4 ms·
The answer to the first interpretation is considerably less than twice the height of the cat, and I think it's an interesting result as well: Assuming a 0.3 me
by liuhenry 9y ago
The answer to the first interpretation is considerably less than twice the height of the cat, and I think it's an interesting result as well:
Assuming a 0.3 meter high cat, you only need ~0.12mm of extra rope (about the width of two sheets of paper) to be able to pull it up at a point and let the cat through.
- davidmanescu 9y agoFollow-up question - after extending the rope by 0.12mm and pulling it up to allow the cat to pass under it, how far away are each of the two points where the rope lifts off the ground?
- mamon 9y agoHaven't done any calculations but I would guess half of the Earth circumference? because if you want to minimize the length of the rope then you have to maximize distance between those two points.
- tromp 9y agoAs the article explains, these points are about sqrt(2Rh) = sqrt(2 x 6400km x 0.3m) = sqrt(3840000m^2) ~ 1.96km away from the cat.
- davidmanescu 9y agoDid a back of the envelope trigonometry calculation as an approximation, seems to check out :)
- mannykannot 9y agoThat would be the same as the 'how far away is the horizon' calculation at the top of this thread, right? (assuming weightless rope.)
- jwilk 9y agoHere's the calculation: https://upload.wikimedia.org/wikipedia/commons/2/21/GeometricDistanceToHorizon.png https://upload.wikimedia.org/wikipedia/commons/2/21/Geometri... R ≈ 6.4E6 h = 0.3 d = √(2Rh + h²) ≈ 2.0E3 γ ≈ sin γ = d/(R + h) ≈ 3.1E-4 extra rope length = 2 × (d − γR) = 2R × (tan γ − γ) ≈ 2R × γ³/3 ≈ 0.13E-3