Originally posted by DejectionWell, ok, that is a theoretical result. I'd like to see someone try to successfully use that to bridge even 5 units of length (2.5 units from each side).
Idk. I know it's infinite for cards. You can experiment with cards if you like. Get a stack of cards, with length unit 2. Then push the top card out 1 unit, the next 1/2, the next 1/3. ect.
Since 1+1/2+1/3.... is infinity, then the length is infinite.
The website above shows a practical success of a three penny distance, using some counterbalance techniques. Counterbalance with your cards would probably make bridge building much easier.
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Originally posted by doodinthemoodIt does. The easy way to look at it is this:
1+1/2+1/3+1/4... never reaches 9.
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + ...
> 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + ...) + ...
= 1 + 1/2 + 1/2 + 1/2 + 1/2 + ...
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Originally posted by wolfgang59Easy there, Frank Lloyd Right...I'm pretty sure he meant 7 coins... 🙄
thats a minimum of 226 coins (I think)
7 will take ... a MINIMUM of 1655 coins (I think) and a steady hand!
🙄
OK! Found the glitch in my calculations. I get the harmonic series now, too. More specifically, the position "L" of the furthest edge of the top coin is given by:
L = sum(i=1...n) 1/(2i) = (1/2) * sum(i=1...n) 1/n
This series is divergent, so the theoretical length of the bridge is infinite.
Originally posted by PBE6😲
Easy there, Frank Lloyd Right...I'm pretty sure he meant 7 coins... 🙄
OK! Found the glitch in my calculations. I get the harmonic series now, too. More specifically, the position "L" of the furthest edge of the top coin is given by:
L = sum(i=1...n) 1/(2i) = (1/2) * sum(i=1...n) 1/n
This series is divergent, so the theoretical length of the bridge is infinite.
5 COINS is not worth posting about!!! I assumed a 5 coin width!