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S99TE14-0710

Ballscrews ( complete program ) - 14370 25

Fig. 4.23 shows the terms for a feed system operated by ballscrew. The formula for motor drive torque is given below :(a) Common transmission (to convert rotary motion to linear motion) T F
b × l
a = 2πη
M19 1 T
a = Drive torque for common transmission (kgf-mm) F
b = Axial load (kgf) F
b = F
bm + μ × W (for horizontal motion) l = Lead (mm)η
1 = Mechanical efficiency (0.9~0.95, Ref. M3)W = Table wight + Work piece weight (kgf)∆μ = Friction coefficient of table guide way (b) Reverse transmission (to convert linear motion to rotary motion) T F
b × l × Î·
2 c = 2Ï€
M20 η
2 = Mechanical effciency (0.9~0.95, Ref. M4) T
c = Torque for reverse transmission (kgf-mm) (c) Motor drive torqueFor normal operation : T N
1 M = ( T
a + T
b + T
d ) × N
M21 2 T
M = Motor drive torque (kgf-mm) T
b = Friction torque of supporting bearing (kgf-mm) T
d = Preload drag torque (kgf-mm, Ref. M2) N
1 = Number of teeth for driver gear N
2 = Number of teeth for driven gear For acceleration operation : T’a = JαT’a
M22 : Motor drive torque during acceleration (kgf)J : System inertia (kgf-mm-sec
2 )α : Angular acceleration (rad/sec
2 )α = 2 πN
dif 60 t
a N
M23 dif = rpm
stage2 − rpm
stage1 t
a = acceleration rising time (sec)
2 2 2 2 2 J = J N
1 12g
N 1 1 M + J
G1 + J
G2 D N Wg l N N
s 2 + W 2 N
M24 2 + 2Ï€ N
2 = Motor inertia + Equivalent gear inertia + Ballscrew inertia + Load inertia (Fig.4.23) W
S : Ballscrew weight (kgf) D
N : Ballscrew nominal diameter (mm) g : Gravity coefficient (9800 mm/sec
2 ) J
2 M : Inertia of motor (kgf-mm-sec ) J
2 G1 : Inertia of driver gear (kgf-mm-sec ) J
2 G2 : Inertia of driver gear (kgf-mm-sec )