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

' y9 m% C# y1 c6 w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
, t: f8 A7 @" d5 Z' m" z1 V求当x=-1.361时  y=?  
>> syms x y
>> 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

5 ?' M. c2 U0 @& C(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)
, p3 G0 _# I4 F0 I* X2 o* u* {5 I" V
f =
9 i! O( k! n; ?
(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)
* S) G0 c, W+ O, L1 J( V$ U
. t5 I; h+ X) m" R! f2 S% W0 [*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368

( F$ }0 v8 p' ~: f1 c1 |7 m$ y% ]+ s>>  subs(f, x, -1.361)
' L& _- U; \" T& [7 `
7 X6 ~/ g$ G/ u) B7 f1 k8 E  tans =

(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin

, g) P+ \% w7 ?& O; \(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -
' H. _, o) H0 M7 x2 o
. n6 W& M- S8 B0 A5 ^. B: y(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368


0 Z! A. Y8 L4 a7 O, u* Z>>  a=0; b=1;
  T8 \. y/ `( l0 T& n5 h* aeps1=1e-8;eps2=1e-8;
0 p( E; l7 i- Y  v" j3 b N=300;
* N$ S1 M" b) ]0 g% {& mf=@ (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);
Hfun=@Bisection;
0 x+ G2 |: Z- y" N[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);
9 s% _' V. H3 j$ ^( g
运行结果
k ,a ,b ,x, f
1 b$ p  [3 Q) V' j1 j) k3 g( Q& R  1, 0.000000000, 1.000000000, 0.500000000,-352805.622314164,
  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,
  ]% _% _, B) K. X8 G  5, 0.062500000, 0.125000000, 0.093750000,13458.364371693,
* {. t' R" j) d$ O$ U  6, 0.093750000, 0.125000000, 0.109375000,4890.124291139,
  7, 0.109375000, 0.125000000, 0.117187500,430.092693351,
) W8 C; m5 }2 z# z7 w  8, 0.117187500, 0.125000000, 0.121093750,-1843.759437944,
  9, 0.117187500, 0.121093750, 0.119140625,-703.184646645,
; ~* W3 J& O  q2 }$ E 10, 0.117187500, 0.119140625, 0.118164063,-135.633470273,
) y% K6 R6 \6 A& Q 11, 0.117187500, 0.118164063, 0.117675781,147.457778641,
12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
  L  q% }, g1 ?. E: P' j3 {- \ 13, 0.117919922, 0.118164063, 0.118041992,-64.817881139,
14, 0.117919922, 0.118041992, 0.117980957,-29.420780403,
5 K( s: X* ?) l5 ~) i/ f 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,
) B! z$ U3 J. k! b2 z* V, J! h 18, 0.117927551, 0.117935181, 0.117931366,-0.665885649,
9 W8 `$ y9 |0 z- Y" a 19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
( C1 c6 L* A2 w& P0 { 20, 0.117929459, 0.117931366, 0.117930412,-0.112953030,
21, 0.117929459, 0.117930412, 0.117929935,0.163512627,
22, 0.117929935, 0.117930412, 0.117930174,0.025279853,
/ ~' i4 L  u- _$ }: W0 l, ? 23, 0.117930174, 0.117930412, 0.117930293,-0.043836575,
7 ]3 {8 o8 \9 I9 O+ C# o 24, 0.117930174, 0.117930293, 0.117930233,-0.009278357,
0 N+ J  ?3 O6 |6 `+ D1 H+ M 25, 0.117930174, 0.117930233, 0.117930204,0.008000749,
, Q$ |8 f) M- I: d; Q4 E 26, 0.117930204, 0.117930233, 0.117930219,-0.000638804,
# I5 Y2 b; m% U4 T6 F 27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
>>

" ?$ n, |' A# @& ~% n结论 当x=-1.361时  y=0.117930219
# ~" A, g. a' K  Y1 y8 l按此方法   可依次  得出150对x与y的值

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
) \2 u! M9 R* Q& O, N, a# s2 p+ P算法？

二分法        
还有一个程序
5 x8 n% B; c; X8 c% Bisection.m
function[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
fprintf('k ,a ,b ,x, f\n')
, k% ^! x8 Y: }$ h% cfor k=1:N
    x=(a+b)/2;
    f_value=f(x);
8 P: H0 H2 E; m    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
/ n2 _: b- L6 ]2 \: d7 `$ E      ,k ,a ,b ,x, f_value)
    if abs(f_value)< eps1||0.5*(b-a)<eps2
    return
6 c/ E" A3 q: e; z  _9 v6 U# R& v    else
        if f(x)*f(a)<0
          b=x;
4 a6 I1 w; i, Q; g: u5 d        else
          a=x;
        end
# A9 i% c4 a( y2 _- _        if k== N
# j# J; o. h! _          warning ('算法超出最大迭代数!')   
! a5 o8 \/ a" E        end
    end

shouce

单车居士 发表于 2015-6-4 12:41
& Z4 v9 s: y$ G* \西交大学生？？问邢老师

这点问题   自己就可以解决

shouce

明月山河 发表于 2015-6-4 19:12
1 Y' v: b6 _1 t. D你这个是代数方程？

  ?5 w3 O' _& _8 S是的呀      不过matlab写出来的方程  括号很多  不好看   

! ^% t7 Q7 G% ?- h5 f6 l# V

shouce

明月山河 发表于 2015-6-4 19:12
) k- x$ t" v. v/ k你这个是代数方程？

是超越方程      


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