Thursday, November 5, 2009
Elektor Foundation Award  A woman's business
Considering women and engineering don't go together too well even in Western Europe, how does it compare to Turkey? 'Our country does not exactly stimulate women to choose engineering', says Fatma Zeynep. However, she does see great potential in the allurement of the possibilities and job opportunities involved. At the same time we see an increase of companies focusing on engineering and electronics.
Fatma Zeynep is an example of how education, networking and fervour can fuel a life of studying and working. Shortly after graduating she was offered a job at the Nuclear Electronics Institute of Ankara, a position which made her a respected colleague, researcher and speaker both home and abroad. Additionally she has always been active in developing new products for education. Based on the Motorola 6800 she developed the first programming set, to be followed by many more. In 1998 she started her own company in the area of computers, electronics and assembly (BETI). Thanks to her own knowledge and that of her students she has always been able to choose the best of the best when it comes to hardware. BETI is still being approached for the special assembly of computer systems; 'on demand' in other words. She also still teaches at the university of Ankara. Education, interest, network  three 'drives' which brought Fatma Zeynep Köksal into the malepopulated world of electronics. And she occupies a special place.
Along with her company BETI, her lectureship at the Ankara University she has also started a company focused on Nuclear Electronics Measurement (Nemo) and we will certainly be hearing more of her. Her enthusiasm for electronics and the way she connects this with her students is definitely Worth an Award.
WH
Wednesday, October 21, 2009
Elektor Award – Will this robot rescue the Flemish technical education?
Yes! according to the creator of the robot. Bart Huyskens has been busy developing a number of robots which can be used for educational purposes, and he's noticed a significant increase of interest amongst students because of it.
The amount of students has doubled in schools where robots are being used in technology lessons, and the classrooms are filling up again.
This is good news for the Flemish Ministry of Education. A few years ago they started putting up Regional Technical Centres where (with government funding) education and companies work together on new initiatives, which are meant to attract more students towards technical education. But its not only for students; teachers can get extra afterschool schooling on new technologies.
Bart has his hands full developing the robots. With the help of sensors they can speak to students, follow a pattern across classrooms, play and dance to music all of which keeps the technohungry crowd pleased. Thanks to Technical Centre funding there are now 14 robots available for use to the students of St. Jozefinstituut in Schoten, and two more will start touring with the TechnoTrailer.
The students will be able to program the robots for different kinds of utilities. The robot itself runs on Eblock technology and speaks Flowcode, but is fluent in Flemish as well.
Bart's enthusiasm has played a big part in the success of the robots. In the past couple of years he has developed the concept, managed to sell it and make a fullfledged product. More importantly: he has managed to interest new groups of students towards electronics, and that's certainly Worth an Award!
Bart Huyskens is hereby nominated for the Elektor Foundation Award 2009.
The granting of this international award will take place during Elektor Live! on the 21 of November in Eindhoven, Holland.
For more information:
 on the Technical Centres can be found at www.rtcantwerpen.be (NL and Flamish)
 on the Elektor Foundation Award: www.elektorfoundation.org
Thursday, October 15, 2009
Radio to the Rescue
Emergency services (such as the police force and fire brigade) know the vital importance of good communication. In order to avoid chaos and unnecessary casualties, it is essential to know where help is needed and what kind of help is needed. This is often easier said than done, as can be seen only too clearly from the situation with the C2000 emergency services radio system in the Netherlands. It works well in theory, but in practice it’s a different story. Emergency aid in event of a major disaster often requires creativity and unusual actions.
In 1953, the North Sea broke through the dikes and submerged large portions of the southern Netherlands. Nothing was spared: people, animals, and buildings all fell victim to the merciless flood waters. A certain Mr Hossfeld was caught in the middle of this catastrophe. Taking his son with him, he plunged into the icecold water, and fortunately they managed to swim to a house where they could enter through an open window and climb onto the roof. The next day they were brought to safety. After they reached dry land, they found that the emergency services were desperately short of communication equipment. Everything had literally been swept away, and the town of Zierikzee was totally cut off.
