# Thread: Maximizing fixed circles in a fixed rectangle

1. ## Maximizing fixed circles in a fixed rectangle

Not sure if this fits here or elsewhere, but I'll try this...

I'm putting together a project where we're buying some rather expensive plastic sheets and making circles out of it.

We need to make roughly 18" diameter circles (actually, because of the router width and some errors, it works out to 18 1/4" or 18.125" per circle).

We've got a few choices of width/length sheets. I'm not sure how many we can get per sheet, obviously the more the merrier.

I know there's packing theories out there, etc, but haven't been able to adapt any of them for our purpose. I tried hexagonal packing etc, but have been failing miserably.

I was told the most we can squeeze on a 48" x 120" sheet is 14 circles, I'm pretty sure that's correct. If we only needed 18" we may be able to fit 17 circles, but, that little bit makes a big difference.

The other two sizes we can choose from are 60" x 120" and 96" x 240" (the bigger the sheet, the more expensive obviously).

Can anyone point me in the direction of an algorithm or something to calculate the number of circles we can fit? I've searched for a program online and found nothing...

2. Originally Posted by barwick11
I was told the most we can squeeze on a 48" x 120" sheet is 14 circles,
That's correct; but with "packing".

3. I wrote a long lengthy response to explain exactly what I was thinking, and found the solution in there...

Two circles of diameter X that are touching have X distance between their centers. If a third circle were placed on top of those two circles, the distance between all their centers Z would be Z = X (an equilateral triangle). The vertical distance between the centers of these circles (H) would be able to be calculated such that X^2 = H^2 + (1/2 X)^2 (in other words, A^2 + B^2 = C^2)

If the two bottom circles were separated, same story, except now it's X^2 = H^2 + (1/2 (X + Y)) ^2

4. Originally Posted by barwick11
I wrote a long lengthy response to explain exactly what I was thinking, and found the solution in there...

Two circles of diameter X that are touching have X distance between their centers. If a third circle were placed on top of those two circles, the distance between all their centers Z would be Z = X (an equilateral triangle). The vertical distance between the centers of these circles (H) would be able to be calculated such that X^2 = H^2 + (1/2 X)^2 (in other words, A^2 + B^2 = C^2)

If the two bottom circles were separated, same story, except now it's X^2 = H^2 + (1/2 (X + Y)) ^2
Ya, that's correct.
But, real-life (you need to "cut" out them circles) almost dictates
NO packing...else how you gonna "slip" yer cutting scissors in there

5. Originally Posted by Wilmer
Ya, that's correct.
But, real-life (you need to "cut" out them circles) almost dictates
NO packing...else how you gonna "slip" yer cutting scissors in there
Router. Actually probably a Waterjet. I actually need 450mm circles (17.7 inches or so), so I already compensated for that with 18 1/4" shapes on the template). I was planning on them using a 3/8" bit originally, but a Waterjet will probably work better.