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As stated in the title, I want to find an expression or a way to calculate the minimum time to go from one point of a path to another when the path is given and acceleration is restricted.

Thus far, I have tried to consider a simplified situation where the path can be represented as a differentiable function $f: [0, l]\to\mathbb{R}$. Here $(0,f(0))$ and $(l, f(l))$ represent the ending points of the path. Also, for now I consider the situation where the initial speed is $v_0=0$.

Let the maximum allowed acceleration be $a_{\max}$.

I have tried to construct a suitable expression to calculate the total time and then minimise it (most likely with Euler-Lagrange) but so far I haven't been able to construct an expression I can manipulate.

We could consider the time to go through an infinitesimal arc length and integrate that: \begin{align*} dt&=\frac{ds}{v(x)}\\ dt&=\frac{\sqrt{1+f'(x)^2}}{v(x)}dx\\ T&=\int_0^l\frac{\sqrt{1+f'(x)^2}}{v(x)}dx. \end{align*} However, minimising this proves to be difficult, or at least I don't seem to get a grasp on it. First of all we'd have to somehow tie the acceleration here: intuitively it should always be at its maximum value so $a(x)=a_{\max}$. Secondly, you'd have to somehow include the limiting factor of $v(x)$ as its maximum value at a specific point is $v_{\max}=a_{\max}\left|\frac{\left(1+f'(x)^2\right)^{3/2}}{f''(x)}\right|$ (i.e. considering that the instantaneous axis of rotation comes from the radius of curvature).

Anyways, any idea on how to proceed, or is there a better way to approach this? Of course, when this simplified case is solved, the next step would be to change the initial speed. Then to give a restriction at the end i.e. $v(l)=v_l$. Furthermore, to make this result useful in some way we'd have to generalise this to 2D parametric curves and it would also be interesting to consider non-differentiable curves as well, but this goes way beyond the initial question.

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  • $\begingroup$ Have you considered adding $a(x)=$constant as a constraint? Your unknown is $v(x)$, so you could define a new integral $T'=T+\lambda a$ and then extremize this problem instead. $\endgroup$
    – paul230_x
    Commented Dec 27, 2021 at 19:18
  • $\begingroup$ @KP99 Hmm, sounds reasonable but the problem here is that $|\vec{a}|=\text{constant}$ but $\vec{a}$ not necessarily. Of course, I could look at the tangential component instead but I am not sure how to tie things together in this case. Because now $v(x)$ is dependent on $a(x)$ in some way (and for some reason I have trouble relating those [units not adding up]). $\endgroup$
    – 110112345
    Commented Dec 27, 2021 at 20:08
  • $\begingroup$ Well you could express $a(x)=v'(x)\frac{v(x)}{\sqrt{1+f'(x)^2}}$. But I am not that familiar with constrains in these contexts. As in I do not know how to setup the correct equation unless the constraint happens to be an integral with respect to $x$ with the same bounds as the original and that integral has a constant value. $\endgroup$
    – 110112345
    Commented Dec 27, 2021 at 20:28

1 Answer 1

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With a maximum magnetude for the acceleration vector, you are limited to a maximum velocity for any curve in the path (a = $v^2$/R). Anticipating this you can accelerate for a while, but then must decelerate before reaching the curve. At each point, one component of acceleration must keep you on the path while the other can adjust the speed.

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  • $\begingroup$ But this doesn't still give us a way to find an expression to find the minimum time. Because you cannot achieve "maximum" speed everywhere as you'd still have to adjust for "future" changes. A good example of this would be a straight line followed by a bend. The "maximal" speed on the straight line would be $\infty$, but we still have to decelerate for the bend. So, how could we find a general way to calculate the optimal acceleration then? $\endgroup$
    – 110112345
    Commented Dec 31, 2021 at 14:38
  • $\begingroup$ The maximum speed on a straight line would be determined on the speeds at each end, and the maximum allowed + and – acceleration. That can be determined with linear kinematics. Finding the time to traverse an irregular curve would get complicated. $\endgroup$
    – R.W. Bird
    Commented Jan 1, 2022 at 15:22

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