How see-saws should be made April 13, 2007Posted by dorigo in games, mathematics, personal, science.
A see-saw is one of the toys you frequently find in children parks and anywhere small children need outdoor entertainment. It is usually built as a long log of wood with a support in the middle and seating for the kids on both ends.
I was looking at my kids playing on one such thing last week, while enjoying a hot sun at the Rifugio Gallo Cedrone, (2200 m above sea level), in val Pusteria, and it just dawned on me that those things are not built the way they should.
My intuition told me that the see-saw should not be perfectly balanced when unloaded, but biased in advance, if the aim is to provide the maximum number of pairs of kids with perfect fun. In fact, when you play with a see-saw and your playmate is heavier or lighter than you, there is arguably less fun, since more effort is needed to perform those ups and downs. A biased see-saw should allow more odd pairs to enjoy the game.
I had forgotten the matter, but yesterday I read a post on http://kea-monad.blogspot.com/ where Kea discusses a simple geometric problem, totally unrelated to the one of designing a see-saw. However, both puzzles belong to a whole class of mathematical problems whose solution involves finding a minimum in a simple function. So I decided I would demonstrate that my intuition was correct.
Today, armed with paper and pencil and ten minutes of free time, I did solve the problem. You want to minimize the squared difference of total weight on the two ends of the see-saw by applying a bias x, assuming two kids with weight w1 and w2 (and posing w1>w2) are sitting with w2 on the side of x. That is, you want to find the value x which corresponds to a minimum of f(x) = (x+w2-w1)^2.
Further assume that w1 and w2 come from a flat distribution of minimum m and maximum M. This may not correspond to the reality of pairs of kids in a playground, but it is a reasonable ansatz – which can be changed anyway. If you do that, the problem reduces to that of a simple integration on w2 from m to w1 of (x+w2-w1), followed by another integration on the result with w1 variable from m to M.
Confused ? It is a problem for junior high school students. The solution is simple: the best value of the bias x is just (M-m)/3. That is, if you assume kids will play on the see-saw if they have a weight between 20 and 80 pounds, the see-saw would do them a better service if it came biased by 20 pounds on one side!