Mr Hossfeld (now 83 years old) did what he could and must do: using a few radio valves (EL3, EL6 and 807) and some coils made by winding wire around a bottle, he put together a transmitter that could deliver 10 watts of power to a 15metre longwire antenna. This was enough to make contact with the outside world (and in a manner of speaking, it was the spiritual ancestor of the C2000 system). For five days and nights, a team of four people constantly manned the PAoZRK transmitter to coordinate assistance activities for Zierikzee.
Radio amateurs such as Mr Hossfeld played a vital role in the initial hours and days of the 1953 floods. Many lives were saved as a result of their efforts.
Our objective with the Elektor Foundation Award is to pay tribute to events such as these: people who managed to make a difference with their knowledge and efforts. Mr Hossfeld is one of the candidates for this award.
More information on the Elektor Foundation Award can be found at http://www.elektorfoundation.org/.
Thursday, October 8, 2009
Old Measurements Never Die
It’s a hygrometer. Nothing special, you might say, but the measuring method is pleasantly interesting. The instrument has two thermometers: one dry and the other wet. The difference in their temperatures is a measure of the relative humidity, which you can read from a table. If only everything were this simple.
I was reminded of this instrument by our preparations for a weather station project, which have been underway for a while now, and our work on a new design for the CO2 meter published in the January 2008 issue. We plan to upgrade the original design with a temperature sensor and a humidity sensor. Naturally, we intend to use a readymade humidity sensor for this, but the old instrument still has a certain charm, perhaps due to its simplicity or the selfevident operating principle. It seems like you learn something from this instrument. The operating principle is somewhat similar to that of a differential amplifier.
Who knows, maybe one of you will put together an electronic version? Or am I simply making things unnecessarily complicated again?
Friday, October 2, 2009
Worth an Award
It happened during an exhibition in
Now, it is not customary (or indeed our expectation) that visitors to Elektor booths at exhibitions hug the staff members, but this man had a reason. He told a story that anyone from
That particular visitor is by no means the only person expressing his feelings in relation to Elektor. We receive letters, visits and emails from people from all over the world who feel a close connection with the magazine. Sometimes with ideas, projects and suggestions; sometimes with criticisms – these are all indications of connectivity.
It sometimes takes my words away. We make, to the best of our ability, a magazine about electronics and surreptitiously the magazine does more than you suspect. People become fascinated with electronics and get busy, begin to study, make discoveries, have their work published, start their own manufacturing company or become an instructor. What interests has this magazine created and what things have come about in the nearly 35 years of its existence?
To give some substance to this curiosity we decided to launch an International Award. This Award is for Elektor readers who have in one way or another accomplished something special; an extraordinary discovery, a piece of fundamental research, a component or new circuit, a new design or application....
Send us with your stories. Who deserves and Award and why? Come with the anecdotes, bring the fascination to life! The Award will be presented on 21 November 2009 during Elektor Live! Elektor Live is an electronics handson event and will be held in the old Philips exhibition building in
The Award is an initiative of the Elektor Foundation (www.elektor.com/foundation). On this page you can find more information about the categories of the Award and the objectives of the Elektor Foundation.
At award2009@elektor.com we look forward to receiving your suggestions for candidates, or a good story or an exciting bit of history.
Wisse Hettinga
WHYTRICITY
It seems like everything that's wireless these days is called wisomething, including witricity. If I understand right, this stands for “wireless electricity”, which means wireless power transfer. Designers everywhere, from MIT to Intel, are busily devising methods to eliminate separate power cables for individual devices.
This makes me wonder: why wi?
In the first place, each of these cables is usually connected to a transformer in an AC adapter. Transformers are not exactly leadingedge technology. The earliest descriptions date back to Faraday in the early decades of the 19th century. If we look more closely at how a transformer works, we usually see a primary winding and a secondary winding fitted on a magnetic core. That’s wireless power transfer.
