Equation for trapzoidal motion in motors

In a trapezoidal motion, velocity is constant for some time, then increases or decreases linearly, then is constant again. That doesn't describe the motor's motion as shown in your graph.In summary, your graph does not represent the motor's motion, and you can't determine the total distance traveled from the graph.
  • #1
baby_1
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15
Hello,
For a motor, we need to calculate the traveled distance and proper timing for its motion. Here is the trapzodial motion of our motor( we need that our motor traveled 20 degree in 2 seconds). (where t2=2t1)
77.png


we know that the traveled distance is surface of our velocity diagram. So:

[itex] \frac{1}{2}at_{1}+V_{0}t_{2}+\frac{1}{2}at_{1}=\Theta_{T}=>(V_{0}=at_{1},t_{2}=2t_{1} )=>t_{1}=\sqrt{\frac{\Theta_{T}}{3a}}[/itex]
for having t1=.5s @ theta=20 degree-> a=27.
by solving this equation we find t1=.49 and t2=.99 which seems be correct.

However, By calculating this equation for theta=90 degree (a=27) =>t1=1.29, t2=2.58. I calculated traveled distance:
[itex] \Theta_{1}=\frac{1}{2}at_{1}=22.46 degree[/itex]
[itex] V=at=>V_{0}=at_{1}=> \Theta_{2}=V_{0}t_{2}=at_{1}t_{2}=89.9 degree[/itex]
[itex] \Theta_{3}=\frac{1}{2}at_{1}=22.46 degree[/itex]
[itex] \Theta_{1}+\Theta_{2}+\Theta_{3}>90[/itex]

what is my mistake?
 
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  • #2
I believe your post has some typos but anyway:
$$t_1=\sqrt{\frac{\Theta_T}{3a}}$$ looks correct to me. But when I apply it for ##\Theta_T=90## and ##a=27## I get ##t_1=1.05##. ##t_2=2.1##
 
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  • #3
baby_1 said:
Summary:: Trapzoidal motion

Hello,
For a motor, we need to calculate the traveled distance and proper timing for its motion. Here is the trapzodial motion of our motor( we need that our motor traveled 20 degree in 2 seconds). (where t2=2t1)
Do you mean that the motor rotated 20 degrees in 2 seconds?
Your problem description is very confusing. The motor is not traveling on a trapezoidal path. This is a graph of velocity vs. time.
baby_1 said:
View attachment 263146

we know that the traveled distance is surface of our velocity diagram. So:

[itex] \frac{1}{2}at_{1}+V_{0}t_{2}+\frac{1}{2}at_{1}=\Theta_{T}=>(V_{0}=at_{1},t_{2}=2t_{1} )=>t_{1}=\sqrt{\frac{\Theta_{T}}{3a}}[/itex]
for having t1=.5s @ theta=20 degree-> a=27.
by solving this equation we find t1=.49 and t2=.99 which seems be correct.
From what you already said, total time is 2 seconds, and t2 = 2t1, so t1 = 0.5 sec. and t2 = 1.0 sec.
baby_1 said:
However, By calculating this equation for theta=90 degree (a=27) =>t1=1.29, t2=2.58. I calculated traveled distance:
[itex] \Theta_{1}=\frac{1}{2}at_{1}=22.46 degree[/itex]
[itex] V=at=>V_{0}=at_{1}=> \Theta_{2}=V_{0}t_{2}=at_{1}t_{2}=89.9 degree[/itex]
[itex] \Theta_{3}=\frac{1}{2}at_{1}=22.46 degree[/itex]
[itex] \Theta_{1}+\Theta_{2}+\Theta_{3}>90[/itex]

what is my mistake?
From your graph, at the start acceleration a is constant and positive, reaching a maximum velocity of v0 in the first 1/2 second. Velocity v is constant for the next full second. At t = 1.5 second, the motor decelerates from v0 to 0, so a is negative.
 
  • #4
I am really appreciative of your response.
Thanks Delta2 , I see that I calculated wrongly and it works fine.
Thanks Mark44 , but I didn't calculate the traveled distance via velocity equation I calculated the traveled distance via surface of the diagram where the value is positive every where.
 
  • #5
baby_1 said:
Thanks Mark44 , but I didn't calculate the traveled distance via velocity equation I calculated the traveled distance via surface of the diagram where the value is positive every where.
What does the diagram represent? You didn't label the axes, but the horizontal axis is time, and the vertical axis appears to be velocity (##v_0## is marked on it).

If this is a graph of velocity vs. time, it doesn't make any sense to calculate the total distance traveled base on the length of the three segments that make up the graph.

As I mentioned before, in the first half second, velocity appears to be increasing from 0 to ##v_0##. In the next second, velocity is constant, and in the last half second, velocity is decreasing from ##v_0## back to 0. This means that that acceleration a is negative in the last half second.

Also, the motor is not moving trapezoidally, at least if I understand what your graph is supposed to represent.
 

Related to Equation for trapzoidal motion in motors

1. What is the equation for trapezoidal motion in motors?

The equation for trapezoidal motion in motors is typically represented as s(t) = s0 + v0t + 1/2at2, where s(t) is the position of the motor at time t, s0 is the initial position, v0 is the initial velocity, and a is the acceleration.

2. How is trapezoidal motion used in motors?

Trapezoidal motion is commonly used in motors to control the speed and position of the motor. It involves ramping up the motor's speed to a desired velocity, maintaining that velocity for a period of time, and then ramping down the speed to come to a stop at a specific position.

3. What are the advantages of using trapezoidal motion in motors?

There are several advantages to using trapezoidal motion in motors. It allows for precise control of the motor's speed and position, resulting in smoother and more accurate movements. It also helps to reduce wear and tear on the motor, as it avoids sudden changes in speed and direction.

4. Can trapezoidal motion be used in all types of motors?

Yes, trapezoidal motion can be used in a variety of motors, including DC motors, stepper motors, and servo motors. However, the specific implementation may vary depending on the type of motor and its capabilities.

5. Are there any limitations to using trapezoidal motion in motors?

One limitation of trapezoidal motion in motors is that it may not be suitable for applications that require very high speeds or very precise positioning. In these cases, other motion control techniques such as sinusoidal or profiled motion may be more appropriate.

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