Brain Teaser No 1

Does this look like an elegant solution? Who in their right mind would buy 4 Sony CDPs......??

BTW, I began to generate cubic equations the other night in connection with the ladder problem. But then found I couldn't find roots to these at all. However, they did look sort of symetrical. I had started with a couple of parametric equations with 'theta' (angle between ladder and ground) as the common parameter and then forcing a development via sin^2 + cos^2 = 1 to give an equation, which happened to be cubic. The symetry in the co-efficiants were 1;3;3;1. Will try to put a bit more time in later this week to develop the idea.

Diagrams were looking more like squares inside squares. At present the skew-wiff squares are side L. Must try making the normal squares L and the skew-wiff ones (1+d+1/d+1) etc.

Seems like the champagne is safe today Bam!

Cheers

Don

Cheers

Don

Marcus wouldn't take part unless Steven was allowed to take part
Steven wouldn't be in the party if Vik was.
Blzebub wouldn't be in the party if both Marcus and Steven were in it.
Vik wouldn't be in the party if Mick was allowed in
Mick will join the party with any of the others
Joe won't be in the party if Marcus is, unless Vik is in it too.

Which four members visited the factory? and what was each one's item of choice?

BTW, any similarity between these fictitious characters and members of this Forum bearing similar but equally fictitious names are entirely in-coincidental.

Cheers

Don

now don, how are you getting on with the ladder?

enjoy

ken

Hmmmmm!

I feel that around Xmas I had climbed about 8 rungs of this ladder (which is a 10 rung ladder).

Last weekend, I felt confident that an attempt on the last two rungs was in sight and went for it!. I then found a simple error in generating my cubic equation. All of a sudden, it seemed as if the rungs on this rickety ladder were snapping like an opening zip and I was heading downwards with my chin and nose ricocheting of the remains of each rung. The biggest ricochet being reserved for that blooming 1m square crate, just as I slumped to the ground!

At the moment, my head is going round in circles!! Ah ha!! Now there's a new idea!!

Back to the drawing board.

How's everybody else doing on this one?

Ken?, Omer?, Paul? Duncan? Martin? anybody???

Cheers

Don

I posted some sin/cos relations recently. Nice and simple, except that when you further process them you end up with familiar looking square roots of square roots.

(I'd thought that the idea of laying the ladder flat on the ground and whirling the walls and floor around it, so that the wall/floor junction describes a circle, had promise. What would the curve traced by the corner of the crate look like?)

I can't wait for BAM to draw the curtains!

Paul

Not my strong point. The 'imaginary' part always seemed 'elusive' to me! Seem to recal them being useful in electricity, and can't imagine (sorry) how to use them in ladders...but then, this might be where that lucky break lies!

Meanwhile, whilst doodling (or should I say using Duncan's method of "proof by excessive writing down") I began to wonder what diameter circle could be inscribed, just touching the wall, ground and ladder, so that each was tangential to the inscribed circle.

Of course, (well its really a conjecture at present!) the circle wouldn't generally touch the ladder at the point (1,1), except in the unique case of the shortest ladder, when its diameter would be 0.586m (=2-root2). And in the case of an infinitly long ladder, the circle would be 0.5m dia and touch the ladder 0.5m above the ground. I think!

Said I was going round in circles.

Cheers

Don

My thoughts also....in fact I can't conceive of it being otherwise....but equally, I can't recall or visualize any substitutions that would lead to a different AND more simple form.

I have also tried a few "wild" shots at "simple" equations eg d=cosh(1/L)+e^(1/L) etc etc.... BTW don't try that one, 'cos I only just made it up as I was typing and it would be funny if it worked !!

Anyway, quite often, there ISN'T any general substitution that can be used. The only substitution is unique and can only be obtained by tackling the problem from a different angle. Once the substitution is thus 'proven', you could use it elswhere....except, in many such cases nobody has found any further use for the substitution.....so it becomes unimportant, and is forgotten about. Until people like Bam turn up!!!

Going round in circles again!

Cheers

Don

Naim requires a speaker cable that consists of two parallel multi-strand conductors, separated by a fixed distance and encased in flexible black plastic. Furthermore, Naim requires that you print the Naim logo on the sleeve and chevrons to indicate the correct orientation of the cable when in use. Being a good salesman you simply agree to all of this whilst remaining po-faced.

