

FORMULAE For Face milling | |||
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![]() *Divide by 1000 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 | ||
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(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. |
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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) | ||
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(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. |
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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 cutting time when 100mm workpiece is machined at 1,000min-1, with feed=0.2mm/rev? (Answer) First, calculate the cutting length per min. from the feed and spindle speed. l=f×n=0.2×1000=200(mm/min) Substitute the answer above into the formulae. Tc=lm÷l=100÷200=0.5(min) 0.5×60=30(sec.) The answer is 30 seconds |
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Tc(min) | Cutting Time | |
vf(mm/min) | Table Feed per Min. | ||
L (mm) | Total Table Feed Length (Workpiece Length: l+Cutter Diameter: D1) | ||
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(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. |
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