Next assignment: what is a similar trick to cut a dodecagon?

Octagons not good enough? ?Oy. ?Not this week, I’m supposed to be loading a truck with our belongings and they’re not even all boxed yet. ?Misjudged how soon my big strong son would be done with his (now online) schoolwork and could help me.

There does have to be some theoretical plane of the blade that would let you use the same trick — i.e., tilt up the corner of the stick to define the desired fence distance. ?(And then having set the fence, reset the blade to 30 degrees off vertical and make your cuts.)

Call the half-dimension of the square “r” (since it will be the radius of your finished cylinder.)

You want to end up with 12 flats, each of them square to the radius at one of the 12 hours of the clock (and at distance r from the middle)

Assuming your blade tilts to the left (otherwise, once you’ve drawn this, hold it in a mirror):

Your desired initial cut will form the 8:00 flat, so it will be be 30deg off vertical, crossing the 8:00 o’clock radius at a distance of r from the center

(This makes it hit the left side of the square at 8:30 but that was a distraction, no help in the trig)

More usefully, it hits the bottom at 7:00 forming a 120deg corner there between the flats of 6:00 and 8:00 because you won’t have cut the flat for 7:00 yet

Draw a radial line to that 7:00 corner: from center of the square to 7 o’clock on the bottom side of the square

Your desired 8:00-flat cut is a line from there upward to left at 30deg off vertical, so the fence distance starts from that 7:00 point and reaches to the lower right corner (at 4:30 so to speak)

Distance from 7:00 to 6:00 measured along the bottom side is r tan 30deg = ((sqrt3)/3) r

Distance from 6:00 to the 4:30 corner is r

so distance from 7:00 to 4:30 corner is r(1+ (sqrt3)/3 ) ?(or r(1+ tan30deg)

Your initial (fence-setting) blade tilt needs to cock the bottom of the stick upward so that with its width of 2 r as hypotenuse, it spans a horizontal distance equal to the fence setting. ?Whatever that tilt is from horizontal, the blade needs to be set at that angle from vertical.

So you want a tilt of arccos ( r(1+ (sqrt3)/3 ) / 2r ) and the r’s cancel and I get 37.94 degrees.

Please retrace my steps and see if I’m awake yet.

BTW I would have to think more whether this non-45 scheme will tolerate radiused corners the way I mentioned that the octagon trick would (I suspect not) -- so at least for now I’d do this one using sharp square-cut corners.