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Re: [Phys-l] Coalesing drops


(delayed by trip to birthday dinner a day away, not mine.)

Yes! You're correct. I forgot about the buoyant force, which is negligible in this case, and the g force is ~ to the cube of the radius, while the drag is only prop. to the square.***'_law

doesn't apply.

hyperphysics has done the work for me:

for 2 mm and 5 mm drops their terminal speeds are v. ~ 6.4 and 10.1 m/s.

I cheated and found in Eisberg and Lerner (I've seen it elsewhere also.) the formula for the position (time) for v^2 drag.

The formulae, as I graphed for two and 5 mm D drops, is on my site.

Here's the generic:

X = (1/gamma) (Ln[exp(sqrt(gamma*g)*t )+ exp(sqrt(gamma*g)*t)]/2)

gamma = drag factor / m = 0.5* density (air) * cross section(Pi * R^2) * coefficient (sphere ~0.5) / density * [(4/3) Pi * R^3]

this reduces to ~ 2.25E-4 / r

I pray no error. The equations for 2 mm and 5 mm diameter are part of the titles of the graphs. They were generated by the function feature in Kaleidograph. I extracted the data it created and subtracted the positions. As one can see if dropped ~ 10 cm apart after v.~ 600 ms the five mm one will catch up. This is after falling about five feet.

*** what happens when ones mind's eye model is incomplete.

bc no experience w/ dropping, so doesn't know if practical.

On 2010, Oct 08, , at 19:25, LaMontagne, Bob wrote:

Just curious - what do you mean by "The smaller will catch up and likely coalesce instead of shatter."? My understanding is that the fall speed of droplets is proportional to the square root of their radii. Don't larger drops catch up with smaller ones?

Bob at P