imensions 10% interest rate Construction drawing indicating lbs/sq inch Maxwell photo

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1 D 4 imensions m model grandfather clock? (ideally being filmed) Weather Map showing measurement in millibars model battle ships English Channel Model l t g model car model in wind tunnel Maxwell photo 10% interest rate Elephant burning model Lucas models Construction drawing indicating lbs/sq inch surface tension bubble small illustration of largest elephant photopotatoes priced per kilo

2 The units of all the variables in the physical world can be created out of only three primitive dimensions: mass M, length L, and time T. 45 Dimensionality only characterizes variables. Units are needed to quantify them. Maxwell and link to info about him In science and engineering, physical variables have two properties: a quality or dimensionality, and a quantity expressed in term of units. Dimensionality is characteristic of the object, characteristic, or event. Velocity, for example, has dimensionality distance divided by time. Distance and time are the primitive dimensions making up the dimensionality of velocity. To express a specific value of velocity, we could say 100km/hr or 60mi/hr or 88ft/sec all express the same quantity of velocity, but in different units. Units are simply measuring scales of a quality or dimension. km/hr, mi/hr, ft/sec are different scales for measuring distance and time. Note that each of these sets of units has the same dimensionality. To express the value of a physical variable one cannot just say that the velocity is what? 45 is a meaningless quantity unless its units or measuring scale are identified. What is interesting is that the units of all variables in the physical world can be created out of only three primitive dimensions: mass M, length L, and time T. This observation was first made by Maxwell in (Usually, variables like temperature and electric charge are called primitive, but they also can be expressed in terms of M, L, and T.) This feature of the natural world helps us predict functional relationships between variables without having to know the physics behind their relationship. Here are a few examples of familiar variables with their characteristic dimensionalities: length velocity Familiar Variables Dimensionality only characterizes variables. Units are needed to quantify them. From a dimensional point of view, velocity could be expressed in any number of ways. No one flinches when we say km/hr, mi/hr, ft/sec. But suppose we would have stated a velocity in terms of yards/day. These would be arbitrary, but dimensionally correct, units. If arbitrary units were used for everything, working with expressions containing several variables would be a nightmare. The solution is to define a set of standard units that are always to be used. This constitutes a system of units. It would be nice if there were only one such system, but, alas, there are several. L L/T force ML/T 6/1/0

3 All terms in a physically consistent equation must reduce to identical primitive dimensions. You re probably aware of the more important systems cgs, MKS, and SI. In the first two systems, the letters stand for the scales of the primitive dimensions: centimeters, grams, seconds and Meters, Kilograms, Seconds. In the last, SI is the abbreviation for the International system of units, a system that is being adopted by the world science community as a single standard. We will primarily use SI units in this course. However, engineering does not always use SI units. Many of its unit conventions come from historical use. One of the advantages of using a system of units is that more complicated variables can be expressed as derived units, i.e., units that are in fact a collection of primitive units. In principle, we would have to say that the force on a object is 50 kilogram-meters per second squared. But since force is such an important variable, we refer to it in terms of the derived unit N, the Newton. And one Newton equals one kilogram-meter/second squared. Here s a partial list of other derived units in the SI system: 3 air conditioner Here are the seven primitive quantities of the SI system: Primitive Quantities of the SI System Primitive Quantities of the SI System a) b) length mass meter kilogram m kg a) force newton N mkg/s b) energy joule J m kg/s c) pressure pascal Pa kg/(ms ) d) power watt W m kg/s 3 c) time second s d) elec. current ampere A e) f) luminous intensity amt of substance candela mole cd mol There are many others, but these few will give you an idea of what s going on. Other systems have their own derived units. g) temperature degree K 6/1/0

