Make a Perfect Dodecahedron: What Angle to Cut Plywood?

[arydberg]

I can cut out 5 sided pieces of plywood on my table saw but then i need to put an angle on the side of each pentagon. My question is what angle do I use so the pieces fit toghter to a solid? How do i find it. I suppose the dot product is used somehow.

 

[DaveC426913]

Try this:
https://en.wikipedia.org/wiki/Table_of_polyhedron_dihedral_angles

 

[ProfuselyQuarky]

I can cut out 5 sided pieces of plywood on my table saw but then i need to put an angle on the side of each pentagon. My question is what angle do I use so the pieces fit toghter to a solid? How do i find it. I suppose the dot product is used somehow.

Wow, such a great idea to make a wooden dodecaheron! Once you've got it down, surely you must create the entire set of Platonic solids! The dot product is used to derive both the general equation of the plane and formula for finding dihedral angles.

The formula for finding such an angle between two planes (or in this case two pieces of wood) is $\theta=\cos^{-1}(\frac{AE+BF+CG}{\sqrt{A^2+B^2+C^2}\sqrt{E^2+F^2+G^2}})$ where the equations of these two planes in general form are $Ax+By+Cz=0$ and $Dx+Ey+Fz=0$. Now, the only thing left to do is determine the values of $A$, $B$, $C$, $D$, $E$, and $F$. For math problems, this information is usually given in some way or another.

I'd like to know how to determine these values when doing something IRL, like for something @arydberg is doing?

 

[arydberg]

Try this:
https://en.wikipedia.org/wiki/Table_of_polyhedron_dihedral_angles

That's what I want. But it still leaves another question. How are these angles determined?

Thanks

 

[The Bill]

Pick a vertex. Consider only the faces of the polyhedron which share that vertex. Now, imagine cutting one edge and laying the resulting shape out flat, like a dressmaker's pattern. Now, consider how much angle each edge would have to be folded through to recreate the original polyhedron's vertex. That is, how much angle is required to rotate the cut edges enough so that they coincide as one edge again. You can use the angle of edge bending to compute the dihedral angle.

Also, if you know them you can use the geometric properties of the polyhedron, such as the relation between edge length and the distance from the center of the polyhedron to the center of a face. Then simply consider a triangle with one vertex at the polyhedron's center, one at the center of a face, and the third at the center of one of that face's edges.

 

[arydberg]

Pick a vertex. Consider only the faces of the polyhedron which share that vertex. Now, imagine cutting one edge and laying the resulting shape out flat, like a dressmaker's pattern. Now, consider how much angle each edge would have to be folded through to recreate the original polyhedron's vertex. That is, how much angle is required to rotate the cut edges enough so that they coincide as one edge again. You can use the angle of edge bending to compute the dihedral angle.

Also, if you know them you can use the geometric properties of the polyhedron, such as the relation between edge length and the distance from the center of the polyhedron to the center of a face. Then simply consider a triangle with one vertex at the polyhedron's center, one at the center of a face, and the third at the center of one of that face's edges.

Pick a vertex. Consider only the faces of the polyhedron which share that vertex. Now, imagine cutting one edge and laying the resulting shape out flat, like a dressmaker's pattern.

The angles of the pentagon measure 108 degrees , three of them add to 324 degrees. The three small angles measure (360 - 324 )/3 or 12 degrees.

Now, consider how much angle each edge would have to be folded through to recreate the original polyhedron's vertex. That is, how much angle is required to rotate the cut edges enough so that they coincide as one edge again. You can use the angle of edge bending to compute the dihedral angle.

How do you do this?

Also, if you know them you can use the geometric properties of the polyhedron, such as the relation between edge length and the distance from the center of the polyhedron to the center of a face. Then simply consider a triangle with one vertex at the polyhedron's center, one at the center of a face, and the third at the center of one of that face's edges.

And where do you use the 12 degree angles?

 

[The Bill]

What level of mathematics education have you had? If you think about what I've said and draw some diagrams of the way the faces fit together flattened in the plane and assembled as a polyhedron, you should be able to work this out.

I'm assuming you know geometry at least at the level of a high school level axiomatic geometry class. I'm trying to give you enough insight to work out the solutions for yourself, so you'll know how to derive the answers to similar problems in the future.

 

[arydberg]

What you said was implied. Yes it is the way to begin but i know that. I asked for an answer.

Right now it appears the best way is to do this is to start with cube. That way the answers are known. ps I am 74.

 

[DaveC426913]

ps I am 74.

Centimetres??

 

[The Bill]

Here's a site which shows sine helpful calculations: http://thales.math.uqam.ca/~rowland/investigations/polyhedra.html

Their second formula relates a polyhedron's volume to its edge length and apothem. Combine that with their calculation for the volume of the dodecahedron given its edge length. That let's you derive an expression for the apothem given the edge length of a dodecahedron.

Then, pick a face of the dodecahedron. Make a right triangle with one vertex at the center of the face, one vertex at the center of an edge of that face, and the third vertex at the center of the dodecahedron.

Since each face is a regular pentagon, the inradius of the face is s/(2tan(π/5)). Now you have a right triangle, and know two of its side lengths. In this case, the inverse tangent of their ratios will give you the angle of this triangle at the point where the faces meet.

Since the dihedral angle of a polyhedron is defined to be the internal angle where the faces meet, it is just double the angle you calculated in the previous paragraph.

 

[arydberg]

Centimetres??

Years. You implied I was in high school.

 

[DaveC426913]

You implied I was in high school.

Not I.

 

[The Bill]

Years. You implied I was in high school.

No, I assumed for the sake of discussion that you know geometry of at least a high school level. I never implied or stated anything about what you are doing currently.

 

[arydberg]

No, I assumed for the sake of discussion that you know geometry of at least a high school level. I never implied or stated anything about what you are doing currently.

Never mind. I will figure it out myself.