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

force of drag 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 = v²*and so

*v = 1*.

For a larger bubble of radius

*r = 2*,

*2³ = 2²v²*, or

*8 = 4v²*, or

*2 = v²*and 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.