Jail Break: CR4 Challenge (08/19/08)

Deleted

I've given you a GA, as yours is the first of its type and I agree the principle. However, the question says magnetised, which that does not necessarily imply that the bar does not have (say) a north pole at each end and a south pole at the centre, for example. Then the unmagnetised bar may not be of a magnetic material...

Maybe try both against his keyring, belt buckle, or glasses frame?

In any case, John should win his appeal, as 3 can be a perfectly satisfactory approximation for some problems. He can then claim compensation for false imprisonment and traduction of reputation.

Well it does say two steel bars, most common steel materials are magnetic...

I would be interested to see how a bar, whose length significantly exceeds its diameter, could be magnetised in any other way then with the poles at each end?

The statement that the jail is of plastic only, indicates that the presence of ferromagnetic material other than the bars in non-existant.

As for the approximation of pi to 3 I agree, he would have good grounds to claim false imprisonment!!!

John.

Agreed that the probable intention is that both bars are of ferromagnetic material.

However, given that this is the land of engineering, the problem would not allow John to claim his right to practice if it was as easy as your default condition.

Assuming that the bar is long enough, it is perfectly straightforward to magnetise a bar so that it has identical poles at each end and a different one in the middle.
If the bar was magnetised by stroking it with a magnet, you would stroke from the centre to each end alternately.
If it was magnetised by placing it in wire solenoids, you could simply use a pair of very similar solenoids, one surrounding each end, with the current passing through both in series, and the direction of current creating identical poles in the middle.

Having said all that, it leaves use with four possibilities for the problem:
First, that the land of engineering has a shortage of jail accommodation, so they make the task as easy as possible (your solution)
Second, that both bars are of ferromagnetic material, and they make the task slightly more difficult by making the magnetisation pattern of the magnetised bar suitably complex.
Third, that they make the task more difficult (but still solvable) by having making the non-magnetised bar of non-ferromagnetic steel, but the magnetised bar is still capable of acting as a compass.
Fourth: either that the authorities believe in severe punishment for its own sake; or that John's use of ∏=3 could have resulted in the failure of a significant project, and that the powers that believe that they can use the punishment as a corrective - John needs to understand what tools are suitable for what task. For either of these situations, they might provide one symmetrically-magnetised bar and one non-ferromagnetic bar. I believe that problem to be unsolvable unless John managed to smuggle some ferromagnetic items into the cell with him (I would add cufflinks, ferrules on shoelaces, and eyelets on his shoes to the list of possible illicit tools that I gave earlier)

Okay, correct me if I am wrong, but I am assuming that the magnetized bar is magnetized end to end and you are thrown into a room with no metal (either on you or in the room).

So, lay both bars cross ways over each other at a 90° angle on the floor. The amount of attraction between two bars in contact with each other will be at a minimum in this configuration. Now take the top bar and roll it to one end of the bar under it while keeping the top bar's center along the long axis of teh bar underneath it.

If the amount of magnetic force between the two bars increases, then the bar underneath is the magnetized bar. If the amount of attraction between the two bars does not change appreciably, then the top bar is the magnetized bar.

Knowing which bar is which will allow you to exchange the unmagnetized bar for a get out of jail card.

Since the one bar is magnetized and not a permanet magnet, if John was to hit both bars together would he not demagnetize the bar. Then both bars would be demagnatized and it would not matter which one he gave them.

This is absolutely the best answer. This is the caveman methodology of engineering. I hope the Gieco caveman isn't reading this blog!!! If you are it's not personal it's professional.

But it might not get John released from jail if the authorities can tell which bar was magnetised originally (the bars don't even have to be the same length, after all).

Plus this is an all-plastic environment, so it is likely to be impossible to provide the sort of shock that will de-magnetise it

I'm assuming firstly that John is not totally naked in this plastic cell

With this assumption, he should tear away some threads from his clothes and suspend the bars at their COG point such that they are oriented horizontally. The magnetised bar will then always line up in a North-South direction as the bar will act like a compass needle.

The unmagnetised bar will show no such preference for it's alignment.

Very good. One problem... How do you know which way is north?

We are only told that the rod is magnetised, not that is has different poles at the two ends. This would not work if the magnetised rod had one pole in the middle and the opposing pole at each end. Similarly, there could be magnetised objects near the cell that locally cancel the Earth's magnetic field.
Unlike Electroman's answer, I can't see how it may be adapted to either of those situations.

It would probably be glaringly obvious very quickly intuitively...
I was just playing with two screwdrivers one of which had a magnetised tip...it was obvious which it was...although not necessarilly easy to explain why... running the tip of one along the shaft of the other and vice versa, one could just feel it..._{(or maybe Bastet told me)}
Del

I have to agree with Del, you can feel the force change as one passes the unmagnetised rod over the magnetized one.

