Lathe threading kinematics A 2 mm pitch screw thread is to be cut on a lathe whose lead screw pitch is 6 mm. If the spindle runs at 60 rpm, what speed must the lead screw run to generate the required pitch (assume simple gear train)?

Introduction / Context:In thread cutting on a lathe, the carriage is driven by the lead screw through a gear train. For a given spindle revolution, the carriage must advance by an amount equal to the desired thread pitch on the workpiece.

Given Data / Assumptions:

• Desired workpiece pitch Pw = 2 mm.
• Lead screw pitch Ps = 6 mm (advance per lead screw revolution).
• Spindle speed Ns = 60 rpm.
• Simple gear train with no additional leads or quick-change gearbox anomalies.

Concept / Approach:For each spindle revolution, carriage advance must equal the workpiece pitch: advance per spindle rev = Pw. Carriage advance = (lead screw revs per spindle rev) * Ps. Set these equal to solve for the necessary lead screw speed.

Step-by-Step Solution:Let Nls = lead screw rpm. Then advance per spindle rev = (Nls / Ns) * Ps.Set (Nls / Ns) * Ps = Pw.Rearrange: Nls / Ns = Pw / Ps = 2 / 6 = 1 / 3.With Ns = 60 rpm → Nls = (1/3) * 60 = 20 rpm.

Verification / Alternative check:Dimensional logic check: larger lead screw pitch requires proportionally lower lead screw speed to achieve a smaller work pitch.

Why Other Options Are Wrong:

• 10 rpm (A) gives advance = (10/60)*6 = 1 mm per spindle rev → 1 mm pitch.
• 120 rpm (C) and 180 rpm (D) would grossly overfeed, making 12 mm or 18 mm pitch equivalents.
• 30 rpm (E) yields (30/60)*6 = 3 mm pitch, not 2 mm.

Common Pitfalls:Inverting the ratio Ps/Pw; forgetting to base advance on per spindle revolution.

Final Answer:20 r.p.m.