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
求当x=-1.361时  y=?  
. z4 i6 ?9 m7 y: k9 O- R>> 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
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(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)
3 q! Y. h2 a( X  K. x/ C% W. r
- o# q+ c) o: ]5 Q( ?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)

( ]$ \# w" _1 J  R*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368
7 z0 L1 `3 x+ S% q+ W" q
>>  subs(f, x, -1.361)

ans =

(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin
9 N- C2 N1 C. k; V! a
3 p& j; y+ P4 Q3 t, c(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -

(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368
2 J6 ~( M) s. l9 ]0 L3 {, w1 [

  s1 U9 |' |0 o) O7 \>>  a=0; b=1;
eps1=1e-8;eps2=1e-8;
N=300;
9 U2 a! P, V9 ^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);
( I1 h! \( L6 Q- cHfun=@Bisection;
- X# {2 R2 N, E# }4 ~5 F[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);

运行结果
k ,a ,b ,x, f
  1, 0.000000000, 1.000000000, 0.500000000,-352805.622314164,
! P) `, j7 y. L- _( ?  2, 0.000000000, 0.500000000, 0.250000000,-92968.331400711,
7 k$ w6 v5 x( r; ?' g4 }4 g  3, 0.000000000, 0.250000000, 0.125000000,-4146.780462183,
  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,
  9, 0.117187500, 0.121093750, 0.119140625,-703.184646645,
9 @+ P9 o- k( u0 U, \: n 10, 0.117187500, 0.119140625, 0.118164063,-135.633470273,
11, 0.117187500, 0.118164063, 0.117675781,147.457778641,
" A9 A7 B3 g, P$ ]! E 12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
13, 0.117919922, 0.118164063, 0.118041992,-64.817881139,
14, 0.117919922, 0.118041992, 0.117980957,-29.420780403,
9 [: f3 [2 o( i  d: P3 k. I5 F 15, 0.117919922, 0.117980957, 0.117950439,-11.724903562,
+ @$ K- k8 H$ H$ F4 R* H. R- f 16, 0.117919922, 0.117950439, 0.117935181,-2.877633532,
17, 0.117919922, 0.117935181, 0.117927551,1.545834384,
2 z8 W* v1 m% _ 18, 0.117927551, 0.117935181, 0.117931366,-0.665885649,
19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
7 w8 n- Z" i: I- e! J4 n 20, 0.117929459, 0.117931366, 0.117930412,-0.112953030,
% c- R: }. {( B& q5 d+ k" f- i! \ 21, 0.117929459, 0.117930412, 0.117929935,0.163512627,
! {* L. l; K$ L2 y# L 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,
25, 0.117930174, 0.117930233, 0.117930204,0.008000749,
26, 0.117930204, 0.117930233, 0.117930219,-0.000638804,
! o$ ^2 T* }3 P 27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
>>
" ]( b- l3 T& T: |' ]
& S' O. f. ?5 N5 `( j" e结论 当x=-1.361时  y=0.117930219
4 k. p$ v% x6 r) r) A按此方法   可依次  得出150对x与y的值
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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
% Bisection.m
function[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
' |+ h2 a& H: E9 l2 Tfprintf('k ,a ,b ,x, f\n')
for k=1:N
3 o$ f, z6 D' [8 Z; h. B    x=(a+b)/2;
    f_value=f(x);
% V* u, v8 J4 T6 D3 z8 j    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
8 o* X! k' g* O& M9 M/ j2 T& e3 w1 {. m      ,k ,a ,b ,x, f_value)
    if abs(f_value)< eps1||0.5*(b-a)<eps2
8 ^$ q& H/ f* G/ K1 j    return
) S; i8 ^: e1 m( m1 e) Z( ^    else
0 V9 P7 r9 M& [$ v        if f(x)*f(a)<0
0 r/ u/ L3 U" f  O6 W  c4 [          b=x;
4 \- o( E! T& @4 }" G( x+ U        else
! H6 q; M2 X, q; F1 ?          a=x;
' R. X$ B0 b9 A: `; n6 `        end
        if k== N
( @2 |# i( f: O' [2 r6 M) L2 c5 e2 y          warning ('算法超出最大迭代数!')   
. y4 K2 ~7 ?$ h. t8 [        end
$ n; \: ]: S4 Z9 j" T    end
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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
' w2 U7 a6 h* B8 F) b

shouce

明月山河 发表于 2015-6-4 19:12
你这个是代数方程？

$ m+ Y7 K9 ?2 `3 n  K6 P是的呀      不过matlab写出来的方程  括号很多  不好看   
& X* v4 h4 }$ i' a- J6 h

shouce

明月山河 发表于 2015-6-4 19:12
你这个是代数方程？

是超越方程      

* C4 j! ^) I! N. J' A# C% {
7 Z. ^4 m3 O# i& O; ]7 z2 ?( v[chāo yuè fāng chéng
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