Mortar-less brick constructions

fatphil · 2005-12-15, 15:24

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?

xilman · 2005-12-15, 15:47

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

axn · 2005-12-15, 19:07

Five, counting the base brick

fatphil · 2005-12-16, 08:16

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

rogue · 2005-12-16, 13:36

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 ?

Greenbank · 2005-12-16, 13:41

Things without uniform weight distribution.

HTH.

fatphil · 2005-12-16, 14:46

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.

mfgoode · 2005-12-18, 11:14

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 ‘Prime obsession’
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 it’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 d’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.

2005-12-15, 15:24	#1
fatphil May 2003 7×37 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?

2005-12-18, 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 ‘Prime obsession’ 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 it’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 d’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.

2005-12-15, 19:07	#3
axn Jun 2003 2³·683 Posts	Five, counting the base brick

2005-12-16, 08:16	#4
fatphil May 2003 259₁₀ 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, 13:41	#6
Greenbank Jul 2005 386₁₀ Posts	Things without uniform weight distribution. HTH.

2005-12-16, 14:46	#7
fatphil May 2003 7·37 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.