Thanks. I'd not met that before. I wonder how it is derived.

Bob

]]>Let p = The number of edges at each face

and q = The number of faces meeting at each vertex

and θ the dihedral angle.

Now,

]]>

Go to

http://en.wikipedia.org/wiki/Dodecahedron

Section titled "Cartesian Co-ordinates"

The diagram shows which points are which, so that should make the task a lot easier.

Bob

]]>Take pairs of points, using the co-ordinates, and find the distance between them.

How???

Where do I get them from??

Good question. I don't know how a person could work them out from scratch. But you can test that they are correct like this:

Draw and label a Schlegel diagram for the solid.

http://en.wikipedia.org/wiki/Schlegel_diagram

Take pairs of points, using the co-ordinates, and find the distance between them. (Best on a spreadsheet with a distance formula)

When you find a pair that are the minimum distance apart they must be adjacent on the Schlegel diagram, so pick a suitable pair of letters and assign them to the co-ordinates. Keep doing this until you have labelled them all.

Then you can choose points for two adjacent faces and do the cosine rule trick. That's basically what I did for post 30. If you look carefully at my screen shot of the spreadsheet, you'll see I have shrunk some columns. That has hidden all my distance apart calculations.

Bob

]]>I used the coordinates from the Wiki page and worked out which points are which on the diagram. Below I have the icosadodecahedron and part of my spreadsheet for determining the coordinates.

Now, how did the wiki work out the co-ordinates without knowing the dihedral angles?

]]>Can anyone help me or point me in the right direction to find the angles needed in order to build a wood model of an icosidodecahedron, truncated icosahedron, and/or the 3V form of an icosahedron. I found Soapy Joes posting on the subject and could not gather how to figure out the angles from his discussions. I would like to be able to derive the angles so that I can later construct the rest of da vincis polyhedra.

thanks

Jesse

I suggest that you purchase the 'MAHEMATICAL MODELS' book by H.Martyn Cundy and

A.P.Rollett go to the inter-net and find a second hand version. Less than $20. ( 1961 )

If you are so inclined, you can learn about Miller Indices, vector analysis, and analytical

geometry. If you are well versed in those subjects, then you can generate your own angles.

I have been making models since 1960 and have a truncated hexacontrahedron in the mathematics archives of the Smithsonian Institute.

]]>Been quite a while since I last looked at this forum...

I have made a dozen or so wood models ...

How do I post pictures to this forum?

Would Jesseherring contact me direct at [hidden]

Soapy

]]>thank you i appreciate it, i will keep my fingers crossed

]]>You're welcome; it's been interesting to explore this. Post back if you want anything clarified.

Bob

]]>i need to sit down and take a good look at this, i cannot thank you enough for your help

jesse

]]>wondering if it would be possible to somehow get a message to SoapyJoe regarding his adventures with similar problem?

jesse

]]>Ok. Here's the complete vector approach.

(i) I used the coordinates from the Wiki page and worked out which points are which on the diagram. Below I have the icosadodecahedron and part of my spreadsheet for determining the coordinates.

(ii) then I made my formula. See the middle diagram

Suppose AHG is one triangle and AHJZX is one pentagon. Let 'E' be the midpoint of AH. (note this E is not the coordinate labelled E in my sheet. I was running short of letters!)

The angle between EG and EZ is the required dihedral angle.

I used the 'dot product' formula for calculating an angle between two vectors.

Top line is the dot product between the vectors. Bottom is the product of their lengths.

These calculations are also shown on the sheet along with a conversion into degrees.

Bob

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