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\title{\ifnum\TeacherEdition=1 TEACHER VERSION \fi\\Cooking a Turkey}
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%\else STUDENT VERSION \fi\\Cooking a Turkey}





\markboth{Cooking a Turkey}{Cooking a Turkey}

\author{Michael A. Karls \\
Department of Mathematical Sciences\\ 
Ball State University\\
Muncie IN 47306 USA\\
mkarls@bsu.edu
}


\makeStitlePDFLaTex


\begin{abstract}
The goal of this project is to investigate several models for the cooking time for a turkey based on weight, test these models with data obtained from heating curves for turkeys of various weights, and develop a new model to fit this data. 
\end{abstract}




\section*{SCENARIO DESCRIPTION}


A dilemma facing many cooks during the holiday season is determining the time needed to safely cook a turkey. Guidance that can be found in sources such as cookbooks, newspaper articles, and cooking websites is often based on the following USDA approximate cooking times for a whole turkey by weight\footnote{Note that in this paper, ``weights'' are to be considered as masses, so technically, one ``pound'' is assumed to represent one ``pound-mass'' in the avoirdupois system of units, \cite{poundmass}.  Typically, when one refers to an item's weight in pounds in commercial use, what is actually implied is the item's mass in pound-mass units.  For simplicity, throughout this paper, we will use the term ``weight in pounds'' to represent an object's mass in pound-mass units and ``weight in kilograms'' to represent the object's mass in kilograms.  Note that one kilogram = 2.20462 pounds.} provided in Table 1.  Oven temperature is 325 \degree F (163 \degree C) and a turkey is considered safely cooked when it reaches an internal temperature of at least 165 \degree F (73.9 \degree C), \cite{USDA2015}.

\begin{table}[!ht]
    \centering
    \begin{tabular}{|l|l|}
    \hline
        %Unstuffed & ~ \\ \hline
        4 to 8 pounds (breast) & 1 1/2 to 3 1/4 hours \\ \hline
        8 to 12 pounds & 2 3/4 to 3 hours \\ \hline
        12 to 14 pounds & 3 to 3 1/4 hours \\ \hline
        14 to 18 pounds & 3 3/4 to 4 1/4 hours \\ \hline
        18 to 20 pounds & 4 1/4 to 4 1/2 hours \\ \hline
        20 to 24 pounds & 4 1/2 to 5 hours \\ \hline
    \end{tabular}
    \caption{USDA approximate cooking times for unstuffed turkeys, \cite{USDA2015}.}
    \label{table1}
\end{table}

Suppose we wish to find a more accurate means to determine cooking time for a turkey based on weight.  In particular, is there a formula one can use to determine cooking time as a function of weight?  In ``The Physics of Christmas'',\cite{Christmas}, author Roger Highfield addresses this question: 
\begin{displayquote}
``In 1947 H. S. Carslaw and J. C. Jaeger went some way toward solving this seasonal problem in a reference work entitled Conduction of Heat in Solids, \cite{CarslawJaeger}. They came up with equations that govern heat transfer in a uniform sphere. Little did they know that they had provided an excellent starting point for a theory to find the correlation between the radius of a turkey and its cooking time.

To ensure that the mathematics did not get out of hand, they made a few simplifying assumptions. In the Christmas context these can be described as follows. First, the oven must be maintained at a constant temperature throughout. Second, the thermal diffusivity -- a measure of how fast heat passes through the turkey -- is independent of temperature and time.  Finally, and most important, the turkey must be extremely plump, so much so as to be spherical.

The time for heat to diffuse through the tissue so that the center of this spherical turkey reaches a certain temperature is proportional to the square of the radius of the turkey. Then we have to figure out the size of the idealized bird. By assuming that the spherical turkey has the same mass, M, as the real one, we can calculate the radius of the required spherical turkey. To do this we can resort to a well-known formula that says the mass of a sphere is proportional to the cube of its radius.

By combining these two formulas, the radius disappears from the calculation, and we are left with a cooking time, t, proportional to M to the power of two-thirds. This would, of course, hold true only if the turkey was a perfect sphere. [Physicist Peter] Barham points out that a cookbook's simple algorithm of twenty minutes per pound plus twenty more minutes for a small turkey, or fifteen minutes per pound plus fifteen minutes for a large bird, is an equally good approximation for a real turkey.''
\end{displayquote}

The same type of formula has been suggested by Pief Panofsky, Stanford Linear Accelerator Center (SLAC) Director Emeritus, \cite{Panofsky}. As accounted by Pief Panofsky's grandson:
\begin{displayquote}

``[t]here was a point in time when my grandfather %[SLAC Director Emeritus] 
... WKH `Pief' Panofsky was not satisfied with the cooking times for turkeys of `30 minutes per pound','' Panofsky wrote. ``This is of course reasonable, because the time a turkey should be cooked is not a linear equation.''
So Pief Panofsky derived an equation based on the ratio between the surface area and mass of a turkey. Cooking time for a stuffed turkey in a 325 \degree F oven is given by
$$t = \frac{W^{2/3}}{1.5}$$
where t is the cooking time in hours and W is the weight of the stuffed turkey, in pounds. The constant 1.5 was determined empirically.''
\end{displayquote}

