Ah — cut with the blade notching into a sacrificial fence! ?Simple but a new one on me. ?So your cutoff is on the
same side as the fence. ?That lets you cut way down on irregularities due to any curvature of the pole, because your fence is right at the corner you’re cutting off. ?And takes way less math.
But I’d certainly stand aside, and maybe also have a gloved helper pull the cutout on through once its end reached the blade and got severed. ?Or at least have an outfeed support that kept the cutout from prying its tail end up against the blade. ?A lot of the happen-before-you-know-it accidents occur when the support of the workpiece or tool suddenly changes — the moment when the cutoff becomes severed, the moment when a drill breaks through the far side, etc.
Yes, the cutout is small but has no less length, so it’s got no less mass per unit of end area than the main workpiece does — it could spear you with about as much frontal pressure as a bigger stick of the same length. ?That stick that took out Hank’s tempered-glass window was less than a 1x1, and less than a foot long.
Crispin
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I counted sheep last night too (well sitting at the dining table) and came up with a couple of constants to use to cut the 12 faced blank.? The lateral distance from the fence to the blade is about .211 times the side of the square and the vertical distance from the table to where the blade intersects the fence is .366 times the side of the square.? Of course you only need one of those if you've got your blade at 30 degrees and the lateral one is most practical.? I usually measure as close as I can and then do a few test cuts, just barely touching the sample wood to the spinning blade.? However, this does use what you probably consider the more dangerous fence/blade configuration, but I have found (so far) that if the blade is embedded into the sacrificial fence, the cutoff is less likely to ride up and kick out.? ?And the relatively small piece of cutoff doesn't pose as much of a danger as when the blade could get hold of the whole blank.? In theory.? But I always stand to the side, out of the line of fire, at any angle of the blade.
I'll try to draw it up and show the equations later but it's simple enough.??
Very good on the 38 degree trick.? I'll go through it and see if I can follow, but will need to draw that too.? I suppose the possibilities for this trick are endless.? Perhaps not practical but endless just the same.
Thanks!,
James
On Thu, May 21, 2020 at 9:28 AM crispin_m_miller <
crispinmm@...> wrote:
Back to you, James,
Of course, once you’ve got the fence-distance formula of ?"1+ (sqrt3)/3 ) r” ?i.e. 1.577 r , then if you know “r" (half the width of your blank), you can also just do the multiplication and set the fence.? Depending on your particular fence and your particular tilt-reading scale, I don’t know which approach would be more accurate. ?
But if you set the fence by measuring tape, you have to deal with a tilted blade, and if you set it by a built-in scale, you need the blade tilt motion to be centered exactly at the table surface.? So the lay-against-the-blade trick does bypass both of those uncertainties, and also the measuring and multiplication tasks.? You do need a blade with teeth enough that a number of them (of the relevant tooth-set direction) fall within the face of the blank, for the blank to rest against the blade accurately.
cm
On May 21, 2020, at 8:33 AM, crispin_m_miller via
??<
crispinmm@...> wrote:
(Then remember to reset your blade to 30deg to make your actual cuts…)
On May 21, 2020, at 8:28 AM, crispin_m_miller via
??<
crispinmm@...> wrote:
Good morning, James —?
On May 20, 2020, at 10:16 PM, crispin_m_miller via
??<
crispinmm@...> wrote:
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.)
OK, used this question as my substitute for counting sheep last night and saw a simpler way to frame it. Woke up now and punched it in.
Can look for a time on the road to prepare an intelligible sketch, but I get that the initial blade tilt for setting the fence should be practically 38 degrees — if I have it right, the trig is
angle = arccos ( ( 1+ (sqrt 3)/3 ) / 2) = 37.94 deg
constructed by (looking at the square blank endwise)
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.
OK, gotta cook brek, wave it under the kid’s nose to wake him up, and we gotta carry boxes
Be careful with your saw
Crispin
