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From: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson)
Newsgroups: comp.soft-sys.matlab
Subject: Re: trapezoidal triangular motion calcs
Date: Fri, 1 Feb 2008 02:59:26 +0000 (UTC)
Organization: National Research Council Canada - Conseil national de rechereches Canada
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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.
-- 
   "I was very young in those days, but I was also rather dim."
   -- Christopher Priest