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Drop-Cutter

Offset ellipse, part 2

More on the offset-ellipse calculation, which is related to contacting toroidal cutters against edges(lines). An ellipse aligned with the x- and y-axes, with axes a and b can be given in parametric form as (a*cos(theta) , b*sin(theta) ). The ellipse is shown as the dotted oval, in four different colours.

Now the sin() and cos() are a bit expensive the calculate every time you are running this algorithm, so we replace them with parameters (s,t) which are both in [-1,1] and constrain them so s^2 + t^2 = 1, i.e. s = cos(theta) and t=sin(theta). Points on the ellipse are calculated as (a*s, b*t).

Now we need a way of moving around our ellipse to find the one point we are seeking. At point (s,t) on the ellipse, for example the point with the red sphere in the picture, the tangent(shown as a cyan line) to the ellipse will be given by (-a*t, b*s). Instead of worrying about different quadrants in the (s,t) plane, and how the signs of s and t vary, I thought it would be simplest always to take a step in the direction of the tangent. That seems to work quite well, we update s (or t) with a new value according to the tangent, and then t (or s) is calculated from s^2+t^2=1, keeping the sign of t (or s) the same as it was before.

Now for the Newton-Rhapson search we also need a measure of the error, which in this case is the difference in the y-coordinate of the offset-ellipse point (shown as the green small sphere, and obviously calculated as the ellipse-point plus the offset-radius times a normal vector) and where we want that point. Then we just run the algorithm, always stepping either in the positive or negative direction of the tangent along the ellipse, until we reach the required precision (or max number of iterations).

Here's an animation which first shows moving around the ellipse, and then at the end a slowed-down Newton-Rhapson search which in reality converges to better than 8 decimal-places in just seven (7) iterations, but as the animation is slowed down it takes about 60-frames in the movie.

I wonder if all this should be done in Python for the general case too, where the axes of the ellipse are not parallel to the x- and y-axes, before embarking on the c++ version?

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