螺桿轉子刀具設計（2） matlab 解法

shouce

螺桿轉子刀具設計 數學模型

0 m3 r) d7 [& R6 m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
# M3 C. J2 V  W8 b" J. g求當x=-1.361時  y=?  
>> syms x y
. D$ O: J; y; M>> 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

( b8 T' N; v7 j) S1 N(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)

' `6 {, W0 ]$ N+ e# w( @& y# jf =
- \( ~) ?+ r$ E7 c! y
(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)
9 A' K2 `$ E5 x6 q
1 o/ }- E/ J4 ^: t- @; K: ~4 l( l*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368
6 f& o$ Q  [5 v  g( }
>>  subs(f, x, -1.361)
8 X1 k, K" s$ Y5 H3 q1 C6 W& M0 }
ans =

+ v/ }7 a; l0 A& M(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin

, f5 E' X% L' U(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


>>  a=0; b=1;
7 k* k8 C0 \; n' s3 K1 o! o  eeps1=1e-8;eps2=1e-8;
N=300;
, D+ q, f% R0 gf=@ (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);
! n, F; y7 [! j9 u# RHfun=@Bisection;
[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);

3 f+ H3 Q! [' _: ]運行結果
k ,a ,b ,x, f
# k, l) Y% m5 m, R3 V  1, 0.000000000, 1.000000000, 0.500000000,-352805.622314164,
  2, 0.000000000, 0.500000000, 0.250000000,-92968.331400711,
0 Y. I# @% V# W. d* X% G1 G  3, 0.000000000, 0.250000000, 0.125000000,-4146.780462183,
. r5 ^7 C! b  `. W9 Y9 Z  4, 0.000000000, 0.125000000, 0.062500000,29178.936016433,
4 r8 B: t! i$ a: D4 w' `  5, 0.062500000, 0.125000000, 0.093750000,13458.364371693,
6 K0 l. }$ r& m  6, 0.093750000, 0.125000000, 0.109375000,4890.124291139,
% d7 N: L$ A5 S+ J) G, C  7, 0.109375000, 0.125000000, 0.117187500,430.092693351,
; x( K  H9 k5 j* R, W  8, 0.117187500, 0.125000000, 0.121093750,-1843.759437944,
! A) e# r2 I, R$ u  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,
+ w" W) [0 \) O. J" E 12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
13, 0.117919922, 0.118164063, 0.118041992,-64.817881139,
: a5 C' Q9 C; E8 d* ?0 W 14, 0.117919922, 0.118041992, 0.117980957,-29.420780403,
. s% o" R8 t4 q2 d. c 15, 0.117919922, 0.117980957, 0.117950439,-11.724903562,
# n2 |. V. p6 W8 y6 |# x 16, 0.117919922, 0.117950439, 0.117935181,-2.877633532,
17, 0.117919922, 0.117935181, 0.117927551,1.545834384,
" l0 c" z; G6 B  _, t: ]- a+ c0 l: [ 18, 0.117927551, 0.117935181, 0.117931366,-0.665885649,
19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
20, 0.117929459, 0.117931366, 0.117930412,-0.112953030,
  Z6 R- m) N; J5 L2 q 21, 0.117929459, 0.117930412, 0.117929935,0.163512627,
0 s9 W. c! [0 j" U$ v 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,
27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
3 l* y5 w) H2 Y" K. ]>>

; {% L4 ?% c+ f* L. d1 C" L" k結論 當x=-1.361時  y=0.117930219
* g* J. T9 m7 j; o8 ]0 W按此方法   可依次  得出150對x與y的值
* V$ l0 o. l) q+ J

單車居士

西交大學生？？問邢老師

明月山河

你這個是代數方程？

cosxuan

算法？

shouce

cosxuan 發表于 2015-6-4 19:45
算法？

二分法        
6 F% T8 U1 p$ O' r還有一個程序
% 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
) w/ s9 x% U- @, F' n8 e4 D    x=(a+b)/2;
2 U9 `. G0 }* B6 n    f_value=f(x);
    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
      ,k ,a ,b ,x, f_value)
    if abs(f_value)< eps1||0.5*(b-a)<eps2
    return
( ?% T+ \* f, H8 y& x8 I    else
$ ^- j; I: T0 P, t        if f(x)*f(a)<0
          b=x;
: T7 _) ]% s: u0 a7 g  u        else
4 y6 E+ |+ ]! |, [8 ]/ Y: _, D          a=x;
" C! \' I8 m* q. T+ |        end
" q- S3 a+ q9 [( z. ]( c2 }5 F; J        if k== N
; j9 T9 r) m% u  S          warning ('算法超出最大迭代數!')   
9 w, f* K& j* d        end
    end
) |) t6 d! o5 r3 ~' j

shouce

單車居士 發表于 2015-6-4 12:41
西交大學生？？問邢老師

這點問題   自己就可以解決
6 }9 m8 s6 t; u+ Q/ Z8 g

shouce

明月山河 發表于 2015-6-4 19:12
) Q5 i9 J2 P' q6 E1 }" ~5 y你這個是代數方程？

是的呀      不過matlab寫出來的方程  括號很多  不好看   

, v8 h. ]; M/ C% B* Q% [6 X2 I

shouce

明月山河 發表于 2015-6-4 19:12
! [, C: ~$ P) w* _+ \  x你這個是代數方程？

是超越方程      


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