Stepper Motors: Sizing and Selection

1. operating speed in steps/second
2. torque (oz.-in.)
3. load inertia (lb-in.²⁾
4. required step angle
5. time to accelerate (ms)
6. time to decelerate (ms)
7. type of drive to be used
8. size and weight -- suitable to application

Much of this information will be provided from application specifications, such as the size and weight considerations, step angle, and the operating speed. Other information must be calculated. Below, several formulas are provided to assist you with these calculations.

Torque (ounce-inches)

T = Fr

where F = force in ounces
r = radius in inches

Load inertia (I = Moment of inertia (lb-in.²)

I (lb-in.²) = Wr²/2 for a disk

I (lb-in.²) = Wr2/2 (r²₁ + r²₂) for a hollow cylinder

where W = weight in pounds

r = radius in inches of solid cylinder or disk
r₁ = inner radius of hollow cylinder
r₂ = outer radius of hollow cylinder

The formula for equivalent inertia to overcome friction in the system and
enough torque to start or stop all inertia loads is:

T = lα/24

where T is torque in ounce-inches
I is the moment of inertia in lb-in.²
α is angular acceleration in radians per second²
1/24 is the conversion factor for converting gravitational units (in.-sec²) to units of mass (lb-in.²)

The formula for calculating the torque required to rotationally accelerate an
inertia load is:

T = 2 x I₀(ω'/t) x (πφ/180) x (1/24)

where T = torque in ounce-inches
I₀ = inertial load in lb-in.²
π = 3.1416
φ = step angle in degrees
ω' = step rate in steps per second
t = time in seconds

Exercise:

Calculate the torque required to accelerate a load that has inertia of 8.2 lb-in.², a step of 1.4 degrees, and an acceleration from 0-1000 steps per second in 0.7 second.

Solution:

T = 2 x 8.2 x (1000/0.7) x (π x 1.4/180) x (1/24)
Torque = 30.75 ounce-inches