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Assume a spherical bubble (I always wanted to say that) of radius r. Its force of buoyancy is a function of its volume (ie, ~r³, and its ...sorry, you cannot view external links. To see them, please
REGISTER or LOGIN is a function of its cross-sectional area and the square of its speed (ie, ~r²v²). Terminal velocity (ie, constant velocity) is reached when force of buoyancy equals force of drag, or r³ = r²v². So, using this simplified math, for a bubble with a radius r = 1, 1³ = 1²v², or 1 = v²and so v = 1. For a larger bubble of radius r = 2, 2³ = 2²v², or 8 = 4v², or 2 = v²and so v² = 2, or v = √2 = 1.414... So, actually, larger bubbles should rise faster than smaller ones, the speed being proportional to the square root of its radius. Assuming spherical bubbles, of course. You can also simplify the original equation, r³ = r²v², down to r = v², or v = √r, which shows the relationship more clearly. Someone check my assumptions and math please.
This actually changes the taste of the champagne; drinking the same champagne from a narrow, deep glass tastes different from a wider, shallow glass.