You have a meeting with your production supervisor and start to flesh out the detailed specification. How many strands are needed per conductor? What should the separation be between the conductors? How thick should the PVC sleeving be? All questions are quickly resolved and all is going well until the supervisor asks you: "how do we decide which direction to put the chevrons in?".

You hadn't thought about this. Your gut instinct is that it doesn't matter as long as there are chevrons and that is what the client requires. However, your supervisor is an engineer and a bit of stickler for process and far too anal to understand the compromises of the wider business environment. So you are going to have to fob him off with some technical rather than marketing reasoning.

The supervisor senses your hesitation and glares at you. He says the only possible directional aspect is the direction that the wire has been drawn through the extruder. He bluntly asks you to define, relative to the chevron direction, which wire orientation you want for each of the strands within each conductor, the direction of twist of the strands within each conductor (clockwise or anti-clockwise when viewed in the direction of the chevrons) and the relative orientation of the two conductors.

What answer should you give the supervisor in order to achieve the best sounding cable for Naim Audio?

BAM
(PS All employees of Naim are allowed to participate in this puzzle)

The effect of the chevrons is to ease or constrain the transmission of musical nuance, a 'dielectric focussing mechanism' in layman's terms.

So don't worry about the copper, just stamp the damned chevrons.

Paul

To paraphrase an old post from Julian Vereker (I have it at work but not here): -

It doesn't seem to matter which way the wire is extruded, the directionality is established at the point the insulation is applied. This means one can easily mark the insulation for direction. 'If I can see a cable being manufactured I can tell you which way round it will sound best' - he never gave away that info though

Naim have never been able to measure anything asymmetrical in the cable, it's an effect that's easy to hear, hard to explain, and as yet impossible to measure.

The current 'why' theory seems to be related to an annealing effect on the copper as the hot insulation is applied.

I can easily demonstrate asymmetry and directionality in microwave cables, but not in audio ones...

Andy.

Andy.

quote:

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Would you mind either elaborating some more or putting a smilie on your post so I know whether you are taking the piss or not?

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I find 'smilies' make a serious impact (for the worse!) on sound quality. It's probably why Steven Toy is wrong....

Paul

Andy.

Please! This is getting even worse.

You said it is easy to hear, Andy. That's something. Under what conditions did you hear the difference - blind trial?

Mr Tibbs, you old cynic! Is this why Naim use a movable band on interconnects rather than a printed label? Ooo. Imagine the opportunities for mix-ups at the connector fitting stage when the coax is unmarked.

quote:

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You said it is easy to hear, Andy. That's something. Under what conditions did you hear the difference - blind trial?

---

That bit should have been in quotes, but since it was paraphrased from memory I didn't bother - those were JV words (almost) not mine.

I've never tried the 'speaker cable both ways round, trusting Naim(!) that it makes a difference, and if it doesn't it doesn't matter anyway, since it is the best cable I've used on my systems for conencting the 'speakers - it was hugely better than the earlier NACA4 it replaced.

For the volumes produced it's sensibly priced too.

I have though heard differences in interconnects, under non-scientific sighted tests, when trying the silver cable I now use. My only defence, not a scientific one, is I'm not generally in the habit of wasting my time on stuff that's irrelevant, bearing in mind I've nothing to gain from it.

The same applies to Naim, they could just as easily leave the cable unmarked for direction, it wouldn't affect sales one jot, bearing in mind the loyalty of the customer base, and it's recommended use owing to the amplifier-matched electrical properties.

Andy.

cheers

Nigel

I mentioned a few posts back, that I had 'extended' Bam's ladder into a 'box'. I drew a floor, a ceiling and two walls, each of equal length (d + 2 + 1/d). I put a 1m square 'crate' in each corner. I drew 4 ladders each of length L, each toe, d from the crate, each top 1/d from the crate. The top of the first ladder marked the toe of the next, around the 4 sides of the room. This gave me a picture of one box inside another, the inner box being of side L and skew to the outer box, side (d + 2 + 1/d). I then drew a third box inside the L box. The 3rd box was 'concentric' with the outside box, but of side (d + 1/d). The inner box touched each crate at the same point that each ladder touched each crate. Distance between the inner box and outer box = 1m. Hope you're all with me so far?

