The easiest way to design a linear-motion system is to add components one by one. Then, step-by-step calculations relate input to power dissipated moving a load in a specific amount of time.Linear systems drive everything from inexpensive seat movers in passenger vehicles to precision laser cutters and CNC machines. All move loads through a linear distance in a specific amount of time.
The second and third steps of the approach outlined here are to choose the motor type and size.
One approach for designing a linear system is to list basic requirements, add components one by one, and define every force interaction along the way.
Define the problem
To begin a linear design, determine the mass of what moves and how fast it goes from A to B. We work in SI units, as they eliminate multiple conversion constants and can always convert back into English units. For example, assume we’re choosing a motor for a simple linear-rail mechanism that moves a load:
Load's mass = 9 kg
Mass orientation: Vertical
A-to-B move distance and time = 200 mm in 1.0 sec
Dwell time = 0.5 sec
Move profile = 1/3-1/3-1/3 trapezoid–controlled acceleration and deceleration
Rotary-to-linear conversion = TFE-coated leadscrew Ø8 mm and 275-mm long
Load support: Linear ball rail and TFE-coated guide rails with a friction coefficient μ = 0.01
Overall size: Limited to the smallest volume possible
Drive architecture: Must be simple, as this is a cost-sensitive application
Drive control: Four-quadrant operation with encoder feedback
Drive power supply = 32 Vdc, 3.5 Arms, 5.0 Apeak maximum output
Worst-case ambient temperature = 30°C
Because force = ma (where a = acceleration due to gravity = 9.81 m/sec2), the 9-kg mass lifted against gravity requires a force of 88 N.
Sizing: Just the beginning
This article explains how to size a motor for a relatively simple single-axis linear-motion application. What don’t we cover here?
Sizing motors for complex designs. The motor-sizing principles we outline are applicable to X-Y tables and multiaxis pick-and-place machines. However, every axis in these designs requires independent analysis of load demands.
Choosing a safety factor so the machine lasts for its intended life. A design’s number of useful cycles depends on motor size, as well as the machine’s mechanical elements including the gearbox and leadscrew assembly.
Accounting for positioning accuracy, resolution, repeatability, maximum roll, pitch, and yaw. Only linear-motion systems that account for these fully meet application requirements.
1. How much power is needed to move the load in the required time?
Calculating minimum power output to translate the load provides a starting point for specifying the rest of the system's components. For our example, this is the average power needed to lift the 9 kg from A to B in 1 sec.
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where P = power, W; F = force, N; S = linear distance, m; and t = time, sec.
Note that power calculated here is less than peak power (or instantaneous power during the move profile) to accelerate and decelerate. Likewise, power calculated here doesn’t factor in extra power to overcome system losses such as friction. We’ll calculate the motor-shaft power for that in a later step.
Pick the motor
2. What motor technology is best for this application?
As outlined in our original parameters, the final design must be inexpensive and have simple drive architecture. Stepper motors satisfy both of these requirements. However, minimizing this machine’s overall volume is also important, so a stepper isn’t recommended: The 17.64-W minimum power requirement at the load (not including system losses and instantaneous peak power) would necessitate a large stepper. A brushless motor solves the problem of design footprint, but adds cost and complicates the drive architecture.
The third option — a dc-brush gearmotor with an in-line planetary gearhead — provides a small footprint, simplified drive, and relatively low cost. Adding a leadscrew for rotary-to-linear conversion keeps gearmotor output speed at around 1,000 rpm, which reduces generated heat at the leadscrew and nut-thread interface.
Gearmotor output
3. What’s the velocity, reflected inertia, and reflected load at the gearmotor output shaft (acting as the leadscrew input)?
Step one: Calculate the peak linear velocity of the application with its 1/3-1/3-1/3 motion profile:
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where vpk = peak linear velocity, m/sec.
Step two: Calculate the minimum pitch needed to keep the leadscrew speed at about 1,000 rpm:
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where pmin = minimum leadscrew pitch, m.
For one typical product, the closest pitch in an 8-mm screw diameter is 20.32 mm.
Step three: Calculate the peak shaft speed of the leadscrew (in rad/sec) for a linear velocity of 0.3 m/sec:
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The leadscrew for stepper motor we select is TFE coated, 275-mm long, 8 mm in diameter with a 20.32-mm pitch, and paired with a freewheeling nut. Assume the leadscrew efficiency, ηs, is 86% and its inertia, Js, is 38.8 × 10-7 kg-m2.
Step four: Determine the total reflected inertia, JT, back from the load to the leadscrew shaft:
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where JL = reflected load inertia, kg-m2; m = mass, kg; and p = leadscrew pitch, m.
Step five: Determine the shaft torque needed to accelerate the load inertia Ta:
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