hi Agnishom,

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

Bob

hi Agnishom,

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

hi Agnishom,

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

jesseherring wrote:

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.

Hi

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

I sent an email about this to "Soapy Joe", hopefully it will reach him.

bob

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

jesse

hi jessehherring

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.