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发表于 2015-12-25 10:33:48
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回复 2# 若枫
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老大:+ l+ @+ s( D& P4 Q% M
感谢您的回复!以下是我的方法,这样的确出来两个G代码。' f& H2 n* f M6 ~: Q' I' m
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if drillcyc$ = 9,
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pdrlcommonb
5 d/ r2 S4 m. F1 v/ h peck1st =peck1$
6 Z: ?0 G+ }+ n; Q/ E if dwell$ = 0,pcan1, pbld, n$, *sgdrlref, "G1181", pxout, pyout, pfzout,
+ w; E* W t! B5 ` prdrlout, dwell$, *peck1st, *peck2$, *feed, strcantext, e$( V& y* K: L7 M' m) S" B4 o
else, pcan1, pbld, n$, *sgdrlref, "G1182", pxout, pyout, pfzout,
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' s' I( U: H; [" h) O' m( `但使用以下的方法出不来一个钻孔循环两个不同的G代码。上面的"G1181”用原来的*sgdrill出来的是G812 E5 x, x1 @6 M# I; y- Q, [ E, v: ^
sg85 : "G85" #bore #1 - no dwell
( W: B! o: Z- ]/ L, h: Tsg85d : "G89" #bore #1 - with dwell
$ X, ^, J) k- X3 r0 usg86 : "G86" #bore #2 - no dwell
2 D# `; _6 c$ f( ^( p0 dsg86d : "G86" #bore #2 - with dwell
+ W; A8 u+ r: O/ y2 _sgm1 : "G76" #fine bore - no dwell0 J; P3 f. A1 H/ k3 p1 `
sgm1d : "G87" #fine bore - with dwell/ H+ ~& k0 b8 ^' j/ O, ]
sgm2 : "G84" #rigid tap - right hand5 o- F* h9 R; N4 t) W
sgm2d : "G74" #rigid tap - left hand
7 R" ]6 F; n0 J6 Ssgc9 : "G183" #custom #9 - no dwell
) b/ ]' c. U& u' r. q/ Z& _sgc9d : "G184" #custom #9 - with dwell- X% [7 C @4 }7 \
sgc10 : "G185" #custom #10 - no dwell$ L; g, q" r' a' M+ y
sgc10d : "G186" #custom #10 - with dwell
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3 L# l8 k l4 W4 y' D请指点一二,谢谢!
8 E! _- K$ m# o, \3 Q6 w最终我要出25个左右钻孔循环,30多个G代码。 |
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