Re: free fall data

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

I convinced myself this morning, by using Excel, that in order to 
demonstrate that g is approximately constant during the fall the
accuracy of measuring distances must be of the order of 0.1%. The
conclusion was reached by analysing "perfect data". Thus I am 
sceptical that a physics teacher can count on nearly equal values 
of g from consecitive frames. Can somebody contradict me by saying
that "all values of g, at least six of them, are nearly always 
identical to within 5% or better"? This may happen by chance but 
not too often. Let me continue after showing the message to which 
I am responding. 

> .... Here is video analysis data for a falling softball. Least 
> squares on the velocity data gave -9.83 +- 0.3 m/s/s. While the 
> video was captured at 30 FPS, only every other frame was analyzed, 
> and a different finite derivative is used than in your message. 
> For example, the velocity at t = 0.2 seconds is calculated:
>
>      v = (0.504 - 0.909)/(0.267 - 0.133) = -3.022 m/s
>
> The difference between this value and the one in the table are 
> because of truncation in the table. 
>
>   t(s)	 Y1(m)	 Dy1(m)	  Vy1(m/s)  Ay1(m/s/s)	
> 0.00E+00	1.134	 0.0			
> 6.70E-02	1.039	 -0.095	  -1.677		
> 1.33E-01	0.909	 -0.225	  -2.338     -10.051	
> 2.00E-01	0.725	 -0.409	  -3.024     -9.672	
> 2.67E-01	0.504	 -0.63	  -3.634		
> 3.33E-01	0.238	 -0.896			
>
> Things you might try to improve your measurement error are: ....

Good suggestions; I will keep them in mind in the lab this week.
And I promis to share the data. The reason I am sceptical is based 
on the following consideration. The accuracy of y must be between 
0.1 and 1% according to my Excel analysis. Everybody can verify this; 
just create ideal data and use the spreadsheet to see what happens 
to individual g when small changes are made in the corresponding 
values of y. 
 
What accuracy can be expected? Suppose that the vertical distance 
of 0.9 meters (see table above) covers all 480 pixels of a frame.
The distance errors are probably 2 pixels at each end. This leads
to an error of 4 pixels for each y (about 1% of the frame's height, 
h). Individual values of y are fractions of h and percentage errors 
on them would be typically 5% or more. I know that two consecutive
errors of 5% in y can lead to differences in g which are enormous.

And I suspect that 480-pixel resolution, used above, is not possible 
when a VHS tape is used as a source. If I recall correctly, the vertical 
resolution of what is recorded on tape is something like 240 lines. If 
this is true then 480 pixels, used to digitize a frame, are illusionary; 
what was lost (when analog pictures are magnetically recorded on tape) 
can not be recovered through digitalization. The effective vertical 
resolution would thus be 240 pixels and an error of 4 pixels would 
correspond to 10% error in a typical value of y. That is why I was 
asking about the side-wise filming. The horizontal resolution may 
possible be limited by the digitizing board only (640 pixels per line). 
Does this make sens to those of you who are familiar with technology
of video-recording?
                                                   Ludwik Kowalski