Re: free fall data

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

Ludwik,

It seems to me that the best way of analyzing this type of data is simply
to fit the raw (d vs. t) data to a function of the expected form.  These
days simple programs abound which will perform regressions to a large
number of possible functions and give you estimates of the uncertainties
in the parameters. I don't see any reason why students, even in an
algebra-based course, shouldn't be able to understand what a regression
does and even how it can get estimates for the uncertainties in the
parameters.

I used a simple six year old freeware program for the Mac (Curve Fit 0.7)
to fit your data to a second order polynomial. Interpreting the
coefficient of the second order term as half the acceleration, it gives 

  a_free fall = (9.84 +/- .46) m/s^2.

Two comments:  1) The discrepancy between this result and the accepted
value is surprisingly small given the size of the uncertainty. 2) The data
point at t = .73333 s is very obviously out of line with the rest of the
data.  When it is eliminated the regression gives

  a_free fall = (10.16 +/- .05) m/s^2.

It's easy and instructive to interpret these facts as indicating 1) that
there is very likely one bad data point, 2) that the data is otherwise
highly consistent with the hypothesis of constant acceleration, and 3)
that you have a significant discrepancy with the accepted value.

Now the task is to determine if that significant discrepancy is the result
of a systematic error or a local gravitational anomaly. ;-)

John

On Sun, 14 Sep 1997, LUDWIK KOWALSKI wrote:

> Here are free fall data I took from the DSON017 movie which came on the
> same CD-ROM as VideoPoint. Called "Vertical Ball Toss" it was recorded
> at 30 fps. There are 29 frames but only the 2nd half, the ball on the 
> way down is analysed here. "Insignificant" digits were used to avoid 
> additional rounding errors (in that sense digits are very significant).
>
>   time(s)   dist(m) speed(m/s) g(m/s^2)
>
>   0.36666   0.000                     AVERAGES
>   0.40000   0.015     0.450
>   0.43333   0.047     0.960     15.30          As you can see, the
>   0.46666   0.088     1.230      8.10   11.1   values of g fluctuate
>   0.50000   0.140     1.560      9.00          between 5.4 and 18.7
>   0.53333   0.198     1.740      5.40
>   0.56666   0.271     2.190     13.50   10.8   The mean acceleration is
>   0.60000   0.359     2.640     13.50          10.66 (or 10.08 if 15.3
>   0.63333   0.453     2.820      5.40          and 18.72 are ignored).
>   0.66666   0.557     3.120      9.00    9.6   
>   0.70000   0.677     3.600     14.40          The mean is very good
>   0.73333   0.897     3.900      9.00          but the constancy of g
>   0.76666   0.948     4.224      9.72    9.3   is not demonstrated. I
>   0.80000   1.099     4.530      9.18          the most simple way of
>   0.83333   1.265     5.001     14.13          calculating v and g.
>   0.86666   1.437     5.157      4.68    12.5
>   0.90000   1.630     5.781     18.72          Even averages of 3 values
>                                                fluctuate by more than 5%
> ...

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