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Sometimes I'll tune into YouTube livestreams a few minutes late, but I want to watch the whole thing and catch up with the live feed, so I'll start from the beginning at 2x speed or 1x speed. I'd like to find a way to calculate exactly how much time it will take for me to reach the live feed.

So if I start watching 20 minutes behind at 2x speed, after 20 minutes of watching, I'll be 10 minutes behind, after 10 minutes, I'll be 5, etc.

I would guess this would look something like this

$ total = \frac{20}{2} + \frac{10}{2} + \frac{5}{2} + ... $

How would you create a general equation for this? Could you use summation, product, or something like a Taylor series? I recall an old VSauce Video about SuperTasks which feels relevant here as well.

Tessa Painter
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First, analogize to chasing a piece of driftwood that is in a stream, whose current is going $1$ mile per hour. Assume that you are running parallel to the driftwood, along the bank of the stream, at $x$ miles per hour. Also assume that you start out $k$ miles behind the driftwood.

Since you are gaining $(x-1)$ miles, for ever hour that you run, and since you start out $k$ miles behind the driftwood, it will take you $~\dfrac{k}{x-1}~$ hours to catch the driftwood.

The analogy is apt. Assume that you start out $k$ minutes behind the show. Also assume that you watch the show at $r$ times the normal speed. For example, $r=2$ implies that you are watching the show at twice the normal speed.

This means that every minute that the show progresses by $1$ minute, you progress by $r$ minutes. This implies that you are catching up to the metaphoric driftwood, at a rate of $(r-1)$ minutes per minute.

Note, that here, the distance to catch up is measured in minutes, rather than (for example) miles.

So, you are catching up by $(r-1)$ minutes, for every minute that you watch, and the driftwood starts out $k$ minutes ahead.

So, it will take you $~\dfrac{k}{r-1}~$ minutes to catch up.

user2661923
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  • Interesting special case is when watching at 2x speed, it will take you k minutes to catch up if you're k minutes behind, which to me seems intuitively incorrect for some reason – jonnybolton16 Aug 01 '23 at 09:27
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Say you are $t$ minutes behind and you start to watch the stream at $m$x speed.So after $\frac{t}{m}$ time you will be $\frac{t}{m}$ behind ,after $\frac{t}{m^2}$ time you will be $\frac{t}{m^2}$ behind and so on

So total time$=\frac{t}{m}+\frac{t}{m^2}+\frac{t}{m^3}+.......=\frac{\frac{t}{m}}{1-\frac{1}{m}}=\frac{t}{m-1}$

Anecdote

This is a really interesting way to visualise the Zeno's Dichotomy paradox which you can find here

Amit Kumar Basistha
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    This is the Achilles and the tortoise paradox, not the dichotomy paradox. – Sandejo Jun 14 '22 at 19:35
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    Achilles and the tortoise paradox is similar to the dichotomy one to be fair as both accounts to the fact that a task with infinite steps can indeed be completed – Amit Kumar Basistha Jun 14 '22 at 19:57
  • The paradox idea is cool, thanks! Presumably the proof that the paradox argument is false is that it assumes the tortoise will always be moving relative to Achilles, however there will come a point when Achilles reaches the tortoise and the relative motion of the tortoise becomes 0 thus he stops moving and Achilles can pass – jonnybolton16 Aug 01 '23 at 09:21
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Say you are consuming video at the rate $m$ (seconds per second, in other words a dimensionless quantity; double speed for instance gives $m=2$). At the same time new video is produced at the rate$~1$. Therefore the amount of unconsumed video decreases as the rate $m-1$ (assuming $m>1$; when $m\leq 1$ one would prefer saying it increases at the rate$~1-m$, although technically that is an equivalent statement). If $t$ is the time you were behind (in other words the amount of unconsumed video) initially, then it takes $t/(m-1)$ (an amount of time) to decrease that to $0$. For $m<1$ this gives a negative amount, and for $m=1$ a division by $0$; in either of these last two cases, the reality is that you will never catch up.

No sophisticated summation techniques are required. If, when driving behind another vehicle you observe the gap between you is closing at a constant speed $v$, and the distance is $l$ initially, then you don't need to go through Zeno's paradox or even do any computation involving your speed to know that you will hit that vehicle in time $l/v$ unless you hit the breaks in time.

Marc van Leeuwen
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