Article 075 - A Solution for the Delian Doubling of the Cube Problem.

A Solution for the Delian Doubling of the Cube Problem.

Historical Context.

The people of Delos suffering from a plague that had apparently been sent by Apollo consulted the Delphi Oracle as to how they might remove the plague.

The Oracle at Delphi suggested that they should double the size of the alter of Apollo which itself was a regular cube.

The problem with this solution is that it destroys the symmetry of the original cube and so would cause further offense to their deity.

Further if the side dimensions are simply doubled then the surface area increases by a factor of 4 and the volume increase by a factor of 8.

The solution needed is to achieve a cube, of constructible units, with double the volume of the original.

Volume of original cube alter            = L x b x h  = 1 x 1 x 1                    = 1

Volume of new cube alter required    = L x b x h  = 1.26 x 1.26 x 1.26       = 2

Volume to be removed from centre

of first cube                                     = 0.26 x 0.26 x 0.26 = 0.017576

This leaves the original cube as

hollow and with a wall thickness of    = 1 – 0.26 / 2           = 0.37  

Now form a new cube of material

from the original and the removed

internal material                               = L x b x h = 1.26 x 1.26 x 1.26       = 2

The final alter cube is therefore constructed out of no new material and so retains its original dimensional qualities and also increases in volume to achieve the Delphi Oracles requirements.

1.259 = the cube root of 2.

1.26 x 1.26  x 1.26 as constructible units creates a volume of 2.0003 which is within the tolerance range of the period in question between 428 BC to 347 BC

This solves the ‘Delian Problem and makes the alter of Apollo constructible at twice its original volume.

Ian K Whittaker

Website: https://sites.google.com/site/architecturearticles

Email: iankwhittaker@gmail.com

07/01/2014

14/10/2020

343 words over 1 pages