Path: news.mathworks.com!not-for-mail
From: "Mark " <medwar19.nospam@hotmail.com>
Newsgroups: comp.soft-sys.matlab
Subject: Re: trapezoidal triangular motion calcs
Date: Sun, 3 Feb 2008 01:34:01 +0000 (UTC)
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"Roger 
Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote 
in message <fnu80l$5vl$1@fred.mathworks.com>...
> roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) wrote in 
message <fnu1ue
> $6ul$1@canopus.cc.umanitoba.ca>...
> > In article <fnu13k$5ur$1@canopus.cc.umanitoba.ca>,
> > Walter Roberson <roberson@ibd.nrc-cnrc.gc.ca> wrote:
> > 
> > >d[total] = d[acceleration] + d[deceleration] + 
d[flat] =
> > >  1/2 * a * t1^2 + 1/2 * a & t1^2 + a * t1 * t2 =
> > >  a * t1^2 + a * t1 * t2 =
> > >  a * t1 * (t1 + t2)
> > >
> > >But for lowest acceleration that gets you there in the 
available time tL,
> > >then t1 + t2 = tL, and d[total] = D, so
> > 
> > Oh, an error! tL is not t1 + t2: tL is 2*t1 + t2 and 
thus t2 is tL-2*t1
> > 
> > >d[total] = a * t1 * tL
> > 
> > That should be,
> > 
> > d[total] = a * t1 * (t1 + tL - 2*t1) = a * t1 * (tL - 
t1)
> > 
> > >D = a * t1 * tL
> > 
> > That should be
> > 
> > D = a * t1 * (tL - t1)
> > 
> > a = D / (t1 * (tL - t1))
> > 
> > The denominator is maximized at t1 = tL/2, so 'a' is 
minimized at
> > t1 = tL/2, and a = D / (tL^2/4) so a = 4 * D / tL^2
> > 
> > The rest of the conclusion is the same as before: 
accelerate half way
> > then decelerate. No lower acceleration will get you 
there on time,
> > and no higher acceleration is needed.
> ---------
>   Mark's aim in both his articles is that of minimizing 
the maximum velocity, 
> not the acceleration.  To minimize the acceleration in 
this case produces a 
> maximum velocity of 2*d/t which is greater than that 
possible using greater 
> acceleration (unless the maximum allowed acceleration is 
only just sufficient 
> to get there in time.)
> 
>   I quote him: "The question is knowing the basic 
equations for motion how 
> do I calculate the minimum velocities needed to achieve 
my timing?" and "I'm 
> looking to minimise the maximum velocity of the mech 
whilst making the 
> distance in time."
> 
> Roger Stafford
> 

Hello,

Thanks for your time and method of dealing with 
acceleration. I'm interested in writing a script that will 
allow me to put any path and calculate a bunch of motion 
parameters including the minimum velocity, accleration and 
jerk that can be used to get round the path in a specific 
time.

This is a common robotics problem that has proven harder to 
solve that I expected.

Once again, I appreciate your time and insight.

Regards,

Mark.