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MILLING TITANIUM  Table 14 Auxiliary Data for Face and Peripheral Milling When AE<90°



                                                                                              0.95
                                                                                 0.85
                                                        0.65
                                                                     0.75
                        Ae/d
                                           0.55
                                                        107.5°
                                                                                              154°
                        AE
                                           95.7°
                                                                     120°
                                                                                 134.4°
                                                                     60°
                        AE/2
                                           47.8°
                                                        53.7°
                                                                                 67.2°
                                                                                              77°
                                                                                              0.974
                                           0.741
                                                                                 0.921
                                                                     0.866
                                                        0.806
                        Sin (AE/2)
                        Example
                        Find chip thickness for a deep square shoulder in a titanium part is roughly machined by an 80
                        mm (3 in) diameter indexable extended flute cutter that is operated with the following data:
                        - width of cut ae=60 mm (2.25 in),
                        - feed fz=0.15 mm/tooth (.006 ipt).
                        Basing on Fig. 13, case a), and Table 12, average chip thickness hm=fz×sin(AE/2) and hmax=fz.
                        Due to ae/d=60/80=0.75, from Table 14 AE=120° and sin(AE/2)=0.866.
                        So, hm=0.15×0.866=0.13 (mm) and hmax=0.15 mm.
                        Note. Estimating hm with the use of 2nd method (Table 12) gives the result as below:
                        hm=fz/2×[√2/2+cos((AE-90°) /2)]
                        =0.15/2×(√2/2+cos15°)=0.125 (mm)
                        Similarly, for 3 in. Dia. milling cutter:
                        ae/d=2.25/3=0.75, sin(AE/2)=0.866, hm=.006×0.866=.005 (in), hmax=.006 in.
                        Note. Estimating in accordance with the 2 method (Table 12) will give
                                                    nd
                        hm=.006/2×(√2/2+cos15°)=.005 (in)

                         Unfavorable, unstable, heavy and heavy-duty examples
                         The above terms may be simply explained with everyday life analogies:

                         Imagine that you are driving along a road. If your road is horseshoe and undivided,
                         covered with stones or pitted –  you are driving in unfavorable conditions.

                         If you are driving in a car carrying two bicycles on an upright mount
                         on the car roof – you are driving in unstable conditions.
                         A large-size truck transporting a caterpillar excavator is a typical heavy vehicle.

                         If the driver of a light pickup car, which is loaded to maximal capacity, decides to
                         overtake the truck, the pickup car motor will work in heavy-duty conditions.



                            If the cutting edge angle of a milling tool is not 90° (Fig. 14), the equations (2) and (2a)
                         change to
                          hm=fz×√(ae/d) ×sinχ    (4)
                         and
                          fz=hm×√(d/ae)×1 /sinχ    (4a)
                         correspondingly

                         Here χ – cutting edge (entering) angle

                         In the same manner, the cutting edge angle should be taken into account in
                         calculating hm and hmax with the use of the equations in Tables 11 and 12.






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