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