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
& {1 D0 @3 ^5 f0 M求当x=-1.361时  y=?  
>> syms x y
! @4 D, ]2 q8 i/ w7 ^6 R" y2 j>> 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)

f =
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(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)

# x# K0 x& ^8 O5 L*sin(x) - (6184732290414159*y*cos(x)*cos(y))/34359738368 + (6184732290414159*y*sin(x)*sin(y))/34359738368

>>  subs(f, x, -1.361)
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+ p0 X  ?# S' C3 b$ L% Eans =

9 d  R0 K/ g& P(3359995*sin(1361/1000)*cos(y))/2 - (2999995*cos(1361/1000)*sin(y))/2 - (7215543768789811*sin(1361/1000))/4294967296 + 75000*sin

) f: y5 T$ }: I: }! p, M(1361/1000)^3 - 75000*cos(1361/1000)*sin(1361/1000)*cos(y) - (6184732290414159*y*cos(1361/1000)*cos(y))/34359738368 -

5 v' C  y# e$ q1 z: C& n. ~# V(6184732290414159*y*sin(1361/1000)*sin(y))/34359738368
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>>  a=0; b=1;
7 L4 o8 E2 O$ F8 Q5 reps1=1e-8;eps2=1e-8;
! p+ }" V' t4 H 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);
Hfun=@Bisection;
* N5 a) O3 @# c3 j[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);
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0 v. P" B8 H- C( H! H( c运行结果
k ,a ,b ,x, f
2 }+ E, c0 \, n  1, 0.000000000, 1.000000000, 0.500000000,-352805.622314164,
  2, 0.000000000, 0.500000000, 0.250000000,-92968.331400711,
& g- H$ k. H8 \" Z  3, 0.000000000, 0.250000000, 0.125000000,-4146.780462183,
  4, 0.000000000, 0.125000000, 0.062500000,29178.936016433,
" c6 I4 O3 a. J5 {  5, 0.062500000, 0.125000000, 0.093750000,13458.364371693,
* }' Z  ^8 p) a: k# ]8 {  6, 0.093750000, 0.125000000, 0.109375000,4890.124291139,
  7, 0.109375000, 0.125000000, 0.117187500,430.092693351,
4 c: b- T2 V/ {' J& l+ I  R  8, 0.117187500, 0.125000000, 0.121093750,-1843.759437944,
# _$ f, A( b6 ~; a. k" C" g5 q  9, 0.117187500, 0.121093750, 0.119140625,-703.184646645,
3 ^# Z* H! O' i7 L 10, 0.117187500, 0.119140625, 0.118164063,-135.633470273,
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,
& \; d; s& e: V8 @4 R 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,
& m, p2 Y1 ?, Y 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,
- v2 M4 d& L0 T( W+ w: @  t 22, 0.117929935, 0.117930412, 0.117930174,0.025279853,
" ^: w( l) G7 \) g5 _, E 23, 0.117930174, 0.117930412, 0.117930293,-0.043836575,
24, 0.117930174, 0.117930293, 0.117930233,-0.009278357,
8 V) X- Q- X$ v' B* e5 C 25, 0.117930174, 0.117930233, 0.117930204,0.008000749,
' l! G- Y( r- f0 t 26, 0.117930204, 0.117930233, 0.117930219,-0.000638804,
27, 0.117930204, 0.117930219, 0.117930211,0.003680972,
>>
* Q# P1 z9 r+ ]. ]
结论 当x=-1.361时  y=0.117930219
: ~6 [- ~" m7 _6 {3 v% H! }% h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
, Q2 F; ^9 `& c1 K4 k$ H/ w0 B算法？

" z+ n9 }! M2 \& g3 I6 X! H二分法        
5 C+ q7 H9 e! D; z  j2 w) d; }2 g还有一个程序
% Bisection.m
* h5 }! w% l' I  e8 U: h) H  qfunction[k,x,f_value]=Bisection(f,a,b,eps1,eps2,N)
% p% y9 l% {$ F7 d9 `. Xfprintf('k ,a ,b ,x, f\n')
3 i* Q% D/ v& @6 Sfor k=1:N
4 r" |# Z& L3 S0 X    x=(a+b)/2;
$ A/ B* C$ U5 ^4 T    f_value=f(x);
5 A" Q+ W* x2 ^) {3 T9 w6 S    fprintf( '%3d, %10.9f, %10.9f, %10.9f,%10.9f,\n'...
      ,k ,a ,b ,x, f_value)
  T: ^/ c4 E) p( }- y; ^/ p    if abs(f_value)< eps1||0.5*(b-a)<eps2
# R- r4 v8 i4 R" Y  V    return
$ ~' G8 u, V5 J5 Q( x1 p    else
        if f(x)*f(a)<0
          b=x;
! \+ q( ^8 u# x  ^8 L        else
- K4 V. b- c: K: o          a=x;
) B5 Y. H, ~9 U* N; `- Y        end
        if k== N
          warning ('算法超出最大迭代数!')   
        end
- U/ R; L2 P0 z- u2 p& c    end
6 o5 B* n: H" Z+ k7 M9 o

shouce

单车居士 发表于 2015-6-4 12:41
7 m1 x3 g8 `7 {2 |( F4 W/ K西交大学生？？问邢老师

* U* p- S' c- w8 T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？

  g5 T# T" v" b是的呀      不过matlab写出来的方程  括号很多  不好看   

& w6 `7 k1 ~- s# l$ D

shouce

明月山河 发表于 2015-6-4 19:12
4 D. @: `% a% V" m你这个是代数方程？

; \- v) ]* w" @* f) B
3 C: e/ S  M) N7 P9 V% F5 ^: m是超越方程      
9 \% U# N3 S4 G( T( K3 e% f
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8 N2 p0 e2 e9 Z- R' f6 E[chāo yuè fāng chéng
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