Shoelace Clock problem

October 19, 2005 1:55 PM

I have a problem that asks what is the smallest time interval that you can accurately measure with a shoelace that burns at a non-constant rate, matches and a pair of scissors. I think this is just one of a whole family of problems like this. Where can I find other problems like this? Help me find the solution to this problem.

Complete statement of the problem: You are given some matches, a shoelace, and a pair of scissors. The shoelace burns like a fuse at a non-constant rate and takes 60 minutes to burn. However, it has a symmetry property: the burn rate at a distance x from the left end is the same as the burn rate at the same distance x from the right end. What is the smallest time interval that you can accurately measure with this shoelace?

posted by noether to sports, hobbies, & recreation (43 comments total)

Would it be cheating to tie the scissors to the end of the shoelace, ignore the matches, and make yourself a pendulum?
posted by wanderingmind at 2:05 PM on October 19, 2005

i can get 15 minutes
posted by andrew cooke at 2:07 PM on October 19, 2005

sorry that wasn't very useful was it? i didn't read the question. apologies again...
posted by andrew cooke at 2:08 PM on October 19, 2005

trying to be more useful: the strange thing about the question is the detail about it burning equally from both ends. that implies that there's some twist related to symmetry (at least to my mind, although in retrospect i don't know how to explain that intuition more clearly).
posted by andrew cooke at 2:14 PM on October 19, 2005

Here's an idea to help get to the solution:

Cut the shoelace in half. Burn one half of it from the middle out. If both ends burn out at the same time, you've got 15 minutes (60/2/2).

Cut the half remaining into half itself. Start it burning from the middle. If both sides burn out at the same time, you've got 7.5 minutes (15/2).

Etc.
posted by lorrer at 2:19 PM on October 19, 2005

"Chop the rope in half and simultaneously ignite both left ends. (Without reorienting either section, of course.) If they both burn out at the same time, we have measured 30 minutes. If not, instantaneously chop the unburnt portion of rope in half and ignite the non-burning left end. And so on, ad infinitum...

Since, at any given time, we have two pieces of rope burning in the same direction, once the entire rope burns out, we have measured 30 minutes!

If we treat this as a logic problem, and conceptualise the burning rope as a line segment with point(s) moving along it, this approach offers some advantages over Evan's solution. Firstly, as already mentioned, it does not implicitly assume a unidirectional burn rate. This appeals (to me) aesthetically. Secondly, it can immediately be generalised to yield a way of measuring any integer division of 60 minutes. "
posted by null terminated at 2:20 PM on October 19, 2005

Roll the shoelace into a ball.
Stand the Scissors on end on the ground.
Hold the shoelace ball as high as the top of the scissors.
Drop the shoelace.
Wait for it to hit the ground.

That's probably the shortest _repeatable_ time interval you can measure with the shoelace.
posted by Crosius at 2:21 PM on October 19, 2005

Okay, first what we know for sure.

Cut the rope in half and we have two 30min. timers.

Take one of them and light it on both ends. It will finish burning in 15 minutes.

Note where long the length of the rope the two burning ends meet. Mark this spot on the unburned 30min rope and cut it at that point.

We now have two 15 minute ropes remaining. (I think -- maybe I'm making a faulty assumption here).

Take one of them and burn on both sides. This will take 7.5 minutes. Mark the spot, etc., etc., etc.

I think we can do this an infinite number of times if we ignore things like scissor's accuracy and rope's particle makeup and all sorts of other things I don't know enough about. . .
posted by nobody at 2:22 PM on October 19, 2005

On preview: yes.
posted by nobody at 2:22 PM on October 19, 2005

Hmm, my previous suggestion will not work if, as implied, the burn rate along the sholace is different at any given point, but symmetrical.

In that case, you can still burn it from the middle out, giving you 30 minutes.

You could also do this: cut the string in half. Each half is 30 minutes. Start one string burning from the outside end, and the other from the side that was formerly the middle. Lay them so the outside end of one is on the same side as the inside end of the other, and they are paralllel. When t he burnt part of each reaches that of the other, you've got 30 minutes. cut that portion off. The strings you have left together measure another 30 minutes, but all the points along them could have different burn rates. So I'm not sure what you could do with the string thereafter, and this is a lot more complicated than just burning it from the middle out, plus you lack the averaging potential of making sure the ends burn out simultaneously.
posted by lorrer at 2:27 PM on October 19, 2005

nobody put it together, thanks nobody!
posted by lorrer at 2:28 PM on October 19, 2005

My gut tells me that with perfect measurement and cutting, you can get arbitrarily accurate times using limits. This isn't very rigorous, but here are my thoughts.

