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Contents

Section 1.1 - Introduction

Section 1.2 - Scaling of area and volume

Section 1.3 - Scaling applied to biology (optional)

Section 1.4 - Order-of-magnitude estimates

Section 1.5 - Summary

Section 1.1 - Introduction

Section 1.2 - Scaling of area and volume

Section 1.3 - Scaling applied to biology (optional)

Section 1.4 - Order-of-magnitude estimates

Section 1.5 - Summary

Why can't an insect be the size of a dog? Some skinny stretched-out cells in your spinal cord are a meter tall --- why does nature display no single cells that are not just a meter tall, but a meter wide, and a meter thick as well? Believe it or not, these are questions that can be answered fairly easily without knowing much more about physics than you already do. The only mathematical technique you really need is the humble conversion, applied to area and volume.

Area can be defined by saying that we can copy the shape of interest onto graph paper with 1 cm \(\times\) 1 cm squares and count the number of squares inside. Fractions of squares can be estimated by eye. We then say the area equals the number of squares, in units of square cm. Although this might seem less “pure” than computing areas using formulae like \(A=\pi r^2\) for a circle or \(A=wh/2\) for a triangle, those formulae are not useful as definitions of area because they cannot be applied to irregularly shaped areas.

Units of square cm are more commonly written as \(\text{cm}^2\) in science. Of course, the unit of measurement symbolized by “cm” is not an algebra symbol standing for a number that can be literally multiplied by itself. But it is advantageous to write the units of area that way and treat the units as if they were algebra symbols. For instance, if you have a rectangle with an area of \(6 \text{m}^2\) and a width of 2 m, then calculating its length as \((6\ \text{m}^2)/(2\ \text{m})=3\ \text{m}\) gives a result that makes sense both numerically and in terms of units. This algebra-style treatment of the units also ensures that our methods of converting units work out correctly. For instance, if we accept the fraction

\[\begin{equation*}
\frac{100\ \text{cm}}{1\ \text{m}}
\end{equation*}\]

as a valid way of writing the number one, then one times one equals one, so we should also say that one can be represented by

\[\begin{equation*}
\frac{100\ \text{cm}}{1\ \text{m}} \times \frac{100\ \text{cm}}{1\ \text{m}} ,
\end{equation*}\]

which is the same as

\[\begin{equation*}
\frac{10000\ \text{cm}^2}{1\ \text{m}^2} .
\end{equation*}\]

That means the conversion factor from square meters to square centimeters is a factor of \(10^4\), i.e., a square meter has \(10^4\) square centimeters in it.

All of the above can be easily applied to volume as well, using one-cubic-centimeter blocks instead of squares on graph paper.

To many people, it seems hard to believe that a square meter equals 10000 square centimeters, or that a cubic meter equals a million cubic centimeters --- they think it would make more sense if there were \(100\ \text{cm}^2\) in \(1\ \text{m}^2\), and 100 \(\text{cm}^3\) in \(1\ \text{m}^3\), but that would be incorrect. The examples shown in figure b aim to make the correct answer more believable, using the traditional U.S. units of feet and yards. (One foot is 12 inches, and one yard is three feet.)

Based on figure b, convince yourself that there are 9 \(\text{ft}^2\) in a square yard, and 27 \(\text{ft}^3\) in a cubic yard, then demonstrate the same thing symbolically (i.e., with the method using fractions that equal one).

(answer in the back of the PDF version of the book)◊ Solved problem: converting \(\text{mm}^2\) to \(\text{cm}^2\) — problem 10

◊ Solved problem: scaling a liter — problem 19

◊

How many square centimeters are there in a square inch? (1 inch = 2.54 cm) First find an approximate answer by making a drawing, then derive the conversion factor more accurately using the symbolic method.

Great fleas have lesser fleas

Upon their backs to bite 'em.

And lesser fleas have lesser still,

And so ad infinitum. -- *Jonathan Swift*

Now how do these conversions of area and volume relate to the questions I posed about sizes of living things? Well, imagine that you are shrunk like Alice in Wonderland to the size of an insect. One way of thinking about the change of scale is that what used to look like a centimeter now looks like perhaps a meter to you, because you're so much smaller. If area and volume scaled according to most people's intuitive, incorrect expectations, with \(1\ \text{m}^2\) being the same as 100 \(\text{cm}^2\), then there would be no particular reason why nature should behave any differently on your new, reduced scale. But nature does behave differently now that you're small. For instance, you will find that you can walk on water, and jump to many times your own height. The physicist Galileo Galilei had the basic insight that the scaling of area and volume determines how natural phenomena behave differently on different scales. He first reasoned about mechanical structures, but later extended his insights to living things, taking the then-radical point of view that at the fundamental level, a living organism should follow the same laws of nature as a machine. We will follow his lead by first discussing machines and then living things.

