I am trying to determine the viscosity of a fluid. Therefore, I let a sphere of known mass m and radius r fall (sink) into the fluid. I then measure the time it takes to sink/travel a given height.

Then there are the following forces having a effect on the sphere:

$F_g = mg$ and

$F_a = \frac{4 \rho \pi r^3 g}{3} $ (buoyant force)

with $g = 9.81 \ ms^{-2}$ and $\rho$ being the density of the fluid.

The result is a motion downwards (sinking), hence I get friction $F_r$.

Assuming this friction gets that strong to have no resulting force on the sphere (i.e. its velocity remains constant), I can use the following equation:

$F_g - F_a - F_r = 0$

According to Stokes Law, the friction of a moving sphere in a fluid with viscosity n is:

$F_r = 6 \pi r n v$ where v is the constant velocity.

Using these equations, I can determine the viscosity:

$n = \frac{g}{6 \pi r} (m - \rho \frac{4}{3} \pi r^3)\frac{1}{v}$

with $v = \frac{s}{t}$.

Question about Ladenburg correction

The fluid is inside of a cylinder, so I cannot use Stokes Law "as it is" because the walls of the container add more friction.

Hence I have to apply the so called Ladenburg correction with a correction factor

$f = (1 + \frac{(2.1 r)}{R}) (1 + \frac{(3.3 r)}{H})$

When / where do I have to apply this factor?

Question about extrapolation

As I am doing this with spheres of different radiuses, I get multiple values for the viscosity:

radius/mm | viscosity/(kg/(m*s))
1.0       | 0.594
2.0       | 0.608
2.5       | 0.631

Extrapolation to $r = 0 \ mm$ gives a value $n_0 = 0.575$.

I do not understand: Why do I need this extrapolation?

  • $\begingroup$ I am not familiar with the Ladenburg correction, but judging by the form of the equation (going to 1 for infinite cylinder radius and height) I think you just have to multiply $F_r$ with it $\endgroup$
    – Michiel
    Commented Mar 10, 2013 at 15:19
  • $\begingroup$ I also do not understand the extrapolation. I would assume that if the calculations are correct, you would get viscosities which are independent of the radii of the spheres $\endgroup$
    – Michiel
    Commented Mar 10, 2013 at 15:20
  • $\begingroup$ Yep...the viscosity should come out same for all radii. $\endgroup$
    – Cheeku
    Commented Mar 10, 2013 at 23:12
  • $\begingroup$ Well, by fitting a straight line to the viscosity values, I get a non null slope ( > 0). It seems as if I should apply the correction factor to the extrapolated value n_0 at some radius r, which gives my a corrected value for this radius. Example: r = 10 mm, R = 100 mm, H = 100 mm. So f = 1.21 * 1.33. Hence i would use (0mm|n_0) as a first point, and (10mm|n_0 * f) as second point. Between thos points, I have to draw the corrected (less slope, but not null) line/graph. That is what I have to do, but I don't understand it :( $\endgroup$ Commented Mar 10, 2013 at 23:18

2 Answers 2


The correction needs to be applied to the terminal velocity you obtain :- $v$.$$v_{corrected} = v_{measured} L$$

, where $L$ is Ladenburg correction

Here is the link. Though the correction expression is different due to different conditions, but the concept is same.

Also, after applying the Ladenburg correction, your viscosity should become constant for all radii.

  • $\begingroup$ Unfortunately, this is not the case. I understand why the viscosity should be independent (so in this case constant) from the radius. Applying the factor f to each of the value pairs above (which means dividing the values by f, according to the equation) gives me a (extremly good fitting) linear graph with a negative slope (which is nonsense). This is what I did, but got marked by my corrector. (Asking the corrector about it leads to: "I dont know, just do it this way" ... where "this way" referes to my first comment to the question) $\endgroup$ Commented Mar 10, 2013 at 23:25
  • $\begingroup$ @Can you mention the actual values(v,r,R,H) for above three cases? It would help me to help better. :) $\endgroup$
    – Cheeku
    Commented Mar 10, 2013 at 23:28
  • $\begingroup$ @DanielOertwig Unfortunately enough, this is really weird. I would try out things myself. I find this really interesting. Well, assuming the table you gave in question is correct. I just need the R and H for the cylinder. :) $\endgroup$
    – Cheeku
    Commented Mar 10, 2013 at 23:36
  • $\begingroup$ R = 23.25 mm and H = 800 mm. $\endgroup$ Commented Mar 11, 2013 at 10:26
  • $\begingroup$ @DanielOertwig I am getting values of viscosity around 0.51 for all the three radii, doing division of viscosity by Ladenburg correction. It works. What values did you get? I suspect a calculation error. P.S. If you're using a plotting program to find the slope, the program might scale the graph to fit the screen and make it look a negative slope. The error over all the three values from 0.51 is around 0.001, which is less than least count I guess. $\endgroup$
    – Cheeku
    Commented Mar 11, 2013 at 14:45

I see an error in the Lindeburg correction formula presented by you. Instead of 2.1r it should be 2.4r. I found in an article of Smoluchowski, that the coefficient (1+2.4r/R) was proved by Landenburg in the paper: R. Ladenburg, Ann. d. Phys. 23, p.447 (1907). The error is probably responsible for the non-constant value of viscosity.


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