Recently, in the paper ``Physics in turkey cooking: Revisit the Panofsky formula'',\cite{PanofskyRevisited},  Yihei ``Jenny'' Jin, Lisa R. Wang, and Jian Jim Wang, showed how Panofsky’s formula can be derived using essentially the same ideas as suggested by Highfeld.  They arrived at the following ``modified Panofsky formula''
\begin{equation}\label{ModifiedPanofskyFormula}
    t = \left[
    \frac{\left(\frac{3}{4 \pi \rho}\right)^{2/3}}{\pi^2 \cdot \alpha}
    \ln\left\{
    \frac{2(T_h-T_0)}{T_h-T_c}
    \right\}
    \right]\cdot W^{2/3}.
\end{equation}
The parameters in their model are \textit{turkey density} $\rho$, \textit{oven temperature} $T_h$, \textit{initial turkey temperature} $T_0$, desired \textit{turkey center temperature} $T_c$, and \textit{turkey thermal diffusivity} $\alpha$, with $$\alpha = \frac{k}{\rho c},$$ where $k$ and $c$ are the \textit{turkey thermal conductivity} and \textit{turkey specific heat}, respectively.
From (\ref{ModifiedPanofskyFormula}), one sees that instead of a ``Panofsky coefficient'' of $1.5$, the ``modified Panofsky coefficient'' would be:

\begin{equation}\label{ModifiedPanofskyCoefficient}
    P = \frac{3600 \cdot \pi^2 \cdot \alpha}{\left(\frac{3}{2.20462 \cdot 4\pi \rho}\right)^{2/3}
    \ln\left\{
    \frac{2(T_h-T_0)}{T_h-T_c}
    \right\}}.
\end{equation}

\noindent Note that for the modified Panofsky formula, (\ref{ModifiedPanofskyFormula}), weight is in kilograms and time is in seconds, while for the modified Panofsky coefficient, (\ref{ModifiedPanofskyCoefficient}),  weight is in pounds and time is in hours.  Unlike Panofsky's model, in which the coefficient's value of $1.5$ depends on a fixed oven temperature and unknown initial turkey temperature, we see that the modified coefficient $P$ is affected by several additional factors, including initial turkey temperature, $T_0$, and desired turkey center temperature, $T_c$. 

%\newpage

\section*{STUDENT ACTIVITIES}

\begin{enumerate}
    \item  From Figure 1, which reproduces Figure 2 in \cite{HeatingCurves},
    one can measure the times to heat turkeys of weights 12 lb (5.4 kg), 15 lb (6.8 kg), 18 lb (8.2 kg), 21 lb (9.5 kg), and 24 lb (10.9 kg), to a safe temperature of 165 \degree F (73.9 \degree C (Celsius)).  This can be done by hand or using tools to more accurately measure data from these curves, such as the free Web Plot Digitizer, \cite{WebPlotDigitizer}.    
    Table 2 provides values for model parameters, found from \cite{ChangCarpenterToledo}, \cite{PanofskyRevisited}, and \cite{HeatingCurves}.   
    % Using these parameter values, what do each of the models (Panofsky, Modified Panofsky, and Barham's cookbook algorithm (Cook's 15 and 20 min)) predict for turkey cooking times for each of these weights?
    % Compare these model predictions to the the actual heating times for each turkey,  
    % %in \cite{HeatingCurves}, 
    % as well as USDA approximate cooking times found in Table 1.
    
%\begin{figure}[H]
\begin{figure}[ht]
\centering
\fbox{\includegraphics[width=0.9\textwidth]{Ref6 Fig2.png}}
\caption{Rate of heating for unstuffed turkeys of different weights [Figure 2 from \cite{HeatingCurves}].}
\label{figure0}
\end{figure}
    
    %\begin{table}[H]
    \begin{table}[!ht]
    \centering
    \begin{tabular}{|l|l|}
    \hline
        %Model Parameters & ~ \\ \hline
        Density $\rho$ (${\rm kg/m^3}$) & 1050 \\ \hline
        Thermal Conductivity $k$  (W/(m K)) & 0.464 \\ \hline
        Specific Heat $c$ (J/(kg K)) & 3530 \\ \hline
        Thermal Diffusivity $\alpha = k/(c \rho)$ (${\rm m^2/s}$) & $1.25185\times 10^{-7}$ \\ \hline
        Oven Temperature $T_h$ (\degree C) & 163 \\ \hline
        Initial Turkey Temperature $T_0$ (\degree C) & 2.8 \\ \hline
        Turkey Center Temperature $T_c$ (\degree C) & 73.9 \\ \hline
    \end{tabular}
    \caption{Model parameters.}
    \label{table2}
\end{table}

Using these parameter values, what do each of the models (Panofsky, Modified Panofsky, and Barham's cookbook algorithm (Cook's 15 and 20 min)) predict for turkey cooking times for each of these weights? Compare these model predictions to the the actual heating times obtained from Figure 1 for each turkey,  
%in \cite{HeatingCurves}, 
as well as USDA approximate cooking times found in Table 1.


% \begin{table}[!ht]
%     \centering
%     \begin{tabular}{|l|l|}
%     \hline
%         Weight & Actual Time to \\
%         W (kg) & 73.9 \degree C (hr) \\
%         \hline
%         5.4 & 3.13 \\ \hline
%         6.8 & 3.52 \\ \hline
%         8.2 & 3.97 \\ \hline
%         9.5 & 4.46 \\ \hline
%         10.9 & 5.03 \\ \hline
%     \end{tabular}
%     \label{table3a}
%     \caption{Actual heating times for turkeys of different weights.}
% \end{table}


