Formula for average width of a circle

Question

I'm trying to make a costing calculator for engraving a circle on acrylic using a laser. The laser can move at 150mm/sec, and lightly cuts lots of parallel lines spaced 0.05mm apart. If I engrave a circle of, say, 20mm radius, I want to know how long that will take. To do this I think I need the average width of a circle (expressed as a function of the radius), multiplied by the number of 0.05mm-spaced lines that would fall within the circle, (also a function of the radius). This should give me the total length of cut, which divided by the speed, will give me the overall time in seconds. But I'm stuck on the formula for the average width of a circle. Can anyone help me with this? Regards - Dave

Answer 1

You can think of the lines being $0.05\,\text{mm}$ wide, hence with length $\ell$ covering an area of $\ell\cdot0.05\,\text{mm}$. This shall cover the circle area of $\pi r^2$, so you need a length $$ \ell \approx \frac{\pi r^2}{0.05\,\text{mm}},$$ hence a time of $$ \frac{\ell}{150\,\text{mm}/\text s}\approx\frac{\pi r^2}{7.5\,\text{mm}^2}\,\text s,$$ so for $r=20\,\text{mm}$ aproximately(!) $167\,\text{s}$. This formula will work for any shape, just replace $\pi r^2$ with the area. However, some inaccuracy is hidden in the $\approx$, coming from navigation times between separate line segments (these will typically be in the range of about half the perimeter of the shape - here $\pi r\approx 63\,\text{mm}$, adding $\approx 0.4\,\text{s}$) and possibly curved line segments (which cover an area just slightly deviating from $\ell\cdot 0.05\,\text{mm}$), acceleration times, and quantisation or rounding errors (the widened lines will typically cover a polygonal shape instead of a round circle, and the exact difference even depends on the "phase" you start the covering with.

Answer 2

The area of the circle, in terms of the radius, is $\pi r^2$. So, dividing by the ‘height’, we have $\frac{\pi r^2}{2r} = \frac{1}{2} {\pi r}$ for the average width of the circle.

The number of $0.05$mm-spaced lines that will fit into a circle of radius $r$ is $\lfloor \frac{2r}{0.05}\rfloor + 1$.

So the total length of cut will be $\frac{1}{2} {\pi r}(\lfloor \frac{2r}{0.05}\rfloor + 1)$, and the overall time for engraving will be this number divided by the speed, as you say.

To engrave your circle of 20mm radius with a laser moving at 150mm/sec and cutting lines spaced 0.05mm apart, the time taken will be $\frac{\frac{1}{2} \pi \cdot 20 (801)}{150} \approx 168$ seconds.

As Cuspy Code points out, this figure will be an underestimate and a better model would take into account the time taken for the laser to move between cutting parallel lines.

Answer 3

Presumably, the cuts fill the circle down to the center, and the successive radii are $0.05k$ for $k$ from $1$ to $n=400$.

Hence using the triangular numbers, the total length is $0.05\dfrac{n(n+1)}22\pi$, to which you add the transverse moves. Hence

$$25216\,mm$$

traveled in

$$168.1\ s.$$

To this you must probably add the time for $800$ changes of direction.