Was it really necessary to invoke Brownian motion (zig-zagging)
in a simple description of precipitation without rotation? Probably
not. Proportionality between the terminal velocity (vt) and the mass
of a particle (m) alone would be sufficient to explain what happens,
qualitatively. That is how precipitation is actually explained in
Chapter 9 of Serway and Faughn (p 294). They derive the formula
vt = (m*g/k) * (1 - ro1/ro2)
where ro1 if the density of fluid ro2 is the density of particles.
For spherical particles k would be 6*Pi*eta*r (Stokes formula),
where r is the radius and eta is the viscosity coefficient.
A quantitative centrifuge problem could thus be presented to
students. They would substitute 9.8 for g in a stationary vessel
or r*w^2 in a rapidly rotating vessel (when r*w^2>>9.8).
But not a single precipitation or centrifuge problem can be
found at the end of Chapter 9. Why is it so? I suspect that
they tried and found that the estimated "sedimentation times"
are orders of magnitude shorter than what is actually observed
for very small particles. This would be a good indication that
Brownian motion does play a significant role, especially when
g=9.8. But this is only a guess. Does anybody know the mass
of a red cell and the ro1/ro2 ratio for blood?
Ludwik Kowalski