# Viscosity and Drag

1. Feb 23, 2012

### Vaal

The linear drag coefficient is a function only of object size and medium viscosity. Multiply sources say viscosity of air is independent of pressure and density. How can the linear drag be independent of density of the medium? It seems like less dense medium should mean fewer collision and in turn less drag force. What am I missing here?

2. Feb 23, 2012

### roboticmehdi

the coefficient may be independent of density but the drag force depends on density. F=0.5 * density * speed^2 * drag coefficient * area. So yes, in denser medium there is higher drag force. ( source: wikipedia, i took the formula from there )

3. Feb 23, 2012

### Vaal

4. Feb 23, 2012

It is dependent on density. Stokes' Law shows drag explicitly as a function of terminal velocity, and that terminal velocity is a function of the density. It even shows this in the article you linked.

5. Feb 23, 2012

### Vaal

The article I linked may have not been the best example, I think it is referring to the force at terminal velocity (notice that a general v is no were in the expression). Classical Mechanics by Taylor explicitly says drag force is b*v and in problem 2.2 says b is given by 3pi*n*D where d is diamante of a sphere and n is the viscosity.

6. Feb 23, 2012

Right well in the case of Stokes flow, the velocity is so low that it is assumed that all drag is viscous in nature, which means that the density has no part in it. It only affects the terminal velocity. With such a low flow velocity, viscous effects are much, much more significant than pressure and density effects.

Consider the nondimensional Navier-Stokes equations:

$$\frac{\partial u_i^{*}}{\partial t^{*}} + u_j^{*}\frac{\partial u_i^{*}}{\partial x_j^{*}} = -\frac{\partial p^{*}}{\partial x_i^{*}} + \frac{1}{\textrm{Re}}\frac{\partial^2 u_i^{*}}{\partial x_j^{*} \partial x_j^{*}}$$

Stokes flow assumes that $\textrm{Re} \ll 1$, so the dissipation term, $\frac{\partial^2 u_i^{*}}{\partial x_j^{*} \partial x_j^{*}}$, will be an order of magnitude greater than the pressure term. In other words, the forces on the object in question are going to be dominated by viscosity with negligible contribution by pressure/density effects.

7. Feb 23, 2012

### Vaal

When you point out the mathematics I guess it does make sense. I don't have a lot of dynamics experience and I guess I am just having trouble intuitively getting used to the idea that the number of interactions (which is proportional to the density) would not change the viscosity.

Thanks for your help.