Challenge Questions

Ooops, I see that all my percentages are a factor of 100 greater than what they should be.

This means my last attempt is in fact accurate to 0.0000085% - could this be correct or have I made another silly mistake here ???

Looks like you nailed it down pretty good...

Well, good job.

GA, I think - I'll add a significant figure to your error calculation (just to show that I checked): about 8.49x10^-6%.

On reading the challenge, I did a quick estimate of the number of different values we might expect from a combination of up to 18 resistors; it's certainly many millions. I would also expect the organisation of the values to be chaotic, with large numbers of results in some regions and significant holes in between. So a proportionate accuracy that is sub parts in 10⁷ should not be such a great surprise - although may be a surprise that it lies on one of the more 'natural' constructional paths.
I think it would be very difficult to perform a systematic search, because most of the structures lie in branches that behave far less regularly than the one that you investigated. However, just from curiosity, I'll look at a couple of the more complicated (but still well-behaved) topologies to see if I can find one that comes close.

Gets my vote.

Did you solve this based on the expansion or did you try various combinations to home in on the correct value?

I think I managed to simply home in.

My methodology was, firstly knowing that 3 and 1/7 is larger than pi, I knew that I had to somehow decrease the value of one of what I call the "parallel 7" resistors. It doesn't take much trial and error to get to where I got.

The big question is : Is there a more accurate combination of the resistors - I'm inclined to think there is, but I haven't found it yet. Back to the drawing board!!!

Great answer, not only because it is remarkably close, but also because it is remarkably well-presented.

Good Answer! I do believe this is the closest to Pi that can be obtained with 18 resistors. However, you can get the same value with fewer resistors. How many less? Maybe that should be the next half of this challenge. How few resistors can be used to get a resistance of 355/113?

Thank you all who gave a GA, but Jim35848 has posed a great follow-on question.

I have looked at this for a lomg time and I cannot find how you can get a value of 355/113 with less than 18 resistors.

Anyone?

Arithmetic is certainly a pig, but I churned through it for the hell of it and yes you are correct. It does produce the fraction 355/113.

Well done with 16 resistors and GA from me.

Did you use MathLab or some other software to come up with this circuit?

No, I didn't use any software - except SPICE* to check that I hadn't made a glaring error.

I confined my investigations to "A" shapes plus series resistors at the top. I also restricted the search to regions where the presence of the cross-bar made a relatively small difference to the total resistance**. Add a bit of number theory, and it didn't take all that long.

To my mind, the scary bit is that even 14 resistors allows a large number of possible cross-connected topologies, and these are capable of providing very fractions with remarkably long demoninators (sic). So I would not be surprised if someone finds even smaller networks that achieve 355/113. I would also anticipate that there exist <=18-resistor configurations that are more accurate [possibilities might include 312689/99532 (error~9x10^-10%), 1146408/364913 (error~5x10^-11%) or even 5419351/1725033 (error~7x10^-13%)]; on the other hand, without some tools that I don't know about*** to narrow the search, I would expect a very long run-time for any programme to generate these improved networks.

*the free version of SIMetrix (default settings are OK for these moderate accuracies).
**Ideally, I was searching for a region where we could insert the next terms of the continued fraction without excessive difficulty, but it turns out that your original arrangement is still less sensitive than my arrangement (even accounting for the extra two resistors that are already there)
***I mean something equivalent to what Stern-Brocot does for generating fractional approximations

Hello Maths_Physics_Maniac:

Your name says it all!........If you cannot solve this then no one can!

Your answer is very clever.........And knowing only basic Electronics, I am still checking it. But looks ok to me.

Take care and have a wonderful holiday!...............

Very good answer. And very well presented as stated.

Agree!

If the resistors are carbon, you can file them down to get ∏.

Mildly humorous, but no G.A. from me because the terms of the question effectively exclude carbon. Plus, when I last tried adjusting a solid carbon resistor it simply fell apart (not to mention that anything that is perfect is bound to self-destruct the moment you try to modify it).

