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
2 k$ v4 [+ Q9 l+ J
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
6 M* j4 D& @6 O0 R+ A4 t求当x=-1.361时  y=?  
; l  P2 c/ }5 v( T4 B5 u$ `>> 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

: o& j: M. I4 ?8 k3 S& \(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)

5 n* \( Q" t; n  `0 v  l$ ?f =

(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
: ]1 W0 w, B/ H  _! {; F7 S- }
' f- x! L! g2 Z( c7 x>>  subs(f, x, -1.361)
2 T/ `- E3 D, R8 o7 a  R
ans =
: Q: e$ \+ S; Z
(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin
  J* ^3 l  P( z
(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -

) S, y  V) r) S2 e9 p: Z7 ]- C(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368
, H2 M1 K. L* W
+ I0 M/ {- T- ~9 ?  V+ T4 r
>>  a=0; b=1;
eps1=1e-8;eps2=1e-8;
N=300;
f=@ (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;
5 E. c/ @+ P: {3 g2 q[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);
! C/ o1 w# \- l" _( u) l6 h; Q* W
" {) ~* v  i7 |+ f) {/ F运行结果
k ,a ,b ,x, f
1 f% }/ P; Z- V4 S3 c  1, 0.000000000, 1.000000000, 0.500000000,-352805.622314164,
: U. `0 U* P! s/ y  2, 0.000000000, 0.500000000, 0.250000000,-92968.331400711,
, [& Y' U' b: n- n) T! C" L  3, 0.000000000, 0.250000000, 0.125000000,-4146.780462183,
5 H+ O. C( F+ a  ]# Z3 u  4, 0.000000000, 0.125000000, 0.062500000,29178.936016433,
  5, 0.062500000, 0.125000000, 0.093750000,13458.364371693,
# @' W, N" h8 n* p  6, 0.093750000, 0.125000000, 0.109375000,4890.124291139,
  7, 0.109375000, 0.125000000, 0.117187500,430.092693351,
4 {, a6 ^% d# c' \4 W, l& S" S! T  8, 0.117187500, 0.125000000, 0.121093750,-1843.759437944,
2 X+ N2 y: j$ p8 N  9, 0.117187500, 0.121093750, 0.119140625,-703.184646645,
6 M1 O! }7 B6 u, v, t. O! K# m9 ? 10, 0.117187500, 0.119140625, 0.118164063,-135.633470273,
11, 0.117187500, 0.118164063, 0.117675781,147.457778641,
12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
6 K& L: m. Y+ k. o# {" `' O 13, 0.117919922, 0.118164063, 0.118041992,-64.817881139,
9 j5 v" M7 ]; C' g- A" l2 i+ w 14, 0.117919922, 0.118041992, 0.117980957,-29.420780403,
: U# i$ I6 Z8 t( E: O5 R  ]; y 15, 0.117919922, 0.117980957, 0.117950439,-11.724903562,
16, 0.117919922, 0.117950439, 0.117935181,-2.877633532,
: Z8 A! [) e. {" J& G 17, 0.117919922, 0.117935181, 0.117927551,1.545834384,
: l( ^7 q  f1 ?6 P* }2 m 18, 0.117927551, 0.117935181, 0.117931366,-0.665885649,
) Z4 {/ n6 w, B" i 19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
20, 0.117929459, 0.117931366, 0.117930412,-0.112953030,
2 z9 l# M$ y1 S9 o8 D+ E6 D+ @% v# @ 21, 0.117929459, 0.117930412, 0.117929935,0.163512627,
: I/ M0 o. M7 l( C$ ]+ \& V" E# c 22, 0.117929935, 0.117930412, 0.117930174,0.025279853,
23, 0.117930174, 0.117930412, 0.117930293,-0.043836575,
24, 0.117930174, 0.117930293, 0.117930233,-0.009278357,
6 M/ V# K8 }5 u4 x 25, 0.117930174, 0.117930233, 0.117930204,0.008000749,
26, 0.117930204, 0.117930233, 0.117930219,-0.000638804,
27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
% u% O6 @4 C, h# ?3 v1 W3 b- }' m>>
$ J' |6 g; t$ V3 h) _2 x
结论 当x=-1.361时  y=0.117930219
$ _4 ?2 f; z9 _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
$ q/ C( _9 U+ D算法？

二分法        
( `, I3 c6 {" q: C: |7 m还有一个程序
% Bisection.m
function[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
' G$ w, k+ F4 U+ S5 G9 Z8 xfprintf('k ,a ,b ,x, f\n')
for k=1:N
    x=(a+b)/2;
% `0 c  x, g- u5 F    f_value=f(x);
    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
: A  V$ q7 T8 t6 A      ,k ,a ,b ,x, f_value)
) f/ d0 u* ^  e  i' p    if abs(f_value)< eps1||0.5*(b-a)<eps2
    return
7 t: c6 L2 M, `. a0 \! }, x    else
        if f(x)*f(a)<0
) o& }9 t. S* A3 n          b=x;
( E2 Y" _* O7 S. g        else
          a=x;
) `- ~: K; ?. y, C, j: b7 B5 p        end
; [3 F; f- q9 U, k        if k== N
# i: c* T+ k5 |4 F+ C          warning ('算法超出最大迭代数!')   
, c3 F4 G2 g+ ?0 z! ~# V        end
, J, B- i1 e9 Y+ U0 O2 z    end

shouce

单车居士 发表于 2015-6-4 12:41
西交大学生？？问邢老师

) d0 w+ n2 A$ z4 S' C5 L0 [; L8 s这点问题   自己就可以解决

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   

8 N8 o- W/ R2 a9 y: `

shouce

明月山河 发表于 2015-6-4 19:12
; y' s. d" ]: [+ Y! F你这个是代数方程？

& F: N$ r% V  ~: Z
, z; o* v# Z/ @0 ]* w是超越方程      
# ^/ Y) \3 Y0 G6 n: ]* X! R
7 u2 }9 J: \0 c4 s* u, A, A
) u9 P. S6 R$ ^) p, A. J) U[chāo yuè fāng chéng