Sign in to follow this  
Followers 0
frumpydolphin

Movement of a point over time question

12 posts in this topic

Background

- System of points a, b, c and q

- a, b, and c exert a push on q moving it to the end of a vector that is 2 times the vector from a, b, or c to q.

- a, b, and c do this at the same time, after each cycle the x and y values of q are the average of the end points of the vectors.

- This happens n times

 

First derivation

The final x value is the sum of the averages of all the vectors endpoints divided by the number of iterations so n.

Same goes for y

 

 

I further thought why keep it at intervals when it can be a constant force.

From this I derived

The integral from 1 to n of the average of the averages of the vector endpoints in respect to the x value of q.

Same for y

Hope I wasn't too vague with it but am I at least close to right?

Also how do I insert equations?

This wasn't homework by the way don't rage at me.

 

0

Share this post


Link to post
Share on other sites

Ya, so there is a point(q) starting with coordinates x​q and yq​. Three other points a, b, and c have there x and y coordinates labelled in the same manner​. A, b, and c exert a "push" on q each second suppose, and move it to the endpoint of a vector 2 times the magnitude of vector a,b, or c to q. The direction remains the same and the origin of the vector is still a,b or c. The average of the endpoints created by these vectors, 1 for a, 1 for b, and1 for c, gives q's position after 1 second. After n times though q's position would be the average of all these averages.

I derived the sum to n from 1 of the small averages divided by the number of seconds, so n, to get the final x and y values of q. (really need to figure out how to use equations)

Hope that helps but that is only the first part.

0

Share this post


Link to post
Share on other sites

 

 

A, b, and c exert a "push" on q each second suppose, and move it to the endpoint of a vector 2 times the magnitude of vector a,b, or c to q.

All at the same time or 3 separate cases?

0

Share this post


Link to post
Share on other sites

 

 

The direction remains the same and the origin of the vector is still a,b or c.
The direction of what? Origin of what? Origin a,b, or c - very ambiguous.
0

Share this post


Link to post
Share on other sites

Geez I suck at this, a, b, and c act as the origin of the vectors, the direction is the direction of the vector from a, b, or c towards q

0

Share this post


Link to post
Share on other sites

Your overall description is confusing. a,b,c are different vectors with differing origins and directions. Your description talks about one origin and one direction. I can't put it together.

0

Share this post


Link to post
Share on other sites

Try to describe the scenario using equations. It should clarify for yourself what you intend.

0

Share this post


Link to post
Share on other sites

I understand it as follows:

You start with a vector [math]\vec{q}(0)[/math], and (simultaneously) apply forces [math]\frac{2}{3} (\vec{a}-\vec{q}(0))[/math], [math]\frac{2}{3} (\vec{b}-\vec{q}(0))[/math] and [math]\frac{2}{3} (\vec{c}-\vec{q}(0))[/math]. You repeat [math]n[/math] times. That means, in the n-th step you apply forces [math]\frac{2}{3} (\vec{a}-\vec{q}(n-1))[/math], [math]\frac{2}{3} (\vec{b}-\vec{q}(n-1))[/math] and [math]\frac{2}{3} (\vec{c}-\vec{q}(n-1))[/math] on [math]\vec{q}(n-1)[/math] to get [math]\vec{q}(n)[/math].

You can give an explicit formula for the result: [math]\vec{q}(n)=\vec{q}(0) + \frac{2n}{3} (\vec{a}+\vec{b}+\vec{c})[/math]. All the intermediate terms cancel out nicely.

Edited by renerpho
0

Share this post


Link to post
Share on other sites

I'm not entirely sure about the correct sign (the question is not clear about that). So it is [math]\vec{q}(n)=\vec{q}(0) \pm \frac{2n}{3} (\vec{a}+\vec{b}+\vec{c})[/math], depending on the direction in which the force is acting.

I will leave vector arrows aside from now on; the following refers to vectors in [math]R^{2}[/math].

 

The continuous system can be modelled analytically, too. Note that [math]q=q(t)[/math].

You have [math]F(t,q)=q \pm \frac{2}{3} (a+b+c)[/math]. From Newton's second law, we have [math]F(t,q)=m \cdot \ddot{q}[/math], where [math]m[/math] is the mass of the test particle in [math]q[/math]. So the differential equation for the system becomes [math]m \cdot \ddot{q} - q = \pm \frac{2}{3} (a+b+c)[/math].

The two equations to solve are

[latex]\begin{pmatrix} m \ddot{x}-x \\ m \ddot{y}-y \end{pmatrix}=\pm \frac{2}{3} \begin{pmatrix} a{_x}+b{_x}+c{_x} \\ \ a{_y}+b{_y}+c{_y} \end{pmatrix}[/latex].

The solution of this system is

[latex]\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} v{_1} \cdot e^{\frac {t}{\sqrt{m}}}+v{_2} \cdot e^{-\frac {t}{\sqrt{m}}}+v{_3} \\ w{_1} \cdot e^{\frac {t}{\sqrt{m}}}+w{_2} \cdot e^{-\frac {t}{\sqrt{m}}}+w{_3} \end{pmatrix}[/latex], with constants [math]v{_i},w{_i}[/math] depending on the initial choice of [math]a,b,c,q(0)[/math].

 

The question does not mention the mass of the test particle. But as it is based on forces, the mass is an important factor for the continuous system.

Edited by renerpho
1

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now
Sign in to follow this  
Followers 0