Threaded Insert for Cutter SNGU

Product features and specifications

Basic specifications and features of the product

• Originally designed double-sided insert
• Sudden breakage / chip welding suppression function
• Smooth discharge of cutting chip

Drawing

Recommended cutting conditions

Recommended cutting condition table

Dry cutting conditions

Wet cutting conditions

Precautions for cutting conditions

*1) Please set the cutting conditions according to the operating environment by referring to the above table.
*2) Wet cutting is recommended when the finished surface is important. (The lifespan is shorter than that of the dry cutting.)

Applicable work material

Selection support information

Compatible parts

* Tightening Torque (N⋅m): TPS4R=3.5

Selection support information

Cutting speed (vc)

* The purpose of division by 1,000 is to change the unit from mm to m

vc (m/min): Cutting speed
DC (mm): Cutter diameter
π (3.14): Pi
n (min^-1): Spindle revolutions per minute

(Example) When cutting with the axis rotational speed of 350 min^-1 and the cutter outer diameter of ø125, ^{what is the cutting speed?,} (Answer) Substituting π = 3.14, DC = 125, n = 350 into the formula,
vc = π · DC · n/1,000 = 3.14 × 125 × 350/1,000 = 137.4 m/min
Accordingly, the cutting speed is 137.4 m/min.

Feed per tooth (fz)

fz (mm/tooth): Feed per tooth
z: Insert number
vf (mm/min): Table feed per minute
n (min^-1): Main axis rotational speed (Feed per revolution f = z × fz)

(Example) When the main axis rotational speed is 500 min^-1, the number of inserts is 10, and the table feed ^{is 500 mm/min, what is the feed per tooth?,} (Answer) Substitute the above figures into the formula,
fz = vf/z × n = 500/10 × 500 = 0.1 mm/tooth
Accordingly, the feed per tooth is 0.1 mm/tooth.

Table feed (vf)

vf (mm/min): Table feed per minute
fz (mm/tooth): Feed per tooth
z: Insert number
n (min^-1): Spindle revolutions per minute

(Example) When the feed per tooth is 0.1 mm/tooth, the number of inserts is 10, and the main axis rotational speed is
500 min^-1,what is the table feed?, ^{(Answer) Substitute the above figures into the formula,} vf = fz × z × n = 0.1 × 10 × 500 = 500 mm/min
Accordingly, the table feed is 500 mm/min.

Cutting time (Tc)

Tc (min): Processing time
vf (mm/min): Table feed per minute
L (mm): Total table feed length (work material length (l) + milling machine diameter (DC))

(Example) What is the cutting time required for finishing the surface of cast iron (FC200) block with 100 mm width and 300 mm length, when the face milling machine diameter is ø200, the number of inserts is 16, the cutting speed is 125 m/min, and the feed per tooth is 0.25 mm (Main axis rotational speed is 200 min^-1)
(Answer) Calculate table feed per minute,
vf = 0.25 × 16 × 200 = 800 mm/min
Calculate total table feed length, L = 300 + 200 = 500 mm
Substitute the above answers into the formula,
Tc = 500/800 = 0.625 (min)
0.625 × 60 = 37.5 (sec) It takes approx. 37.5 seconds.

Cutting power (Pc)

Pc (kW): Cutting power
ap (mm): Depth of cut
ae (mm): Cutting width
vf (mm/min): Table feed per minute
Kc (MPa): Specific cutting force
η: (Machine coefficient)

(Example) What is the cutting power required for milling tool steel at a depth of 2 mm cut, a cutting width of 80 mm, and a table feed of 280 mm/min, based on the conditions of a cutting speed 80 m/min, cutter diameter ø250, 12 inserts and machine coefficient 80%.

(Answer) First, calculate the main axis rotational speed to obtain feed per tooth of the milling machine.
n = 1,000 vc/πDC = 1,000 × 80/3.14 × 250 = 101.91 min^-1
Feed per tooth fz = vf/z × n = 280/12 × 101.9 = 0.228 mm/tooth
Substitute into the formula,
Pc = 2 × 80 × 280 × 1,800/60 × 10^6 × 0.8 = 1.68 kW