\item The equation that governs heat transfer in a sphere centered at the origin, \cite{CarslawJaeger}, \cite{PanofskyRevisited}, \cite{Polking2003}, is the \textit{three-dimensional heat equation}
\begin{equation}\label{3D}
    \Delta u = \frac{1}{\alpha}\frac{\partial u}{\partial t}
\end{equation}
 where $u = u(x,y,z,t)$ is the temperature of the sphere at any point $(x,y,z)$ in the sphere's interior for time $t>0$ and the \textit{Laplacian} operator applied to $u$, $\Delta u$, is given by
 \begin{equation}\label{LaplacianRectangular}
     \Delta u = u_{xx} + u_{yy} + u_{zz}
 \end{equation}
in rectangular coordinates.
If we assume that $u$ is radially symmetric, i.e. $u = u(r,t)$, for 
\begin{equation}\label{requation}
    r = (x^2 + y^2 + z^2)^{1/2},
\end{equation}
then using the chain rule, the Laplacian becomes
\begin{equation}\label{LaplacianRadial}
    \Delta u = u_{rr} +\frac{2}{r} u_r,
\end{equation}
so we can write equation (\ref{3D}) as
\begin{equation}\label{RS}
    u_{rr}(r,t) +\frac{2}{r} u_r(r,t) = \frac{1}{\alpha} u_t(r,t).
\end{equation}

We assume that while the turkey cooks, the surface of the spherical turkey has a constant temperature $T_h$, that of the oven, and at time zero, the turkey has constant initial temperature of $T_0$ at every point in the turkey's interior. We also assume that the temperature at the center of the turkey remains finite during cooking.  Mathematically, these conditions can be described by \begin{equation}\label{BCright}
    u(R,t) = T_h, \ \ \ \ {\rm for} \ \ t>0,
\end{equation}
\begin{equation}\label{BCleft}
    u(0,t) < \infty, \ \ \ \ {\rm for} \ \ t>0,
\end{equation}
and
\begin{equation}\label{IC}
    u(r,0) = T_0, \ \ \ \ {\rm for} \ \ 0<r<R,
\end{equation}
where $R$ is the radius of the sphere.

Show that with the substitution 
\begin{equation} \label{substitution}
    v(r,t) = r u(r,t),
\end{equation}
equation (\ref{RS}) becomes the \textit{one-dimensional heat equation}
\begin{equation}\label{1D}
    v_{rr}(r,t) = \frac{1}{\alpha} v_t(r,t),
\end{equation}
with boundary conditions (\ref{BCright}), (\ref{BCleft}) and initial condition (\ref{IC}) becoming
\begin{equation}\label{BCright_revised}
    v(R,t) = R T_h, \ \ \ \ {\rm for} \ \ t>0,
\end{equation}
with
\begin{equation}\label{BCleft_revised}
   v(0,t) = 0, \ \ \ \ {\rm for} \ \ t>0,
\end{equation}
and
\begin{equation}\label{IC_revised}
  v(r,0) = r T_0, \ \ \ \ {\rm for} \ \ 0<r<R.  
\end{equation}

\item Using separation of variables, one can show that the solution\footnote{If you are familiar with the separation of variables technique for solving partial differential equations, use it to get this solution! For an introduction to or review of this technique, especially in the context of the heat equation, see \cite{PanofskyRevisited}, \cite{GroundWater2020}, \cite{Polking2003}, or \cite{Powers2006}.} to the one-dimensional heat equation \textit{initial value--boundary value problem (IVBVP)} (\ref{1D}) -- (\ref{IC_revised}) is:
\begin{equation}\label{1Dsolution}
    v(r,t) = r T_h + \sum_{n=1}^{\infty} B_n \sin \left(\frac{n \pi}{R}r\right) e^{-\frac{\alpha n^2 \pi^2}{R^2}t},
\end{equation}
where
\begin{equation}\label{Bn's}
    B_n = 
\frac{2}{R}
\int_0^R (r T_0 - r T_h) \sin \left(\frac{n \pi}{R}r\right) \ dr = -\frac{2R(T_h-T_0)(-1)^{n+1}}{n \pi}.
\end{equation}
With the substitution (\ref{substitution}), it follows from (\ref{1Dsolution}) and (\ref{Bn's}) that the solution to our original IVBVP (\ref{RS}) -- (\ref{IC}) is
\begin{equation}\label{solution}
    u(r,t) = T_h - \frac{2R(T_h-T_0)}{\pi r}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin \left(\frac{n \pi}{R}r\right) e^{-\frac{\alpha n^2 \pi^2}{R^2}t}.
\end{equation}
Using the idea suggested by Roger Highfield, that the mass of a sphere is proportional to the cube of its radius, show how the modified Panofsky formula (\ref{ModifiedPanofskyFormula}) follows from (\ref{solution}), with $T_c = u(0,t)$ the temperature at the center of the spherical turkey at time $t>0$.  Hint:  Compare the size of coefficients of each term in the series -- which is the dominant term?

\item From Figure 1,
%Figure 2 in \cite{HeatingCurves}, 
for each given turkey weight, one can collect temperature data over time.  Find a model that can be used to determine temperature of a turkey of a given weight at any time t.  How well does this model fit the given data?

\item Use this new model to predict the time needed to heat turkeys of weights 12 lb (5.4 kg), 15 lb (6.8 kg), 18 lb (8.2 kg), 21 lb (9.5 kg), and 24 lb (10.9 kg), to a safe temperature of 165 \degree F (73.9 \degree C).  Compare to the results for the models investigated in question 1.

\item Bonus Question:  Repeat the last two questions with data collected from heating curves for stuffed turkeys - see Figure 2, which reproduces Figure 3 in \cite{HeatingCurves}.

\begin{figure}[H]
%\begin{figure}[ht]
\centering
\fbox{\includegraphics[width=0.9\textwidth]{Ref6 Fig3.png}}
\caption{Rate of heating of stuffed turkeys of different weights [Figure 3 from \cite{HeatingCurves}].}
\label{figure9}
\end{figure}
    
\end{enumerate}

%\newpage

\begin{thebibliography}{99}

%\bibitem{Asmar2000} Asmar, N.~H. 2000.  \textit{Partial Differential Equations with Fourier Series and Boundary Value Problems}.  Upper Saddle River NJ:  Pearson Prentice Hall.

