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Nathan Moore 2023-01-24 19:38:12 -06:00
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@ -83,8 +83,8 @@ Useful information: the student has a mass of $80kg$ and is made mostly of water
Here's a possible answer:
equate food energy with calorimetric heating and assume human bodies have the same heat capacity as water, about $1\frac{kcal}{kg\cdot\degC}$. This allows us to calculate the body's temperature increase.
\bea
3000kcals &=& 80kg\cdot1 \frac{kcal}{kg\cdot \degC}\cdot\Delta T \nonumber \\
\Delta T &\approx& +37.5\degC \nonumber
3000kcals &=& 80kg\cdot1 \frac{kcal}{kg\cdot \degC}\cdot\Delta T , \\
\Delta T &\approx& +37.5\degC .
\eea
Students are normally quite surprised at this number. Although wildly unrealistic, $\Delta T \approx +6\degC$ is typically fatal, there is a related phenomena of diet-induced thermogenesis,\cite{meat_sweats} known informally as ``the meat sweats''. Some students connect this calculation to feeling quite hungry after a cold swim in the pool (a similar effect). On a larger scale, discussing what's wrong with this estimate is useful. The main storage mechanism for storing food energy is fat tissue, which the calculation completely ignores. Infants are generally born with little fat, and an infant sleeping through the night often coincides with the baby developing enough fat tissue to store sufficient kcals to make it though a night without waking up ravenously hungry. A related follow-up is that if a person is stranded in the wilderness, they should immediately start walking downstream (ie, towards civilization) as they likely won't be able to harvest an amount of kcals equivalent to what they already have stored on their hips and abdomen.\cite{trout} The contrast of bear hibernation\cite{fat_bear} and songbirds constantly eating through the winter are related connections to investigate.
@ -92,7 +92,7 @@ Students are normally quite surprised at this number. Although wildly unrealist
A more realistic question to follow up with relates to the average \textit{power} given off by a person over a day.
Again, assuming $3000kcal$ is burned over $24 hours$, with useful information: $1 kcal \approx 4200J$ and $1 J/s=1W$.
\be
\frac{3000kcal}{24hours}\cdot\frac{4200J}{1kcal}\cdot\frac{1hour}{3600sec}\approx145W
\frac{3000kcal}{24hours}\cdot\frac{4200J}{1kcal}\cdot\frac{1hour}{3600sec}\approx145W .
\ee
Most students still remember $75Watt$ lightbulbs, but given the spread of LED lighting, ``A person's body heat is two $75W$ light bulbs'' will probably only make sense for a few more years. Desert or cold-weather camping, alone versus with friends, and survival swimming are also examples for students to make sense of this answer. If you can take advantage of other people's waste body heat, you'll sleep more pleasantly and survive longer in cold water.
@ -122,8 +122,8 @@ The other $2$ blocks of energy are transformed into heat and leave the hiker's b
\bea
\frac{1}{3}\cdot600kcal\cdot\frac{4200J}{1kcal}
&=& 80kg\cdot10\frac{Joules}{kg\cdot m}\cdot height \label{eq:bar_chart}\\
height &\approx& 1000 m
&=& 80kg\cdot10\frac{Joules}{kg\cdot m}\cdot height , \label{eq:bar_chart}\\
height &\approx& 1000 m .
\eea
This estimate is again surprising to students. Five trips up the bluff to burn off $\$2$ of saturated fat, sugar, and flour! A nice followup calculation is to imagine a car that can burn a $100kcal$ piece of toast in the engine: from rest, what speed will the toast propel it to? If (again) the engine converts $1/3$ of the energy into motion (kinetic energy), a $1300kg$ Honda Civic will reach a speed of about $15\frac{m}{s}\approx33mph$!
@ -176,14 +176,14 @@ A table from a USDA booklet giving 1917 yields for various farm products.
So, another question using this data. If you want to feed your family of four people potatoes, how much land will you need to cultivate?
Here's an estimate: a family of 4 requires $3000kcal/person$ each day.\cite{calorie_age} If we over-estimate and produce food for the entire year, the family will need about $4.4$ million kcals.
\be
4~people\cdot\frac{3000kcal}{person\cdot day}\cdot\frac{365~days}{year} \approx 4.4 M kcal
4~people\cdot\frac{3000kcal}{person\cdot day}\cdot\frac{365~days}{year} \approx 4.4 M kcal .
\ee
A brief aside for those bored by the simplistic unit conversion: when I ask students to solve problems like these, one undercurrent of conversation is ``Should I divide by 365 or multiply?'' Particularly with online homework systems, checking your answer for reasonability isn't typically graded. Asking the students to reason proportionally with units is a skill that can give meaning to numbers.
From figure \ref{1917_yields} we can estimate $1.9~million~kcals$ per acre of potato production. Again the students might ask, should I multiple $4.4$ and $1.9$ or should I divide them? It can be useful in a class discussion to have the students discuss and vote which of the following two forms will give the meaningful answer.