One of the world's most famous pieces of scientific writing is Galileo's Dialogues Concerning the Two New Sciences. Galileo was an entertaining writer who wanted to explain things clearly to laypeople, and he livened up his work by casting it in the form of a dialogue among three people. Salviati is really Galileo's alter ego. Simplicio is the stupid character, and one of the reasons Galileo got in trouble with the Church was that there were rumors that Simplicio represented the Pope. Sagredo is the earnest and intelligent student, with whom the reader is supposed to identify. (The following excerpts are from the 1914 translation by Crew and de Salvio.)

Sagredo:

Yes, that is what I mean; and I refer especially to his last assertion which I have always regarded as false...; namely, that in speaking of these and other similar machines one cannot argue from the small to the large, because many devices which succeed on a small scale do not work on a large scale. Now, since mechanics has its foundations in geometry, where mere size [ is unimportant], I do not see that the properties of circles, triangles, cylinders, cones and other solid figures will change with their size. If, therefore, a large machine be constructed in such a way that its parts bear to one another the same ratio as in a smaller one, and if the smaller is sufficiently strong for the purpose for which it is designed, I do not see why the larger should not be able to withstand any severe and destructive tests to which it may be subjected.

Salviati contradicts Sagredo:

Salviati:

... Please observe, gentlemen, how facts which at first seem improbable will, even on scant explanation, drop the cloak which has hidden them and stand forth in naked and simple beauty. Who does not know that a horse falling from a height of three or four cubits will break his bones, while a dog falling from the same height or a cat from a height of eight or ten cubits will suffer no injury? Equally harmless would be the fall of a grasshopper from a tower or the fall of an ant from the distance of the moon.

The point Galileo is making here is that small things are sturdier in proportion to their size. There are a lot of objections that could be raised, however. After all, what does it really mean for something to be “strong”, to be “strong in proportion to its size,” or to be strong “out of proportion to its size?” Galileo hasn't given operational definitions of things like “strength,” i.e., definitions that spell out how to measure them numerically.

Also, a cat is shaped differently from a horse --- an enlarged photograph of a cat would not be mistaken for a horse, even if the photo-doctoring experts at the National Inquirer made it look like a person was riding on its back. A grasshopper is not even a mammal, and it has an exoskeleton instead of an internal skeleton. The whole argument would be a lot more convincing if we could do some isolation of variables, a scientific term that means to change only one thing at a time, isolating it from the other variables that might have an effect. If size is the variable whose effect we're interested in seeing, then we don't really want to compare things that are different in size but also different in other ways.

Salviati:

... we asked the reason why [shipbuilders] employed stocks, scaffolding, and bracing of larger dimensions for launching a big vessel than they do for a small one; and [an old man] answered that they did this in order to avoid the danger of the ship parting under its own heavy weight, a danger to which small boats are not subject?

After this entertaining but not scientifically rigorous beginning, Galileo starts to do something worthwhile by modern standards. He simplifies everything by considering the strength of a wooden plank. The variables involved can then be narrowed down to the type of wood, the width, the thickness, and the length. He also gives an operational definition of what it means for the plank to have a certain strength “in proportion to its size,” by introducing the concept of a plank that is the longest one that would not snap under its own weight if supported at one end. If you increased its length by the slightest amount, without increasing its width or thickness, it would break. He says that if one plank is the same shape as another but a different size, appearing like a reduced or enlarged photograph of the other, then the planks would be strong “in proportion to their sizes” if both were just barely able to support their own weight.

Also, Galileo is doing something that would be frowned on in modern science: he is mixing experiments whose results he has actually observed (building boats of different sizes), with experiments that he could not possibly have done (dropping an ant from the height of the moon). He now relates how he has done actual experiments with such planks, and found that, according to this operational definition, they are not strong in proportion to their sizes. The larger one breaks. He makes sure to tell the reader how important the result is, via Sagredo's astonished response:

Sagredo:

My brain already reels. My mind, like a cloud momentarily illuminated by a lightning flash, is for an instant filled with an unusual light, which now beckons to me and which now suddenly mingles and obscures strange, crude ideas. From what you have said it appears to me impossible to build two similar structures of the same material, but of different sizes and have them proportionately strong.

