Some precisations about the "gold number"

PHI in electronics is still triggering a lot of people. Please find Rafaels precisations on the 'Gold Number'

WH

Salutations!

First I have to note that Phi is not the "Golden ratio", as you stated in your article. Exactly it is the "Gold-number", or "Number of Phydias", as it is present in the architecture of the Parthenon, in Athens.

Well, exactly "Phi" is the limit in the ratio of two consecutive numbers in the series of Fibonacci. It is a perfect "irrational" number, and as for "Pi" or "e", it only can be defined by a relation in a series of numbers. In this case it also represents the solution for the "Aurea proportio" in the division of a segment. It seems you think that Phi was firstly defined by Pacioli. In fact it was discussed by Euclid in his "Elements", as you can read in the following articles.
http://en.wikipedia.org/wiki/Euclid
http://www.pauloporta.com/Fotografia/Artigos/epropaurea1.htm

When you consider the ratio between three segments which satisfy that L=a+b, while L/a=a/b, the calculation brings us the formula X²-x-1=0, whose solution is x=(1+5^1/2)/2; but also x=(1-5^1/2)/2. As the second solution is not a real number, then we use the first for the operation in the physical world, and we call it "The Golden number".

Well, the fact is that the Fibonacci series is present in a lot of events, in the nature. It is the factor for the growth of a specimen, for the "Harmonische" (tempered) scale of music, for the aesthetic look of a sculpture... But you also find it in our own body. It is the ratio between the facial segments, between the length of the members... Considering this, and as electronics is only a matter of nature, it is normal than you'll find "Phi" in several effects. For example, this is the ratio in the growth of a Larsen signal, or the dumping factor in a natural attenuation. Actually, whenever you have any event where several (more than three) values of the same magnitude vary in the time, or in a ratio of parameters, in a re-entrant manner, you'll meet the Fibonacci series, hence, the Golden number in the progression.

As you can check, the value is not steady till you consider larger numbers in the series. For instance, as the series is caused by he addition of the precedent pair, the ratio between 2 and 1, 3 and 2, 5 and 3, 8 and 5, is far away from Phi. This is the cause that normal observations don't consider the fact, as the values are limited to a few observations, and the Maxwell theory provides an exact answer to the continuous environment (in the linear equations domain). But really the matter is that Phi and the Fractal geometry have a lot in common; thus, the "Golden ratio" is a simple approximation to the phenomena of "fractal growth".

Then don't ask for "applications" where "phi" is of use; better if you ask for the contrary, which phenomena don't fit the theory. Or if you like, ask for examples of how it is observable in the electronic field.

For instance, whenever you have to compute the impedance values for an attenuator, and you need match commercial values, sure than Fibonacci will come in your help.

Best regards
R. de Miguel

MORE BALLS PLEASE

Thanks for all your replies on the question of the Electronic Ball. There are more answers then I am able to publish on this blog. Below you will find an answer of Ron from New Zealand and Helmut from France. Ron uses LTSpice to get to the answer - Helmut the 'poor mans approach' Thanks guys!

wh

Hi Guys,

I have a solution to your resistor ball problem:

I flattened the ball, drew the schematic and loaded it into LTSpice circuit simulator, running on a laptop. I added a 1000 volt battery to the schematic and connected it to two diametrically opposite points. I ran the simulator in DC mode, and it produced a list of all the nodes with their voltages, and all the components (battery and resistors) with the currents through them. Knowing the applied voltage and the battery current, I calculated the resistance for question one.

Helmut from France wrote

Hi,

Here are some thoughts on the calculation of the equivalent resistance of two adjacent nodes, a subject that attracted less attention than the equivalent resistance of the two opposing nodes. Being more a practical kind of person I am not going to do a calculation that belongs in a theoretical treatise on electronics, instead I will try to find some poor man’s approach that will hopefully result in an acceptable approximation.

Some years ago a similar question was raised in Elektor about an infinite flat resistor grid and I was thinking that maybe it would be possible to reuse some of that. My starting point was (and still is) that the further a resistor is away from the two adjacent nodes, the lesser its contribution to the equivalent resistance will be. Intuitively, this value should be somewhere in between two limiting values, depending on the value 0 (short) or infinite (open) attributed to these far away resistors. So why not take the arithmetic average of the two? This approximation should get better when the number of jumps needed to get to the border nodes of the grid increases. These nodes will be given a value of open or shorted. The easiest case is the one with 0 jumps: short equals 0 ohm, open equals 1 Mohm, average equals 0.5 ohm. Averaging several approximate calculations of the equivalent resistance when the circuit gets more complicated gives this table (for the side of a pentagon):

Jumps	0	1	2
Short circuits	0.00	0.27 (3/11)	0.602
Open circuits	1.00	1.00	0.696
Average	0.50	0.64 (8/11)	0.649
Elektor measurement	0.65	0.65	0.65

With more jumps the calculation becomes complicated! But, with resistors of 2%, it seems that even with just one jump (still within reach of my mental calculation skills) we arrive at a good approximation.