Now let's look at all the modern approaches to obtaining wireless power transfer. Basically, they all amount to reworking the transformer principle, with a primary coil and secondary coil coupled by magnetic induction. The only difference is that people are experimenting with different frequencies and using resonant coils. The last part also sounds a bit familiar – isn’t that how radio broadcasting works, with electromagnetic waves? And let's not forget Tesla, whose enormous Wardenclyffe project was intended to provide wireless power transmission.
So we already have wireless power, but it’s not enough to meet our needs, and furthermore it’s not very efficient. This brings me to my question for you this week: do you see a future for wireless power, or should we start thinking about new forms of power distribution, and what would they be?
WH
your view/repsons:
Ron Wesselman
Monday, September 28, 2009
ON and OFF
Do you think
I’m writing this during the
If you know an interesting way to turn something on or off (people, animals, cars, radios, TVs, etc.), you’re more than welcome to send your ideas to theelectronicball@elektor.com.
http://robotics.me.es.osakau.ac.jp/~taniguti/indexeng.html
I reflashed a router to a linux server. He is doing my webserver, mails, backup server tetc.
He knows
 the time (from the internet)
 day and night to switch on the lights
See the developments http://edimax.geens.nl/001.hardware/020.Patches/111.lichtschakelaar/
P.v.Geens (Holland)
I would think that if Edison used a switch in his experiments and/or apparatus, it would have been a knife switch. In the early part of the century when Edison done most of his work, the knife switch was the most popular type available to both the experimenter and to industry. Therefore, it stands to reason that Edison too would have used this type of switch.

Best Regards,
Technical Support Engineer
2 Mount Royal
Marlborough, MA 01752
(article can be dispatched on demand  thanks, WH)
Friday, September 25, 2009
THE IT GUYS AT AU BON PAIN
In general IT departments don't have very good reputation; when you need them thei're not there and when you don't need them they doing upgrading and copying stuff keeping you from working.
Not with Darrel (r) and George (l): they are always there! They run a kind of DIY IT department in a Au bon Pain, just opposite the big Christian Science Church on Massachusetts Ave, Boston. Darrel knows everything about notebooks  he's got a dozen I think  he tears them all down to the real bits and puts them back together and repairs them. George is the expert in asking how you're doing and giving you permission to leave  don't forget to ask.
George and Darrel are more the hub of a small community of people that like to see each other every morning, checking everything is OK. They have questions on websites, computers, mails, the weather forecast etc. and is an nice example of how IT can bring together people.
Straight down the road is the Hynes Convention centre. The home base for the ESC show. That also draws a crowd. Students and professionals on electronics gather discussing the latest and greatest on embedded electronics. Lots of them also visited the Elektor booth to pick up the latest issue.
The visitors on the ESC and the earlymorningIfirstneedacoffeeDIYITdepartment from George and Darrel were excellent!
WH
Wednesday, September 16, 2009
THE ELECTRONIC BALL QUESTIONS
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.
I repeated this with modified diagrams, to obtain the answers to questions two and three.
Results: I calculated Q1 answer as 1000Volts/0.00647059Amps = 1,545,454 ohms, which agrees with your answer of 1.545 Megohm.
I calculated Q2 answer as 1000/0.0015412 = 648,845 ohms, which agrees with your answer of 0.65 Megohm.
I calculated Q3 answer as 1000/0.00149481 = 668,9891 ohms, which agrees with your answer of 0.67 Megohm.
I think I could claim to have a greater accuracy than other solutions.
I attach the following files:
(1) Three asc files, one for each question, (these are LTSpice simulator files) just in case you have a PC with LTSpice or a compatible circuit simulator, and are able to run them.
(2) Three schematic diagram (pdf) files, one for each question. These are essentially the same except for the measurement connection points.
