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
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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=?  
% o" F9 l0 |! {9 B0 y# f: m>> syms x y
3 x# c; X, K. ?4 ^  U>> 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)

/ E1 H! u6 [  @+ 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)

% n$ o  p8 S* L; _5 X. n; W*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368
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' {1 Y7 W4 k) e2 x1 [. T>>  subs(f, x, -1.361)
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, c/ b$ e8 Z; _8 eans =

(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin
( E$ \' D6 m" y  b2 n
" f# S  g, m" _0 {(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -

0 Y. R3 l, H% |(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368


* `! S! h9 x) F>>  a=0; b=1;
; y5 t! q9 i* n& Z+ h7 Aeps1=1e-8;eps2=1e-8;
  _0 a) f8 m' o5 F 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);
, h& c, _" f  N4 m  W) JHfun=@Bisection;
3 g5 x+ K1 }& o+ }& F5 Y( g# G[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);
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% u" G3 c4 f7 D) U  R* }+ W6 M运行结果
k ,a ,b ,x, f
  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,
, \% U6 j/ S* t$ f; R: l  5, 0.062500000, 0.125000000, 0.093750000,13458.364371693,
  6, 0.093750000, 0.125000000, 0.109375000,4890.124291139,
" j/ G1 `4 S4 O  7, 0.109375000, 0.125000000, 0.117187500,430.092693351,
* N* R, V; K% S# p9 Z/ T/ {  8, 0.117187500, 0.125000000, 0.121093750,-1843.759437944,
4 O4 {8 A' ^: P% a' A  9, 0.117187500, 0.121093750, 0.119140625,-703.184646645,
10, 0.117187500, 0.119140625, 0.118164063,-135.633470273,
# T! p; H7 O6 r5 L5 C1 k& ]! L 11, 0.117187500, 0.118164063, 0.117675781,147.457778641,
0 d$ F2 @/ r; _# c, c  o7 | 12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
# r0 d/ X" ?6 y 13, 0.117919922, 0.118164063, 0.118041992,-64.817881139,
; C0 E* H+ N- b. d9 S$ F$ n! @ 14, 0.117919922, 0.118041992, 0.117980957,-29.420780403,
  N% [! v/ l) }- R( [* d) c 15, 0.117919922, 0.117980957, 0.117950439,-11.724903562,
16, 0.117919922, 0.117950439, 0.117935181,-2.877633532,
3 g/ ]- q  n8 l0 C! v2 _# S 17, 0.117919922, 0.117935181, 0.117927551,1.545834384,
1 p) |; u; z- p0 P. H1 j  W 18, 0.117927551, 0.117935181, 0.117931366,-0.665885649,
- B8 Z6 x8 s$ R$ W, b 19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
5 }- `: O% s. I  D; I 20, 0.117929459, 0.117931366, 0.117930412,-0.112953030,
1 E6 V, w8 t) B) G 21, 0.117929459, 0.117930412, 0.117929935,0.163512627,
22, 0.117929935, 0.117930412, 0.117930174,0.025279853,
9 D2 x5 C3 M1 C) ~% W% x 23, 0.117930174, 0.117930412, 0.117930293,-0.043836575,
: y6 @- l& z8 \ 24, 0.117930174, 0.117930293, 0.117930233,-0.009278357,
% x7 c3 L) q0 e) M8 }2 V, P 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,
>>
* |9 I/ v' E$ |6 `5 R* q
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        
6 \7 o5 P% g) X2 o4 k4 q, x还有一个程序
- X1 C4 U# l  b! f( _% Bisection.m
' z( N6 U  Z7 I1 _" Nfunction[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
fprintf('k ,a ,b ,x, f\n')
for k=1:N
# k9 r" U6 f: @8 j3 c3 p    x=(a+b)/2;
    f_value=f(x);
    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
& k3 D- i3 O8 f- G& @      ,k ,a ,b ,x, f_value)
: H" L& I1 Z: V1 L    if abs(f_value)< eps1||0.5*(b-a)<eps2
    return
' e) x# C, Z, X4 ~6 ^    else
& @* M) F# n- E% N: g        if f(x)*f(a)<0
; w4 r/ [( M2 k& H' A% E          b=x;
        else
          a=x;
$ E6 G, B8 ~# J1 P        end
8 C, a  Q; v5 @' f+ t; s        if k== N
3 T0 M8 N& ~/ M) N8 J: X          warning ('算法超出最大迭代数!')   
# k/ Y) @, ^8 h0 |8 s- I        end
" _! ~# g/ V1 m    end
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shouce

单车居士 发表于 2015-6-4 12:41
- _, N6 V$ ~& r% g% y$ `: f9 h6 g西交大学生？？问邢老师

/ ]# S( l$ V( W1 O% Q" c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   
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shouce

明月山河 发表于 2015-6-4 19:12
( j5 j/ e/ g% D- [8 ~你这个是代数方程？

是超越方程      
2 S: [1 Y: N% E

& m. h6 k" W( S0 o- L- s6 P: z[chāo yuè fāng chéng
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