The nature of the application under consideration dictates what information is required to properly select a motor. For example, operating at a fixed speed will have a different demand than operation under servo conditions where the duty cycle (duration on versus duration off) may be required. In general, three parameters will determine motor selection: (1) peak torque requirement, (2) RMS torque requirement, and (3) speed of operation.
Peak torque, Tp is the sum of the torque due to acceleration of inertia, TJ, load, TL, and friction, TF:
TP = TJ+ TL + TF * ( Equation 1 )
*Other factors contribute to the overall torque requirement. The values of these factors are typically more difficult to assess. They are taken into consideration by employing a “rule-of-thumb” safety margin: 20% of the calculated torque value.
Looking at the separate components, the torque due to inertia is the product of load (including motor rotor) inertia and load acceleration:
TJ = JL+M Xα (α = acceleration) ( Equation 2 )
The torque due to the load is defined by the configuration of the mechanical system coupled with the motor. The mechanical system also determines the amount of torque required to overcome friction in a given application. These systems will be described later in this document.
Root-Mean-Square or RMS torque is a value used to approximate the average continuous torque requirement of an application. It is a statistical approximation described by the following equation:
( Equation 3 )
Where t1 is the acceleration time, t2 is the run time, t3 is the deceleration time, and t4 is the dwell time in a move.
Speed of operation is also dictated by the configuration of the mechanical system that is coupled with the motor shaft and by the type of move that is to be affected. For example, a single speed application would require a motor with a rated operating speed equal to the average move speed. A point-to-point positioning application would require a motor with a rated operating speed higher than the average move speed. (The higher operating speed would account for acceleration, deceleration, and run times of the motion profile, resulting in an average speed equal to the move speed). Figures 1A & 1B relate rated operating speed to average move speed for point-to-point positioning move profiles.
( Figures 1A & 1B )
This section presents conversion factors and physical characteristics of motion that are utilized in the sizing and selection of motors. The information provided is a technical basis for the calculations discussed later in this document.
Inertia is a very important consideration during acceleration and deceleration of loads. Because belts, pulleys, gears, sprockets, drive shafts, driven shafts, etc., are typically utilized in power transmission applications, it is appropriate to review inertias of cylindrical objects. Figure 2 illustrates two objects rotated about the cylinder axis and equations describing the corresponding rotational inertias.
( Figure 2 )
Inertia can be calculated whether or not the weight is known. The equations below provide the necessary elements required for each scenario.
For objects of known weight, W, substituting W/g (g = acceleration of gravity) for m:
( Equation 4 )
( Equation 5 )
If weight is unknown but volume, V, and material density, ƿ, is known, substituting Vƿ/g for m:
( Equation 6 )
( Equation 7 )
Table 1 shows the densities of commonly used materials. The values shown in the table should be substituted for p, in Equations (6) and (7), when calculating inertias.
( Table 1 )
1.1 Aluminum disk:
Equation ( 6 ) shows that the inertia of a solid cylinder of length L, radius R, and density p may be determined as follows:
J = (0.0041 Sec²/in)R⁴Lp = (0.0041 Sec²/in)(4”)⁴(0.300”)(1.54oz/in.ᶾ) = 0.4849 oz-in Sec²
1.2 Plastic chuck:
J = (0.0041 Sec²/in)(2.5”)⁴(0.5”)(0.64 oz./in.ᶾ) = 0.051 oz-in Sec²
1.3 Motor rotor: To be included later
Step 2 – Acceleration Calculation:Units of Radians/Sec² should always be used in acceleration rate calculations, since “Radian” is a unitless number. The product of acceleration and inertia will cancel out, resulting in units of torque. Acceleration rate = operating speed/acceleration time:
3,600 RPM/9.55=377 Rad/Sec. and Acceleration Rate = (377 Rad/Sec) / (2 Sec) = 188 Rad/Sec²
Step 3 – Peak Torque Calculation: Peak torque is the sum of the torque due to inertia, load, and friction. Since load (usually windage) and friction torque in applications such as disk certifiers are often negligible, the motor may be selected based on the product of inertia and acceleration alone:
Tƿ = (inertia) (acceleration rate) = (disk inertia + chuck inertia + rotor inertia) (acceleration) = (0.4849 + 0.051 + rotor, oz.in. Sec²) (188 Rad/Sec²)
Ignoring rotor inertia for now:
Tp = (0.5359 oz-in Sec²) x (188 Rad/Sec²) = 100 oz-in
Step 4 – Motor Selection: A motor with a peak torque rating higher than the value calculated in Step 3 should be selected. Again, referring back to Table4, Model DIN34-20 has a peak torque rating of 120 oz-in, it also has a rotor inertia of 7.7 X 10-ᶾ oz-in Sec²; a value that is negligible compared to the total load inertia. (If the rotor inertia was comparable to or greater than the load inertia, it would be necessary to proceed with motor selection iterations, plugging the rotor inertia value in the inertia-times-acceleration-rate equation to insure a proper peak torque requirement.)