(3) Three “Results” files (jpg), being a small portion of the simulator dc mode readouts. The only results of importance here are the very last entry, "I(V1)" on each sheet, which is the battery current. The applied voltage is 1000 volts, so the resistance is obtained from these two figures.
(4) A plot (pdf) of the “Transient” simulation for Q1 only. This is not necessary for the solution, but it does give an answer without any manual calculation. In this simulation, LTSpice plots the applied voltage and the battery current; it also calculates and plots the resistance. So you get three lines on the graph, blue for the voltage with a scale second from the left, green for current with a scale on the right, and red for resistance, with a scale on the far left. And as you can see, the resistance is midway between 1.54545 Megohm and 1.54546 Megohm, confirming the manual calculation. (Being dc, the graph doesn’t look much, but it also works for ac and transients.)
Regards
Ron Badman (ZL1AI)
Friday, August 14, 2009
Some precisations about 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 "Goldnumber", 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 X^{2}x1=0, whose solution is x=(1+5^{1/2})/2; but also x=(15^{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 reentrant 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
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
Friday, July 10, 2009
CORRECTION
The right formule must be ((1 + SQRT5)/2)
Thanks for the correction.
Kind regards,
WH
Thursday, July 9, 2009
THE GOLDEN APPROXIMATION
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 logspiral? 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
Wir haben die Potentiometer mit RelaisundWiderstandsreihen ausgetauscht und das Ganze hat mal hier für ziemlich aufsehen und Kleinfurore gesorgt, als ich die RReihen 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 E24Näherungswerten zum Schluss fast exakt den goldenen Schnitt reproduziert. Wenn man alle RWerte 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 NichtÜberraschung.
Aber ja, echt schräg wie wenig Phi eigentlich in der Elektronik vorkommt...
sowas wie ein PhiPoti wär vielleicht mal ne Sache ;)
YOU WANT ANOTHER CRAZY ANSWER?
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
Monday, June 29, 2009
THE ELECTRONIC BALL
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 AA', BB', CC', DD' and EE' 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 5folded symmetry seen above reducing to a simple vertical symmetry. However this 2symmetry 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 11' 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 22', 33', 44', 55' and 66'.
This is the new equivalent circuit
We are close to dividing circuit into two halves, we just have to cope with the two branches 77' and 88'.
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 loopcurrents 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=R1V
then we “only” have to invert R matrix to get R1 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 allzeroesandoneone known term vector we only need R1(13,13) to calculate I13. So we have:
R1(13,13)=0.323529412
and then since V=1V and R=1Mohm
I13=R1(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
.
Tuesday, June 23, 2009
UFF!
Nehmen wir einmal an, wir legen entlang jeder Naht eines Fußballes einen Widerstand von einem Megaohm und löten alle Verbindungspunkte zusammen. Wie hoch ist nun der Widerstand zwischen zwei gegenüberliegenden Punkten?"
Obwohl ich leider erst sehr spät auf dieses Rätsel stieß, reizte es mich natürlich, da mitzurätseln. Ein besonderer Anreiz ist, dass im ElektorLabor anscheinend auch noch gerätselt wird und die Anwort nicht schon vorher existiert.
Für einen HobbyElektroniker wie mich ist es eher unwahrscheinlich, dass von einem einzigen Widerstandswert gleich 90 Stück paratliegen. Somit war  auch im Hinblick auf eine folgende Berechnung  die Suche nach einer Vereinfachung des Problems geboten.
Symmetriebetrachtungen
Der Widerstand soll zwischen zwei gegenüberliegenden Punkten  also Ecken  gemessen werden. An jeder Ecke treffen sich zwei Sechsecke und ein Fünfeck. Somit gibt es für die  willkürlich ausgewählte  Ecke eine Linie, um die das Gebiet spiegelsymmetrisch ist.
SymmetrieLinie
Diese Linie bildet insgesamt einen Kreis um den Ball herum, der diesen in zwei gleich Hälften teilt. Somit genügt es, den Widerstand einer Hälfte zu ermitteln und diesen dann  wegen der Parallelschaltung mit der anderen Hälfte  im Wert zu halbieren.
