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Ant on a stretchy rope puzzle

Дата публикации: 29-05-2026 07:36:03



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  • Thread starter Thread starter DaveC426913
  • Start date Start date Mar 14, 2026
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Here is my take on it:

Answer: Yes.

More and more of the rope is stretching out *behind* the ant. That halves the expansion the ant needs to deal with.

After one second, the ant has covered (1cm/(2km/2)) = 1/100,000th of the distance to the far end of the rope.
After two seconds the ant has covered (2cm/(3km/2)) = 1/75,000th of the distance to the far end of the rope.
After three seconds the ant has covered (3cm/(4km/2)) = 1/66,666th of the distance to the far end of the rope.

The ant is definitely making progress.

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As a spreadsheet:
1773533798345.webp

I guess I haven't proven that column G reaches one in a finite number iterations....

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I would say no. If you think of the end of the rope as the destination then for every second forward the ant is (1km - 1cm) further away.

Basically, the ant is effectively traveling away from the end of rope destination.

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Let the band have initial length ##L##, with one end at rest and the other moving at constant speed ##V##. Thus at time ##t## the band has length ##L+Vt##, and a point at distance ##x## from the stationary end has speed ##\frac x{L+Vt}V##. The ant's speed relative to the rubber is ##u##, so its speed relative to the stationary end is $$\frac{dx}{dt}=u+\frac{xV}{L+Vt}$$Maxima says that this is satisfied by $$\frac xL=\frac uV\left(1+\frac{Vt}L\right)\ln\left(1+\frac{Vt}L\right)$$meaning that the ant reaches ##x=L+Vt## when$$\begin{eqnarray*}
\frac Vu&=&\ln\left(1+\frac{Vt}L\right)\\
t&=&\frac LV\left(e^{V/u}-1\right)
\end{eqnarray*}$$Plugging in ##L=10^3\,\mathrm{m}##, ##V=10^3\,\mathrm{ms^{-1}}##, and ##u=10^{-2}\,\mathrm{ms^{-1}}## this becomes ##t=e^{10^5}-1\approx 10^{43400}\,\mathrm{s}##.

Bear in mind that the universe is less than ##10^{18}\,\mathrm{s}## old. So there is a solution to the maths (assuming I didn't make any mistakes), but it's not realistic.

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Edit: only just saw @Ibix' solution. Seems to be even a very similar choice of variable names.

The rope starts at length L and grows at rate u.
If at time t the ant is x from the far end of the rope then in further time dt the whole rope grows from ##L+ut## to ##L+ut+udt##, and the rope ahead of the ant grows in the same ratio: from ##x## to ##x\frac{L+ut+udt}{L+ut}##. That would make the ant ##x(1+\frac{udt}{L+ut})## from the far end. Against that, the ant moves on ##vdt##, so
##\dot x=x\frac{u}{L+ut}-v##.

I get ##x=(L+ut)(1-\frac vu\ln(\frac{L+ut}L))##, so ##x=0## at ##t=\frac Lu(e^{v/u}-1)##.

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I don't agree. Plugging the numbers into my formula, after 1s, ##\frac xL=10^{-5}\times 2\ln 2\approx 1.4\times 10^{-5}##, which is barely 0.007% of the length of the rope at that time.

The ant is barely moving with respect to the rope, and one tiny bit of the rope is barely moving with respect to the next tiny bit, so it would be quite surprising if the ant moved very far from its end in such a short time.

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DaveC426913
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I don't agree. Plugging the numbers into my formula, after 1s, ##\frac xL=10^{-5}\times 2\ln 2\approx 1.4\times 10^{-5}##, which is barely 0.007% of the length of the rope at that time.

The ant is barely moving with respect to the rope, and one tiny bit of the rope is barely moving with respect to the next tiny bit, so it would be quite surprising if the ant moved very far from its end in such a short time.

Right. Yeah. You're right.

The ant is 1/100,000th of the way along the rope, so only 1/100,000th (~1cm) of the total 1km expansion occurs behind it.

Oops.

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Ah. There it is.

"What we need to do is think about the ant's position as a fraction of the length of the rope. The above reasoning shows that this fraction is always increasing, but this is not yet enough. (The ant might asymptotically approach some fraction of the rope and never come close to reaching the target point.)"

This is how I went about intuitively resolving it (as a fraction) but I quickly realized it doesn't prove that it actually reaches the end.

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Here's the animation from the wiki page:
Ant_on_a_rubber_rope_animation.gif

The effect is vastly downgraded, as the rope's expansion is on the same order of magnitude as the ant's pace, but still...

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Here's the animation from the wiki page:
View attachment 370239

The effect is vastly downgraded, as the rope's expansion is on the same order of magnitude as the ant's pace, but still...

This reminds me of the expanding universe for some reason.
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It reminds me of the horizontal escalators in airports. Some people stand still while others continue walking but it feels like running.

I’ll have to consider my answer in light of these new insights. Thank you all for showing me error of my ways.

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It reminds me of the horizontal escalators in airports. Some people stand still while others continue walking but it feels like running.

I’ll have to consider my answer in light of these new insights. Thank you all for showing me error of my ways.

The worst part is when people stand on both sides of the moving walkway and theres no method to walk around them, and the connecting flight is only 30 minutes from boarding. Sometimes it is quicker to use the regular floor.
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I come up behind them and politely say "Excuse me". They always move.

If I am irritated, as I pass I say "Walk left; stand right."

And if they give me any guff, I say "New to the Big City are we?"

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I come up behind them and politely say "Excuse me". They always move.

If I am irritated, as I pass I say "Walk left; stand right."

And if they give me any guff, I say "New to the Big City are we?"

I do the same. Most of the time it works
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Here's the animation from the wiki page:
View attachment 370239

The effect is vastly downgraded, as the rope's expansion is on the same order of magnitude as the ant's pace, but still...

You can notice from the animation that the ant absolutely moves at an increasing speed, while the rope end absolutely moves at a constant speed.

By the way, 8,9×1043421 years is a long period, and the ant will definitely die before it reaches the rope end.

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