Overhang
tc...@lsa.umich.edu
of the American Mathematical Monthly entitled "Overhang." Earlier versions
of this work may be found, for example, on Zwick's homepage.
http://www.cs.tau.ac.il/~zwick/
The topic is the old problem of stacking bricks on a table so that they
hang over the edge by an arbitrarily large amount. The standard solution
involves a harmonic series, but the authors show that one can do much
better if one allows more than one brick at each level.
One thing that I mentioned to Paterson and Zwick (though apparently too
late for them to include it in their Monthly article) is that there is,
in fact, at least one place in the literature where the possibility of
more than one brick on each level is explicitly considered. I refer to
the book "Ingenious Mathematical Problems and Methods" by L. A. Graham
(Dover 1959), specifically the problem about the "Bridal Arch." In the
solutions section, Graham reproduces three proposed solutions to the
problem sent in by his readers, all of which have more than one brick
per level. (I wish I had a scanner---perhaps someone else can scan in
the relevant pages and post them on the web somewhere?)
However, Graham dismisses all three solutions as incorrect "models of
graceful instability" without any analysis! I asked Paterson and he
said that at least some of these solutions might be correct, but he did
not actually check them in detail. Presumably, Graham simply assumed
that those other solutions could not be correct. Can anyone here either
vindicate them, or show that they are wrong?
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
William Elliot
> Mike Paterson and Uri Zwick have a very nice article in the current issue
> of the American Mathematical Monthly entitled "Overhang." Earlier versions
> of this work may be found, for example, on Zwick's homepage.
>
> http://www.cs.tau.ac.il/~zwick/
>
> The topic is the old problem of stacking bricks on a table so that they
> hang over the edge by an arbitrarily large amount. The standard solution
> involves a harmonic series, but the authors show that one can do much
> better if one allows more than one brick at each level.
>
Are the bricks the usual red clay fired 2" x 4" x 8" bricks of uniform
density or are they lite concrete blocks 6" x 6" x 12" hollow out with
holes or are they ornamental bricks 12 " x 12" x 4" hollow out
symmetrically with holes?
Can the bricks be stacked anyway one chooses or must they stacked like a
portion of a wall, level by level? Can be the bricks be Leggos?
Is there a known maximal or most efficient geometrical configuration for
laying the bricks?
> One thing that I mentioned to Paterson and Zwick (though apparently too
> late for them to include it in their Monthly article) is that there is,
> in fact, at least one place in the literature where the possibility of
> more than one brick on each level is explicitly considered. I refer to
> the book "Ingenious Mathematical Problems and Methods" by L. A. Graham
> (Dover 1959), specifically the problem about the "Bridal Arch." In the
> solutions section, Graham reproduces three proposed solutions to the
> problem sent in by his readers, all of which have more than one brick
> per level. (I wish I had a scanner---perhaps someone else can scan in
> the relevant pages and post them on the web somewhere?)
>
Ascii Art?
The Last Danish Pastry
news:49614960$0$307$b45e...@senator-bedfellow.mit.edu...
Done...
http://thelastdanishpastry.blogspot.com/2009/01/bridal-arch.html
--
Clive Tooth
http://www.shutterstock.com/cat.mhtml?gallery_id=61771
tc...@lsa.umich.edu
William Elliot <ma...@rdrop.remove.com> wrote:
>Are the bricks the usual red clay fired 2" x 4" x 8" bricks of uniform
>density or are they lite concrete blocks 6" x 6" x 12" hollow out with
>holes or are they ornamental bricks 12 " x 12" x 4" hollow out
>symmetrically with holes?
Graham explicitly says they're 2x4x8.
>Can the bricks be stacked anyway one chooses or must they stacked like a
>portion of a wall, level by level? Can be the bricks be Leggos?
They can be stacked any which way, but they're perfectly smooth with flat
surfaces and sharp edges, so they're not Legos.
>Is there a known maximal or most efficient geometrical configuration for
>laying the bricks?
