- #1

baby_1

- 159

- 15

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:

[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?