First, if you cut it into halves A and B, every subsegment of A has a corresponding segment in B with the same burn time.

Now cut up one of the halves into an arbitrary number of sections (n) of equal length, and start burning them all at the same time. Whenever one of those sections burns out, note what fraction of the other section remains. This gives a relationship between the burn time of each subsection.

Since we also know that the sum of burn times of all subsections must equal 30 minutes, that gives us enough information to calculate the burn time of each subsection.

Even with one half of our shoelace gone, there are still n subsections in the other half, and we know how long it would take for each of those burn. As n increases, the accuracy of our timing grows.
posted by Eamon at 2:30 PM on October 19, 2005

Would it be cheating to tie the scissors to the end of the shoelace, ignore the matches, and make yourself a pendulum?

How does that help? You can tell that the pendulum swings with a constant period, but you don't have a way of figuring out what that period is, as far as I can tell.

I can get 7.5 minutes. I don't know if that's minimal.

ROT13:
Phg gur fubrynpr va unys (ol sbyqvat vg bire naq phggvat vg ng gur zvqcbvag). Lbh abj unir gjb ynprf juvpu jvyy ohea va bar-unys ubhe, vs yvg sebz bar raq. Fvzhygnarbhfyl yvtug bar cvrpr ng bar raq, naq gur bgure cvrpr ng obgu raqf. Gur bar gung oheaf ng obgu raqf jvyy ohea va svsgrra zvahgrf. Jura gur cvrpr yvg ng obgu raqf oheaf bhg, yvtug gur frpbaq raq bs gur bgure cvrpr. Gur gvzr sebz yvtugvat gur frpbaq raq bs gung cvrpr, gb gur gvzr jura gung cvrpr vf shyyl ohearq, vf frira naq n unys zvahgrf.

On preview: nobody's solution only works down to 7.5 minutes, as there is no guarantee that the two 15-minute ropes are identical; they may not even be the same length!
posted by DevilsAdvocate at 2:30 PM on October 19, 2005

1) Use the shoelace as a reference length
2) Hold the scissors shoelace length from the ground
3) Drop scissors.

Additionally, you can continue to fold shoelace in half and cut to get smaller and smaller segments, as far as you feel comfortable that you'll still be able to hold and drop the scissors from.

A pendulum might be a good idea as well. If you knew how long the shoelace actualy was, you could convert those units into minutes and seconds.
posted by delmoi at 2:32 PM on October 19, 2005

nobody's solution is a much more elegant version of mine. Way to go!
posted by Eamon at 2:32 PM on October 19, 2005

note what fraction of the other section remains

How do you intend to precisely measure that?

I just realized I can improve on my previous solution and get 3.75 minutes:

Hfr abobql'f zrgubq gb trg gjb svsgrra-zvahgr ebcrf, gura hfr gubfr va gur zrgubq V qrfpevorq va zl cerivbhf fbyhgvba vafgrnq bs gur gjb guvegl-zvahgr ebcrf.
posted by DevilsAdvocate at 2:37 PM on October 19, 2005

On Preview: I'm not reading the obfuscation but I think we're in agreement Devil's Advocate.

Nobody's solution doesn't work past 15 minutes. Picture your lengths of rope

[------------------]

[------------------]

30 minutes each

[-----------] [----]

We burn one in 15 minutes and it marks above. Cut the other rope along those sections

We're left with:

[-----------]

[---]

Each will finish burning in 15 minutes.

You can't perform the same trick you did the first time because now the ropes are different lengths.

However, it should be noted that if you light one of these ropes on fire at BOTH ends, it will burn out in 7.5 minutes.

This is the smallest time quantum that is obvious at the moment. Sorry I can't solve this definitively :).
posted by onalark at 2:42 PM on October 19, 2005

DA on checking your encryption I agree on the 3.75 minutes.
posted by onalark at 2:46 PM on October 19, 2005

DevisAdvocate: We have 2n subsections of shoelace. n have been burned, but n are fine at this point. You can approximate (with arbitrary precision) the relative length of the burnt pieces by folding/cutting any one of the unburnt pieces in half an arbitrary number of times.
posted by Eamon at 2:48 PM on October 19, 2005

other puzzles of the same flavour involve finding quantities of liquid using beakers of known volume in combination. however, i've no idea how you describe the similarity in a way that will let you find other variations.

best i can do with the (readable) help above is 3.75, too. there's a different first step, which is to burn the two pieces at one end each, but use opposite ends, note when/where the flames "meet" (if the halves placed parallel), and extinguish them. but the variation doesn't buy you anything, that i can see.
posted by andrew cooke at 2:57 PM on October 19, 2005

Inspired in part by Eamon, I think I've found a way around the problem with his method that will allow you to measure an arbitrarily small interval of time (if you have perfect timing, and don't mind performing a potentially infinite number of tasks in a finite time).