For this, you should see Paterson and Zwick's paper, that I linked to.
>Ascii Art?
O.K., so here's my attempt to reproduce (the critical portion of) one of
the proposed solutions. Does this one work? That is, is it in equilibrium?
Is it a stable equilibrium? If it works, then how does it compare to the
overhang you would get with the standard solution?
_ _ _ _ _ _ _ _ _
| |_ _ _ _| | |
| |_ _ _ _| _|_ _ _ _|
| |_ _ _ _|_ _| |
|_|_ _|_ _ _ _|_ _ _ _|
| | | | | | |
| | |_| |_|_ _ _ _|
| | | | | |
|_|_|_|_ _|_ _ _ _|
|_ _ _ _| |
|_ _ _ _|_ _ _ _|
| | | |
| |_|_ _ _ _|
| | |
|_ _|_ _ _ _|
| | |
|_|_ _ _ _|
|_ _ _ _|
|_ _ _ _|
William Elliot
> William Elliot <ma...@rdrop.remove.com> wrote:
Oh dang. Please don't clip the problem. As it's not included, I'm not
responsible for any errors caused from not having the problem statement
close at hand to review. (I compose off line.)
> Graham explicitly says they're 2x4x8.
>
> They can be stacked any which way, but they're perfectly smooth with flat
> surfaces and sharp edges, so they're not Legos.
>
Pile 2" x 4" x 8" bricks on a table so that they stick out as far as
possible. Are the bricks of uniform density?
>> Is there a known maximal or most efficient geometrical configuration for
>> laying the bricks?
>
> For this, you should see Paterson and Zwick's paper, that I linked to.
>
>> Ascii Art?
> O.K., so here's my attempt to reproduce (the critical portion of) one of
> the proposed solutions. Does this one work? That is, is it in
> equilibrium? Is it a stable equilibrium? If it works, then how does it
> compare to the overhang you would get with the standard solution?
>
What is there to compare other than the extend of overhang?
Nothing, IIRC. There no additional criteria were given.
This is complex problem because the torque off the edge of the table has
to be balanced with torque on the table. In addition, the bricks that
support other extended bricks at each level have to be counter balanced.
The counter balancing will in turn add more torque to the whole system
What's the standard solution?
> _ _ _ _ _ _ _ _ _
> | |_ _ _ _| | |
> | |_ _ _ _| _|_ _ _ _|
> | |_ _ _ _|_ _| |
> |_|_ _|_ _ _ _|_ _ _ _|
> | | | | | | |
> | | |_| |_|_ _ _ _|
> | | | | | |
> |_|_|_|_ _|_ _ _ _|
> |_ _ _ _| |
> |_ _ _ _|_ _ _ _|
> | | | |
> | |_|_ _ _ _|
> | | |
> |_ _|_ _ _ _|
> | | |
> |_|_ _ _ _|
> |_ _ _ _|
> |_ _ _ _|
> --
I assume the last three lines are the table top.
Let's look at the top.
_ _ _ _ _ _ _ _ _
| |_ _ _ _| | |
| |_ _ _ _| _|_ _ _ _|
| |_ _ _ _|_ _| |
|_|_ _|_ _ _ _|_ _ _ _|
| | | | | | |
| | |_| |_|_ _ _ _|
| | | | | |
|_|_|_|_ _|_ _ _ _|
I remove the brick that don't come into play where the "table top"
for the bottom is 1/4 brick to the left of the right end of the right
most brick. Torque by level: left right.
_ _ _ _ _ _ _ _
|_ _ _ _| | |
|_ _ _ _| _|_ _ _ _| 0 10
|_ _ _ _|_ _| |
|_ _|_ _ _ _|_ _ _ _| > 1 6
| | | | | |
| |_| |_|_ _ _ _| 15 3
| | | | |
|_|_|_ _|_ _ _ _| 3 1
Ok, this portion is stable.
tc...@lsa.umich.edu
William Elliot <ma...@rdrop.remove.com> wrote:
>Oh dang. Please don't clip the problem. As it's not included, I'm not
>responsible for any errors caused from not having the problem statement
>close at hand to review. (I compose off line.)
