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Help finding the equation of a line to a moving point

  1. Sep 10, 2014 #1
    "In a private project of mine, I have come across a challenging problem.

    I have 3 points: A, B and C.

    A is the motionless point of origin and C is the orbital point of destination. B is the linear transmission between the two; it must intersect with the changing position of C after leaving A, accounting for it's continued movement. The issue is, I'm not sure how to calculate whether or not B will meet C given certain values

    The equation attached to C should be familiar to most of you; it is the equation of a circle. (Cx-h)2+(Cy-k)2 = r2

    At the core of my problem is that C is a point moving along this circle at a set rate (call the rate 't')

    B, however, is the problem point; B also moves, however it moves from A's a/b to C at a set rate ('s').

    This gives us two extra variables:

    t = the rate of motion of C's orbit
    s = the rate of transmission of B

    There are some additional considerations to keep in mind. The line of transmission B moves on (B*) cannot be curved; it must remain linear. I need to figure out how to calculate the equation of this line from A to C given 's'.

    So far I have attempted treating the equation of a circle as one point in an equation for a line, however this gives me a range of lines, and from this point I don't know where to go.
    y= (xk+√(r2-(Cx-h)2)-b)/(h+√(r2-(Cy-k)2)-a)+b-(ka+√(r2a2-x2a2+2a2xh-h2)-ab)/(h+√(r2-(y-k)2)-a)

    But this would just give me the range of values given I know the current point of C. does anyone know how I can factor in the travel of C and the movement of B?
    Last edited: Sep 11, 2014
  2. jcsd
  3. Sep 11, 2014 #2


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    The circle [itex](Cx- h)^2+ (Cy- k)^2= r^2[/itex] can be written in parametric form as [itex]Cx= r cos(\theta)+ h[/itex], [itex]Cy= r sin(\theta)+ k[/itex] where "[itex]\theta[/itex]" is, geometrically, the angle the line from (h, k) to (Cx, Cy) makes with the horizontal line Cy= k. If the point is moving around the circle with linear speed "v" (I will not use "t"- I want to reserve that for "time"), then, since the circle has radius r, so circumference [itex]2\pi r[/itex] the point (Cx, Cy) will go completely around the circle in [itex]2\pi r/v[/itex] seconds so [itex]2\pi[/itex] radians in [itex]2\pi r/v[/itex] seconds- the angular velocity is v/r radians per second. That means that [itex]\theta= vt/r[/itex] radians in t seconds and the equations of the orbit can be written as [itex]Cx= r cos(vt/r)+ h[/itex], [itex]Cy= r sin(vt/r)+ k[/itex].

    Now, the straight line from the center of the circle, (h, k) to the point (Cx, Cy) has slope (Cy- k)/(Cx- h) and, since y= k when x= h, equation y= [(Cy- k)/(Cx- h)](x- h)+ k. Replacing Cx and Cy with the formulas above, that is
    [tex]y= \frac{r sin(vt/r)}{r cos(vt/r)}(x- h)+ k= tan(vt/r)(x- h)+ k[/tex].

    That is the equation of the line which is what you asked for. I'm not sure what you mean by "B is moving", presumably along the line from (h, k) to (Cx, Cy). The length of that line is always r so at speed s, B will take time r/s to go from (h, k) to (Cx, Cy). If you want to "lead" the point so that B and C get to the same point on the orbit at the same time, you will have to replace "t" in the above equation with "t+ r/s". That would make it
    [tex]y= \frac{r sin(vt/r+ v/s)}{r cos(vt/r+ v/s)}(x- h)+ k= tan(vt/r+ v/s)(x- h)+ k[/tex].
  4. Sep 11, 2014 #3
    Thank you HallsofIvy for a very detailed response.
    I apologise for my terrible ability to convey what it is I am trying to do, I will try using an example case to explain where the problem I see with this is.

    let A = (0,0)[origin]
    let C = h=7, k=5, r=1
    (all other factors are fine)

    it appears like you read my post as A= (h,k) meaning A is inside the circle. Were it inside the circle, I probably would have consider it a might easier. However in this case A=(0,0) so it is outside the circle, my main point of issue.

    As for B, I have no problems with your interpretation of B, thank you.

    Thank you,
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