## Sunday, September 21, 2008

### Determining the proper angles to cut pyramid faces

I have built a few projects that involved the construction of a hollow pyramid by assembling triangular faces. As the construction progressed, I quickly realized that the proper miter and bevel angles of each pyramid face are not trivial. ie They are not 45*, 22.5*, 60*, or any other common construction number. The reason is that the faces of the pyramid meet each other in a compound angle, thus making it difficult or impossible to cut the pieces without first calculating the miter and bevel using geometry.

I made this diagram in Google SketchUp with the dim_angle.rb plugin (http://www.crai.archi.fr/RubyLibraryDepot/Ruby/em_geo_page.htm) . The angles are as follows:

VA = Vertex angle
SA = Slope angle
HA = Hip angle
DA = Dihedral angle (measured in a plane that is orthogonal to the pyramid's sloping edge)
MGA = Miter gauge angle

When building pyramids, the VA and SA are inputs to the following equations. HA is an intermediate step, and the MGA and DA are outputs. These two output angles indicate how the miter and bevel adjustments must be set on the saw to cut the pyramid faces.

For a pyramid with a square base, the VA is 45*. The SA is chosen by the designer to give the pyramid a desired aspect ratio. High SA values, would make the pyramid very tall relative to the base. For this example, let's choose 60*. The first calculation is for MGA:

VA= 45
SA = 60
MGA = arctan(tan(VA)/cos(SA))
MGA = arctan(tan(45)/cos(60)) = 63.43 (not a "common" angle at all!)

Now, the HA is not really useful for setting the saw, but makes the final equation easier to manage.
HA = arctan(sin(SA)*tan(MGA)*cos(VA))
HA = arctan(sin(60)*tan(63.43)*cos(45)) = 50.77

DA = 2(90-(arctan(cot(VA)*tan(HA)*cos(HA)))
DA = 2(90-(arctan(cot(45)*tan(50.77)*cos(50.77) = 104.48

Depending how your saw measures bevel cuts, the proper bevel angle setting is given by just this part of the equation:

arctan(cot(VA)*tan(HA)*cos(HA))
arctan(cot(45)*tan(50.77)*cos(50.77) = 37.76

So, set your saw's miter gauge to the MGA (63.43) and the bevel to 37.76, and cut away. I'm sure you math guys that really loved the "fun" trig identities will have a more compact form for these equations, so feel free to post them.

1. Hi, I've tried to build a 5 sided pyramid using this formula but it didn't all meet up, I took the VA for a pentagon as 108/2. I'll email you the excel sheet I used if could take a look for me?

thanks, Will.

2. Will, I've never tried a five-sided pyramid, but it should work. Are you building the pyramid from paper or other thin material so that the thickness and bevel is negligible?

Scroll down to N-sided box:
http://jansson.us/jcompound.html

If you enter "4-sided" and 60* from horizontal, the MGA and DA angles agree with my calculations. Let me know how it goes.

3. VA= 54
SA = 45

So I took VA from 108/2 as for a pentagon. I used 45 for SA as this is the largest angle I can cut with my circular saw.
My outputs were

MGA = 62.8
DA = 24.5

I tested my spreadsheet using your inputs and the outputs matched yours.

I'm building these out of 3/4" plywood. Unfortunately I just tried it out and 5 pieces do not all marry up.
I will try a four sided one using your dimensions next to see if that matches up ok.
Not having much luck getting this sorted!! Haha

thanks, Will.

4. Will, are you having problems with the miter angle or the bevel angle of the joint? The DA is sometimes not helpful for setting the bevel angle on your saw. At the end of my blog post, I indicate the part of the last equation will give the bevel, which is probably what your saw should be set to. Try doing the calculations for a four-sided pyramid with a very high SA (eg 89*, almost a rectangular box). The bevel should be 45*, now increase to 5-sided, still with an 89* SA. The bevel angle should be 36*, try cutting just a simple, unmitered bevel joint with 36* (or 90-36 = 54, depending on your saw).

5. Okay, I'm an art metals student. I'm trying to make an octahedron out of 18 gauge bronze sheet. None of this makes any sense to me, but that could be because I can't access the website you mentioned and I'm terrible at math. I do not have a machine or anything of the sort to make the beveled edges. I have a protractor, a pair of dividers, a ruler, and a caliper. I just want to know what angle the bevel should be around each equilateral triangle. Or perhaps another way to put it, if the metal is about 1/16" thick, how far in do I make the interior triangle to file down to to make the bevel around the sides? Any information would be awesome and appreciated.
~Steve (kungfuguru88@yahoo.com)

1. I'm sure this is too late to matter, but here goes anyway...

Your ocahedron can be thought of as eight pyramids, with each face of the octahedron being the base of a 3-sided pyramid with VA=30, and MGA=45, and SA being the included angle of the beveled edges.

From Ben's first equation, you'd get SA=acos(tan(VA)/tan(MGA)) = 54.7 degrees.

The inside edges would be cot(54.7)*1/16" = 1/sqrt(2) * 1/16" or 0.7071/16" in from the outside edges.

6. Hi. My name is Bob and I am a hobbyist and amature photographer. In reference to Will's 5 sided pyramid: I found the site below which hosts a calculator for an N-sided pyramids. It yielded a dihedral angle of 121.3°. If my read is correct that the bevel angle is 180° - DA / 2, then the bevel angle is 29.35°

http://jansson.us/jcompound.html

7. The DA calculation can be simplified. Specifically the section inside the arctan calculation.

Sorry typos this is being typed on my iPad.

To make the angle calculations easier, don't use the cot and just divide by the tan of VA instead. Also tan(HA) x cos(HA) can just be sin(HA) since the tangent is defined to be the sine over the cosine.

The arctan calculation just becomes sin(HA)/tan(VA)

Thanks for the great tutorial. I have been struggling with this and confirmed your math.

8. If you insert the expression for MGA into that for HA, it reduces easily into a simpler expression:

HA = arctan(sin(SA)*tan(MGA)*cos(VA))
= arctan(sin(SA)*tan(VA)/cos(SA)*cos(VA))
= arctan(tan(SA)*sin(VA))

9. I guess while I'm at it...

bevel angle = 90-DA/2 = asin(sin(SA)*cos(VA))

10. Good article Ben !
My 2 bits: The dihedral angle (DA) for a 4 sided pyramid can also be expressed in terms of the height (H) and base length (B) of the pyramid:
DA = arcCos( -1 /( 1 + (2H /B)² ))

11. Hi Ben , Im a aluminium fabricator and a customer wants me to make 3 sided pyramid in round tubing and im have a hard time cutting the double ended mitres every corner is 60 degrees on the flat plane got any clues its going to be tig welded so i can get away with a bit. Thanks Ross .

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