Survival time of Princess Kaguya in an air-tight bamboo chamber
SSurvival time of Princess Kaguya in an air-tight bamboo chamber
Akio Inoue a , Hiroyuki Shima b, ∗ a Faculty of Environmental and Symbiotic Sciences, Prefectural University of Kumamoto, 3-1-100, Tsukide, Higashi-ku, Kumamoto 862-8502, Japan b Department of Environmental Sciences, University of Yamanashi, 4-4-37, Takeda, Kofu, Yamanashi 400-8510, Japan
Abstract
Princess Kaguya is a heroine of a famous folk tale, as every Japanese knows. She was assumed to be confined in a bamboo cavitywith cylindrical shape, and then fortuitously discovered by an elderly man in the forest. Here, we pose a question as to how long shecould have survived in an enclosed space such as the bamboo chamber, which had no external oxygen supply at all. We demonstratethat the survival time should be determined by three geometric quantities: the inner volume of the bamboo chamber, the volumetricsize of her body, and her body’s total surface area that governs the rate of oxygen consumption in the body. We also emphasize thatthis geometric problem shed light on an interesting scaling relation between biological quantities for living organisms.
1. Introduction
Imagine being trapped in a closed room with no window,chimney, or hole in the wall to allow the flow of air. Would yousurvive for long or remain safe in that situation?The most serious and imminent danger associated with en-trapment in a sealed room is the inevitable development of oxy-gen deficiency disease. Oxygen is one of the key elements re-quired for the sustenance of life, as experimentally confirmedin the 1770’s by Joseph Priestley [1]. The author demonstratedthat mice could survive in closed containers as long as they con-tained plants that emit oxygen through photosynthesis; other-wise, the mice could not survive without the plants. This maybe true for all mammals including humans [2]. By incorpo-rating oxygen into the body, humans create approximately 30molecules of the energy source adenosine triphosphate (ATP)from one molecule of glucose [3, 4]. Therefore, once all theoxygen in a sealed room is consumed by an entrapped individ-ual, the body can no longer produce energy without which itwould be di ffi cult to survive. An interesting mammalian excep-tion is a rat that lives underground called the naked mole rat[5]. In the anoxic state, this rat generates energy for survival bya mechanism that is di ff erent from ordinary oxygen-based en-ergy production, and, consequently, can survive for 18 minutesin an oxygen-deprived state and does not experience consider-able cellular damage . Of course, it would not be feasible to“recreate” all mammals as naked mole rats to avoid the risk ofoxygen deficiency. Therefore, let us consider how long a humanbeing would survive in a sealed room with no external oxygensupply. ∗ To whom correspondence should be addressed:Hiroyuki Shima (Email: [email protected]) It was reported in Ref. [6] that in emergencies, the naked mole rat has beenreported to increase the production of the saccharide fructose in its body, whichsuggests that it uses fructose, instead of glucose, as an energy source for thesurvival of relevant tissues such as the brain and heart. Figure 1: Cartoon of the situation where a girl was discovered by an old man.The folk tale states that the old man found a small girl sitting inside a bamboochamber that shined like gold. He thought the girl must be a god damsel and,so, he then took her home, named her Princess Kaguya, and nurtured her.
In the present article, we proposed a Gedanken experimentthat provides a strategy for developing a viable solution to thequestion and problem posed. Our argument is based on a fic-tional mysterious girl “Princess Kaguya”, who is the heroineof one of the oldest and most famous folk tales in Japan [7].In the tale, an old man working as bamboo cutter discovereda miniature girl inside a glowing bamboo shoot [8]; see Fig. 1.Believing her to be a divine presence, he and his wife decided toraise her as their own child. When the girl came of age [9], she
Preprint submitted to Forma December 19, 2017 a r X i v : . [ phy s i c s . pop - ph ] D ec s conferred with the formal name “Princess Kaguya”, whichincorporates the Japanese term Kaguya derived from the lightand life she radiated. The legendary story of Princess Kaguya,which has been handed down through generations [10] was an-imated in 2013, released worldwide, and was finally nominatedfor a 2015 Academy Award for a feature length animation film[11].However, we were not concerned with the entire legendarystory, but we focused on her entrapment in the sealed space.As mentioned above, Princess Kaguya was assumed to havebeen discovered inside the stalk of a glowing bamboo plant (seeFig. 2), which was completely air-tight. This poses the ques-tion of how long Princess Kaguya had survived in the bamboochamber. We will see below that solving this problem may fa-cilitate our understanding of the interplay between metabolismand respiration [13, 14]. In addition, it will be revealed that thisproblem can be easily extended to situations on di ff erent scales,and is not limited to a “miniature girl situation”; it thus providesan excellent intellectual exercise for considering the allometricrelationship of living matter with respect to oxygen.
