# 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

JL = 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 JM of the motor,

if a 400W motor is selected, JM = 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, JM = 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):

JM > 6 [kg·cm2]

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

Torque required to overcome friction:

Tf = 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 Tf, and the maximum torque should be greater than Tf + Ta.

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

JW = M * (PB / 2π)²

= 200 * (2 / 6.28)²

= 20.29 [kg·cm²]

The rotational inertia of the screw

JB = MB * DB² / 8

= 40 * 25 / 8

= 125 [kg·cm²]

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

Tf = 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

TA1 = 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

TA2 = JB * α / η

= JB * (N * 2π / 60 / t1) / η

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

= 10.903 [N·m]

Total torque required for acceleration

TA = TA1 + TA2 = 12.672 [N·m]

4. Selection of Servo Motor

Rated torque of the servo motor

T > Tf and T > Trms

Maximum torque of the servo motor

Tmax > Tf + TA

Finally, the ECMA-E31820ES motor was selected.

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Author

#### Shane

##### Founder of MachineMFG

As the founder of MachineMFG, I have dedicated over a decade of my career to the metalworking industry. My extensive experience has allowed me to become an expert in the fields of sheet metal fabrication, machining, mechanical engineering, and machine tools for metals. I am constantly thinking, reading, and writing about these subjects, constantly striving to stay at the forefront of my field. Let my knowledge and expertise be an asset to your business.

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