CHAPTER 15 APPENDIXES

(1) Max. traveling speed (v)

If the reduction ratio is 1/1 and the rotation speed of the motor shaft is 3000 r/min

v = (3000/60) × 10×(1/1) = 500 mm/s

(2) Load inertia converted to motor axis (JL)

 Screw (J1) Suppose Ø20 and 500 mm in length.

πρ

J1 = 32

L

1000

D1

1000

4

× GL2

π× 7.85 × 103

=

32

500

1000

20

1000

4

× ( 1/1)2

= 0.6 × 10-4 kg m2

 Moving parts (J2) Suppose a transfer mass of 20 kg.

J2 = W

1

2π

×

BP

1000

2

× (GL)2

= 20

1

2π

×

10

1000

2

× ( 1/1)2

= 0.5 × 10-4 kg m2

JL = J1 + J2 = 1.1 × 10-4 kg m2

(3) Load torque converted to motor axis (TL)

Suppose a transfer mass of 20 kg, friction coefficient (µ) of 0.1 and machine efficiency (η) of 0.9.

TL =

(µW + F) × 9.81

2πη

= (0.1 × 20 + 0) × 9.81

2π × 0.9

= 0.03 Nm

BP

1000

10

1000

× GL

× (1/1)

BP : Screw lead [mm]

D : Screw dia. [mm]

GL : Deceleration ratio

JL : Load inertia converted to motor

shaft [kgm2]

L : Screw length [mm]

TL : Load torque converted to motor

shaft [Nm]

W : Transfer weight [kg]

μ : Friction factor

15

Capacity Selection Calculation 15-25