2. Assumed condition
Our discussion will be based on the hypotheses enumeratedbelow. • The space where Princess Kaguya was trapped had cylin-drical shape and was a completely sealed room with noinflow and outflow of air. • The height of the Princess is assumed to be 9.10 cm [7]and not to change while she was trapped. • The Princess remained sitting quietly in the space untilher destined discovery by the old bamboo cutter. • Her body produced energy only by breathing the oxygeninside the enclosed space.
3. Way of thinking
Suppose that the inner volume of the bamboo chamber inwhich the Princess was trapped is V inn [L]. The chamber wasfilled with air and initially, the oxygen concentration could beassumed to be 21 %, similar to that of ambient air. For humansin general (including Princess Kaguya), the oxygen concentra-tion of exhaled breath is 16 %. Figure 3 provides a visual il-lustration; the outer large square boundary represents the innerchamber that encloses the flesh air with 21 % in oxygen con-centration at the initial state. The total air inside the boundaryis divided into those confined in many small square domains,each of which represents the amount of air that Princess inhalesand exhales by one breath. For simplicity, every exhaled breath Figure 2: (Left) Photo of a bamboo forest. (Right) Cross section of a bambooculm segment. In the culm, a long cavity inside the cylindrical woody section isdivided into numerous chambers by sti ff diaphragms. Reprinted from Ref. [12]with permission. does not di ff use but remain in the same position. As a re-sult, the oxygen concentration in a small domain (highlightedby gray) is displaced by 16 % following each breath. Everysingle breath, Princess inhales flesh air in di ff erent small do-mains in order to intake a constant amount of oxygen. Even-tually when she breaths half the volume of the total air in thechamber, a net oxygen concentration wil be 18.5 % at the finalstate (see the bottom of Fig.3). This value is nearly the same asthe lower limit under which oxygen deficiency occurs. Takinginto account the volume V Pr that Princess’ body occupies in thechamber, therefore, ( V inn − V Pr ) / Q air be the amount of air that Princess intakes per unittime. From the above argument, the viable time T for Princessreads T = ( V inn − V Pr )2 Q air . (1)In the following discussion, we first evaluate the Princess-relatedparameters: Q air and V Pr in Sections 3.2 and 3.3, and then ad-dress the estimation of the bamboo-related parameter V inn inSection 3.4. If the di ff usion of exhaled breath is taken into account, the net oxygenconcentration in the inner chamber will decrease with time and Princess willbreathe fast by actively moving diaphragm and muscles needed breathing. Thismay cause the increase in the consumed energy per hour and thus result ina decline of Princess’ survival time compared with that obtained under thedi ff usion-free condition. ...
21 %21 % 21 %21 %21 %21 % 21 %21 % 21 %21 % 21 %16 %21 %16 % 21 %21 %21 % 21 % 21 % 21 %
21 % =
21 %21 % 21 %16 %21 %21 % 21 %21 % 21 %21 % 21 %21 %21 %16 % 21 %21 %21 % 21 % 21 % 21 %21 % 21 %16 %21 % 21 % 21 % 21 %21 %16 % 21 %21 % 21 % ==
16 %16 %16 % 16 % 16 %16 %16 % 16 %
16 %21 %
Onebreath OnebreathManybreathsInitial state Final state
Figure 3: Schematic diagram of the time variation in the oxygen concentration inside a bamboo chamber.
Based on the previous determination, one breath by the Princessconstitutes an oxygen intake corresponding to only 5 % of thetotal volume of air she inhaled, which could be expressed asfollows: Q oxy [L / h] = Q air [L / h] × . . (2)Here Q oxy is the amount of oxygen that the Princess intakes perhour, which she uses for energy production. Hereafter, L / h isthe unit of Q oxy and Q air . Human cells use oxygen to create theenergy source ( i.e., ATP) through an oxygen-based chemicalreaction to fuel cell functions. The amount of energy producedby the intake of 1 L oxygen is estimated to be 4.83 kcal / L [15],under the assumption that the respiratory quotient of the hu-man body is 0.82. As a result, when the oxygen uptake rate is Q oxy [L / h], the following applies: E p [kcal / h] = .
