Science > Physics > Path of a particle moving through 2 points with minimum Jerk
| Topic: |
Science > Physics |
| User: |
"X" |
| Date: |
16 Apr 2005 06:07:41 AM |
| Object: |
Path of a particle moving through 2 points with minimum Jerk |
Let's say at a particle (mass m) passes through the point A=(x1,y1)
with velocity v1.
If the particle is to pass through the point B=(x2,y2), What is the
path that it has to travel in order to experience minimum jerk (da/dt)
at each point of it's trajectory between A and B?
either an explantion or the name of the topic does this comes under
would be nice.
(There are no external forces gravity etc are present)
Thank you
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| User: "zigoteau" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
16 Apr 2005 05:25:10 PM |
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(X) wrote in message news:<377e671d.0504160307.414fcf40@posting.google.com>...
Hi, X,
Let's say at a particle (mass m) passes through the point A=(x1,y1)
with velocity v1.
If the particle is to pass through the point B=(x2,y2), What is the
path that it has to travel in order to experience minimum jerk (da/dt)
at each point of its trajectory between A and B?
What's a? Acceleration? If so, do you mean minimum |da/dt|?
either an explantion or the name of the topic does this comes under
would be nice.
(There are no external forces gravity etc are present)
The general topic is Calculus.
If I have correctly understood your problem, the aim is to apply at
each instant the force required to minimize jerk. The force can be
applied in any way.
My gut feeling is that the solution to the problem would have constant
jerk j.
It only makes sense if you specify that the acceleration starts off
being zero, otherwise you could choose the value which would satisfy
the boundary conditions with zero jerk.
In 1D you have
a = jt
v = v1 + j*t^2/2
x2 = v1*t + j*t^3/6 (there is no loss of generality in putting x1=0)
j = 6*x2/t^3 - 6*v1/t^2
x2 and v1 are given, t is an arbitrary constant, and the aim is to
minimize j.
The extremum of x2*u^3 - v1*u^2 occurs when u=2/3*v1/x2, ie a value
of -4/27*v1^3/x2^2
You'll have to prove/disprove my gut feeling, and then extend to
higher dimensions.
Cheers,
Zigoteau.
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| User: "Sbharris[atsign]ix.netcom.com" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
16 Apr 2005 03:36:57 PM |
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If there are no forces acting, the object will necessarily continue to
travel at velocity v1, right? Along a straight line, right?
(Galleleo's law and Newton's first law). If point x2,y2 happens to be
on its path, then it will drift though with no acceleration at all at
any time, so a = 0 and da/dt is obviously 0.
You didn't frame this problem very well.
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| User: "XGR131" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 03:21:51 AM |
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You changed my question, v1 is not neccessirily in direction of AB.
but your answer to your own question is correct though.
Thanks
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| User: "Sbharris[atsign]ix.netcom.com" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
16 Apr 2005 03:37:08 PM |
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If there are no forces acting, the object will necessarily continue to
travel at velocity v1, right? Along a straight line, right?
(Galleleo's law and Newton's first law). If point x2,y2 happens to be
on its path, then it will drift though with no acceleration at all at
any time, so a = 0 and da/dt is obviously 0.
You didn't frame this problem very well.
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| User: "Sbharris[atsign]ix.netcom.com" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
16 Apr 2005 03:37:16 PM |
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If there are no forces acting, the object will necessarily continue to
travel at velocity v1, right? Along a straight line, right?
(Galleleo's law and Newton's first law). If point x2,y2 happens to be
on its path, then it will drift though with no acceleration at all at
any time, so a = 0 and da/dt is obviously 0.
You didn't frame this problem very well.
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| User: "Gregory L. Hansen" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 08:54:40 AM |
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In article <377e671d.0504160307.414fcf40@posting.google.com>,
X <XGR131@gmail.com> wrote:
Let's say at a particle (mass m) passes through the point A=(x1,y1)
with velocity v1.
If the particle is to pass through the point B=(x2,y2), What is the
path that it has to travel in order to experience minimum jerk (da/dt)
at each point of it's trajectory between A and B?
either an explantion or the name of the topic does this comes under
would be nice.
(There are no external forces gravity etc are present)
Thank you
Oh, I think I know how to solve it, assuming you can pass through point B
in any direction and don't need to stop, and you can change your speed.
The jerk is constant. If it weren't constant, then a smaller jerk at one
time must be made up by a larger jerk at another time. And integrating
polynomials is easy, so it almost selves itself.
I would rotate and translate the coordinate system so the particle passes
through point A=(0,y1) at t=0 with velocity v=(v_x0, 0), and finishes at
point B=(x2,0). The jerk vector is then
j = (-j_x, -j_y)
And just integrate, plugging in initial conditions.
x(t) = v_x0 t - 1/6 j_x t^3
y(t) = y1 - 1/6 j_y t^3
x(t) is hard to invert because it's cubic, but inverting y(t) is easy, and
you can eliminate t from x to get x(y). And you can invert the rotation
and translation to get back to the original conditions. The math will
continue for pages.
It just occured to me that we could also have j=(j_x*f(t),j_y*g(t)), with
f^2+g^2=1, constant magnitude but a direction that changes arbitrarily in
time.
It's hard to solve these sorts of things without just doing one special
case or another.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
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| User: "Gregory L. Hansen" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 08:08:07 AM |
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In article <377e671d.0504160307.414fcf40@posting.google.com>,
X <XGR131@gmail.com> wrote:
Let's say at a particle (mass m) passes through the point A=(x1,y1)
with velocity v1.
