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There is a body moves along an eight figure path and there is a laser beam at the origin that rotating with rotation rate $w$.
How to get the points that originate each rotation from the intersection between the moving body path (eight figure) and the rotating beam.

Note that the laser beam is used for nothing but only as time reference.

Given:

  1. the equation that describes that path:
    $(x-h)^4 = a^2((x-h)^2-(y-k)^2)$
    where $(h,k)$ = (50,70)
  2. The start point of the curve $(X_S,Y_S)$
    where $(X_S,Y_S)$ = (30,70).

  3. The speed of the moving body ($V$=150 m/s).

  4. The speed of rotation of the laser beam $w= 36$ deg/sec.

  5. The value of a = 10.

Required:

The points on the curve that came from the intersection between the beam line and the motion path (yellow points on the attached figure).

Please note that:
I do not know any information about how many points came from the intersection between the body path and the beam, for sure it depends on the body speed and the rotation rate of the beam.

Here is a figure for more illustration:

enter image description here

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    $\begingroup$ @AdrianKeister when the moving beam start to rotate, it will hit the body which moves in an eight figure path, i want to get the points of hits(between the line and the body) or in other words (the points of intersection between the beam and the path of the body(eight figure)). those points will appear every rotation because the beam hit the body once per complete rotation $\endgroup$
    – AAEM
    Sep 18, 2018 at 17:16
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    $\begingroup$ Does the beam hit the body at point $(X_S,Y_S)$ initially (i.e., at time $t=0$)? Are we given the value of $a$? $\endgroup$ Sep 18, 2018 at 17:23
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    $\begingroup$ Let $\big(r(t),\theta(t)\big)$ denote the polar coordinates of the particle at time $t$. Thus, the laser hits the particle at time $t$ if and only if $$\theta(t)\equiv wt+\phi_0\pmod{2\pi}\,,$$ where $\phi_0$ is the initial phase of the laser (namely, the laser's direction at time $t$ is given by the equation $\phi(t)=w t+\phi_0$). From your description, $\phi_0=\dfrac{\pi}{2}$. Thus, you will need to solve for $\theta(t)$, which does not look very fun. $\endgroup$ Sep 18, 2018 at 17:34
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    $\begingroup$ And I would expect that there are finitely many intercepts if and only if the period $T_l:=\dfrac{2\pi}{w}$ of the laser beam's rotation, and the period $T_p:=\dfrac{L}{V}$ of the particle in the figure-eight path satisfy $\dfrac{T_p}{T_l}\in\mathbb{Q}$. Here, $L$ is the arc length of the figure-eight curve (which looks very ugly, i.e., $L\approx 6.09722\,a$, so I expect infinitely many intercepts). However, I do not have the will power to try to solve for $T_p$ exactly, or to find out what the intercepts are. $\endgroup$ Sep 18, 2018 at 17:41
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    $\begingroup$ How is this substantially different from your previous (and more general) question? $\endgroup$
    – amd
    Sep 18, 2018 at 23:58

1 Answer 1

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I haven't solved the question, but I have an approach you can try.

You have the starting point and the speed and the curve. So you should be able to get the position of the body as a function of time.

Solve speed = 150 m/s = $ \sqrt{ \frac{dx}{dt}^2 + \frac{dy}{dt}^2 }$ and $ 4(x-h)^3 \frac{dx}{dt} = a^2[ 2(x-h)\frac{dx}{dt} - 2(y-k)\frac{dy}{dt}] $

to get $\frac{dx}{dt}$ and $\frac{dy}{dt}$. You also have the initial position $(x_0,y_0)$.

So now you have the position of the body as a function of time as $x = x_0 + \int x(t)$ and $y = y_0 + \int y(t)$.

For the next part, try converting to polar coordinates by using $x = r cos \theta$ and $y = r sin \theta $.
This way, you will have the angle the position vector of the object makes with the x-axis at the origin as a function of time.
Equate that to the laser's angle and you get your theta values. This can further give you the times and therefore the position of the body.

Hope this helps.

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  • $\begingroup$ it would be better to provide complete answer $\endgroup$
    – AAEM
    Sep 30, 2018 at 10:12
  • $\begingroup$ The complete answer would require a lot of calculation, which would require time. Also, you haven't given the value of 'a' in the equation of the path. Check out this link to work with this kind of paths. $\endgroup$ Oct 1, 2018 at 1:35
  • $\begingroup$ I will try to solve but i need to know the following : (1)what do you mean by "to get dx/dt and dx/dt" i think it is a typo? (2) after simplifying the equation to be function only in {x,y and (dy/dt)}, how can i calculate dy/dt? (3) what is x(t) and y(t) represent for? (4) how can your approach substitute for the antenna rotation (Omega)? $\endgroup$
    – AAEM
    Oct 1, 2018 at 10:26
  • $\begingroup$ 1) Yes, it is. Sorry about that. $\endgroup$ Oct 1, 2018 at 17:02
  • $\begingroup$ 2) The resulting differential equation is time consuming to solve. I'm not exactly sure how to solve the one that you get. 3) They are the x and y co-ordinates of the body i.e. the position of the body as a function of time from the starting time when the initial position is given.4) Once you have the position as a function of time, you can simply use $\theta = tan^{-1} \frac{y(t)}{x(t)} $ to get the angle subtended at the origin as a function of time. You already know the laser beam's angle as a function of time $ \theta = 36t$. So equating both will give you the intersection points. $\endgroup$ Oct 1, 2018 at 17:13

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