\bibitem{CarslawJaeger} Carslaw, H.~S. and J.~C. Jaeger. 1959. \textit{Conduction of Heat in Solids, second}, Oxford: Oxford University Press.

\bibitem{ChangCarpenterToledo} Chang, H.~C., J.~A. Carpenter, and R.~T. Toledo. 1998. ``Modeling Heat Transfer During Oven
Roasting of Unstuffed Turkeys,'' \textit{Journal of Food Science}, Volume 63 (2), pp. 257-261.
\url{https://ift.onlinelibrary.wiley.com/doi/epdf/10.1111/j.1365-2621.1998.tb15721.x}
Accessed 7 July 2022.

\bibitem{Christmas} Highfield, Roger.  1998. \textit{The Physics of Christmas}, Boston, MA:  Little, Brown and Company. 

\bibitem{PanofskyRevisited} Jin, Yihei ``Jenny'', Lisa R. Wang, and Jian Jim Wang. 2021. ``Physics in turkey cooking: Revisit the Panofsky formula,''  AIP Advances \textbf{11}, 11531. 
\url{https://aip.scitation.org/doi/pdf/10.1063/5.0067811} Accessed 4 March 2022.

\bibitem{GroundWater2020} Luther, Kenneth, Michael Karls, and Jeremy Christman. 2020. ``9-002-T-GroundWaterFlow,'' \url{https://www.simiode.org/resources/8072}.  Accessed 4 March 2022.

\bibitem{HeatingCurves} Martinez, J.~B., A.~J. Maurer, and L.~C. Arrington. 1984. ``Heating Curves During Roasting of Turkeys'' \textit{Poultry Science}, Volume 63 (2), pp. 260-264. \url{https://www.sciencedirect.com/science/article/pii/S0032579119463414?via\%3Dihub}. Accessed 4 March 2022.

\bibitem{Polking2003} Polking, J.~C. 2003. \textit{An Introduction to Partial
Differential Equations in the
Undergraduate Curriculum Lecture 12:  Heat Transformation in the Ball}. \url{https://www.math.hmc.edu/~ajb/PCMI/lecture12.pdf}.  Accessed 4 March 2022.

\bibitem{Powers2006} Powers, D. 2006. \textit{Boundary Value Problems, Fifth Edition}, San Diego, CA: Elsevier Academic Press.
\url{https://books.google.com/books?id=IFkjYFEGTbEC&lpg=PA135&pg=PA135\#v=onepage&q&f=true}.  Accessed 17 January, 2020.

\bibitem{USDA2015} USDA Food Safety and Implementation Service. 2015. \textit{Let's Talk Turkey—A Consumer Guide to Safely Roasting a Turkey}.
\url{https://www.fsis.usda.gov/food-safety/safe-food-handling-and-preparation/poultry/lets-talk-turkey-roasting}.  Accessed 4 March 2022.

\bibitem{WebPlotDigitizer} Rohatgi, Ankit. 2021. \textit{Web Plot Digitizer}.  Version 4.5. Pacifica, California, USA. \url{https://automeris.io/WebPlotDigitizer}. Accessed 2 March 2022.

\bibitem{poundmass} Weisstein. Eric W. 1996--2007. \textit{Pound-Mass}.  Eric Weisstein's World of Physics. \url{https://scienceworld.wolfram.com/physics/Pound-Mass.html}. Accessed 4 March 2022.

\bibitem{Panofsky} Workman, Shawne.  2008. ``The Panofsky Turkey Constant,'' \textit{SLAC Today}. \url{https://today.slac.stanford.edu/a/2008/11-26.htm}. Accessed 4 March 2022.


\end{thebibliography}

\newpage

\teacher{
\section*{NOTES FOR TEACHER}

\begin{enumerate}
    \item From the heating curves in 
    Figure 1,
    %Figure 2 in \cite{HeatingCurves}, 
    one can measure the actual times to heat turkeys to a temperature of 165 \degree F (73.9 \degree C).  As stated in the Student Version, this can be done by hand or using tools to more accurately measure data from these curves, such as the free Web Plot Digitizer, \cite{WebPlotDigitizer}. 
    %We provide the actual heating times we obtained in the second column of Table 3.
    Table 3 and Figure 3 compare these actual heating times to the estimated heating times for the three models (Panofsky, Modified Panofsky, and Cookbook (Cook's) algorithm estimates for 15 minutes per pound and 20 minutes per pound), respectively.  Since we aren't sure which turkey weights correspond to a ``large'' vs. a ``small'' turkey, we apply the Cookbook algorithm for both 15 minutes per pound and 20 minutes per pound to all given weights.
    
\begin{table}[!ht]
    \centering
    \begin{tabular}{|l|l|l|l|l|l|}
    \hline
    %Weight W (kg) & Time to 73.9 \degree C (hr) & Panofsky (hr) & Modified Panofsky (hr) & Cook's 15 (hr) & Cook's 20 (hr) \\ \hline
    Weight & Time to & Panofsky (hr) & Modified & Cook's & Cook's \\ 
    W (kg) & 73.9 \degree C (hr) & & Panofsky (hr) & 15 (hr) & 20 (hr) \\ \hline
    
        5.4 & 3.13 & 3.47585 & 3.29914 & 3.22624 & 4.30165 \\ \hline
        6.8 & 3.52 & 4.05325 & 3.84719 & 3.99785 & 5.33047 \\ \hline
        8.2 & 3.97 & 4.59206 & 4.3586 & 4.76947 & 6.35929 \\ \hline
        9.5 & 4.46 & 5.0654 & 4.80788 & 5.48597 & 7.31463 \\ \hline
        10.9 & 5.03 & 5.55157 & 5.26933 & 6.25759 & 8.34345 \\ \hline
    \end{tabular}
    \caption{Estimated cooking times in hours for each model.}
    \label{table3}
\end{table}
    
\begin{figure}[ht]
\centering
\fbox{\includegraphics[width=1\textwidth]{Figure3.eps}}
\caption{Comparison of turkey cooking models.}
\label{figure1}
\end{figure}

For each model, the estimated cooking time to reach 165 \degree F is greater than the measured time, with RMSE of 0.53 hr, 0.30 hr, 0.83 hr, and 2.43 hr for Panofsky, Modified Panofsky, Cook's estimate (15 min per lb), and Cook's estimate (20 min per lb), respectively.  Of these estimates, the Modified Panofsky formula appears to be the best predictor, however we are still getting an average overestimate of about eighteen minutes for cooking time. We note that the actual cooking times align with the USDA approximate cooking times found in Table 1, so each model also overestimates the expected cooking times for each weight found in Table 1.  