In other words, this specific experiment, using things like wooden planks that have no intrinsic scientific interest, has very wide implications because it points out a general principle, that nature acts differently on different scales.

To finish the discussion, Galileo gives an explanation. He
says that the strength of a plank (defined as, say, the
weight of the heaviest boulder you could put on the end
without breaking it) is proportional to its cross-sectional
area, that is, the surface area of the fresh wood that would
be exposed if you sawed through it in the middle. Its
weight, however, is proportional to its volume.^{1}

How do the volume and cross-sectional area of the longer plank compare with those of the shorter plank? We have already seen, while discussing conversions of the units of area and volume, that these quantities don't act the way most people naively expect. You might think that the volume and area of the longer plank would both be doubled compared to the shorter plank, so they would increase in proportion to each other, and the longer plank would be equally able to support its weight. You would be wrong, but Galileo knows that this is a common misconception, so he has Salviati address the point specifically:

Salviati:

... Take, for example, a cube two inches on a side so that each face has an area of four square inches and the total area, i.e., the sum of the six faces, amounts to twenty-four square inches; now imagine this cube to be sawed through three times [with cuts in three perpendicular planes] so as to divide it into eight smaller cubes, each one inch on the side, each face one inch square, and the total surface of each cube six square inches instead of twenty-four in the case of the larger cube. It is evident therefore, that the surface of the little cube is only one-fourth that of the larger, namely, the ratio of six to twenty-four; but the volume of the solid cube itself is only one-eighth; the volume, and hence also the weight, diminishes therefore much more rapidly than the surface... You see, therefore, Simplicio, that I was not mistaken when ... I said that the surface of a small solid is comparatively greater than that of a large one.

The same reasoning applies to the planks. Even though they are not cubes, the large one could be sawed into eight small ones, each with half the length, half the thickness, and half the width. The small plank, therefore, has more surface area in proportion to its weight, and is therefore able to support its own weight while the large one breaks.

You probably are not going to believe Galileo's claim that this has deep implications for all of nature unless you can be convinced that the same is true for any shape. Every drawing you've seen so far has been of squares, rectangles, and rectangular solids. Clearly the reasoning about sawing things up into smaller pieces would not prove anything about, say, an egg, which cannot be cut up into eight smaller egg-shaped objects with half the length.

Is it always true that something half the size has one
quarter the surface area and one eighth the volume, even if
it has an irregular shape? Take the example of a child's
violin. Violins are made for small children in smaller size
to accomodate their small bodies. Figure i shows
a full-size violin, along with two
violins made with half and 3/4 of the normal length.^{2}
Let's study the surface area of the front
panels of the three violins.

Consider the square in the interior of the panel of the full-size violin. In the 3/4-size violin, its height and width are both smaller by a factor of 3/4, so the area of the corresponding, smaller square becomes \(3/4\times3/4=9/16\) of the original area, not 3/4 of the original area. Similarly, the corresponding square on the smallest violin has half the height and half the width of the original one, so its area is 1/4 the original area, not half.

The same reasoning works for parts of the panel near the edge, such as the part that only partially fills in the other square. The entire square scales down the same as a square in the interior, and in each violin the same fraction (about 70%) of the square is full, so the contribution of this part to the total area scales down just the same.

Since any small square region or any small region covering part of a square scales down like a square object, the entire surface area of an irregularly shaped object changes in the same manner as the surface area of a square: scaling it down by 3/4 reduces the area by a factor of 9/16, and so on.

In general, we can see that any time there are two objects with the same shape, but different linear dimensions (i.e., one looks like a reduced photo of the other), the ratio of their areas equals the ratio of the squares of their linear dimensions:

\[\begin{equation*}
\frac{A_1}{A_2} = \left(\frac{L_1}{L_2}\right)^2 .
\end{equation*}\]

Note that it doesn't matter where we choose to measure the linear size, \(L\), of an object. In the case of the violins, for instance, it could have been measured vertically, horizontally, diagonally, or even from the bottom of the left f-hole to the middle of the right f-hole. We just have to measure it in a consistent way on each violin. Since all the parts are assumed to shrink or expand in the same manner, the ratio \(L_1/L_2\) is independent of the choice of measurement.

It is also important to realize that it is completely unnecessary to have a formula for the area of a violin. It is only possible to derive simple formulas for the areas of certain shapes like circles, rectangles, triangles and so on, but that is no impediment to the type of reasoning we are using.