Phg gur ebcr vagb a (a zhfg or rira) frtzragf, naq yvtug bar raq bs rnpu fvzhygnarbhfyl. Jurarire bar frtzrag tbrf bhg (juvpu unq orra oheavat sebz bayl bar raq), yvtug gur frpbaq raq bs n fgvyy-oheavat frtzrag. Vs n frtzrag juvpu unq orra oheavat sebz obgu raqf tbrf bhg, yvtug n frtzrag juvpu vf nyernql oheavat fbzrjurer va gur zvqqyr bs gung frtzrag (perngvat gjb arj oheavat frtzragf; obgu oheavat ng obgu raqf vs gur bevtvany unq orra oheavat ng obgu raqf; bar oheavat ng obgu raqf naq bar ng bar raq vs gur bevtvany unq orra oheavat bayl ng bar raq). Abgr gung ng nal tvira gvzr, gurer ner a oheavat raqf. Guhf, gur gvzr sbe gur ragver ebcr gb ohea vf fvkgl zvahgrf qvivqr ol a.

On preview: OK Eamon, I see that you can measure the unburned length as a fraction of the whole, but I'm not sure where you get from that: let's say when segment 1 is fully burned, segment 2 is 3/4 burned. What then? Since the rope doesn't burn evenly, you can't conclude that the total burning time for segment 2 is 4/3 that of segment 1.
posted by DevilsAdvocate at 2:59 PM on October 19, 2005

Wow. I close a browser window to get some work done, and when I return I find I took something an infinite number of steps too far. Indeed. My answer only works to 7.5.
posted by nobody at 3:01 PM on October 19, 2005

I love these problems!

Since the original question also asked where to find more, I'll point out that [wu:riddles] is a wonderful cache of them (at least for a broad understanding of "problems like this"). That site sometimes offers hints, but doesn't give solutions.
posted by gorillawarfare at 3:06 PM on October 19, 2005

Finally figured out the ROT13 bit. Brilliant, D.A.
posted by nobody at 3:20 PM on October 19, 2005

DA I follow your logic for the infinitely small time step and it appears sound to me.
posted by onalark at 3:25 PM on October 19, 2005

Using null terminated's link, I found a link to the Shoelace Clock problem in a book for Martin Gardiner. The solution is on page 64. Their solution gives 3.75 minutes as the answer. I think the solution is similar to DAs.
posted by noether at 3:27 PM on October 19, 2005

You could just drop the scissors from some height x. You measure X by burning the laces.
posted by rdr at 3:56 PM on October 19, 2005

How does that help? You can tell that the pendulum swings with a constant period, but you don't have a way of figuring out what that period is, as far as I can tell.

Couldn't you build some kind of Foucalt's Pendulum contraption? I very hazily suspect that the period must be related to the rate of the earth's rotation or something...but I could be totally wrong about that.
posted by juv3nal at 4:15 PM on October 19, 2005

I can't see better than 3.75
Cut the rope in half. You have two identical ropes of 30 mins.
Start burning one rope at both ends. Measure where the flame meets.
Cut the second rope at that point. You have two different ropes of 15 mins.
Start burning one of those two ropes at both ends. At the same time, start burning the second rope at one end.
When the both-end rope burns out, stop the second rope burning. You have a rope which will burn for 7.5 mins.
Burn this rope at both ends. When the flames meet, you've measured 3.75 mins.
posted by seanyboy at 4:39 PM on October 19, 2005

Assuming the pendulum will keep swinging...

Cut the rope in half.
Tie the scissors to one end of rope 1.
Swing the rope from left to right.
Start burning the second rope at both ends
count the number of swings it takes to burn the second rope.
Divide 15 mins by the number of swings.
That's your smallest measurable time instance.
posted by seanyboy at 4:47 PM on October 19, 2005

As to where you can find other problems like this.
How would you move mount fuji has problems like this and examines scenarios for solving them.
posted by seanyboy at 4:52 PM on October 19, 2005

I like seanyboy's answer, but it doesn't take into account the natural slowdown of the swinging rope. I'm sure some physicists in here could provide an equation describing how long a pendulum will swing unattended before coming to a stop. I'd guess the various single-swing durations would decrease logarithmically, and of course their sum would add up to the 15 minute total, the question would be whether or not the smallest swing's duration can be expressed as a fraction of that 15 minutes using the above formula.
posted by odinsdream at 6:20 PM on October 19, 2005

