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\begin{document}

\newpage
\title{\ifnum\TeacherEdition=1 TEACHER VERSION \else STUDENT VERSION \fi\\Falling Coffee Filters}





\markboth{Falling Coffee Filters}{Falling Coffee Filters}








\author{Brian Winkel, Director SIMIODE\\
Chardon OH 44024 USA\\
BrianWinkel@simiode.org
}

\makeStitlePDFLaTex


\begin{abstract}
We are given data on the time and position of a stack of coffee filters as it falls to the ground. We attempt to model the falling mass and we confront the different resistance terms and models.
\end{abstract}




\section*{SCENARIO DESCRIPTION}

\begin{figure}[h]
\centering
\includegraphics[width=8.0 cm]{3-16-VernierAndFilters.eps}

\vspace*{-4.9 true cm}
\parbox{7.4 true cm}{\caption{\label{fig:3-16-VernierAndFilters}  Vernier Software \& Technology Motion Detector and stack of coffee filters.}}

\end{figure}


We are given data on time and position of a free falling stack of coffee filters.  The  data offered in Table~\ref{table:data} is from an experiment to model a falling stack of 12 commercial coffee filters.  Each coffee filter has  mass  0.01908 kg for total stack mass $m = 12\times 0.01908 = 0.22896$ kg. The data was collected with a Vernier Software \& Technology Motion Detector on 20 February 2010 at the United States Air Force Academy CO where the acceleration due to gravity is $g = 9.78976 m/s^2$. The Motion Detector, along with a stack of coffee filters, is depicted in Figure~\ref{fig:3-16-VernierAndFilters} while a plot of the data from Table~\ref{table:data} is offered in Figure~\ref{fig:3-16-DataPlot}.





\begin{table}[h]\label{table:data}
\begin{center}
\small
\begin{tabular}{|r|r||r|r|} \hline  \hline
\multicolumn{1}{|c|}{Time (s)} & \multicolumn{1}{|c||}{Distance (m)}& \multicolumn{1}{|c|}{Time (s)} & \multicolumn{1}{|c|}{Distance (m)}\\ \hline \hline
0.00 & 0.0000& 0.22 & 0.3619\\ \hline
0.02 & 0.0228& 0.24 & 0.4050\\ \hline
0.04 & 0.0478& 0.26 & 0.4500\\ \hline
0.06 & 0.0730& 0.28 & 0.4931\\ \hline
0.08 & 0.1054& 0.30 & 0.5440\\ \hline
0.10 & 0.1369& 0.32 & 0.5927\\ \hline
0.12 & 0.1710& 0.34 & 0.6429\\ \hline
0.14 & 0.2063& 0.36 & 0.6937\\ \hline
0.16 & 0.2426& 0.38 & 0.7458\\ \hline
0.18 & 0.2810& 0.40 & 0.7991\\ \hline
0.20 & 0.3197& 0.42 & 0.8556\\ \hline  \hline

\end{tabular}

\parbox{12 true cm}{\caption{\label{table:data}  Data on the time and position of a stack of coffee filters as it falls to the ground. We measure the distance falling positively downward and say $y(t)$ is the distance fallen in meters at time $t$ in seconds, i.e. $t$ is time in seconds from release and $y(t)$.  is distance in meters fallen from release.}}
\end{center}

\end{table}


\begin{figure}[h]
\centering
\includegraphics[width=13cm]{3-16-DataPlot.eps}


\parbox{7.4 true cm}{\caption{\label{fig:3-16-DataPlot}  A plot of the data from Table~\ref{table:data}.}}

\end{figure}
We will make a mathematical model of this phenomenon and estimate the parameters in the model. We shall also compare the model's predictive power with the actual data. 




We have actually taken our observed data and adapted it as follows to produce the data in Table~\ref{table:data}. In order to estimate the initial velocity (for the equipment only offers position with averaging for its internal estimate of velocity) we use the first and third data point $(t_1,y_1)$  and $(t_3,y_3)$  where $t_i$ is time in seconds of the  $i^{\rm th}$ observation and  $y_i$ is the distance fallen at the $i^{\rm th}$   observation  and form the symmetric difference about the second observation of our collected data as an approximation of initial velocity, calling this variable, AveVelInitial:

$${\rm AveVelInitial} = \frac{y_3-y_1}{t_3-t_1} \, . $$ 
					
In our case this was ${\rm AveVelInitial} = 1.10446$ m/s.  We then start our time at $t = 0$ at the second observation of our collected data and note our distance fallen here is 0 m, shifting all other time and distance observations accordingly, with an initial velocity of 1.10446 m/s.

\begin{itemize}
  \item  [a)] Write down some assumptions about a controlled environment in which you might collect such data.
 \item  [b)] With these assumptions, construct a Free Body Diagram (FBD)  and indicate all the forces acting on the mass of the stacked coffee filters.  
 \item  [c)] Using Newton' s Second Law of Motion which says that the change in momentum is equal to the sum of all the external forces acting on the mass of the stacked coffee filters, construct a mathematical model for determining the velocity, $v(t) = y'(t)$, of this object. Here $y(t)$ is the distance the coffee filters have fallen from rest in meters at time $t$ in seconds. 

		$$d/dt(m*v (t)) = m y''(t) =  \rule{6 true cm}{.01cm} \, . $$

 \item  [d)] Solve your differential equation for $v(t)$ or $y(t)$.

