Re: Air Resistance

[LUDWIK KOWALSKI <KOWALSKIL@alpha.montclair.edu>]

> Date: Wed, 10 Dec 1997 13:43:27 -0500 (EST)
> From: "James W. Wheeler" <jwheeler@eagle.lhup.edu>
> Subject: Re: Air resistance
>
> Question:  What is the volume and density of the falling object?  Is
> Flotation an important effect?
> 	F = -gV(density object-density air) -cv^n

The object was "WILSON OFFICIAL SIZE-WEIGH" ball but it was poorly inflated.
The weight was 0.550 kg, the diameter was 9.5 inches (not 9, as reported
earlier). This gives volume V=0.0588 m^3 and density=9.35 kg/m^3. Room 
temperature probably 25 C. I am puzzled by your first sign, you probably 
did not mean the sign of m*g and the sign of the air resistance term, c*v^n,
to be the same. But you are right that the buoyancy correction should be
made when R is evaluated from a. I will do it and  report a better value 
of n, probably tonight. 

With new diameter the empirical formula from a book (see below) gives:

     v --> 2.5 m/s   3.0    3.5     4.0     4.5     5.0 
     R --> 0.080 N  0.115  0.157   0.205   0.260   0.320 

Not dramatically different from our experimental data. Air resistance
is clearly not negligible at the velocities of this experiment. Note
that v=5 m/s is about 11 mi/hr.

The  reference for the empirical formula: "Analytical Mechanics" by 
G.R. Fowles and G.L. Cassiday, Fifth Edition, 1993, page 62. It is the
usual c1*v+C2*v^2 description. For a sphere of diameter D the empirical
values are:  c1=1.55*10^-4*D and c2=0.22*D^2, all in SI units.