Mathematics and Oenology: Exploring an Unlikely Pairingoenology and in food and wine pairing. We introduce and study some partial differential equations for the correct deﬁnition

Mathematics and Oenology: Exploringan Unlikely Pairing

Lucio Cadeddu, Alessandra Cauli, and Stefano De Marchi

Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Maths and Wine-Related Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Barrel Volume Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2The Mathematics of Wine Aging: Arrhenius and Eyring Equations . . . . . . . . . . . . . . . . . . . 5Optimal Wine Storage Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9The Influence of the Heat Flow in the Temperature Equation . . . . . . . . . . . . . . . . . . . . . . . . 12The Optimal Depth for a Wine Cellar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15The Temperature Equation at the Optimal Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17A Qualitative Study of the Depth of a Wine Cellar Based on the ChosenReference Period and Soil Conditions While the Temperature Is Changing . . . . . . . . . . . . . 18

What’s Food and Wine Pairing? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21The Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Geometrical Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Matching Algorithm (MA) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Implementation Details and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

More Recent Investigations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

L. Cadeddu (�)University of Cagliari, Cagliari, Italye-mail: [email protected]

A. CauliPolitecnico di Torino, Turin, Italye-mail: [email protected]

S. De MarchiUniversity of Padova, Padova, Italye-mail: [email protected]

© Springer Nature Switzerland AG 2019B. Sriraman (ed.), Handbook of the Mathematics of the Arts and Sciences,https://doi.org/10.1007/978-3-319-70658-0_67-2

1

http://crossmark.crossref.org/dialog/?doi=10.1007/978-3-319-70658-0_67-2&domain=pdf

mailto:[email protected]



https://doi.org/10.1007/978-3-319-70658-0_67-2

2 L. Cadeddu et al.

Abstract

The aim of this chapter is to discuss some applications of mathematics: inoenology and in food and wine pairing. We introduce and study some partialdifferential equations for the correct definition of a wine cellar and to thechemical processes involved in wine aging. Secondly, we present a mathematicalmethod and some algorithmic issues for analyzing the process of food and winepairing done by sommeliers.

Keywords

Wine · PDE · History of mathematics · General applied mathematics

Introduction

Mathe matics is hidden everywhere and it surprises us with its several andfascinating applications. It is one of the oldest sciences that have developed inthe course of human history, constantly evolving and resolving practical problemsand deeply influencing our daily lives. In this chapter, an introduction is given toseveral applications of mathematics to solve problems arising when dealing withwine: aging, storing, building of wine cellars, and so on. In particular, applicationsof partial differential equations to the correct definition of a wine cellar and to thechemical processes involved in wine aging are given, as well as some historicaldetails about wine-inspired problems, such as measurement of wine barrel internalvolume. A mathematical method and some algorithmic issues for analyzing thetechnique of food and wine pairing are discussed. The approach is based on thecomparison of the area of two planar polygons after some affine transformations. AMATLAB package, which analyzes the overlap of these planar polygons, has beendeveloped for this purpose and is here presented.

Maths and Wine-Related Problems

Barrel Volume Calculations

In 1613, Johannes Kepler married, at the age of 42, the 24-year-old SusannaReuttinger, after considering 11 different candidates over 2 years (this problemwas later formalized as the “marriage problem” or the “secretary problem”). Thewedding was celebrated in Linz, Austria. To celebrate the event, Kepler bought abarrel of wine but questioned the method the wine merchant used to measure thevolume of the barrel and thus determine the selling price. Upset by the apparentlyincorrect merchant’s method, Kepler decided to study the problem of how todetermine the correct volume of a wine barrel and, moreover, find the optimal

Mathematics and Oenology: Exploring an Unlikely Pairing 3

Fig. 1 Horizontal sheets in afull barrel

proportions that maximize internal volume. He formalized the problem in his “newstereometry (solid geometry) of wine barrels” (Kepler 2018).

The scheme of a wine barrel (see Fig. 2) shows how the wine merchant calculatedthe internal volume and thus the price of the wine. The merchant inserted a stickthrough the tap hole till the opposite edge of the lid of the barrel (at the upper leftin the picture). The length of the stick d determined the volume of the barrel andhence the selling price. This method outraged Kepler, who saw that a narrow, highbarrel might have the same d as a wide one and would indicate the same wine price,though its volume would be ever so much smaller.

To determine the volume of a wine barrel accurately, Kepler thought of the winein a full barrel, or of any solid body, as made up of numerous thin horizontal sheetsarranged in thin layers and treated the volume as the sum of the volumes of theseleaves (see Fig. 1). In the case of a wine barrel, each of these leaves was, at leastapproximately, a cylinder (Klein 2004, p. 209). Clearly, he had just introduced thenotion of integration by cylindrical shells (more or less in the same fashion ofCavalieri’s theorem), which was still to come. Incidentally, Kepler’s calculationsgave – approximately – the same results the wine merchant used to determine thecost of the barrel! At the same time, Kepler found the optimal proportions thatmaximized the internal volume of the barrel.

Nowadays, the problem of optimization of that internal volume can be easilysolved by means of simple differential calculus. Considering a cylinder of height hand diameter 2r, it is possible to determine h as a function of the length d of the stickthe merchant used to determine the volume (see Fig. 2).

Applying Pythagoras’ theorem r can be determined as a function of d and h, more

precisely: r2 = d2

4 − h2

16 . The volume of the cylinder is V = πr2 h. Substitutinginto this formula this expression of r2 in terms of h and d, it is easily seen thatV = π

4 d2h − π16h3, i.e., a cubic polynomial in h which attains its maximum at

h = 2d√3.

4 L. Cadeddu et al.

Fig. 2 Wine merchant’scalculation scheme

Fig. 3 Barrel volume V function graph as a function of h, only h > 0 and V > 0 are admissible

For simplicity’s sake, choosing d = 1 in order to draw the cubic function

V = π16

(4d − d3

)which attains its maximum at d = 2

√3

3 and the resulting

maximum volume of the barrel is 4π√

33 . This volume changes very slowly near

its maximum (see Fig. 3); hence, small changes of h do not generate large changesof V. In other words, Kepler discovered that Austrian barrel shape was very closeto the optimal one, that which maximizes internal volume and small variations of d


were negligible. This couldn’t be completely accidental; perhaps internal volumeshad been empirically calculated to obtain this result. Kepler calculation showedthat the volume of cylinders of equal diagonals d does not vary significantly in aneighborhood of the maximum. Finally, the merchant’s method was fairly accurateand honest. The length of the stick d determines the volume of the barrel in asatisfactory way, provided the proportions satisfy the 3h2 = 4d2 requirement, asfound on Austrian barrels.

