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- Formulas for Drilling
FORMULAS FOR DRILLING
CUTTING SPEED (vc)
- vc (SFM)
- :Cutting Speed
- D1 (inch)
- :Drill Diameter
- π(3.14)
- :Circular Constant
- n (min-1)
- :Rotational Speed of the Main Spindle
*Unit transformation (from "mm" to "m")
(Problem) :
What is the cutting speed when main axis spindle speed is 1350min-1 and drill diameter is .500inch ?
(Answer):
Substitute π=3.14, D1=.500inch, n=1350 into the formulae
vc=π× D1×n÷12=3.14×.500×1350÷12=176.6SFM
The cutting speed is176.6SFM.
(Problem) :
What is the cutting speed when main axis spindle speed is 1350min-1 and drill diameter is .500inch ?
(Answer):
Substitute π=3.14, D1=.500inch, n=1350 into the formulae
vc=π× D1×n÷12=3.14×.500×1350÷12=176.6SFM
The cutting speed is176.6SFM.
FEED OF THE MAIN SPINDLE (vf)
- vf (inch/min)
- :Feed Speed of the Main Spindle (Z axis)
- fr (IPR)
- :Feed per Revolution
- n (min-1)
- :Rotational Speed of the Main Spindle
(Problem):
What is the spindle feed (vf) when feed per revolution is .008IPR and main axis spindle speed is 1350min-1?
(Answer):
Substitute fr=.008, n=1350 into the formulae
vf = fr×n = .008×1350 = 10.8inch/min
The spindle feed is 10.8inch/min.
What is the spindle feed (vf) when feed per revolution is .008IPR and main axis spindle speed is 1350min-1?
(Answer):
Substitute fr=.008, n=1350 into the formulae
vf = fr×n = .008×1350 = 10.8inch/min
The spindle feed is 10.8inch/min.
DRILLING TIME (Tc)
- Tc (min)
- :Drilling Time
- n (min-1)
- :Spindle Speed
- ld (inch)
- :Hole Depth
- fr (IPR)
- :Feed per Revolution
- i
- :Number of Holes
(Problem):
What is the drilling time required for drilling a 1.2inch length hole in alloy steel at a cutting speed of 165SFM and feed .006IPR ?
(Answer):
Spindle Speed n =(165×12)÷ (.59×3.14) = 1068.8min-1
Tc =(1.2×1)÷ (1068.8×.006) = .187
= .187×60≒11.2 sec
What is the drilling time required for drilling a 1.2inch length hole in alloy steel at a cutting speed of 165SFM and feed .006IPR ?
(Answer):
Spindle Speed n =(165×12)÷ (.59×3.14) = 1068.8min-1
Tc =(1.2×1)÷ (1068.8×.006) = .187
= .187×60≒11.2 sec
CUTTING SPEED (vc)
*Unit transformation (from "mm" to "m")
- vc (m/min)
- Cutting Speed
- D1 (mm)
- Drill Diameter
- π (3.14)
- Pi
- n (min-1)
- Rotational Speed of the Main Spindle
(Problem)
What is the cutting speed when main axis spindle speed is 1350min-1 and drill diameter is 12mm ?
(Answer)
Substitute π =3.14, D1=12,n=1350 into the formulae
What is the cutting speed when main axis spindle speed is 1350min-1 and drill diameter is 12mm ?
(Answer)
Substitute π =3.14, D1=12,n=1350 into the formulae
vc =(π ×D1×n)÷1000 =(3.14×12×1350)÷1000=50.9m/min
The cutting speed is 50.9m/min.
The cutting speed is 50.9m/min.
FEED OF THE MAIN SPINDLE (vf)
- vf(mm/min)
- Feed Speed of the Main Spindle(Z axis)
- f (mm/rev)
- Feed per Revolution
- n (min-1)
- Rotational Speed of the Main Spindle
(Problem)
What is the spindle feed (vf) when feed per revolution is 0.2mm/rev and main axis spindle speed is 1350min-1?
(Answer)
Substitute f=0.2, n=1350 into the formulae
What is the spindle feed (vf) when feed per revolution is 0.2mm/rev and main axis spindle speed is 1350min-1?
(Answer)
Substitute f=0.2, n=1350 into the formulae
vf = f×n = 0.2×1350 = 270mm/min
The spindle feed is 270mm/min
The spindle feed is 270mm/min
DRILLING TIME (Tc)
- Tc (min)
- Drilling Time
- n (min-1)
- Spindle Speed
- ld (mm)
- Hole Depth
- f (mm/rev)
- Feed per Revolution
- i
- Number of Holes
(Problem)
What is the drilling time required for drilling a 30mm length hole in alloy steel at a cutting speed of 50m/min and feed 0.15mm/rev ?
(Answer)
Spindle Speed
What is the drilling time required for drilling a 30mm length hole in alloy steel at a cutting speed of 50m/min and feed 0.15mm/rev ?
(Answer)
Spindle Speed
n=(50×1000)÷(15×3.14)=1061.57min-1
Tc=(30×1)÷(1061.57×0.15)=0.188
=0.188×60=11.3sec.
Tc=(30×1)÷(1061.57×0.15)=0.188
=0.188×60=11.3sec.