Manufacturing Calculations: Speeds, Feeds and Forces

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1. Cutting speed and spindle speed
2. Machining time for turning
3. Material removal rate in turning
4. Milling feed rate
5. Material removal rate in milling
6. Drilling time
7. Cutting force and power
8. Blanking and punching force
9. Bend allowance
10. Taylor tool life equation
From CAD model to machining plan

These are the sums a machinist, a process planner or a design engineer does before a part is made: how fast to spin it, how long it will take, how much power the machine needs and how big a press must be. Shop practice mixes units, so each formula states the units it expects. Diameters are in millimetres, cutting speed in metres per minute and spindle speed in revolutions per minute.

Turning. The bar spins at N rev/min, so its surface passes the tool at the cutting speed V = πDN. The tool moves along the bar by the feed f every revolution and removes a layer as deep as the depth of cut d.

1. Cutting speed and spindle speed

V = \frac{\pi D N}{1000} \qquad\Longleftrightarrow\qquad N = \frac{1000\,V}{\pi D}

Symbol	Meaning	Unit
$V$	Cutting speed (surface speed)	m/min
$D$	Diameter of the workpiece (turning) or of the cutter (milling, drilling)	mm
$N$	Spindle speed	rev/min

Assumptions: $V$ comes from the tool maker’s data for the tool and work material. The 1000 converts millimetres to metres.

Worked example

A 50 mm steel bar is turned with a carbide insert at a recommended cutting speed of 150 m/min.

N = \frac{1000 \times 150}{\pi \times 50} = 955\ \text{rev/min}

AnswerN ≈ 955 rev/min (set the nearest available speed)

2. Machining time for turning

t = \frac{L}{f\,N}

Symbol	Meaning	Unit
$t$	Cutting time for one pass	min
$L$	Length of cut, plus approach and overrun	mm
$f$	Feed	mm/rev
$N$	Spindle speed	rev/min

Assumptions: constant feed and speed through the pass. Multiply by the number of passes and add handling time for the full cycle.

Worked example

Turn 120 mm of the bar from example 1 at a feed of 0.2 mm/rev.

t = \frac{120}{0.2 \times 955} = 0.63\ \text{min}

Answert ≈ 38 s per pass

3. Material removal rate in turning

MRR = 1000\,V\,f\,d

Symbol	Meaning	Unit
$MRR$	Material removal rate	mm³/min
$V$	Cutting speed	m/min
$f$	Feed	mm/rev
$d$	Depth of cut (radial)	mm

Assumptions: depth of cut small compared with the diameter. MRR sets how much power the cut needs and how fast roughing goes.

Worked example

Turning at V = 150 m/min, f = 0.2 mm/rev and d = 2 mm:

MRR = 1000 \times 150 \times 0.2 \times 2 = 60\,000\ \text{mm}^3/\text{min}

AnswerMRR = 60 cm³/min

4. Milling feed rate

v_f = f_z\,z\,N

Symbol	Meaning	Unit
$v_f$	Table feed rate	mm/min
$f_z$	Feed per tooth (chip load)	mm/tooth
$z$	Number of teeth (flutes)	none
$N$	Spindle speed	rev/min

Assumptions: $f_z$ comes from the cutter maker’s data; too low a chip load rubs instead of cutting and wears the tool quickly.

Worked example

A 4-flute end mill runs at 3000 rev/min with a chip load of 0.05 mm/tooth.

v_f = 0.05 \times 4 \times 3000 = 600\ \text{mm/min}

Answerv_f = 600 mm/min

5. Material removal rate in milling

MRR = a_e\,a_p\,v_f

Symbol	Meaning	Unit
$a_e$	Width of cut (radial engagement)	mm
$a_p$	Depth of cut (axial)	mm
$v_f$	Table feed rate	mm/min

Assumptions: steady cutting with constant engagement.

Worked example

The end mill from example 4 cuts 6 mm wide and 10 mm deep.

MRR = 6 \times 10 \times 600 = 36\,000\ \text{mm}^3/\text{min}

AnswerMRR = 36 cm³/min

6. Drilling time

t = \frac{L + 0.3D + c}{f\,N}

Symbol	Meaning	Unit
$L$	Depth of the hole	mm
$0.3D$	Length of the drill point (118° point angle)	mm
$c$	Approach and breakthrough clearance, typically 1–2 mm	mm
$f$	Feed	mm/rev
$N$	Spindle speed	rev/min

Assumptions: a through hole with a standard 118° drill; for a blind hole, drop the breakthrough allowance.

Worked example

Drill a 10 mm through hole in a 25 mm steel plate with an HSS drill at 25 m/min and 0.15 mm/rev.