Of course the general technique works reasonably well with any type of film resistor. Use a small ceramic-tipped drill or an air abrader and you'll get quite stable results.

You'll have quite a task adjusting the resistor to beat the 0.85-parts in 10-million precision that has been achieved so far, but if you can trim one of these resistors to within 0.05% of 1.4991486Ω you'll be able to beat that precision using a hybrid technique. Now, where did you put that precision resistance bridge?

I gave him a GA because he was a lot more sophisticated than I; I would just stick a big soldering iron on 'em. (plus, you already had your votes).

And, I do love a sneaky guy.

Filing carbon comp resistors can and has been done but since carbons are lousy resistors to begin with, you're wasting your time. I can do this with precision wire wound resistors which can easily be 'adjusted' to very tight tolerances, something that the world of film resistors cannot do despite claims to the contrary.

if n >1

u must divide n to 1 and then u must divide thats division to 1 and then u must divide the last division to 1 and so on ... and u will connect the 1ohm resistors(number of the first division) in series , and then u will connect the " amount of the second division 1 ohm resistors " in parallel to the first group and so on..

if n < 1

u must divide 1 to n and then u must divide thats division to n and then u must divide the last division to n and so on. and u will connect the 1 ohm resistors (number of the first division) in parallel, and then u will connect the "amount of the second division 1 ohm resistors" in series to the first parallel 1 ohm resistors group... and so on ...

Here's a somewhat more 'natural' network that gives R=355/133 using 17 resistors. That's more resistors than the wilder arrangement I posted previously; on the other hand, the error can be further reduced using a smaller total number of resistors than either M_P_M's arrangement (posts ## 2 & 3) or the wild one (post # 16).

I suspect that the number can be further reduced by incorporating more of the parallel elements from post #1, so I'll give that a whirl when I take my next break.

You are a Master in topological arranjaments of 1 ohm rezistor !

Thanks for demonstrations !

Most flattering, but before I succumb I think I should wait for jim35848 to show us it how few resistors are really needed. I won't be surprised if he comes up with something that not only uses fewer resistors but is also simpler than any of my arrangements!

BTW, I think I have demonstrated that my suggestion above is a non-starter (i.e. that incorporating more than one of the resistors from the group of seven at the base into the complex network will not give any advantage over what I already have).

If I remember correctly, π is approximately 3.14159265... With 18 resistors you can get to 3.1411776..., which is an error of 0.0004162 or 0.0132%. With 10 resistors you can get "only" 3.142857, which is an error of 0.040%.

1. Build a parallel/series network with 9 parallel branches. The first 6 branches are single resistors. Branch 7 is two in series. Branch 8 is three in series. Branch 9 is four in series. This network will have a perfect value of 0.1411776 Ω.

2. Place your remaining three resistors in series with this network, to give a final resistance of 3.1411776... Ω.

It was fun--John M.

Posts 2 and 16 give a very much more accurate answer (0.00000849%) with 18 and 16 resistors respectively. Even with 16 resistors, there are multiple ways to achieve this - which suggests we should be shooting for yet a smaller number. Plenty more fun to be had, then - assuming that you have the time.

This is a serious question, not a sophistic comment.

We all know there is no such thing as, say, an integer meter, i.e. 1 meter doesn't exist, only 1.00 m or 1.00000000 m and so on. So, in the reverse case, is there such a thing as ∏ Ω, or only something arbitrarily close?

Your question reminds me of the recent cube-a-sphere/square-a-circle thread. As you know, "squaring a circle" is one of the classic "things that are impossible" in the field of Euclidian geometry (i.e, using only a compass and straight edge). Trisecting an angle is another.

Someone unacquainted with the rules of Euclidian geometry would say: "It is patently obvious that I can trisect an angle: start with a 99 degree angle, divide by 3, and layout three 33 degree angles, using a protractor."

