TRIBONACCI CONSTANT* = Swannies Constant PLATINUM RATIO* in Swannies Theorems
The Fibonacci Constant = 1.6180... = The Golden Ratio is defined as the constant ratio to wich adjacent Fibonacci numbers tend and is the well known Golden Ratio.
The Tribonacci Constant = 1,8393... = A No Name Ratio, up till now, likewise can be defined as the ratio to wich adjacent Tribonacci numbers tend, but was never given a Precious Metal name, most likely because Platinum and Silver had already been adopted for use elsewhere and the other precious shiny/hard ones were not well enough known. The (variable) Platinum named ratio of Economics has however indeed been partly duplicated and appears as a (fixed/constant) Platinum Ratio = 3,5616.... Swannie was unaware of these applications /too lazy to properly check and indeed also called the Tribonacci Constant 1,8393... the Platinum Ratio and now has no wish to revamp/change the history of his past efforts. Where ever Platinum Ratio is refer to on this site you may prefer to read Tribonacci Constant or to boost an ego read Swannies Ratio! He at least applied/devised this ratio to easy to understand, elementary maths models to be enjoyed by all as art!?* Note also that Fibonacci ('one' son of Nacci) never discovered "his" constant and that no Nacci named Tribonacci ('three sons of Nacci?) ever lived. Why then double honour or is it fourfold honour for the Naccis?))) * Except for Snub cubes.
Swannies Theorem with 27 + 64 + 125 = 216 (Plato's cubes) gives with the same kind of exercise as above but adding the last three values to give the next one, the following sequence with the ratios between one number and the previous one shown below, settling to the fourth decimal, to 1,8393.
27 ; 64 ; 125 ; 216 ; 405 ; 746 later 4631 ; 8516 ; 15665 ; 28812 ; 52993
2,37.. 1,95.. 1,72.. 1,87.. 1,84.. 1,8389. 1,8395 1,8393 1,8393
Will call it THE PLATINUM RATIO* = 1,8393 (to 4th dec.). Note : It is a ratio between Volumes, related to Swannies First Theorem.
Now Plato's 27 + 64 + 125 = 216, that is 3^3 + 4^3 + 5^3 = 6^3, the start of the sequence, gave nice Swannies models like the stainless steel one but 8516 + 15665 + 28812 = 52993, that is 20,4211^3 + 25,0213^3 + 30, 6566^3 = 37,5612^3, gives a near perfectly proportioned PLATINUM MODEL.
To see how these models compare first find the dimensions of a platinum tetrahedron of the near same size to the former. So start off also with 3^3 = 27 and multiply with the Platinum Ratio P = 1,8393 as follows:
27 + 27P + 27PP = 27PPP
27 + 49,6611 + 91,3417 = 168,0047
3^3 + 3,6757^3 + 4,5036^3 = 5,5179^3
It thus remained quite close to Plato's 3^3 + 4^3 + 5^3 = 6 ^3 start of the sequence but gives a more compact model with perfectly proportioned prisms. Again please note that these values represent Volumes – the volumes of Prisms stuck on the faces of a trirectangular tetrahedron in accordance with Swannies First Theorem. Also, the values 3 or 3,6757 etc does not represent any linear dimension of either a tetrahedron or a pyramid, they have to be calculated by the same method shown on the Equations Page as follows:
You have 27 ; 49,6611 ; 91,3417 ; 168,0047. Let the sides of the tetrahedron meeting at right angles (at the vertex) be a, b and c, from short to long. (See diagram below if you wish) Therefore the areas of the rectangular triangles are equal to ½ab, ½ac and ½bc. Now square these areas (De Gua and Swannie) and set each equal to the numerical value of its prism's volume in accordance with the theorem:
¼ a^2 b^2 = 27 (1)
¼ a^2 c^2 = 49,6611 (2)
¼ b^2 c^2 = 91,3417 (3)
To simplify take the square roots on both sides. Then :-
½ a c = 7,0470 (2)
½ b c = 9,5573 (3)
c = 1,3562 b
Substitute c in (3) ½ b (1,3562 b) = 9,5573
b^2 = 14,0942
b = 3,7542
Substitute b in (3) ½ (3,7542) c = 9,5573
c = 5,0915
Substitute b in (1) ½ a (3,7542) = 5,1962
a = 2,7682
Using Pythagoras's side^2 + side^2 = hypotenuse ^2 and a, b and c, the sides of the base
of the tetrahedron for:-
a and c to be e = 5,7954
and for b and c to be f = 6,3259
Below on the right: Diagram in black of an opened up and labbled Platinum trirectangular tetrahedron. The red represents the near same dimensions of the Plato's steel model's tetrahedron.