"you can feel the force change as one passes the unmagnetised rod along the magnetized one"

GA, I think. Assuming the unmagnetised rod is itself ferromagnetic, I would take this to be an improved expression of Electroman's and Del's answers

I detect a little sarcasism in that remark. I know it is not scientific but one has to use what he has got at hand.

No sarcasm intended - in case it dispels your doubt, my GA rating was not just in the comment, but also given to your post.
But I think John needs a more comprehensive strategy - hopefully my next posting.

I disagree with the argument that having two north poles would completely defeat the center position methode. It would only give the magnetised bar two locations that are magnetically neutral about 1/4 of the way from either end instead of just the center.

You may not realise it, but you are not disagreeing with the content of post #13; this was about why the compass method proposed by MPM would not necessarily work.

But it also demonstrates that you cannot know where the neutral locations are until you test for them, so the method of poking each rod in turn to the centre of the other is inadequate - you need to find them; the simplest method being the one indicated by Del and described in sufficiently more detail by Rick Lee

Of course, symmetrical end-to-end magnetisation was just an example. If I wanted to make this a real test of ingenuity I would provide a not-very-strongly magnetised bar, with the field running across the width/thickness, that had a pair of north poles on two opposing surface regions and a similar pair of opposing south poles.

Over to you.

Further to the suspended north pointing rod.

It is assumed that the plastic sell is fixed and that there are no N painted on the wall.

If the rod moves he has the answer but if it does not he must first do another test of swinging it through 90deg.

I actually think that the answer should be in the value of pi.

What he should do is suspend one of the bars from the tip and swing it like a pendulum. the rotation of the earth will have the swing changing direction with the maximum prescribing a circle. The magnitude and deflection of each timed swing can be measured and pi can be calculated.

The normal bar will swing concentric and will prove the value of pi and the magnetized one will not swing true.

He will have to wait pi days before positive proof - Why? - I don't know.

Alternatively I can send him a GOOZ card.

I was taught that the true test for magnetism was demonstrating repulsive force between like poles. Attractive force can be caused by unlike poles or an induced pole in a magnetic but non-magnetized material

Therefore John should hold one bar (A) vertically and stroke it top to bottom several times with either tip of the other bar (B)

He should then place the tip with, which he has been stroking bar A, to the upper end of bar A. If he feels a repulsive force he has magnetized A with B, and he should hand in bar A

If he feels an attractive force, bar B was not a magnet and the should hand in bar B.

Unfortunately he has a 50-50 chance of magnetising the non-magnetic bar with this method. He will not get out of jail if he hands his captors a magnetised bar, even if that bar was unmagnetised to begin with, so it is not really a good way to have a 100% certainty of being released.

I still think suspending the bars horizontally and using the earth's magnetic field (Or indeed any local magnetic fields for that matter) will work.

The reason I believe it will work is that firstly you do not need to know which way North is, all you need to do is perform the experiment at 2 arbitrary orientations in the plastic cell that are 90 degrees to each other. The magnetised bar will change direction at one or both of these orientations whereas the unmagnetised bar will hold the orientations without twisting.

Fyz,

I believe that even if what you suggested is true, I.E. that the bar could be magnetised with similar poles at both ends and an opposite pole in the middle, the magnetised bar will still twist from one of the experimental positions. Such a bar might want to line up in an East-West direction perhaps???

What do you guys think?

If you had said that it is unlikely the magnetised bar would be perfectly symmetrically magnetised, I would agree. Equally, I'm certain that I could ensure sufficient symmetry that the "set" of any thread that was strong enough to support the bar would dominate over magnetic pointing to better than 0.1-degrees.

The bigger problem with a symmetrically magnetised bar is the interaction between it and a ferromagnetic non-magnetised bar in what is (presumably) a relatively small space. Either of these will affect the other in roughly equal amounts, and far more than would the Earth's magnetic field.

In the end, I think that John needs to go through sequential hoops: I'll put these in a reply to the original question.

It's hard to think of less complicated than electroman's answer. He would be out in π minutes

Your method would probably take π hrs (I believe your method works)

Mine would need π² hours (I believe the jailer knows which bar was originally unmagnetized on sight)

By the way I never suggested that the bar be magnetized N-SS-N or vice-versa, I simply magnetized it N----S and stayed faithful to my highschool physics teacher by looking for repulsion.