\item This activity provides a nice application of the chain rule along with a change of coordinates to transform a three-dimensional problem into a one-dimensional problem followed by a substitution to transform the resulting PDE into the familiar one-dimensional heat equation.
We provide some details for both the initial coordinate transformation and substitution.  

Following \cite{Polking2003}, to show that (\ref{LaplacianRadial}) follows from (\ref{LaplacianRectangular}), we can apply the chain rule to $u(r,t)$ using (\ref{requation}).  Differentiating with respect to $x$,
\begin{align}
    u_x(r,t) &= u_r(r,t)r_x \nonumber \\
             &= u_r(r,t)\frac{x}{r} \nonumber
\end{align}
\begin{align}
    u_{xx}(r,t) &= u_{rr}(r,t)r_x^2 + u_r(r,t)r_{xx} \nonumber \\
                %&= u_{rr}(r,t)\frac{x^2}{r^2} + u_r(r,t)\frac{(1)(r)-(x)\left(\frac{x}{r}\right)}{r^2} \nonumber \\
                &= u_{rr}(r,t)\frac{x^2}{r^2} + u_r(r,t)\frac{r^2-x^2}{r^3} \label{uxx}
\end{align}
Similarly, differentiation with respect to $y$ and $z$ yield
\begin{align}
    %u_y(r,t) &= u_r(r,t)\frac{y}{r} \\
    u_{yy}(r,t) &= u_{rr}(r,t)\frac{y^2}{r^2} + u_r(r,t)\frac{r^2-y^2}{r^3} \label{uyy} \\
    %u_z(r,t) &= u_r(r,t)\frac{z}{r} \\
    u_{zz}(r,t) &= u_{rr}(r,t)\frac{z^2}{r^2} + u_r(r,t)\frac{r^2-z^2}{r^3} \label{uzz}
\end{align}
Substituting (\ref{uxx}), (\ref{uyy}), and (\ref{uzz}) into (\ref{LaplacianRectangular}), and using (\ref{requation}), we see that (\ref{LaplacianRadial}) follows.

To show that the substitution (\ref{substitution}) transforms the radially symmetric heat equation (\ref{RS}) into the one-dimensional heat equation (\ref{1D}), differentiate (\ref{substitution}) with respect to $r$ and $t$.
\begin{align}
    \frac{\partial}{\partial{r}}\left[v(r,t)\right] &=     \frac{\partial}{\partial{r}}\left[r u(r,t)\right] \nonumber \\
    &= u(r,t) + r u_r(r,t) \label{vr}
\end{align}
\begin{align}
    \frac{\partial^2}{\partial{r^2}}\left[v(r,t)\right] &=     \frac{\partial}{\partial{r}}\left[u(r,t) + r u_r(r,t) \right] \nonumber \\
    &= 2 u_{r}(r,t) + r u_{rr}(r,t) \label{vrr}
\end{align}
\begin{align}
    \frac{\partial}{\partial{t}}\left[v(r,t)\right] &=     \frac{\partial}{\partial{t}}\left[r u(r,t)\right] \nonumber \\
    &= r u_t(r,t) \label{vt}
\end{align}
Multiplying (\ref{RS}) by $r$, we see that
\begin{equation}\label{RSr}
    ru_{rr}(r,t) + 2 u_r(r,t) = \frac{1}{\alpha}r u_t(r,t).
\end{equation}
% Substituting (\ref{vrr}) and (\ref{vt}) into (\ref{RSr}) and using (\ref{substitution}) with  (\ref{BCright})--(\ref{IC}) yields (\ref{1D}) and (\ref{BCleft_revised})--(\ref{IC_revised}).
Substituting (\ref{vrr}) and (\ref{vt}) into (\ref{RSr})  yields (\ref{1D}).  Finally, using (\ref{substitution}) with  (\ref{BCright})--(\ref{IC}) leads to (\ref{BCleft_revised})--(\ref{IC_revised}).

\item The solution to IVBVP (\ref{1D}) -- (\ref{IC_revised}) via separation of variables along with Fourier series techniques provides an excellent introduction to or review of these ideas.  A detailed solution can be found in \cite{GroundWater2020} as well as many texts that introduce these techniques, 
%including \cite{Asmar2000}, 
such as \cite{Powers2006}, which also includes a nice derivation of the one-dimensional heat equation, (\ref{1D}).  Reference \cite{PanofskyRevisited} provides another derivation of the one-dimensional heat equation and an outline of the separation of variables solution of (\ref{1D}), and \cite{Polking2003} specifically addresses applying separation of variables to the radially symmetric version of the three-dimensional heat equation, (\ref{RS}).  %\textit{If students are not familiar with partial differential equations, the heat equation, or separation of variables, we recommend introducing these concepts before starting this project!}

