minimum and maximum acceleration of the parameter s for the time-optimal path-constraint trajectory planning -(lagrangian mechanics-Ordinary differantial equations) lagrangian formulation of dynamics what's the Fill "1-2-3 "?
minimum and maximum acceleration of the parameter s for the time-optimal path-constraint trajectory planning -(lagrangian mechanics-Ordinary differantial equations) lagrangian formulation of dynamics what's the Fill "1-2-3 "?
Related questions
Question
minimum and maximum acceleration of the parameter s for the time-optimal path-constraint trajectory planning -(lagrangian mechanics-Ordinary differantial equations) lagrangian formulation of dynamics
what's the Fill "1-2-3 "?
![OPTIONS=odeset ('MaxStep',0.01);
4
ml=5;
m2=3;
6 -
Il=0.1;
7 -
12=0.05;
8 -
rl=0.2;
u min=[-100; -40];
u max=[100; 40];
9 -
10 -
abc (ml, m2,I1,12,rl,u min,u max);
LU (a,b, c,u_min, u_max);
11
[a,b, c] =
[L_max, U_min,u_L,u_U] =
12 -
13 -
q_initial=[3*pi/4;sqrt (2)];
q_final-[pi/4; sqrt (2)];
14 -
s_initial=[0 0];
s final=[1 o);
f L=e (t,s) [s (2) ;L (s, sdot) ];
f_U=@ (t,s) [s (2);U (sdot, sdotdot)];
if U_min (0,0) <0
15 -
16 -
17 -
18 -
19 -
20
message='error'
21
return
22
end
t_final=1;
[ttl,sl]=ode45 (f_U, [0, t_final],s_initial,OPTIONS); $Run s forward
23 -
24 -
25 -
[tt2,s2]=ode45 (f L, [t final,0],s final,OPTIONS); $Run s backward
26 -
[tmin, smin, imin,jmin, distance, tleft] =
findswitch (ttl,sl,tt2, s2); Find the intersection of forward and backward runs
%3D
[ttl,sl]=ode 45 (f U, [0, $FILL1%],s initial,OPTIONS); *Run s forward again with accelaration this time up to the switch
[tt2,s2]=ode45 (f_L, [tmin, tmin+$FILL2],$FILL3%,OPTIONS); $Run s forward with deccelaration after the switch
27 -
28
29
$FILL3: s initial will be different
30
subplot (2,2,1)
31
plot (sl (:,1),sl (:,2))
32
hold on
33
plot (s2 (:,1),s2 (:, 2))
34
xlabel ('s')
ylabel ('ds/dt')
legend ('acceleration','decceleration', 'Location','south')
35
36](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7611a395-318b-4577-b358-b0b6f732a783%2F09ddff1e-5bc7-4d5d-bb72-2a5b22cbde96%2Fsnfhar_processed.png&w=3840&q=75)
Transcribed Image Text:OPTIONS=odeset ('MaxStep',0.01);
4
ml=5;
m2=3;
6 -
Il=0.1;
7 -
12=0.05;
8 -
rl=0.2;
u min=[-100; -40];
u max=[100; 40];
9 -
10 -
abc (ml, m2,I1,12,rl,u min,u max);
LU (a,b, c,u_min, u_max);
11
[a,b, c] =
[L_max, U_min,u_L,u_U] =
12 -
13 -
q_initial=[3*pi/4;sqrt (2)];
q_final-[pi/4; sqrt (2)];
14 -
s_initial=[0 0];
s final=[1 o);
f L=e (t,s) [s (2) ;L (s, sdot) ];
f_U=@ (t,s) [s (2);U (sdot, sdotdot)];
if U_min (0,0) <0
15 -
16 -
17 -
18 -
19 -
20
message='error'
21
return
22
end
t_final=1;
[ttl,sl]=ode45 (f_U, [0, t_final],s_initial,OPTIONS); $Run s forward
23 -
24 -
25 -
[tt2,s2]=ode45 (f L, [t final,0],s final,OPTIONS); $Run s backward
26 -
[tmin, smin, imin,jmin, distance, tleft] =
findswitch (ttl,sl,tt2, s2); Find the intersection of forward and backward runs
%3D
[ttl,sl]=ode 45 (f U, [0, $FILL1%],s initial,OPTIONS); *Run s forward again with accelaration this time up to the switch
[tt2,s2]=ode45 (f_L, [tmin, tmin+$FILL2],$FILL3%,OPTIONS); $Run s forward with deccelaration after the switch
27 -
28
29
$FILL3: s initial will be different
30
subplot (2,2,1)
31
plot (sl (:,1),sl (:,2))
32
hold on
33
plot (s2 (:,1),s2 (:, 2))
34
xlabel ('s')
ylabel ('ds/dt')
legend ('acceleration','decceleration', 'Location','south')
35
36
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps
