Selecting the right motor is critical to ensuring stable, efficient, and safe performance in linear motion systems. In vertical applications, where a ball screw is installed upright and driven by an AC motor, the design must consider not only the required thrust and speed but also factors such as load holding, back-driving risk, and system reliability. This case provides a typical example of motor selection for a vertically mounted ball screw combined with an AC motor drive, offering practical guidelines for engineers during system design and optimization.

1. Specifications and Operating Conditions of the Drive Mechanism

1.1 Structure and Configuration

As shown in the diagram, the worktable moves up and down using a ball screw, and the driving power comes from an AC geared motor equipped with an electromagnetic brake.

Additional Note: There are many practical examples using this type of structural design, such as lift devices at both ends of vertically arranged double-deck conveyor lines, lifting worktables, top-loading feeding devices, robotic arms, and so on.

1.2 Required Specifications

1.3 Operating Conditions (Working Conditions)

Motor Power Supply: Single-phase 220V, 50Hz

Operating Time: Intermittent operation for 5 hours per day

Operating Conditions: Repeated starting and stopping; load must be maintained when stopped.

2 Determining the Reduction Ratio of the Gearbox

2.1 Output Shaft Speed of the Gearbox:

【Formula Understanding】The formula is relatively easy to understand, where 60 is the conversion factor to change the time unit from seconds to minutes.

2.2 Reduction Ratio of the Gearbox

For the motor with electromagnetic brake (4-pole type), the rated speed at 50Hz is between 1200 and 1300 [r/min].

Based on this, the reduction ratio is selected as $i = 9$.

【Formula Understanding】The motor speed is determined according to the requirements and the relevant motor catalog information.

3 Calculation of Required Torque (TM)

3.1 Load in the Operating Direction

3.2 Pre-load of Ball Screws

Considering the safety factor $S_f = 2$.

Based on the previous calculation results: Reduction ratio $i = 9$, Load torque $T_L = 0.86 \, \text{[N·m]}$, select a gearbox and motor with electromagnetic brake that meet the allowable torque of the gearbox.

Referring to the relevant catalog, "Allowable Torque with Gearbox," the initially selected motor is 4RK25GN-CW2ML1, and the gearbox is 4GN9KF.

3.4 Required Torque (T_M)

Convert the load torque to the value on the motor output shaft to calculate the required torque $T_M$

The transmission efficiency of the gearbox 4GN9KF.

The starting torque of the previously selected motor 4RK25GN-CW2ML1 is 160 [mN·m], which is sufficient to meet the required torque of 118 [mN·m]. Therefore, the system can start. Additionally, it needs to be confirmed whether the electromagnetic brake can hold the gravity load when stopped. Here, it is assumed that the load is the same as the previously calculated load torque.

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3.5 Required Torque on the Motor Output Shaft to Maintain Load (T'M)

= 0.0956 [N·m] = 95.6 [mN·m]

The static friction torque of the electromagnetic brake part of the previously selected motor 4RK25GN-CW2ML1 is 100 [mN·m], which is sufficient to meet the required torque to maintain the load of 95.6 [mN·m].

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4 Confirmation of Load Moment of Inertia (J)

4.1 Moment of Inertia of the Ball Screw

= $0.993 \times 10^{-4}$ [kg·m²]

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4.2 Moment of Inertia of the Workbench and Workpiece

[kg·m²]

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4.3 Load Moment of Inertia of the Gearbox Output Shaft (J₀)

$J₀ = J_s + J_M = 0.993 \times 10^{-4} + 0.286 \times 10^{-4} = 1.28 \times 10^{-4} \, \text{[kg·m²]}$

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4.4 Moment of Inertia Determination

The allowable load moment of inertia $J_G$ for the gearbox 4GN9KF with a reduction ratio of 9, according to the product catalog, is determined using the following formula:

[kg·m²]

Since $J < j < J_G$, the load moment of inertia is within the allowable value, so it can be used. There is still torque margin available, and the moving speed can be confirmed by the no-load speed (approximately 1470 r/min).

= 13.6 [mm/s]

$N_M$ - Motor Speed

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【Formula Understanding】The speed of the motor after reduction by the gearbox, multiplied by the lead of the ball screw, gives the distance the workbench moves per minute. The factor 60 is used to convert the time unit from seconds to minutes.

The above confirmation results all meet the specifications. Therefore, the selected motor is 4RK25GN-CW2ML1, and the gearbox is 4GN9KF.

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5 Summary

Overall Calculation Approach and Steps:

Determine the Reduction Ratio of the Gearbox: This is the most basic requirement to ensure the workbench moves at the desired speed, and the motor must meet this requirement.

Calculate the Required Torque: This serves as the basis for the preliminary selection of the motor and gearbox. The required torque provides enough force to move the load. As the name suggests, "required" is a necessary condition but not necessarily a sufficient condition.

Confirm the Load Moment of Inertia: Verify the preliminary selected motor and gearbox. The entire drive system is calculated as a complete system, and the system’s moment of inertia is compared with the allowable values for the motor (and gearbox) output shaft to ensure it meets the requirements.

In summary:
Speed (reduction ratio) → Drive capability (required torque) → Motion confirmation (moment of inertia).