If the particle is to pass through the point B=(x2,y2), What is the
path that it has to travel in order to experience minimum jerk (da/dt)
at each point of it's trajectory between A and B?
either an explantion or the name of the topic does this comes under
would be nice.
(There are no external forces gravity etc are present)
Thank you
That's an interesting problem. To minimize the acceleration you would
simply follow a circular arc with the largest possible radius. But that
would introduce an infinite jerk when you suddenly go from uniform motion
to accelerated motion around the circle. Following a quarter of a cosine
would be better.
The problem definition could be tightened up. Will the particle move at
constant speed, or can its energy change? You might find a lower jerk if
it can slow down. Should the particle be moving in any particular
direction when it goes through point B, or does it simply have to go
through point B?
This seems like a calculus of variations problem. Not quite the sort
of "Find the curve that minimizes the time for a particle to slide from
point A to point B" problem that physics students do once or twice in
their academic careers, but maybe similar after a suitable change of
variables is made. I don't know how to solve it off-hand.
--
"No one need be surprised that the subject of contagion was not clear to
our ancestors."-- Heironymus Fracastorius, 1546
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| User: "andy everett" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
16 Apr 2005 07:59:52 PM |
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It goes straight and slows very very slowly and travels a long distance,
comes to a stop and reverses its direction of travel very very slowly
and heads back to point B.
Let da/dt go tend to zero --> dv/dt will be small --> long distance
traveled.
I think you are missing some facts to make the problem interesting?
X wrote:
Let's say at a particle (mass m) passes through the point A=(x1,y1)
with velocity v1.
If the particle is to pass through the point B=(x2,y2), What is the
path that it has to travel in order to experience minimum jerk (da/dt)
at each point of it's trajectory between A and B?
either an explantion or the name of the topic does this comes under
would be nice.
(There are no external forces gravity etc are present)
Thank you
.
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| User: "XGR131" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 03:17:27 AM |
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True,
Also i forgot to mention that the particle should remain within a
rectangle where its
diagonal is AB before reaching B.
Thanks
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| User: "zigoteau" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 04:00:58 AM |
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andy everett <vze2qxq3@verizon.net> wrote in message news:<cyi8e.9437$ok4.7574@trndny07>...
Hi, Andy,
It goes straight and slows very very slowly and travels a long distance,
comes to a stop and reverses its direction of travel very very slowly
and heads back to point B.
Good point.
Let da/dt go tend to zero --> dv/dt will be small --> long distance
traveled.
I think you are missing some facts to make the problem interesting?
The real-world problem would be: you are scooting merrily along in a
spaceship at velocity v, when suddenly you see the planet Krypton out
of the porthole located at the point (x,y,z), and you decide you want
to stop there. How fast can you do it without killing the passengers
or spilling their coffee? There would be constraints on |a| as well as
on |da/dt|, and you might also want to match velocities with the
planet Krypton. Rather than doing it in minimum time, you might want
to minimize the fuel consumption.
It's actually a problem in linear(-ish) programming in an
infinite-dimensional vector space.
It brings to mind the problem of how to get a lift as fast as possible
from an initial floor to a destination. As I remember the problem,
|da/dt| was not a consideration, but |a| was. The solution is to
accelerate as hard as permissible up to the halfway point, then slam
on the brakes as hard as permissible for the other half of the
journey.
Cheers,
Zigoteau.
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| User: "XGR131" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 07:06:46 AM |
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I would need it to pass through another point, I am trying to find out
what is the smoothest path that will pass thrugh n points while keeping
the jerk to minimum at each point of journey. the only condition
required is that the length of path between each two points to be less
than or equal to |x1-x2|+|y1-y2|
Thanks
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| User: "XGR131" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 07:12:24 AM |
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I meant to say between every each two points that the particle has to
pass through
the length of path travelled is less than or equal to |x1-x2|+|y1-y2|;
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| User: "John C. Polasek" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 05:00:23 PM |
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On 16 Apr 2005 04:07:41 -0700, (X) wrote:
Let's say at a particle (mass m) passes through the point A=(x1,y1)
with velocity v1.
If the particle is to pass through the point B=(x2,y2), What is the
path that it has to travel in order to experience minimum jerk (da/dt)
at each point of it's trajectory between A and B?
either an explantion or the name of the topic does this comes under
would be nice.
(There are no external forces gravity etc are present)
Thank you
I guess you really need an Archimedes spiral instead of the circle.
Mr. Dual Space
If you have something to say, write an equation.
If you have nothing to say, write an essay
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| User: "John C. Polasek" |
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| Title: Re: Path of a particle moving through 2 points with minimum Jerk |
17 Apr 2005 04:12:35 PM |
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On 16 Apr 2005 04:07:41 -0700, (X) wrote:
Let's say at a particle (mass m) passes through the point A=(x1,y1)
with velocity v1.
If the particle is to pass through the point B=(x2,y2), What is the
path that it has to travel in order to experience minimum jerk (da/dt)
at each point of it's trajectory between A and B?
either an explantion or the name of the topic does this comes under
would be nice.
(There are no external forces gravity etc are present)
Thank you
This is a simple problem and it can be solved for zero jerk. I'll give
you the graphics diagram and you solve it.
Put points A nd B on paper, and from point A draw the vector v1.
Now from point A draw the normal to v1 toward B.
Somewhere along this normal put the center of a circle whose radius R
reaches point A. (You can find this by trial arcs struck from A and B
with a little greater radius. The center is inside the lunar zone).
Swing the arc using the "found" point over to point B.
The acceleration with zero jerk will be V1^2/R.
Mr. Dual Space
If you have something to say, write an equation.
If you have nothing to say, write an essay
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