Swannies Theorems, Art in Maths and a follow-up to Pythagoras, De Tinseau and De Gua De Malves.

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TRIBONACCI CONSTANT* =  Swannies Constant PLATINUM RATIO* in  Swannies Theorems

(((* NOTE:- Precious Metals Duplications? : After designing his Golden Rectangular Triangle with the known 1,6180...  Swannie wished to do a similar ratio excercise with a trirectangular tetrahedron, the obvious being adding three figures for the right angle faces to give a fourth for the base, discovering 1,8393... , resulting in what he called his Platinum Tetrahedron. Below and elsewhere thus inadvertantley named THE PLATINUM* RATIO and then related to SWANNIES PLATINUM TRIRECTANGULAR TETRAHEDRON and its PRISMS as combined in a Swannies First Theorem model also mentioned.  Only later he learned about Tribonacci and sorted  it all out as here follows:     20/06/2014:
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 b =  5,1962        (1)
                                                                                                      ½ a c =  7,0470        (2)
                                                                                                      ½ b c =  9,5573        (3)
                                                                                (2) ÷ (1)          c ÷ b =  7,0470 ÷ 5,1962
                                                                                                             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 b is found to be                                                                      d = 4,6644
                                         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:
 A Platinum trirectangular tetrahedron lying on its base on a diagram showing its opened up rectangular faces (light blue). The dimensions of the model is in inches because inches gives a convenient sized diagram and model. All the dimensions are listed below.

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'' -  the well known shape re-appears!   

 A = 9,56 inch^2   ;  B = 7,05 inch^2  ;  C = 5,92 inch^2   ;  D = 12,96 inch^2

(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.

Wishfull thinking:
A Swannies First Theorem (real) Platinum model (also in inch units) to be displayed  next to the Stainless Steel one  seems to remain a dream to come true; unless a platinum (so few tiny units thick ) coated one is OK  and affordable!

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:- If the right angled sides are a, b and c let the volumes be ½ ab times c for the height etc.
For the four prisms the volumes relate as follows: ½ abc + ½ acb + ½ bca  =  3/2 abc  =  the volume of the base
   From de Gua and Swannies First Theorem the volume of a base prism is also (½ ab)^2 + (½bc)^2  + (½ ac)^2 as derived from squaring the area of the base which is of course the square root of it:
√((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 - The tie up with De Tinseau/.. ) has met with unfavourable comment. Maths and its practisioners, being rather snobish regarding the factuality/accuracy of their subject sneer at suchlike practice. So on request I'll also explain to you why I am so indebted to this fairy and how she came to my rescue when the Maths Devil stymied my progress with this theorem! For now I'll only leave you with the statement: "Units may be engaged to numbers but they are not married to them". The essay is of a "Much ado about little" kind.

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