Choosing a Lathe Spindle Speed and Estimating Machining Time
A turner in an ITI workshop must rough-turn a 40 mm diameter mild steel bar, 300 mm long, on a centre lathe whose headstock offers spindle speeds of 63, 90, 125, 180, 250, 355 and 500 rev/min. The recommended cutting speed for the carbide-tipped tool on this steel is 30 m/min. Two millimetres must be removed from the radius of the bar, and the workshop rule limits the depth of cut to 1 mm per pass. Feed is set at 0.2 mm per revolution for roughing. The trainee, wanting to finish quickly, sets the machine to 500 rev/min; within a few minutes the tool tip is glowing and the insert has worn badly, and the surface finish has deteriorated. The instructor now asks for a proper speed and time calculation before the next component is mounted.
1. Calculate the required spindle speed and select one from the available steps. Justify the direction in which you round.
Read the worked answer
N = 1000 V / (π D) = (1000 × 30) / (3.1416 × 40) = 30000 / 125.66 = 238.7 rev/min.
Available steps: 180 and 250. Choosing 250 gives an actual cutting speed of π × 0.040 × 250 = 31.4 m/min, only 4.7% above the recommendation - well inside normal practice. Choosing 180 gives 22.6 m/min, about 25% below, which wastes productivity without extending tool life in any useful way.
So select 250 rev/min. The rounding rule to state explicitly: select the nearest available step, and when the calculated value falls between two steps and tool protection is the priority (hard material, thin or interrupted cut, form tool), round down; here the nearest step is also the safe one. The calculation must use the bar diameter in metres and the cutting speed in m/min, since mixing mm and m is the commonest source of a factor-of-1000 error.
2. Using the chosen speed, determine the number of passes and the machining time for the 300 mm length.
Read the worked answer
Depth of cut per pass is limited to 1 mm, and 2 mm must come off the radius, so 2 passes are required (1 mm + 1 mm), with a final sizing pass if tolerance demands.
Feed rate = feed per revolution × spindle speed = 0.2 × 250 = 50 mm/min. Time per pass = length of cut / feed rate = 300 / 50 = 6 minutes. Two passes = 12 minutes of actual cutting.
Judgement on top of the arithmetic: the calculated 12 minutes is cutting time only. Allowance must be added for approach and overtravel (say 5 mm each end, adding about 0.2 minutes per pass), for returning the carriage, for measuring between passes, and for the fact that a roughing feed of 0.2 mm/rev will be reduced on the finishing pass. A realistic shop-floor estimate is nearer 15–18 minutes. Quoting 12 minutes to a supervisor as the total job time is a planning error, not a calculation error.
3. Evaluate the trainee's choice of 500 rev/min quantitatively, and state a practical rule for the workshop.
Read the worked answer
At 500 rev/min the actual cutting speed is π × 0.040 × 500 = 62.8 m/min, more than double the recommended 30 m/min.
Tool life does not fall linearly with speed. Cutting temperature rises steeply with speed, and carbide inserts typically lose life roughly in inverse proportion to speed raised to a power of three or more, so doubling the speed can cut insert life to a small fraction of normal. That is exactly what the glowing tip and rapid wear show. Secondary damage: higher speed with the same feed raises vibration on a 300 mm slender bar, spoiling finish and risking chatter marks and taper.
The workshop rule: cutting speed is set by the tool and material combination, never by the operator's wish to finish sooner. Time saved on the spindle is paid for several times over in insert changes, scrap and rework. If output must increase, raise depth of cut or feed within their limits, or improve rigidity and coolant - not the cutting speed beyond the recommended value.