Biscuits, Books, Coins and Cards: Massive Hangovers

Have you ever tried to build a high stack of coins? In theory it’s fine: as long as the centre of mass of the coins above each level remains over the next coin, the stack should stand. But as the height grows, it becomes increasingly trickier to avoid collapse.

Ten chocolate gold grain biscuits, with a hangover of about one diameter.

Ten chocolate gold grain biscuits, with a hangover of about one diameter.

In theory it is possible to achieve an arbitrarily large hangover — most students find this out for themselves!  In practice, at more than about one coin diameter it starts to become difficult to maintain balance.

Maximum Overhang

Let the coin radius be our unit of length. Assume the tilt is to the right and take the origin at the right hand edge of the top coin. We build the stack downwards. To achieve the theoretical maximum overhang at each stage, we place the stack with its centre of mass just over the right hand edge of the next coin. For n coins, let C(n) be the horizontal distance from the origin to the centre of mass. Thus, the overhang D(n+1) for n+1 coins is C(n):

[ Overhang for n+1 coins ] = [Centre of mass for n coins ]

If there is just one coin, C(1) = 1. With two coins, it is clear that the upper one may extend by up to one unit before it topples. The centre of mass is half way between the centres of the two coins, 1.5 units from the origin. Thus, if the stack of two coins is placed on a third, the overhang is D(3) = C(2) = 1.5 units.

We proceed by induction to get the general result. The strategy is to place the balanced stack of coins upon a new coin so that the augmented stack remains balanced. The centre of mass of the stack of n coins is a distance C(n) from the origin. The centre of the new coin is displaced C(n)+1 from the origin. The centre of mass of the n+1 coins is the weighted average of the stack of n and the new coin:

C(n+1) = [ n C(n) + (C(n)+1) ] / (n+1) = C(n) + 1/(n+1)

Thus we have the recursive relationship

C(n+1) = C(n) + 1/(n+1).

Since C(1) = 1, we have C(2) = 1+½ , C(3) = 1+ ½ + ⅓ and in general

C(n) = H(n) = 1 + ½ + ⅓ + ¼ + ⅕ + … + 1/n

where H(n) is the n-th harmonic number.

The harmonic numbers increase without limit, but very slowly. The harmonic series is divergent; in principle, there is no limit to the extent of the overhang. However, the rate of increase of H(n) with n is so slow that, in practice, only a limited overhang is possible.

Books and Cards

Below are ten volumes of the Encyclopedia Britannica stacked up with an overhang of about one book-length. Attempts to increase this resulted in collapse of the stack.

Ten volumes of the Encyclopedia Britannica.

Ten volumes of the Encyclopedia Britannica.

The length unit for circular items like biscuits and coins is the radius. For rectangular items, like books and cards, the unit is half the length. The theoretical maximum overhang for the ten volumes is D(10) = C(9) = H(9) units. This is about 1.4 book-lengths.


Thick, heavy books are hard to stack. With card decks it is easier to achieve a large hangover. The image above shows a stack with an overhang of about 1.5 card lengths. The theoretical maximum for 52 cards is H(51) units or H(51)/2 card-lengths, which is about 2.26 card-lengths.

Try this yourself: you can probably do better!

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