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Re: free fall data

THIS WAS POSTED LAST NIGHT BUT I DID NOT RECEIVE IT. REPOSTING:
***************************************************************
Responding to:
... Strong zooming, to cover as many pixels as possible, is
equivalent to filming from a short distance. It may be associated
with sizable paralax errors.

Brian Whatcott <inet@intellisys.net> writes:

This is factually incorrect. A prime method to REDUCE parallax errors
is to place a camera at as large a distance as possible and use the
largest magnification which a telephoto lens can offer - i.e. by zooming.
Visualize the extreme rays from the target, and the angle which they
subtend at the camera; furthest distance = smallest angle.

This is a very important piece of information. I can not visualize the
rays. Can you give us a reference, Brian?
Thanks.
.........................................................................
P.S. My own lab preparation data (free fall) are not good.
The acceleration is fluctuating a lot, as shown below.

23.3 6.03 14.9 and 0.00 ---> mean = 11.0 m/s^2

I repeated the measurements (working with VidePoint on the same frames)
and the following results were obtained:

11.7 6.03 8.91 and 8.91 ---> mean = 8.9 m/s^2

Pixels must be the main source of error, different sequences are obtained
from the same frames. The mean of two means is very good but I am not happy.
I envy those who wrote they can keep fluctuations below 5%. My VHS shutter
was 1/1000.

The class analysis will be limited to the speed-time line approach.
I will say something like this. "g is the slope of the v-t line. Thus to
verify its constancy we must verify that the v(t) curve is a straight line.
It is not a perfect linear graph but we will find the average slope. How
close is it to 9.8?" What else can I do with data on displacements which
are not accurate enough?
Ludwik Kowalski