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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 02:40:19 +0000 (UTC)
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"Mark " <medwar19.nospam@hotmail.com> wrote in message <fntmc9$2ir
$1@fred.mathworks.com>...
> Hi Roger,
> 
> Your assumption was right about the problem. It is simple 
> motion between 2 points. However no mech really achieves 
> infinite acceleration and this is the main question. 
> 
> The mech will accelerate with limited acceleration, then a 
> triangular velocity or trapizodal (if maximum velocity can 
> be reached in time) before limited deceleration. The mech 
> will be stationary at the start and end.
> 
> I'm looking to minimise the maximum velocity of the mech 
> whilst making the distance in time.

  Well, then you should accelerate at the maximum amount allowed until 
reaching a certain maximum velocity, coast at that maximum velocity for a 
period of time, then decelerate at the maximum amount permissible, all in 
such a way as to come back to zero velocity after total time t has elapsed.  
The maximum velocity should be chosen so that the integral under the 
velocity versus time curve is equal to d, the given distance.  In other words, 
the total area under the line v = d/t should equal the total area above it.  
Since the "curve" is actually of trapezoidal shape, choosing this maximum 
velocity is a matter of solving a simple linear equation obtained from the 
triangles and rectangles involved in these areas.  This will minimize your 
maximum velocity attained.

> I read that acceleration should be 1/3 the maximum speed?

  It is a bit arbitrary comparing acceleration with speed.  They are measured 
in different kinds of units and their relative numerical values therefore 
depend on the particular system of units used.

Roger Stafford