Duty cycle check: (100 oz-in)20% = 20 oz-in, well within the continuous stall rating of 50 oz-in.
Step 5 – Winding Selection: The winding should be selected in the same manner as shown in the single speed application example. In this case, the high peak torque requirement necessitates selection of a winding with a low resistance, so that enough current may be drawn to provide 100 oz-in of torque. Winding A appears to be a suitable candidate.
Step 6 – Controller Selection: THigh acceleration rate applications require selection of a controller that has a peak current rating at least as high as the current dictated by motor winding parameters. Winding A, selected in Step 5, will result in a peak current magnitude of 100 oz-in/6.1 oz-in/A, or 16A. Remember, considering motor efficiencies, a controller with a peak rating at least 20% higher than the calculation value, or about 20A, should be selected.
The high acceleration, single speed application also enables the use of two-quadrant control.
A. Point-to-Point Positioning Application:
The point-to-point positioning application is by far the most complex of all applications; not just from the standpoint of system integration and load compensation, but because of the difficulty in estimating an average velocity profile or duty cycle, in practice.
The remainder of this section describes five basic power transmission mechanical configurations. Equations specific to each configuration are provided to enable determination of the motor performance required to affect a point-to-point move. Once determined, move speed, reflected torque, and reflected inertia may be used to calculate the peak torque, RMS torque, and speed of operation motor selection parameters discussed earlier. The section will conclude with an example of a chemical etching/plating process requiring point-to-point positioning capabilities.
JL = Load inertia (oz-in Sec²)
JM = Motor inertia (oz-in Sec²)
JB = Belt inertia (oz-in Sec²)
WL = Load weight (oz.)
FG = Force due to gravity (oz.)
FP = Push/Pull force (oz.)
WB = Belt weight (oz.)
TL = Load torque (oz-in)
TF = Friction torque (oz-in)
Eᶠ = Power transmission efficiency (%/100)
g = Acceleration of gravity (386 in./Sec²)
P = Leadscrew pitch (Revolutions/inch)
µ = Coefficient of friction (%/100)
wM = Rotational velocity (RPM) at motor shaft
w L= Rotational velocity (RPM) at motor shaft
v = Average move velocity (inches/min)
( Figure 5 )
Example: Figure 5 illustrates equipment used in a chemical etching/plating process
Drive = Motor with a 10.1 Reduction gear head (E=0.80), inertia, JGB = 0.04 oz-in Sec², pinion
radius, Rp = 1.5” pinion material = stainless steel, O.5” thick. (Rack and pinion efficiency = 0.90).
Weight = 4,000 oz., D = 24” move on linear bearings; move time = 1 Sec. Velocity Profile:
Trapezoidal for controlled acceleration and controlled decal. Duty cycle = 1 Sec move, 8 sec rest, repeated indefinitely. Vsource = 36V DC.
Step 1 - Break up the analysis into components as follows;
(1) Calculate values of parameters at the pinion
(2) Values reflected to the motor shaft, through the gear head.
Calculation of load parameters at the pinion: Figure 6D describes the equations pertaining to a rack and pinion system.
