Question

Tool Compensation Worksheet 1 - 1.00 — F 1.40 30.96° 31.570 x t R0.25 2.00 + 1.00 22.50° Typ. - 1.00 - 2.00 - 3.00 - 4.00 - SETPT is 1.0 off of LLC of WP @ 100 above -Cutter Dia. = 0.5 -Machine on outside edge, CW direction & 0.100 depth - Program in incremental SETPT

Coul I get help calculating the tool compensation for this part and write a program for it. This is for a Mill.

Answer 1

იუ he p eop၅ ဆ as a poco sJk02 First calculate the all inclined heights By Applying The trigonometry BC wat Bottom 70.4rg 425 20:56 D ke * coso = BK & sino a B * COS 29.1= * CK =Bexsino BC = 1.08 * CK = sin(3.9-S>) x1.00 ICK=0.414 BC El at Right end Top 1.40 ط ر مو برا 231.57 1.63 HG * cose = HL «) sino? La +9= HL - LG = HG sino loso & LG = 1.63x sin (31.57). HG = 1.40 cos 31-57) 1420.8602 49 = 1.63 )

(x) Top Mk I left End I (x) (0 s8 = MI & Coso = 1 TI O sind - M5 an MJ = sinox I J IJ 30-96 TJ sin (30-96) x 1.166 TJ- Cosc 30-96) «MT = 0.5297 IT- 1.166 objects height for 1 Hence tale find the all inclined positioning the tool Movement easy Calculation you can we tand= opp lite si adjaced side (x) substitute the all find outed values (213) 1.6 lit out 0.599 0.86 1.401/ 6.0 6 SETAT S. Pk, BKR S A pkisk PG, 1) is the the starting point let us go in the one clock klise direction (it is our choise to chose direction of Movent) = 1

6 oo xo yo ForMo3. NOI G71 G90 G94 No No 3 400 мо кож Mo 4 Gol 942 002 yo yo s100 Mo5 Gol XlYI No 6 GOI XIy2.401 107 GOI X2 y 3 NO 8 GOI X 3.6 y 3 Nog Gol x 4.75 y 2.14 NIO G02 X5 y 189 NII Gol 85 yi N12 GOI X4 YI NIS GOI X3 70.414 M12 Gol x2 x M13 GOI XIYI N14 940 MOQ, Moy, M20 M21 M30, > M= Block Numby Gol = lineay inty polation G71 = Metric Selection Gor= clockwise direction 990- Absloute programming Mos spindle on Mo4 = spindle off G9u feed in mm Mo7 = Coolent on Goo - Rapid travers MOG - Coolent off M30 = End programme 11. 02