已解决
r2*cosθ2-r1*cosθ1=r3
r1*sinθ1-r2*sinθ2=r4
其中r1、r2、r3为已知量，r4为输入变量，θ1、θ2为输出量，想求出θ1与r4、θ2与r4的两个单独的关系式，两个未知数两个方程感觉应该可以求解吧
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2式平方和可得到θ1-θ2，可是较难得到θ1+θ2的简洁表示 r2²+r1²-2r1r2(cosθ1cosθ1+sinθ1sinθ1)=r3²+r4² cos(θ1-θ2)=(r3²+r4²-r1²-r2²)/2r1r2 θ1-θ2=arcos[(r3²+r4²-r1²-r2²)/2r1r2]=φ(r4) θ1=θ2+φ代入r1*sinθ1-r2*sinθ2=r4 r1(sinθ2cosφ+sinφcosθ2)-r2*sinθ2=r4 (r1cosφ-r2)sinθ2+r1sinφcosθ2=r4 令ψ(r4)=arcos[(r1cosφ-r2)/√{(r1cosφ-r2)²+(r1sinφ)²}/r4] cosψsinθ2+sinψcosθ2=1 sin(θ2+ψ)=1 θ2=2kπ-½π-ψ(r4) θ1=2kπ-½π-ψ(r4)+φ(r4) φ(r4)=arcos[(r3²+r4²-r1²-r2²)/2r1r2] ψ(r4)=arcos[(r1cosφ-r2)/√{(r1cosφ-r2)²+(r1sinφ)²}/r4] |