Sometimes it is inconvenient to write all the equations in terms of ratios, especially when more than two objects are being compared. A more compact way of rewriting the previous equation is

\[\begin{equation*}
A \propto L^2 .
\end{equation*}\]

The symbol “\(\propto\)” means “is proportional to.” Scientists and engineers often speak about such relationships verbally using the phrases “scales like” or “goes like,” for instance “area goes like length squared.”

All of the above reasoning works just as well in the case of volume. Volume goes like length cubed:

\[\begin{equation*}
V \propto L^3 .
\end{equation*}\]

When a car or truck travels over a road, there is wear and tear on the road surface, which incurs a cost. Studies show that the cost \(C\) per kilometer of travel is related to the weight per axle \(w\) by \(C \propto w^4\). Translate this into a statement about ratios.

(answer in the back of the PDF version of the book)If different objects are made of the same material with the same density, \(\rho =m/V\), then their masses, \(m=\rho V\), are proportional to \(L^3\). (The symbol for density is \(\rho\), the lower-case Greek letter “rho.”)

An important point is that all of the above reasoning about scaling only applies to objects that are the same shape. For instance, a piece of paper is larger than a pencil, but has a much greater surface-to-volume ratio.

Correct solution #1: Area scales in proportion to the square of the linear dimensions, so the larger triangle has four times more area \((2^2=4)\).

Correct solution #2: You could cut the larger triangle into four of the smaller size, as shown in fig. (b), so its area is four times greater. (This solution is correct, but it would not work for a shape like a circle, which can't be cut up into smaller circles.)

Correct solution #3: The area of a triangle is given by

\(A=bh/2\), where \(b\) is the base and \(h\) is the height. The areas of the triangles are

\[\begin{align*}
A_1 &= b_1 h_1/2\\
A_2 &= b_2 h_2/2\\
&= (2b_1)(2h_1)/2 \\
&= 2b_1 h_1\\
A_2/A_1 &=(2b_1 h_1)/(b_1 h_1/2)\\
&= 4
\end{align*}\]

(Although this solution is correct, it is a lot more work than solution #1, and it can only be used in this case because a triangle is a simple geometric shape, and we happen to know a formula for its area.)

Correct solution #4: The area of a triangle is \(A= bh/2\). The comparison of the areas will come out the same as long as the ratios of the linear sizes of the triangles is as specified, so let's just say \(b_1=1.00\) m and \(b_2=2.00\) m. The heights are then also \(h_1=1.00\) m and \(h_2=2.00\) m, giving areas \(A_1=0.50\ \text{m}^2\) and \(A_2=2.00\ \text{m}^2\), so \(A_2/A_1=4.00\).

(The solution is correct, but it wouldn't work with a shape for whose area we don't have a formula. Also, the numerical calculation might make the answer of 4.00 appear inexact, whereas solution #1 makes it clear that it is exactly 4.)

Incorrect solution: The area of a triangle is \(A=bh/2\), and if you plug in \(b=2.00\) m and \(h=2.00\) m, you get \(A=2.00\ \text{m}^2\), so the bigger triangle has 2.00 times more area. (This solution is incorrect because no comparison has been made with the smaller triangle.)

Correct solution #1: Volume scales like the third power of the linear size, so the larger sphere has a volume that is 125 times greater \((5^3=125)\).

Correct solution #2: The volume of a sphere is \(V=(4/3)\pi r^3\), so

\[\begin{align*}
V_1 &= \frac{4}{3}\pi r_1^3 \\
V_2 &= \frac{4}{3}\pi r_2^3 \\
&= \frac{4}{3}\pi (5r_1)^3 \\
&= \frac{500}{3}\pi r_1^3 \\
V_2/V_1 &= \left( \frac{500}{3}\pi r_1^3 \right) / \left( \frac{4}{3}\pi r_1^3 \right)
&= 125
\end{align*}\]

Incorrect solution: The volume of a sphere is \(V=(4/3)\pi r^3\), so

\[\begin{align*}
V_1 &= \frac{4}{3}\pi r_1^3 \\
V_2 &= \frac{4}{3}\pi r_2^3 \\
&= \frac{4}{3}\pi \cdot 5r_1^3 \\
&= \frac{20}{3}\pi r_1^3 \\
V_2/V_1 &= \left( \frac{20}{3}\pi r_1^3 \right) / \left( \frac{4}{3}\pi r_1^3 \right)
&= 5
\end{align*}\]

(The solution is incorrect because \((5r_1)^3\) is not the same as \(5r_1^3\).)

