Replies

1. 
   SgianDubh | Oct 03, 2002 04:52pm | #1

   I may be getting the wrong end of the stick here, Doc, but if it's a columnar structure out of something like veneered sheet goods, you are in effect making an eight sided box. If you further plan to mitre all the panels together, then you have 360° /16 mitres  = a 22.5° mitre cut on the end of each panel.  If you have a sliding table saw, start by ripping the panels to the requisite height, and then tilt the blade to 22.5° and square the pieces roughly 6-12 mm (1/4"- 1/2") or so overlength. Then set the length stop on the fence and mitre the other end of each panel. If you are using a US style contractor or  cabinet saw, the procedure is similar in getting the parts square to start with, but you'll have to use the rip fence to work the 22.5° mitre. (In the latter case, if you are using a right tilt saw its often useful to cut a piece of 6 mm (1/4") MDF to sit on the table so that already mitred end doesn't slip under the fence. You are registering from the outside face of each panel.) The mitre's can be reinforced with things like splines or biscuits and the whole assembled with band or webbing clamps.

   A slow setting glue like polyurethane or liquid hide glue might be a good choice at assembly because they give you time to fiddle and align things correctly prior to finally tightening the cramps. If you don't have a table saw of some sort, and your structure is not as I've outlined for, let us know, because the approach will almost certainly need to be different. Slainte.

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1. 
   famousdoctor | Oct 03, 2002 05:01pm | #2

   Thanks for the feedback. I'm using an american table saw, good tip regarding the mdf.  My goal is for the post / pedestal to be appx 8" diameter.  My rough drawings indicate each piece width to be 4". 

   Does this sound accurate to you?

1. 
   theRev | Oct 03, 2002 05:34pm | #3

   Can't help with the math right now, sorry. It's after 11 here and the brain is bit worn down.

   Couple of tricks, though. First, sneak up on your ripped width. It's more work, but the pieces will go together better. If your finish width is 4", then first rip at your angle on both sides to a width of 4 1/4, then reset your fence slightly. You'll find that the blade will leave a somoother finish when it only has to take off a kerf or less of wood per side for a final cut. There will be fewer gaps in your pieces.

   Second, rip a couple extra pieces at slightly wider just in case you need them. Just the slightest change in 22.5 +/- will add up quickly over 16 cuts. You might find that the last piece will probably need to be fudged in width and/or angle to fit nicely. Those extra pieces will come in handy for fitting.

   Clamping (or cramping as Sgian would say) can be a bear. A previous issue of FWW had some good strategies for this. One, lay your pieces face up on your worktable, and tape them all together tightly edge to edge with masking tape. Depends on your project size, but I would say every 4 inches of height should be one row of tape. You might need a third hand for this, though. Then, gently turn it over, and gently roll it up. The tape will keep it together, and you can now check your fit and make any adjustments you might need. When I did mine, after getting the final fit before gluing, I retaped it to get it very tight.

   Slow glue, definitely. Work it on all sides. Roll it up slowly. Then tape it closed and reach for surgical tubing to bind it up tightly.

   If it's a table leg (not an umbrella stand like below), you'll probably want to work in some backing for foot and top attachment before gluing it all up.

   Have fun.

   Scott

   Edited 10/3/2002 10:40:57 AM ET by the rev

   
2. 
   DouglasABaker | Oct 16, 2002 08:59pm | #8

   4" will be too long.

   A Right triangle has sides in the proportions of 1:1:Square Root of 2, or 1:1:~1.41.

   The _outside_ diameter of your octagon, then, will be comprised of a full side plus 2 sides (one on each) coming off it at 45 degrees.  Mathematically, your diameter then = (side length) + 2*(side length)/(sqrt 2).

   For an external diameter of 8" (note that stock size is irrelevant) you need a side length of 3.31" on the long side.  If you want an _internal_ diameter of 8", you need 3.31" on the _short_ side.

   At 4" for the long side, your external diameter would be 9.65".

   Doug

1. 
   fredsmart48 | Oct 19, 2002 04:08pm | #10

   It is nice that you can remember all the formulas. Is there a good book on construct math that explains it.

   On a small projects I use a protector.

   Lay it out the angles you need. In this case 45 deg.

   Mark 4" on both legs.

   Measure from the 4" on both legs that is the length of the board you need.

2. 
   AlanTurner | Oct 04, 2002 04:24pm | #4

   I have seen in the Lee Valley catalogue router bits for cutting birdsmouth "notches" in edges, at the appropriate agnles for 6, 8 12 & 16 sided pieces. As such, they lock together when assembling. Not too expensive. Don't know how they would work on MDF. Would need to work these off of a router table/fence set-up. I haven't yet tried them.

   Has anyone used these yet?

   
   

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   Edited 10/4/2002 10:22:00 AM ET by s4s

1. 
   UncleDunc | Oct 04, 2002 06:46pm | #5

   I've seen those too, and remember thinking it would be silly to buy extra bits when you could get the same effect with a rabbeting bit by tilting the router relative to the table. Then I thought about how to make a tilt adjustment for a router table, and when it appeared that it would require cutting part of the base off the router, I decided a few extra bits would be a cheap and easy solution if you do much of that kind of work.

1. 
   AlanTurner | Oct 04, 2002 07:09pm | #6

   Perhaps tilting the stock via an angled platen would be another way to approach this. A bit less invasive, from the router's perspective.

1. 
   UncleDunc | Oct 04, 2002 08:27pm | #7

   Doh.

   Go away. You're making me look bad. :)

3. 
   mrbird90 | Oct 16, 2002 11:05pm | #9

   If you are working with a circumscribed circle(dia. around points) the long edge will be 3.06" and if you are working to an inscribed circle(radius contacts center of side)  your long edge will be 3.31" Hope this helps.