Finally in my belief, a bar N-SS-N would orientate itself EW in the southern hemisphere; neutrally at the equator; and NS in the northern hemisphere (all directions are of course ideal spherical magnetic)

Electroman's method could mislead if the magnetised rod is NSSN- so it needs checking both ways around.
Rick Lee's modification won't give false positives, and is scarcely slower, so that would be a good start

I think you already acknowledged that your answer would be slower than Electroman's; plus it's no more reliable at identification than Rick Lee's, and awkward authorities might have a tester on hand to see whether you have indeed handed them "the unmagnetised bar".
[I believe that MPM is right to recognise that the terms of release did not say "the bar that was originally unmagnetised", and John is dealing with legal authorities here, even if he is in the land of engineering (quality engineering, anyone?)]

Interresting thought! I have heard some interesting ideas put forth, such as the unmagnetized bar is not ferric and hanging the bar from a thread. Let's toss some other twists into this.

Firstly, I believe that a magnetized bar can not have poles as N-S-N. The magnetic domains in the metal always orient in one dirction. That is what makes it magnetic. I think though the proposer knows this and was just trying to get a response on this.

Next, the unmagnetized bar not being ferric. Good thought but if you look at what other metals it could be, visual inspection eliminates a lot due to major color differences. Brass, Bronze, Gold, Silver, Tin, would be obvious. I doubt they have any Tantalum or Platinum as to the cost. That leaves Titanium, aluminum, and stainless steel. Titanium and aluminum are light weight and easily dismissed. The difference between stainless and ferric metal can be determined though on a close side by side comparision as the ferric bar would show bits of corrosion no mater how clean it might be. So let's rule out a non ferric bar.

Next, the questioner did not classify what he meant by BAR. I am sure everybody immediately thought a nice 2 cm wide round 20 cm long cylinder weighing 400 grams. What if the bars are large ingots weighing 10 Kg or more, I doubt the man is going to have a thread to support that.

I still go with the force idea, if the dissimilar metal bit has been discounted.

I take it you would apply the same reasoning to steel wire? In that case you have discounted the more successful implementations of wire recorders, the precursors of magnetic recording media from tape recorders to hard disc drives.

If you care to look up steel data, you will find that ferromagnetic grades are available with good oxidation resistance (e.g. 20/0), and that not all non-ferromagnetic materials have the highest corrosion resistance. Another trick is to leach the protective oxides from the surface of the higher grade steel, which then leaves a high-grade steel with a surface that will rust.
Do you really suppose that the authorities would provide a test that could be solved on the basis of unwarranted assumptions - particularly when the original "crime" had been of similar type?

Regarding supporting the weight of the bar, there should be enough material in John's unravelled socks to make a multi-strand thread/string/rope that would support many kilograms. He might first try using the arms of his shirt for the NSN test (non-destructively), in case that gives enough sensitivity.
The bigger problem could be finding supports for the suspension - would the plastic chair provide suitable supports? At worst he might have to support the suspended bar from his hands (braced against the wall) and push the other around the floor with his foot.

I suspect John was not really in the land of engineering but rather in the land of mathematics. Good engineers always employ a factor of safety in their designs to account for the uncertainties in such factors as loading conditions, material properties, construction tolerances, corrosion and the like. Sometimes engineers need more precision and accuracy in their results than 5% but usually the conservative factors of safety used would make a 5% error in Π to be insignificant. I'm not saying we shouldn't use the best available values but sometimes worrying about lots of significant digits in an answer in just an indication of an inadequate understanding of the accuracy of the input parameters.

From a purely mathematical viewpoint, 3.14159 is no better an answer than 3 - unless the required precision is spelled out. You either need a precise answer (e.g. sum of series), a definition, a method, or a value to whatever precision you wish - so long as the resolution limit is explicit.

We don't know how difficult the authorities have decided to make the task. For example, they have the option of making the non-magnetised bar from steel that is not ferroelectric.

As a consequence, John needs a multi-stage approach.

If he has managed to retain some ferromagnetic item on his person (that could be one of belt buckle, keyring, shoe-lace ferrule, shoe eyelet, spectacle case or frame...) his best first- option will be to use that to check on the properties of each bar, probably using a method akin to that proposed by Rick Lee in #15.However, the authorities are likely to have removed all ferromagnetic objects from his person, so we can't rely on this being possible. In that case he will need to adopt an approach that could take several steps.

