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- Technical Information
- Reference
- Formulas for Milling

CUTTING SPEED (vc)

- vc (SFM)
- :Cutting Speed
- D1 (inch)
- :Cutter Diameter
- π (3.14)
- :Pi
- n (min
^{-1}) - :Main Axis Spindle Speed

(Problem) :

What is the cutting speed when main axis spindle speed is 350min-1 and cutter diameter is Ø5" ?

**(Answer):**

Substitute π3.14, D=5", n=350 into the formulae.

vc =π×D×n÷12=3.14×5"×350÷12= 457.9SFM

**The cutting speed is 457.9SFM.**

What is the cutting speed when main axis spindle speed is 350min-1 and cutter diameter is Ø5" ?

Substitute π3.14, D=5", n=350 into the formulae.

vc =π×D×n÷12=3.14×5"×350÷12= 457.9SFM

FEED PER TOOTH (fz)

- fz (IPT)
- :Feed per Tooth
- z
- :Insert Number
- vf(inch/min)
- :Table Feed per Min.
- n (min
^{-1}) - :Main Axis Spindle Speed (Feed per Revolution fr=z x fz)

What is the feed per tooth when the main axis spindle speed is 500min

Substitute the above figures into the formulae.

fz =Vf÷(z×n)=20÷(10×500) = .004IPT

TABLE FEED (vf)

- vf(inch/min)
- :Table Feed per Min.
- fz (IPT)
- :Feed per Tooth
- n (min
^{-1}) - :Main Axis Spindle Speed
- z
- :Insert Number

What is the table feed when feed per tooth is .004IPT, insert number is 10, and main axis spindle speed is 500min

Substitute the above figures into the formulae.

vf = fz×z×n = .004IPT×10×500 = 20inch/min

CUTTING TIME (Tc)

- Tc (min)
- :Cutting Time
- vf(inch/min)
- :Table Feed per Min.
- L (inch)
- :Total Table Feed Length (Workpiece Length(l)+Cutter Diameter(D1)

What is the cutting time required for finishing 4" width and 12" length surface of a cast iron (GG20) block when cutter diameter is Ø4", the number of inserts is 16, the cutting speed is 410SFM , and feed per tooth is .01. (spindle speed is 200min

Calculate table feed per min vf=.01×16×200=32inch/min

Calculate total table feed length. L=12+4=16inch

Substitute the above answers into the formulae.

Tc =16÷32= 0.5(min)

The answer is 30 sec.

CUTTING SPEED (vc)

* Divide by 1,000 to change to m from mm.

- vc (m/min)
- Cutting Speed
- D1 (mm)
- Cutter Diameter
- π (3.14)
- Pi
- n(min
^{-1}) - Main Axis Spindle Speed

(Problem)

What is the cutting speed when main axis spindle speed is 350min^{-1} and cutter diameter isø125? (Answer)

Substitute π=3.14, D1=125, n=350 into the formulae.
**vc=(π×D1×n)÷1000=(3.14×125×350)÷1000 =137.4(m/min)**

The cutting speed is 137.4 m/min.

What is the cutting speed when main axis spindle speed is 350min

Substitute π=3.14, D1=125, n=350 into the formulae.

The cutting speed is 137.4 m/min.

FEED PER TOOTH (fz)

- fz (mm/tooth)
- Feed perTtooth
- z
- Insert Number
- vf (mm/min)
- Table Feed per Min.
- n (min
^{-1}) - Main Axis Spindle Speed (Feed per Revolution
**f = z x fz**)

(Problem)

What is the feed per tooth when the main axis spindle speed is 500min^{-1}, insert number is 10, and table feed is 500mm/min?

(Answer)

Substitute the above figures into the formulae.**fz=vf÷(z×n)=500÷(10×500)=0.1mm/tooth**

The answer is 0.1mm/tooth.

What is the feed per tooth when the main axis spindle speed is 500min

(Answer)

Substitute the above figures into the formulae.

The answer is 0.1mm/tooth.

TABLE FEED (vf)

- vf (mm/min)
- Table Feed per Min.
- z
- Insert Number
- fz (mm/tooth)
- Feed per Tooth
- n (min
^{-1}) - Main Axis Spindle Speed

(Problem)

What is the table feed when feed per tooth is 0.1mm/tooth, insert number is 10, and main axis spindle speed is 500min^{-1}?

(Answer)

Substitute the above figures into the formulae.**vf=fz×z×n=0.1×10×500=500mm/min**

The table feed is 500 mm/min.

What is the table feed when feed per tooth is 0.1mm/tooth, insert number is 10, and main axis spindle speed is 500min

(Answer)

Substitute the above figures into the formulae.

The table feed is 500 mm/min.

CUTTING TIME (Tc)

- Tc(min)
- Cutting Time
- vf(mm/min)
- Table Feed per Min.
- L (mm)
- Total Table Feed Length (Workpiece Length:
**l**+Cutter Diameter:**D1**)

(Problem)

What is the cutting time required for finishing 100 mm width and 300 length surface of a cast iron (GG20) block when cutter diameter is ø200, the number of inserts is 16, the cutting speed is 125m/min, and feed per tooth is 0.25 mm. (spindle speed is 200 min^{-1})

(Answer)

Calculate table feed per min vf=0.25×16×200=800mm/min

Calculate total table feed length. L=300+200=500mm

Substitute the above answers into the formulae.

**Tc=500÷800=0.625(min)**

0.625×60=37.5(sec). The answer is 37.5 sec.

What is the cutting time required for finishing 100 mm width and 300 length surface of a cast iron (GG20) block when cutter diameter is ø200, the number of inserts is 16, the cutting speed is 125m/min, and feed per tooth is 0.25 mm. (spindle speed is 200 min

(Answer)

Calculate table feed per min vf=0.25×16×200=800mm/min

Calculate total table feed length. L=300+200=500mm

Substitute the above answers into the formulae.

0.625×60=37.5(sec). The answer is 37.5 sec.