Regards,

Helmut Müller

CORRECTION

Some of you noticed that the formule for PHI was not correct... that is correct!

The right formule must be ((1 + SQRT5)/2)

Thanks for the correction.

Kind regards,

WH

THE GOLDEN APPROXIMATION

Ricardo Oddone Geraldo Snel from Brasil, believes PHI is in the ball and also puts in the Golden Approximation - WH

According to the picture (of the resistorball - WH), Fibonacci numbers probably have something to do with the ball resistance, too.

What about soldered resistors building up a log-spiral? Will the resistance value converge at infinite length? Check log antennas...

Besides "Golden Ratio" and "Golden Equations" (e. g. exp (i*pi) + 1 = 0) we have at least one "Golden Approximation" in electronics:

2^10 ~ 10^3

It follows of course that 1 K ~ 1 k, but also (taking log base 10):

10 log 2 ~ 3 (dB).

EIN PHI POTI

ich bin Lukas und bin an der Entwicklung von einem neuen Equalizer beteiligt, der in dem High-End-Bereich in Studios zum Einsatz kommt.
Wir haben die Potentiometer mit Relais-und-Widerstandsreihen ausgetauscht und das Ganze hat mal hier für ziemlich aufsehen und Kleinfurore gesorgt, als ich die R-Reihen durchgerechnet hab, um eventuell das Konzept zu optimieren, und war heftig überrascht (kann mich noch an ein paar interessante Mails erinnern) als Projektleiter Stefan, der alles mit dem Lötkolben und dem Gehör macht, auf der Suche nach einer angenehm zu schaltenden Widerstandsreihe aus reiner Intuition heraus eine Reihe rausgearbeitet hat, die in den E24-Näherungswerten zum Schluss fast exakt den goldenen Schnitt reproduziert. Wenn man alle R-Werte zusammen- zieht, dann landet man bei einer durchschnittlichen Schrittbreite von fast exakt Phi, was mich doch sehr verwundert hat, da man im Audio- bereich höchstens mit nem log oder zumindest nem Quasilog unterwegs ist, aber Phi ist so "da draussen" - und das gar nicht erst mit Vorsatz, sondern aus reinem Gehör heraus...das war ne ziemliche...man kann fast schon wieder sagen Nicht-Überraschung.

Aber ja, echt schräg wie wenig Phi eigentlich in der Elektronik vorkommt...
sowas wie ein Phi-Poti wär vielleicht mal ne Sache ;)

YOU WANT ANOTHER CRAZY ANSWER?

I’ve been enjoying Elektor very much!

If you want another crazy answer. These problems (calculate the resistance of the ball) somehow always relate to PHI (the golden mean, (1+sqrt(5))/2). So simply taking your measurement I would say the answer is 5/(1+sqrt(5)) or 5/2*PHI (1.5450850). Now if this is the exact answer I will be pleasantly surprised.

-Scott Wurcer

THE ELECTRONIC BALL

First step to be taken is translating that complicated 3-D thing into something easier to handle and draw, ie we have to flatten the ball. Just have a look to the picture below, you can imagine each branch of our graph -the ball- is made of some ideally elastic resistor, then you stretch one of pentagons so that it can surround all the rest. Of course we loose neat angles and branch lengths you can see on the 3-D ball but this is the only way to flatten it. That neat arrangement of polygons can only exist in 3-D space. Nonetheless topology is the same, in other words connections between resistors in the branches is exactly the same. We still have 20 hexagons and 12 pentagons and connections, common branches and “neighbors” are just the same as in the real 3-D thing. Note that the 12th pentagon is “the biggest one” surrounding the hole graph.























This could be done even using one of the hexagons as graph center taking us to the same conclusions.

Next step is answering the question: Is resistance between any two diametrically opposite nodes the same? This cannot really taken for granted as is. However looking at the picture below we can see circular symmetry as 360/5=72° and this means that circuit and hence resistance between any couple of nodes A-A', B-B', C-C', D-D' and E-E' is the same.





















Of course which pentagon is used as center of the graph has no influence and then the same shall apply to any pentagon in the ball. This will include all the nodes in the ball and gives us opportunity to calculate resistance between any couple of nodes above having the same result.




