哪里有計算圓柱度誤差的程序,？

deltabobchen

對呀, 三座標(biāo)可以給你自動評定的呀, 你只要結(jié)果就可以了!

luxiang821

我原來也關(guān)注過一些關(guān)于形位公差評定的數(shù)學(xué)方法,，不過沒學(xué)過編程很想了解實現(xiàn)這些方法的編程算法的一些問題,。不知道有沒有人有興趣談?wù)撓拢�讓我也長長學(xué)問,。

zhuai1003

我也不會編程，這兒有篇測量方面的文章,，與大家共享

胡雷

只是 想知道這方面的計算是怎樣編程的 ,，一定要知道評價的條件,，在把條件列入編程   如有不對地方多多指教

hezhikai

我也在找啊

foreverhua

我會編一點 但是那個程序老是運行部出來  程序如下 請高手指點
8 j! m. n% A9 R* Q5 B: z0 |function z=myfun2(x)
* N  z6 X4 W3 x, R* ^: I9 l; M( E8 Sl=[38.8263,35.4687,25.9806,11.9948,-4.0615,-19.4158,-31.4083,-37.9757,-37.9629,-31.4080,-19.4108,-4.0573,
11.9961,25.9829,35.4653,38.8230,35.4619,25.9784,11.9957,-4.0606,-19.4149,-31.4044,-37.9785,-37.9743,-31.4093,
9 t9 U' k+ ]+ O( ]-19.4135,-4.0569,11.9971,25.9795,35.4636]
m=[0.0014,15.7895,28.8524,36.9273,38.6135,33.6208,22.8189,8.0678,-8.0705,-22.8217,-33.6211,-38.6107,-36.9260,
-28.8503,-15.7902,-0.0005,15.7852,28.8491,36.9245,38.6123,33.6212,22.8174,8.0687,-8.0721,-22.8214,-33.6248,
/ U+ O' v; R4 P9 h" N9 r-38.6129,-36.9251,-28.8459,-15.7876]
n=[-2.0010,-2.9670,-3.9326,-4.8979,-5.8630,-6.8282,-7.7937,-8.7601,-9.7259,-10.6904,-11.6543,-12.6209,-13.5862,
. t" F0 j$ u( v' h-14.5507,-15.5175,-16.4838,-17.4502,-18.4149,-19.3811,-20.3463,-21.3115,-22.2779,-23.2418,-24.2071,-25.1726,
; Q5 g$ F# ]' ~0 @" ]9 T2 _-26.1366,-27.1026,-28.0690,-29.0360,-30.0008]
z=0;
for k=1:30
6 S" `  z$ G% r1 }    r(k)=(sqrt((l(k)-x(3)*n(k)-x(1))^2+(m(k)-x(4)*n(k)-x(2))^2-x(5))^2
1 |0 o& O6 S5 o, s$ [7 w    z=z+r(k)
% v2 k- |; M4 z; O/ M# M+ v" zend


: G3 d3 s/ x5 y' U; ]: K& x
9 W1 T! J1 K6 }+ R2 B% |8 }$ ux0=[0,0,1,1,38.000];
[x,favl,exitflag,output]=fminunc(@myfun2,x0)