On analyzing the fall of an object released in air electromagnetically from rest, I found that the data perfectly fit the following quadratic curve:

$$y = 9.7686 \tfrac{\textrm{m}}{\textrm{s}^2} * \tfrac{t^2}{2} + 1.22618 \tfrac{\textrm{m}}{\textrm{s}} * t + 0.00029 \textrm{m}$$

The data that went into this calculation were captured by a high-speed camera snapping pictures of an object falling in front of a marked background with precisely measured distances, and the coefficient of determination between the above function and the data is $1.00000$.

This equation confuses me, because it doesn't seem to fit any reasonable form. If you solve for the general form of acceleration and then plug into $y = \frac{at^2}{2}$, you should get an equation that can be set equal to the quadratic above to get the constants, which should tell you the exponent on the drag.

Without wind drag, you would expect the position equation to fit $y = \frac{gt^2}{2}$. With constant drag force, which from what I understand isn't typically how wind drag presents, you would expect the equation to fit the same form but with a different $g$ as $y = \frac{at^2}{2}$.

With drag proportional to velocity, expressed as $a = g - cv$, you would expect the position equation to fit $y = \frac{gt^2}{2 (ct+1)}$.

With drag proportional to velocity squared, which is what I was expecting, expressed as $a = g - cv^2$, you would expect the position equation to fit $y = \left[\frac{-1}{2ct^2} ± \sqrt{\frac{1}{4c^2t^4}+\frac{g}{ct^2}}\right]\frac{t^2}{2}$. But if you do the math, the quadratic doesn't fit any of these, because it seems to start with a non-zero velocity.

Why isn't the curve of the fall, which perfectly matches the data, not consistent with any of these forms? The curve is quadratic, but starts with a non-zero velocity, and I have no idea why.

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    $\begingroup$ If the object is released before $t=0$, you would have an initial velocity. You can find a $t^{'}=t+t_0$ that gets rid of the linear term. If $a<g$, you have an upward force somewhere. If it doesn't have the right form for air resistance, then it isn't air resistance. Or maybe it is another error. If your object is in front of the scale, it might be parallax introducing an error in your distance measurements. $\endgroup$ – mmesser314 Feb 10 '17 at 12:45
  • $\begingroup$ @mmesser314 If you do the math, there's too much initial velocity, far more than can be accounted for by releasing before $t=0$. It's quadratic, just not consistent with starting from rest. $\endgroup$ – TheEnvironmentalist Feb 10 '17 at 12:48
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    $\begingroup$ Then perhaps your release mechanism is throwing your object. It is hard to say without seeing your setup. $\endgroup$ – mmesser314 Feb 10 '17 at 12:49
  • $\begingroup$ @mmesser314 Just an electromagnet and an iron bob, and a timer that cuts power to the electromagnet. $\endgroup$ – TheEnvironmentalist Feb 10 '17 at 12:51
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    $\begingroup$ Re I found that the data perfectly fit the following quadratic curve: No you didn't. Except for scientists / engineers who cook the books, there is no such a thing as a perfect fit in experimental sciences. In fact, too good a fit can be a sign of cooking the books. Aside: Your expressions for linear and quadratic drag are incorrect. For linear drag, $y(t) = \frac g {c^2}(ct + \exp(-ct)-1)$ (assuming $y(0)=y'(0)=0$, and positive $y$ is down). For quadratic drag, $y(t) = \frac 1 c \ln(\cosh(\sqrt{cg}\,t))$ (same assumptions). $\endgroup$ – David Hammen Feb 10 '17 at 13:13

It looks like you have started timing $(t=0)$ and measuring distance $(y=0)$ when the ball has reached a velocity of $1.226m/s$ at a time of $1.226ms^{-1}/9.77ms^{-2} \approx 0.125s$ after release. The ball has by then travelled $\frac12\times 1.226ms^{-1}\times 0.125s \approx 77mm$ since leaving the electromagnet.

Perhaps your timer is triggered by the ball passing a light gate located $77mm$ below the electromagnet? This seems to be the simplest explanation.

If your object is an iron ball bearing and the distance is no more than 2m I would not expect much air resistance anyway. Wasn't the purpose of the experiment to measure $g$ rather than air resistance? The value of $g=a$ is given by the $t^2$ term of the quadratic fit. You can ignore the $t^1$ and $t^0$ terms. There is no need to make an adjustment to the time scale in order to get the form $y=\frac12 at^2$.

How many data points did you have? If you only have 3 data points, you will always get a perfect fit to a quadratic formula.

It would be helpful if you described the apparatus, gave dimensions, and posted your data.

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  • $\begingroup$ Very good analysis and reasonable pointers/guesses about the underlying causes - given the absence of more detail. $\endgroup$ – Floris Feb 10 '17 at 23:53
  • $\begingroup$ So we had about 6 data points, and we really should have had more, but we were limited by the markings on our backboard. What I'm wondering is why the velocity at $t_0$ given acceleration $9.77\tfrac{\textrm{m}}{\textrm{s}^2}$ doesn't match the position at $t_0$. This is really what I'm trying to figure out. The velocity is far more than can be accounted for by constant acceleration over the displacement, even though $r^2$ seems like an indicator of a perfect match $\endgroup$ – TheEnvironmentalist Feb 11 '17 at 1:18
  • $\begingroup$ I have not seen your experiment, you have not provided any details, so it is difficult for me to comment. I just made an educated guess. If you measured distance and time starting at the light gate, that is consistent with your equation, which gives $y \approx 0$ when $t=0$. $\endgroup$ – sammy gerbil Feb 11 '17 at 2:54

Just wanted to expand a bit on Sammy's very good answer, in light of David Hammen's comment.

When an object falls under quadratic drag, the velocity as a function of height can be written as

$$v = \sqrt{v_t^2\left(1-e^{\frac{-2g(y_{peak}-y)}{v_t^2}}\right)}$$

Where the terminal velocity $v_t$ is given by

$$v_t=\sqrt{\frac{2 m g}{C_D\rho A}}$$

This tells you that drag becomes important when $\frac{2 g y}{v_t^2}$ is no longer a small number - so when $\frac{C_D \rho A y}{m}$ is not small. It's not hard to see that this will happen when the total mass of the air traversed ($\rho A y$) becomes comparable to the mass of the object falling $m$. That's actually a pretty interesting result.

There are a few things you need to do with your data.

First - ask yourself how many datapoints you had, and how many coefficients you are fitting. Obviously a second order fit through three points is "perfect"

Second - assuming you have enough redundancy in your data, plot the residuals - that is, the difference between the calculated value, and the fit. Sometimes a "almost perfect" fit will prove to have a small systematic error that can be see in the residuals but not the original

Third - as Sammy pointed out, if your fitted equation of motion tells you there is an offset in time and space, then maybe your coordinate system didn't start at exactly y=0, or time didn't start exactly when the object was first released.

Fourth - using the equation above, ask yourself if you expect drag to be significant in this case. At the end of the drop, were you getting anywhere near the terminal velocity? Were you at 1%, 5%, 20% of $V_t$?

Finally - it is often good, when you do an experiment, to try to plot the data in such a way that you expect a straight line. Then, if your hypothesis is incorrect, you will see something that is "not quite straight", or that has an obvious offset; also, often if you plot the right thing, either the intercept or the slope will be a good measure of the thing you are measuring (for example, $g$). I will let you mull that over a bit.

Bottom line - don't "plug numbers in" to a calculation or formula; especially experimental data. It deserves plotting, and careful thought. You spent time doing the experiment; spend equal time analyzing the result and making sure you really understand what the data are telling you.

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