Sizing a Motor to Drive a Load along a Motion Trajectory
To qualify for the given system, the motor must successfully pass all three following tests:
A. The motor must be able generate peak torque for the application
B. The motor must be able to run at the maximum velocity
C. The motor must be able to generate toque without overheating
A. The motor must be able generate peak torque for the application
In order to drive the load along the required motion trajectory, the motor must generate a certain torque
function 𝑇𝐺 . The torque depends on the moment of inertia, J, of the load, required angular acceleration, α and
opposing frictional torque, TF, of the load. The torque is given by the following equation:
𝑇𝐺 = 𝐽𝛼 + 𝑇𝐹
Where:
J = Moment of Inertia of the load, kg-m2
α = Angular Acceleration, m/s2
TF = Opposing Friction of the load, N-m
Given Velocity Profile:
Velocity
0
20
40
60
Time (ms)
The load it needs to drive has the following specification:
a)
b)
c)
d)
e)
Moment of inertia of the load, JL = 10-3 kg-m2
Assume that moment of inertia of the motor, JL = 10-4 kg-m2
Opposing frictional torque, TF =0.2 N-m
Cycle time is 60 ms, means that the motor must complete one revolution in 60 ms.
This process must be repeated every 60 ms or 100ms.
Solution:
We are looking for a motor that will be able to perform above motion trajectory:
In order to accomplish the required move with the specified time, the application peak velocity equals:
𝜔𝑜 =
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
1
1
𝑡𝑎𝑐𝑐𝑒𝑙 + 𝑡𝑠𝑙𝑒𝑤 + 𝑡𝑑𝑒𝑎𝑐𝑐
2
2
Or,
𝜔𝑜 =
2𝜋
. 04
= 157 𝑟𝑎𝑑⁄𝑠
Since the velocity must be reached within the acceleration time 20 ms, the acceleration rate equals:
𝛼=
Since: 1
𝜔𝑜
𝑡𝑎𝑐𝑐𝑒𝑙
𝑟𝑎𝑑
157 𝑠
=
= 7850 𝑟𝑎𝑑/𝑠 2
0.02 𝑠
rev
radian
 0.105
min
sec ond




radian
1


0  157
 1495  rpm

radian 
sec ond  

0.105
sec ond 

Thus, the required torque level can be computed using the following equation:
𝑻𝑮 = 𝑱𝜶 + 𝑻𝑭
Remember that motor has the moment of inertia, JM that is normally in the order of ten to fifty times less
than load’s moment of inertia, JL. J in the above equation is the total J = JL + JM = 10-3+10-4=1.1x10-3N-m
Or,
𝑇𝐺 = (1.1 𝑥 10−3 𝑁­𝑚) (7850
𝑟𝑎𝑑
) + 0.2 𝑁­𝑚
𝑠2
Or,
𝑻𝑮 = 𝟖. 𝟖𝟑𝟓 𝑵­𝒎
Similarly, during the slew and de-acceleration intervals, the generated torque equal 0.2 Nm and -1.056 Nm
respectively.
𝑇𝐺 𝐷𝑒𝑎𝑐𝑐𝑒𝑙 = (1.1 𝑥 10−3 )(−7850) + 0.2 𝑁­𝑚
Or,
𝑇𝐺 𝐷𝑒𝑎𝑐𝑐𝑒𝑙 = −8.435 𝑁𝑚
B. The motor must be able to run at the maximum velocity
Since the velocity 157 radian/sec must be reached within the acceleration time of 20 ms, the acceleration
rate equals:
𝛼=
Since: 1
𝑟𝑎𝑑
157 𝑠
=
= 7850 𝑟𝑎𝑑/𝑠 2
0.02𝑠
𝜔𝑜
𝑡𝑎𝑐𝑐𝑒𝑙
rev
radian
 0.105
min
sec ond