To do that we shall connect a voltage generator between A and A' and calculate current flowing









This connection will unfortunately invalid the 5-folded symmetry seen above reducing to a simple vertical symmetry. However this 2-symmetry will allow us to reduce circuit to be considered.

In fact due this symmetry voltage on node 1 will be the same as that on node 1', hence voltage across branch 1-1' will be zero and no current will ever flow, so we can even remove this branch without modifying operation of the circuit. The same will naturally apply to 2-2', 3-3', 4-4', 5-5' and 6-6'.


















This is the new equivalent circuit






















We are close to dividing circuit into two halves, we just have to cope with the two branches 7-7' and 8-8'.

We first will replace resistor R with two parallel connected 2R resistors and than we will do the same as before, due symmetry current out of X and X' is the same, then current in the horizontal branch is zero, we then can remove it and finally cut our ball in two halves.













Then the remaining of our ball is
























So resistance seen from any two diametrically opposite nodes will be one half of resistance seen from voltage generator in the equivalent circuit below.





















Despite several tries to simplify the circuit above using common series, parallel and star/delta rules on electric nets I did not manage to achieve any further improvement apart from the obvious series connection of couples of resistors (not shown above to make clear belonging polygon).

So I had to go through the safe and hard way, ie solve the circuit using loop-currents method and Kirchhoff's laws. Let's have a look to diagram below where I replaced series resistors with their equivalent and depicted 12 loop currents and our “most wanted” voltage generator current with their (conventional) directions.























Each loop shall satisfy : sum of voltage generator(s) equals sum of voltage drop across resistors:

V=RI

for instance, looking at loop I1 we see we have no voltage generator ie V=0 and then given different directions of currents involved.

0=7RI1 – RI2 – RI3 – RI4 + 3RI13

This , repeated 13 times will give us a set 13 linear equations in 13 unknowns which can be conveniently described and solved using matrix algebra.



This is the R matrix describing the above circuit:

......................7.-1 . -1. -1. 0. 0. 0. 0. 0. 0. 0. 0. 3
....................-1. .6 . -1 ..0 -1 -1 0 0 0 0 0 0 0
....................-1 -1 5 -1 0 -1 -1 0 0 0 0 0 0
....................-1 . 0 -1 6 0 0 -1 -1 0 0 0 0 2
.....................0 -1 ..0 0 5 -1 0 0 -1 0 0 0 0
.....................0 -1 -1 0 -1 6 -1 0 -1 -1 0 0 0
R =............. 0 0 -1 -1 0 -1 6 -1 0 -1 -1 0 0
.....................0 0 0 -1 0 0 -1 5 0 0 -1 0 2
.....................0 0 0 0 -1 -1 0 0 6 -1 0 -1 0
.....................0 0 0 0 0 -1 -1 0 -1 5 -1 -1 0
.....................0 0 0 0 0 0 -1 -1 0 -1 6 -1 2
.....................0 0 0 0 0 0 0 0 -1 -1 -1 7 1
.....................3 0 0 2 0 0 0 2 0 0 2 1 10

then we need the known terms column ie a vector describing voltage generators connected – only one in loop 13 in our case- and a second vector containing our 13 unknowns.
We shall suppose voltage applied is 1V just to make calculations easier, value is not important at all being the circuit linear. We actually need only I13 but unfortunately this cannot be calculated on its own, we need to solve the whole system instead.


..................0 ............I1
..................0.............I2
..................0............ I3
..................0 ............I4
..................0 ............I5
..................0 ............I6
V =........... 0 .... I = .I7
..................0 ............I8
..................0 ............I9
..................0 ............I10
..................0 ............I11
..................0 ............I12
..................1 .............I13

Once we have these powerful instruments problem becomes logically simple, since

RI=V then I=R-1V

then we “only” have to invert R matrix to get R-1 even if this is logically simple requires quite a huge amount of calculations to be done, hence I got MatLab to do this. I will omit inverted matrix here since we only need one of its elements, in fact given the all-zeroes-and-one-one known term vector we only need R-1(13,13) to calculate I13. So we have:

R-1(13,13)=0.323529412


and then since V=1V and R=1Mohm

I13=R-1(13,13) V13 = 0.323529412 . 1 = 0.323529412 A

and hence resistance of one half of the ball shall be

rh = 1/ 0.323529412=3.0909 Mohm

One interesting consideration is that since all numbers in the R matrix are integers and operations involved in calculating its inverse are sums, differences, products and divisions result must be a rational number ie ratio of two integers. It actually looks like

rh =3 +1/11 =34/11 Mohm

Further investigations with MatLab showed that this is true down to its machine precision so we will assume it for true.
So finally taking into account the second half of the ball parallel connected we get


rball =17/11 Mohm = 1.5454 Mohm






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