4 If an equation is not dimensionally consistent, it s wrong, before the physics even gets involved. With a single system of units, the calculations associated with science and engineering problems are least complex. But, when mixed systems of units are used, then the calculations require units conversions and become much more tedious and more susceptible to calculation errors. Consequently there is a compelling reason to describe everything in a single system. We ll get to this topic in a bit, but first let s discuss three fundamental rules of dimensions: 1) The dimension of a mathematical term or expression is the product of the dimensions of the constituent variables. For example, the expression ½ at, where a is accelera- L tion and t is time, has units T = L. That is, the resulting expression has dimension length or T distance. If someone writes the equation distance s = v 0 + ½ at, you don t even need to know the physics behind the equation. It s simply wrong with or without the physics. Why? Because the first and last terms have dimension L and the middle term has dimension L/T. Not possible. This principle affords a quick way to keep you from looking dumb. When you deduce or derive an equation, check the dimensionality of the terms. If they re not the same, you ve done something wrong. Dimensionallyconsistent equations may not be always right, but then it becomes a question of the physics. If an equation is not dimensionally consistent, it s wrong, before the physics even gets involved. Always check the consistency of your terms. 3) Terms may be added together only if the units of each term are the same. 4 ) All terms in an equation must reduce to identical primitive dimensions. You cannot add 3 m to 5 in. and get 8 min. In more colloquial terms, you can t add apples and oranges. What this means is that all terms in a physically consistent equation must reduce to identical primitive dimensions. This is where units conversion comes in. 6/1/0

5 One standard is not necessarily convenient for all disciplines. Units conversion is the process of expressing one measuring scale in terms of another. In this case it concerns the relationship in = 1 m. Each side of this equation has two elements: a numerical quantity and a unit. Note that the dimensionality of each side is the same L. The numbers and the units can be algebraically manipulated separately. So, by dividing both sides of this equation by various factors, one can obtain the following equations from in = 1 m: = 1 m/in; 1 = in/m; 1 in = m. The first two of these equations are dimensionless, which means the terms in these equations are expressed in ratios of units whose dimensionality cancel. Dimensionless variables and equations play an extremely important role in engineering and science, but we re getting ahead of ourselves... So, getting back to our problem of adding 3m to 5 in., how do we do it? The answer is: multiply one of the terms by the quantity one which, of course, does not change the value of the term. So, the distance 3m can be also expressed as 3 m * 1 = 3 m * in/m = m in/m = in. (The m s cancel). Now we have an expression for 3m in terms of the same units as the other term. And, we can add them in + 5 in = in. That s a lot of work just to change from one set of units to another. Suppose I tell you that the floor of a building will currently support a loading of 100 pounds per square foot and I want to reinforce it so that it will support a loading of Pascals. The problem is conceptually trivial, but the details of obtaining an answer is tedious because the units are presented in two different ways. But there are reasons why a single set of units are not used by every one. Let s take pressure. There are probably a dozen standard ways to express pressure p.s.i., mm Hg, Torr, Pascals, bars, etc. Why don t we simply agree to use, say, SI units and eliminate the tedium of all this conversion business? 5 We choose the explicit form for one as the middle equation above. Well, although SI units do provide a way to describe pressure, one standard is not necessarily convenient for all disciplines. 6/1/0

6 For example, a meteorologist is interested in pressure in relation to standard atmospheric pressure. So he uses the bar as a scale of pressure. 1 bar = 1000 millibars = 1 standard atmosphere. With such a scale, he can describe a low pressure system as having a pressure of 984 millibars, and immediately the number is meaningful in terms of standard atmospheric pressure. On the other hand, a construction engineer is interested in how much loading a floor design can support. So he thinks in pounds (force) per square foot. He thinks pounds because that s the everyday way to imagine weight, and square foot because that s how areas of large structures are quantified. The choice of units is also partially dictated by the magnitudes they produce. But using a molecule as a unit of measure is extremely cumbersome, because any practical quantity of substance would contain billions upon billions of molecules. Hence, a more convenient quantity the mole has been invented to express chemical quantity. A mole is that quantity of substance whose weight in grams or pounds or other convenient measure equals the substance s molecular weight. So, to effect the reaction mentioned above, one would take kg-moles of sulfur dioxide (18 kg) plus 1 kg-mole of oxygen (64 kg) to yield kg-moles of sulfur trioxide (19 kg). In engineering, the nuisance of mixed units will remain. So, an engineer must become adept with the algebra of units conversion. And his most-used reference will probably become a table of conversion factors. 6 Construction drawing A materials scientist will often refer to tensile strength of a material in giga-pascals (10 9 Pascals) because the resulting numbers are more convenient. It s a lot easier to communicate the tensile strength of iron as 0.4 giga-pascals rather than 8,600,000 pounds per square foot. photo: engineer looking at a calculator In chemistry, one is concerned with reactions between atoms and molecules. For example, a chemist might write the reaction SO + O =SO 3, which means that two molecules of sulfur dioxide and one molecule of oxygen can be combined to yield two molecules of sulfur trioxide. 6/1/0