Correct solution: The amount of ink depends on the area to be covered with ink, and area is proportional to the square of the linear dimensions, so the amount of ink required for the second “S” is greater by a factor of \((48/36)^2=1.78\).

Incorrect solution: The length of the curve of the second “S” is longer by a factor of \(48/36=1.33\), so 1.33 times more ink is required.

(The solution is wrong because it assumes incorrectly that the width of the curve is the same in both cases. Actually both the width and the length of the curve are greater by a factor of 48/36, so the area is greater by a factor of \((48/36)^2=1.78\).)

Reasoning about ratios and proportionalities is one of the three essential mathematical skills, summarized on pp.532-533, that you need for success in this course.

◊ Solved problem: a telescope gathers light — problem 11

◊ Solved problem: distance from an earthquake — problem 12

◊

A toy fire engine is 1/30 the size of the real one, but is constructed from the same metal with the same proportions. How many times smaller is its weight? How many times less red paint would be needed to paint it?

◊

Galileo spends a lot of time in his dialog discussing what really happens when things break. He discusses everything in terms of Aristotle's now-discredited explanation that things are hard to break, because if something breaks, there has to be a gap between the two halves with nothing in between, at least initially. Nature, according to Aristotle, “abhors a vacuum,” i.e., nature doesn't “like” empty space to exist. Of course, air will rush into the gap immediately, but at the very moment of breaking, Aristotle imagined a vacuum in the gap. Is Aristotle's explanation of why it is hard to break things an experimentally testable statement? If so, how could it be tested experimentally?

The left-hand panel in figure o shows the approximate validity of the proportionality \(m\propto L^3\) for cockroaches (redrawn from McMahon and Bonner). The scatter of the points around the curve indicates that some cockroaches are proportioned slightly differently from others, but in general the data seem well described by \(m\propto L^3\). That means that the largest cockroaches the experimenter could raise (is there a 4-H prize?) had roughly the same shape as the smallest ones.

Another relationship that should exist for animals of different sizes shaped in the same way is that between surface area and body mass. If all the animals have the same average density, then body mass should be proportional to the cube of the animal's linear size, \(m\propto L^3\), while surface area should vary proportionately to \(L^2\). Therefore, the animals' surface areas should be proportional to \(m^{2/3}\). As shown in the right-hand panel of figure o, this relationship appears to hold quite well for the dwarf siren, a type of salamander. Notice how the curve bends over, meaning that the surface area does not increase as quickly as body mass, e.g., a salamander with eight times more body mass will have only four times more surface area.

This behavior of the ratio of surface area to mass (or, equivalently, the ratio of surface area to volume) has important consequences for mammals, which must maintain a constant body temperature. It would make sense for the rate of heat loss through the animal's skin to be proportional to its surface area, so we should expect small animals, having large ratios of surface area to volume, to need to produce a great deal of heat in comparison to their size to avoid dying from low body temperature. This expectation is borne out by the data of the left-hand panel of figure p, showing the rate of oxygen consumption of guinea pigs as a function of their body mass. Neither an animal's heat production nor its surface area is convenient to measure, but in order to produce heat, the animal must metabolize oxygen, so oxygen consumption is a good indicator of the rate of heat production. Since surface area is proportional to \(m^{2/3}\), the proportionality of the rate of oxygen consumption to \(m^{2/3}\) is consistent with the idea that the animal needs to produce heat at a rate in proportion to its surface area. Although the smaller animals metabolize less oxygen and produce less heat in absolute terms, the amount of food and oxygen they must consume is greater in proportion to their own mass. The Etruscan pigmy shrew, weighing in at 2 grams as an adult, is at about the lower size limit for mammals. It must eat continually, consuming many times its body weight each day to survive.

Large mammals, such as elephants, have a small ratio of surface area to volume, and have problems getting rid of their heat fast enough. An elephant cannot simply eat small enough amounts to keep from producing excessive heat, because cells need to have a certain minimum metabolic rate to run their internal machinery. Hence the elephant's large ears, which add to its surface area and help it to cool itself. Previously, we have seen several examples of data within a given species that were consistent with a fixed shape, scaled up and down in the cases of individual specimens. The elephant's ears are an example of a change in shape necessitated by a change in scale.

Large animals also must be able to support their own weight. Returning to the example of the strengths of planks of different sizes, we can see that if the strength of the plank depends on area while its weight depends on volume, then the ratio of strength to weight goes as follows:

\[\begin{equation*}
\text{strength}/\text{weight} \propto A/V \propto 1/L .
\end{equation*}\]

Thus, the ability of objects to support their own weights decreases inversely in proportion to their linear dimensions. If an object is to be just barely able to support its own weight, then a larger version will have to be proportioned differently, with a different shape.

