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

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
; X( u& z# J- @1 [求当x=-1.361时  y=?  
; D- T( U! l* o- x) T: {/ f6 Z% A" s>> syms x y
4 k$ [+ [- a5 R* E- P4 Q+ ]+ h>> 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)
& o- b: Z9 c, ]' d7 a7 H# v4 h
) a% @  m; G' o0 O$ mf =
& p2 {9 i% u. s" F2 m7 X) 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)

/ m; n: N+ G% {5 C/ w*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368

>>  subs(f, x, -1.361)
8 h: @: B6 ]( j" Z& h
  X2 a* o  l' \8 B7 s: ~ans =
- N5 q7 g) A: \4 B4 A$ c2 h9 R0 ~
. D" Y4 @# [; ]+ N1 m2 x: e(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin
4 p6 I; \" X1 L9 Z* _
(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -

% [, a/ j  z4 X2 }! G/ a(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368

5 G3 B5 p% x: R9 s! U# S
>>  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);
4 \8 j0 D+ g6 f  Q$ F2 g" j$ SHfun=@Bisection;
9 l4 c. \/ [8 Z% y[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);
! J( C. N! F, x: r6 o2 ]" q
: `$ F) x4 s0 _运行结果
k ,a ,b ,x, f
  1, 0.000000000, 1.000000000, 0.500000000,-352805.622314164,
( z8 L) W1 h+ g7 h! r, Q2 A  2, 0.000000000, 0.500000000, 0.250000000,-92968.331400711,
/ t. a/ y9 n( u4 M9 ?7 J; u  3, 0.000000000, 0.250000000, 0.125000000,-4146.780462183,
( F3 c8 k# q( y/ m+ F- y  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,
( u$ c# \6 S* O2 c  k6 _9 v  7, 0.109375000, 0.125000000, 0.117187500,430.092693351,
& J8 k. g8 {1 J9 B  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,
. _) y: S6 }+ J* u5 | 11, 0.117187500, 0.118164063, 0.117675781,147.457778641,
: J5 ]+ L% _( X% h* O) E 12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
' Y, M; G1 b' }, s7 x+ s/ H( n 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,
16, 0.117919922, 0.117950439, 0.117935181,-2.877633532,
17, 0.117919922, 0.117935181, 0.117927551,1.545834384,
18, 0.117927551, 0.117935181, 0.117931366,-0.665885649,
: ?! p! |  Z! H# o 19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
0 C+ [+ }% |8 R% C 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,
4 Y1 }; r1 J5 K, q0 a 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,
27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
>>
# _% I( P# v2 i: r
; P0 n& o# s% A' F结论 当x=-1.361时  y=0.117930219
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        
还有一个程序
; Y) \, f1 `9 R$ p2 K( a0 n# K% Bisection.m
function[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
fprintf('k ,a ,b ,x, f\n')
# z) E# J4 M- X4 P! `1 \4 L3 Qfor k=1:N
  F- x. P) k5 [    x=(a+b)/2;
    f_value=f(x);
    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
; n# i- l) v) }5 R      ,k ,a ,b ,x, f_value)
! |- I4 N# S2 O& I6 x- a0 u    if abs(f_value)< eps1||0.5*(b-a)<eps2
. |: z% h+ t# s' b1 T    return
6 R: ?2 w2 q+ q    else
        if f(x)*f(a)<0
          b=x;
/ z. |2 M# C8 J( O        else
) @! D( H# V4 N- A: A9 F: f          a=x;
        end
$ f( E# ]8 t4 ~, ?: R        if k== N
          warning ('算法超出最大迭代数!')   
! {! H9 X4 r: u  S3 w1 N        end
) G6 [" y# q+ q! @    end

shouce

单车居士 发表于 2015-6-4 12:41
西交大学生？？问邢老师

$ y3 E- x* u. @3 X9 a这点问题   自己就可以解决
$ l2 P% f$ s( v. U9 `

shouce

明月山河 发表于 2015-6-4 19:12
! W8 F/ M! B! K- I4 P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
" Q- W6 y2 G7 I# O你这个是代数方程？

, ^, x% D  Y5 }$ [  J* M
是超越方程      


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