7 We should mention one peculiar tradition we have about a very common unit: the kilogram. You can go to a store and ask for a kilogram of potatoes, and you ll be given approximately. pounds of potatoes as weighed on a scale. Are there other ways that we can take advantage of the concept of dimensionality? We ve already seen that we can use it to check for the consistency of equations. The answer is yes. 7 photo- potatoes priced per kilo But a kilogram isn t a unit of weight (force); it s a unit of mass. The Newton is the unit of force. Force and mass don t even have the same units. So, how can we use kilograms in two different ways? It s called dimensional analysis. And dimensional analysis can give insight into the physics that s involved in a problem. What s remarkable is that it can suggest the physics without our having to know the physics. m l t g The answer: we don t. Unfortunately we use the same word to mean two different things: kilogram mass and kilogram force. When we say kilogram we are assuming the listener can differentiate our intent between kilogram mass and kilogram force. In the grocery store, the merchant only thinks weight, so that s how he interprets it. But in science and engineering, sometimes one has to be very explicit in order not to be confusing. A kilogram of weight(force) is a kilogram of mass times the acceleration of gravity 9.8 m/s. So, 1 kg of weight in fact weighs 9.8 N. But we carelessly refer to it as 1 kg. (The same confusion arises with pound, because there is pound mass and pound force.) That s enough about units. Let s get back to dimensionality and the idea that physical variables have intrinsic dimensions associated with them. It s easiest to explain with an example. Suppose you want to determine the period of oscillation of a simple pendulum a mass suspended from a string. You hypothesize that a number of factors may be involved: length of the string, the mass of the weight, and, probably, gravity. This is as much about the physics as you know. A more mathematical way of putting it is that you are expecting that the period t is functionally related to the length l, the mass m, and gravity g; or t= f(l,m,g). What you want to know, of course, is exactly what that relationship is. First, let s write down the dimensionality of each of these variables: period t [T] mass m [M] length l [L] gravity g L T 6/1/0

8 Since t = f(l,m,g), we know that whatever combination of l,m,g we end up with will have to have the same dimensionality as t. Since t has dimension [T], we need to arrange l,m,g in some fashion to produce a final dimension of [T]. The first thing to notice is that mass m is the only variable in the bunch that contains the dimension M. Since M is not contained in the dimensionality of t, and,since there are no other variables which might be combined with mass in such a way that the M s would cancel out, we must conclude that mass cannot possibly be part of the functional relationship. So, we ve just deduced a bit of the physics of the problem: mass doesn t affect the period of the pendulum. The only variables we have left are gravity g and length l. So, we need to form an expression whose dimensionality is [T]. There s only one way: l has dimensionality [ ][ ] T g Therefore, period τ ~ L = T. L l g. That s the relationship we re looking for. It doesn t give us an exact equation, just the relationship between the variables. The exact relationship is 1 l τ =. g So, we ve learned something about the physics of a pendulum through dimensional analysis. Suppose you had a jet of water impinging on a wall. Forgetting about gravitational effects, could you deduce an expression that relates pressure on the wall to variables associated with the water jet maybe, water density r [M/L 3 ] and water velocity v [L/T]? Try it, and see what you come up with. (This is an easy problem.) Let s try a somewhat more difficult problem. Suppose we want to analyze the lift force f ML T of an airfoil, i.e., an airplane wing. We presume it to depend on area A[ L ], velocity v L T, density of air ρ M L 3, and viscosity of air µ M LT. Again, by inspection we can find a combination of variables whose dimensionality is the same as that of force. An answer is f ~ ρ v A. But that s not the only answer. Another one is 1 f ~ µ va. 8 6/1/0

