Domino Overhang

[BMAD]

Let the length of a domino be 1. On the edge of a table, a single domino can extend out a distance of 1/2 before being unbalanced. A stack of two dominoes can extend out a distance of 3/4. How large of an overhang can you create with 5? 10? N? How many dominoes are needed before the top domino is one full horizontal length away from the edge of the table, for a full overhang length of 2? Each domino must be placed in the same orientation (don't rotate the dominoes or stand them on end.)

this can also be done with playing cards

[gavinksong]

Let the length of a domino be 1. On the edge of a table, a single domino can extend out a distance of 1/2 before being unbalanced. A stack of two dominoes can extend out a distance of 3/4. How large of an overhang can you create with 5? 10? N? How many dominoes are needed before the top domino is one full horizontal length away from the edge of the table, for a full overhang length of 2? Each domino must be placed in the same orientation (don't rotate the dominoes or stand them on end.)

this can also be done with playing cards

To answer BMAD's original questions using the harmonic method:

It only takes four dominoes before the top domino is one full horizontal length away from the edge.

With five, you can create an overhang of 137/120 off the edge of the table. With ten, you can create an overhang 7381/5040 off the edge.

For N, the overhang length is sum(1/(2n), 1, N).

Edited August 3, 2013 by gavinksong

[DeGe]

The basic premise of the balance is that:

a) the "system" of dominoes should have its Centre Of Gravity (COG) on the table

b) Any domino on any extremity must have its COG either on the table or on another domino

For 2 dominoes, the 3/4th overhang is possible as follows:

(1 Domino length is 4 units here)

The system of 2 dominoes has its COG at the edge of the table.

For 5 Dominoes, a simple construction could be as follows:

The overhang here is 1 domino.

With this set-up, the COG is at (22x/5x) = 4.4 while the edge is at 5; The whole system can be moved by 0.15 domino length (0.6/4) and still have it stable. The total overhang then is 1.15 domino length.

In fact, a better configuration can be achieved by placing the dominoes as follows:

4 5

3 2

1

The first domino overhangs 0.29 domino length.

The total overhang in this case would be 1.29 domino length.

Edited August 1, 2013 by DeGe

[DeGe]

Total overhang 1.75 domino length

Edited August 1, 2013 by DeGe

[DeGe]

Overhang of 1.75 domino lengths (previous one was not correct)

Edited August 1, 2013 by DeGe

[TimeSpaceLightForce]

5 dominoes

[gavinksong]

In a stack of N dominos, the following recursive requirement must be met:

For any M dominos at the top of the stack, their total center-of-gravity must not be past the edge of the domino/table directly beneath it.

So let N be infinity. Pretend the stack is straight (there is no overhang). Now, let's start pushing out the dominos starting at the top.

(M=1) The topmost domino, since its center of gravity is at half its length, can be pushed past the edge of the domino below it by half its length.

(M=2) Since the first domino extends past the edge of the second domino by 1/2, the top two dominos has its center-of-gravity at 1/2 + (1/2)(1/2) = 3/4 the length of the second domino. Thus, the second domino can be pushed out by 1/4 of its length.

(M=3) Since the center-of-gravity of the first two dominos extends past the edge of the third domino by 1/2, the center-of-gravity of the top three dominos is at 1/2 + (2/3)(1/2) = 5/6 the length of the third domino, so the third domino can be pushed out by 1/6 of its length.

(M=4) Since the center-of-gravity of the first three dominos extends past the edge of the fourth domino by 1/2, the center-of-gravity of the top four dominos is at 1/2 + (3/4)(1/2) = 7/8 the length of the fourth domino, so the fourth domino can be pushed out by 1/8 of its length.

...

...

(For any M) Since the center-of-gravity of the first M-1 dominos extends past the edge of the Mth domino by 1/2, the center-of-gravity of the top M dominos is at 1/2 + ((M-1)/M)(1/2) = 1-(1/2M) the length of the Mth domino, so the Mth domino can be pushed out by 1/2M of its length.

Thus, the total overhang of an infinite stack of dominos is half the sum of the harmonic series (1+1/2+1/3+1/4+1/5+...), which is infinite.

Of course, you wouldn't actually be able to make a infinite overhang with "real" dominos.

Edited August 3, 2013 by gavinksong

[gavinksong]

Um. So this is embarrassing. There is a slight typo in my spoiler for infinite overhang.

Since the center-of-gravity of the first M-1 dominos

lies on the edge of the Mth domino...

[Perhaps check it again]

BMAD posted:

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"How many dominoes are needed before the TOP [my emphasis] domino is one full horizontal

length away from the edge of the table, for a full overhang length of 2?"

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The second diagram of post # 2, and the diagrams of post numbers 3 and 4 of DeGe do not count,

and the diagram of TimeSpaceLightForce in post # 6 also does not count, because none of those

scenarios involve the top domino being the domino that is doing the overhanging away from the

edge of the table.