[Guest]

As I was thinking about the subject, I was reminded of a puzzle* I did in my school days:

Given as many jenga bricks as you like, if you constructed a tower, placing only one brick above another (ie you cannot have more than one brick in a single row) how far could you make the tower project over the edge of a table? How far could you project with n bricks?

With just one brick you can project half a brick over the edge, but by cunningly balancing them is it possible to get further?

* actually a piece of maths coursework

[Guest]

As I was thinking about the subject, I was reminded of a puzzle* I did in my school days:

Given as many jenga bricks as you like, if you constructed a tower, placing only one brick above another (ie you cannot have more than one brick in a single row) how far could you make the tower project over the edge of a table? How far could you project with n bricks?

With just one brick you can project half a brick over the edge, but by cunningly balancing them is it possible to get further?

* actually a piece of maths coursework

I think you could go as much as you wanted to.

dIf you build a diamond centered on the edge of the table. The diamon would have 1 block on the table, two on top of that, three on top of that, 2 on top of that, and 1 on top of that. So you would have 5 blocks high, and 1 and a half bricks extended over the edge of the table.

By building out both left and right sides at the same time, you keep the centre of gravity constantly on the edge of the table. So they blocks should stay ( assuming super steady hand, can place two bricks at a time, no wind etc..

Edit: sorry, this fails to meet the specifications ( I didn't reread the question, and go carried away. Since you can only place one brick on top of another, I would suspect that the tanswer would like between 0.5 and 0.8 of a brick.

a formation simular to

/

/

/-

Edited February 18, 2009 by taliesin

[Guest]

I do not know the lenghts of the bricks but this is the way I can see it working. Placing the bottom brick lenght wise 1/2 the width of a horizontal brick will remain on the table. Place horizontal bricks on top as counterweight. So the Length hanging over the table top should be L - 1/2W

.____

|____|

|____|

|____|__________

|_______________|

__|

__| edge of table

__|

[Guest]

infinitely close to one bricks length. the center of mass of the bricks on top of the projecting brick would have to be on a line that would be perpendicular to the table top and as close to the edge as possible.

[Guest]

i'm confused!!!

[Guest]

the answers given above are correct. As long as the center of gravity remains at the edge of the table and no further, then the mass of blocks can overhang to 1 block long before toppling over.

In this sketch, a linear progression of blocks is represented as a trapezoidal mass whose center of gravity passes along the same axis as one of the diagonals.

Also, the center of gravity of and given mass of blocks must not go beyond the pivot point of the block below the mass. This is why a trapezoid shape works well, slicing it horizontally at and level maintains a center of mass not past the pivot point.

[Guest]

How far could you project with n bricks?

1/2 + 1/4 + ... + 2^-(n-2) + 2^-(n-1)

[Guest]

the answers given above are correct. As long as the center of gravity remains at the edge of the table and no further, then the mass of blocks can overhang to 1 block long before toppling over.

In this sketch, a linear progression of blocks is represented as a trapezoidal mass whose center of gravity passes along the same axis as one of the diagonals.

Also, the center of gravity of and given mass of blocks must not go beyond the pivot point of the block below the mass. This is why a trapezoid shape works well, slicing it horizontally at and level maintains a center of mass not past the pivot point.

That would be true if you had some glue...

[Guest]

1/2 + 1/4 + ... + 2^-(n-2) + 2^-(n-1)

No. Wait. I think that's wrong. The answer is surprising...

1/2 ( 1 + 1/2 +1/3 + ... + 1/(n-2) + 1/(n-1) )

which has no limit. =

[Guest]

i don't think you could make any part of the stack stick out any further than one bricks length minus something a little less than half its width (assuming all bricks are the same dimensions). You can use a stack of however many to hold the extending brick in place but their center of gravity has to be over the table and over the extending brick.

I'm not sure if that makes any since. perhaps i'll make an illustration a little later.

[Guest]

No. Wait. I think that's wrong. The answer is surprising...

1/2 ( 1 + 1/2 +1/3 + ... + 1/(n-2) + 1/(n-1) )

which has no limit. =

could you explain what that equation means in terms of stacking jenga peices? because it seems the way i am visualizing it:

That would be true if you had some glue...

[Guest]

could you explain what that equation means in terms of stacking jenga peices?

It's not to scale, but you get the idea...

EDIT: Oops. Figure appeared twice. Sorry about the size, but I guess you can't remove attachments...

Edited February 18, 2009 by d3k3

[Guest]

ah i see, I was looking at it upside down..

[Guest]

It's not to scale, but you get the idea...

EDIT: Oops. Figure appeared twice. Sorry about the size, but I guess you can't remove attachments...

The man has it. And yes, the surprising result is that there is no limit to the sum, and you can project them unto infinity. A demonstration with half a dozen bricks can easily show you can reach more than one bricks length over the edge. I did have a photo, but that was several computers ago.

The way to construct it is by starting from the top. Place the 1st brick on top of the second, and stick it out as far as possible, over the edge of that brick (1/2 lengths) then take these two and project them as far over a third brick as possible (another 1/4) and then this pile of three bricks will rest 1/6 of a brick over the edge of the next*. The general formula is 0.5 x (1 + 1/2 + 1/3 + ... + 1/n) which tends to infinity as n tends to infinity.

*calculate the centre of gravity of each successive pile of bricks.

EDIT: spoiler inserted

Edited February 18, 2009 by armcie