9 What do we mean by dimensionless? It s a variable or expression whose constituent dimensions cancel. 10% interest rate Which one is right? Both of them! In this case, we don t get a neat, unique solution like the way we did for the pendulum. Instead, we get groupings of variables that are proportional to force. How do we interpret this result? In these two examples, we carried out analyses essentially by inspection. But, inferring physics from almost no knowledge using dimensional analysis gets difficult very quickly when the number of possible parameters increases. Fortunately, there s a methodical way for carrying out the analysis. And this analysis relies on dimensionless variables and dimensionless expressions. But before we can talk more about dimensional analysis, we better discuss the idea of a dimensionless variable. What is meant by dimensionless? It s a variable or expression whose constituent dimensions cancel. We use dimensionless variables all the time, without really thinking about it. There are many other examples: radian angle, index of refraction, coefficient of friction. Anything that is termed coefficient of... is most likely to be a dimensionless parameter. Dimensionless variables are useful because they give us information about a problem that is independent of scale or units. Lets get back to the interest rate example. Suppose a friend of yours comes up to you and says : Wow, I made $00 in interest in my savings account last year. Is that good or bad? What you really need to know to share in his excitement is how much money he had in his account. If he had $0,000, you politely change the subject to the weather. If he had $000, you ask him for the name of his bank. The point is that in many cases, we are interested in knowing a quantity relative to another quantity. And this information is dimensionless. 9 One example is interest rate. Interest rate is a percentage. It s the number of dollars earned divided by the number of dollars invested. This ratio has dimension $/$ = nothing. So, a percentage is a dimensionless variable. 6/1/0

10 There are other advantages in working with dimensionless quantities. First, quantitative results are universal, i.e., they do not depend on units. 10 Here the variables refer to several characteristics. And, recall the pendulum problem we just did. We could have expressed the period of a pendulum in a non-dimensional way as: m l t g Second, if equations are expressed in dimensionless form, it is easy to see which are the important terms. Third, scaling is automatically built-in. For example, if you specify that a battleship has a length to beam ratio of 4, it applies to model battleships and full-size battleships and everything in between. Dimensionless quantities can have many forms. For example, an expression for dimensionless gravity g could be: g = g/g e, where g is gravity on some other planet and g e is gravity on earth. g, then, would be gravity on another planet relative to that of earth. Here, all the variables refer to a single characteristic. But, we might have a dimensionless measure of static pressure in a moving fluid as: p = (p-p 0 ) /(rv ), where p is pressure, p 0 is a reference pressure, r is fluid density, and v is fluid velocity. τ g / l = const. (Incidentally, you re not expected to understand the physics of the material here. That will come in a few years. The purpose of these examples is to illustrate the use of dimensionless variables.) Suppose we have a problem whose outcome z 1 depends on a number of other variables z, z 3, etc. We can describe this relationship mathematically as: z 1 = f(z, z 3,...,z n ). So there s a total of n variables that are interrelated. We can also express this relationship as: h(z 1, z, z 3,..., z n ) = 0. But these z i s can t appear in just any arbitrary way, because, in this last expression, the functional combination of all the z i s must have a dimensionality of zero. This is where dimensional analysis begins to take form. 6/1/0

