Hexagonal Geometry Wood Project

[Reality Believer]

I am doing a woodworking project that involves an interesting geometry problem.

The piece is essentially a vertical hexagonal tube that I want to put a "roof" on. The roof peak is in the center of the hexagon and is 2 inches above the sides. Each side is 5 1/8 inches along the flat, and the hexagon is 8 7/8 inches across the flats.

I want to construct the roof by cutting 6 pie shaped pieces out of 3/4" material so they fit properly to form essentially a hexagonal pyramid.

What is the miter angle and bevel angles that would be required to make the hexagonal pyramid roof fit together seamlessly?

 

[Dr. Trintignant]

I am doing a woodworking project that involves an interesting geometry problem.

The piece is essentially a vertical hexagonal tube that I want to put a "roof" on. The roof peak is in the center of the hexagon and is 2 inches above the sides. Each side is 5 1/8 inches along the flat, and the hexagon is 8 7/8 inches across the flats.

I want to construct the roof by cutting 6 pie shaped pieces out of 3/4" material so they fit properly to form essentially a hexagonal pyramid.

What is the miter angle and bevel angles that would be required to make the hexagonal pyramid roof fit together seamlessly?

Each piece, on the outer surface, is an isosceles triangle with one side of 5.125" length. The other sides form a right triangle with legs of 2" and 5.125", which we can feed into Pythagoras's theorem to give a length of 5.5". So far, so good.

Now, we want to find the miter angle on the 5.125" side (the base). Consider a line drawn from the center of this base up to the peak. That line is the hypotenuse of a right triangle with a height of 2" and width of sin(60)*5.125", or 4.44". Take arctan of 2" over 4.44" and we get 24.25°. This assumes that the top of the hexagon is flat.

The miter angle between sides is trickier. Consider two of the triangular sides. On the line that joins them, there is a point that, when connected with the outside points on the bottom of the sides, forms lines perpendicular to the edge. Those three points form a triangle that we can use to compute the angle.

Each of those lines also forms a right triangle with the base. We know that one of the angles is arccos(5.125"/(2*5.5")) = 62.23°. The hypotenuse of the triangle is 5.125", so the length in question is sin(62.23°)*5.125" = 4.535". This length forms two of the sides of the original triangle we were considering. The last side is actually the diameter of the hexagon, or 8.88".

We can look at half of this triangle, and see that the angle we're interested in is arcsin(8.88"/(2*4.535")) = 78.25°. And that is just miter angle between pieces.

Sorry if this description was confusing (hard to draw pictures here), but I believe it's correct. To summarize:
Length of edges between pieces: 5.5"
Angles at base of pieces: 62.23°
Miter angle at base of pieces (the 5.125" edge): 24.25°
Miter angle between pieces (the 5.5" edges): 78.25°

Hopefully someone else can back me up here .

- Dr. Trintignant

 

[screensnot]

I drew it in AutoCAD and came up with the same numbers (almost). I got the bevels at 24.24° and 78.14°.

 

[Soapy Sam]

Use more glue. Never fails. The Bodge-it-yourself school of design theory.

 

[Skeptical Greg]

Reality Believer,

Is this a project that you were actually expected to do yourself, and receive a grade on ?

 

[Reality Believer]

Reality Believer,

Is this a project that you were actually expected to do yourself, and receive a grade on ?

Thanks Dr. Trintignant for that, and no Diogenes, this is a personal project.

 

[Dr. Trintignant]

I drew it in AutoCAD and came up with the same numbers (almost). I got the bevels at 24.24° and 78.14°.

Yeah--after posting, I decided to run the numbers with a few more sig-figs, and got 78.14 for the second angle (the arcsin amplifies errors somewhat when close to 90°). Close enough for woodworking, in any case. And good to know that my trig skills aren't in as bad shape as I thought .

- Dr. Trintignant

 

[Dr. Trintignant]

Thanks Dr. Trintignant for that, and no Diogenes, this is a personal project.

Glad to help. Trig problems are fun from time to time (somehow, I have more fun with them now than I did back in high school).

- Dr. Trintignant