Here's the link to the main paper on the subject again.
http://www.cs.tau.ac.il/~zwick/
>Pile 2" x 4" x 8" bricks on a table so that they stick out as far as
>possible. Are the bricks of uniform density?
Yes.
>What is there to compare other than the extend of overhang?
That's what I meant; sorry if it wasn't clear.
>What's the standard solution?
One brick per level, where brick number n-1 from the top sticks out 1/n
of a brick length from the nth brick.
Han de Bruijn
> In article <2009010601...@agora.rdrop.com>,
> William Elliot <ma...@rdrop.remove.com> wrote:
>
>>Oh dang. Please don't clip the problem. As it's not included, I'm not
>>responsible for any errors caused from not having the problem statement
>>close at hand to review. (I compose off line.)
>
> Here's the link to the main paper on the subject again.
>
> http://www.cs.tau.ac.il/~zwick/
>
>>Pile 2" x 4" x 8" bricks on a table so that they stick out as far as
>>possible. Are the bricks of uniform density?
>
> Yes.
>
>>What is there to compare other than the extend of overhang?
>
> That's what I meant; sorry if it wasn't clear.
>
>>What's the standard solution?
>
> One brick per level, where brick number n-1 from the top sticks out 1/n
> of a brick length from the nth brick.
http://mathforum.org/kb/message.jspa?messageID=5506453&tstart=0
Han de Bruijn
tc...@lsa.umich.edu
The Last Danish Pastry <cli...@gmail.com> wrote:
>Done...
>http://thelastdanishpastry.blogspot.com/2009/01/bridal-arch.html
Excellent! Thank you very much!
Bill Dubuque
>
> Mike Paterson and Uri Zwick have a very nice article in the current issue
> of the American Mathematical Monthly entitled "Overhang." Earlier versions
> of this work may be found, for example, on Zwick's homepage.
>
> http://www.cs.tau.ac.il/~zwick/
>
> The topic is the old problem of stacking bricks on a table so that they
> hang over the edge by an arbitrarily large amount. The standard solution
> involves a harmonic series, but the authors show that one can do much
> better if one allows more than one brick at each level.
>
> One thing that I mentioned to Paterson and Zwick (though apparently too
> late for them to include it in their Monthly article) is that there is,
> in fact, at least one place in the literature where the possibility of
> more than one brick on each level is explicitly considered. I refer to
> the book "Ingenious Mathematical Problems and Methods" by L. A. Graham
> (Dover 1959), specifically the problem about the "Bridal Arch." In the
> solutions section, Graham reproduces three proposed solutions to the
> problem sent in by his readers, all of which have more than one brick
> per level. (I wish I had a scanner---perhaps someone else can scan in
> the relevant pages and post them on the web somewhere?)
Someone already scanned the whole book. It can easily be located
via the usual ebook databases, e.g. gigapedia.com links to
http://ifile.it/znhvyqk/126631.rar
Rich Grise
> Mike Paterson and Uri Zwick have a very nice article in the current issue
> of the American Mathematical Monthly entitled "Overhang." Earlier
> versions of this work may be found, for example, on Zwick's homepage.
>
> http://www.cs.tau.ac.il/~zwick/
>
> The topic is the old problem of stacking bricks on a table so that they
> hang over the edge by an arbitrarily large amount. The standard solution
> involves a harmonic series, but the authors show that one can do much
> better if one allows more than one brick at each level.
>
Am I allowed to overhang at each end? If so, then wouldn't the answer be
"an arbitrary number"? i.e., say:
____
__|____|__
__|____||____|__
|____||____||____|
|____||____|
|____|
and so on?