In the same way, one can very easily calculate the length of a side of a square that has the same area as that of of a given circle to any reasonable level of precision. So clearly, "Squaring a circle" has a particular meaning that can make the statement "You cannot cannot square a circle" something other than nonsensical. Therefore, I was surprised that so many good answer votes were given to TheVoices' answer and so few given to emc_c's. The Voices answer simply does the algebra, and I am sure the ninth grader posing the question is familiar with both algebra and the formulas for volume and areas of common shapes. Therefore, an algebraic answer would not seem to advance the student's knowledge at all. Emc_c's answer, on the other hand, explains that it is only when using the peculiar rules of Euclidian geometry constructions that that there is any difficulty in squaring a circle (or trisecting an angle), probably somewhat advancing the knowledge of the student.

My CAD system can trisect an angle more accurately than I can bisect one on paper, using Euclidian rules. However, conceptually, I will have precisely bisected the angle on paper, but will have approximated a trisection to some limit I establish in the CAD system. I'd argue that in the Euclidan world, perfect line segments exist in our imaginations, and that what we draw is not inaccurate but instead symbolic, and as a result can be perfectly accurate.

So can we have 1 apple? The apple I hold now is lighter than the same apple of just a minute ago. Certainly anything we can smell can be held partly in our hand and partly in the air around us. .9998 apple seems a little silly though.

When we say 2 pi, we are assuming the 2 is exact, but the pi is not. We assume, I think, that 2 pi is exactly twice 1 pi. (Pi might complain: "Why do I have to be calculated to 5, 8, 10, or 12 places, but you don't even put a decimal point after 1?) Can we measure precisely 2 of anything? How would we know? Even the standard for the meter residing in Paris has its good days and its bad days, so if we can say .999999 m as an indication of as precise a measure of one meter as we can get, the standard itself varies more than the "measurement."

Occasionally, I will use 3.14159 for pi, no matter what the significant digits in the (for example) diameter. Much more frequently, I will use pi to the 32 digits provided by the Windows calulator. Seems ironic that it is easier to use a very precise value rather than a more obvious approximation.

Conceptually, a kilometer is exactly 1000 meters, I think. Conceptually, pi Ω is exactly pi Ω. We cannot measure such a value with high precision, but we know it exists. Emperically calculating pi itself is impossible, because we can't tell if our measured circle is round or if our diameter is really a diameter. But mathematically we can calclulate pi to as many places as desired, with two mathemeticians finding the same 115th digit.

This seems to drifting into an incoherent ramble.

Hans Reichenbach had interesting things to say (see page 28) about distinctions between technical impossibilities and logical impossiblities.

I'm not quite satisfied with the official answer, since it uses only 13 resistors to achieve a certain accuracy. The problem as I understood it was to come as close to pi as possible, and it seems to me that the remaining 5 resistors can be used in some way to improve the accuracy.

Admittedly, all the masterful circuitry and mathematics by MPM, Fyz, and others are far above my comprehension level -- somewhat limited to traditional mechanical items like shafts and gears and bearings *. I'm merely quibbling about the semantics, in case there's anyone still looking in here.

My attempt to award a GA (my very first one) to ba/ael for his limerick 25 has apparently not impressed the powers that be at CR4 -- as lamented in my post 38.

=TeeSquare=

* Since the question is nevertheless an interesting one I tried to see if I can actually calculate the equivalent resistance of any of the circuits provided -- and failed miserably. They contain intermediate branches which cannot be classified as being in series or parallel, and my elementary text books don't deal with such complications. I'm also hopeless at finding any information on the internet.

Just to satisfy my own curiosity though, I tried to figure out from first principles the equivalent resistance of five known resistors in a Wheatstone bridge pattern -- A, B, C, D in cyclic order, with A and B in series, parallel to D and C, and E across the middle (I won't go through the trauma of inserting a diagram here). I got a HUGE expression, which I think could be correct from symmetry considerations, and is probably a standard result in circuit theory.

My numerator was:

ABC+BCD+CDA+DAB + E(A+B)(C+D),

and the denominator was:

(A+D)(B+C) + E(A+B+C+D).

I'll be thankful if someone can confirm whether this is correct. Most of my accessible acquaintances, former classmates and colleagues who have engineering degrees seem to be busy in computer or management or stock market activities, and I can't get them interested in any technical discussion. No profit involved perhaps! =TeeSquare=

Besides using series-parallel combinations, you also need to use a Y-Δ transformation.

