bndcons, odecons]}, Thread[{vars, init}]];

Extract the solution points.

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solpts = Table[{i h, y[i] /. sol[[2]]}, {i, 0, nh}];

Plot the position of the chain with minimal potential energy.

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ListPlot[solpts, ImageSize -> Medium, PlotTheme -> "Marketing"]

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Use FindFit to fit the result to the catenary curve.

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catenary[t_] = c1 + (1/c2) Cosh[c2 (t - c3)];

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fitsol = FindFit[solpts, catenary[t], {c1, c2, c3}, {t}]

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Plot the solution points together with the catenary curve.

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Show[Plot[catenary[t] /. fitsol, {t, 0, 1}, PlotStyle -> Directive[Green, Thickness[0.01]], ImageSize -> Medium], ListPlot[Take[solpts, 1 ;; nh ;; 5], PlotStyle -> PointSize[.02]]]

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