Ever since the earliest days of wine, various containers have been used for storageand maturation. Their capacity was not always known and differed from one countryto the next. For this reason, the “stick” method was so popular. Different barrels alsoshared the same name, adding confusion to storage measurements. The medievalbarrel used in Florence (Italy) contained just 45 l, while the fifteenth-century Englishwine barrel measured 143 l. Nowadays the situation is different as there are only afew standard wine barrels worldwide. The French barrel (Bordeaux or Burgundy)volume varies from 225 to 228 l. The modern “hogshead” stores 300 l, while the“puncheon” has a capacity of 500 l. There are also larger barrels that range from 550to 630 l. Generally, the 225-l wine barrel has more or less become the standard barrelsize for wine maturation. This size became widely adopted more than a century ago,because a barrel of this size could be handled by a single person.

The Mathematics of Wine Aging: Arrhenius and Eyring Equations

Wine aging is a result of oxidation and non-oxidative processes. Complex chemicalreactions involving wine’s sugars, acids, and phenolic compounds (e.g., tannins)can modify the aroma, taste, color, and mouthfeel of the wine in such a way thatit might be more pleasing to the taste. Not all wines can age, though. The ratio ofsugars, acids, and phenolics to water is a key determination of how well a wine canage. For example, the less water in the grapes prior to harvest, the more likely theresulting wine will have a good aging potential. Climate, grape variety, vintage, andviticultural practice are always relevant. Contact with oak either during fermentationor after (the so-called barrel aging) will add more phenolic compounds to the wine.

Wine aging, hence, is a process which involves many chemical reactions takingplace over time. Each of these reactions occurs at a certain rate, and each one isaffected by temperature changes in a different way. Any chemical reaction hasa unique “energy factor,” “activation energy,” or natural energy barrier that mustbe overcome for the reaction to occur. For a chemical reaction to proceed at areasonable (or desirable) rate, the temperature of the system should be high enoughin such a way that there exists an appreciable number of molecules with energyequal to or greater than the activation energy. The term activation energy was firstlyintroduced in 1889 by the Swedish Nobel Prize-winning scientist Svante Arrheniusand the empirical equation of reference is the so-called Arrhenius equation (seeLaidler 1987). He followed the work of Dutch chemist J. H. van’t Hoff (1884). Theequation reads as:

6 L. Cadeddu et al.

ln k = ln k0 − Ea

RT

or, equivalently:

k = k0e− Ea

RT .

This equation links the dependence of the rate constant k of a chemical reactionto the absolute temperature T (in kelvin), where k0 is the pre-exponential factor(constant for small temperature variations), Ea is the activation energy of thechemical reaction (also constant for small temperature variations), and R is theuniversal gas constant. This formula can be easily deduced from the following van’tHoff differential equation which states that the rate of variation of K in time isdirectly proportional to the activation energy Ea and inversely proportional to thesquare of the temperature T multiplied by the universal gas constant R:

dK

dT= Ea

RT 2K.

Now, integrating this equation by a simple separation of variables yields:

∫ k

k0

dK

K=

∫ T

T0

Ea

Rτ 2 dτ

and hence

ln k − ln k0 = − Ea

RT

(assuming Ea

RTis zero at T = T0) and, finally, the exponential form1

k = k0e− Ea

RT .

Arrhenius states that for every 10 ◦C the temperature, as a result of a chemicalreaction, increases between 50% and 200%. Similarly to this “empirical” Arrheniusequation, the so-called Eyring equation describes the variance of the rate of achemical reaction with temperature. It is also known as Eyring-Polanyi equationbecause it was developed almost simultaneously in 1935 by the chemist HenryEyring, the physical chemist Meredith Gwynne Evans, and the polymath MichaelPolanyi. This equation provides insight into how a reaction progresses at themolecular level. Its expression results from transition state theory as follows:

k = kkBT

he− �H‡

RT E�S‡R ,

where:


• k is the reaction rate constant.• κ is the transmission coefficient.• kB is the Boltzmann’s constant.• h is the Planck’s constant.• T is the absolute temperature.• �H‡ is the enthalpy of activation.• R is the gas constant.• �S‡ is the entropy of activation.

According to the transition state theory, the reaction does not occur instantly atthe moment of the collision, but molecules, at the beginning at an infinite distance,when they approach, start to interact: the binding lengths change gradually untila particular configuration is reached, the so-called transition state or activatedcomplex. All these transformations correspond to an increasing potential energyand the activated complex corresponds to the activation energy Ea.

The activated complex is in an energy level higher than that of the reactants andthe products and it is a very unstable state. For this reason, the activated complexexists only for a limited time, and it evolves soon toward the formation of productsor it can recede forming again the initial reactants. The activated complex cannot beisolated.

Consider a bimolecular reaction A + B → C and K = [C][A][B] , where K is the

equilibrium constant.In the transition state model, the activated complex AB is formed:

A + B � AB‡ → C

K‡ = [AB]‡

[A] [B].

The rate of a reaction is equal to the number of activated complexes decomposingto form products. Hence, it is the concentration of the high-energy complexmultiplied by the frequency of it surmounting the barrier.

Rate = v[AB‡

](1)

= v [A] [B] K‡ (2)

The rate can be rewritten:\vspace*{-2pt}

Rate = k [A] [B] . (3)

Combining Eqs. (2) and (3) gives:

8 L. Cadeddu et al.

k [A] [B] = v [A] [B] K‡ (4)

k = vK‡ (5)

where

• v is the frequency of vibration.• k is the rate constant.• K‡ is the thermodynamic equilibrium constant.

The frequency of vibration is given by:

v = kBT

h(6)

where

• kB is the Boltzmann’s constant (1.381 × 10 −23 J/K).• T is the absolute temperature in Kelvin (K).• h is Planck’s constant (6.626 × 10 −34 Js).

Substituting Eq. (6) into Eq. (5):

k = kBT

hK‡. (7)

Equation (7) is often tagged with another term (M 1−m) that makes the units equalwith M as the molarity and m as the molecularity of the reaction.

k = kBT

hK‡

(M1−m

). (8)

The following thermodynamic equations further describe the equilibrium con-stant:

�G‡ = −RT ln K‡ (9)

�G‡ = �H ‡ − T �S‡ (10)

where �G‡ is the Gibbs energy of activation, �H‡ is the enthalpy of activation,and �S‡ is the entropy of activation. Combining Eqs. (9) and (10) to solve for lnK‡

gives:


ln K‡ = −�H ‡

RT+ �S‡

R. (11)

The Eyring equation is finally given by substituting Eq. (11) into Eq. (7):

k = kBT

he− �H‡

RT e�S‡R . (12)

In particular, the difference between Arrhenius activation energy Ea and theactivation enthalpy �H‡ is quite small and numerically close to the accuracyattained in most experiments. These two energies are therefore frequently usedinterchangeably to define the activation barrier of a reaction.