\bea
\frac{4.4 M kcal}{family}\cdot\frac{1 acre}{1.9M kcal} & \textrm{~~or~~}&
\frac{4.4 M kcal}{family}\cdot\frac{1.9M kcal}{1 acre}
\frac{4.4 M kcal}{family}\cdot\frac{1.9M kcal}{1 acre} .
\eea
The choice of operation is difficult to make without seeing the units present, which is again a learning opportunity for the students.
@ -213,9 +213,9 @@ Looking at the map with ImageJ,\cite{imageJ} it seems like the recorded area dev
$16,000~acres$ -- details are given in \ref{appx_imageJ}.
With these assumptions, we could equate the corn energy production from chinampas with the population's yearly food need. Note, in this version of the story, the corn productivity, $P\frac{bu}{acre}$ is treated as an unknown variable.
\bea
Food~production &=& 16,000acres\cdot \frac{2~corn~crops}{year}\cdot P \frac{bu~of~corn}{acre} \nonumber \\
Population~requires &=& 100,000~people\cdot \frac{3000kcal}{person\cdot day}\cdot\frac{365days}{year}\cdot\frac{1lbs~corn}{1594kcal}\cdot\frac{1bu}{56lbs} \nonumber \\
P \approx 38\frac{bu}{acre} &&
Food~production &=& 16,000acres\cdot \frac{2~corn~crops}{year}\cdot P \frac{bu~of~corn}{acre} . \\
Population~requires &=& 100,000~people\cdot \frac{3000kcal}{person\cdot day}\cdot\frac{365days}{year}\cdot\frac{1lbs~corn}{1594kcal}\cdot\frac{1bu}{56lbs} . \\
P \approx 38\frac{bu}{acre} . &&
\eea
This crop productivity is in remarkable agreement with the 1917 USDA yields, $35bu/acre$, which seems to validate the assumed $100,000$ person population of Tenochtitlan. Some references\cite{Chinampas_1964} describe an extensive tribute system that Aztec government required of its subjects, which certainly would have been necessary to support populations on the upper end of historical estimates.\cite{400k}
@ -241,20 +241,20 @@ How much food does the island need?
\bea
food~needed~per~year &=& 8.5\times10^6~people
\cdot \frac{3000}{person\cdot day }
\cdot \frac{365days}{year} \nonumber \\
&\approx& 9.3\times 10^{12} kcals \nonumber
\cdot \frac{365days}{year} , \nonumber \\
&\approx& 9.3\times 10^{12} kcals .
\eea
How much land area, sown in potatoes, would produce this food?
\bea
9.3\times10^{12}kcals /\left(1.908\times 10^6\frac{kcal}{acre}\right) &=& 4.87\times10^6 acres \nonumber\\
&\approx& 19,700 km^2 \nonumber
9.3\times10^{12}kcals /\left(1.908\times 10^6\frac{kcal}{acre}\right) &=& 4.87\times10^6 acres , \nonumber\\
&\approx& 19,700 km^2 .
\eea
How much land area, sown in oats, would produce this food?
\bea
9.3\times10^{12}kcals /\left(1.254\times10^6\frac{kcal}{acre}\right) &=& 7.41 \times10^6 acres \nonumber \\
&\approx& 30,000 km^2 \nonumber
9.3\times10^{12}kcals /\left(1.254\times10^6\frac{kcal}{acre}\right) &=& 7.41 \times10^6 acres , \nonumber \\
&\approx& 30,000 km^2 .
\eea
Summed, $49,700km^2$, these two areas devoted to oats and potatoes are roughly equivalent to the amount of arable land estimated above for Ireland, $54,000km^2$.\cite{arable_percentage}
@ -306,7 +306,7 @@ Within this database, foods are identified by an FDC ID.
An example calculation (implemented in the Jupyter notebook) follows for Corn.
In 2022 the USDA reported an average production of $172.3$ bushels of corn per acre of farmland.
\be
172.3\frac{bu}{acre}\cdot\frac{56lbs~corn}{bu}\cdot\frac{453.6~grams}{lbs}\cdot\frac{365~kcal}{100~grams} = 15,974,657 \frac{kcal}{acre}
172.3\frac{bu}{acre}\cdot\frac{56lbs~corn}{bu}\cdot\frac{453.6~grams}{lbs}\cdot\frac{365~kcal}{100~grams} = 15,974,657 \frac{kcal}{acre} .
\label{example_calculation}
\ee
Obviously the result is only reasonable to two significant figures!
@ -363,7 +363,7 @@ Specifically, to find the area of the two large chinampas areas near Tenochtitla
\centering
\includegraphics[width=\columnwidth]{imageJ_analysis.jpg}
\caption{
Three screen captures showing chinampa areas and the calibration stick used to convert pixel-squared area into $miles^2$. The image being analyzed is available in \cite{Chinampas_1964}.
Three screen captures showing chinampa areas and the calibration stick used to convert pixel-squared area into $miles^2$. The image being analyzed is available online. \cite{Chinampas_1964}
}
\label{imageJ}
\end{figure}