For $t>0$, we can find the temperature at the center of the turkey from (\ref{solution}) by taking the limit as $r \rightarrow 0$, noting that we can distribute the $\displaystyle \frac{R}{\pi}$ coefficient over the sum to get a familiar quantity,
\begin{align}
    u(0,t) &= \lim_{r\rightarrow 0} \left[T_h - \frac{2R(T_h-T_0)}{\pi r}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin \left(\frac{n \pi}{R}r\right) e^{-\frac{\alpha n^2 \pi^2}{R^2}t}\right] \nonumber \\
    &= T_h - 2(T_h-T_0)\sum_{n=1}^{\infty} (-1)^{n+1} \lim_{r\rightarrow 0}\left[\frac{\sin \left(\frac{n \pi}{R}r\right)}{\frac{n \pi r}{R}}\right] e^{-\frac{\alpha n^2 \pi^2}{R^2}t} \nonumber \\
    &= T_h - 2(T_h-T_0)\sum_{n=1}^{\infty} (-1)^{n+1} e^{-\frac{\alpha n^2 \pi^2}{R^2}t}. \label{*}
\end{align}
Now the first term in the series (\ref{*}) dominates, so for $t$ with $u(0,t)=T_c$,
\begin{equation}
    T_c \approx T_h - 2(T_h-T_0) e^{-\frac{\alpha  \pi^2}{R^2}t} \label{**}.
\end{equation}
Solving (\ref{**}) for $t$, 
\begin{equation*}
    \ln \left\{\frac{T_h - T_c}{2(T_h - T_0)} \right\} \approx -\frac{\alpha\pi^2}{R^2} t,
\end{equation*}
from which it follows that
\begin{equation}\label{***}
    t \approx \frac{R^2}{\alpha \pi^2} \ln \left\{\frac{2(T_h - T_0)}{T_h - T_c}\right\}.
\end{equation}
Now, since the density of a sphere of weight $W$ and volume $V$ is 
$$\rho = \frac{W}{V},$$ 
it follows that the weight of a sphere of radius $R$ and density is related by
\begin{equation}\label{29}
    \frac{W}{\rho} = \frac{4}{3} \pi R^3.
\end{equation}
Solving (\ref{29}) for $R$, we find 
\begin{equation}\label{4*}
    R = \left(\frac{3 W}{4 \pi \rho}\right)^{1/3}.
\end{equation}
From (\ref{***}) and (\ref{4*}), we see that
$$
t \approx \left[\frac{\left(\frac{3}{4 \pi \rho}\right)^{2/3}}{\pi^2 \cdot \alpha}\ln\left\{\frac{2(T_h - T_0)}{T_h - T_c}\right\}\right] \cdot W^{2/3}.
$$
This is the Modified Panofsky formula, (\ref{ModifiedPanofskyFormula})!

\newpage

\item For our model, we look at three cases:  initial model, revised models, and further revised models.
%an initial model using given parameters, revised models in which the parameters are varied, and further revised models in which the parameters are replaced with functions of weight.

\begin{enumerate}
    \item For our initial model we use equation (\ref{solution}) with 16 terms (as including more than 16 terms leads to exponential terms that are too small to represent in Mathematica as a normalized machine number, so precision may be lost) to fit model to heating curve data taken from 
    Figure 1.
    %Figure 2 in \cite{HeatingCurves}.  
    For parameters $T_0$, $T_h$, and $\alpha$, we use the given values from Table 2.  Since the temperatures are measured via a thermocouple in breast meat 4.5 cm from the surface of the turkey, we choose $r = R - 4.5/100$ m.  As Figure 4 shows,  our results aren't too great. Graphically, for each weight, there is a poor match between model (curves) and measured data (dots).
    
%\newpage

%\begin{figure}[H]
\begin{figure}[!htbp]
\centering
\fbox{\includegraphics[width=0.38\textwidth]{Figure4.eps}}
\caption{Initial model with given values chosen for parameters $\alpha$, $T_0$, and $T_h$.}
\label{figure4}
\end{figure}
    
\item For our first model revision, treating
$\alpha$ as an unknown parameter, we notice that we can choose $\alpha$ to get a ``good'' graphical fit, with a different value of $\alpha$ for each weight. Using Mathematica's FindMinimum command, to obtain an $\alpha$ value for each weight to minimize the sum of the square errors (SSE) for each weight yields the results shown in Table 4 and Figure 5(a).  Clearly we get a much better graphical fit, with square error ranging from approximately $2644$ (\degree C)$^2$ for $5.4$ kg to $89$ (\degree C)$^2$ for $10.9$ kg, with an RMSE of about $8.15$ \degree C.
    \begin{table}[!ht]
    \centering
    \begin{tabular}{|l|l|l|}
    \hline
        Weight W (kg) & $\alpha$ (m$^2/$sec) & SSE ((\degree C)$^2$) \\ \hline
        5.4 & $7.31954\times10^{-8}$ & 2644.06 \\ \hline
        6.8 & $7.05338\times10^{-8}$ & 1241.15 \\ \hline
        8.2 & $6.16343\times10^{-8}$ & 974.879 \\ \hline
        9.5 & $5.94693\times10^{-8}$ & 564.305 \\ \hline
        10.9 & $5.31524\times10^{-8}$ & 89.1743 \\ \hline
    \end{tabular}
    \caption{First model revision with $\alpha$ chosen for each weight to minimize  SSE for each weight.}
    \label{table4}
\end{table}