As the aging of a wine is the combination of several chemical reactions, theArrhenius and the Eyring equations are needed to study the problem of the averageof the temperature and to understand why wine collectors are so hung up ontemperature.

One of them was Jean-Baptiste Joseph Fourier (1768–1830) who, as any goodFrenchmen, was motivated to carry out his research into the understanding of thepropagation of heat because of his love for wine and food. More precisely, hewas in trouble when trying to keep his bottles of wine at the ideal temperature.During summer, the bottles would get too hot and, hence, change the taste. Duringwinter, the bottles would get way too cold, freeze, and eventually crack. The obvioussolution was to keep the bottles in a cellar with small temperature changes, butunderstanding the perfect depth of a cellar was a trouble. Too deep and it couldbecome too expensive to build and troublesome to manage (e.g., getting wine bottlesin and out); not deep enough, and temperature fluctuations might have been harmfulboth for the aging process and the long-term storage.

For these reasons, in 1804, he started studying the propagation of heat, and in1807 he submitted to the Paris Institute the first paper on the topic, titled “On thePropagation of Heat in Solid Bodies.” The paper caused several objections, raisedby many well-known scholars (viz., Lagrange, Laplace, Boit, Poisson), but he keptworking on it and in 1822 published it as “Théorie analytique de la chaleur” (for arich Fourier’s biography, we refer to O’Connor and Robertson 2003).

Before attacking the problem the Fourier’s way, it is needed to study andunderstand the parameters that affect wine aging and storage.

Optimal Wine Storage Conditions

The physical parameters that play an important role during wine aging processare optimal average temperature, temperature fluctuations, humidity, light, andvibrations.

10 L. Cadeddu et al.

Optimal Average TemperatureAs seen in the previous section, maintaining the right temperature is certainly oneof the most important factors in the preservation of a wine. Studying the averagetemperature to be maintained in a wine cellar and its possible variations in relationto the external environment is of paramount importance. At around 10 ◦C, wine isalready aging very well; increasing the temperature to 20 ◦C makes it age in a fewyears. Increasing the temperature to 30 ◦C, wine ages in just a few months, but it isnot recommended.

Chemical reactions occurring inside a bottle that increases its temperature arequite complicated and proceed at different speeds. In order to stop or control thesechemical reactions, it is recommended to keep the wine cooler within the optimalrange to achieve ideal aging of the bottle during storage. Thus, a wine stored at atemperature of 30 ◦C is of lower quality with respect to a wine stored at 13 ◦C.Indeed, wine experts have found the ideal wine storage temperature to be about13 ◦C in order to get a well-balanced aging. Keeping the wine at lower temperaturessuch as 10 ◦C is not so harmful as long as it does not freeze. However, there arethree drawbacks in keeping the wine at a low temperature until the end of the agingprocess: first of all, it grows (too) old and it is not ideal if you do not drink itright away. Then, it is the best solution if you plan to leave it to future generations,provided the wine can age for a long time. The second drawback is that keeping thewine at a low temperature is expensive. The ground temperature in most of the USAis about 10 ◦C or higher, so maintaining a constantly cooler temperature impliesa continuous artificial cooling of the cellar. The third possible problem that canbe encountered during storage at too low temperatures is the inability to properlyhandle moisture levels. Proper, artificial refrigeration of the cellar tends to dry theair, which is not recommended, as it might affect (see next section C) wine quality.It has been shown that the maximum allowed temperature for storing a wine is about18 ◦C, except for particular delicate wines for which the value should be lower, asit has been already observed. Higher temperatures, above 21 ◦C, might be good toaccelerate the time to get the final product, while a few days at 27 ◦C or a fewhours at 32 ◦C can permanently damage any even rugged wine. Hence, the idealtemperatures for storing wine are 13–16 ◦C for red wines, 10–14 ◦C for white wines,and 10–11 ◦C for rosé wines. A decades-old wine could taste quite young if storedcontinually under these optimal conditions. The basic idea is therefore to keep thetemperature constant at an optimal average value in such a way that the wine agesgradually over time.

Temperature FluctuationTemperature fluctuation is perhaps even more important than the optimal aver-age storage temperature. Using the Arrhenius equation (see section “Maths andWine-Related Problems”), the highest value of the temperature within a cycle speedsup the aging process much more than a low value that instead slows it down. Inaddition, the temperature oscillations move the air inside the bottle and the wineexpands as soon as it is heated. What can happen is that by increasing the pressure,


the bottle breaks. Then, the cork moves slightly outward or a small amount ofcontent pushes the cork. As soon as the bottle cools down, the wine retracts andthe air compresses inside the bottle. Excessive temperature variations bring smallamounts of air to replace the wine. This brings the old bottles to fill up to the brim.Since oxygen, one of the most reactive gases, is the most damaging element forwine, bottles that are subjected to repeated temperature fluctuations tend to losetheir freshness. Minimizing the frequency of temperature oscillations is as importantas minimizing the extent of these oscillations. An acceptable level of temperatureoscillation is about 7 ◦C (average annual oscillation). A change of about 11 ◦C canbecome harmful if it occurs on a daily basis. The idea is that a deep wine cellarrepairs by seasonal variations in temperature and also by the daily changes. Thisleads immediately to the problem of building an appropriate wine cellar.

HumidityThis is the most controversial issue in the whole wine conservation process.The general consensus among experts is that a wine should be kept in a dampenvironment to keep the cork wet and perfectly sealed. Some experts have recentlydealt with this problem by asserting that as long as the bottle is stored with thewine stopper, it remains wet and the humidity seems insignificant for optimalconservation. But other taste experts have indicated that wines kept at an appropriatetemperature but in dry conditions seem to lose their freshness. Moreover, in dryconditions, the top of the cork which is normally not in contact with the wine driesup and therefore becomes smaller in diameter. This slightly loosens the seal andthe air may enter inside the bottle. This, together with the temperature oscillations,favors harmful wine oxidation. So, keeping the wine in too dry conditions is a riskto avoid. On the other hand, controlling humidity inside a cellar is easy and nottoo expensive. As it has been said, humidity is therefore much less critical thantemperature. Storing wine below 50% of relative humidity is harmful, while above80% there are mold risks. The range between 50% and 80% of relative humidityis therefore acceptable. The amount of water that can dissolve in air increases withtemperature. The maximum amount that can be held at a particular temperature is100% relative humidity. It is important to distinguish between humidity and relativehumidity.

LightIt is advisable to keep the wine in dark environments, as light, at least for smallwavelengths, lowers the resistance of the complex of molecules that will createthe main shades of taste during a proper aging process. Glass absorbs most ofthe ultraviolet rays and in particular dark green glass absorbs all the other smallwavelength rays. If you have a good wine in clear bottles, like white wines orChampagne, you should keep them in a dark place. It is also not recommendedto expose the bottles to artificial light.


