螺杆转子刀具设计（2） matlab 解法

shouce

螺杆转子刀具设计 数学模型

" [/ Q9 Z% x! ~3 j9 E. L解如下方程：1679999.7*sin(x)-1679997.5*cos(y)*sin(x)-1499997.5*sin(y)*cos(x)-75000*sin(x)^3-179999.4000005*y*cos(y)*cos(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)=0
求当x=-1.361时  y=?  
>> syms x y
+ ?2 h/ G7 B2 D. x6 Q>> f=1679999.7*sin(x)-1679997.5*cos(y)*sin(x)-1499997.5*sin(y)*cos(x)-75000*sin(x)^3-179999.4000005*y*cos(y)*cos

(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)

1 G. y/ x+ X" k5 D1 [' o5 qf =

3 H$ ^% W( Q& z3 I(7215543768789811*sin(x))/4294967296 - (2999995*cos(x)*sin(y))/2 - (3359995*cos(y)*sin(x))/2 - 75000*sin(x)^3 + 75000*cos(x)*cos(y)
1 [- i2 p# v1 N4 r( c
*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368

>>  subs(f, x, -1.361)

ans =
' ]* w  N8 W/ ^8 h5 N5 Q
9 H3 x0 n% ?3 t& a; J8 }(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin
& H$ J5 [. V+ T. n+ y
(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -
; N, r8 z& Q. Y
$ k2 D8 A, T" E4 Y(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368
& q# o3 D$ Q7 {: d3 M7 ~
1 r1 B3 K; n+ N% Z5 R
0 L8 n, x5 L* x0 A4 V. h9 T0 X>>  a=0; b=1;
eps1=1e-8;eps2=1e-8;
' n) Q/ w! d8 q0 z N=300;
0 y+ f3 r. ]8 H- V8 G, `7 Af=@ (x) ((3359995*sin(1361/1000)*cos(x))/2 - (2999995*cos(1361/1000)*sin(x))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin(1361/1000)^3- 75000*cos(1361/1000)*sin(1361/1000)*cos(x) - (6184732290414159*x*cos(1361/1000)*cos(x))/34359738368 -(6184732290414159*x*sin(1361/1000)*sin(x))/34359738368);
  l0 ~7 M6 T6 D- a7 I  }Hfun=@Bisection;
[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);

/ b% D& y3 e+ F+ Z运行结果
7 J3 c1 a4 D5 V* ck ,a ,b ,x, f
  W" f* |* ]# h8 x$ m% S  1, 0.000000000, 1.000000000, 0.500000000,-352805.622314164,
4 U5 \( g7 ^; @" p. L  2, 0.000000000, 0.500000000, 0.250000000,-92968.331400711,
  3, 0.000000000, 0.250000000, 0.125000000,-4146.780462183,
  4, 0.000000000, 0.125000000, 0.062500000,29178.936016433,
  5, 0.062500000, 0.125000000, 0.093750000,13458.364371693,
  6, 0.093750000, 0.125000000, 0.109375000,4890.124291139,
  7, 0.109375000, 0.125000000, 0.117187500,430.092693351,
  8, 0.117187500, 0.125000000, 0.121093750,-1843.759437944,
  9, 0.117187500, 0.121093750, 0.119140625,-703.184646645,
10, 0.117187500, 0.119140625, 0.118164063,-135.633470273,
11, 0.117187500, 0.118164063, 0.117675781,147.457778641,
, A- X1 \1 C0 b( ?' v 12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
13, 0.117919922, 0.118164063, 0.118041992,-64.817881139,
$ V9 m: |  x8 E2 A- Z 14, 0.117919922, 0.118041992, 0.117980957,-29.420780403,
$ m# D) Q6 [* l, M' d 15, 0.117919922, 0.117980957, 0.117950439,-11.724903562,
16, 0.117919922, 0.117950439, 0.117935181,-2.877633532,
17, 0.117919922, 0.117935181, 0.117927551,1.545834384,
0 z# x: X. s& J: Y6 R* h* K/ P7 E3 q 18, 0.117927551, 0.117935181, 0.117931366,-0.665885649,
( j+ n0 g6 }; B9 A 19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
) ]' L6 \2 c, B8 w6 F1 y! S 20, 0.117929459, 0.117931366, 0.117930412,-0.112953030,
21, 0.117929459, 0.117930412, 0.117929935,0.163512627,
" y2 x  l4 c/ y6 j 22, 0.117929935, 0.117930412, 0.117930174,0.025279853,
3 W  D6 R7 d/ ~4 `% P3 E) _ 23, 0.117930174, 0.117930412, 0.117930293,-0.043836575,
24, 0.117930174, 0.117930293, 0.117930233,-0.009278357,
25, 0.117930174, 0.117930233, 0.117930204,0.008000749,
26, 0.117930204, 0.117930233, 0.117930219,-0.000638804,
+ u, V3 E/ Z  l# o& ~/ e7 g 27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
>>

' V0 a& a1 E" L7 d结论 当x=-1.361时  y=0.117930219
! D. `% P" B# s- M8 O5 \按此方法   可依次  得出150对x与y的值
. K, N: u3 b% ~

单车居士

西交大学生？？问邢老师

明月山河

你这个是代数方程？

cosxuan

算法？

shouce

cosxuan 发表于 2015-6-4 19:45
* R1 H& u& v( M0 |. i1 k算法？

5 n5 [0 A' r" }$ Z: H二分法        
还有一个程序
% Bisection.m
function[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
fprintf('k ,a ,b ,x, f\n')
; [4 N$ t1 d% f( h# t; wfor k=1:N
    x=(a+b)/2;
    f_value=f(x);
9 ~1 J* @0 ~2 g  e1 c! P    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
      ,k ,a ,b ,x, f_value)
: n7 p8 @- x1 w# S4 a  B    if abs(f_value)< eps1||0.5*(b-a)<eps2
    return
    else
        if f(x)*f(a)<0
          b=x;
' @: y. u0 t8 ?6 o# @        else
          a=x;
1 y" x6 P6 v- }+ s9 Y) ?* e- ~        end
        if k== N
          warning ('算法超出最大迭代数!')   
6 S0 [/ e$ n9 h5 o; Y0 D6 c6 N        end
    end
' I7 E& p0 U2 Q( k* ?7 Z6 @

shouce

单车居士 发表于 2015-6-4 12:41
! i2 t  ?! z& c/ m# S/ k/ S6 b西交大学生？？问邢老师

8 ?. M8 x& K2 Q这点问题   自己就可以解决
5 u4 k, W# e* i8 H, r) S

shouce

明月山河 发表于 2015-6-4 19:12
你这个是代数方程？

是的呀      不过matlab写出来的方程  括号很多  不好看   

shouce

明月山河 发表于 2015-6-4 19:12
8 Z+ F1 d& x5 n3 ]3 I7 ]你这个是代数方程？

# ~- h: E7 k0 l2 ?/ U2 x% U- M- V- c
是超越方程      


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