83 [kcal / L] × Q oxy [L / h] , (3)where E p is the rate of energy production following the intakeof oxygen at the rate of Q oxy . This is the energy consumed bythe body for the maintenance of physiological life activities.It is interesting to note that the consumption energy by hu-man and many kinds of mammals is proportional to the totalbody surface area [16]. This is primarily because many mam-mals regulate their body temperature by heat exchange through The respiratory quotient (RQ) is a dimensionless number used in calculat-ing the basal metabolic rate estimated from carbon dioxide production. It is de-fined as the volumetric ratio of the carbon dioxide removed from the body to theoxygen consumed by the body. The RQ value indicates the kinds of nutrientsthat are metabolized: values of 0.7, 0.8, and 1.0 indicate that lipids, proteins,and carbohydrates, respectively, are being metabolized. The approximate RQof a mixed diet is 0.82, which implies that the energy produced following theuptake of 1 L oxygen is 4.83 kcal / L as derived from Ref. [15]. the skin surface and the respiratory tract to maintain a constantbody temperature at an extremely narrow range ( e.g. , 36-37 ◦ Cfor humans). Especially at rest, most of this heat exchangeoccurs through the skin surface. Therefore, it would be easyto determine there is a high degree of correlation between thebody surface area and the energy consumption rate of the hu-man body. The value of the proportionality constant is deter-mined by the degree of active motion of the body, as listed inTable 1.For instance, when a person is sitting still the energy con-sumption per hour per surface area is 50 kcal / (m · h). Accord-ingly, the energy E c consumed per hour can be calculated asfollows: E c [kcal / h] =
50 [kcal / (m · h)] × A [m ] , (4)where A [m ] is the total surface area of the body. In the presentstudy, we assumed that the energy was produced only by breath-ing and, consequently, E c = E p .From Eqs. (2)-(4), we obtain Q air [L / h] =
50 [kcal / (m · h)] × A [m ]4 .
83 [kcal / L] × . . (5)Therefore, the remaining task is the determination of the body Table 1: Metabolic rates for selected activities [13].
Activity Metabolic rate [kcal / (m · h)]Sleeping 35Sitting upright 50Standing 60Walking 140Running 6003urface area ( A ) of the Princess, which will be solved in the nextsubsection. Developing methods for estimating the surface area of thehuman body has long been an important subject in the fieldof biology, physiology, medicine, and engineering [17, 18]. Itis currently accepted that the total surface area A [m ] of thehuman body generally obeys a power law with respect to theheight H [m] and weight M [kg] of the body as provided by[19] A = . × M . × H . . (6)Let us use this power law to evaluate A of Princess. First, weremind that the body height of Princess Kaguya was assumedto be H = .
10 cm [7]. Next, we considered the weight of thePrincess, which was finally determined by assuming a similarrelationship between H and M based on di ff erent scales as ex-plained below. In Japan, the average height ( H ) and weight( M ) of 5-years-old girls are H = .
87 cm and M = . M , is M = M × (cid:32) HH (cid:33) = . × (cid:32) .
10 [cm]107 .
87 [cm] (cid:33) = . × − [kg] . (7)In summary, the final calculation for Princess Kaguya is as fol-lows: A = . (cid:16) . × − (cid:17) . × (cid:16) . × − (cid:17) . = . × − [m ] . (8)In addition, the volume of Princess’ body V Pr can be deducedfrom Eq. (7) as V Pr = . ] , (9)where the specific gravity of the Princess’s body was assumedto be equal to that of water: 1 g / cm . If the bamboo chamber in which the Princess was trappedwas assumed to be cylindrical with no tapering, the inner vol-ume of the chamber V inn could be calculated as follows: V inn = π (cid:96) ( d − w ) , (10)where (cid:96) is the internode length, d is the outer diameter of thecross section, and w is the wall thickness of the bamboo culm.Figure 4 shows the schematics of the three geometric param-eters and that of the total height ( h ) of a bamboo culm. Wehypothesize that the Princess was in a chamber with a height of0.6 m from the ground. This would be a reasonable height foran elderly man to cut with an axe or sickle while stooping. Fur-thermore, since the internode length decreases as it approachesthe ground from above [21], those located lower than 0.6 m in ℓ 𝑤𝑤𝑑𝑑ℎ Figure 4: A bamboo culm with the total height of h and the internode located0.6 m above the ground. The parameters (cid:96) , d , w used in the text are also defined. altitude would tend to have had insu ffi cient volume to enclosethe Princess with a height of 9.10 cm.To quantify V inn of the internode chamber depicted in Fig. 4,we had to specify the species of bamboo in which the Princesswas trapped. There were two possible candidates of bamboospecies, which are listed in Table 2 and commonly called “Madake”and “Hachiku” in Japan [22].These two species have existed in Japan for at least 800years since they were domesticated in the late 9th century to theearly 10th century when the tale of Princess Kaguya is thoughtto have been written. Most of the other bamboo species in Japancan be excluded as potential candidates because they were in-troduced in Japan only a few centuries ago [23] or they wouldnot have had a su ffi cient volume to confine the Princess witha 9.10 cm height. Based on this background, we focused onthe two bamboo species that have existed from ancient times(see Table 2), and actually personally collected several samplesfrom the forest (similar to what the bamboo cutter did) to mea-sure the (cid:96) , d , and w values using Eq. (10). The internode wasmeasured at a height approximately 0.6 m from the ground.Using the measurement values listed in Table 2 we used thefollowing equation for the subsequent calculation: V inn = . × [cm ] for Madake , (11)and V inn =
98 [cm ] for Hachiku . (12) Table 2: Measurement data of the bamboo culm sizes.
