Thread: Octagonal box prototype

Rocker

Nice work, I look forward to seeing more of this style, are you considering a dome lid?

Tony,

I have no idea how I could go about making a domed lid for an octagonal box. I suppose you could laminate each sector; but joining them neatly would be a nightmare. Do you have any suggestions as to how to go about it?

Rocker

Sweet! Man you got on this quick. Already made the jig to cut the fingers and starting to cut the sides! Looks good so far. I am very excited to see what you will do with this. Especially the lid. keep posting pictures.

Hi Rocker, amazing work - both with the jig and the box!!

I recently made a rectangular raised box lid by hand planing the sides until they met with as little gap as possible while keeping the raised look. For a job done completely in the dark, I think it turned out pretty well. But I will have to line the box lid as the underside looks terrible

Attached is the shape of the lid pieces and the side view.

I know this probably doesn't help you much at all Rocker, especially as I can't give you any precise details at all , but I thought it might spark off an idea or two.

cheers

Wendy

Rocker

These images may help? http://www.thedragonslair.com

Originally Posted by rufflyrustic

I recently made a rectangular raised box lid by hand planing the sides until they met with as little gap as possible while keeping the raised look. For a job done completely in the dark, I think it turned out pretty well. But I will have to line the box lid as the underside looks terrible



Wendy

Wendy,

As you know, I tend to shy away from the dark side, so my first reaction was to draw a low-angle octagonal pyramid in 3D in TurboCAD, and then measure the angle between adjacent faces to determine the mitre angle. Drawing the octagonal pyramid is not that hard; but I was defeated by the problem of how to measure the angle between the adjacent faces. So I think I shall have to make a set of eight isosceles triangles with apical angles of 45°, and mitre their long edges at say 86°, then fit them together and see if the resultant pyramid is roughly the desired height; and then, if necessary, modify their mitre angles accordingly.

Correction: I have now nutted out how to measure the interfacial angle, and the mitre angle between adjacent faces of the octagonal pyramid. If the sides of the octagon are 62 mm, and the height of the pyramid is 20 mm, then the mitre angle is 84.34°. You have to admit that TurboCAD is the bee's knees - try doing that in Sketchup.

Rocker

There are some pics in this thread of the jig I use to cut triangles with bevelled sides on the tablesaw.
Cutting triangles

Originally Posted by Rocker

You have to admit that TurboCAD is the bee's knees - try doing that in Sketchup.

Rocker

I just did this task in Sketchup and found it quite easy. For an upward tilt of each octant of 5.7 degrees the mitre angle is 87.8 degrees.

Some trigonometry would also allow the angle to be calculated using sines and cosines. Anyone want to try it?

Update: tried reproducing Rocker's TurboCad dimensions and got a corresponding angle of "~84.5 deg" The Sketchup angle output is limited to 0.1 degrees resolution. If you can measure to 0.01 degrees, then TurboCAD can give you more precision.

Oops, I underestimated Sketchup. Zenwood, Skew was asking in another thread for a formula to give the mitre angle of a regular pyramid, and your name was mentioned as a maths guru. Can you give a general formula for the mitre angle, if the length and number of sides and the pyramid height are known?

Rocker

Turned out to be more interesting than I thought. The attached explains how to calculate the mitre angle for an octagonal pyramid made from eight equal facets. (No more than high-school maths is required.) It would be quite easy to generalise this to the case of an n-agonal pyramid (give me time).

What was the link to that other thread with Skew's question?

Zenwood,

Many thanks for that very detailed treatment of the problem. There still remains, though, the question of the size of the apical angle of the inclined faces of the pyramid. I had forgotten, when I first posed this question, that this apical angle becomes progressively less than 45°, as the pyramid's height increases. I suggested, in the other thread - http://www.woodworkforums.ubeaut.com...ad.php?t=39586 - that perhaps someone could come up with a couple of simultaneous equations, linking this apical angle and the mitre angle to the length and number of sides, and the height of the pyramid. Maybe that should have been just linking the apical angle and mitre angle to the number of sides and the slope of the faces, since the angles are the same whatever the scale of the pyramid.

Rocker

Rocker, you may shy away from the Darkside , but I have to confess that I shy away from the 'calcuated physics/mathematics' in woodworking .

cheers
Wwendy

So how do you make each side of an octo box precisely the same length and mitre angles

Echidna,

Making the sides of the box the same length is not too hard, and the angle between the sides (135&#176 is determined by the jig used to cut the fingers. Cutting the triangles for a pyramidal lid is a bit harder, but armed with Stuart's calculator in the other thread, you can determine the apical angle and the mitre angle, if you input the slope angle of the faces of the pyramid, rather than its height. Actually cutting the triangles safely probably requires a dedicated wide-kerfed crosscut sled, with some auxiliary wedges to give the required apical angle of the triangles.

Rocker