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本帖最后由 caiyingkai 于 2010-6-6 16:41 编辑
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- `3 W! {% S) ^* r* Z( q4 e( ?0 g处理出来的程式放机器上跑+Z超程?5 S' Z5 x& C8 n; y4 s% z
这程式怎么超程的? b2 M5 b5 \1 R: n A4 J* G
发那科系统!* d0 r0 |6 v7 |
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如下1 n6 _! e/ D; S
N0010 G00 G40 G49 G17 G80 G90
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N0030 T01 M06
+ a" {1 b. @& zN0040 G00 G90 G54 X0.0 Y0.0& Q9 Y( h( F( h& G L
N0050 M08& a8 z! N! `6 y3 _. A
N0060 S1800 M03
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N0090 Z24.0 ^' L/ N: P1 v V) E
N0100 G01 Z21. F1000.1 q; G: [3 v8 E. u& M, T
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N0190 Z36.5
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N0260 G00 Z37.51 A3 m* v9 h6 U5 _
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, t6 ?' R% o. J1 `" ?; f: c还有就是出来一个报警020
5 o- Y, d8 p! q' F020 圆弧插补中,起始点和终点到圆心的距离的差大于876号参数指定的 & J, Y3 m i: F9 r& [$ \
数值。) y2 S5 `7 }3 r9 t. m
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这个是什么意思呢!我的程式为
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N0090 Z19.- |0 { x2 C& L# l( F- _8 @
N0100 G03 X-.459 Y-1.265 I.5 J0.0 F800.
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9 a! u6 k8 w8 xN0120 X0.0 Y-1.5 I.265 J.4599 b, V/ B/ G. u/ p! N
N0130 X.266 Y-1.507 I0.0 J1.5- n% X% x, V. G; ?5 N6 b [9 x
N0140 X.523 Y-1.438 I-.266 J1.507
. C* j; r* B& ` ]N0150 X.75 Y-1.299 I-.523 J1.4389 f, ~0 ~" f. E! q3 d2 `# ]
N0160 X1.238 Y-.899 I-.75 J1.299/ C6 c9 e9 s/ t' j: f
N0170 X1.497 Y-.318 I-1.238 J.899
; N, E% E5 {( y/ iN0180 Y.318 I-1.497 J.318
0 U, m( X) i* n$ wN0190 X1.238 Y.899 J-.3188 |& N8 `; S) h6 L: N: J
N0200 X.75 Y1.299 I-1.238 J-.899) Z( R! r6 V4 y- l9 N" j
N0210 X.16 Y1.522 I-.75 J-1.299' y! _9 D, T0 ]% D' T
N0220 X-.473 Y1.455 I-.16 J-1.5224 o9 [3 T4 u$ C0 ?# B1 Z
N0230 X-1.024 Y1.137 I.473 J-1.455- i, p' X/ V% {. e& L7 d/ j
N0240 X-1.398 Y.622 I1.024 J-1.137 |
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