哪里有計(jì)算圓柱度誤差的程序,？

deltabobchen

對(duì)呀, 三座標(biāo)可以給你自動(dòng)評(píng)定的呀, 你只要結(jié)果就可以了!

luxiang821

我原來(lái)也關(guān)注過(guò)一些關(guān)于形位公差評(píng)定的數(shù)學(xué)方法,，不過(guò)沒(méi)學(xué)過(guò)編程很想了解實(shí)現(xiàn)這些方法的編程算法的一些問(wèn)題,。不知道有沒(méi)有人有興趣談?wù)撓拢�讓我也長(zhǎng)長(zhǎng)學(xué)問(wèn),。

zhuai1003

我也不會(huì)編程，這兒有篇測(cè)量方面的文章,，與大家共享

胡雷

只是 想知道這方面的計(jì)算是怎樣編程的 ,，一定要知道評(píng)價(jià)的條件,，在把條件列入編程   如有不對(duì)地方多多指教

hezhikai

我也在找啊

foreverhua

我會(huì)編一點(diǎn) 但是那個(gè)程序老是運(yùn)行部出來(lái)  程序如下 請(qǐng)高手指點(diǎn)
function z=myfun2(x)
2 Q0 P/ X) n8 T8 a; Nl=[38.8263,35.4687,25.9806,11.9948,-4.0615,-19.4158,-31.4083,-37.9757,-37.9629,-31.4080,-19.4108,-4.0573,
+ q# y0 m& a+ T; p. Z7 k11.9961,25.9829,35.4653,38.8230,35.4619,25.9784,11.9957,-4.0606,-19.4149,-31.4044,-37.9785,-37.9743,-31.4093,
-19.4135,-4.0569,11.9971,25.9795,35.4636]
: I6 ^+ e0 d, ]7 @9 H/ Im=[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,
( m" H( b  {4 H5 T-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,
. D# n) P* |) v! y( ~& I$ c-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,
0 o, Q3 p3 X* }" a' B" o-14.5507,-15.5175,-16.4838,-17.4502,-18.4149,-19.3811,-20.3463,-21.3115,-22.2779,-23.2418,-24.2071,-25.1726,
, N. A" I; {: u$ _$ `  D-26.1366,-27.1026,-28.0690,-29.0360,-30.0008]
4 P. S: X! ]! B5 M: Z' n% tz=0;
for k=1:30
    r(k)=(sqrt((l(k)-x(3)*n(k)-x(1))^2+(m(k)-x(4)*n(k)-x(2))^2-x(5))^2
4 |9 V0 _+ {) u, E3 p    z=z+r(k)
" ^. Q* w2 N# J3 w( dend

* Z- F2 A- @  z3 \7 l
$ \; F, _4 Z* m& h' R/ y- j& [  M- a
x0=[0,0,1,1,38.000];
[x,favl,exitflag,output]=fminunc(@myfun2,x0)