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My professor gave us an assignment in which we have to find the cubic equation for a 3-DOF manipulator. The end effector is resting at A(1.5,1.5,1) and moves and stops at B(1,1,2) in 10 seconds. How would I go about this? Would I use the Jacobian matrix or would I use path planning and the coefficient matrix to solve my problem. I'm assuming coefficient matrix but I am not given the original position in angle form. I was only taught how to use path planing when the original angles are given.

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    $\begingroup$ I would use inverse kinematics to determine the motions of each joint (start and end positions). Then use path planning to set the desired motion profiles for each axis over those ten seconds. $\endgroup$ – SteveO Dec 11 '15 at 4:59
  • $\begingroup$ Does it have to be in joint space? $\endgroup$ – 50k4 Dec 11 '15 at 5:49
  • $\begingroup$ Good comment @50k4. You could plan the trajectory in Cartesian space, and then use inverse kinematics repeatedly throughout the 10 sec to stay on that path. The nice thing about that approach is you can keep the end effector on a linear path during the trajectory. However, it requires repeated inverse kinematics in the motion loop, which the earlier approach does not need. $\endgroup$ – SteveO Dec 11 '15 at 7:59
  • $\begingroup$ Well, it depends on the exercise, I mean if it is a theoretical exercise about trajectory planning, solving the inverse kinematics seems to be an unneeded overhead...Cartesian curves might be enough for the exercise... $\endgroup$ – 50k4 Dec 11 '15 at 10:52
  • $\begingroup$ Welcome to robotics Jimenemex. On stack exchange, questions asking for homework help must include a summary of the work you've done so far to solve the problem, and a description of the difficulty you are having solving it. Please edit your question to add this information. $\endgroup$ – Mark Booth Dec 11 '15 at 14:30
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I guess you want to find a cubic polynomial for the end effector. You have 3 coordinates for your points A and B, from your question is not clear if they are $x,y,z$ or $x,y,\theta$. Anyway, I'll show here the procedure for $x$, and you can repeat it for the other two coordinates.

Given the cubic parametric form $x = a_0+a_1 t + a_2 t^2 + a_3 t^3$ ($*$), you want to find the parameters $a_i$.

At $t=0$ you know that $x = 1.5$, so from ($*$) $a_0 = 1.5$.

At $t=10$ you know that $x = 1$, so from ($*$) $a_0+10a_1 + 100a_2 + 1000a_3 = 1$.

Let's differentiate the polynomial: $\dot{x} = a_1 + 2a_2 t + 3a_3 t^2$ ($**$).

At $t=0$ you know that $\dot{x} = 0$ (rest condition), so from ($**$) $a_1 = 0$.

At $t=10$ you know that $\dot{x} = 0$ (stop condition), so from ($**$) $a_1 + 20a_2 + 300a_3 = 0$.

So you have:

$a_0 = 1.5$

$a_1 = 0$

$a_0+10a_1 + 100a_2 + 1000a_3 = 1 \Rightarrow 1.5 + 100a_2 + 1000a_3 = 1$

$a_1 + 20a_2 + 300a_3 = 0 \Rightarrow 20a_2 + 300a_3 = 0$

From the last equation you get $a_2 = -15a_3$, substituting in the second last you get $a_3 = 0.001$, and substituting $a_3$ back you get $a_2 = -0.015$.

So you have all the $a_i$ coefficients to replace in $(*)$ and you can get the solution $x(t) = 1.5 - 0.015t^2 + 0.001t^3$. You can do the same for the other 2 coordinates changing the initial and final conditions.

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  • $\begingroup$ Can you explain how you got to the final equation for x from using the previous two? $\endgroup$ – Jimenemex Dec 11 '15 at 16:26
  • $\begingroup$ I edited the question adding more details. Basically, you have a system of four equations and four unknowns. $\endgroup$ – charles Dec 14 '15 at 8:56

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