Resistance

[BigMoosie]

I have recently been studying the mathematics behind projectile motion through air resistance. We are given problems where the resistance force is proportional to v² or v or something else (simple) containing v. I know that shape has alot to do with it but what is normally the function used for approximating air resistance for say, a spherical object. Or are such functions so complex that it would be impossible to solve without experimental data?

[timo]

The function used depends on the substance your are passing through and the velocity you´re using. For slow velocities, a function ~v might be sufficient. For higher velocities, you´ll need v² terms. Generally, all those function should better be regarded as an approximation only valid within a certain range. Computing friction for realistic scenarios is a complicated task and afaik an industry / branch of research in itself: The friction is not calculated by simple terms like above, but instead a full simulation which also includes turbulences and stuff like that is run. I don´t know how good these simulations are these days but I doubt that anyone would let a plane transport passengers when its aerodynamical properties have only been calculated on a computer and not really tested in an experiment - the computer simulation could at most be used to reduce the amount of experimenting because effects can already be estimated from the simulation. Another interesting thing in this context is that such experiments are not scale-invariant. You cannot just model your new plane with a size reduced by a factor of 100 and expect it to have the same aerodynamic properties as the real plane - a reason being that you cannot scale down the size of an air molecule by a size of 100.

 

In short: A realistic calculation can be arbitrarily complex, I think. For simple things like shooting someone in the back, effective models like ~v² can be sufficient.

[YT2095]

yeah, that sort of thing requires Reynolds numbers and laminar flow calcs Aaaargggghhhhh!

[Bignose]

For a single sphere travelling in an infinite fluid, the drag coefficient has simple expressions in the high velocity and low velocity regions, but not intermediate. To know what region you are in, you need to know the particle Reynolds number: Rep = Udr/m

 

where U=particle's velocity

d = diameter of particle (sphere)

r = density of the fluid

m = viscosity of the fluid

Rep is a dimensionless number (so all the units of U,d,r & m have be converted so that they cancel)

 

Then, the drag coefficient, CD, can be written as:

 

For Rep<0.3, CD=24/Rep

For 0.3<Rep<500, CD=(24/Rep)*(1.0 + 0.15*Rep^0.687)

For Rep>500, CD=0.44

 

This is just one of several correlations that exist, most of them varying in their representation of the intermediate region. The low Rep region, known as the Stokes flow region, can be analytically calculated, whereas the high Rep very turbulent region has been confirmed experimentally many, many times.

 

with the drag coefficient, you can write the drag force:

 

Fd = (pi/8)*CD*d^2*r*|V-U|*(V-U)

 

where V is the fluid velocity (Note the absolute values!)

 

So, you see, when the flow speed is low, because of the Reynolds number dependence in CD, the drag force is proportional to the velocity difference to the first power. But, when the speed is high, because CD becomes just a constant, the drag is proportional to the velocity difference squared.

 

An iterative procedure can be used: Guess a velocity, use that guess to calculate a CD, then calculate the velocity for that CD. If the calculated velocity and guessed velocity are the same, you are done, but normally, you have to use that newly calculated velocity as the next guessed velocity and perform the cycle a few times for the desired accuracy.

 

Then, of course, all the other difficulties in calculating drag: non-spherical particles, the presence of other particles, the effects of the walls of the container the fluid is in, just to name a few.

[BigMoosie]

This is much more complicated than I was expecting, thanks so much guys, your help is as always very valuable to me.