Thanks,
Rich
tc...@lsa.umich.edu
Rich Grise <ri...@example.net> wrote:
>Am I allowed to overhang at each end? If so, then wouldn't the answer be
>"an arbitrary number"? i.e., say:
>
> ____
> __|____|__
> __|____||____|__
> |____||____||____|
> |____||____|
> |____|
>
>and so on?
You are allowed to hang over both ends, and your proposal here is a
plausible one. However, it is not correct. If you work it out in detail,
you will find that your configuration loses equilibrium eventually,
I think when n=5 (if your picture above is n=3). On the other hand,
modifications of your idea *do* work. Again, see Zwick's homepage.
Grover Hughes
> the proposed solutions. Does this one work? That is, is it in equilibrium?
>
> _ _ _ _ _ _ _ _ _
> | |_ _ _ _| | |
> | |_ _ _ _| _|_ _ _ _|
> | |_ _ _ _|_ _| |
> |_|_ _|_ _ _ _|_ _ _ _|
> | | | | | | |
> | | |_| |_|_ _ _ _|
> | | | | | |
> |_|_|_|_ _|_ _ _ _|
> |_ _ _ _| |
> |_ _ _ _|_ _ _ _|
> | | | |
> | |_|_ _ _ _|
> | | |
> |_ _|_ _ _ _|
> | | |
> |_|_ _ _ _|
> |_ _ _ _|
> |_ _ _ _|
> --
> Tim Chow tchow-at-alum-dot-mit-dot-edu
Yes, the above configuration is stable, and in fact, at least 3 of the
bricks shown are redundant. They are the 3 "on-edge" ones which are
lying one above the other in 3 separate levels. The bottom 2 bricks
are also redundant and can be considered as representing the
supporting structure for the tower of bricks. After removing the 3
bricks I mentioned above, this leaves 22 bricks to reach an overhang
of 2.
I have found what I believe to be the smallest number of bricks needed
to reach an overhang of exactly two L, and I wonder if others of you
have found this same value, which is 15. If you can beat this, I'd be
interested. The stacking method I found follows a pattern which
applies to other overhangs in increments of L/2, where L is the length
of a single brick or block. The formula for the number of bricks
required to reach a specified overhang in terms of half-lengths is
easily found from consideration of the results for the first few such
overhangs.
Here's my stack for the 2l overhang:
____ Let each brick measure 1 x 2 x 4 inches and weigh
1 pound.
|_15_|
|_14_|
|_13_|____
|_11_|_12_|____ _________
|__8_|__9_|_10_|__|____4____|
|__6_|__7_|__|____3____|\p <- there is a void between
bricks 7 and 3,
|__5_|___|____2____|\q <- also one between bricks 5
and 2.
__________|____1____|\r
///////////////|\t
///////////////| Sum of moments about p = 0 i.e., brick 4 is balanced
///////////////| Sum of moments about q = w4 * 2 - w10 * 2
/////////////| = 1*2 - 1*2 = 0 , so
balanced.
////////| Sum of moments about r = w3 * 2 + w4 * 4 - (w7 + w9
+ w12)*2
= 2 + 4 -6 = 0 , so
balanced.
Sum of moments about t = (w2 + w3 + w4 + w7 + w9 +
w10 + w12) * 2
-(w5 + w6 + w8 + w11 + w13
+ w14 + w15)*2
= 0, so balanced.
Grover Hughes
Grover Hughes
oops! I rechecked the balance of the above configuration, and found
that only two (2) of the "on-edge" bricks I mentioned can be
discarded. Hope I got it right this time!
I also believe that the lower left-hand corner brick, another "edge-
on" one, can be removed with no ill results.
Regards to all,
Grover Hughes
tc...@lsa.umich.edu
Grover Hughes <ghu...@cei.net> wrote:
>oops! I rechecked the balance of the above configuration, and found
>that only two (2) of the "on-edge" bricks I mentioned can be
>discarded. Hope I got it right this time!
Good job!
I've circulated this question on the math-fun mailing list; hopefully
there will be some responses there.