The first rough guess from looking at frame 20 (the dropping cap) goes as follows: The grid is 6x8 inches, so the cap is dropped from about 7.5 vertical grid spacings or 150 cm. With a bit of squinting and guessing, I can see 28 images of the cap on the way down. Using s = 1/2gt2, I get t = 0.55 sec, whichleads to about 0.02 seconds between flashes, or about 50 Hz.

The next try is a bit more detailed: Using a graphics program (the GIMP) I read off the pixel positions of the top and bottom of the cap dropped in frame 20. They can be well-determined in 19 positions: 
pos#  top   bot   avg
------------------------
 1    127   141   134	
 2    137   151   144	
 3    148   162   155	
 4    162   176   169	
 5    177   191   184	
 6    193   208   200.5	
 7    213   228   220.5	
 8    234   248   241	
 9    257   272   264.5	   Click on the
10    283   297   290	   picture to see
11    309   324   316.5	   the full size --->
12    338   352   345	
13    368   381   374.5	
14    399   413   406	
15    434   447   440.5	
16    469   482   475.5	
17    506   516   511	
18    546   552   549	
19    586   592   589
I then did a fit to the data in the first and last column with the following equation:

(ypx-c) = a*(tstrobe-b)2      (1)

where ypx is the position of the falling cap in pixels, and tstrobe is the time measured in intervals between flashes of the strobe light. Parameters b and c are arbitrary offsets, and the one we want is parameter a, since we want to get to the equation for a falling object:

ycm = 1/2gtsec2      (2)

with y in cm and t in seconds, and g=981 cm/s2 is the acceleration of gravity. (see this NASA site for more on this)

The results of the fit are:
a = 0.814
b = -5.35
c = 109.8
so I write the simplified (ignoring b and c) equation as:
ypx = 0.814 tstrobe2      (3)
Now let's convert from pixels to cm. On the image, the location of the horizontal lines of the grid are at 710, 810, 912, 1013, 1113 and 1214 pixels. These rungs are spaced 8 inches apart, so I get 4.96 pixels/cm. With this, equation (3) turns into:
ycm = 0.165 tstrobe2      (4)
Now we have to convert from strobe counts tstrobe to time tsec in seconds. If the conversion factor is d, I can write
ycm = 1/2g(d*tstrobe)2      (5)
Comparing (4) and (5), I get d = 0.0183, which is the time between flashes in seconds. Inverting this, the strobe ran at 54.4 Hz. The setting on the strobe scale scale was 106.7 Hz. [ Obviously, I lost a factor of 2 somewhere here...]

On frame 25, the white (masking) tape stripe on the bottle is visible in 3 locations (apart from the launch position at 1149), at pixels 354, 817, and 1053. The difference between the first and last points, 1053-354 converts to 141.5 cm, and 2 strobe intervals are 0.0366 sec, so the average velocity between these points was 38 m/s or 137 km/h or 87 mph.

The next approximation I can make on the back of an envelope is to assume that there is uniform acceleration from the launch at 1149 pixels (from the top of the picture) at the moment of launch at t=0 through the point at 1053 pixels at an unknown time t1 to the point at 354 pixels, which is 2 strobe clicks later. I have 2 unknowns, namely t1 and the acceleration a, and 2 equations. There is a solution with t1 = 0.22 seconds and a = 80g. Quite a blast!

Click on the
picture to see
the full size --->

Last update 22 Oct 2003 - HvH
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