% \newpage

%\item 
%Next, 
For our second model revision,
we treat $T_h$ and $T_0$ as parameters that are unknown, along with parameter $\alpha$ and repeat the last case, choosing a set of parameters $\alpha$, $T_0$, and $T_h$ for each weight that minimize SSE for each weight.  We get even better results, which are shown in Table 5 and Figure 5(b).\footnote{For each turkey weight, we see that parameter $T_0$ is greater than measured $T_0$ (initial turkey temperature) and parameter $T_h$ is less than measured $T_h$ (oven temperature).
Why might this be the case?  One possibility is that the turkey is not a solid sphere, rather has a hollow inside (if unstuffed),
so the hollow interior may be warmer than the actual turkey, which makes the turkey's initial temperature higher than measured $T_0$ for the points closer to the sphere's center.  Since the turkey is much colder than the oven temperature, the temperature at the outer surface interface will be most likely be less than the measured oven temperature $T_h$.}
Notice that SSE is significantly reduced with an RMSE of about $1.41$ \degree C.  The only drawbacks to this model are that the parameters $\alpha$, $T_0$, and $T_h$ depend on the specific weights for which we have measured data and the model can only be used for these specific weights.

\begin{table}[!ht]
    \centering
    \begin{tabular}{|l|l|l|l|l|}
    \hline
        Weight W (kg) & $\alpha$ (m$^2/$sec) & $T_0$ (\degree C) & $T_h$ (\degree C) & SSE ((\degree C)$^2$)  \\ \hline
        5.4 & $1.53163\times10^{-7}$ & 5.2988 & 94.8857 & 13.8725 \\ \hline
        6.8 & $1.22326\times10^{-7}$ & 4.66593 & 104.993 & 54.6895 \\ \hline
        8.2 & $1.00452\times10^{-7}$ & 5.19228 & 110.75 & 48.4685 \\ \hline
        9.5 & $8.213\times10^{-8}$ & 6.79055 & 120.594 & 10.4006 \\ \hline
        10.9 & $5.90282\times10^{-8}$ & 3.74536 & 146.171 & 37.002 \\ \hline
    \end{tabular}
    \caption{Second model revision with $\alpha$, $T_0$, and $T_h$ chosen for each weight to minimize SSE for each weight.}
    \label{table5}
\end{table}

%\vspace*{-.1 true cm}

%\begin{figure}[H]
\begin{figure}[htbp]
  \centering
  \subfloat[First model revision with $\alpha$ chosen for each weight to minimize  SSE for each weight.]{\fbox{\includegraphics[width=0.38\textwidth]{Figure5a.eps}}}
  \hfill
  \subfloat[Second model revision with $\alpha$, $T_0$, and $T_h$ chosen for each weight to minimize SSE for each weight.]{\fbox{\includegraphics[width=0.38\textwidth]{Figure5b.eps}}}
  \caption{Revised models with parameters chosen for each weight to minimize SSE for each weight.}
  \label{figure5}
  \end{figure}
  
\newpage
  
\item Suppose we wish to apply the model to a turkey of \textit{any weight}.  A natural question to ask is ``Could there be a relationship between the parameters $\alpha$, $T_0$, and $T_h$ and the weights of the turkeys?'' Plotting each parameter vs. weight, we see that there is essentially a linear relationship between each parameter value and the weights.
\footnote{The increased outer surface area for larger turkeys may account for the increase in parameter $T_h$ at the outer interface as turkey weight increases.  Also, since turkey density $\rho$ and specific heat $c$ decrease as temperature increases, and turkey thermal conductivity $k$ increases as temperature increases, \cite{ChangCarpenterToledo}, \cite{PanofskyRevisited}, it follows that thermal conductivity $\alpha$ increases as temperature increases.  Since smaller turkeys take less time to heat than larger turkeys, we'd expect that smaller turkeys would have higher average temperatures for longer periods of time than larger turkeys, leading to a decrease in thermal diffusivity $\alpha$ as turkey weight increases. This higher average temperature for smaller turkeys may also affect the initial temperature in the same way, leading to a decrease in parameter $T_0$ for larger turkeys, due to higher cavity temperatures in the smaller turkeys.} 
    
%\begin{figure}[H]
\begin{figure}[ht]
\centering
\fbox{\includegraphics[width=1\textwidth]{Figure6.eps}}
\caption{Parameters $\alpha$, $T_0$, and $T_h$ vs. weight W.}
\label{figure6}
\end{figure}

%\newpage

For our third model revision,
%Fitting 
fitting linear functions of weight $W$ to the choices of $\alpha$, $T_0$, and $T_h$ from the last model (second revision), that minimize SSE for each weight, we get a model that is a function of weight only, that can be applied to \textit{any turkey weight}!  Figure 7 (a) and Table 6 show that we get an excellent graphical fit with the trade-off from our last model of slightly higher SSE and an RMSE of about $2.75$\degree C.
    

    
Finally, 
as a fourth model revision, 
in order to reduce SSE, we construct a model that has six parameters, via new linear functions for $\alpha$, $T_0$, and $T_h$, namely  $\alpha(W) = m_{\alpha} W + b_{\alpha}$, $T_0(W) = m_1 W + b_1$, and $T_h(W) = m_2 W + b_2$.  Choosing values for these six parameters $m_{\alpha}$, $b_{\alpha}$, $m_1$, $b_1$, $m_2$, and $b_2$ that minimize the square errors for all of the weights, we get a model that works for all weights $W$ with essentially the same graphical fit as our third model revision, significantly smaller SSE, and an RMSE of about $1.92$ \degree C.  Table 7 and Figure 7(b) outline these results.

