Servo Motor Sizing: A Step-by-Step Guide for Mechanical Engineers

Case One

Given:

• Disc mass M=50 kg
• Disc diameter D=500 mm
• Maximum disc speed 60 rpm

Please select the servo motor and reduction gear, component schematic as follows:

Calculating the moment of inertia for the disc rotation

J_L = MD²/8 = 50 * 50² / 8 = 15625 [kg·cm²]

Assuming a gear reduction ratio of 1:R, the load inertia reflected on the servo motor shaft is 15625/R².

According to the principle that the load inertia should be less than three times the rotor inertia J_M of the motor,

if a 400W motor is selected, J_M = 0.277 [kg·cm²],

then: 15625 / R² < 3*0.277, R² > 18803, R > 137,

the output speed = 3000/137 = 22 [rpm],

which does not meet the requirement.

If a 500W motor is selected, J_M = 8.17 [kg·cm²],

then: 15625 / R² < 3*8.17, R² > 637, R > 25,

the output speed = 2000/25 = 80 [rpm],

which satisfies the requirement.

This type of transmission has minimal resistance, so torque calculations are ignored.

Case Two

Given:

• Load weight M = 50 kg
• Synchronous belt wheel diameter D = 120 mm
• Reduction ratio R₁ = 10, R₂ = 2
• Friction coefficient between load and machine table µ = 0.6
• Maximum motion speed of load: 30 m/min
• Time for load to accelerate from rest to maximum speed: 200ms

Ignoring the weight of each conveyor belt wheel,

What is the minimum power requirement for a motor to drive such a load?

The schematic diagram of the component is as follows:

1. Calculating the load inertia reflected on the motor shaft:

JL = M * D² / 4 / R₁² 

   = 50 * 144 / 4 / 100

   = 18 [kg·cm²]

According to the principle that load inertia should be less than three times the motor rotor inertia (JM):

J_M > 6 [kg·cm²]

2. Calculating the torque required to drive the motor load:

Torque required to overcome friction:

T_f = M * g * µ * (D / 2) / R₂ / R₁

= 50 * 9.8 * 0.6 * 0.06 / 2 / 10

= 0.882 [N·m]

Torque required for acceleration:

Ta = M * a * (D / 2) / R₂ / R₁

= 50 * (30 / 60 / 0.2) * 0.06 / 2 / 10

= 0.375 [N·m]

The servo motor’s rated torque should be greater than T_f, and the maximum torque should be greater than T_f + T_a.

3. Calculating the required motor speed:

N = v / (πD) * R₁

= 30 / (3.14 * 0.12) * 10

= 796 [rpm]

Case Three

Given:

• Load weight M = 200 kg
• Screw pitch PB = 20 mm
• Screw diameter DB = 50 mm
• Screw weight MB = 40 kg
• Coefficient of friction µ = 0.2
• Mechanical efficiency η = 0.9
• Load movement speed V = 30 m/min
• Total movement time t = 1.4 s
• Acceleration and deceleration time t₁ = t₃ = 0.2 s
• Resting time t₄ = 0.3 s

Please select the servo motor with the minimum power that meets the load requirements,

The component diagram is as follows:

1. Calculation of Load Inertia Converted to the Motor Shaft

Load inertia of the weight converted to the motor shaft

J_W = M * (PB / 2π)²

= 200 * (2 / 6.28)²

= 20.29 [kg·cm²]

The rotational inertia of the screw

J_B = M_B * D_B² / 8

= 40 * 25 / 8

= 125 [kg·cm²]

Total load inertia

JL = JW + JB = 145.29 [kg·cm²]

2. Calculation of Motor Speed

Required motor speed

N = V / PB

= 30 / 0.02

= 1500 [rpm]

3. Calculation of Torque Required to Drive the Motor Load

The torque required to overcome friction

T_f = M * g * µ * PB / 2π / η

= 200 * 9.8 * 0.2 * 0.02 / 2π / 0.9

= 1.387 [N·m]

Torque required when the weight is accelerating

T_A1 = M * a * PB / 2π / η

= 200 * (30 / 60 / 0.2) * 0.02 / 2π / 0.9

= 1.769 [N·m]

Torque required when the screw is accelerating

T_A2 = JB * α / η

= J_B * (N * 2π / 60 / t₁) / η

= 0.0125 * (1500 * 6.28 / 60 / 0.2) / 0.9

= 10.903 [N·m]

Total torque required for acceleration

T_A = T_A1 + T_A2 = 12.672 [N·m]

4. Selection of Servo Motor

Rated torque of the servo motor

T > T_f and T > Trms

Maximum torque of the servo motor

T_max > T_f + T_A

Finally, the ECMA-E31820ES motor was selected.