Species
Phyllostachys Phyllostachysbambusoides nigra
Sieb. et Zucc. var. henonis (Madake) (Hachiku) h [m] 13.9 8.3 (cid:96) [cm] 20 20 d [cm] 6.2 3.8 w [cm] 0.74 0.654 . Results Substituting the results obtained thus far into Eq. (1), we ar-rived at the conclusion that the viable time for Princess Kaguyacould be calculated as follows: T = .
16 [h] = . , (13)and T = .
041 [h] = . . (14)These are the solutions to our questions on the viable time forhuman survival in the bamboo chamber.It should be noted that in both cases above, the viable timewas found to be less than 10 min, which appears to be surpris-ingly short considering that the Princess was allegedly rescuedby the old man who was passed by the bamboo she was trappedin by chance. In fact, the situation allowed of no delay; the lifeof the Princess would have been in danger if the old man hadnot rescued her promptly. This observation would create theimpression that she was an extremely fortunate or blessed girl.Alternatively, it could be proposed that she was an “ancestor”of the naked mole rat and was endowed with extreme toleranceagainst oxygen deprivation; the acceptable notion would ulti-mately be based on the readers’ discretion or decision.
5. Biophysical scaling relation
From the viewpoint of biophysics, the shortness in T givenby Eqs. (13) and (14) is regarded as an important consequenceof an allometric relation deduced from Eq. (1). Note that in theright-hand side of Eq. (1), the numerator ( i.e. , V inn − V Pr ) scalesin cubic manner to the length dimension, and the dominator( i.e., Q air ) scales in square manne,, respectively. Therefore, theviable time T was proportional to the characteristic length ofthe system. Specifically, the smaller the body size and trap-ping space, the shorter the survival time of the live organismtherein, although the ratio of the both sizes remains unchanged( i.e. even when the geometric similarity holds). In this con-text, we considered the viable time problem to be an allometricproblem, which would be instructive in acquiring a better un-derstanding of the scaling relation between biological quanti-ties.The above discussion can be clarified by the informationillustrated in Fig. 5, which presents a simplified situation inwhich a cube with length of side λ is confined in a cubic cavitywith length of side L . Using the inner cube to simulate a livingorganism, we could then obtain the equation V inn = L , V Pr = λ , Q air ∝ λ , (15)which implied that T ∝ L − λ λ . (16)If L /λ is fixed as a constant c , the following equation is ob-tained: T ∝ λ ( c − λ ∝ λ. (17) Figure 5: Cubic-shaped closed space with a side length of L that contains acubic-shaped living organism with λ as the linear dimension. Equation (17) reveals that the viable time is proportional tothe characteristic body length λ when the geometric similarity L /λ ≡ c is satisfied.
6. Summary
In this study, we estimated the time for which Princess Kaguyacould have survived in the completely airtight bamboo cham-ber. In the analysis, we used several basic concepts associatedwith the metabolic rate and geometric relationship of the hu-man body, which are both frequently encountered in the ordi-nary course of biophysics. Based on the assumed sizes of thePrincess’ body and bamboo chamber, we concluded that the du-ration was less than 10 min, which was shorter than what wouldbe intuitively expected. The short duration is attributable to theallometric scaling, which implies that the viable time durationis linearly scaled to the characteristic length scale of the system.As a final remark, we emphasize that the problem addressedis not limited to a fictitious “miniature-girl” situation, but canbe extended easily to di ff erent situation in the body size, theamount of activity, and the number of persons trapped in asealed space. The following questions could be reasonably pro-posed for further investigations, which will provide good exer-cises for the study of scaling relation between biological quan-tities. • Consider a situation where a normal-sized person is trappedin an airtight safe made of steel. Using a self-determinedsize for the safe, estimate the time that this person had tobreathe. • Discuss a solution similar to that described above, in which100 spectators are trapped in an airtight concert hall. • Consider the breathing-time problem for a normal-sizedperson who moves actively in a sealed room using thenumeric data listed in Table 1.
Acknowledgments
The authors acknowledge Prof. Eiichi Yoshimura for help-ful comments on the estimation of basal metabolic rate, and5rof. Motohiro Sato for constant discussions about bamboo sci-ence. This work was supported by JSPS KAKENHI GrantNumbers JP 25390147 and 16K14948.
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