An Introduction to Swannies Platinum Trirectangular Tetrahedron and its Prisms
The GOLDEN RIGHT-
The Golden Ratio 1,6180 (to 4 th dec.) is covered by Google. Presented here, however, is a somewhat different approach to it and also a Golden Right-
`The conventional Fibonachi sequence starts with 1 + 1 = 2, then 1 + 2 = 3 etc. always adding the last two values to give the next one. Here is shown part of the sequence and below it the ratios between a value and the previous one fluctuating up and down and settling down to reach, to the fourth decimal, the Golden Ratio:
1 ; 1 ; 2 ; 3 ; 5 ; 8 ; 13 ; 21 ; later 144 ; 233 ; 377
1,000 2,0000 1,5000 1,6666 1,6000 1,6250 1,6154 1,6081 1,6180
Staying with positive numbers, including fractions, you can however start with any two numbers, including your month and day of birth, any way round. Elsewhere I mentioned the 4, 7, 11 ... Eu du Cologne sequence! You always end up with a 1,6180 ratio between one and the previous one. First of all one might think of adding a length unit to the numbers and compare linear properties ( Yes there are such examples) . However with the triangles that now follows the numbers are used to represent area units and later on with the tetrahedron they represent volumetric units.
A sequence starting with 9, 16 and then 25 is of course the well known 3^2 + 4^2 = 5^2 used to illustrate the Theorem of Pythagoras. Partly copying the above exercise we get by addition:-
1,7778 ,5625 1,6400 1,6098 1,6181 1,6180
Now 9 +16 = 25 or 3^2 + 4^2 = 5 ^2 gives you a nicely numbered Pythagorean right angled triangle. However 453 + 733 = 1186 or 21,2838^2 + 27,0740^2 = 34,4384^2, the last three figures above gives you a (very near) Golden right-
To see how this Golden right-
9 + 9G = 9GG
9 + 14,5623 = 23,5620
3^2 +3,8160^2 = 4,8540^2
Still close to its 3^2 + 4^2 = 5^2 origen but now perfectly proportioned!
Below on the Right: The existing steel model (prisms mounted on tetrahedron) with cubes of the same volume as the prisms shown next to it, the number of cubic inches of the three smaller cubes adding up to the 216 cubic inches of the large cube below. The exercise that follows on the next page gives a very similar model, as it is closely related; just more compact, and lets say, better proportioned .