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A product is initially at rest on a conveyor belt:

Product on Conveyor

The initial conditions of the product can be described as follows:$$x_i=0$$ $$v_i=0$$ $$a_i=0$$$$j_i = j⋆ $$.

The product will be moved forward distance of exactly Δx (known) to position it under the conveyor belt:

Product on Conveyor Moved

When the movement is complete, the product will be sprayed for a few moments, so it must come to a complete rest at this point:

Product on Conveyor Sprayed

The conveyor, which must move the product into position, will complete the movement in the minimum time possible.

The conveyor's control system will apply a constant jerk of j⋆ (known), -j⋆ (known) or zero.

However, the conveyor must obey these limits:

  • It has a max velocity of v⋆ (known),
  • a max acceleration of a⋆ (known),
  • a max deceleration of d⋆ (known).

Assume the product does not slide or slip on the conveyor.

How long will it take the product to complete the movement of Δx?

Side Note:

At first glance, it seems as though the problem can be solved by a simple application of the kinematic equation for one-dimensional motion with constant jerk (and its derivatives): $$x = x_0 + v_0 t + \frac{1}{2} a_0 t^2 + \frac{1}{6}jt^3$$ However, the limits v⋆, a⋆ and d⋆ pose a major problem. For some values of Δx, the conveyor may not even reach max velocity, max acceleration, or max deceleration. Therefore, there will be 8 possible scenarios in which different combinations of these limits are reached. To assist with this part of the problem, I have created a decision chart that illustrates the different possible scenarios:

Conveyor Limit Decision Chart

As Δx gets smaller and smaller, the possibilities range from all three limits reached...

All limits reached

... to no limits reached...

No limits reached

... making it hard to describe the time passed during the movement with a single function. I am stumped at this point- how do I predict which limits will be met and which will not?

Luminaire
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    What do you mean by "acceleration" and "deceleration"? Assuming this problem is in more than 1 dimension, acceleration is typically defined as a vector quantity, and deceleration becomes difficult to define such that it is distinct from acceleration. – Kajelad Jul 19 '17 at 21:50
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    @Kajelad Yes defining acceleration differently than deceleration in multiple dimensions would be tricky. This problem is just in 1 dimension. To keep things simple, acceleration is defined as the increase in magnitude of velocity over time, and deceleration is defined as the decrease in the magnitude of velocity over time. – Luminaire Jul 19 '17 at 22:01
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    Shall the object also be finally still, i.e. still at its destination? – md2perpe Jul 19 '17 at 22:41
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    Do you have more information about the magnitude of your known parameters? Also, what have you tried so far? – Andrei Jul 21 '17 at 11:34
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    As I understand the question, we want to find a twice-differentiable function $x:[0,T]\to\mathbb R$ satisfying the conditions $$\begin{gather} x(0)=x'(0)=x''(0)=0, \ x(T)=X,\ x'\le v,\ d\le x''\operatorname{sgn}(x')\le a, \ x'''\in{-j,0,j} \text{ wherever $x'''$ is defined,} \end{gather}$$ and such that $T$ is as small as possible. (I've renamed $x^*$ to $x$ and $x$ to $X$.) Is this interpretation correct, @Luminaire? –  Jul 21 '17 at 15:39
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    @Rahul I think you also have $x'(T)=x''(T)=0$. In addition, $d<0$, and I can think you can safely assume that $x'\ge 0$ if $X>0$ – Andrei Jul 21 '17 at 18:21
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    @Andrei: You're right, I missed that part of the question (and I should have written $-d$ for the lower bound). I agree that $x'\ge0$ should be true in the optimal solution, but I don't think we should assume it as part of the question. –  Jul 21 '17 at 19:11
  • @md2perpe- yes, it must be finally still. I updated the question to explain why! @Andrei- I have tried applying the kinematic equation for one-dimensional motion with constant jerk. I am able to solve the problem by making certain assumptions- I.e. the conveyor is able to meet max velocity, max acceleration and max deceleration. However, determining which assumptions are safe to make has stumped me. @Rahul yes that's exactly right- and what has stumped me. We must determine what x''' will look like in order for the movement to take the minimum amount of time. – Luminaire Jul 24 '17 at 18:57
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    Nice edit! I hope the question will get enough reopen votes now. –  Jul 24 '17 at 19:09
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    Hypothesis: The jerk as a function of time takes the values $j^,0,-j^,0,-j^,0,j^$, in that order, for consecutive time intervals between times $0\le t_1\le t_2\le\dots\le t_7$. That is, the jerk is $j^*$ between time $0$ and $t_1$, $0$ between $t_1$ and $t_2$, and so on. Some of those intervals may be degenerate (for example, you could have $t_1=t_2$). I believe the ending conditions and the kinematic constraints can be expressed as linear equations and inequalities on the variables $t_1,\dots,t_7$, after which minimizing $t_7$ becomes a linear programming problem. –  Jul 24 '17 at 20:49
  • @Rahul I went and looked up "linear programming problem," and it sounds exactly like what I am facing! I am very interested in learning how to solve this as a linear programming problem. I tried expressing the kinematic constraints as linear inequalities, but I ran into a problem- the velocity constraint is not linear! $$\v_{\star}> \frac{1}{2}j(t_1-t_0)^2 + a_\star(t_2-t_1)+ a_\star(t_3-t_2) + \frac{1}{2}j(t_3-t_2)^2$$. Did I express the kinematic constraint wrong, or is there a way to make this expression linear? – Luminaire Jul 24 '17 at 23:32
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    No, you're right, the velocity constraint isn't linear. My bad... –  Jul 25 '17 at 00:21

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