Since the data on the cockroaches seemed to be consistent with roughly similar shapes within the species, it appears that the ability to support its own weight was not the tightest design constraint that Nature was working under when she designed them. For large animals, structural strength is important. Galileo was the first to quantify this reasoning and to explain why, for instance, a large animal must have bones that are thicker in proportion to their length. Consider a roughly cylindrical bone such as a leg bone or a vertebra. The length of the bone, \(L\), is dictated by the overall linear size of the animal, since the animal's skeleton must reach the animal's whole length. We expect the animal's mass to scale as \(L^3\), so the strength of the bone must also scale as \(L^3\). Strength is proportional to cross-sectional area, as with the wooden planks, so if the diameter of the bone is \(d\), then

\[\begin{align*}
d^2 &\propto L^3 \\
\text{or}
d &\propto L^{3/2} .
\end{align*}\]

If the shape stayed the same regardless of size, then all linear dimensions, including \(d\) and \(L\), would be proportional to one another. If our reasoning holds, then the fact that \(d\) is proportional to \(L^{3/2}\), not \(L\), implies a change in proportions of the bone. As shown in the right-hand panel of figure p, the vertebrae of African Bovidae follow the rule \(d\propto L^{3/2}\) fairly well. The vertebrae of the giant eland are as chunky as a coffee mug, while those of a Gunther's dik-dik are as slender as the cap of a pen.

◊

Single-celled animals must passively absorb nutrients and oxygen from their surroundings, unlike humans who have lungs to pump air in and out and a heart to distribute the oxygenated blood throughout their bodies. Even the cells composing the bodies of multicellular animals must absorb oxygen from a nearby capillary through their surfaces. Based on these facts, explain why cells are always microscopic in size.

◊

The reasoning of the previous question would seem to be contradicted by the fact that human nerve cells in the spinal cord can be as much as a meter long, although their widths are still very small. Why is this possible?

It is the mark of an instructed mind to rest satisfied with
the degree of precision that the nature of the subject
permits and not to seek an exactness where only an
approximation of the truth is possible. -- *Aristotle*

It is a common misconception that science must be exact. For instance, in the Star Trek TV series, it would often happen that Captain Kirk would ask Mr. Spock, “Spock, we're in a pretty bad situation. What do you think are our chances of getting out of here?” The scientific Mr. Spock would answer with something like, “Captain, I estimate the odds as 237.345 to one.” In reality, he could not have estimated the odds with six significant figures of accuracy, but nevertheless one of the hallmarks of a person with a good education in science is the ability to make estimates that are likely to be at least somewhere in the right ballpark. In many such situations, it is often only necessary to get an answer that is off by no more than a factor of ten in either direction. Since things that differ by a factor of ten are said to differ by one order of magnitude, such an estimate is called an order-of-magnitude estimate. The tilde, \(\sim\), is used to indicate that things are only of the same order of magnitude, but not exactly equal, as in

\[\begin{equation*}
\text{odds of survival} \sim \text{100 to one} .
\end{equation*}\]

The tilde can also be used in front of an individual number to emphasize that the number is only of the right order of magnitude.

Although making order-of-magnitude estimates seems simple and natural to experienced scientists, it's a mode of reasoning that is completely unfamiliar to most college students. Some of the typical mental steps can be illustrated in the following example.

\(\triangleright\) Roughly what percentage of the price of a tomato comes from the cost of transporting it in a truck?

\(\triangleright\) The following incorrect solution illustrates one of the main ways you can go wrong in order-of-magnitude estimates.

Incorrect solution: Let's say the trucker needs to make a $400 profit on the trip. Taking into account her benefits, the cost of gas, and maintenance and payments on the truck, let's say the total cost is more like $2000. I'd guess about 5000 tomatoes would fit in the back of the truck, so the extra cost per tomato is 40 cents. That means the cost of transporting one tomato is comparable to the cost of the tomato itself. Transportation really adds a lot to the cost of produce, I guess.

