Gilson: Polyhedra - A New Approach

For general discussion of polyhedra, not necessarily Stella-specific.
User avatar
oxenholme
Posts: 83
Joined: Wed Jan 16, 2008 5:32 am
Location: North West England

Post by oxenholme » Sun Nov 04, 2012 9:42 pm

removed
Last edited by oxenholme on Mon Nov 05, 2012 4:24 pm, edited 1 time in total.

User avatar
Ulrich
Posts: 165
Joined: Tue Jan 29, 2008 8:08 am
Location: Germany
Contact:

Post by Ulrich » Mon Nov 05, 2012 11:07 am

If the lateral pentagons are regular and their edge length is 1, Stella tells me that the other edge length is 4.69217462142686, the acute angle is 29.8878°:

Image

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

User avatar
oxenholme
Posts: 83
Joined: Wed Jan 16, 2008 5:32 am
Location: North West England

Post by oxenholme » Mon Nov 05, 2012 4:35 pm

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.

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

User avatar
guy
Posts: 86
Joined: Mon Feb 11, 2008 10:30 am
Location: England
Contact:

Post by guy » Mon Nov 05, 2012 10:03 pm

Looks really cool. Congratulations.

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

Post by Peter Kane » Tue Nov 06, 2012 9:18 am

A great ruler & compass & skewer achievement! If only Euclid had been fond of kebabs, he might have produced an Elements 2.

Well done!

Pete K

kohli149
Posts: 1
Joined: Mon Jul 27, 2020 10:12 am

Re: Gilson: Polyhedra - A New Approach

Post by kohli149 » Tue Jul 28, 2020 9:19 am

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 !
Get complete information about etagovt here.

Post Reply