radian  
1

0  157
 1495  rpm


radian 
sec ond  

0.105
sec ond 

Because  
d 

dt
t
Conversion:
3000 rev/min= 50 rev/ s
1 Rev is 2π Radians=3.1428 x 2= 6.284 rad/rev
Therefore Motor Peak Velocity,
50
𝑟𝑒𝑣
𝑠
= 50
𝑟𝑒𝑣
𝑠
𝑥 6.284
𝑟𝑎𝑑𝑖𝑎𝑛𝑠
𝑟𝑒𝑣
𝑟𝑎𝑑𝑖𝑎𝑛𝑠
= 314.2 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
Thus, motor must be capable of running at 1495 rpm.
Required Power for the Motor:
Thus the Required motor power:
𝑃=
𝑛𝑇
𝑊𝑎𝑡𝑡𝑠
9.55
Or,
𝑇=
𝑃 = (1495 𝑟𝑝𝑚).
30
( 𝜋 )𝑃
𝑛
8.835 𝑁𝑚
= 1383.352 𝑊𝑎𝑡𝑡𝑠
30/𝜋
P = 1.383 kW
Another important Equation:
∆𝑛 =
9.55𝑇∆𝑡
𝐽
Where:
∆𝑛 = Change in speed in rev/min
𝑇 = 𝑇𝑜𝑟𝑞𝑢𝑒 𝑖𝑛 𝑁. 𝑚
∆𝑡 = 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑜𝑓 𝑡𝑖𝑚𝑒 𝑑𝑢𝑟𝑖𝑛𝑔 𝑤ℎ𝑖𝑐ℎ 𝑡ℎ𝑒 𝑡𝑜𝑟𝑞𝑢𝑒 𝑖𝑠 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑖𝑛 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝐽 = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑖𝑛 𝑘𝑔. 𝑚2
9.55 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑜 𝑡𝑎𝑘𝑒 𝑐𝑎𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡𝑠[ 𝑒𝑥𝑎𝑐𝑡 𝑣𝑎𝑙𝑢𝑒 =
30
]
𝜋
A. The motor must be able to generate toque without overheating
This 3rd test which is to determine if the motor can generate the peak torque without overheating is a little
more involved. Temperature rise in the motor is proportional to the RMS value of the torque. It is necessary
to compute the RMS value and compare it with the related continuous torque.
The RMS value of a periodic function is:
𝑇𝑅𝑀𝑆 = √
1 𝑇 2
∫ 𝑇 𝑑𝑡
𝑇 0 𝐺
Graphically,
8.835 N-m
Torque
0.2 N-m
-8.435 N-m
60 ms
𝑇𝑅𝑀𝑆 = √
1
1
(8.8352 𝑥. 02 + 0.22 𝑥. 02 + (−8.435)2 𝑥. 02) = √
(2.985 𝑁­𝑚) = 7.053 𝑁­𝑚
. 06
. 06
Continuous torque available from a motor is generally given by the motor manufacturer. This continuous
torque available from a motor is the TRMS value and safe limit to run at that torque without overheating.
Based on the above criteria we found a motor having following specification:
Manufacturer: Sure Servo, Model# SVM -220, Medium inertia motor
Parameter
Value
𝑅𝑎𝑡𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 𝑜𝑢𝑡𝑝𝑢𝑡(𝑊𝑎𝑡𝑡)
2000
𝑅𝑎𝑡𝑒𝑑 𝑇𝑜𝑟𝑞𝑢𝑒 𝑁­𝑚)
9.4
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑇𝑜𝑟𝑞𝑢𝑒𝑁­𝑚
Resistance)
23.5
𝑇𝐶 (Continuous Torque) N-m
9.4
𝑇𝑃 (Peak Torque)
23.5 𝑁. 𝑚
𝜔𝑝 (Peak Angular Velocity)
3000 𝑟𝑝𝑚 = 314 𝑟𝑎𝑑/𝑠
𝐾𝑡 (Torque Constant)N-m/A
0.77𝑁. 𝑚/𝐴
𝐽𝑀 (Moment of Inertia)
27.8 𝑥10−4 𝐾𝑔. 𝑚2
𝜌(Armature Resistance) Ω
0.6 Ω
𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑇𝑖𝑚𝑒 𝐶𝑜𝑛𝑠𝑡𝑎𝑛 )
ms
𝐴𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝐼𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒, 𝑚𝐻
10.1
Rated Speed (rpm)
2000 𝑟𝑝𝑚
6.1