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This story seems to almost be out of science fiction (or The Prisoner) but the pictures don't lie. The water people in Los Angeles are so desperate to save what water is left in some reservoirs that they are essentially putting a roof on it by filling the lakes with black plastic balls:
Facing a long-term water crisis, officials concerned with preserving a reservoir in Los Angeles hatched a plan: They would combat four years of drought with 96 million plastic balls.
On Monday, Mayor Eric Garcetti of Los Angeles arrived at the 175-acre Los Angeles Reservoir to release the final installment of the project: 20,000 small black orbs that would float atop the water. [...]
Mr. Garcetti said that the dark balls would help block sunlight and UV rays that promote algae growth, which would help keep the city's drinking water safe. Officials also said the balls would help slow the rate of evaporation, which drains the water supply of about 300 million gallons a year. The balls cost $0.36 each and are part of a $34.5 million initiative to protect the water supply.
This is an ingenious way of reducing evaporation. Perhaps the state's aqueduct system can also be filled with these balls to stop evaporation there.
HughPickens.com also submitted this article just minutes later!
(Score: 0) by Anonymous Coward on Thursday August 13 2015, @09:04AM
You seem to have done your calculation with a square. But the closest packing of balls is not a square lattice, but a hexagonal one. And indeed, the balls are likely to locally form an approximation of that hexagonal lattice.
The hexagon has the area of six equilateral triangles, each having a height of r. Since for an equilateral triangle, the height is sqrt(3)/2 times the side length, and the area is half the product of side length and height, we get
Ahexagon = 6 × 1/2 × 2r/sqrt(3) × r = 2 sqrt(3) r² ≈ 3.4 r²
Clearly 2 sqrt(3) is much closer to π than 4 is.