The problem is that the human brain is not very good at estimating area or volume, so it turns out the estimate of 5000 tomatoes fitting in the truck is way off. That's why people have a hard time at those contests where you are supposed to estimate the number of jellybeans in a big jar. Another example is that most people think their families use about 10 gallons of water per day, but in reality the average is about 300 gallons per day. When estimating area or volume, you are much better off estimating linear dimensions, and computing volume from the linear dimensions. Here's a better solution to the problem about the tomato truck:

As in the previous solution, say the cost of the trip is $2000. The dimensions of the bin are probably 4 m \(\times\) 2 m \(\times\) 1 m, for a volume of \(8\ \text{m}^3\). Since the whole thing is just an order-of-magnitude estimate, let's round that off to the nearest power of ten, \(10\ \text{m}^3\). The shape of a tomato is complicated, and I don't know any formula for the volume of a tomato shape, but since this is just an estimate, let's pretend that a tomato is a cube, 0.05 m \(\times\) 0.05 m \(\times\) 0.05 m, for a volume of \(1.25\times10^{-4}\ \text{m}^3\). Since this is just a rough estimate, let's round that to \(10^{-4} \text{m}^3\). We can find the total number of tomatoes by dividing the volume of the bin by the volume of one tomato: \(10\ \text{m}^3/10^{-4}\ \text{m}^3=10^5\) tomatoes. The transportation cost per tomato is $$2000/10^5$ tomatoes=$0.02/tomato. That means that transportation really doesn't contribute very much to the cost of a tomato.

Approximating the shape of a tomato as a cube is an example of another general strategy for making order-of-magnitude estimates. A similar situation would occur if you were trying to estimate how many \(\text{m}^2\) of leather could be produced from a herd of ten thousand cattle. There is no point in trying to take into account the shape of the cows' bodies. A reasonable plan of attack might be to consider a spherical cow. Probably a cow has roughly the same surface area as a sphere with a radius of about 1 m, which would be \(4\pi (1\ \text{m})^2\). Using the well-known facts that pi equals three, and four times three equals about ten, we can guess that a cow has a surface area of about \(10\ \text{m}^2\), so the herd as a whole might yield \(10^5\ \text{m}^2\) of leather.

Its torso looks like it can be approximated by a rectangular box with dimensions \(10\ \text{m}\times5\ \text{m}\times3\ \text{m}\), giving about \(2\times10^2\ \text{m}^3\). Living things are mostly made of water, so we assume the animal to have the density of water, \(1\ \text{g}/\text{cm}^3\), which converts to \(10^3\ \text{kg}/\text{m}^3\). This gives a mass of about \(2\times10^5\ \text{kg}\), or 200 metric tons.

The following list summarizes the strategies for getting a good order-of-magnitude estimate.

- Don't even attempt more than one significant figure of precision.
- Don't guess area, volume, or mass directly. Guess linear dimensions and get area, volume, or mass from them.
- When dealing with areas or volumes of objects with complex shapes, idealize them as if they were some simpler shape, a cube or a sphere, for example.
- Check your final answer to see if it is reasonable. If you estimate that a herd of ten thousand cattle would yield \(0.01\ \text{m}^2\) of leather, then you have probably made a mistake with conversion factors somewhere.

\(\propto\) — is proportional to

\(\sim\) — on the order of, is on the order of

{}

Nature behaves differently on large and small scales. Galileo showed that this results fundamentally from the way area and volume scale. Area scales as the second power of length, \(A\propto L^2\), while volume scales as length to the third power, \(V\propto L^3\).

An order of magnitude estimate is one in which we do not attempt or expect an exact answer. The main reason why the uninitiated have trouble with order-of-magnitude estimates is that the human brain does not intuitively make accurate estimates of area and volume. Estimates of area and volume should be approached by first estimating linear dimensions, which one's brain has a feel for.

**1**. How many cubic inches are there in a cubic foot?
The answer is not 12.(answer check available at lightandmatter.com)

**2**. Assume a dog's brain is twice as great in diameter as a
cat's, but each animal's brain cells are the same size and
their brains are the same shape. In addition to being a far
better companion and much nicer to come home to, how many
times more brain cells does a dog have than a cat?
The answer is not 2.

**3**. The population density of Los Angeles is about
\(4000\ \text{people}/\text{km}^2\). That of San Francisco is about
\(6000\ \text{people}/\text{km}^2\). How many times farther away is the average
person's nearest neighbor in LA than in San Francisco?
The answer is not 1.5. (answer check available at lightandmatter.com)

**4**. A hunting dog's nose has about 10 square inches of active
surface. How is this possible, since the dog's nose is only
about 1 in \(\times\) 1 in \(\times\) 1 in = 1 \(\text{in}^3?\) After all, 10 is
greater than 1, so how can it fit?