( Figure 6 )
As previously discussed, Jp = (0.0041 Sec²/in)(1.5”)⁴(0.5”)(4.48 oz./inᶾ) = 0.046 oz-in Sec² - i.e. negligible.
Step 2 – Calculations of total parameters at motor shaft: The gear box equations shown in Figure 6B may be used for these calculations.
Step 3 – Calculation of peak torque requirement:Earlier, we discussed Tp = TJ + TL + TF = (JL+M)α + TL + TF = (0.33 oz-in Sec² + JM) α + TL + TF, but a = (2,300 RPM/9.55)/t1 = (241 Rad/Sec)/0.333 Sec = 723 Rad/Sec², and Tp = (0.00 oz-in Sec² + J) (723 Rad/Sec²) + 33 oz-in + 0 oz-in = 272 oz-in, assuming motor rotor inertia is small relative to reflected load inertia.
Step 4 – Calculation of RMS torque requirement: equation (3) provides the desired value as follows:
Step 5 – Application of 20% safety margins:
1.2 X Tp = 1.2 (272 oz-in) = 326 oz-in
1.2 X TRMS = 66 oz-in X 1.2 = 79 oz-in
Step 6 – Motor and winding selection: Referring to Table 4, it appears Model DIN34-36 meets the peak and RMS torque requirements. Looking at winding A:
Vsource – (Kb) () = 36V – (0.099 V/Rad/Sec)(241 Rad/Sec) = 12 V
Current available to produce torque = V/RM = 12V/0.36Ω - 33A
Current required to drive the load: peak=(326 oz-in)/(14 oz-in/A)=23A
RMS=(79 oz-in)/(14 oz-in/A)=6A
Conclusion: Winding A is a good match.
Step 7 – Comparison of reflected load inertia to rotor inertia: We saw that the reflected load inertia was calculated at a value of 0.33 oz-in Sec². The DIN34-36 has a rotor inertia of 0.022 oz-in Sec². The ratio is therefore 0.33/0.022 = 15, which is slightly higher than the recommended value. The system may perform adequately under these conditions. If not, a solution may be utilization of a motor with higher rotor inertia. Another solution may be direct coupling of the gear head to the motor shaft, thereby increasing the total shaft inertia almost to the level of the reflected load inertia. Electrical modifications to the control circuitry may also help reduce the effects of inertia mismatches. However, this is a solution usually employed “after the fact.”
Step 8 – Controller Selection: This application requires a 36V DC rated drive with continuous and peak current outputs of 6A and 23A, respectively.
In most of the equations provided above, friction was estimated from appropriate coefficients of friction. In practice, it is desirable to measure actual friction levels whenever possible. Friction is a parameter that is often under-estimated in value, but is paramount to proper motor selection. (Measured friction values would replace µW in all of the preceding equations.)
The data, specifications and electrical parameters presented in this guide illustrate typical applications, are for reference only and are subject to change without notice. Although efforts have been made to insure the accuracy of the information given, nothing herein is intended or should be construed as a warranty of the performance or design of BEI Kimco products. Product and data warranties are described solely in BEI Kimco contractual documents.
Peak Torque (Tp) – The torque that can be produced for 10 seconds without exceeding the maximum allowable winding temperature.
Continuous Stall Torque (Tcs) – This is the amount of torque that can safely be produced over an indefinite period of time under a stalled rotor condition. This value is measured with the motor mounted on an aluminum plate (6” x 6” x 1.8”) heat sink, to the maximum allowable temperature of the windings.
Moment of Inertia (Jm) – The moment of inertia of the rotating member, equal to Wk²/g where W = weight (oz), k = radius of gyration (in.) and g = acceleration due to gravity (386 in/Sec²).
Motor Constant (Km) – A figure of merit of the motor. The higher the value for a given volume of motor, the more powerful the motor.
Electrical Time Constant (TE) – This value is equal to L/R (inductance divided by resistance). It is also equal to the time it takes for the current to reach 63% of its steady state value when the winding is energized by a step input of voltage.