VibrationsFinally, wine should be stored in a vibration-free environment. It has been carefullyverified that this is not so important for wines that do not emit deposits. For thosethat produce sediment, the most harmful aspect is the agitation of that deposit beforeserving the wine. Hence, if the conservation area is subject to small vibrations, it ispreferable to leave the wine for a couple of weeks in a quiet place before serving it.The same attention should be paid whenever an aged red wine is moved.

The Influence of the Heat Flow in the Temperature Equation

The heat equation is of paramount importance in several scientific fields. Inmathematics, the heat equation is the prototype for any parabolic partial differentialequation. In financial mathematics, it is needed to solve the Black–Scholes partialdifferential equation in option pricing models. In probability theory, this equationis related to the study of Brownian motion (random walks) via the Fokker–Planck equation. A more general version of the heat equation, called the “diffusionequation,” is related to the study of chemical diffusion and other related processes,as those which arise in wine aging. The simplest form of the heat flow density isgiven by Fourier’s Law, which states that the time rate of heat transfer througha material is proportional to the negative gradient in the temperature and to thearea, at right angles to that gradient, through which the heat flows. Along thez axis (see picture below), we have qs(t) = −k ∂T

∂z, where k is the coefficient of

thermal conductivity and it is always positive as is T = T(t, z), the temperaturewhich varies, in time t, along the z axis. All materials obey to this law if, at leastat first approximation, the material is not heated excessively and quickly. The heatflow qs(t) is actually a vector and it is assumed that this is propagated along thez axis only. As seen in the previous paragraph, wine storage in underground cellarsrequires proper temperature control. The idea is that a good condition of the soilrepairs wine from seasonal and daily variations in temperature. Clearly, very deepcellars have this advantage, but their construction is very expensive, and they arenot very practical to use and maintain in good condition. Therefore it is relevantto ask how the emission of heat propagating through the surface of the cellar(represented along the t axis of the figure) affects the temperature function. Consideras a reference a semi-infinite homogeneous semi-space, isolated, that is, withoutheat sources inside. Thus the heat flow can be represented through the surface alongthe axis t. Indicate with z the depth of the cellar. Solving the heat equation meansfinding the precise relationship between temperature T, the depth z of the wine cellar,and the time t (Fig. 4).

The goal is to find the best depth of a wine cellar. Formulating the problem withthe heat diffusion equation in the form:

∂2T

∂z2− 1

k

∂T

∂t= 0, (13)


Fig. 4 Find z0, the optimal depth!

where the temperature is a function of the space and time that is denoted by T(t, z)and k = ρc

kis the diffusion coefficient of the Earth. Assume also that the heat flow

qs(t) is a periodic function, so that the initial condition is expressed by:

−k∂T

∂z= F0e

−iωt , (14)

where k is the coefficient of thermal conductivity.Hence, it is needed to solve the following second-order differential system:

{∂2T∂z2 − 1

k∂T∂t

= 0

− k ∂T∂z

= F0e−iωt

, (15)

whose solution, obtained by straightforward application of Fourier series techniques(for more details refer to Bornemann (2004), Cadeddu and Cauli (2018), Siegfriedand Marcuson (2010), and Taler (2006)), is of the form:

T (z, t) = Af (z)e−iωt . (16)

In fact, substituting the expression of T(z, t) in the heat diffusion equation (13),yields:

Ae−iωt

[(d2f (z)

dz2

)− (−iω)

kf (z)

]= 0


which thus becomes an ordinary equation of the second order:

d2f (z)

dz2+ iω

kf (z) = 0

that has a general solution of the form:

f (z) = ae

√− iω

k + be−

√− iω

k .

Since the exponential decay of the temperature oscillation, when the depth growsis relevant, choose a = 0 and consider only the second term of f (z), which must bereplaced in the expression (16) of the temperature, which provides:

T (z, t) = Af (z)e−iωt = Ae−

(1−i√

2

)√ωkze−iωt . (17)

By separating the oscillatory part:

T (z, t) = Ae−√ω2k

zei(√

ω2k

z−ωt)

. (18)

It remains to determine the constant A to obtain the particular solution of theproblem (15) considered. Derive the quantity (18) with respect to z:

∂T

∂z= −Ae−iωt

(1 − i√

2

) √ω

ke−

(1−i√

2

)√ωkz

and impose the boundary (initial) condition (14):

kA

√ω

k

(1 − i√

2

)e−iωt = F0e

−iωt

and

kA

√ω

ke−iωt−i π

4 = F0e−iωt ,

from which the value of the constant is obtained:

A = F0

k

√k

ωei π

4 .

Thus, the desired solution, which expresses the temperature under the action ofthe heat flow that crosses the surface, is:


T (z, t) = F0

k

√k

ωe−√

ω2k

zei(√

ω2k

z)−ωt+ π

4 (19)

By comparing the equations of the temperature wave (19) and the heat flow(14), it is readily seen that the oscillatory parts differ from π

4 . This means thatthe temperature anomaly due to the heat flow on the surface, for each given depth,returns the oscillation period on the surface of 1/8 of the oscillation period whichappears in the temperature wave equation. Equation (19) completely solves the heatdiffusion problem since it shows the precise relationship between temperature T, thedepth z of the cellar, and the time t. Given T and t, one can find z and vice versa.This will be done in the next paragraph.

The Optimal Depth for a Wine Cellar

The aim is to estimate the depth z of a wine cellar to reduce the temperature anomaly,so that the oscillation is negligible.

Suppose that the temperature variation at the surface is described by the wave lawT(0, t) = T0 + A0 cos ωt, where T0, A0, and ω are positive constants, in particular,ω = 2π

trepresents the rate of change of the temperature corresponding to the initial

known value of the temperature at the surface, T0. For example, taking as referenceperiod of storage the year, ω = 2π rad

365days = 2π rad31536000 s = 1.99 × 10−7rad/s. The

problem is finding the optimal depth z0 of the wine cellar or establishing the winerysize for different values of ω. Then, formulate the problem in the case where thewave describing the temperature at the surface is of the form T(0, t) = T0 + A1cosω1 t + A2cosω2 t, where ω1 is the temperature variation frequency correspondingto the period of 1 year and ω2 is the frequency of the temperature variation withrespect to the 1-day period. Let T = T0 + U. Then the model of the initial problemis formulated by the following second-order system:

⎧⎨

⎩

Ut − kUzz = 0, 0 < z < ∞, −∞ < t < ∞U (0, t) = A0 cos ωt, −∞ < t < ∞| U (z, t) |< C, 0 ≤ z < ∞, −∞ < t < ∞

,

whose solution is of the form:

U (z, t) = ℜA0e−αzei(ωt−αz) = A0e

−αz cos (ωt − αz) ,

where α = (ω2k

) 12 . Notice that the solution U(z, t) must have the average surface

temperature T0 added to it in order to obtain the expression of the temperaturefunction. Observe that κ depends on the nature of the soil: it can vary by a factor of5 or more between a wet and a dry soil. For more precise calculations (see Turcotteand Schubert 2002), assume an average value of κ = 2 × 10−7 m2/s while assumingthat the Earth’s temperature fluctuates around 20 ◦C. Impose the condition αz = π


to obtain the optimal depth of the cellar for a period of a year:

z0 = π

α= π

(2k

ω

) 12 ≈ 4.45 m.