John's first step could be to adopt the method proposed by Rick Lee in #15- pass the tip of each object along the other; the one in his hand that displays the more pronounced zeroes in the attraction is the non-magnetised bar.
This method was chosen for its speed, but would fail if either the magnetisation was too weak to feel (they only said "one bar is magnetised"), or if the second bar was not ferromagnetic

To accommodate a lightly-magnetised bar, John could then hang each of the bars in turn from a pair of threads* taken from his clothing (thank you Maths-Physics Maniac). He would then touch the tip of the free bar at various points along the length of the hung bar, and pull it away. Movement of the bar would indicate the level of attraction; it's basically the same method post #15, but far more sensitive.
If this doesn't work, John has to assume that the interaction between the two bars is too small to be used for this test.

All that seems to remain (subject to someone finding a method that works when this does not...)) is the use of a single thread to see whether either bar behaves as a compass needle. This still won't necessarily work, as the authorities might have chosen to provide bars that are respectively non ferro-magnetic and symmetrically magnetised (alternatively, there may be magnetic objects near to the cell that cancel the Earth's magnetic field).
In this case, John will have no recourse other than to contemplate some other interesting problem for the remaining 3 (≈∏) days.

So, if the last experiment doesn't identify the correct bar, John should hand one at random to his captors - 50% probability of release is better than none.

*If the authorities remove all suitable clothing, John may still have some body hair to provide suspension - but in the worst-case his experiments may take more than 0.14159 days.

If the bars are the right shape and size, maybe John can balance each on it's end. The one that falls over is magnetic and trying to align itself with the Earths magnetic field. I reckon he'd suffocate and die in a sealed plastic room anyway.

While in prison John should not find the unmagnetised bar and hand it over to his captor(sic) because if he does he will not get out on "Good engineering" i.e. find out that pi is 3.41159.

He needs to stay in prison for the stipulated period of 3.14159 days, get a receipt for the time served and get out "on good engineering". I am sure in the land of engineering (where I live) prisons are places for luxurious resting. Therefore, there is nothing to be gained by testing for magnetisation of the 2 steel bars and handing one of the bars to the captor.

In the land of engineering (where I live) spelling and minor grammatical errors are universally ignored provided that the meaning is not compromised; this is necessary because, being engineers, approximately 70% of the population is dyslexic.

However, this posting indicates that the province where John dropped his doo-doo appears to be so badly organised (or possibly strapped for cash) that the engineering policeman who captured him also has to serve as gaoler. I will therefore be standing at the next election - my platform: better to distribute and organise national resources.

There being no restriction on the number of tests or attempts that John is allowed, a slight extension of Electroman's solution in post 2 plus the insights provided by Del the Cat and RickLee in posts 14 & 15 is all that is necessary. There's nothing new in what follows -- I've merely attempted to spell things out in pedantic fashion (also simplistic compared to the comprehensive strategy spelt out by Physicist in post 26).

Firstly some assumptions which are not explicitly stated in the problem:

1. The steel bars are essentially straight and of reasonable size (say a few inches to a foot long) with a small cross section, presumably round or rectangular, hence light enough and convenient for John to hold in his hand.

2. We consider only conventional ferromagnetism as commonly observed in steel (other than certain stainless or austenitic varieties).

3. The 'strength' of magnetisation is sufficient for any attractive/repulsive forces between the bars to be noticeable when John holds them in his hands.

4. Presence or absence of earth's magnetic field or any such external influence is irrelevant to the problem, and there is no ferromagnetic item other than the two steel bars in the plastic prison.

Now, allowing for any convoluted configuration in the magnetised bar (as suggested by Physicist? in post 8 and later), it would be reasonable to suppose that two or more 'poles' exist and can be located along its length (or even across its width?). John can try the following procedure for a quick release from his incarceration:

Hold bar A firmly in one hand or along an edge of a table say, and suspend bar B horizontally by holding it lightly between fingers at the centre, allowing it to pivot.

Keep one end of B a slight distance away from A, with their lengths at right angles to each other, and move it along A from one end to the other.

If A is the magnetised bar, then the presence of a pole anywhere along its length can be sensed by the tendency of B to swing as the end passes by in its vicinity.

If B enjoys a smooth run, it is presumably the magnetised bar, but there is no bar on John interchanging the bars and repeating the test. If he is still unable to resolve the problem, he better head for the nearest bar after pi days.

I will leave it to the geniuses out there in CR4 to cope with the situation where the sadists have magnetised a bar across its width.

As an engineer I strongly endorse what jim35848 has stated in post 23. I have elsewhere bemoaned the regressive progress from slide rules to calculators displaying umpteen meaningless and distracting digits by default! I hold that a great majority of practical problems in the world of engineering require only a slide rule and sketchpad. The advent of calculators and computers has resulted in diminished capacity for thinking and conceptualising, or so it seems to me.

But I have no arguments with Kris' exact pound of flesh (post43)

=TeeSquare=