Water clocks puzzle
Mar. 1st, 2006 01:10 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
jackPuzzle: I have two identical cylindrical lemonade bottles I punched small holes into the sides of, just above the bottom1. I find that when I fill them each with water the first empties in 225 seconds and the second more quickly in 75 seconds. Without any other form of measurement, how do I use them to time 25 seconds?
Q. That's just one of those hourglass/numerology puzzles. If one hourglass empties in time p minutes, and the second in time q, then calculate p-1 mod q, and you know pp-1=1 (mod q) so pp-1=1+nq. Run glass Q n times and glass P p-1 times; when the first has finished you have exactly one minute before the second does.
A. If you think that's easy, try it.
Q. It's a trick, isn't it.
A. Probably.
Q. Tilt one of the bottles...
A. If you like. They *are* cylindrical, so you *can* halve the amount of water by tilting them until the water only just covers the base, if you think it'll help.
Q. I could use a...
A. No other equipment is necessary. You can use additional vessels of unknown size if you want to.
Q. How quickly did you solve it?
A. Actually, I made the puzzle up myself last night.
Q. Are you sure your solution works?
A. I may be mistaken. In which case I apologise, and promise to feel very embarassed.
Q. Do I need to use much maths for this?
A. Some. Nothing not taught at A-level IIRC.
Q. Can I find a different solution?
A. I don't think so, but try; it might be better.
[-1] That's not a footnote, that means 'inverse of p'
[1] Really.
Q. That's just one of those hourglass/numerology puzzles. If one hourglass empties in time p minutes, and the second in time q, then calculate p-1 mod q, and you know pp-1=1 (mod q) so pp-1=1+nq. Run glass Q n times and glass P p-1 times; when the first has finished you have exactly one minute before the second does.
A. If you think that's easy, try it.
Q. It's a trick, isn't it.
A. Probably.
Q. Tilt one of the bottles...
A. If you like. They *are* cylindrical, so you *can* halve the amount of water by tilting them until the water only just covers the base, if you think it'll help.
Q. I could use a...
A. No other equipment is necessary. You can use additional vessels of unknown size if you want to.
Q. How quickly did you solve it?
A. Actually, I made the puzzle up myself last night.
Q. Are you sure your solution works?
A. I may be mistaken. In which case I apologise, and promise to feel very embarassed.
Q. Do I need to use much maths for this?
A. Some. Nothing not taught at A-level IIRC.
Q. Can I find a different solution?
A. I don't think so, but try; it might be better.
[-1] That's not a footnote, that means 'inverse of p'
[1] Really.
no subject
Date: 2006-03-01 05:57 pm (UTC)I suppose I should apply IB fluid dynamics to what would happen if you made the hole 3rd of the size, but then I'd have to charge you :)
no subject
Date: 2006-03-01 06:10 pm (UTC)A1. That wasn't the question asked.
A2. If you want me to provide a rationale, pretend I happen to have a 225 second egg timer. I can't use that to measure any intermediate time.
A3. I can 75 seconds with the water, I want to time 25 ;)
I suppose I should apply IB fluid dynamics to what would happen if you made the hole 3rd of the size,
I thought of you two, actually, when I was working out the rate of flow. I calculated that the velocity expelled should be proportional to the square root of the height of water, and the time taken to the square root of the initial height -- but I wasn't sure if you'd left simple non-turbulent flows like behind :)
but then I'd have to charge you :)
*flourishes red flag*