Bringing Newton’s Second Law of Motion into the Motion Control World.
For any linear motion control application, the solution to the problem requires at a minimum three points of data:
- How far do I need to move?
- How heavy is the payload?
- How fast do I need to get there?
These three points of data can be fed into Newton’s Second Law of Motion stating that the amount of acceleration is dependent upon the force acting on it and the object’s mass.
$$acceleration=\frac{force}{mass}$$
The equation above is the basic building block of all motion profile acceleration calculations. You can utilize this equation to determine the maximum theoretical accelerations possible by any motor or actuator if you know the mass of the moving object, and the amount of force the motor generates. There are several opposing forces that affect the maximum acceleration such as friction, drag, and hysteresis. Factoring in the losses, the maximum continuous accelerations that can be achieved through direct drive linear motion are between 5-10 G’s for closed loop position control applications and 10-20 G’s for open loop sinusoidally oscillating applications.
While there are an infinite number of motion profiles, there are three profiles that define acceleration best for most linear motion applications.
- triangular
- sinusoidal
- trapezoidal
The triangular motion profile, also known as a saw tooth profile, is the simplest as it assumes constant acceleration and deceleration through the motion profile. This model is best to understand the basic requirements in your motion profile. Below you can find the equation for calculating the acceleration with a triangular motion profile.
$$acceleration\;in\;G's=\frac{4\times displacement}{(time\:to\:complete travel)^2\times acceleration\:due\:to\:gravity}$$
example
To get the acceleration necessary to move one inch in 0.050 seconds, you plug this into the equation above to get the following:
$$4.15\:G's=\frac{4\times1\:inch}{(0.050\:seconds)^2 \times386\frac{inches}{seconds^2}}$$
A trapezoidal motion profile is derived from the triangular profile, as it assumes a constant acceleration until a desired velocity is achieved, and then maintains a constant velocity for a time. In this case, you will need to know two of three variables: Time to reach the target constant velocity, the targeted constant velocity, or distance available to reach the targeted constant velocity. The basic trapezoidal profile can be seen below.
- If the two known variables are time and distanceto achieve the constant velocity:
- then the calculation for the acceleration necessary to accelerate to a constant velocity can be determined with the formula below:
$$acceleration\:in\:G's=\frac{2\times distance\:available\:to\:reach\:the\:targeted\:constant\:velocity\:}{(time\:to\:reach\:the\:targeted\:constant\:velocity)^2\times acceleration\:due\:to\:gravity}$$
example
To get the acceleration necessary to move to a constant velocity after one inch in 0.050 seconds, you plug this into the equation above to get the following:
$$2.07\:G's=\frac{2\times 1\:inch}{(0.050\:seconds)^2\times 386\frac{inches}{second^2}}$$
- If the two known variables are velocity and distance to achieve the constant velocity:
- then the calculation for the acceleration necessary to accelerate to a constant velocity can be determined with the formula below:
$$acceleration\:in\:G's=\frac{target\:constant\:velocity^2}{distance\:available\:to\:reach\:the\:targeted\:constant\:velocity\times acceleration\:due\:to\:gravity}$$
example
To get the acceleration necessary to move to a constant velocity of 50 inches per second in one inch, you plug this into the equation above to get the following:
$$6.48\:G's=\frac{(50\frac{inches}{second})^2}{1\:inch\times 386\frac{inches}{second^2}}$$
- If the two known variables are velocity and time to achieve the constant velocity:
- then the calculation for the acceleration necessary to accelerate to a constant velocity can be determined with the formula below:
$$acceleration\:in\:G's=\frac{2\times target\:constant\:velocity}{time\:to\:reach\:the\:targeted\:constant\:velocity\times acceleration\:due\:to\:gravity}$$
example
To get the acceleration necessary to move to a constant velocity of 50 inches per second in 0.050 seconds, you plug this into the equation above to get the following:
$$5.18\:G's=\frac{2\times 50\frac{inches}{second}}{0.050\:seconds\times 386\frac{inches}{second^2}}$$
This type of profile is typically found in long travel applications where a part needs to achieve a constant velocity. Once you determine the acceleration in any of these applications you can also calculate the missing variable whether it be distance, time, or velocity.
The third common motion profile is typically utilized in oscillatory systems, which is a sinusoidal motion profile. These profiles have a constantly changing acceleration to ensure a smooth oscillation between two end points. To calculate the acceleration with this motion profile, it is necessary to know two of three variables: Frequency of oscillation, displacement of oscillation, or maximum velocity.
- If the two known variables are frequency of oscillation and displacement in the oscillatory system:
- then the calculation for the acceleration necessary to oscillate can be determined with the formula below:
$$acceleration\:in\:G's=\frac{2\times \pi^2\times frquency\:of\:oscillation^2\times displacement}{acceleration\:due\:to\:gravity}$$
example
To get the acceleration necessary to oscillate one inch at 20 Hertz, you plug this into the equation above to get the following:
$$20.45\:G's=\frac{2\times \pi^2 \times 20Hz^2 \times 1\:inch}{386\frac{inches}{second^2}}$$
- If the two known variables are maximum velocity and frequency of oscillation in the oscillatory system:
- then the calculation for the acceleration necessary to oscillate can be determined with the formula below:
$$acceleration\:in\:G's=\frac{2\times \pi\times frquency\:of\:oscillation\times maximum\:velocity}{acceleration\:due\:to\:gravity}$$
example
To get the acceleration necessary to oscillate at 20 Hertz with a maximum velocity of 50 inches per second, you plug this into the equation above to get the following:
$$16.28\:G's=\frac{2\times \pi\times 20Hz\times 50\frac{inches}{second}}{38\frac{inches}{second^2}}$$
- If the two known variables are maximum velocity and displacement in the oscillatory system:
- then the calculation for the acceleration necessary to oscillate can be determined with the formula below:
$$acceleration\:in\:G's=\frac{2\times maximum\:velocity^2}{displacement\times acceleration\:due\:to\:gravity}$$
example
To get the acceleration necessary to oscillate at 20 Hertz with a maximum velocity of 50 inches per second, you plug this into the equation above to get the following:
$$12.95G's=\frac{2\times (50\frac{inches}{second})^2}{1\:inch\times 386\frac{inches}{second^2}}$$
No matter the motion profile the basic solutions for calculating acceleration will give an understanding for acceleration you are trying to get out of a system. With this understanding, any user can verify the best technology to suit their desired motion control application.