Counting Cards with a Harmonic Series
Richard Green demonstrates the use of a harmonic series (H(n) = 1 + 1/2 + 1/3 + … + 1/n) to count how far a deck of cards will overhang from the base card. Richard discusses the card’s center of mass, and found that the maximum displacement is 2.26 card widths for 52 cards. Furthermore, he found that if a deck had a million cards, the total stable overhang would only be 14.39.
Please post in the comments your favorite use of a harmonic series!
Originally shared by Richard Green
Massive Hangovers and the Harmonic Series
This picture shows a stable deck of 52 cards in which the top card overhangs the bottom card by about 1.5 card widths. In theory, it is possible to do even better so that the overhang is almost 2.26 card widths. The reason for this has to do with centres of mass and the harmonic series.
In order to understand where the number 2.26 comes from, it is helpful to keep track of the horizontal displacement of each card. We will measure horizontal distances from an origin at the rightmost edge of the top card, in such a way that each card is 2 units wide. The centre of mass of the top card is then 1 unit to the left of the origin.
The overhang of a stack of cards is the horizontal distance between the rightmost edge of the top card and the rightmost edge of the bottom card. In a stack of two cards, the way to create the largest overhang is to put the centre of mass of the top card above the rightmost edge of the lower card. The combined centre of mass of these two cards will then be half way between their individual centres of mass; in other words, 1.5 units to the left of the origin.
Now consider creating an overhanging stack with n+1 cards by placing an overhanging stack with n cards on top of a single card. (This is probably not a good way to create a stack in practice, but it is mathematically helpful.) In order to maximize the overhang, the cards should be positioned so that the centre of mass of the top n cards is directly above the rightmost edge of the bottom card. If we define C(n) to be the horizontal distance between the centre of mass of the n-card stack and the origin, this shows that the size of the overhang for n+1 cards is equal to C(n).
The centre of the bottom card of the new n+1 card stack is at a distance of C(n)+1 from the origin. The centre of mass of the new n+1 card stack is thus given by the weighted formula
C(n+1) = (nC(n) + C(n)+1)/n+1,
which simplifies to C(n+1) = C(n) + 1/(n+1).
Since C(1) = 1, we can solve this to show that C(n) is given by the sum of the first n terms of the harmonic series, H(n) = 1 + 1/2 + 1/3 + … + 1/n. A famous property of this series is that it diverges to infinity. For example, if you were prepared to make n large enough, you could add up enough terms to make H(n) larger than 1000, or any other large number you care to specify. However, the divergence is very slow, and adding up the first million terms of the series only gives a total sum of about 14.39.
Going back to the skewed stack of 52 playing cards, we now know that the size of the overhang for 52 cards is equal to H(51), which works out at around 4.5188. However, the width of a card is two units, so in terms of card widths, the maximum overhang is about 2.2594, which is where the figure of 2.26 comes from.
Since the series H(n) diverges to infinity, it is possible in theory to stack objects in this way so that the overhang is arbitrarily large. However, the thicker and heavier the objects become, the harder it is to achieve large overhangs.
This post is based on a post in the blog ThatsMaths by Peter Lynch, who is a Professor of Meteorology at University College Dublin. The blog post also shows a photograph of a stack of ten biscuits (cookies) and a stack of ten volumes of the Encyclopædia Britannica.
Peter Lynch’s original blog post, Biscuits, Books, Coins and Cards: Massive Hangovers, can be found here: http://thatsmaths.com/2014/06/12/biscuits-books-coins-and-cards-massive-hangovers/
Here’s an online harmonic series calculator by Jim Carlson which you can use to calculate values of H(n) for various n: http://www.math.utah.edu/~carlson/teaching/calculus/harmonic.html
And your point is…?
Marcie, I find you lack of curiosity disturbing.
This is great. When I try it for friends, though, I can only get to 2.25.
Since the series inside the parentheses doesn’t converge, you can get to any amount of overhang if you just have enough cards. In theory, that is.
Exactly, Bill Tyler!
??? No role for length of the cards ?? ..
V Good info..
This is not optimal. It is a mistake to think that the optimal stack is optimal at every level. I don’t remember how to improve it but am certain it can be improved. I believe the furthest-projecting card should not be the top one.
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