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標題: 螺桿轉子刀具設計（2） matlab 解法 [打印本頁]

作者: shouce 時間: 2015-6-4 11:00
標題: 螺桿轉子刀具設計（2） matlab 解法
螺桿轉子刀具設計數學模型

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
求當x=-1.361時  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

(x)+179999.4000005*y*sin(y)*sin(x)+75000*cos(y)*cos(x)*sin(x)

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)

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

>>  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

(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;
eps1=1e-8;eps2=1e-8;
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;
[k,x,f_value]= feval(Hfun, f,a,b,eps1,eps2,N);

運行結果
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,
  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,
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,
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,
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,
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,
>>

結論當x=-1.361時  y=0.117930219
按此方法可依次  得出150對x與y的值

作者: 單車居士 時間: 2015-6-4 12:41
西交大學生？？問邢老師

作者: 明月山河 時間: 2015-6-4 19:12
你這個是代數方程？

作者: cosxuan 時間: 2015-6-4 19:45
算法？

作者: shouce 時間: 2015-6-5 08:37

cosxuan 發表于 2015-6-4 19:45
) R- N. h; g& t算法？

二分法
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
x=(a+b)/2;
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
else
      if f(x)*f(a)<0
      b=x;
      else
      a=x;
      end
      if k== N
      warning ('算法超出最大迭代數!')
      end
end

作者: shouce 時間: 2015-6-5 08:40

單車居士發表于 2015-6-4 12:41 ) G; r7 F! t: s
西交大學生？？問邢老師

這點問題自己就可以解決

作者: shouce 時間: 2015-6-5 08:46

明月山河發表于 2015-6-4 19:12
! g; b( T# c1 }$ H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

作者: shouce 時間: 2015-6-6 18:55

明月山河發表于 2015-6-4 19:12
/ Q i* }! {, f6 q* D @你這個是代數方程？

是超越方程

( J4 j9 P2 Z! S- s8 B7 Q[chāo yuè fāng chéng

作者: 設計者AF 時間: 2015-6-6 20:19
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