Is a broken clock right twice a day?

Question

I was thinking about this expression, and I wondered if it holds true when the clock is slow. I can imagine a slow clock which is not right at all in the span of 12 hours—imagine a clock which ticks 5 minutes every 12 hours, which points to 11:59 at exactly 12:00. 12 hours later it is 12:00 and the clock is pointing to 12:04. The clock won't be right for another 5 minutes or so, and will have gone more than 12 hours without being right.

Is there an upper bound on the amount of time that a clock which moves at a slow but constant rate will spend being wrong before it is right? What would be a good way to model this? How should I tag this question?

Answer 1

If the bad clock runs wrong by a factor $k\in\mathbb{R}_+$, for example $k=2$ means that it runs at double speed, then the first time the bad clock is correct is $(12\ \mbox{hours})/|1 - k|$ after the start (where the two clocks agree). By choosing $k$ very close to $1$, you can make that time span as long as you desire (so no, there's not an upper bound).

Answer 2

I'm more familiar with the saying as "a stopped clock is right twice a day", which is obviously correct for a standard 12 hour analog clock (with a possible exception involving leap-seconds).

As for your question, what I would consider to be the key insight is that, if you have two clocks running at (different) constant speeds, $\tau$, $\omega$, then the difference between them also changes at a constant rate, $|\tau-\omega|$. It follows immediately that, if the first clock is correct and working properly, then $\tau=1$, and the difference of the other from the correct time, $t$, is given by $|1-\omega|t$.

Since clocks are periodic, if the broken clock is correct at $t=0$, it will also show the correct time at any $t$ such that $|1-\omega|t=(12hours)\times k$, or $t=\frac{(12 hours)\times k}{|1-\omega|}$ (for $k\in\mathbb{Z}$). The period of that, $T$, is clearly given by $T={12hours\over|1-\omega|}$, and, since $\omega$ can be made as close as you choose to $1$, $T$ can be made as close to $\infty$ as you wish.

That also makes intuitive sense. If a clock loses or gains time very slowly - say a second per year - then it also takes a very long time for that error to be noticeable, and a much, much longer time before that error accumulates to be 12 hours.

Meanwhile, a clock that's running at a high enough speed (forward or backward) will display every time multiple times per second, including the correct time, so, at least in theory, for any short, finite time interval, $\delta t$, you can describe a clock that runs continuously at a fixed speed, is correct at some instant during the interval $[t,\delta t)$ for any time $t$, and which is completely useless in practice.

Answer 3

Let's let the 'correct' time be modeled by $f=x\pmod {12}$. If we have another clock moving at a constant rate, then that's going to look like $f=\alpha\cdot x+\beta\pmod{12}$, for some $0<\alpha<1$ and $0\leq\beta<12$. $\alpha$ represents the 'slowed' tick rate and $\beta$ the time offset. Is this a good enough model? You can more easily solve $x\equiv\alpha\cdot x+\beta\pmod{12}$.

Edit: or see that this equation has no solution, as André points out.

Answer 4

Use $12$ hours as time unit, and denote real time by $t$. The time shown by the clock is $$f(t):=\lambda t+ c\ ,$$ where $\lambda>0$ is a constant. When the clock is gradually getting behind this $\lambda$ is a trifle smaller than $1$. The clock shows the correct time whenever $f(t)-t\in{\mathbb Z}$, i.e., $$(\lambda-1) t+c\in{\mathbb Z}\ .$$ The time interval $\Delta t$ between two such incidences is $$\Delta t={1\over |\lambda -1|}\ .$$ An example: When the clock is $1$ minute per day behind then $|\lambda-1|={1\over 24\>\cdot\> 60}$. It follows that $\Delta t=1440$, which corresponds to $720$ days.