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[matlab] 快速求离散点刀具坐标程序

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发表于 2015-11-26 21:28:44 | 显示全部楼层 |阅读模式
本帖最后由 shouce 于 2015-11-26 21:34 编辑
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% examlpe7.m% t& @+ }  N& T. M& |6 v. ^1 @; i
a=0; b=2;  ]+ I/ V- T# k) h& Z
eps1=1e-12;eps2=1e-12;
) ?$ h; l3 }& R/ v$ w, C N=300;
  }7 P: T+ n9 F, x9 G* \A=[-0.104020146        35.0002164        -0.004160847];
7 ?7 F& Z2 p6 \5 ?/ z[m,n]=size(A);
  N8 z9 j4 N. o/ D4 Mfor i=1:m
( }+ _( T. i. Y0 q1 na1=A(i,1);b1=A(i,2);m1=A(i,3);
9 C" h  ~, G4 t9 T: D* z( y+ Tend# w; S' q' j( e* H+ W! O
f=@ (x) ((285+60*tan(45*pi/180)-b1*sin(x)-a1*cos(x))*(a1+b1*m1)+(60)^2*x*(sin(x)-m1*cos(x))-60*285*tan(45*pi/180)*(m1*sin(x)+cos(x)));
* I4 T! b6 q: J. j; HHfun=@Bisection;
5 N0 \" r. D/ x, ^1 T, N( k  p% O[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);5 J  j4 w3 F( u# }- q
- j- f, Y' u4 K% h* ?

5 x  F% f. N9 Z  ^1 q+ m/ L% A 为离散点坐标及导数  中心距  285  左旋  螺旋角45度     在二分法中增加一部些程序,方便了很多/ ]9 S. S1 z: F& J+ O: i9 A0 W
9 k+ `+ ~( ]3 R1 J/ o! [; I4 L
部分运行结果:0 L8 `: [0 s+ k4 o
35, 1.303240794, 1.303240794, 1.303240794,0.000000896,! e+ C2 H1 a/ M1 a2 i6 [
36, 1.303240794, 1.303240794, 1.303240794,0.000000279,
* M/ h; j  H8 t4 H6 k0 V 37, 1.303240794, 1.303240794, 1.303240794,-0.000000030,
/ F5 x  u  s9 n 38, 1.303240794, 1.303240794, 1.303240794,0.000000124,( U5 R5 W0 `. A- r- e) }: I8 i! R: _' ~
39, 1.303240794, 1.303240794, 1.303240794,0.000000047," l1 ^# b. x6 q0 G0 g  w
40, 1.303240794, 1.303240794, 1.303240794,0.000000009,' H' r- P7 Y: T6 L. s# o6 n, y
41, 1.303240794, 1.303240794, 1.303240794,-0.000000011," o! ?7 F. [( J
$ a+ i9 c' _0 q: r; N

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