11 l t The basis of dimensional analysis comes from what s called the Buckingham Pi Theorem. In 1914, Buckingham (an American physicist) proved that there are limited ways that you can group variables together to form non-dimensional mathematical expressions. Specifically, if you have a problem in which there are n variables, and, contained in those n variables are N primitive dimensions; then, there are n-n possible dimensionless groupings of variables. A problem consisting of n variables z i in a relationship: h(z 1, z, z 3,..., z n ) = 0, 11 First, the groupings of the z i s into the dimensionless variables p i tells us how the original variables must appear in an expression that describes the problem. Second, by expressing the problem in dimensionless variables (the p i s), we have reduced the number of variables in the problem from n to N-n. (Incidentally, by convention, the notation for the dimensionless variables is the letter p.) The relationship between the p i s can also be expressed as p 1 = f (p,...p n-n ). Let s see how this all works. We ll first return to the pendulum problem. m g can be expressed as a relationship: j(p 1, p,...p n-n ) = 0, First, we have n = 4 original variables: period t, length l, gravity g, mass m. wind over airfoil where the p i s are dimensionless groupings of the z i s. The theorem basically says that a problem of n variables z i can be reduced to a problem of n-n variables p i, where the p i s are dimensionless and are constructed from the original z i s. Within these 4 variables are N = 3 primitive dimensions: [M], [L], [T]. This means we can form n N = 1 dimensionless group. So we begin with a problem There are two benefits in this. h(t, l, g,m)=0; and we know we ll end up with an expression j(p 1 ) = 0. 6/1/0

12 l t m g wind over airfoil The only question that remains is how to form the dimensionless p 1 out of the original variables t, l, g, m. All expressions must have dimensionality M a L b T g, i.e., the primitive units to some power. The powers can be positive, negative, zero, or fractional. What we re looking for is some combination of t, l, g, m, such that that combination has dimensionality M 0 L 0 T 0 ; that is, no dimension. The most general expression involving the original variables is t a l b g c m d. And this expression will have dimensionality [T] a [L] b [LT - ] c [M] d. But we want this dimensionality to equal [M] 0 [L] 0 [T] 0. Therefore, reorganizing the expression for the original variables, we have [M] d [L] b+c [T] a-c = [M] 0 [L] 0 [T] 0. Since this equation can be true only if the powers of M, L, T on the left-hand side each equals zero, we get two equations and three unknowns. But that s OK. Since we re interested in an expression that involves t, we want to set a = 1. With that value for a, we conclude that c = ½, and b = -½. 1 Consequently, the general expression t a l b g c m d can now g be explicitly specified as τ. This is our one dimensionless variable p for this l problem. So, a statement of the character of the period of a pendulum is ϕ τ g = 0. l This says that behavior of a pendulum is determined by a single dimensionless variable g τ, made up of the initial l variables t, g, l. This is equivalent to saying that from the power or [M]: d = 0 from the power of [L]: b + c = 0 from the power of [T]: a c = 0. τ ~ l g. We have deduced that d = 0, i.e., there is no role for mass m in the period of a pendulum. That leaves us with Now, let s do the airfoil problem we glossed over before. 6/1/0

13 13 wind over airfoil The variables are: a b c d e F A v ρ µ a c d e T 3 ML T force or lift F ML, wing area A [L ], T velocity v L, air density ρ M, T 3 L and air viscosity µ M. LT We have n = 5 original variables and N = 3 primitive variables, which means that we should expect an expression with two dimensionless groups: ϕ (π 1, π ) = 0, or π 1 = φ (π ). As before, a general expression containing the original variables is To find the powers a,b,c,d,e which will make the expression dimensionless we write b L M M ( a+ d + e) ( a+ b+ c 3d e) ( a c e) ( L ) = M L T = M L T L LT Solving separately for the powers of M, L, T, we obtain a + d + e = 0 a + b + c -3d - e = 0 -a - c - e = 0 Three equations with five unknowns. Since we want to find an expression for lift F, we can set a = 1. To eliminate µ, we ll also choose e = 0.. This gives one of the dimensionless groups as We know that there s another dimension group to be obtained. This time, we ll eliminate the lift by setting a = 0; and we ll make sure to include the viscosity by setting e = 1. Now choose a = 0 to eliminate F, and c = 1 to include v. This gives Finally, π = FA v ρ µ = 1 π = φ π = ρ v A µ = ( π ) or ( ρ v A) Note that our expression for π is the reciprocal of the one we deduced before. Dimensional analysis doesn t care. If an expression is dimensionless, so is its reciprocal. What s important is that we now have an expression for lift as F = ρv F µ 1 ρva F µ = φ ρv A ρva µ Aφ ρva It depends explicitly on the parameters ρ, v, and A, and on some function of the dimensionless group π. Actually, the groupings π 1 and π occur frequently enough in aeronautics that they ve been given names: π 1 is called the lift coefficient ; and the reciprocal of π is called the Reynolds number /1/0