N = \frac{1000 \times 25}{\pi \times 10} = 796\ \text{rev/min}, \qquad t = \frac{25 + 3 + 2}{0.15 \times 796} = 0.25\ \text{min}

Answert ≈ 15 s per hole

7. Cutting force and power

F_c = k_c\,f\,d, \qquad P = \frac{F_c\,V}{60}

Symbol	Meaning	Unit
$F_c$	Main cutting force	N
$k_c$	Specific cutting force of the material	N/mm²
$f$ , $d$	Feed and depth of cut	mm/rev, mm
$P$	Power at the tool	W
$V$	Cutting speed (÷ 60 converts m/min to m/s)	m/min

Assumptions: $k_c$ depends on the material and the chip thickness; about 2000 N/mm² is typical for medium carbon steel and 700 N/mm² for aluminium alloys. Divide by the machine’s efficiency (often 0.75–0.85) for motor power.

Worked example

For the turning cut in example 3 (f = 0.2 mm/rev, d = 2 mm, V = 150 m/min) in steel:

F_c = 2000 \times 0.2 \times 2 = 800\ \text{N}, \qquad P = \frac{800 \times 150}{60} = 2000\ \text{W}

AnswerP = 2 kW at the tool, about 2.5 kW at the motor

8. Blanking and punching force

F = L\,t\,\tau_s

Symbol	Meaning	Unit
$F$	Force to cut through the sheet	N
$L$	Length of the cut edge (perimeter)	mm
$t$	Sheet thickness	mm
$\tau_s$	Shear strength of the sheet, roughly 0.8 × tensile strength	N/mm²

Assumptions: flat punch and die with normal clearance. Choose a press with 20–30 % margin above this force.

Worked example

Blank a 50 mm diameter disc from 2 mm mild steel with a shear strength of 300 N/mm².

F = (\pi \times 50) \times 2 \times 300 = 94\,200\ \text{N}

AnswerF ≈ 94 kN (about 9.6 tonnes)

9. Bend allowance

BA = \theta\,\frac{\pi}{180}\,(R + K\,t)

Symbol	Meaning	Unit
$BA$	Length of the neutral axis through the bend	mm
$\theta$	Bend angle	degrees
$R$	Inside bend radius	mm
$K$	K-factor: position of the neutral axis as a fraction of thickness, 0.33–0.5	none
$t$	Sheet thickness	mm

Assumptions: the K-factor depends on the material, the tooling and the radius-to-thickness ratio; take it from your press brake’s test bends or the CAD sheet-metal table. The flat length is the sum of the straight flanges plus the bend allowances.

Worked example

A 90° bend in 2 mm steel with an inside radius of 2 mm and K = 0.44:

BA = 90 \times \frac{\pi}{180} \times (2 + 0.44 \times 2) = 4.52\ \text{mm}

AnswerBA = 4.52 mm

10. Taylor tool life equation

V\,T^{\,n} = C

Symbol	Meaning	Unit
$V$	Cutting speed	m/min
$T$	Tool life	min
$n$	Tool life exponent: about 0.1 HSS, 0.2–0.3 carbide, 0.4–0.6 ceramic	none
$C$	Cutting speed for 1 minute of tool life	m/min

Assumptions: the constants hold for one tool and work material pair and a fixed feed and depth. The equation shows why a small cut in speed gives a large gain in tool life.

Worked example

A carbide tool has n = 0.25 and C = 400 m/min. Compare tool life at 200 and 150 m/min.

T = \left(\frac{C}{V}\right)^{1/n}: \qquad \left(\frac{400}{200}\right)^{4} = 16\ \text{min}, \qquad \left(\frac{400}{150}\right)^{4} = 51\ \text{min}

AnswerSlowing down by 25 % makes the tool last about three times longer

From CAD model to machining plan

Most of these numbers come straight off the part drawing: diameters and lengths for turning, hole depths for drilling, perimeters for blanking and bend radii for sheet metal. When we model parts in CAD, we check that radii, hole sizes and bend radii match standard tooling, so the shop can make them with these standard calculations and no special tools.

Key points

Start from the cutting speed recommended for the tool and material, then calculate spindle speed.
Feed and depth of cut set the removal rate, the force and the power.
Blanking force is perimeter × thickness × shear strength; leave a margin on press capacity.
Tool life rises steeply as cutting speed falls.

Manufacturing calculations: machining, sheet metal and tool life

1. Cutting speed and spindle speed

2. Machining time for turning

3. Material removal rate in turning

4. Milling feed rate

5. Material removal rate in milling

6. Drilling time

7. Cutting force and power

8. Blanking and punching force

9. Bend allowance

10. Taylor tool life equation

From CAD model to machining plan

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