Thread: How do you produce a smooth curve transaction between two velocities

Hi, I have depicted the problem above.
I have a machine that is travelling to point A at a speed of 600mm/sec
I need to reduce the speed over the fixed distance (200mm) and fixed time (1.5 sec) to the velocity at point B (8mm/Sec)
I realise this will be a variable acceleration to solve the movement.
But how do I go about solving this to give me points to plot on the curve.
Thank you.

Is it allowed for the machine to stop and move (a little) in the opposite direction provided all other conditions are satisfied? Is this graph acceptable?

Originally Posted by emakarov

Is it allowed for the machine to stop and move (a little) in the opposite direction provided all other conditions are satisfied? Is this graph acceptable?

Sorry emakarov the move has to consistent in the same direction, no -velocities.
I am also looking to keep the acceleration or in this case (deceleration) to a minimum.

Originally Posted by calltronics

I am also looking to keep the acceleration or in this case (deceleration) to a minimum.

With this requirement, if you want to find an optimal solution, this may be a problem from optimal control, which is a pretty advanced area. If you want to just find some solution, then it is possible to search for the velocity v(t) in the form of a polynomial. We have three restrictions on v: v(0) = 600, v(1.5) = 8 and . Since a quadratic polynomial is determined by three coefficients, there is a single quadratic function v(t) that satisfies these requirements, but unfortunately v(t) becomes negative at some point. It is possible to look at polynomials of higher degrees: then solutions will not be unique, so by tweaking the coefficients it may be possible to make v(t) always positive.

It looks to me like you want the curvature to be continuous through the three curves. That means you want specific values for the function at A and B, specific values for the derivative (so there is no "corner") at those points, and specific values for the second derivative (so the curvature is continuous). That is that you have six conditions. You could fit a fifth degree polynomial, of the form which has 6 coefficients to determine.

Originally Posted by HallsofIvy

It looks to me like you want the curvature to be continuous through the three curves. That means you want specific values for the function at A and B, specific values for the derivative (so there is no "corner") at those points, and specific values for the second derivative (so the curvature is continuous). That is that you have six conditions. You could fit a fifth degree polynomial, of the form which has 6 coefficients to determine.

Yeh, I am on the same track as HallsofIvy, at least I find comfort in not been alone on this solution. -

I don't think the OP said that the second derivative should be continuous. Yes, in the technical sense, "smooth" means infinitely differentiable, but informally it is enough to have continuous first derivative. For example, f(x) = x² for x ≥ 0 and -x² for x < 0 is pretty smooth. If this is so, then distance can be a cubic polynomial. On the other hand, there are additional requirements that velocity is nonnegative and that the maximum of acceleration is minimal...

Sorry emakarov, I assumed that the picture of the curve would act as a depiction of the "Ideal" solution showing only positive velocity. The addition of minimal acceleration is I think unlikely to be calculated and be more of a result to the solution.

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