【答案】应助回帖

10楼: Originally posted by cooooldog at 2016-03-22 08:59:16
楼主还在不？

这个问题简单，只须用Newton法反向迭代求解就可以了

只是牛顿法需要求解的方程组表达式对应于曲线围成的面积的求解函数。

楼主如果能冒个泡出来，免费求解。

sweety   hsd3521...

ClearAll["Global`*"];
R = 4878/100;
z1 = 6;
r = R/z1;
z2 = z1 - 1;
e = 705/100;
f = r/e;
re = 126/10;
θ = ArcTan[Sin[z1 τ]/(f + Cos[z1 τ])] - τ;
φ = ArcSin[f Sin[θ + τ]] - θ;
ψ = z1/(z1 - 1) φ;

(* 两个参数方程 *)
curve01 = {(R - r) Sin[τ] + e Sin[z2 τ] -  re Sin[θ], (R - r) Cos[τ] - e Cos[z2 τ] +  re Cos[θ]} // Simplify;

(*定子*)
ParametricPlot[curve01,{τ,0,2 π},Exclusions\[Rule]None,MaxRecursion\[Rule]15,PlotPoints->
500,PlotStyle->Red]


(*转子*)
curve02 = {curve01[[1]] Cos[φ - ψ] -  curve01[[2]] Sin[φ - ψ] - e Sin[ψ],
    curve01[[1]] Sin[φ - ψ] +   curve01[[2]] Cos[φ - ψ] - e Cos[ψ]} //Simplify;

base=ParametricPlot[curve02,{τ,0,2 π},Exclusions->None,MaxRecursion->15,PlotPoints->500,PlotStyle->Blue]; Show[base]

τp = (y - x) /.FindRoot[(curve02 /. τ -> x) == (curve02 /. τ -> y), {x, Pi/20}, {y, 2 Pi}];

arc = Table[curve02, {τ, 0, τp/10, .00001}];

Show[base,Graphics[{{Red,Line[{curve02 /. τ -> 0, {0, 0},curve02 /. τ -> τp/10}], Line@arc}}]]

τp = 5 Pi/3;
5 (NIntegrate[-First[curve02] D[Last@curve02, τ] +  Last[curve02] D[First@curve02, τ], {τ, 0, τp/10}, Method -> "LocalAdaptive", MaxRecursion -> 100]) // NumberForm[#, 15] &

curveArea[r_]:=Module[{z1,z2,e,f,re,R,θ,φ,ψ,τ,curve01,curve02},
z1=6;
R=r z1;
z2=z1-1;
e=705/100;
f=r/e;
re=126/10;
θ=ArcTan[Sin[z1 τ]/(f+Cos[z1 τ])]-τ;
φ=ArcSin[f Sin[θ+τ]]-θ;
ψ=z1/(z1-1) φ;
curve01={(R-r) Sin[τ]+e Sin[z2 τ]-re Sin[θ],(R-r) Cos[τ]-e Cos[z2 τ]+re Cos[θ]};curve02={curve01[[1]] Cos[φ-ψ]-curve01[[2]] Sin[φ-ψ]-e Sin[ψ],curve01[[1]] Sin[φ-ψ]+curve01[[2]] Cos[φ-ψ]-e Cos[ψ]};
(5 (NIntegrate[-First[curve02] D[Last@curve02,τ]+Last[curve02] D[First@curve02,τ],{τ,0,τp/20,τp/10},MaxRecursion->100,Method->"LocalAdaptive"]))
]

FindRoot[curveArea[x] == 6557.19, {x, 7.5}, Evaluated -> False,Method -> {"Newton", "StepControl" -> "TrustRegion"}] // AbsoluteTiming

r=8.1294128480878

11楼: Originally posted by Mr__Right at 2016-03-22 20:57:03
边学习，边回答这个问题。

工具：用Mathematica 解决。求解这类问题，Mathematica首屈一指。

首先是描述两条曲线的代码：


ClearAll;
R = 4878/100;
z1 = 6;
r = R/z1;
z2 = z1 - 1;
e = 705/100;
...

10楼: Originally posted by cooooldog at 2016-03-22 08:59:16
楼主还在不？

这个问题简单，只须用Newton法反向迭代求解就可以了

只是牛顿法需要求解的方程组表达式对应于曲线围成的面积的求解函数。

楼主如果能冒个泡出来，免费求解。

sweety   hsd3521...

11楼: Originally posted by Mr__Right at 2016-03-22 20:57:03
边学习，边回答这个问题。

工具：用Mathematica 解决。求解这类问题，Mathematica首屈一指。

首先是描述两条曲线的代码：


ClearAll;
R = 4878/100;
z1 = 6;
r = R/z1;
z2 = z1 - 1;
e = 705/100;
...

12楼: Originally posted by 丫头丫头2014 at 2016-05-17 10:49:39
在呢 不知道还可以问吗
...