Weitere Vereinfachungen Zeichnet man besagten Kreis in eine Skizze ein, so ergibt sich folgendes Bild.
SymmetrieKreis teilt den Ball in zwei Hälften
Der Teil außerhalb des Kreises gehört zur Rückseite des Balls; somit kann diese Skizze natürlich nicht perspektivisch richtig sein, was für die folgende Betrachtung aber auch nich sein muss. Auch die 1MOhmWiderstände sind hier nicht eingezeichnet; jede Kante soll einen Widerstand darstellen.
Der Kreis spaltet zwei Widerstände, die an den AnschlussEcken liegen, der Länge nach. Einige weitere Widerstände werden quer durchgeschnitten. Da beide BallHälften exakt gleich sind, besteht keinerlei Grund dafür, dass von einer Hälte zur anderen ein Strom fließen könnte. Die quer durchgeschnittenen Widerstände können also einfach weggelassen werden, und die beiden längs gespaltenen Widerstände werden durch doppelt so hohe ersetzt, also 2MOhm.
An dieser Stelle habe ich dann den Lötkolben angeheizt und einen Ball gebaut, der nur auf einer Seite aus Widerständen besteht und ansonsten  quasi als Dummy  nur aus einfachen Drähten. An den Trennstellen wurden die Drähte unterbrochen und mit Schrumpfschlauch wieder verbunden, ohne dass eine elektrische Verbindung entsteht.
Mit Spannung den Widerstand gemessen  das Messgerät zittert mit und zeigt Werte zwischen 3.03 und 3.15 MOhm, das ist im Mittel 3.08 MOhm. Wird berücksichtigt, dass der Widerstand für den kompletten Ball zu ermitteln ist, so ergibt sich ein Wert von 1.54 MOhm. Das bestätigt mit dem im Rätsel schon angegebenen ElektorMesswert 1.545 MOhm meine Überlegungen bis jetzt ja recht gut! :)
Jetzt kann man im nächsten Schritt gleich diese Parallelschaltung berücksichtigen: Die zwei gespaltenen Widerstände bekommen wieder 1 MOhm und alle anderen 500 kOhm.
Einige Widerstände entfallen, das bleibt übrig
Jetzt gibt es einige Ecken, an denen sich nicht mehr 3, sondern nur noch 2 Widerstände treffen. Diese können daher jeweils zu einem Widerstand zusammengefasst werden, also 1 MOhm und an zwei Stellen 1.5 MOhm. An allen anderen Stellen bleibt es bei 500 kOhm.
Strom für eine Ecke und Gesamtstrom
Nochmals Symmetrie
Das verbleibende Gebilde aus 0.5 MOhm, 1 MOhm und 1.5 MOhmWiderständen ist drehsymmetrisch. Es ist also nicht notwendig, Berechnungen für alle Ecken anzustellen, sondern es reicht eine Hälfte, z.B. nach Unterteilung an der eingezeichneten Grenzlinie.
Der Gesamtstrom durch das Maschenwerk ist die Summe der Ströme, die diese GrenzLinie überschreiten. Natürlich könnte man zur Berechnung des Gesamtstroms auch andere Grenzen ziehen, z.B. direkt unter der Ecke U und die Ströme durch die beiden von dort ausgehenden Widerstände zusammenzählen. Aber der skizzierte Weg passt besser zu der Symmetriebetrachtung. ;)
Spannung an den Ecken des Fußballs
Jetzt hat sich die Anzahl der Ecken, über die Berechnungen auszuführen sind, auf 11 reduziert. Für die Berechnung benutze ich die Gesetzmäßigkeit, dass nirgends Strom aus dem Nichts entstehen kann, also der von einer Ecke ausgehende Strom immer Null ist. Wird an das Gesamte Gebilde oben eine Spannung +U angelegt und  der Symmetrie wegen  unten eine Spannung U, dann lassen sich die Spannungen an den übrigen Ecken mit 'A' bis 'K' bezeichnen und bereichnen.