Below on the left:
Dimensions (to second decimal only to simplify and as higher accuracy can in any case not be attained in table top models) of above tetrahedron and related Swannies prisms:
Sides: a = 2, 77'' ; b = 3,75'' ; c = 5,09'' ; d = 4,66'' ; e = 5,80'' ; f = 6,3259''
(Note: (i) b/a = c/b = 1,3562 (ii) The shape of the smallest triangle C: 'Up' a, b and d by
multiplying each with 1,0837 and get 3'', ≈ 4'' and ≈5'' -
Faces: A = 9,56 inch^2 ; B = 7,05 inch^2 ; C = 5,92 inch^2 ; D = 12,96 inch^2
Prisms (not shown): on A = 91,34 inch^3 ; on B = 49,66 inch^3 ; on C = 27 inch^3 ;
and on D = 168,00 inch^3 with ratio between them 1,8393 – the Platinum Ratio
So all in all there is not much difference in shape between the tetrahedron and the model of Stainless Steel and its more proportional relative the Platinum one.
BELOW : A VARIATION ON THE FIRST THEOREM
Above on the left: Prisms: Small base, more height; larger base, less height; same volume.
Above on the right: Level, a property of the Variation? Contents (ask granny) won't spill on release of handle?
A VARIATION on First Theorem. See model above.
The Volumes of the prisms on the right angled faces of the trirectangular tetrahedron are made equal as follows:-
For the four prisms the volumes relate as follows: ½ abc + ½ acb + ½ bca = 3/2 abc = the volume of the base
√((½ ab)^2 + (½bc)^2 + (½ ac)^2) which times height = 3/2 abc its above defined volume. Simplifying this gives:
height of base prism= 3abc/√(a^2b^2 + a^2c^2 + b^2c^2)
Example: Building a model with dimensions a = 3cm , b = 4cm and c = 5cm , the model shown above, you have (with Pythagoras) all you need as defined underlined and shown above except the height of the base prism which calculated with the given equation gives 6,49cm to the second decimal.
Likewise to this example there can there be a = b = c and more variations.
Q: Lets say you have a solid (aluminium) model of a "Variation" as dealt with above. Guess or find its centre of gravity. Any easy/simple way of doing it?
Herewith A DETAILED REHASH OF THE above VARIATION -
because whilst playing with the layout made me wonder if some of it (of repetitive nature) can not perhaps be set to music.
About models of PRISMS on the faces of Trirectangular Tetrahedrons with right angle sides a,b and c and heights, say in cm, to the tune of what follows:
For First Theorem For its VARIATION
Area(cm^2) x Height (cm)* Area(cm^2) x Height (cm)
Volume say the right angled prism with smallest base: 1/2ab x 1/2ab 1/2ab x c
Volume say the right angled prism with biggest base: 1/2bc x 1/2bc 1/2bc x a
Volume of inbetween based right angled prism: 1/2ac x 1/2ac 1/2ac x b
Volume of base prism = above added: 1/4(a^2b^2 + b^2c^2 + a^2c^2) 3/2 abc
( De Gua and First Theorem) (Also added as defined)
Area of base of base prism = √ (of its above volume) (De Gua and First Theorem reversed)
= 1/2√(a^2b^2 + b^2c^2 +a^2c^2)
But for the Variation the Area above x a Height = Volume of the base prism as defined = 3/2abc
1/2√(a^2b^2 + b^2c^2 +a^2c^2) x a Height = 3/2abc
Therefore Height of base prism = 3/2abc / (1/2√(a^2b^2 + b^2c^2 +a^2c^2))
= 3abc / √(a^2b^2 + b^2c^2 +a^2c^2)
Building a Variation Model with a,b and c known, you still need to use Pythagoras to find the sides of the base prism and above equation to find its height.
Note: (i)*According to the First Theorem only the numerical values of the product is (obviously) used together with a linear, cm in this case, unit for height.
(ii) A special set of values is of course when a = b = c, which simplifies all above in an interesting way, should you care to check .
Q: Lets say you have a solid model of a "Variation" as dealt with above. Guess or find its centre of gravity. Any easy/simple way of doing it, even perhaps only for (ii) above?
Need to Defend the Maths Fairy: The Maths Fairy making a cm dissappear in the First Theorem (see page:Pyth.&box -