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Topic: Finger Gauges (Read 1826 times) |
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James Fingas
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Finger Gauges
« on: Nov 22nd, 2002, 11:45am » |
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Willy Wutang is making a robot, and he has bought himself a set of 24 finger gauges. For those of you who have never seen a set of these, each gauge is a strip of metal about 1/2" wide and about one foot long. Each is a different thickness (0.001" up to 0.024"). Each has a hole in one end, and they are fastened together with a nut and bolt, in order of thickness.
Willy's calculations indicate that, with the flame thrower attachment extended, he must adjust the spark plug to be 0.086" from the jellied gasoline nozzle. He pulls out his trusty finger gauge set, and realizes they only go up to 0.024". How can he accurately set the spark plug distance?
If you've figured out how Willy measured 0.086", what is the smallest measurement (in thousandths of an inch) that Willy can't make using his set of feeler gauges? What is the smallest measurement (in thousandths of an inch) that Willy can't approximate to +/-0.001"?
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Jeremiah Smith
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Re: Finger Gauges
« Reply #1 on: Nov 22nd, 2002, 11:52am » |
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Well, he could just whip out the .023, .022, .021, and .020 gauge ones. Since they're all adjacent, you can set them against each other, and use them as a single gauge, which is .086" thick. I don't know what the smallest measurement he can't do is, though. Maybe I'll do that later.
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| « Last Edit: Nov 22nd, 2002, 11:52am by Jeremiah Smith » |
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S. Owen
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Re: Finger Gauges
« Reply #2 on: Nov 22nd, 2002, 12:52pm » |
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Essentially we're asking what the smallest number is that cannot be expressed as the sum of consecutive integers between 1 and 24.
I got the answer quickly just by brute force, but here is a bit of the insight you could use to get it by hand. Maybe this can be extended into a more elegant solution?
Notice what the sum of n consecutive integers is, mod n. If n is odd, it will be 0. If n is even, it will be n/2. Clearly Willy can measure thicknesses between 1 and 24 thousandth of an inch. So you can start checking 25, 26, .... The result above helps you get the answer pretty quickly.
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Garzahd
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Re: Finger Gauges
« Reply #3 on: Nov 24th, 2002, 5:47pm » |
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Also a nice problem: Say the gauge has an infinite number of consecutive thicknesses of metal (it doesn't quite fit in Willywutang's garage, I know). Then what measurements are unreachable, and why?
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Chronos
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Re: Finger Gauges
« Reply #4 on: Nov 24th, 2002, 6:50pm » |
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Garzahd, I'm not quite sure what you're asking. I can get any integer thickness just by pulling out that single finger, and there's no way I can get any non-integer thickness. That seems too trivial. Are you also requiring that each measurement use two or more fingers? In that case, the only positive integers which are unattainable are powers of 2
This probably also works for the set of 24, within reasonable limits, so I'm going to guess that the smallest integer you can't get with the set of 24 is 32.
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S. Owen
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Re: Finger Gauges
« Reply #5 on: Nov 24th, 2002, 7:41pm » |
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Indeed that is the answer - yeah, I think you have to consider the "infinite case", and then guestimate that the result will hold for 24. That's as rigorous as I can come up with, at least.
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Jeremiah Smith
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Re: Finger Gauges
« Reply #6 on: Nov 24th, 2002, 10:27pm » |
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So no power of 2 can be expressed as the sum of consecutive integers less than itself? Or have I read this wrong?
You know, I find that result somewhat depressing and I don't know why.
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James Fingas
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Re: Finger Gauges
« Reply #7 on: Nov 25th, 2002, 9:27am » |
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Let's say that Willy gets frustrated with his finger gauges (because they can't measure some thicknesses). He undoes the screw holding them together, and decides to rearrange them so that they can make any measurement from 0.001" to 0.300" in increments of 0.001".
Is this possible? If so, what order should Willy put them in? If not, why not?
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towr
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Re: Finger Gauges
« Reply #8 on: Nov 25th, 2002, 10:47am » |
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[0.001 0.003 (0.004-0.024) 0.002] is the only way to get 0.296 - 0.300 (the bit between () is not ordered yet). But there is no way to get 0.295, since at best you could put the 0.004 and the 0.002 next to each other and leave them out of the bunch, which gives you 0.294
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| « Last Edit: Nov 25th, 2002, 10:48am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Chronos
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Re: Finger Gauges
« Reply #9 on: Dec 1st, 2002, 9:52pm » |
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jeremiahsmith: A proof of that result. First, let's prove that none of those numbers is attainable.You have n consecutive gagues. n is either even or odd. If n is odd, then let t be the thickness of the middle gague. Then the total thickness is n*t, and since n is odd, this cannot be a power of 2. Alternatively, n might be even. In that case, let the sum of the two middle gagues be T. Since T is the sum of two adjacent numbers, it must be odd. The total thickness in this case is T*(n/2). Again, this has an odd factor, so it cannot be a power of 2.
Now, let's show that all other numbers are attainable.If the total thickness is not a power of 2, then it has an odd factor. Assuming that we're allowed negative numbers, we can use either of the methods referred to above, and we're guaranteed to find an odd set of gagues and an even set of gagues (probably several, if our number has more than one odd factor). If we're not allowed to use negative numbers, then we can still always get an answer..., the proof being left as an exercise for the student.
(edit: Dangit, I can never get the color right first try)
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| « Last Edit: Dec 1st, 2002, 9:56pm by Chronos » |
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HappyFunBall
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I have no idea how to do the invisible text, so here's a spoiler warning.
The first 2 parts were already answered (32 is the first size which can't be measured). For the last part, you have to find the first consecutive 3 integers which can't be measured. A little insight and brute force reveals these to be 157-159. So 158 is the first width that can't be at least approximated +/- 1.
Back to the first part, note that the sum of k consecutive integers beginning at n is kn + (k^2 - k)/2. Setting equal to 86 (our target) and rearranging gives: 2n + k - 1 = 172/k. Since the left side must be an integer, the right side must be also. The only values of k that fit in the constraints of the problem and divide 172 are 1, 2, and 4. Checking each of these quickly reveals that the only answer that fits is k = 4, n = 20.
I used similar reasoning to help me with the last part.
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