## Gilson: Polyhedra - A New Approach

metric data:

28 facelets, of 2 types; 70 edgelets; 44 vertices, of 4 types.

Radius 4.83391, Diameter 9.66782

Facelet type 0: area 1.72048

Edge 0-0: Joins 1-1, Length 1, Dihedral 142.09 deg, Mitre 18.9551 deg

Face angle 108 deg

Edge 0-1: Joins 0-1, Length 1, Dihedral 152.938 deg, Mitre 13.5311 deg

Face angle 108 deg

Edge 0-2: Joins 1-0, Length 1, Dihedral 142.09 deg, Mitre 18.9551 deg

Face angle 108 deg

Edge 0-3: Joins 1-2, Length 1, Dihedral 142.09 deg, Mitre 18.9551 deg

Face angle 108 deg

Edge 0-4: Joins 0-4, Length 1, Dihedral 152.938 deg, Mitre 13.5311 deg

Face angle 108 deg

Facelet type 1: area 6.68965

Edge 1-0: Joins 0-2, Length 1, Dihedral 142.09 deg, Mitre 18.9551 deg

Face angle 135.234 deg

Edge 1-1: Joins 0-0, Length 1, Dihedral 142.09 deg, Mitre 18.9551 deg

Face angle 135.234 deg

Edge 1-2: Joins 0-3, Length 1, Dihedral 142.09 deg, Mitre 18.9551 deg

Face angle 119.822 deg

Edge 1-3: Joins 1-4, Length 4.69217, Dihedral 137.658 deg, Mitre 21.1709 deg

Face angle 29.8878 deg

Edge 1-4: Joins 1-3, Length 4.69217, Dihedral 137.658 deg, Mitre 21.1709 deg

Face angle 119.822 deg

Ulrich

Vive le difference!

I didn't realise Stella could "do" this one.

It is quite amusing seeing all the data correct to umpteen decimal places.

I divided the 360 degrees into 14, and used a couple of spring bow compasses to produce the net for the equatorial pentagons - not measuring any dimensions anywhere.

I pushed a skewer through the centre, and used the ruler as a rigid flat object to determine the height of the 7 sided pyramid. I then took the first and second actual measurements - the long pentagons being 59.3mm in height and 29.7mm in width. Again I used a couple of spring bow compasses to determine the vertices of the long pentagons.

I actually quite like the proportions of the end result.

I didn't realise Stella could "do" this one.

It is quite amusing seeing all the data correct to umpteen decimal places.

I divided the 360 degrees into 14, and used a couple of spring bow compasses to produce the net for the equatorial pentagons - not measuring any dimensions anywhere.

I pushed a skewer through the centre, and used the ruler as a rigid flat object to determine the height of the 7 sided pyramid. I then took the first and second actual measurements - the long pentagons being 59.3mm in height and 29.7mm in width. Again I used a couple of spring bow compasses to determine the vertices of the long pentagons.

I actually quite like the proportions of the end result.

Last edited by oxenholme on Tue Nov 06, 2012 7:17 am, edited 1 time in total.

- Peter Kane
**Posts:**90**Joined:**Sun Oct 25, 2009 11:50 am**Location:**S.E England

### Re: Gilson: Polyhedra - A New Approach

I may be wrong, but I think that you can start by drawing a pentagon that has similar angles to the 14 pentagons around the equator of the figure. I don't think that you need to be precise, so long as you make it symmetrical. Make 14 of these and arrange them into a ring. The remaining pentagons that make up the upper and lower pyramids are them defined by the sides of the equatorial pentagons, plus a fairly arbitrary length for the longer sides.

Don't forget to publish a pic !

Don't forget to publish a pic !

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