Based on this drawing and previously posted formulae :-

Outer box is side [(L^2 + 1)^0.5] + 1 = (d + 2 + 1/d)
Inner box is side [(L^2 + 1)^0.5] - 1 = (d + 1/d)
Middle box (skew) is side L = [(1+d)^2 + (1+1/d)^2]^0.5

The ratio between the areas of the boxes for any given L is constant ie area of inner box to area of L^2 box = area of L^2 box to area of outer box. When L=2*sqrt (2) this ratio is 0.5 (or 2, depending which way up you take the ratio) and tends towards 1.0 as the L increases towards infinity. The curve looks a bit like a rectangular hyperbola.

The 3 boxes can be drawn for any value of L using the [(L^2 + 1)^0.5] - 1 formula etc you don't need to calculate d first. I have drawn the 3 boxes both with the outer/inner boxes parallel to the paper edge and also with the middle L box parallel to the paper edge…… Just in case one gave me more inspiration…….it didn't!!!!!!!

I also noticed that in any right angled triangle with sides a,b and h (h=hypotenuse) that if you draw a circle radius r, centred on the hypotenuse and the circle just touches the other two sides, then 1/a + 1/b = 1/r . If r = 1 which it does in our case, then 1/a + 1/b =1.

As I said….just a couple of thoughts. might inspire somebody towards a solution!!

Cheers

Don

Not as intersting as the snaic charmer!!

X = INT(L-1)

In other words, take the ladder length, deduct 1, then round the answer down ie ignore the remainder.

The shortest ladder length occurs when X=1, in which case L=square root of 8, or approx 2.8. This gives the maximum value of L-X of approx 1.8, rounded down to 1

As the ladder length increases, and tends to infinty, then X also tends to infinity and L-X tends to 1. This happens very quickly eg if X=10, L=11.05, and L-X=1.05

The formula works precisely in situations where X values are integers (and ladder values are not). I acknowledge that it is not precise where L values are integers (and X values are not).

Nonetheless, is it neat enough???

STEVE D

I now wonder whether the solution is something to do with calculus. Perhaps BAM could confirm this is the right track!

STEVE D

Glad to see you're still trying for a solution, however, I'm having a bit of difficulty with your proposed X = Int(L-1)

First, this is presumably with the ladder almost horizontal, (until L gets to be less than about 5m)

Second, it doesn't work too well with intermediate values of L such as L=10.5m. Using the above formula gives X = 9. An accurate value in this case would be 0.105914m or 9.4416m depending on whether the ladder is nearly vertical or horizontal.

In fact, your formula works slightly better without the 'Int' bit, especially at larger values of L. Eg if L=10.5 then your modified formula gives X=9.5 which compares reasonably well with 9.44. To get the 'other' value for X you would find the reciprocal of 9.5 (which = 0.1053), and which compares reasonably well with 0.1059.

Of course both this and the earlier versions fall down completely once L gets less than about 5m. Have you tried plotting the curves of L v X (or d as I called it) using the 'accurate' formula, and then tried to find a more simple formula to descibe the curve?

I still can't find either an algebraic way of simplifying the 'neat' formula that was posted a few weeks back or an alternative starting picture. Might try the library again later today.

Cheers

Don

I think it should have referred to a conjecture rather than a theorem, because Fermat never actually published his proof. The theorem stated in effect that for positive integer values, a cube could not be represented by the sum of two other cubes, and generally any number to the power 'n' could not be represented by the sum of two other numbers also to the power 'n', where 'n' was >2. Now this conjecture wasn't invented by Fermat, but had been written down in a book titled 'Arithmetica'? a few centuries earlier. Fermat writes in his copy of the book words to the effect that "I have a truly marvellous proof of this conjecture, but the margin of this book isn't big enough to squeeze it in" You bet the margin wasn't big enough! It took Andrew Wiles and couple of mates, over a decade, and a few PhDs, to get a solution and it wasn't elegant by any stretch of the imagination.

Now perhaps you can see some sort of relationship here with Bam's ladder problem. Bam says he has an elegant solution. The rest of us feel like we deserve some kind of PhD for the effort in getting a practical and 'neat' solution. Now I'm reluctant to take this analogy much further because Fermat's notes in the margin only came to light after he had died. This meant nobody could get him to reveal his 'truly marvellous proof' - if indeed he ever had one or, worse still had one that could stand up to scrutiny.

Now I don't want to worry Bam, but wouldn't it be a good idea to lodge your elegant ladder solution with an independent arbitrator such as Paul S, just in case something dreadful happened before we found the answer……..but please don't let this put you off revealing your identity at the hifi show tomorrow….we are all looking forward to that last pint with you….oooops did I say last pint?

Cheers
Don