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Constant speed
04-13-2016, 03:29 PM (This post was last modified: 04-13-2016 03:30 PM by Martin Hepperle.)
Post: #11
RE: Constant speed
(04-11-2016 03:53 PM)Tugdual Wrote:  ...
Thanks Martin, sounds like a good answer but I struggle with further conclusions.
First, at constant speed, I would assume that ds/dt = 0 so dx = 0?
Other question is how do you move from your expressions to a parametric equation for x(t) and y(t) over the time?

Almost ... constant speed would be \(v = {ds \over dt} = const.\) not zero.
where: \(ds\) ... distance and, \(dt\) ... time interval.

Using a given (constant) speed \(v\) and a (user selected) time step \(dt\) you would determine \(ds = v \cdot dt\) and then use the equation above to find \(dx\).
You would start at e.g. time \(t=0\) at position \(x=0\), calculate \(dx\) for the given speed, march to \(x=x+dx\) and plot \(f(x)\) at the new point (for time \(t=t+dt\). Then repeat for the next time step.

My simple equations will work as long as the function is monotonic \(y=f(x)\). For a parametric curve (e.g. a circle) one would have to rewrite with the parameter \(t\) or \(s\) (arc length) to obtain \(dx\) as well as \(dy\) as a result.
This is left as an exercise to the inclined reader ... ;-)

Martin
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Messages In This Thread
Constant speed - Tugdual - 04-04-2016, 10:33 PM
RE: Constant speed - PANAMATIK - 04-05-2016, 07:01 AM
RE: Constant speed - Tugdual - 04-05-2016, 07:19 AM
RE: Constant speed - Dave Britten - 04-05-2016, 11:14 AM
RE: Constant speed - Tugdual - 04-05-2016, 06:04 PM
RE: Constant speed - Dave Britten - 04-06-2016, 11:16 AM
RE: Constant speed - Tugdual - 04-07-2016, 06:13 AM
RE: Constant speed - Dave Britten - 04-07-2016, 11:16 AM
RE: Constant speed - Martin Hepperle - 04-11-2016, 02:01 PM
RE: Constant speed - Tugdual - 04-11-2016, 03:53 PM
RE: Constant speed - Martin Hepperle - 04-13-2016 03:29 PM
RE: Constant speed - Tugdual - 04-13-2016, 06:30 PM
RE: Constant speed - Martin Hepperle - 04-18-2016, 07:27 AM



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