I know that this is a very simple question, but I am stuck at the very last part of this process and can't find the solution elsewhere (I figured I'd find it on this site, but I didn't see it).

I have an object that is colliding with a circle and I need it to deflect properly, like this: examples of real-world deflection
I know the coordinates of the center of the circle and the object when it is on the circle's perimeter. I know the direction that the object is traveling on contact and can calculate the direction to the center (pointing inwards).

From similar questions, I know that the tangent line is perpendicular to the radius line I calculate. But, I'm not sure where to go after that. I need to calculate the new direction of the object in degrees, but my idea $\theta = \theta + 2(radiusline - \theta)$ , where $radiusline$ is the vector pointing towards the center, is inaccurate.

What is the proper formula for this deflection?

  • $\begingroup$ use the tangent to calculate the normal at the point of contact, and the angle the direction vector makes with it. The reflection vector is then simple to calculate. $\endgroup$ – K. Rmth Dec 8 '13 at 21:07
  • $\begingroup$ I have the direction vector. Is the tangent vector (since it points in a direction) equal to the direction plus 90 or minus 90? The normal at the point of contact is just the line to the radius, isn't it? $\endgroup$ – person27 Dec 8 '13 at 21:33
  • $\begingroup$ Aren't mixing two things up? The direction vector applies to the moving object whilst the tangent vector applies to the circle. Using the normal implies the benefit of simplicity; the normal vector(direction towards the circle centre) at a point of incidence $(b,c)$ is just $\vec N=-b\underline{\hat i}-c\underline{\hat j}$ $\endgroup$ – K. Rmth Dec 8 '13 at 21:52
  • $\begingroup$ I don't understand what the carot over the i is, what the i and j stand for, or what the underscores are. I have the math education and capacity of your average high-school student. You could say I'm definitely not mathematically-minded. $\endgroup$ – person27 Dec 8 '13 at 21:52
  • $\begingroup$ I will write an answer to this. Hopefully it will no longer be greek to you then.$\hat i$ is the unit vector in the positive x-direction (horizontal) and $\hat j$ is likewise for the y-direction (vertical). $\endgroup$ – K. Rmth Dec 8 '13 at 21:53

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$$\begin{align} \vec N & = \text{normal at point of incidence}=-a \hat{\mathbf{i}}-b \hat{\mathbf{j}} \\ \vec V & = \text{incident vector}=u \hat{\mathbf{i}}+v \hat{\mathbf{j}} \\ \vec R & = \text{reflected vector}=c \hat{\mathbf{i}}+d \hat{\mathbf{j}} \\ \end{align}$$ $$\begin{align} \text{using}\ \vec R =\vec V -2\vec N(\vec V \cdot \vec N)&={u \choose v}-2{-a \choose -b}\left[{u \choose v}\cdot {-a \choose -b}\right]\\ &\\ &={u \choose v}-2(au-bv){-a \choose -b}\\ &={u+2a^2u-2abv \choose v+2a^2u-2b^2v} \equiv {c \choose d}\\ \end{align}$$ Hence $\vec R=(u+2a^2u-2abv)\hat{\mathbf{i}}+(v+2a^2u-2b^2v)\hat{\mathbf{j}}$


Another way of doing this would be to get the tangent that runs through the point of collision and reflect off of this. The tangent is perpendicular to the line OP where O is the center of the circle and P is the point of collision.

  • $\begingroup$ Oh he referenced That question. I figured this way out by myself, but why can't he use this method it seems so much simpler what are the advantages for k rmth method? $\endgroup$ – dylan Oct 9 '17 at 16:51

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