- 01 Nov '05 20:09 / 1 editSuppose you have n circular coasters, a table, and a lot of free time. What is the maximum table overhang you can acheive by stacking the coasters on the edge of the table? You may assume that each of these coasters:

1) has unit mass;

2) has unit diameter;

3) is weighted so that its centre of gravity runs directly through its geometric centre; and

4) is infinitely strong, so that is does not deform under stacking stresses.

To get started, what is the maximum table overhang for:

a) 1 coaster?

b) 2 coasters?

c) 3 coasters?

d) 10 coasters?

And what is the maximum table overhang you can acheive given an infinite number of coasters? (Drinking games will never be the same again!!) - 01 Nov '05 21:23

I don't know the answer for 1 coaster. It would involve the coaster being placed with it's center of gravity exactly on the edge of the table and then calculating how much of the coaster hangs over.*Originally posted by PBE6***Suppose you have n circular coasters, a table, and a lot of free time. What is the maximum table overhang you can acheive by stacking the coasters on the edge of the table? You may assume that each of these coasters:**

1) has unit mass;

2) has unit diameter;

3) is weighted so that its centre of gravity runs directly through its geometric centre; and

4) is ...[text shortened]... n acheive given an infinite number of coasters? (Drinking games will never be the same again!!)

For the other questions, you need to know the size of the table. Also, if you stack a coaster on a coaster, does the top coaster count as overhanging if the one below it also does? If so, then you can stack infinitely many coasters on top of one another. As long as there are no air currents or earth movements or settling of the table's joints the stack should be stable...maybe quantum uncertainty might destabilize it at some point. - 01 Nov '05 21:39

I think you answered one of your own questions. You would indeed place the first coaster with its centre right on the edge, so half of the coaster would overhang (total overhang = 1/2). You don't need to know the size of the table to determine the overhang for subsequent coasters, but you can imagine the table to be a semi-infinite slab (you can place the coasters as far back as you want to, if you need to).*Originally posted by AThousandYoung***I don't know the answer for 1 coaster. It would involve the coaster being placed with it's center of gravity exactly on the edge of the table and then calculating how much of the coaster hangs over.**

For the other questions, you need to know the size of the table. Also, if you stack a coaster on a coaster, does the top coaster count as overhangi ...[text shortened]... oints the stack should be stable...maybe quantum uncertainty might destabilize it at some point.

In answer to your other question, the overhang is equal to the distance from the edge of the table to the furthest edge of the furthest coaster hanging off the table. For instance, if you had 1 coaster hanging 1 inch over the edge, and you placed another coaster on top hanging 1.2 inches past the first, the total overhang would be 1 inch + 1.2 inches = 2.2 inches. If you placed another coaster on top hanging 2.5 inches past the second, the total overhang would be 2.2 inches + 2.5 inches = 5.7 inches, etc...

For the purposes of this question, you can assume there ain't nothin' that's gonna shake these coasters loose. - 02 Nov '05 10:36It seems to me that the maximum overhang is 1/2. If you placed a second coaster on top of the first but with greater overhang it would topple both of them. Is that not correct?

And when you refer to 2 coasters, 3 coasters, etc...are you assuming they are all stacked on top of each other? I thought you meant that they could be placed next to one another and the overhangs would add together...thus 2 coasters would have a summed overhang of 1. It seems from context that you didn't mean this though. If you placed them next to one another the overhang would still be 1/2 right?

The reason I think 1/2 is the maximum overhang is that a coaster with 1/2 overhang would be barely balanced on the table. Any slight point weight on the coaster that was over the overhang would be enough to topple it. If a coaster above the bottom had greater overhang, it's center of mass would be over the bottom one's overhang and would destroy the delicate balance of the one below and topple it. - 02 Nov '05 15:32

The maximum overhang is greater than 1/2. I won't give away the solution, but consider what happens if you were to construct the coaster tower backwards (placing coasters underneath instead of on top). It's possible to reverse the construction process you come up with this way, and build the "leaning tower" by judicious placement of the base coasters. Balance is definitely the main issue here.*Originally posted by AThousandYoung***It seems to me that the maximum overhang is 1/2. If you placed a second coaster on top of the first but with greater overhang it would topple both of them. Is that not correct?**

And when you refer to 2 coasters, 3 coasters, etc...are you assuming they are all stacked on top of each other? I thought you meant that they could be placed next to one a ...[text shortened]... he bottom one's overhang and would destroy the delicate balance of the one below and topple it. - 02 Nov '05 20:38

