mike paterson

Mike Paterson

Overhang

Academic Sponsors’ Day, MSRI, March 2, 2012

Peter WinklerUri Zwick

Yuval PeresMikkel Thorup

A Crow Problem:

How long does it take to drive off the crow?

-n 0 1-1-2 2 n

Simple random walk: about n2 throws.

One way to see that: consider the probability distribution of crow’s location; its variance

goes up by 1 after each throw.

A new problem, brought to MSRI in spring ’05

by Zwick: the crow comes back…

…at night!Now what---your first stone will hit the crow and dislodge him, but after that

you’re increasingly unsure where he is.

You can certainly get him off the wallin order n3 throws, and you certainly

still need at least n2 . Which is the truth?

Theorem: Order n3 throws are necessary.

Proof: Uses two different potential functions, each for the wrong problem.

An unusual case of two wrongs making a right.

-n 0 n

The overhang problemHow far off the edge of the table

can we reach by stacking n identical blocks of length 1?

“Real-life” 3D version

Idealized 2D version

Back in time with the overhang problem…

John F. Hall, Fun with Stacking Blocks, Am. J. Physics December 2005.Martin Gardner - Scientific American’s “Mathematical Games” column, 1969.

J.G. Coffin – Problem 3009, American Mathematical Monthly, 1923. George M. Minchin, A Treatise on Statics with Applications to Physics, 6th ed. (Clarendon, Oxford, 1907), Vol. 1, p. 341.

William Walton, A Collection of Problems in Illustration of the Principles of Theoretical Mechanics 2nd ed. (Deighton, Bell, Cambridge, 1855), p. 183.

J.B. Phear, Elementary Mechanics (MacMillan, Cambridge, 1850), pp. 140–141.

The classical solution

“Harmonic Stack”

Using n bricks we can get an overhang

of

Is the classical solution optimal?

Apparently not. How can we improve the construction?

Inverted pyramids?

Claimed to be stable in Mad About Physics, by Chris Jargodzki and Franklin Potter,

but…

They are unbalanced, when the number of layers exceeds 2.

Diamonds?

The 4-diamond is balanced…

But the 5-diamond is …

not.

What really happens?

What really happens!

Why is this unbalanced?

… and this balanced?

Equilibrium

F1 + F2 + F3 = F4 + F5

x1 F1+ x2 F2+ x3 F3 = x4 F4+ x5 F5

Force equationMoment equation

F1

F5F4

F3

F2

Forces between bricksAssumption: No

friction.All forces are vertical.

Equivalent sets of forces

Balanced StacksDefinition: A stack of bricks is balanced iff there is an admissible set of forces under which each brick is in equilibrium.

1 1

3

How can we tell if a stack is balanced?

Checking for balance

F1 F2 F3 F4 F5 F6

F7 F8 F9 F10 F11 F12

F13 F14 F15 F16

F17 F18

Equivalent to the feasibilityof a set of linear

inequalities:

Stability and Collapse

A feasible solution of the primal system gives a set of balancing

forces.A feasible solution of the dual

system describes an infinitesimal motion that decreases the potential

energy.

Small optimal stacks

Overhang = 1.16789Bricks = 4

Overhang = 1.30455Bricks = 5

Overhang = 1.4367Bricks = 6

Overhang = 1.53005Bricks = 7

Small optimal stacks

Small optimal stacks

Small optimal stacks

Overhang = 2.14384Bricks = 16

Overhang = 2.1909Bricks = 17

Overhang = 2.23457Bricks = 18

Overhang = 2.27713Bricks = 19

Support and counterweight bricks

Support

set

Counter-weights

These examples are “spinal”: support stack has only one brick per level, so overhang increases

with height.Spinal stacks can achieve overhang S(n) ~ log n.

100 bricks example

But are spinal stacks optimal?No! When # bricks reaches 20 . . .

Support set is not spinal.

Overhang = 2.32014, slightly exceeding S(20).

Optimal weight 100 construction

Overhang = 4.20801

Bricks = 47Weight = 100

Brick-wall constructions

Brick-wall constructions

“Parabolic” construction

5-stack

Number of bricks: Overhang:Stable!

Thus: n bricks can achieve an overhang of order n1/3 ...

an exponential improvement over the order log n overhang

of spinal stacks.

Mayan from 900 BC---no keystone

Yes! Argument is based on the idea that laying bricks is like stoning crows.

Each additional brick…

spreads forces the same way that throwing a stone (at night) spreads the crow’s probability

bar.

The Upper BoundIs order n1/3 best possible??

In particular, a stack of only n bricks cannot overhang by more than 6n1/3 brick

lengths. The parabolic construction gives overhang (3/16)1/3 n1/3 ~ .572357121 n1/3, so we have the order right but the constant is off by an

order of magnitude. Simulations suggest that the constant

can be improved by adjusting the shape of the brick wall construction…

A generalized version of the “stoning crows” analysis shows that it takes order n3 bricks to

get the stack to lean out by n .

“Vases”

Weight = 1151.76

Bricks = 1043

Overhang = 10

“Vases”

Weight = 115467.

Bricks = 112421

Overhang = 50

“Oil lamps”

Weight = 1112.84

Bricks = 921

Overhang = 10

giving overhang of about 1.02 n1/3 .

How about using the third dimension?Our upper bound proof makes no use of the fact

that bricks cannot overlap in space! Hence, the 6n1/3 bound applies even in 3D, as long as there are no non-vertical forces.

However, the constant can be improved in space by skintling,

Effectively increasing the brick length to (1+w)1/2 .

Open problems● What is the correct constant in the maximum

overhang, in the rectilinear case? In the general 3-dimensional case?

● What is the asymptotic shape of “vases”?● What is the asymptotic shape of “oil lamps”?● What is the gap between brick-wall

constructionsand general constructions?

● Can the proof be extended to cover non-vertical forces (if, indeed, they are possible for 3D bricks)?

● How much friction is needed to change the 1/3 exponent for overhang?

Thank you for your attention. Happy stacking…