Catenary

悬链线数学表达式的证明

  


  如右图,设最低点A处受水平向左的拉力H,右悬挂点处表示为C点,在AC弧线区段任意取一段设为B点,则B受一个斜向上的拉力T,设T和水平方向夹角为θ,绳子的质量为m,受力分析有:   Tsinθ=mg;   Tcosθ=H,   tanθ=dy/dx=mg/H,   mg=ρs,, 其中s是右段AB绳子的长度,ρ是绳子线重量密度,代入得微分方程dy/dx=ρs/H;利用弧长公式ds=√(1+dy^2/dx^2)*dx;所以s=∫√(1+dy^2/dx^2)*dx;   所以把s带入微分方程得dy/dx=ρ∫√(1+dy^2/dx^2)*dx/H;…..(1)   对于(1)设p=dy/dx微分处理   得 p’=ρ/H*√(1+p^2)……(2)   p’=dp/dx;   对(2)分离常量求积分   ∫dp/√(1+p^2)=∫ρ/H*dx   得ln[p+√(1+p^2)]=ρx/H+C,即asinhp(反双曲正弦)=ρx/H+C   当x=0时,dy/dx=p=0;带入得C=0;   整理得asinhp=ρx/H 另祥解: (ln[p+√(1+p^2)]=ρx/H);   p=sh(ρx/H) (1+p^2=e^(2ρx/H)-2pe^(ρx/H)+p^2);   (p=[e^(ρx/H)-e^(-ρx/H)]/2=dy/dx);   y=ch (ρx/H)* H / ρ (y=H/(2ρ)*[e^(ρx/H)+e^(-ρx/H)] );   令a=H/ρ: y=a*cosh (x/a)   (y=a[e^(x/a)+e^(-x/a)]/(2)= a*cosh(x/a))。

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About yhtian

I am an academic researcher working on Offshore Geotechnical Engineering. My blog aims to write down some work related trivial things and tricks about software, programming. It is basically a memo for me to back up some thoughts and small details. But I am more than happy if someone would visit and discuss.
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