14 14 surface tension bubble Now, you can try one. Find an expression for the period of oscillation of a small drop of liquid under the influence of surface tension. Assume no influence of gravity and a restoring force dependent only on surface tension. (It s a phenomenon you may have seen on the news. Sometimes astronauts are interviewed in space and there are a few drops of liquid that are loose inside the cabin. These drops often exhibit a strange undulation as they drift around. That s what this problem is about.) Your variables are period of oscillation t [T], surface tension s [MT - ], density r [ML -3 ], and diameter [L]. We ve given a few examples related to mechanics and aeronautics, but the technique is useful in almost all branches of engineering and science. For example, it has been used in a very clever way to show how physical factors present limitations on life itself. Read the excellent essay Exposing Life s Limits with Dimensionless Numbers by Steven Vogel in Physics Today, Nov 1998, pp -7. This brings us to another topic of dimensional reasoning: scaling, modeling, and similarity. The question is: if we learn something about the behavior of one physical system, can we apply our knowledge of that system to a similar one, say, twice as large? Answer: τ ~ 3 ρd σ Can we, for example, measure the drag on a model car and infer what the drag will be on a full-size car? The dimensional analysis we ve presented here only scratches the surface. There are many books on the subject, and there are many strategies for approaching problems. As you can see, dimensional analysis is a powerful tool to give insight into the physics of problems. Although we ve presented dimensional analysis as a fairly straightforward procedure, there are definitely subtleties that can make this tool an art. Can we put a model bridge in a wind tunnel and predict what oscillations might occur on a real bridge in a strong cross wind? Fortunately, the answer is yes. We can take results from one system and apply them to a similar system. But the definition of similar is tricky. 6/1/0

15 model grandfather clock? (ideally being filmed) There are at least three different kinds of similarity between two objects or processes: 1) Geometric similarity ) Kinematic similarity 3) Dynamic similarity includes force scale similarity, i.e., equality of Reynolds number (inertial/viscous), Froude number (inertial/buoyancy), Rossby number (inertial/coriolis), Euler number (inertial/ surface tension) Geometric similarity is simple: Do they look the same? Are the linear dimensions proportional between one system and the next? Are the angles the same? Kinematics is involved with motion of a system the rate at which an object moves from one point to another. Kinematic similarity requires that velocities are proportional from one system to another, so that time scales are preserved. Suppose you make a ¼ size model of a grandfather clock (with a pendulum) that you want to use in a film production. Will it appear to perform as a real one? No. From what we now know about pendulums, the shorter pendulum will oscillate more rapidly, thus not exhibiting kinematic similarity. 15 For kinematic similarity, the period of the pendulum of the model would have to be the same as that of the full size clock, i.e., the velocities would have to be proportionately smaller. (If we really wanted to use this model clock in your film, what would we have to do to provide kinematic similarity? Think about the filming location...) Dynamics involves forces. For two systems to be dynamically similar, the equations that determine the behavior of the system must be identical. If these equations are made dimensionless, the forces acting on the system can be made to appear as ratios of forces, e.g., inertial forces vs. viscous forces, or inertial forces vs. surface tension. So dynamics similarity requires that these ratios of forces are the same. The difficulty is that often there are many different force ratios governing the dynamics of a system, and all of them simultaneously cannot be matched in a model. But, just as often, one group of forces are much more important than others, and models can provide very good approximate information about their full sized counterparts. 6/1/0

