# Torque and Force Requirements

## Torque is required to both overcome friction in the nut, and accelerate the motor and load to the required top speed.

Similarly, in linear motor applications, force is required to overcome friction in the ways, cable bending forces, and to accelerate the moving mass of the stage and the user payload. In general, calculations of this sort make the most sense (and are a lot easier) when the MKS system of units is employed. The MKS unit of torque is the Newton-meter (N-m), and the corresponding unit of rotational inertia is kilogram-meters squared (kg-m2).

##### Rotary Motor Example

The load which must be accelerated consists of the manual positioning knob, the motor rotor, the flexible shaft coupling, the leadscrew, the moving portion of the positioning table, and the user payload. The rotational inertia (J) of the knob used on DOVER positioning tables is 6.3 x 10-6 kg-m2 while the helical shaft coupling has an inertia of 2.2 x 10-6 kg-m2. The rotor inertia varies with motor frame size and length. Specific values of the rotor inertia for six standard motors are as follows:

### Rotor Inertia

##### Stepping Motors:

17 frame, 0.16 N-m (23 oz-in) holding torque:3 x 106kg-m2
23 frame, 0.38 N-m (50 oz-in) holding torque:11 x 106kg-m2
23 frame, 0.71 N-m (100 oz-in) holding torque:23 x 106kg-m2

##### Servo Motors:

40 mm square brushless:5.65 x 106kg-m2
50 mm square brushless:38.1 x 106kg-m2
Brush servomotor:26.1 x 106kg-m2

Depending on size, our tables can be provided with either ~12 mm or ~18 mm outer diameter (OD) leadscrews, whose rotational inertias are 2.7 x 10-6 and 1.2 x 10-5 kg-m2 per 100 mm of travel, respectively. The process of determining the required torque for a given application begins by adding the rotary inertia of the knob, motor rotor, coupling, and leadscrew. The user payload mass and the moving mass of the positioning table must then be summed, and converted into an equivalent rotary inertia, via the following formula:

### Rotational Intertia of Table & Load

J: rotational inertia, in kg-m2
m: total moving mass, in kg
e: screw efficiency

The efficiency of the leadscrew is typically 0.6 for our leadscrews with anti-backlash nuts, and 0.9 for ballscrews. The moving masses of single-axis stages, as well as the upper and lower axis moving masses for X-Y tables, are listed with the specifications for each table. Finally, the total rotational inertia is converted to a torque which, when summed with the friction torque, equals the total required torque. The frictional torque of DOVER positioning tables is held between 0.03 and 0.06 N-m for 12 mm OD leadscrews, and between 0.06 and 0.09 N-m for 18 mm OD screws. While the nut can be adjusted to lower torque values, this can reduce its otherwise excellent (1-2 micron) repeatability, and decrease the axial stiffness. Due to the presence of a lubricant film, the friction between leadscrew and nut increases with rpm, as well as at lower temperatures.

As an example, consider our XYL-1515-SM, a 300 mm x 300 mm (12″ x 12″) travel xy table. It is supplied with a standard motor, and has an upper axis moving mass of 8.6 kg. We will assume a screw lead of 5 mm (0.005 m), a leadscrew diameter of 18 mm, and a user payload of 23 kg. This load must be accelerated at 2 meters per second squared. To begin, we sum the rotational inertias of the respective components:

### Sum of Rotational Inertias

J Knob = 6.3 x 106 kg-m2
J Coupling = 2.2 x 106 kg-m2
J Motor = 23 x 106 kg-m2
J Leadscrew = 12 x 106 kg-m2 x (300mm/100mm) = 36 x 106 kg-m2

Total Rotational Inertia = 1.0 x 10-4 kg-m2

Note that the rotational inertia of the leadscrew is greater than that of the payload.

The formula to convert rotational inertia to torque is as follows:

### Torque

T = J x A / Lead
T: torque, in N-m
J: rotational inertia, in kg-m2
A: acceleration, in m/s2