Setting CNC
cutting speeds
Run a cutter too fast and it burns out in seconds; too slow and the job takes all afternoon. You will use compound measures, inverse proportion and a real-life graph to set a machine up properly.
The job
Read the spec like an engineerAshfield's machine shop is milling a slot in a mild steel bracket. The programmer has to choose a cutting speed and a cutter diameter, and from those the machine's spindle speed and feed rate follow.
- Spindle speed N = (1000 × Vc) ÷ (π × d), in rev/min
- Feed rate Vf = N × fz × z, in mm/min
- The cutter has z = 4 teeth and a feed per tooth of fz = 0.08 mm
- The machine's spindle tops out at 4000 rev/min and its feed at 3000 mm/min
- Carbide tooling in mild steel wants Vc between 100 and 200 m/min for decent tool life
Your question: which combinations of cutter diameter and cutting speed keep the tool happy without asking the machine for more than it can give?
Explore the drawing
Turn the dial, pick a cutterNotice the shape of the curve: halve the diameter and the spindle speed doubles. That is inverse proportion, and it is why small cutters are the ones that hit the machine's ceiling first.
Worked solution
Open one step at a timeTake a 25 mm cutter at Vc = 120 m/min and work the numbers through.
Vc is how fast the edge of the tool travels through the metal, in metres per minute. The spindle speed N is how fast the whole cutter turns.
Every revolution drags 78.54 mm of cutting edge through the work. The 1000 in the formula converts metres into millimetres so the units agree.
Inside the machine's 4000 rpm ceiling. Passes.
Use the π key, not 3.14. Over a multi-step question the difference compounds, and mark schemes are written around the exact value.
Each of the 4 teeth takes a 0.08 mm bite every revolution.
Well inside the 3000 mm/min feed limit.
Feed rate is a compound measure — distance per unit time — so it behaves exactly like speed.
Decimal minutes are not minutes and seconds. 0.368 min is 22 seconds, not 36.8. Multiply by 60; never read the decimal part as seconds.
The curve is N against d for a fixed Vc. It is a reciprocal graph, because d sits on the bottom of the fraction.
Halving the diameter doubles the speed every time. The curve never touches the horizontal axis, and it climbs without limit as d approaches zero — which is exactly why the 6 mm cutter cannot be run at this cutting speed.
You can still use the 6 mm cutter if you drop Vc. Make Vc the subject and put the machine's ceiling in.
But the tooling wants at least 100 m/min for decent tool life. So the 6 mm cutter simply does not suit this machine and this material. That is a legitimate engineering answer: sometimes the maths tells you to change the plan.
Your turn
Answers checked instantlyWork out the spindle speed for a 20 mm cutter running at Vc = 90 m/min, to the nearest rev/min.
A 3-tooth cutter runs at 1432 rev/min with a feed per tooth of 0.10 mm. Work out the feed rate, to the nearest mm/min.
The machine is limited to 4000 rev/min. Work out the highest cutting speed possible with an 8 mm cutter, to 1 decimal place.
Exam-style question
Full mark schemeA slot 250 mm long is milled with a 4-tooth, 16 mm cutter. The cutting speed is Vc = 140 m/min and the feed per tooth is fz = 0.06 mm.
Work out how long the cut takes, in seconds, to 1 decimal place.
Mark scheme — 5 marks
- N = (1000 × 140) ÷ (π × 16) M1
- N = 2785 rev/min A1
- Vf = 2785.2 × 0.06 × 4 = 668.4 mm/min M1
- time = 250 ÷ 668.4 = 0.374 min M1
- 0.374 × 60 = 22.4 seconds A1
Accepted answer: 22.4 s (22.3–22.5 for sensible rounding). The two habitual errors are forgetting to multiply by the number of teeth, and leaving the answer as 0.4 minutes. Read the question’s units back to yourself before you write the final line.
Produced for EAL engineering learners.
Maps to GCSE Mathematics (9–1) — compound measures, inverse proportion, and interpreting real-life graphs.