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Suppose that we have a body that will move over a curve (for example a parabolic curve). The equation of that curve is : $$ y+k=(x+h)^2 $$ Where (h,k) are the (x,y) of the vertex. Also suppose that the body moves with speed (v) m/s and the position of the start point of the path is known for example:$$(y_s, x_s)$$

Please note that the position is calculate when a laser machine beam which rotates at speed 36 degree/sec, hits the body. How can one calculate the (x,y) of the body positions(E,B,C,E,F,G) along the path each laser beam hit.

I have performed the required but in a particular case (circular path at any center). But in our case it is different and I want the concept of the solution to be applicable to any function. Thanks in advance.

Here is a figure for more clarification: enter image description here

AAEM
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  • What exactly can the laser measure? Only that it currently hits the body or also the distance or other measures? Or is the laser not for measuring but you only want to use it as time reference? – koalo Sep 17 '18 at 10:41
  • @koalo the laser measures no thing, i want to use it as a time reference, i have edited the question to illustrate that point – AAEM Sep 17 '18 at 10:49
  • A parabola has no center. Did you mean to say that $(h,k)$ is the vertex? – amd Sep 18 '18 at 01:01
  • @amd so what is point (h,k) called? – AAEM Sep 18 '18 at 07:01
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    It’s the vertex, as I wrote already. – amd Sep 18 '18 at 07:37
  • @amd yes i mean that – AAEM Sep 18 '18 at 07:38
  • @amd but the answer below is not clear for me, please if you have a simpler one please post it – AAEM Sep 18 '18 at 07:38
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    It looks like you basically want a multiple of the arc length parameterization, but I don’t understand the function of the rotating laser. The figure you’ve added doesn’t really clarify anything. – amd Sep 18 '18 at 07:45
  • @amd the rotating laser is used as a reference for the required point position, i.e. i want the position of the body when it is hit by the laser – AAEM Sep 18 '18 at 07:46
  • Even for something as simple-looking as a parabola, there’s no explicit formula for the arc length parameterization, which requires inverting the arc length function for the curve. For an ellipse, there’s not even a nice closed form for the arc length itself. On top of that, you’re trying to solve for all of the times when the line defined by the rotating beam intersects the object’s position. I suggest you explore numerical solutions to this. – amd Sep 18 '18 at 08:10
  • @amd "you’re trying to solve for all of the times ", this is not true , i am trying to find only the position of intersection between the laser beam line and the path of the object – AAEM Sep 18 '18 at 08:21
  • And that occurs at a particular set of times. – amd Sep 18 '18 at 19:53

2 Answers2

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You can parameterize any equation to make it a function of time. Some are easier, so of course, others require more effort. In the case of a parabola expressed as $y+k=(x+h)^2$, let $x(t)=t$ and $y(t)=t^2+2ht+h^2-k$. Using the parametric equation, we can know where the body is at any time t. As written, $t=0$ is at the vertex. If we wanted to start elsewhere, we could let $x=t-p$ where p is the period of the cycle (i.e. p=36 sec). $$\left( \begin{array}{c} x(t)=t-p\\ y(t) = t^2+2ht-2pt+h^2-2hp+p^2-k \end{array}\right).$$ Suppose for example that $p=36$ and we let $x=t-5p$, then $y(t)=(t-5\cdot 36)^2 + 4(t-5\cdot 36)+1$ and $t$ can range from 0 to 360, but you are only interested in $t$ when it is $0, 36, 72 ...$ So we build a graph that shows the points calculated at those times.enter image description here

Narlin
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  • please give an example to make it more clear, for example: – AAEM Sep 17 '18 at 12:08
  • OK. I edited to add an example – Narlin Sep 17 '18 at 12:27
  • In the example $h=2$, $k=3$ – Narlin Sep 17 '18 at 12:37
  • Why in the answer you said x = t−p, and what do you mean by x=t−5p? and why you said "let", is this means that we can "let" it to be any value? so why you selected x=t−5p or just as an example? and why not just let x(t) = t – AAEM Sep 17 '18 at 12:41
  • This is just an example. By letting $x=t-5p$, we have set the function so that $a(0)$ is five periods before the vertex. You could just as easily use $x=t$ and then deal with negative time. – Narlin Sep 17 '18 at 12:46
  • but if we used the suggestion that $$x=t$$, the final equation will be $$y=t^2+4t+1$$, and let t =0, we expect that y also equals to 0 but here it is equal to 1?? – AAEM Sep 18 '18 at 08:31
  • also please correct this statement if it is wrong"by doing the above calculations, you the x axis will be considered as a t axis on which the body will appear every p seconds" – AAEM Sep 18 '18 at 08:38
  • why you do not answer? – AAEM Sep 18 '18 at 16:56
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The laser beam is effectively a family of rays parameterized by time, which can be represented by the rotating unit vector $$\mathbf L(t) = (\cos\omega t, \sin\omega t),$$ where $\omega$ is the angular velocity of the ray. For simplicity, I’m assuming that the laser starts off pointing in the positive $x$-direction at $t=0$. You can always add a constant phase shift $\delta$ or offset the time if this is not the case. If you can come up with a time parameterization $\mathbf\gamma(t)$ of the object path, then the problem becomes one of finding the solutions to $\mathbf\gamma(t)=c\mathbf L(t)$, $c\ge0$.

In theory, finding this time parameterization is straightforward. Every regular parameterized curve can be reparameterized by arc length: if $\mathbf\gamma$ is regular, then the arc length function $$s(\lambda) = \int_a^\lambda \|\mathbf\gamma'(u)\|\,du$$ is an increasing differentiable function, so has a differentiable inverse $\lambda = \lambda(s)$. Setting $s=vt$ in this arc length parameterization of the curve then has the property that the object moves with constant speed $v$, as required. (You can verify this by an application of the chain rule.)

In practice, however, the arc length function for most curves doesn’t have a nice inverse, and in fact the arc length itself might not be expressible in terms of elementary functions. For instance, for your parabola, the arc length function derived in the answers to this question doesn’t have a closed-form inverse. Indeed, as Will Jagy notes in a comment to that question, a parabola is one of the few curves for which the arc length integral has a closed form, however ugly. Since we have $s=vt$, you could try solving $$c\left(\cos\left(\frac\omega vs(\lambda)\right),\sin\left(\frac\omega vs(\lambda)\right)\right) = \mathbf\gamma(\lambda)$$ for some convenient parameterization of the parabola, but that doesn’t look very promising, either. In general, you’re going to have to resort to numeric methods to solve this problem.

amd
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  • so how one can use a numeric method to solve this problem, or i should start a new thread? – AAEM Sep 19 '18 at 16:24