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
3 r2 Z/ n4 z7 g/ }" E/ f
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
+ Y9 t* L* X( G# i% a4 p( q求当x=-1.361时  y=?  
>> 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
* w3 I" J. y5 d8 e9 z" V( N! S$ M
(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)
: \) @% _7 S( k# I0 |: N
f =
* C: O) j6 c3 d/ m  M
6 K$ v/ K  @; U- k- 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)

*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368

& f$ s6 M) @$ V* D- Q>>  subs(f, x, -1.361)
. m5 h- w/ I) H* {1 Q; r
ans =

3 F. X! D8 v  d. V& q7 m(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin

' q; t( B  X, Z' y) q- R8 I' r4 n(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -
. |9 m* E8 S& [
(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368


>>  a=0; b=1;
* ~% C" k8 y6 v# V) Reps1=1e-8;eps2=1e-8;
4 s& i8 W. B; t- M. }/ B8 L N=300;
0 J+ M- `: Q# k. Z# _$ @$ p% wf=@ (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);
8 x& R# R$ h4 jHfun=@Bisection;
, |+ U& H7 S: o( q[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);
- w* h" A! H) \* W; d
运行结果
) h/ ]3 I% t. e" B7 `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,
3 C" P- q* \8 |  4, 0.000000000, 0.125000000, 0.062500000,29178.936016433,
2 M) t* d2 E5 c0 _# l. {; L  5, 0.062500000, 0.125000000, 0.093750000,13458.364371693,
! ~7 s, k/ |) [0 J" f( h1 v* x  6, 0.093750000, 0.125000000, 0.109375000,4890.124291139,
, {. g) W$ _/ r" B7 Y4 n+ ^  7, 0.109375000, 0.125000000, 0.117187500,430.092693351,
  8, 0.117187500, 0.125000000, 0.121093750,-1843.759437944,
! p' O8 V% ^& Z: U2 |+ U+ D" ~! R  9, 0.117187500, 0.121093750, 0.119140625,-703.184646645,
10, 0.117187500, 0.119140625, 0.118164063,-135.633470273,
' S( r1 W. ~* |) d3 g1 I  R! ~7 j 11, 0.117187500, 0.118164063, 0.117675781,147.457778641,
12, 0.117675781, 0.118164063, 0.117919922,5.969190901,
13, 0.117919922, 0.118164063, 0.118041992,-64.817881139,
1 N' j5 |2 P5 @% G 14, 0.117919922, 0.118041992, 0.117980957,-29.420780403,
15, 0.117919922, 0.117980957, 0.117950439,-11.724903562,
# G3 f* x$ H  g% \" N9 r 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,
19, 0.117927551, 0.117931366, 0.117929459,0.439977849,
20, 0.117929459, 0.117931366, 0.117930412,-0.112953030,
21, 0.117929459, 0.117930412, 0.117929935,0.163512627,
1 y5 ?8 |% j/ g' {& {" n9 h 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,
3 T& @9 a. M4 |2 f$ T 25, 0.117930174, 0.117930233, 0.117930204,0.008000749,
) t  z8 V. C0 Z 26, 0.117930204, 0.117930233, 0.117930219,-0.000638804,
27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
5 v* T8 T" j" N2 `>>

/ x" S- v: }% \9 }; m结论 当x=-1.361时  y=0.117930219
8 Q3 G7 i) p- c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？

) d7 Z$ }' r/ b' W二分法        
还有一个程序
" c0 T: |7 H9 T) u# u7 y9 X0 y) n% Bisection.m
2 s- v9 W- B4 l, w# |( Q$ t+ zfunction[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
fprintf('k ,a ,b ,x, f\n')
for k=1:N
    x=(a+b)/2;
" i) C; r( q; C. E0 S    f_value=f(x);
" S, w  k* a3 w( N$ _6 T# g    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
      ,k ,a ,b ,x, f_value)
5 X" `$ X  d) o- ?  s8 I    if abs(f_value)< eps1||0.5*(b-a)<eps2
- r/ w) k  t  w. L$ A7 ], T4 w5 V5 ~    return
; V: k/ k3 F9 @# p- n6 D    else
. q. Q- f3 a2 A" f5 E        if f(x)*f(a)<0
+ E( D7 e3 k) u! v% U& w9 J2 b          b=x;
        else
          a=x;
        end
: O3 {, {* L7 Q) E        if k== N
          warning ('算法超出最大迭代数!')   
        end
    end
" U; O1 |! m/ L% n. N, ?( _

shouce

单车居士 发表于 2015-6-4 12:41
, d$ G: s  ~2 ?% C6 ~7 c西交大学生？？问邢老师

, G* H+ j0 S5 ~6 w' [* S: F- D这点问题   自己就可以解决

shouce

明月山河 发表于 2015-6-4 19:12
+ w( F6 I6 n+ f4 H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
" c6 l6 H; K: h: a7 v. C5 x  o" c你这个是代数方程？

3 Y$ t% w2 D! U1 i3 |3 x  n
是超越方程      
1 y4 u' v3 Q" L

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