mersenneforum.org Mortar-less brick constructions
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 2005-12-15, 15:24 #1 fatphil     May 2003 3·7·11 Posts Mortar-less brick constructions How many bricks are required to build a structure such that one of the bricks overhangs at least one brick-length beyond the base of the structure, assuming nothing apart from normal gravity holds the structure together. (By an overhang of at least one brick-length, I mean that the nearest part of that brick to the base is horizontal distance >epsilon from the base, and the furthest part is horizontal distance >1+epsilon, where 1 is the brick's length, it's longest dimension.) What about an overhang of 2?
2005-12-15, 15:47   #2
xilman
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Quote:
 Originally Posted by fatphil How many bricks are required to build a structure such that one of the bricks overhangs at least one brick-length beyond the base of the structure, assuming nothing apart from normal gravity holds the structure together. (By an overhang of at least one brick-length, I mean that the nearest part of that brick to the base is horizontal distance >epsilon from the base, and the furthest part is horizontal distance >1+epsilon, where 1 is the brick's length, it's longest dimension.) What about an overhang of 2?
Ah, another old favourite.

It's the minimum i such that the harmonic series sum
exceeds 2 (resp. 3).

The sum reaches 2.083333333333333333333333333 at i=4 and
3.019877344877344877344877344 at i=11.

To get an overhang of ten, you need 33617 bricks, I believe.

Paul

 2005-12-15, 19:07 #3 axn     Jun 2003 34·5·13 Posts Five, counting the base brick
 2005-12-16, 08:16 #4 fatphil     May 2003 E716 Posts The base brick indeed is counted, and with that you have the right answer. It's pretty difficult to achieve with imperfect bricks. Here's me missing by 1, but able to get plenty of extra overhang from the extra "brick": http://fatphil.org/maths/bricks.jpg
2005-12-16, 13:36   #5
rogue

"Mark"
Apr 2003
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Quote:
 Originally Posted by fatphil The base brick indeed is counted, and with that you have the right answer. It's pretty difficult to achieve with imperfect bricks. Here's me missing by 1, but able to get plenty of extra overhang from the extra "brick": http://fatphil.org/maths/bricks.jpg
What are those ?

 2005-12-16, 13:41 #6 Greenbank     Jul 2005 2×193 Posts Things without uniform weight distribution. HTH.
 2005-12-16, 14:46 #7 fatphil     May 2003 3·7·11 Posts The thicker plastic casing is heavier than the thin aluminium, and so I was at a _disadvantage_ that way round. I just tried with some fairly uniform 'plastic': http://fatphil.org/maths/cards.jpg and it was much easier; I managed to hit the goal of 5.
 2005-12-18, 11:14 #8 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22×33×19 Posts Mortar-less brick constructions This problem is better explained by using a pack of standard cards which is more practical. The problem is beautifully treated in the very first chapter by John Derbyshire in his excellent book โPrime obsessionโ The total overhang is with just 4 bricks (not counting the base card). The total over hang is 1/2 + 1/4 +1/6 + 1/8 = 25/24. Yes your are right Paul its half the harmonic series. Some interesting facts he gives are: For 51 cards the total overhang is a shade less than 2.2594065907334. For 100 cards itโs a tad less than 2.58868875887. For a trillion cards we have a bit more than 14.10411839041479 overhang. The total over hang increases very slowly as the number of cards goes up but never comes to an end as the harmonic series is divergent See proof by Nicole dโOresme. mathworld.wolfram.com/HarmonicSeries.html - 32k - Cached - Similar pages Mally. P.S. 'Prime Obsession' is a must read for anyone interested in the RH.

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