Full report ( Ball Lead Screw Application ) .fr

Ball/Lead screw specifications. Diameter. DB. = 15 [mm]. Total length. LB. = 800 [mm]. Lead (ptich). PB. = 5 [mm/rev]. Efficiency η. = 87 [%]. Material. Steel ρ. =.
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- given information -

Load and linear guide

20 [kg]

Total mass of loads and table

m

=

Friction coefficient of the guide

µ

=

Diameter

DB

=

15 [mm]

Total length

LB

=

800 [mm]

Lead (ptich)

PB

=

Efficiency

η

=

87 [%]

ρ

=

7900 [kg/m3 ]

TB

=

0.05

Ball/Lead screw specifications

Steel

Material Breakaway torque of the screw

5 [mm/rev]

1 [N·m]

External force FA

=

0.1 [N]

Transmission belt and pulleys or gears Primary pulley (gear)

Secondary pulley (gear)

pitch circle diameter (PCD)

Dp1

=

[mm]

Dp2

=

[mm]

mass

mp1

=

[kg]

m p2

=

[kg]

thickness

Lp1

=

[mm]

Lp2

=

[mm]

=

[kg/m3 ]

ρ

=

[kg/m 3]

material

ρ

Mechanism Placement Mechanism angle

α

0 [°]

=

Other requirement(s) Is it necessary to hold the load even after the power supply is turned off?



NO

Is it necessary to hold the load after the motor is stopped, but not necessary to hold after the power supply is turned off?



YES

Operating conditions Variable speed operation

Operating speed

V1

=

0 [mm/s]

V2

=

150 [mm/s]

Stopping accuracy Stopping accuracy

∆l

=

S·F

=

0.005 [mm]

Safety factor Safety factor

1.5

10/11/2016 17:41

Full report ( Ball Lead Screw Application )

http://www.orientalmotor.com/support/frBallLeadScrew.html

- calculated result -

Load Inertia JW

= m (( PB × 10-3 ) / 2π )2

20 × ((

= JS

=

1.268e-5 [kg·m2 ]

=

3.140e-5 [kg·m2 ]

=

4.407e-5 [kg·m2 ]

= ( π / 32 ) ρ ( LB × 10-3 ) ( DB × 10-3 )4 = ( 3.14 / 32 ) ×

JL

5 × 10-3 ) / ( 2 × 3.14 ))2

7900

800

×(

× 10-3 ) × (

15

× 10-3 )4

= JW + JS =(

1.268e-5

+

3.140e-5

)

Required Speed Vm1

= V1 ( 60 / PB )

0

= Vm2

× ( 60 /

5

)

=

0.000e+0 [r/min]

5

)

=

1800 [r/min]

= V2 ( 60 / PB )

150

=

× ( 60 /

Required Torque T

= ( Ta + TL ) ( Safety Factor )

0.000

=(

1.159

+



1.5

=

1.739 [N·m]

=

9.900 [N]

=

1.159 [N·m]

=

3.600e-1 [deg]

Load Torque F

= FA + ( m × 9.8 ) × ( sinα + µcosα )

0.1

= TL

20

+(

× 9.8 ) × ( sin

0

+

0.05

× cos

0

)

= (((( F ( PB × 10-3 )) / 2π ) × 1.1 ) + TB ) ( 1 / ( η × 0.01 ))

9.900

= ((((

×(

5

× 10-3 )) / 2 × 3.14 ) × 1.1 ) +

1

)×(1/(

87

× 0.01 ))

Required Stopping Accuracy ∆θ

= ∆l ( 360° / PB ) =

0.005

× ( 360 /

5

)

Other requirement(s) Holding Torque

- end of the report -

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10/11/2016 17:41