Question

Problem 2: Consider a LSPB trajectory of the following form 0 stst 4o (a) Determine the LSPB trajectory that will satisfy the following constraints: to 0 (second) t 2 (second) q(tr)--20° V-# 15 (deg/sec) (b) Also find the position, velocity, and acceleration when t 1 (second) (c) Sketch both the trajectory and velocity profiles.

Answer 1

to maintain continuity:

$q_0+\frac{\alpha}{2}t_b^2 = \frac{q_f+q_0-Vt_f}{2}+Vt_b$

200 +00-15 * 2 0,012 +15 tb

$\alpha t_b^2 =-10+30t_b$

velocity

$\alpha t_b = V = 15$

to maintain continuity:

out t = tf-t

$\frac{q_f+q_0-Vt_f}{2}+V(t_f-t_b) = q_f - \frac{\alpha}{2}t_f^2 +\alpha t_f (t_f-t_b)- \frac{\alpha}{2}(t_f-t_b)^2$

20 0 15 2 +15(2-t6) = 200-a22 + a * 2(2-t6)--(2-t6)2

200-2a 4a-2ct,--(2-6)2 -5 + 15(2-6)

$-5+15t_b + 2\alpha - 2\alpha t_b- \frac{\alpha}{2}(4+t_b^2-4t_b) = 0$

$-5+15t_b - \frac{\alpha}{2}t_b^2 = 0$

$\Rightarrow \alpha t_b^2 = -10+30t_b$

But from the velocity equation:

tb-15/a

$\Rightarrow \alpha*(15/\alpha)^2 = -10+30*(15/\alpha)$

(225/a) =-10+ (450/a)

$\alpha = 22.5$

$\Rightarrow t_b = 2/3$

So, the LSPB is given by:

$\begin{cases} (22.5/2)t^2 & \text{ if } 0<t\leq 2/3 \\ -5+15t & \text{ if } 2/3 \leq t \leq 4/3 \\ -25+45t-(22.5/2)t^2 & \text{ if } 4/3\leq t \leq 2 \end{cases}$

b)

at t = 1:

$\dot q = 15deg/s$

eq s

c)

(22.5/2)2 if 2/3 < t < 4/3 -25+45t-(22.5/2)2 if 4/3 t 2

İf 0 < t < 2/3 if 2/3 < t < 4/3 if 4/3 StS2 22.5t 45 - 22.5t

trajectory

20 18 16 14 12 10 0 268 10 12 1416 18 20 2 0

velocity:

15 10 0 2 4 6810 12 14 16 1820