Re: Air resistance

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

On 06 Dec 1997 09:56:16 - John Mallinckrodt 

> ... I can (as with the previous data) obtain *excellent* numerical 
> fits to your d vs. t data (i.e., rms errors in d less than 1 least 
> sig fig) with drag exponents (i.e., the n in F_drag = b v^n) anywhere 
> from 1/5 to 5.  
There must be a misunderstanding somewhere. I am reposting the data 
at the end of this message. Just plot R=f(v) on the log-log paper
and get n="the best slope". Without doing this I can say that n=2 
(plus or minus of up to 20%). Here is my reasoning. The region in
which good data were collected spans from about 2.49 to 4.63 m/s 
Thus (v2/v1)=1.8. The corresponding ratio of air resistaces, R2/R1 
is 3.0. What is n when 1.8^n=3? It is close to 2.

For your values on n=5 and n=0.2 the force ratios would be 19 and 1.1,
respectively. The experimental data are not perfect but they are 
certainly not so bad as to accomodate your range of n. And this has 
nothing to do with the value of b (=m*r) in your R=b*v^n formula.

                                                  Ludwik Kowalski
.......................................................................

> With do = d at the first time, vo = v at the first time, r = b/m where 
> m = mass of falling opbject and, F_drag = bv^n, I get the following 
> (all in SI > units) 
>
>    r        n     do       vo      g     rms deviation in d values
>   0.18     0.2   0.608    1.776   9.8             .0009
>   0.023    2.0   0.6078   1.772   9.8             .0007
>   0.0004   5.0   0.6079   1.763   9.8             .0008
>
> The values of the best fit parameters will surely depend to some small
> extent on the numerical method used.  I used a simple "predictor-
> corrector" type method. I could have obtained marginally better fits by
> allowing g to be a free parameter as well, but that was clearly pushing
> the data too hard.
  **********************************************************************
    REPOSTING OF THE FALLING BALL DATA. mass=0.55 kg, diameter 9 inches.

Time zero corresponds to the moment at which the instrument is activated;
the object is usually released about one second later. Values of R were 
obtained form R=0.55*(9.8-a) formula. Use it to get more points for a
serious "best fit" analysis and poste your results. I will do this now. 

      t(s)         d(m)         v(m/s)        a(m/s^2)         R(N)

     1.400        0.608         1.773           9.63
     1.425        0.655         2.013           9.65
     1.450        0.708         2.254           9.65
     1.475        0.768         2.494           9.64           0.09
     1.500        0.835         2.734           9.60
     1.525        0.904         2.975           9.61           0.10
     1.550        0.983         3.215           9.59
     1.575        1.066         3.454           9.54           0.14
     1.600        1.155         3.691           9.48
     1.625        1.251         3.927           9.44           0.20
     1.650        1.352         4.162           9.38
     1.675        1.458         4.395           9.33           0.26
     1.700        1.573         4.632           9.31           0.27
     1.725        1.692         4.865           9.33           
     1.750        1.816         5.035           7.83
     1.775        1.945         4.971           1.27
*************************************************************************