1
$\begingroup$

i currently calculating the total time travel of a servo motor moving from point A to point B.

I having a distance with 1000m with speed 10m/s , acc = 2m/s^2 and dcc=2m/s^2, what is the equation for this type of question ? I have use the trapezoid equation which is time taken = (Acc+Dcc)*(max.dis/max.speed) to solve. it this the correct equation ? if I have a large acceleration or deceleration, what will be the equation ?

$\endgroup$
0
$\begingroup$

I assume $10 \mathrm{ms}^{-1}$ is the maximum velocity. Then the calculation is as follows with the usual symbols $s(t),v(t), t, s(0)=0, v(0)=0$ for distance, velocity, time, and $a=2 \mathrm{ms}^{-2}$. $$v=at=2t,\;s=\frac{a}{2}t^2=1\times t^2$$ After 5 s and 25 m the maximum velocity is reached. Since the deceleration has the same magnitude as acceleration the deceleration phase is analogous to the acceleration (takes 5s and needs 25m). Therefore the distance with constant max. velocity is $1000-2\times 25=950$m and needs $950/10=95$s.

Therefore the total time is $95+2\times 5 = 105$ s.

Your formula cannot be correct, because you suppress the acceleration and deceleration phases, it would give $4\times 1000/ 10 = 400$ s

For general equations let $v_m$ the max. velocity, $D$ the total distance, $t_a, s_a$ the time and distance for the acceleration phase and $t_d, s_d$ the time and distance for the deceleration phase. We have

$$t_a=\frac{v_m}{a},\;s_a=\frac{a}{2}t_a^2=\frac{v_m^2}{2a}$$ $$t_d=\frac{v_m}{d},\;s_d=\frac{d}{2}t_d^2=\frac{v_m^2}{2d}$$

Assuming $s_a+s_d<=D$ you have for the total time $T$

$$T=t_a+t_d + \frac{D-s_a-s_d}{v_m}$$ $$T=\frac{v_m}{a}+\frac{v_m}{d} + \frac{D}{v_m}-\frac{v_m}{2a}-\frac{v_m}{2d}$$

$$T=\frac{1}{2}\left(\frac{v_m}{a}+\frac{v_m}{d}\right) + \frac{D}{v_m}$$

Large $a,d$ values are no problem, but small may invalidate the condition $s_a+s_d<=D$

$\endgroup$
  • $\begingroup$ thank for the equation and explanation, i will do the testing with the timer and actual mechanism. $\endgroup$ – user1804079 Sep 3 '15 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.