The shape of the table doesn't matter either, provided it's at least 1/2 coaster diameter wide at some point.. It could be shaped like a starfish, but if you balance the coaster right on the point it still balances, as will all the other coasters if you set them up properly. For the purposes of this question let's call it a square table with sides 5 coaster diameters long.*Originally posted by slippytoad***It is very important to mention whether the table is round or rectangular (or some other odd shape of a table). A coaster hangs off a round table differently than it would off the straight edge or even corner of a rectangular table.** - 02 Nov '05 22:17

Hmm. Interesting point. One could pin down a single coaster with high overhang with a stack of coasters with little or no overhang. How would you calculate the maximum overhang though? I have no idea how this could be done.*Originally posted by PBE6***The maximum overhang is greater than 1/2. I won't give away the solution, but consider what happens if you were to construct the coaster tower backwards (placing coasters underneath instead of on top). It's possible to reverse the construction process you come up with this way, and build the "leaning tower" by judicious placement of the base coasters. Balance is definitely the main issue here.** - 03 Nov '05 16:11

I'll give you a hint. The first coaster can hang over the edge by 1/2 a diameter (so it balances right on its centre of gravity). Now, imagine a coaster popping up right underneath the first one, so that the first coaster hangs over the second one by 1/2 a diameter, and the edge of the second coaster is flush with the edge of the table. If you move the second coaster along (with the first one stuck on top, moving as well), eventually you'll hit a balance point past which the whole structure will fall. Call the distance moved "x", set up your equation balancing the moments (force x distance) on either side of the fulcrum (the edge of the table), and solve for "x".*Originally posted by AThousandYoung***Hmm. Interesting point. One could pin down a single coaster with high overhang with a stack of coasters with little or no overhang. How would you calculate the maximum overhang though? I have no idea how this could be done.** - 04 Nov '05 16:14 / 2 editsI think it is infinite.

If you imaging a counterweighted protruding stack

..___

..___

..___

___ ___

...___ ___

___ ___ ___

...___ ___

___ ___

...___

.....*

with the table edge at *, coasters at ___, and space at "." , then this stack can be made as big as you like, each protruding coaster is sandwiched by two others top and bottom and you can add as many coasters to the counterweight on the top as you need to balance it.

in fact, why not make a symmetrical stack:

.......___

.......___

.......___

.....___ ___

...___ ___ ___

___ ___ ___ ___

...___ ___ ___

.....___ ___

........___

.........*

We can now see that this stack will easily balance on the table, the force on the top coaster will puch it directly upwards but we can stack as many coaster as we needon top of it to cancel that upward force. The diamond of the stack can be made arbitrarily large as long as our coasters are infinitely stiff.

I suppose there might be some limit as the top coaster went out of the earth's gravitational field, but I don't think this problem was supposed to take gravity fall off into account. - 04 Nov '05 17:26

Stacking the coasters so that there is more than one coaster per level (as in your diamond configuration) is not allowed in this question, although I agree that your configuration would give an infinite overhang for an infinite number of coasters. The idea is to simply stack one coaster on top of another, and see what the maximum overhang will be. The answer may surprise you...*Originally posted by iamatiger***I think it is infinite.**

If you imaging a counterweighted protruding stack

..___

..___

..___

___ ___

...___ ___

___ ___ ___

...___ ___

___ ___

...___

.....*

with the table edge at *, coasters at ___, and space at "." , then this stack can be made as big as you like, each protruding coaster is sandwiched by two others top and bottom and yo ...[text shortened]... ional field, but I don't think this problem was supposed to take gravity fall off into account. - 04 Nov '05 21:00

It seems to me like it should be 1 coaster overhang if given your constraint. If I calculate it will the answer still surprise me?*Originally posted by PBE6***Stacking the coasters so that there is more than one coaster per level (as in your diamond configuration) is not allowed in this question, although I agree that your configuration would give an infinite overhang for an infinite number of coasters. The idea is to simply stack one coaster on top of another, and see what the maximum overhang will be. The answer may surprise you...** - 07 Nov '05 19:09 / 2 edits

One final gasp for this thread before it dies forever. The answers for the questions as posed are:*Originally posted by PBE6***Leisurleysloth has PMed me the correct answer. I'm inclined to get him to post his solution. Any objections?**

a) 1/2 coaster diameter (CD) = 0.5 CD

b) 3/4 CD = 0.75 CD

c) 11/12 CD ~= 0.92 CD

d) 7381/5040 CD ~= 1.46 CD

And the overhang of the stack increases without bound as the number of coasters increases, so there is no maximum overhang (it's infinite). Surprising!