SME
Manufacturing Processes · Problem type
Milling: table feed and removal rate
Use this for “An 80 mm cutter with 6 teeth runs at 120 m/min and 0.1 mm/tooth, 40 mm wide and 3 mm deep. Find the spindle speed, table feed and removal rate.”
Diagram
v = πDN = π × 0.08 m × 477 rpm = 120 m/min. fᵣ = Nnₜfₜ = 286 mm/min.
MRR = wdfᵣ = 34.4 cm³/min.
Depth and feed are enlarged in the detail.
This is an example. Type your own values to start a new problem.
v (m/min)
=
π
×
D (mm)
×
N (rpm)
fᵣ (mm/min)
=
N (rpm)
×
nₜ
×
fₜ (mm/tooth)
MRR (cm³/min)
=
w (mm)
×
d (mm)
×
fᵣ (mm/min)
Formulas
v
=
π
×
D
×
N
1,000
120
=
π
×
80
×
477.4648
1,000
fᵣ
=
N
×
nₜ
×
fₜ
286.4789 = 477.4648 × 6 × 0.1
MRR
=
w
×
d
×
fᵣ
1,000
34.3775
=
40
×
3
×
286.4789
1,000
Assumptions
•
v is the speed at the cutter’s rim, so N = v ÷ (πD) with D in meters.
•
Each of the nₜ teeth bites fₜ on every turn; the cutter spans the width w.
•
The block moves at the table feed fᵣ; 1 cm³ = 1000 mm³.
Step-by-step
WE KNOW
D = 80 mm, nₜ = 6, v = 120 m/min, fₜ = 0.1 mm/tooth, w = 40 mm, d = 3 mm
FIND
spindle speed (N), table feed (fᵣ), removal rate (MRR)
Step 1 · Find spindle speed (N)
Use v = π × D × N ÷ 1,000
The cutting speed in mm each minute, shared over one circumference πD.
N
=
1,000
×
v
π ×
D
N
=
1,000
×
120
π × 80
N
=
120,000
80π
N
=
477.4648
rpm
Step 2 · Find table feed (fᵣ)
Use fᵣ = N × nₜ × fₜ
Every
tooth
bites
fₜ,
nₜ
teeth
each
turn,
N
turns
each
minute.
fᵣ = 477.4648 × 6 × 0.1
fᵣ = 2,864.7888 × 0.1
fᵣ = 286.4789 mm/min
Step 3 · Find removal rate (MRR)
Use MRR = w × d × fᵣ ÷ 1,000
Width × depth × table feed in mm³ each minute; ÷ 1000 turns mm³ into cm³.
MRR
=
40
×
3
×
286.4789
1,000
MRR
=
120
×
286.4789
1,000
MRR
=
34,377.468
1,000
MRR = 34.3775 cm³/min
Check
Put the numbers back in. Both sides should match.
120
=
π
×
80
×
477.4648
1,000
✓
286.4789 = 477.4648 × 6 × 0.1 ✓
34.3775
=
40
×
3
×
286.4789
1,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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