Though, just from a practical standpoint, I doubt a swinging shoelace would keep swinging unattended for the requisite 15 minutes. Combine this technique with some of the others expressed above, like those that give you the ability to measure down to 3.5 minutes, and you might have something.
posted by odinsdream at 6:21 PM on October 19, 2005

The period of the pendulum's oscillation is nearly independent of the size of the swing. Guessing how long it will take to stop swinging is nearly impossible, but you can just give the pendulum an occasional push without any substantial problems. It won't be perfectly accurate, but then again, we're comparing this to measuring time with a burning shoelace.
posted by yarmond at 8:20 PM on October 19, 2005

Couldn't you build some kind of Foucalt's Pendulum contraption? I very hazily suspect that the period must be related to the rate of the earth's rotation or something

It depends on both the rate of the earth's rotation and your latitude. A Foucault pendulum will complete a rotation in 23 hours 56 minutes (1 sidereal day) if it's at a pole, and won't rotate at all at the equator. More generally, if you're at latitude L, it will complete a rotation in a time equal to 1 sidereal day*|csc L|. Good idea, if you know your latitude, or have some way of determining it.
posted by DevilsAdvocate at 9:39 PM on October 19, 2005

One could stick the scissors in the ground, and use them as the center point of a circle. Using the shoelace (or a portion of it) as a constant radius, the matches can be laid out in a nice circle.

Instant sun-dial.

Assuming a sunny day, we should be able to measure any length of time: provided we look closely enough.
posted by aladfar at 10:11 PM on October 19, 2005

aladfar: Damn you, that was precisely my thought. And if you use the width of the shoelace tip as the space between the matches poked into the ground, you can get even distribution. From there you can work out the time measurement from the length of the day.
posted by Kickstart70 at 11:41 PM on October 19, 2005

Wait, so long as you have matches, use those as timers too. Cut the rope in half, leaving two pieces, each good for 30 min. Light one, and light matches as it burns. As each match burns out, light the next (maybe lay them end to end on the ground). This lets you calculate the time per match (30 min / n matches). You can even repeat the process with the other piece of rope and average the two numbers. It isn't perfectly precise, but it should be a much smaller amount of time. This does assume you get as many matches as you want, and that matches are more uniform than rope in this theoretical universe.
posted by lorimt at 11:49 PM on October 19, 2005

Damn you people don't know much about math.

Allowing for the fact that I have a few degrees in science related disciplines, as well as a graduate degree in nuclear physics, allow me to point out that this is remarkably similar to half-life problems. But I'll get to that later.

First, nobody's solution above is false. He has assumed, incorrectly, that the shoelace burns at a constant rate that increases. Not true. It's possible that the rate both increases and decreases. Therefore, all you can really say is that the midpoint on the shoelace is the 30 minute mark.

I'm going to ignore tricky solutions that involve things like dropping them and the like, since they also have problems based on the non-linear burn rate.

Consider the problem in it's most simple state: half the shoelace burns in 30 minutes. What else do you know for a fact? Nothing! That's it. Walk away!

KFJ
posted by kungfujoe at 3:15 AM on October 20, 2005

Other things we know for a fact.
1) If we light a 30 minute shoelace at both ends, it'll burn out in 15 minutes.
2) Shoelace 2 is identical to shoelace 1.

I don't know much about math, but I have a degree in STFU from the school of hard knocks.
posted by seanyboy at 3:42 AM on October 20, 2005

Allowing for the fact that I have a few degrees in science related disciplines, as well as a graduate degree in nuclear physics, allow me to point out that this is remarkably similar to half-life problems.

When all you have is a hammer, everything looks like a nail.

First, nobody's solution above is false.

Gee, and you're only the fourth person in this thread to point that out. (nobody himself was the third.)

He has assumed, incorrectly, that the shoelace burns at a constant rate that increases.

No, he never makes that assumption. His solution is wrong, but not for that reason.
posted by DevilsAdvocate at 6:31 AM on October 20, 2005

I think I've been trolled. "A constant rate that increases?" From someone who alleges to have an advanced degree in physics?
posted by DevilsAdvocate at 6:35 AM on October 20, 2005

Since the rope doesn't burn evenly, you can't conclude that the total burning time for segment 2 is 4/3 that of segment 1.

Ah, nice catch. I guess I could argue that as the length of each segment decreases, so does the variation in their burning rates. But I don't think it's worth going there.
posted by Eamon at 7:37 AM on October 20, 2005