    \begin{table}[!h]
    \centering
    \begin{tabular}{|l|l|l|l|l|}
    \hline
        Weight W (kg) & $\alpha(W)$ (m$^2/$sec) & $T_0(W)$ (\degree C) & $T_h(W)$ (\degree C)& SSE ((\degree C)$^2$)  \\ \hline
        5.4 & $1.49481\times10^{-7}$ & 5.34458 & 91.6808 & 133.756 \\ \hline
        6.8 & $1.26117\times10^{-7}$ & 5.24009 & 103.752 & 65.5593 \\ \hline
        8.2 & $1.02752\times10^{-7}$ & 5.1356 & 115.823 & 220.288 \\ \hline
        9.5 & $8.10568\times10^{-8}$ & 5.03857 & 127.033 & 62.4845 \\ \hline
        10.9 & $5.76924\times10^{-8}$ & 4.93408 & 139.104 & 147.296 \\ \hline
    \end{tabular}
    \caption{Third model revision with linear functions of weight $W$, $\alpha(W)=2.39601\times10^{-7} - 1.66888\times10^{-8} W$, $T_0(W)=5.7476 - 0.0746347 W$, and $T_h(W)=45.1199 + 8.62239 W$, fit to the choices of $\alpha$, $T_0$, and $T_h$ from last model (second revision) that minimize SSE for each weight.}
    \label{table6}
\end{table}
    
    \begin{table}[!ht]
    \centering
    \begin{tabular}{|l|l|l|l|l|}
    \hline
        Weight W (kg) & $\alpha(W)$ (m$^2/$sec) & $T_0(W)$ (\degree C) & $T_h(W)$ (\degree C)& SSE ((\degree C)$^2$) \\ \hline
        5.4 & $1.36091\times10^{-7}$ & 7.23203 & 98.7194 & 38.6273 \\ \hline
        6.8 & $1.1985\times10^{-7}$ & 5.87067 & 105.344 & 58.5903 \\ \hline
        8.2 & $1.03609\times10^{-7}$ & 4.5093 & 111.968 & 79.4516 \\ \hline
        9.5 & $8.85283\times10^{-8}$ & 3.24517 & 118.119 & 30.1645 \\ \hline
        10.9 & $7.22875\times10^{-8}$ & 1.88381 & 124.743 & 100.595 \\ \hline
    \end{tabular}
    \caption{Fourth model revision with revised linear functions of weight $\alpha(W)=1.98734\times10^{-7} - 1.16006\times10^{-8} W$, $T_0(W)=12.483 - 0.972405 W$, and $T_h(W)=73.1688 + 4.7316 W$,
    chosen to minimize SSE for all weights.}
    \label{table7}
\end{table}
\newpage

\begin{figure}[H]
  %\begin{figure}[htbp]
  \centering
  \subfloat[Third model revision with linear functions of weight $W$ for $\alpha$, $T_0$, and $T_h$, fit to the choices of $\alpha$, $T_0$, and $T_h$ for each weight that minimize SSE for each weight.]{\fbox{\includegraphics[width=0.38\textwidth]{Figure7a.eps}}}
  \hfill
  \subfloat[Fourth model revision with linear functions of weight $W$ for $\alpha$, $T_0$, and $T_h$, chosen to minimize SSE for all weights.]{\fbox{\includegraphics[width=0.38\textwidth]{Figure7b.eps}}}
  \caption{Further revised models with linear functions of weight for each parameter.}
  \label{figure7}
  \end{figure}
    
\end{enumerate}

\item For our latest model (fourth revision) we can also estimate the cooking time for each given weight (found by solving numerically for the time $t$ at which the turkey breast (location of the thermocouple) reaches $73.9$ \degree C. These estimated cooking times are listed in Table 8.
\begin{table}[!ht]
    \centering
    \begin{tabular}{|l|l|l|}
    \hline
        Weight W (kg) & Actual Time to 73.9 \degree C (hr) & Latest Model (hr) \\ \hline
        5.4 & 3.13 & 3.24977 \\ \hline
        6.8 & 3.52 & 3.58626 \\ \hline
        8.2 & 3.97 & 3.99686 \\ \hline
        9.5 & 4.46 & 4.50452 \\ \hline
        10.9 & 5.03 & 5.29348 \\ \hline
    \end{tabular}
    \caption{Estimated cooking times in hours for latest model.}
    \label{table8}
\end{table}

Figure 8 provides a graphical comparison between our latest model, actual cooking times, and each of the turkey cooking models.  The figure shows that our latest model is most accurate overall, especially for the middle three weights, with close agreement with the modified Panofsky model for weights closer to 5.4 kg or 10.9 kg.  We find that 
for our latest model, the RMSE is $0.135$ hr, or about $8.10$ min, with an average overestimate of about six minutes.



\begin{figure}[ht]
\centering
%\includegraphics[width=0.75\textwidth]{ModelComparisonAll.eps}
\fbox{\includegraphics[width=1\textwidth]{Figure8.eps}}
\caption{Comparison of turkey cooking models to latest model (dashed curve).}
\label{figure8}
\end{figure}

\item Data for cooking stuffed turkeys can be measured from Figure 2
%Figure 3 in \cite{HeatingCurves}
and compared to the models.  This is an exercise left for the students! 

\end{enumerate}

\noindent \textsc{Final Remark:} \textit{If students are not familiar with partial differential equations, the heat equation, or separation of variables, we recommend introducing these concepts before starting this project!}

%\noindent Note:  Tables were converted from Mathematica or Excel to LaTeX via this free web-based converter: \url{https://tableconvert.com/excel-to-latex}.


%\newpage

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\bibitem{HeatingCurves} Martinez, J.~B., A.~J. Maurer, and L.~C. Arrington. 1984. ``Heating Curves During Roasting of Turkeys'' \textit{Poultry Science}, Volume 63 (2), pp. 260-264. \url{https://www.sciencedirect.com/science/article/pii/S0032579119463414?via\%3Dihub}. Accessed 4 March 2022.

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\bibitem{poundmass} Weisstein. Eric W. 1996--2007. \textit{Pound-Mass}.  Eric Weisstein's World of Physics. \url{https://scienceworld.wolfram.com/physics/Pound-Mass.html}. Accessed 4 March 2022.

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\end{thebibliography}
}





\end{document}
