Equation for trapzoidal motion in motors

[baby_1]

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)

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

$\frac{1}{2}at_{1}+V_{0}t_{2}+\frac{1}{2}at_{1}=\Theta_{T}=>(V_{0}=at_{1},t_{2}=2t_{1} )=>t_{1}=\sqrt{\frac{\Theta_{T}}{3a}}$
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:
$\Theta_{1}=\frac{1}{2}at_{1}=22.46 degree$
$V=at=>V_{0}=at_{1}=> \Theta_{2}=V_{0}t_{2}=at_{1}t_{2}=89.9 degree$
$\Theta_{3}=\frac{1}{2}at_{1}=22.46 degree$
$\Theta_{1}+\Theta_{2}+\Theta_{3}>90$

what is my mistake?

 

[Delta2]

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 $t_1=1.05$. $t_2=2.1$

 

[Mark44]

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.

View attachment 263146

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

$\frac{1}{2}at_{1}+V_{0}t_{2}+\frac{1}{2}at_{1}=\Theta_{T}=>(V_{0}=at_{1},t_{2}=2t_{1} )=>t_{1}=\sqrt{\frac{\Theta_{T}}{3a}}$
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.

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

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.

 

[baby_1]

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.

 

[Mark44]

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.