Es werden 11 Gleichungen mit 11 Unbekannten aufgestellt. Ein Beispiel ist in obige Skizze eingezeichnet: Die von Ecke D ausgehenden Ströme sind (DU)/1.5MOhm + (DK)/1MOhm + (DC)/0.5MOhm (alle unbezeichneten Linien sind 0.5MOhmWiderstände). Diese Summe muss 0 sein, woraus sich die Gleichung 11D2U3K6C=0 ergibt. Setzt man immer wenn U vorkommt, diesen Teil auf die rechte Seite des Gleichheitszeichens, so lautet dieses Beispiel 11D3K6C=2U. Entsprechende Gleichungen werden für alle 11 betrachteten Ecken aufgestellt und in eine Matrix eingesetzt.
Matrix der 11 Gleichungen mit 11 Unbekannten
Die Gleichung für die Ecke D bzw. den Spannungswert D ist hier in der 4. Zeile zu sehen: Jeweils in den Spalten C, D, K und U sind die Koeffizienzen 6, 11, 3 und 2 eingetragen.
Lösen der 11 Gleichungen per MatrixDiagonalisierung
Gibt man diese Tabelle in ein Tabellenkalkulationsprogramm ein und beherrscht dieses MatrixBerechnugen, dann kann man durch Invertieren der Matrix A1 bis K11 und Multiplikation mit dem Vektor U den Ergebnisvektor errechnen. Es ergibt sich:
A = U * 0,5735294117647060
B = U * 0,3382352941176470
C = U * 0,2647058823529410
D = U * 0,3382352941176470
E = U * 0,1911764705882350
F = U * 0,1176470588235290
G = U * 0,1764705882352940
H = U * 0,0735294117647059
I = U * 0,1176470588235290
J = U * 0,0147058823529412
K = U * 0,0441176470588236
und mit der Formel
R = 1MOhm * U / (E + 2F + 2H + 2J + K)
ist schließlich der gesuchte Widerstand
R = 1,54545454545455 MOhm
Lösen der 11 Gleichungen  ganz genau!
Dieses Resultat sieht ja schon sehr genau aus, und man kann erahnen, dass es sich um einen periodischen Dezimalbruch ..545454.. handelt.
Nachdem aber in dem Rätsel die Frage nach der genauesten Lösung gestellt ist, möchte ich im folgenden eine Lösung finden, die möglichst nicht an Genauigkeit noch überboten werden kann. Also machte ich mich daran, die MatrixRechnung von Hand durchzuführen, und zwar so, dass nur mit ganzen Zahlen gerechnet wird, also Brüche zunächst noch nicht ausdividiert werden.
Das wurde doch recht mühselig, weswegen hier nur ein paar Zwischenergebnisse gezeigt werden sollen.
Umformen, damit rechts oben nur Nullen
Jetzt rechts oben nur Nullen
Es wäre einfacher, wenn in der Diagonalen nur Einsen stünden, doch ohne Brüche geht das nicht.
Umformen, damit links unten nur Nullen
Jetzt kommen schon gemein große Zahlen vor.
Geschafft!
Die Zahl 68 spielt hier offensichtlich eine besondere Rolle. Teilweise könnte man noch kürzen, aber so ist es schöner. Jetzt können die Lösungen direkt abgelesen werden: 68 A = 39 U, 68 B = 23 U u.s.w.
Jetzt diese Ergebnisse in die Formel
R = 1MOhm * U / (E + 2F + 2H + 2J + K)
eingesetzt, ergibt den Widerstandswert
R = 1MOhm * 68 / (13 + 2*8 + 2*5 + 2*1 + 3) = 1MOhm * 68 / 44
und, so weit wie möglich gekürzt,
R = 1MOhm * 17 / 11
Uff!
Bernhard Foltz
.
Thursday, June 4, 2009
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