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
# }1 O% b% ?6 o/ }& x+ j0 k
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
! K7 F' {7 f1 G: C1 e+ j求当x=-1.361时  y=?  
8 U7 r! S4 j1 }  ?>> syms x y
' ?2 C, u& f* k2 o% 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

" N. [: }2 J# G(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)

5 X0 u/ T5 @6 [6 r" of =

(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)

*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368
; |% ~/ U+ s" w
# Z. Z3 g) L1 p( ~+ }& ?5 ]>>  subs(f, x, -1.361)

ans =
' M- w' }4 ?6 r3 `0 P3 a
" }/ X4 ?* i: k5 x+ J8 S4 I. a; Q! n(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin
% W( E' C$ V& l6 N; f" D3 I7 z# z; w0 |
5 O7 g4 U& n/ o(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -
! ?0 @! x" H2 S5 a9 w% x" T
1 J1 T3 l# B4 e$ T" ^0 ?# y- n(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368
( ]/ r" y3 p; m9 z' I
; ?2 N/ u2 s( Q2 _- t! u+ z8 ?* J
>>  a=0; b=1;
8 G) T+ @$ g5 l4 P' i+ b- K8 weps1=1e-8;eps2=1e-8;
N=300;
7 }: p0 [8 d6 u8 jf=@ (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;
[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);
8 u* U& b5 g8 K# S
运行结果
k ,a ,b ,x, f
8 S5 i$ m8 p$ d# ]  1, 0.000000000, 1.000000000, 0.500000000,-352805.622314164,
! i& O# S5 c3 {& m  2, 0.000000000, 0.500000000, 0.250000000,-92968.331400711,
  3, 0.000000000, 0.250000000, 0.125000000,-4146.780462183,
- a3 C3 v$ i% [  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,
: M- D8 e& `: [$ z  D  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,
2 i' a6 j* C( V$ ^* N  @ 12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
$ y; E6 d, i6 O* k! d+ I( t5 _ 13, 0.117919922, 0.118164063, 0.118041992,-64.817881139,
14, 0.117919922, 0.118041992, 0.117980957,-29.420780403,
15, 0.117919922, 0.117980957, 0.117950439,-11.724903562,
$ O' `8 j- G- A) }+ n& r" M; s 16, 0.117919922, 0.117950439, 0.117935181,-2.877633532,
" i3 X2 C. d1 Q+ M  t2 q( f 17, 0.117919922, 0.117935181, 0.117927551,1.545834384,
0 z& d+ o3 e0 z# e 18, 0.117927551, 0.117935181, 0.117931366,-0.665885649,
1 H# p. e0 N% t; H9 W$ E# s% o% d 19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
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,
$ z9 Y- B( z" N 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,
( r9 ~5 w: p% _7 ~$ Q 26, 0.117930204, 0.117930233, 0.117930219,-0.000638804,
27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
1 ]% Y  k' E2 F9 h& c>>

结论 当x=-1.361时  y=0.117930219
' j: P9 i* S% |5 L1 C" M. 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
算法？

二分法        
8 c1 @- `' {" R# d3 {* t$ v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')
for k=1:N
* A0 p% [8 d  @    x=(a+b)/2;
    f_value=f(x);
" C: e: k8 z( j$ L: T/ g    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
9 R2 r/ j+ B$ [' A- {  f      ,k ,a ,b ,x, f_value)
    if abs(f_value)< eps1||0.5*(b-a)<eps2
    return
# a2 k) ~1 W' ^# o7 ?    else
. U$ [, Y4 Y" N. x6 I) x7 T        if f(x)*f(a)<0
+ @: C+ v9 O  G. `) ]) L8 @6 \          b=x;
        else
% l* {5 |) z% `3 [: }          a=x;
        end
        if k== N
          warning ('算法超出最大迭代数!')   
8 I" b& m2 L* D& \3 G        end
' h* `) V: ]* a4 s5 s/ z, f. L    end
6 Q( q$ T6 t7 X, ~  z& E( E6 M6 W; V

shouce

单车居士 发表于 2015-6-4 12:41
西交大学生？？问邢老师

这点问题   自己就可以解决
" J6 ~$ f% [6 N) G5 A- E

shouce

明月山河 发表于 2015-6-4 19:12
- @. E2 f  m; I7 T% y你这个是代数方程？

' r2 L2 P' ^2 |6 E8 s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
" n2 [* N' `0 h0 O$ u4 T$ x, q  Q你这个是代数方程？

( \; Y0 B; a. A是超越方程      


" l: A. Y2 o) Y" Z8 g" l5 Y$ A[chāo yuè fāng chéng
0 D% w' a1 Y/ d/ R& b6 }) ]