If the amplitude of the annual fluctuation is assumed to be 15 ◦C, things getbetter because a damping factor of 10 is now sufficient. That is, choose z so thate−αz = 0.1 or

z0 ≈ 3.2 m.

However, considering as period of aging 1 day = 86,400 s, instead of 1 year,yields ω = 2π

86,400 = 7.27 × 10−5rad/s and the optimal depth necessary to damp theoscillations of the temperature is equal to:

z0 = π

α= π

(2k

ω

) 12 = π(0.004)

12

(86, 400

2π

) 12 ≈ 23.3 cm.

So, the optimal depth z0 is linked to the conditions of the wine cellar previouslydescribed, and in particular z0 depends primarily on ω, i.e., on the period of agingand on the diffusion coefficient of the Earth κ .

In the case in which the temperature at the surface is of the form T(z, t) = T0 +A1 cos ω1 t + A2 cos ω2 t, putting T = T0 + U + V, the model system of theproblem becomes:

⎧⎨

⎩

Ut − kUzz = 0, Vt − kV zz = 0, 0 < z < ∞, −∞ < t < ∞U (0, t) = A1 cos ω1t, V (0, t) = A2 cos ω2t, −∞ < t < ∞| U (z, t) | < C, | V (z, t) | < C, 0 ≤ z < ∞, −∞ < t < ∞

,

(20)

Like in the previous case, one gets:

U (z, t) = ℜA1e−α1zei(ω1t−α1z) = A1e

−α1z cos (ω1t − α1z) , α1 =(ω1

2k

) 12

V (z, t) = ℜA2e−α2zei(ω2t−α2z) = A2e

−α2z cos (ω2t − α2z) , α2 =(ω2

2k

) 12

from which the temperature at the depth z and time t is:

T (z, t) = T0 + A1e−α1z cos (ω1t − α1z) + A2e

−α2z cos (ω2t − α2z) ,

with


ω1 = 2π

3.15 × 107 = 1.99 × 10−7 rad/s

and

α1 = α2√365

=(ω1

2k

)1/2.

So, choosing the reference period of the year, the optimal depth of:

z0 = π

α1= π

(2k

ω1

)1/2

≈ 4.45 m.

is found.

The Temperature Equation at the Optimal Depth

Wishing to write the equation of the temperature function T(z0, t), at the optimaldepth z0, its oscillations at the variation of this depth can be highlighted andtherefore it can be determined the best one for building the cellar, taking into accountthe so-called skin effect occurring in the first layer of the Earth surface where thetemperature function decreases with depth at a factor of 1

e.

Consider the inside temperature of the cellar T = 1eT0 = 1

eAe−iwt , being

T0 = T(0, t) = Ae−iωt, the temperature on the outer surface. By matching this

value to the temperature function found earlier, T (z, t) = Ae−

(1−i√

2

)√ωkze−iwt , and

resolving with respect to z, it is easily found the value of the optimal depth (alreadycalculated):

z0 = π

√2k

ω

which depends exclusively on ω and κ .Now rewrite the equation of temperature (19) according to the optimal depth z0

found:

T (z0, t) = F0

k

√k

ωe− z

z0 ei(

zz0

−ωt+ π4

)

(21)

The choice of the optimal depth z0 that has been calculated leads to twoadvantages. First, a reduction by a factor of 23 in the amplitude wave of thetemperature function is obtained while the temperature oscillation is reduced to amere 1 ◦C with a correct design of the winery. Secondly, the phase of the wave ofthe temperature function at these depths is exactly the opposite of the phase of the


same at ground level. This is the desirable effect to dampen excessive temperaturefluctuations and limit the mechanisms of heat transfer, such as the opening of thedoor of the cellar, which pushes the temperature of it well beyond the optimalstorage value of 13 ◦C.

A Qualitative Study of the Depth of a Wine Cellar Basedon the Chosen Reference Period and Soil Conditions Whilethe Temperature Is Changing

In order to study how the depth of the cellar needs to be modified according to thechosen time for the aging of the wine and the soil conditions in which the cellar islocated, choose three main reference time periods, and examine the optimal depthsin considering the oscillations which is subjected to the temperature during 1 dayor 1 year or under intense and persistent cold conditions, i.e., for the three differentvalues of ω:

• ω = 7.27 × 10−5 rad/s, for a daily period• ω = 1.99 × 10−7 rad/s, for an annual period• ω = 1.99 × 10−12 rad/s, for a period of hundreds of thousands of years

Also consider three ground conditions, that is, analyze the cases of a clay, sandy,and rocky soil. In this regard, observe the following table:

Considering 1 day as a reference period and examining the oscillations of theoutside temperature on the surface according to time, the trend is the following:

It can be observed, therefore, that during the hottest hours of the day, thetemperature undergoes strong oscillations, reaching at midday a considerablevariation of 12◦.

Regarding the regulation of the depth z0 of the wine cellar, consider 1 dayas a reference period for temperature fluctuations, and taking into account soilcharacteristics, no major changes are necessary since in order to damp dailyfluctuations, it would be necessary, for a rocky terrain, the optimal depth of only20 cm, as shown in the following table:


Surface temperature for daily oscillations

Sur

face

tem

pera

ture

osc

illat

ion

(kel

vin)

Times (hours)0

0

2

4

6

8

10

12

5 10 15 20 25

Choosing a year as a wine aging period, depending on the characteristics of thesoil, the optimal depths vary between 1.4 and 3.8 m, as shown in the followingtable:

In order to understand how important is adjusting temperature oscillation bymeans of the depth of the cellar, consider an extreme situation of an overly longaging period, with regard to the third ω value which has been considered, of about100,000 years. It should be noted that, in this case, a proper adjustment of the depthof the cellar would be impracticable for the same soil characteristics, since it would


be needed the building of a cellar deeper than 1 km, as confirmed by the followingtable:Now, consider the three main types of soil as a reference and choose a wineproduction area in the world, where the average annual temperature is 13 ◦C andthere are substantial variations in temperature during the considered yearly agingperiod. Look at the respective charts for the optimal depths in the three differentchosen soil conditions for an annual value of ω = 1.99 × 10−7 rad/s.