16 Examples in dynamic similarity are most frequently found in the engineering field of fluid dynamics. So a model ship on a model ocean will look unrealistic because the waves that are created are smooth. 16 Fluid dynamics is the study of the behavior of liquids and gases, under the influence of forces. It plays a fundamental role in many areas of engineering, such as chemical engineering, hydraulics, environmental engineering, heat transfer, biomedical engineering, aeronautics, to name a few. Dynamic similarity is often difficult to obtain in smaller models. In the motion picture industry, models are often used instead of the real thing especially when it comes to blowing things up. Geometric and kinematic similarity are almost always matched: the scenes are in believable proportion, and things move in scaled proportion, e.g., a model car takes as long to pass a model house as a real car takes to pass a real house. But scenes that draw attention to dynamic similarity are another story. The burning of a small scale house does not look realistic, nor does the wave pattern of a model ship moving through a model ocean. Why? Similarly, the flames from a small model house will have much less complexity than those from a burning real house. One of the decisions that must be made by a special effects producer in today s movie industry is how big a model is necessary to produce the visual dynamics of the real things. Bigger models cost more. But, it won t be too long before digital special effects replace models...at least in the movies. One would think that geometric similarity might always be desirable in a physical model. But that is not the case. Sometimes the physical system simply does not model well in correct proportions. An example is a physical model of the English Channel near Grenoble, France. The real Channel is approximately 40km wide, 100 km long and 0.1 km deep. The model is approximately 10m wide, 5 m long and 0.1 m deep. The dynamics of the model does not match the real situation. Large scale flows with high velocities produce many scales of motion. The motion is usually turbulent or chaotic. Small scale flows with lower velocities are smoother or laminar. 6/1/0

17 17 English Channel model A proportional depth would have been 0.05m deep too shallow to permit the multilayer flow patterns in the English Channel to appear. What makes this model especially interesting is that the whole model rotates as a phonograph turntable. A significant force affecting the flow in the English Channel comes from the Coriolis force an apparent force due to the earth s rotation. In order for the model to take that force into account, the model needs to be rotated at a scaled rate. (In contrast, a 6000 m model of the Chesapeake Bay was created. The Bay also is strongly affected by Coriolis force, but the model was stationary and did not take it into account. As a result, the model failed to generate realistic flow patterns and was abandoned.) However, in doing so, we will have increased its weight by a factor of (L) 3 /L 3 = 8, because weight is proportional to the cube of any linear dimension. But the ability of an elephant to support himself depends on the strength of his leg bones. And, here, strength increases by the cross sectional area, say R. Therefore, if his height doubles, his bones would need a cross sectional area of 8R, i.e., his bones would need a radius of R, not R the geometric increase. Using this reasoning, one can also calculate the size of the largest possible elephant at some point the legs would extend beyond the body. Scaling does not always relate to the visual aspects of models. It can also be used to determine structure when other criteria are involved. And now, maybe you can imagine why small animals like insects can get by with very slender legs. Let s take the example of the elephant. Suppose you are a bio-engineer and you have discovered how to create animals of any size. What should an elephant twice as tall as a conventional elephant look like? insect with skinny legs! (daddy long legs ^^) The idea is this: if we simply increase all the geometric dimensions by, we will, of course, have a geometrically similar elephant of height, say L. 6/1/0

18 18 Dimensional reasoning is an extremely important tool in engineering. It provides a way for apparently disparate data, variables, and ideas to fit together into a single theoretical framework. It can keep us from looking silly when writing equations; it can help us to produce physics; it can help us reduce the complexity of systems by showing what s important and what s not; and it can help us to understand the world using models. Become comfortable with the ideas of dimensional reasoning. This topic will give you more insight into science and engineering per study hour, than any other subject. 6/1/0