2 Attachment(s)

How do you produce a smooth curve transaction between two velocities

Attachment 24510

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.

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

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?

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

Quote:

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.

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

Quote:

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 $\displaystyle \int_0^{1.5}v(t)\,dt=200$. 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.

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

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 $\displaystyle ax^5+ bx^4+ cx^3+ dx^2+ ex+ f$ which has 6 coefficients to determine.

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

Quote:

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 $\displaystyle ax^5+ bx^4+ cx^3+ dx^2+ ex+ f$ 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. -

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

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...

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

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.

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

I admire the valuable information you offer in your message. This message are obviously from your personal experience. It’s a authentic and fantastic for anyone. Daily Jobs Ads and Tenders Ads I will remain awhile for your message. Really I am impressed with your message content. Keep up the great work. mcitp examsI will bookmark your blog and have my friends check up here often. I am quite sure they will learn lots of new stuff here than anybody else . I respect you from the core of the heart. Thanks for sharing this information