Rxy = R1 + R2 = Rc(Ra + Rb)/(Ra + Rb + Rc)

Two similar equations can be written ofr Ryz and Rzx and then solved for Ra, Rb, and Rc or R1, R2, and R3.

R1 = RbRc/(Ra + Rb + Rc)

R2 = RaRc/(Ra + Rb + Rc)

R3 = RbRc/(Ra + Rb + Rc)

or

Ra = (R1R2 + R2R3 + R3R1)/R1

Rb = (R1R2 + R2R3 + R3R1)/R2

Rc = (R1R2 + R2R3 + R3R1)/R3

Regarding the extra five resistors: While I don't have proof that 355/113 is as close as you can come to Pi with 18 resistors, I'd be surprised if there is a closer solution.

Thanks,

Jim

Hi, and thanks for alerting me to the more basic (and much simpler!) star-delta transformation, which would presumably make my 'Wheatstone bridge' formula effectively redundant when simplifying a network of resistances in practice.

There seems to be an oversight in your figure -- the labels Rb and Rc should be interchanged. I tried deriving the 'Rxy' formula with which you started, and got a slightly different expression. I think we need to maintain cyclic symmetry of the symbols (all clockwise).

Regards, =TeeSquare=

Nice to know someone is reading this. Yes, I switched Rb and Rc. Corrected figure below.

Happy holidays,

Jim

First, your expression (=?) is correct. Jim has covered some methodology.

I agree that it seems surprising that we can do so well with just 13 resistors and no better even if we have 5 additional resistors to play with. Equally, I believe that the "challenge" was based on the 13-resistor version, and the extra 5 resistors were provided to allow solutions using simpler topologies.

Now, I believe the situation is that we don't actually know whether (or not) there is a more precise solution with up to 18 resistors. What we do know is that there is no solution that can be derived any simple way from any of the "good" solutions that have been presented so far; the reason that we know this is that any simple addition of the five available resistors will reduce the resistance by at least 12-uOhm, compared with the existing 270-nOhm error.
I have reason to believe that an exhaustive search has been made of the "regular" topology - by which I mean resistors that can be built by series or parallel connections of other resistors that were themselves built in the same way. And there will be more than 2¹⁸ of these (and less than 2²⁰). I think it likely that the Wheatstone Bridge topology was also included.
On the other hand, I think your workings on the Wheatstone topology illustrate why writing a programme to handle the full range of topologies would be a nightmare (Jim's methods do help - but things still get difficult when you have more than a couple internal nodes).

Finally, I tend to agree with Jim that a closer approach is not all that likely. As mentioned previously, the "regular" topology would give at most a million arrangements with different values. Perhaps there might be up to 10-Million of the more complex topologies, but there would still be a lot less than a 100k of them that would give values between 3 and 4 Ohms. If we were to spread them uniformly, that would give a spacing of about 10-uOhm - and we already have a value within 0.25-nOhm. Even if this is part of a cluster, a density-factor of 20 is quite severe.
By the way, it is all too common for there to be several apparently different arrangements that give the same value. For example, 1-ohm can be made with 1 resistor, four resistors (two ways: 1/2+1/2 or 2||2), or six resistors (2/3+1/3, or 1.5||3). And we can see that this doesn't imply other arrangements come particularly close - the closest with six resistors is 10/11, which is close to the average for the range 1 ... 2.

Thank you for the explanation. Also to Jim for the added insights. I can't say I really followed it all, but in a qualitative way I've got the sense of what you've written, and it has added to my understanding (of why many issues of a mathematical nature will continue to remain beyond my grasp!). It's good to get away from my nuts and bolts now and then, and wander a bit in wonderland.

Guess it's time to wish all of you good folk out there a Merry Christmas. I still think ba/ael is one GA vote short for his poem in #25, but the workings of CR4 are beyond me! Season's greetings to everyone. =TeeSquare=