Linear Servo Motor

Linear Shaft Motor Sizing

One of the most straightforward tasks in the design of a linear motion system is to specify a motor and drive combination that can provide the force, speed and acceleration that is required by the mechanical design.

This is all too often the most overlooked aspect of the linear motion system design. Often, the engineer will select a motor, and then design a system around it. This approach leads to improperly sized motors for the application, making this the most costly aspect of the system, not only from the perspective of the initial purchase cost, but also due to service maintenance and energy costs.

Linear Shaft Motor Sizing Application Note:

This application note is intended to help walk you through the steps preformed by the NPA Smart sizing software. This will assist you in understanding the process and calculation of even more complicated motion systems. The steps that will be covered include:

  1. Define the operation conditions and motion profile
  2. Calculate forces required by your system
  3. Selecting the proper Linear Shaft Motor for your application
  4. Selecting the proper linear servo driver for your application


Motor Sizing Example

Let’s assume you want to move horizontally a mass of 6kg point-to-point over a distance of 100 mm (X) in 160 msec, including settling time (Tm) to +/- 1 micron. Total travel is 400mm, and a dwell time of 200msec is needed after each move.


Move Profile
We will assume an estimated settling time of 10msec (Ts).
The move cycle time (Tc) is 160 + 200 = 360msec
Using previous move formula:
T (msec) = Tm – (Ts)
T (msec) = 160 – 10 = 150msec
We will assume an efficient trapezoidal profile (1/3, 1/3, 1/3)
Acceleration needed here (see previous move formula):
A = (4.5)*(0.1*0.152)
A = 20m/sec2 (about 2 “g”)
V = (1.5)*(0.1/0.15)
V = 1m/sec
The acceleration and deceleration time becomes (150/3)= 50msec
The time at constant speed is (150/3) = 50msec
We can estimate the acceleration force of the load only (see previously mentioned formula) at 2g*9.81*6kg = 117N.
Based on this we can select S350T (peak force = 592N, continuous force = 148N) assuming a coil mounting plate of 1kg.
Total moving mass: 6kg (load) + 1kg (plate) + 1.9kg (coil mass) = 8.9kg
|Coil resistance = 20.2ohm, Coil Force constant 99N/Ap, Thermal Resistance 2.4°C/W, Back Emf 33Vp/m/sec,
Inductance p-p 33mH, Electrical cycle length 120mm
We assume a good set of linear bearings with μ=0.005 and 20N of friction.
Friction Force: Ff (N) = 8.9*9.81*[sin(0) + 0.005*cos(0)] + 20 = 20.4N
Inertial Force: Fi (N) = 8.9*20 = 178N
Total Acceleration Force: F1 (N) = 178 + 20.4 = 198.4N
Total Constant Velocity Force: F2 (N) = 20.4N
Total Deceleration Force: F3 (N) = 178 – 20.4 = 157.6N
Total Dwell Force: F4 (N) = 0N
RMS Force: Frms (N) = √[{198.42*0.05)+(20.42*0.025)+(157.62*0.05)/0.36]
Frms (N) = 94.7N
RMS Current: Ica = 94.7/99 = 0.96 Amp rms
Peak Current: Ipa = 198.4/99 = 2 Amp rms
Motor Resistance Hot: Rhot = 20.2 * 1.423 = 28.7Ω
Voltage due B EMF: Vbemf = 33 * 1 = 33Vac
Voltage due I*R: Vir = 1.225 * 28.7 *2 = 70.32Vac
Voltage due Inductance: VL = 7.695 * 1 * 33 * 2 / 120 = 4.23Vac
Bus Voltage needed: Vbus = 1.15 * √ [(33 + 70.3)2 + 4.232] = 118.8Vac


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