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本帖最后由 qinyuanmin 于 2012-3-17 16:24 编辑 $ G% v y" G/ P+ g# n T
( B4 e7 y S9 o7 X+ y5 x
" `7 z! @7 ^4 H& } U5 y8 v6 d%
, r! a( @; u- }1 B( z(PROGRAM NAME: 031f8)1 H' y4 e5 Y8 Y. Z4 h, B
(D=32.0, R=6.0)! O7 e5 v) e- H3 @, q9 M
G40 G17 G49 G69 G80 G90 G54: v% e9 g- n R. D: ^* L
G91 G28 Z0.0& S$ m5 C* r1 P/ J& R) d: j, S
S1000 M03" a: e# m, N5 H% \; {1 t
G0 G90 X-68. Y-91.1 Z3 }# q9 J; ], J+ p2 j
Z100.
8 b$ w1 h7 `1 O. EG98 G81 X-68. Y-91. Z-69.9 R3. F200.9 J& h$ J+ [- C1 x
Y91. R3.# L# g" ?# { z
G80+ f% ] C( V ~5 f8 S6 _/ R
G0 Z100. (第1步完,↓第2步的转速2500没有): O2 g( Y! b- U0 }' k' B# N
G0 X-93.299 Y105.875
+ ~0 ~1 ?; J8 e# q2 ?Z100.) F4 Z2 b6 Z* O6 s& n+ k
Z1.% u# Q. a! m" ]& O8 G$ w* z' o
G1 X-91.812 Y106.01 Z.948 F3500.
" P W% G; U1 m; dX-77.208 Z.4384 U0 R, C9 D' ^4 b4 c# j' f1 O
X-76.191 Y107.51 Z.375, K, J2 @: E5 |7 p
X-73.178 Y111.109 Z.211/ h6 I4 Q! S9 K" y$ d+ Y7 X- `
X-69.772 Y114.341 Z.0477 q# z8 K* I/ G! {, I
X-66.021 Y117.163 Z-.117
7 E* E$ x6 O3 }X-64.47 Y118.1 Z-.185 A8 `" Y( Q# A% g" ?1 ^( a; t
同一把刀,2步程序转速不一样,一起后处理出来,第2步默认了第一步的转速,求高手解答,具体在后处理构造器什么地方修改可恢复?" W/ k# a, e9 m
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