07Lecture-G-Code-cir - F3.0 Example 5(on past year’s quiz Start point = X2.0 Y2 Centre point = X2 Y1 End point = X3.0 Y1.0 R-method N10 G91 G02

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ENGR3190U: Manufacturing and Production Processes Lecture 07 Circular Interpolation , Feeds and Speeds - Notes Example 2 Start point = X-1.0 Y-2.5, Centre point = X.5 Y-2.5, End point = X4.0 Y-2.5 R-method N10 G90 G03 X4.0 R2.5 F3.0 IJ - method N10 G90 G03 X4.0 I2.5 F3.0 Example 3 Start point = X5.2 Y-5.05, Centre point = X5.2 Y-3.8, End point = X3.95 Y-3.8 R-method N10 G90 G03 X3.95 Y-3.8 R-1.25 F3.0 IJ - method N10 G90 G03 X3.95 Y-3.8 J1.25 F3.0 Example 4 Start point = X-1.5 Y0, Centre point = X0 Y0, End point = X0 Y1.5 R-method: N10 G90 G02 X0 Y1.5 R1.5 F3.0 IJ – method: N10 G90 G02 X0 Y1.5 I1.5
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Unformatted text preview: F3.0 Example 5 (on past year’s quiz) Start point = X2.0 Y2., Centre point = X2. Y1., End point = X3.0 Y1.0 R-method: N10 G91 G02 X1.0 Y-1.0 R1.0 F3.0 IJ – method: N10 G91 G02 X1.0 Y-1.0 J-1.0 F3.0 Example Consider a 3/8” – Two Flute HSS end mill used to cut (roughing) an aluminum part. Calculate the feeds and speeds. Use simplified formula for RPM. CS = 240 RPM = 4*240/.375 = 2560 therefore set S2560 in program Feed per tooth = .013 0.005/1” = x/.375 -> x=(0.005)(.375)/1 = .001875 F = (2560)(2)(.001875) = 9.6 set F9.6 in program 1...
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This note was uploaded on 12/09/2010 for the course MECH ENG mech301 taught by Professor Yang during the Spring '10 term at UOIT.

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