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Re: Air resistance



Ludwig,
> Please help to decide between the two positions taken:
>
>1) The data shown below demonstrated that n is between 1.6 and 2.4
>2) The data shown below demonstrated that n is between 0.2 and 5.0

Fascinating discussion, great ideas for experiments, but neither position
below satisfies me. I believe that the data show little more than that the
object is falling but not in free fall.

My concerns: You give 16 data points and then use velocity data that uses
15 point averaging. If I want to really check the results I rely on the
actual raw data--namely the time and position. This means that I casn only
generate 2 velocities with 15 point averaging.

What does MacMotion really mean by averaging? Do they do a least squares
fit to 15 points or what?

Plotting position versus time in Excel resuilts in a curved line--some
acceleration is obvious. A second order polynomial fit to the data gives
d = 4.7454 t^2 - 11.495 t + 7.3998, suggesting a constant acceleration of
9.49 m/s^2, as noted earlier by John.

More meaningful data for me would involve complete data for a broader time
range extending from close to the initial drop until close to terminal
velocity.


John,
Precisely what are you fitting? d versus t? With what function? Did you
integrate to get d-t formula for different resistance functions or what?


The data were collected with the Vernier motion detector at the rate
of 40 samples per second, averaging=15. The first 4 colums are output
from the Mac Motion software. ...

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 0.11
1.525 0.904 2.975 9.61 0.10
1.550 0.983 3.215 9.59 0.12
1.575 1.066 3.454 9.54 0.14
1.600 1.155 3.691 9.48 0.18
1.625 1.251 3.927 9.44 0.20
1.650 1.352 4.162 9.38 0.23
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