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From: "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid>
Newsgroups: comp.soft-sys.matlab
Subject: Re: trapezoidal triangular motion calcs
Date: Fri, 1 Feb 2008 04:43:01 +0000 (UTC)
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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