 \item  [e)] If you choose to solve the equation in $v(t)$, the velocity, then you can integrate the velocity function you obtained, $v(t)$, above to get position function,  

$$ y(t) = \int_0^t v(T)\, dT \, . $$

Or you may solve the differential equation in (c) for $y(t)$, the position, and use that solution.

 \item  [f)] Estimate the parameters  in your model.
 \item  [g)] Compare your solution of your model to the data offered above.
 \item  [h)] Write a paragraph description of the entire process in (a) - (g) and indicate how confident you are of the various aspects of this process.
 \item  [i)] What other variations might you consider for this model?
\end{itemize}  

\teacher{
\section*{NOTES FOR TEACHERS}

We review.  We measure the distance falling positively downward and say $y(t)$ is the distance fallen in meters at time $t$ in seconds.

From a Free Body Diagram and Newton's Second Law, which says that the change in momentum ($d/dt(m v (t))   = m*y''(t))$ is equal to the sum of ALL external forces acting on the body, we formulate the general model from the fact that the only external forces are that of gravity acting on the mass (i.e.  $m*g$) and the force in the opposite direction of resistance due air which we presume is proportional to some power of the velocity (i.e.  $k  y'(t)^r$). We obtain  differential equation (\ref{eq:intromodel}):

\begin{equation}\label{eq:intromodel}
m y''(t) = m g - k  y'(t)^r\, . 
\end{equation}  

We attempt to approximate $K = k/m$, the specific coefficient of air resistance,  and $r$ in the differential equation with three models:

\begin{enumerate}
  \item  $r = 1$ (linear resistance), (often this value for $r = 1$ is presumed in simple modeling efforts)
  \item  $r = 2$ (quadratic resistance), ( this value for $r = 2$ is presumed when there is turbulence behind the object), and 
  \item  $r$, (general resistance). 
\end{enumerate}  
We divide both sides of the first differential equation model by $m$  and identify $K = k/m$, calling $K$  the specific coefficient of air resistance, to get to the second differential equation model without mass, $m$, appearing in (\ref{eq:genmodel}).

\begin{equation}\label{eq:genmodel}
 y''(t) =  g - K  y'(t)^r 
\end{equation} 

\subsection*{Analyses}

We offer complete analyses for all three cases, $r = 1, 2$ and in general, with two approaches in the latter case (a) a grid in the parameter space $(K,r, SSE(K,r))$ over which we compute the sum of square errors for each set of parameters $(K,r)$ (some 160,000 of them) and select the parameters which minimize the sum of square errors, and (b) a gradient approach in the parameter space $(K,r, SSE(K,r))$ which converges to the same result as our grid examination.

We offer our results in a Mathematica notebook 3-16-Mma-T-FallingCoffeeFilters-TeacherVersion.nb and an accompanying pdf file 3-16-Mma-T-FallingCoffeeFilters-TeacherVersion.pdf for those who do not have access to Mathematica.

\subsection*{Summary of Analyses}


We are using the following general differential equation to model our falling coffee filters:

\begin{equation}
 y''(t) =  g - K  y'(t)^r 
\end{equation} 
Here  $K = k/m$ and $r$  are the parameters we need to estimate in the differential equation with general power resistance.
  
Of the 3 cases we examined: for $r = 1$ (linear resistance), $r = 2$ (quadratic resistance), and $r$, (general resistance) we found the general resistance power to be the best fit - by far!   

In the case for $r = 1$ (linear resistance) we find the constant $k/m = K = 3.0804$ and  our minimum sum of square errors is $SSE = 0.00041054$.

In the case for $r = 2$ (quadratic resistance) we find the constant $k/m = K = 1.69014$ and  our minimum sum of square errors is $SSE = 0.0029966$.


In the general resistance model, using our grid approach, we found   $K = 3.95$,  $r = 0.55$, and our minimum sum of square errors is $SSE = 0.0000241985$.


In the general resistance model, using our gradient approach,    we found   $K = 3.93142$,  $r = 0.560844$, and our minimum sum of square errors is $SSE = 0.0000239852$.

\begin{figure}[h]
\centering
\includegraphics[width=13cm]{3-16-GradientResultPlot.eps}


\parbox{8.9 true cm}{\caption{\label{fig:3-16-GradientResultPlot}  A plot of the best result model (in terms of smallest sum of square error) over the original data from Table~\ref{table:data}.}}

\end{figure}


Thus the parameters obtained through the gradient search seem to offer the lowest sum of square errors and thus become our model of record.  This means for 12 coffee filters with $k/m =  K = 3.93142$ and $m = 0.22896$ that our specific coefficient of air resistance, $k$, (independent of mass) is found by solving $k/0.22896 = 3.93142$ or $k = 0.900138$ when $r = 0.560844$. We offer a plot of this best fitting model over the original data in Figure~\ref{fig:3-16-GradientResultPlot}.

\subsection*{Issues}

There are many proposed models for falling bodies with resistance, some from first principles and others empirical in an attempt to just fit the data.  We do not direct the students in this modeling scenario, but rather let them produce a model with which they are comfortable. We are also interested in strategies for estimating parameters in models and we offer several in   3-016-Mma-T-FallingCoffeeFilters-TeacherVersion.nb and an accompanying pdf file 3-016-Mma-T-FallingCoffeeFilters-TeacherVersion.pdf.

We have found that following through the modeling process by estimating parameters from a model reinforces the power of differential equations to model the real world.
}




\end{document}
 