Temperature depending on depth and time: clay soil

Dep

th (

Met

ers)

Temperature variations in time (kelvin)–4

–12

–10

–8

–6

–4

–2

0

–2–3 –1 10 2 3 4

Choosing the optimal wine conservation conditions which have been describedin this section, from the charts provided, notice that for a clay soil, the optimal depthis over 6 m; for a sandy soil, it is almost 8 m; and, finally, for a rocky soil, which issubjected to strong variations in temperature, it should imply a depth of more than20 m.

Finally, in order to keep the temperature of the wine conservation environmentconstantly equal to the optimal temperature of 13 ◦C, the depth of a wine cellar iscrucially important for the aging process as it is able to cope with the oscillationsalso daily of the temperature function. The following pic shows a chart thathighlights the importance of correct sizing of a wine cellar even in persistently coldclimates throughout all the period. The different values in meters can be seen in


Temperature depending on depth and time: sandy soil

Dep

th (

Met

ers)


–12

–10

–8

–6

–4

–2

0

–2–3 –1 10 2 3 4

Temperature depending on depth and time: rocky soil

Dep

th (

Met

ers)


–30

–35

–40

–25

–20

–15

–10

–5

0

–2–3 –1 10 2 3 4

the graph, in order to adjust the depth of the cellar in an attempt to mitigate theoscillations of the temperature function that occur during the annual aging process.For wine cellars located almost on the surface, where the temperature function takesup more than 10◦ below 0, one can observe how, by modifying the depth of the cellarand increasing it, it is easier to attenuate the frequent temperature oscillations.

What’s Food and Wine Pairing?

The discussion starts by explaining what is the process of food and wine pairing.“Food and wine pairing is often cast as a rather mysterious science, but in truth itis actually quite simple, and the experimentation involved is great fun. People often


–30

–35

–40

–502007 2007.5 2008 2008.5 2009 2009.5 2010

0.25 m1 m1.5 m2 m3 m5 m7 m10 m

–45

–25°C

–20

–15

–10

–5

Attenuation of temperature oscillations with depth0

thought that the purpose for a wine is for it to be drunk with food in a situation whereboth complement each other, and it still amazes how often a rather humble wine willsynergize with a food match in a profound way.” These words give a perfect idea ofwhat is aimed by the process of matching food and wine.

Mathematically speaking, one can formalize the process as an algorithm thatgeometrically can be modelled as the comparison of the area of two polygonsobtained by connecting the values indicated in the sommelier graph depicted inFig. 5). This diagram is somehow reminiscent of a dart board highlighting the moreimportant characteristics of the food and those of the wine that we would like tounderstand when tasting a food and smelling a wine. How does one compare thesepolygons? What kind of information does one wish to get from this comparison?What follows a pairing algorithm based on the technique that sommeliers use bytasting the food and the wine is presented and tested on some classical food andwine pairing.

On reading the chapter, one should be convinced of the importance of choosingthe proper wine for every dish: the proper wine exalts the dish and its preparation.

The Graph

Firstly the graph that the Italian Association of Sommeliers (called AIS) uses tofind the wine that better match a given food is presented (see Fig. 5). There are twodifferent sets of descriptors, called characteristics, that are pointed out in the graphwith different characters.


SWEETNESS

SOFTNESS

sweetnesstendency

fattiness

ALCOHOLLEVEL

TANNINCONTENT

sapiditypersistence taste-aroma

ACIDITY

EFFERV.

SAPIDITY

greasiness

iuiciness

AROMA INTENSITY

I.A.P.

spiciness

aromatic

bitterness tendency

acidity tendency

sweetness

Fig. 5 The graph for food and wine pairing used by the AIS association

• Words in normal size correspond to the food characteristics evaluated usingthe natural number from 0 to 10. These characteristics are considered in coun-terclockwise order and are greasiness, juiciness, sweetness tendency, fattiness,sapidity/bitterness tendency/acidity tendency/sweetness, and persistence taste-aroma/spiciness/aromatic.

• Words in capitals correspond to the wine characteristics again using the valuesfrom 0 to 10. As above, these are considered in counterclockwise order,opposite to the circles center, as follows: alcohol level, tannin content, sapid-ity/effervescence, acidity, aroma intensity/IAP (intense aromatic persistence),and sweetness/softness

Notice:When one has a group of characteristics (such as those for the food sapid-

ity/bitterness tendency/acidity tendency/sweetness) and gives to each of them avalue, the one reported in the graph is the highest that corresponds to the mostperceptible characteristic of the group.


To be more precise:

1. The value 0 is used when a characteristic is absent.2. The values in the range 1–3 are used when a characteristic is barely perceptible.3. The values in the range 4–6 are used when a characteristic is better perceptible

than before, but not clearly.4. The values 7 and 8 are considered when a characteristic is clearly perceptible.5. Finally, the values 9 and 10 are given when a characteristic is perfectly

perceptible.

These values are represented in the graph of Fig. 5 by using 11 concentriccircles of radii 0 up to 10, intersected by lines emanating from the inner circle andconnecting the value of the characteristics.

In Fig. 6 is presented a slightly different form of the original graph (asthat in Associazione Italiana Sommeliers 2001) that gives a good idea of whatItalian sommeliers should handle when performing food and wine pairing. Somecharacteristics of the food the wine pairing process are obtained by opposition whileothers by accordance.

It is time to look to the mathematical point of view. Fixing a Cartesian coordinate(x, y) system, the graph has essentially three main symmetries: one bottom-up thatcorresponds to the horizontal line x = 0 describing a matching by accordance andthe other two describing a matching by opposition, corresponding to the bisectionlines y = x and y = − x.

(a) Consider the bottom-up line, parallel to x = 0. In the bottom part, there are thecharacteristics of the food that should agree with the corresponding ones of thewine along the same side (left, sapidity, bitterness tendency, acidity tendency,sweetness; right, persistence taste-aroma, spiciness, aromatic), on the top of thegraph for the wine. There is one exception: the sapidity, bitterness, and aciditytendency of the food have to be considered as opposed to the sweetness andthe softness of the wine. As a simple example, if the food has aromatic 6 andspiciness 5, the wine should have aroma intensity and/or IAP with almost thesame values.

(b) The second symmetry is along the bisection line y = x. Here the matching ismade by opposition. For example, a food which is juicy with value 8 must matcha wine with similar values of alcohol level and/or tannin content.

(c) The third symmetry is along the bisection line y = −x. The matching is hereagain by opposition. For example, a food which is fat should be matched with awine with good percentage of acidity and effervescence or sapidity. The reasonis simple: effervescence and acidity have the effect of polish from fat in themouth.

