6 ms·

Is that a puzzle? Isn't it just 2πr + d = 2π(r + d/2π)?

marcosdumay 4y ago
Yes, to both. It's very counter intuitive, and very easy to solve.
drran 4y ago
When stretched, the length of the rope will be two straight lines plus part of the circle. The length of the segment can be calculated using a formula for horizon. Then we can create an equation: (s arc distance to horizon for h)*2 + delta = (d straight Distance to horizon for h)*2, or d-s-delta/2=0. d=sqrt(h(2R+h)) km/m (see https://en.wikipedia.org/wiki/Horizon#Approximation https://en.wikipedia.org/wiki/Horizon#Approximation ) s=R*arccos(R/(R+h)) (see https://en.wikipedia.org/wiki/Horizon#Arc_distance https://en.wikipedia.org/wiki/Horizon#Arc_distance ) sqrt(h(2R+h))-R*arccos(R/(R+h))-delta = 0 6 foots is about 2 meters. sqrt(h(2*6371000+h))-6371000*arccos(6371000/(6371000+h))-2 = 0 h=306m
- gshubert17 4y ago
  I think I found a small typo, where you have delta (instead of delta/2) in the third line. I believe it should be sqrt(h(2R+h))-R * arccos(R/(R+h))-delta/2 = 0 This can be solved for h given delta numerically, such as with a spreadsheet. Use caution here because this formula leads to a loss of significant digits from cancellation of nearly identical terms. For example, given R = 6378000 meters and trying h = 193 meters: the first (d) term is 49618.0 meters and the second (s) term is 49617.0 meters. The difference is 1.0 meter, which is the half delta we wanted. But there's been a loss of 4 significant digits (because the first 4 digits in each value are the same: 4961). The smaller the arc distance, the smaller the angle, the worse the relative error becomes. It's possible to analytically factor out sqrt(2hR) from both the d and s terms. In the first case, d = sqrt(2Rh+h^2) = sqrt(2Rh) * (1+h/4R) using Taylor series for (1+h^2/2Rh)^0.5 In the second case, I can derive one formula, but the coefficients aren't the same as my numerical best fit, which is s = sqrt(2Rh) * (1-5h/12R) Combining the two series versions for d and s, gives us this for delta, delta = 4h/3 * sqrt(2h/R) Finally, solving for h given delta, take the cube root of h^3: h^3 = (9R/32) * delta^2 Thus for a delta of 2 meters, I get height h = 192.88 meters.
- texaslonghorn5 4y ago
  I believe this is overcomplicating the problem, the answer is just r' = (2pi*r + 6ft)/2pi - r = 6ft/2pi ≈ just under 1ft
  - drran 4y ago
    Just by folding the rope you will be able to reach 6ft/2 = 3ft. Check your formula, please.
- deleted 4y ago
  [deleted]
quantum_magpie 4y ago
That is for a smooth sphere, which the world is neither.

Yes, to both. It's very counter intuitive, and very easy to solve.

When stretched, the length of the rope will be two straight lines plus part of the circle. The length of the segment can be calculated using a formula for horizon. Then we can create an equation: (s arc distance to horizon for h)*2 + delta = (d straight Distance to horizon for h)*2, or d-s-delta/2=0. d=sqrt(h(2R+h)) km/m (see https://en.wikipedia.org/wiki/Horizon#Approximation https://en.wikipedia.org/wiki/Horizon#Approximation ) s=R*arccos(R/(R+h)) (see https://en.wikipedia.org/wiki/Horizon#Arc_distance https://en.wikipedia.org/wiki/Horizon#Arc_distance ) sqrt(h(2R+h))-R*arccos(R/(R+h))-delta = 0 6 foots is about 2 meters. sqrt(h(2*6371000+h))-6371000*arccos(6371000/(6371000+h))-2 = 0 h=306m

gshubert17 4y ago
I think I found a small typo, where you have delta (instead of delta/2) in the third line. I believe it should be sqrt(h(2R+h))-R * arccos(R/(R+h))-delta/2 = 0 This can be solved for h given delta numerically, such as with a spreadsheet. Use caution here because this formula leads to a loss of significant digits from cancellation of nearly identical terms. For example, given R = 6378000 meters and trying h = 193 meters: the first (d) term is 49618.0 meters and the second (s) term is 49617.0 meters. The difference is 1.0 meter, which is the half delta we wanted. But there's been a loss of 4 significant digits (because the first 4 digits in each value are the same: 4961). The smaller the arc distance, the smaller the angle, the worse the relative error becomes. It's possible to analytically factor out sqrt(2hR) from both the d and s terms. In the first case, d = sqrt(2Rh+h^2) = sqrt(2Rh) * (1+h/4R) using Taylor series for (1+h^2/2Rh)^0.5 In the second case, I can derive one formula, but the coefficients aren't the same as my numerical best fit, which is s = sqrt(2Rh) * (1-5h/12R) Combining the two series versions for d and s, gives us this for delta, delta = 4h/3 * sqrt(2h/R) Finally, solving for h given delta, take the cube root of h^3: h^3 = (9R/32) * delta^2 Thus for a delta of 2 meters, I get height h = 192.88 meters.
texaslonghorn5 4y ago
I believe this is overcomplicating the problem, the answer is just r' = (2pi*r + 6ft)/2pi - r = 6ft/2pi ≈ just under 1ft
- drran 4y ago
  Just by folding the rope you will be able to reach 6ft/2 = 3ft. Check your formula, please.
deleted 4y ago
[deleted]

quantum_magpie 4y ago

That is for a smooth sphere, which the world is neither.