Question

1115 LU ULJU III 18.3 A tapered surface is turned on an automatic lathe. The workpiece is 550 mm long with mini- mum and maximum diameters of 100 mm and 200 mm at opposite ends. Automatic controls on the lathe permit the surface speed to be maintained at a constant value of 200 m/min by adjusting the rotational speed as a function of workpiece diameter. Feed = 0.25 mm/rev, and depth of cut = 3.0 mm. The rough geom- etry of the piece has already been formed, and this operation will be the final cut. Determine (a) the time required to turn the taper and (b) the rotational speeds at the beginning and end of the cut (assume cutting starts at the smaller diameter).

Answer 1

200 mm V = 200m/min feed - 0.25 mm/rev. d = 3.0 mm. - 550 mm ) At the beginning TID, N = 200 60 60 = ITX (100) x 10-3 XN,= 200 -> N N, = 200x103 - = 200 x 10 = 2000 pm Toon T At finish V = TTD₂N₂ : 2 - STX (200), 10-3x N₂=200 7 N₂ = 1000 rpm tan x = (200-100): 10% Taper L,X=> 50 2:10.30 Machining time = 1 1 1 1 ? we assume single papp . Since, roughly the Surface dimensions have been attained. Neue feed 0.25 mm/rev. 4 Axial travel per rev. Now, for 500mm 550 rev = 2200 rev. 0.25 We assume arg. Turning speed / Turning dia Arg. dia - 200 +100 - 150 mm. Now, Tm= I - TIL Darg - ITX (500) x (100) = 0.432min f.N 1000.f.V 1000 x 3 x 200 0 2 Aus- 25.92 sec. -0-432 X 60 = 25.92 see