In Fig. 6 is shown the graph, similar to that can be found in Italian books forsommeliers (cf., Associazione Italiana Sommeliers 2001), where the characteristics


SWEETNESS

SOFTNESS

sweetnesstendency

fattiness

ALCOHOLLEVEL

TANNINCONTENT


ACIDITY

EFFERV.

SAPIDITY

greasiness

iuiciness

AROMA INTENSITY

I.A.P.

spiciness

aromatic

bitterness tendency

acidity tendency

sweetness

Fig. 6 The graph used by sommeliers to match food and wine by accordance or opposition

have been identified by opposition with the colors red and black and the ones byaccordance with the colors green and magenta.

Example 1In Fig. 7 the graph for matching of a slice of S. Daniele ham, Italian prosciuttocrudo (blue polygon), and a red wine from Sicily, DOC Nero d’Avola, year 2002,14% (red polygon), is presented.

One question arises: “given a food, which characteristics should a wine have forthe optimal match (or the best possible)?”

Since our problem consists of the “comparison” of two polygons inscribed in11 circumscribed circles, the comparison is made by analyzing the intersectionsof their areas and how much they overlap. Hence, the best matching problem canbe performed, modulo a roto-translation, by looking at how much the polygonsoverlap! The more they overlap, the better the matching is.


SWEETNESS

SOFTNESS

sweetnesstendency

fattiness

ALCOHOLLEVEL

TANNINCONTENT


ACIDITY

EFFERV.

SAPIDITY

greasiness

iuiciness

AROMA INTENSITY

I.A.P.

7

7

4

5

7

6

77

5

6

6

spiciness

aromatic

bitterness tendency

acidity tendency

sweetness

Fig. 7 This graph corresponds to the match of a slice of ham and a glass of Nero d’Avola (Sicilianwine), aged 2002

Geometrical Issues

The vertices of the polygons are identified as follows:

• Put the common center of the circles at the origin.• Every polygon has vertices at the points

(xs, ys) = (k cos θs, k sin θs) ,

where k ∈{0, . . . , 10} and θ s are the corresponding angles.• On looking at the graph used by Italian sommeliers in Fig. 5, for the wine

characteristics, one can consider the angles θ1 = π /2 − δ/2, θ2 = π /2 + δ/2,θ3 = π + δ, θ4 = π + 2δ, θ5 = −2δ, θ6 = − δ with δ = π /12.


• Similarly for the food. Food characteristics are at the polygon vertices that, bysymmetry with respect to the center of the circles, have angles αs − π + θ s,s = 1, . . . , 6, where the θ s are the angles of the wine’s characteristics.

Let

W = {(xs, ys) , s = 1, . . . , SW }F = {(us, vs) , s = 1, . . . , SF }

be the two polygons. Applying Green’s theorem (cf., Kaplan 1991) to the regionenclosed by the polygons, we can easily compute AW and AF , i.e., the signed areaof the polygons. The vertices must be ordered clockwise or counterclockwise: ifthey are ordered clockwise, the area has negative sign.

The wine polygon has vertices

AW = 1

2

SW∑

i=1

(xiyi+1 − xi+1yi) , (22)

where xSW+1 = x1 and ySW+1 = y1 (cf., Bockman 1989). Equation (22) canbe interpreted as the product (crossed) of the two column array formed by thecoordinates of the vertices. One has also to consider the location of the centroidof each polygon, i.e., the geometrical analogue of the center of mass. The generalformula is well-known. For the sake of completeness, the coordinates of the centroidfor the food polygon are:

υF = 1

6AF

SF∑

i=1

(ui + ui+1) (uivi+1 − ui+1vi) (23)

υF = 1

6AF

SF∑

i=1

(vi + vi+1) (uivi+1 − ui+1vi) (24)

Matching Algorithm (MA)

Let start by constructing the graph of Fig. 5 and the corresponding polygons. In eachpolygon, the two moment lines corresponding to the centroidal principal momentsabout axes parallel to the Cartesian ones are indicated. Then one finds the centroid,as described above.

The matching algorithm (MA) can be summarized as follows:

1. Find the equation of the moment lines2. Apply a roto-translation to the food polygon (with the aim to make the polygons

having the same centroid)


3. Analyze the area of the difference

E (W,F) = (W ∪ F) \ (W ∩ F) , (25)

to check if the wine matches the food.Mathematically speaking, we have to face the question of finding the minimum

of E (W, F ).From linear programming, each (convex) polygon represents a planar region that

can be represented as A x ≤ b. Hence, W := A1x<b1 and F := A2x<b2 will bethe polygons of the wine and the food, respectively.

Moreover, let r(ϕ) = eiϕ the complex exponential function and let F1 :=A2x<b2 be the food polygon obtained either by a reflection or by a rotation ofangle ϕ (obtained by multiplying each polygon vertex by r(ϕ)) followed, possibly,by a translation of a vector t. After these transformations, the union of the polygonscan be rewritten as

W ∪ F1 = (Bx<c

) ∪ m∪j=1

(Cj x<dj

). (26)

where the first term is W ∩ F1 which gives the (percentage of) overlapping and thesecond represents the union of the small polygons not belonging to the intersection,that is, the difference described by the function (25). Thus, the minimum of (25) isobtained when m = 0 (if possible), which is equivalent to say that W∪F1 = W∩F1.This means that both the wine polygon and the transformed food polygon overlapperfectly.

As regards to the optimal rotation, recalling the fact that the angles of thecharacteristic lines of the food are the same as the ones of the wine modulo π ,in order to maximize the overlap, a heuristic strategy consists in performing sixrotations, which correspond to the angles of the six lines of the wine characteristicswith the constant δ halved. The angle which maximizes the overlap can then beconsidered the “optimal” one for the rotation.

Implementation Details and Examples

The implementation was done in MATLAB by means of the function polygeomby H. J. Sommer and the function PolygonClip from the toolbox Polygonclipper by S. Hölz, both downloadable at Matlab Central FileExchange. For details on MATLAB, please refer to the monograph (Highamand Higham 2000).

By the function polygeom, we can compute the areas, the centroids, theperimeters, and the moment lines of the polygons. The user can then decide which


Table 1 Three examples of wine and food matching. The second and third column give the areaof the original polygons. The fourth column gives the area of E (W, F ) after reflection. The sixthcolumn the area E (W, F ) after roto-translation by the optimal angle (in radians) of the fifthcolumn. The last column is the difference between the values in columns 6 and 4

Example Area W Area F Area R Opt. angle AreaR − T

Diff.

1 65.5 34.8 34.8 3.1743 32.5 2.32 57.7 38.6 37.8 π 38.6 1.23 54.1 52.6 43.0 π 44.0 1.0

transformation applies to the food polygon (or reflection plus translation or roto-translation).

