- 08 Nov '07 07:26

Well, 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).*Originally posted by Dejection***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. - 08 Nov '07 19:43 / 1 edit

So far i've made it up to 5 but I've got the shakes after too many beers last night. I might be able to get to 7 when my hands settle down.*Originally posted by PBE6***What is the longest single-penny width bridge you can make by stacking pennies on top of each other on the edge of a table?** - 08 Nov '07 21:03 / 1 edit

Easy there, Frank Lloyd Right...I'm pretty sure he meant 7*Originally posted by wolfgang59***thats a minimum of 226 coins (I think)**

7 will take ... a MINIMUM of 1655 coins (I think) and a steady hand!

*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. - 08 Nov '07 21:31
*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!