The function PolygonClip has been then designed to calculate the overlap oftwo polygons. Then, by the MATLAB function polyarea, one is able to computethe area of their intersections and compare it with those of the original polygons.

In Table 1, the comparison of both strategies in three examples is presented. Theexamples correspond to food and wine pairing examples found at the web page ofthe Italian Association of Sommeliers (see http://www.sommelier.it).

In Fig. 8, the shorthand notation for the characteristics has been adopted: initialsin lower case or in upper case. The values corresponding to the characteristics areeasily deducible by just counting the circle to which the vertex belongs to. When acharacteristic is absent, the corresponding vertex is not drawn. This is obtained withthe software just described applied to Example 1.

The first example corresponds to the match of an oyster with a sparklingwine, similar to French Champagne, called Franciacorta Brut DOCG (Italy). Thecorresponding diagram and the polygons after reflection and roto-translation areshown in Fig. 8. Notice that in this case, the wine polygon has bigger area than thatof the food polygon and some characteristics of the food are not balanced by thoseof the wine. For example, the effervescence of the wine has a value 8 which is twicethe value of the sweet tendency of the oyster. This means that the matching is notharmonic.

In the second example, the matching is between a dish consisting of potato andonion pudding with a Chardonnay Colli Orientali del Friuli DOC vintage 1995 (awine from the region Friuli, in the northeastern part of Italy).

The third example is the matching of a smoked herring fillet with onions and ared wine (from the region Marche, east-center part of Italy) called Lacrima di Morrod’Alba DOC, vintage 1996. In this case, the match is almost perfect. Sommelierssimply say “the matching is harmonic,” even if the polygons are not completelyoverlapped. The reason is that it is really difficult to find the complete overlap. Whenthe overlap is the best possible, as in this case, sommelier concludes the matchingis harmonic. A final comment concerns the so-called optimal rotation angle whenroto-translation has been applied. In all examples, the optimal angle is π or nearlyπ radians. This should not be too surprising, since the polygons are constructed on(characteristics) lines which differ of π radians.

http://www.sommelier.it


–10

–10

–8

–8

–6

–6

–4

–4

–2

–2

0

0

2

2

4

4

6

6 st=4

Wine Character

Food CharacterSO=6 AI=6, IAP=5

J=6

A=6

E=8

pta=2, a=4sa=6

8

8

10

10

o

AL=6

–8

–6

–4

–2

0

2

4

6

8

–10 –8 –6 –4 –2 0 2 4 6 8 10–4

–2

0

2

4

6

8

–6 –4 –2 0 2 4 6 8

–8

–6

–4

–2

0

2

4

6

8

–10 –8 –6 –4 –2 0 2 4 6 8 10–4

–2

0

2

4

6

8

–6 –4 –2 0 2 4 6 8

Fig. 8 Matching of oyster with a glass of Franciacorta Brut DOCG (Italy). First row: the diagram.Second row: the polygons after reflection. Third row: the polygons after “optimal” roto-translation


More Recent Investigations

Fourier’s work has been recently applied to spectroscopic analysis for monitoringthe quality of grape and wine. Fourier-transform infrared (FT-IR) spectroscopyis a nondestructive analytical technique (Boulet et al. 2007; Moreira and Santos2004) that provides structural information on molecular features of a large rangeof compounds. This is a technique which is widely applied in the food industry,though acceptance of its use in the grape and wine industry has been relatively slowand mainly restricted to large wineries. Direct spectroscopic measurement is veryuseful in this context since it allows for quick and simultaneous analysis of severalcompounds which are of paramount importance in grape and wine quality analysis:alcohols, sugars, acidity, glycerol, phenolic compounds, and sulfur dioxide.

For a more recent application of mathematical models to wine tasting andanalysis, we refer to De Marchi (2007) where wine is interpreted as a chaoticdynamical system.

Conclusion

In this chapter, partial differential equations have been applied to modelize (andsolve) problems involved in wine making and storing (chemical reactions duringaging, correct “natural” design of a wine cellar) and a simple application of basiccalculus for the determination of the internal volume of a wine cellar, as intuitivelydeveloped by Kepler without the aid of derivatives, have been showed. Moreover,by the help of a purposely developed MATLAB package which analyzes the overlapof polygons, a mathematical method and some algorithmic issues for studying thetechnique of food and wine pairing have been discussed.

References

Associazione Italiana Sommeliers (2001) Abbinamento cibo-vino Ed. AIS (Associazione ItalianaSommeliers), Milan, Italy

Bockman SF (1989) Generalizing the formula for areas of polygons to moments. Am Math Mon96(2):131–132

Bornemann F (2004) In the moment of heat. In: The SIAM 100-digit challenge: a study in high-accuracy numerical computing. SIAM, Philadelphia

Boulet JC, Williams P, Doco T (2007) A Fourier transform infrared spectroscopy study of winepolysaccharides. Carbohydr Polym 69:89–97

Cadeddu L, Cauli A (2018) Wine and maths: mathematical solutions to wine–inspired problems.Int J Math Educ Sci Technol 49:459–469

De Marchi S (2007) Mathematics and wine. Appl Math Comput 192(1):180–190Higham DJ, Higham NJ (2000) Matlab guide. SIAM, PhiladelphiaHoff JH (1884) Etude de dynamique chimique, Oxford University, F. Muller & CompanyKaplan W (1991) Green’s theorem. In: Advanced calculus, 4th edn, Section 5.5. Addison-Wesley,

Reading, pp 286–291Kepler J (2018) Eberhard Knobloch Nova Stereometria Dolorium Vinariorum/New Solid Geom-

etry of Wine Barrels: Suivi De Accessit Stereometriae Archimedeae Svpplementvm/a Supple-ment to the Archimedean Solid Geometry Has Been Added. Les Belles Lettres


Klein F (2004) Elementary mathematics from an advanced standpoint. Arithmetic, algebra,analysis. Dover, Mineola

Laidler KJ (1987) Chemical kinetics. Harper and Row, New YorkMoreira JL, Santos L (2004) Spectroscopic interferences in Fourier transform infrared wine

analysis. Anal Chim Acta 513(1):263–268O’Connor J, Robertson EF (2003) Joseph Fourier. MacTutor history of mathematics archive,

University of St AndrewsSiegfried M, Marcuson R (2010) The wine cellar problem. Periodic heating of the surface of the

Earth. Geodynamics SIO 234Taler J (2006) Solving direct and inverse heat conduction problems. Springer, Berlin/New YorkTurcotte DL, Schubert G (2002) Geodynamics. Cambridge University Press, Cambridge

Documents

Mathematics and Oenology: Exploring an Unlikely Pairingoenology and in food and wine pairing. We introduce and study some partial differential equations for the correct deﬁnition