**6**. In a computer memory chip, each bit of information (a 0
or a 1) is stored in a single tiny circuit etched onto the
surface of a silicon chip. The circuits cover the surface of
the chip like lots in a housing development. A typical chip
stores 64 Mb (megabytes) of data, where a byte is 8 bits.
Estimate (a) the area of each circuit, and (b) its linear size.

**7**. Suppose someone built a gigantic apartment building,
measuring 10 km \(\times\) 10 km at the base. Estimate how tall the
building would have to be to have space in it for the entire
world's population to live.

**8**. A hamburger chain advertises that it has sold 10 billion
Bongo Burgers. Estimate the total mass of feed required to
raise the cows used to make the burgers.

**11**. (solution in the pdf version of the book) Compare the light-gathering powers of a 3-cm-diameter
telescope and a 30-cm telescope.

**12**. (solution in the pdf version of the book) One step on the Richter scale corresponds to a
factor of 100 in terms of the energy absorbed by something
on the surface of the Earth, e.g., a house. For instance, a
9.3-magnitude quake would release 100 times more energy than
an 8.3. The energy spreads out from the epicenter as a wave,
and for the sake of this problem we'll assume we're dealing
with seismic waves that spread out in three dimensions, so
that we can visualize them as hemispheres spreading out
under the surface of the earth. If a certain 7.6-magnitude
earthquake and a certain 5.6-magnitude earthquake produce
the same amount of vibration where I live, compare the
distances from my house to the two epicenters.

**13**. In Europe, a piece of paper of the standard size,
called A4, is a little narrower and taller than its American
counterpart. The ratio of the height to the width is the
square root of 2, and this has some useful properties. For
instance, if you cut an A4 sheet from left to right, you get
two smaller sheets that have the same proportions. You can
even buy sheets of this smaller size, and they're called A5.
There is a whole series of sizes related in this way, all
with the same proportions. (a) Compare an A5 sheet to an A4
in terms of area and linear size. (b) The series of paper
sizes starts from an A0 sheet, which has an area of one
square meter. Suppose we had a series of boxes defined in a
similar way: the B0 box has a volume of one cubic meter, two
B1 boxes fit exactly inside an B0 box, and so on. What would
be the dimensions of a B0 box? (answer check available at lightandmatter.com)

**15**. According to folklore, every time you take a breath, you
are inhaling some of the atoms exhaled in Caesar's
last words. Is this true? If so, how many?

**16**. The Earth's surface is about 70% water. Mars's diameter
is about half the Earth's, but it has no surface water.
Compare the land areas of the two planets.(answer check available at lightandmatter.com)

**17**. (solution in the pdf version of the book) The traditional Martini glass is shaped like a cone
with the point at the bottom. Suppose you make a Martini by
pouring vermouth into the glass to a depth of 3 cm, and then
adding gin to bring the depth to 6 cm. What are the
proportions of gin and vermouth?

**18**. The central portion of a CD is taken up by the hole
and some surrounding clear plastic, and this area is
unavailable for storing data. The radius of the central
circle is about 35% of the outer radius of the data-storing area.
What percentage of the CD's area is therefore lost? (answer check available at lightandmatter.com)

**19**. The one-liter cube in the photo has been marked off into smaller
cubes, with linear dimensions one tenth those of the big one.
What is the volume of each of the small cubes?(solution in the pdf version of the book)

**21**. Estimate the number of man-hours required for building the Great Wall of China. (solution in the pdf version of the book)

**22**. (a) Using the microscope photo in the figure, estimate the mass of a one cell of the
*E. coli* bacterium, which is one of the most common ones in the human
intestine. Note the scale at the lower right corner, which is 1 \(\mu\text{m}\). Each of
the tubular objects in the column is one cell.
(b) The feces in the human intestine are mostly bacteria (some dead, some alive),
of which *E. coli* is a large and typical component. Estimate the number of bacteria
in your intestines, and compare with the number of human cells in your body, which is
believed to be roughly on the order of \(10^{13}\). (c) Interpreting your result from part b,
what does this tell you about the size of a typical human cell compared to the size of
a typical bacterial cell?

**23**. A taxon (plural taxa) is a group of living things. For example, *Homo sapiens* and
*Homo neanderthalensis* are both taxa --- specifically, they are two different species within
the genus *Homo*. Surveys by botanists show that the number of plant taxa native to
a given contiguous land area \(A\) is usually approximately proportional to \(A^{1/3}\). (a) There are 70
different species of lupine native to Southern California, which has an area of about
\(200,000\ \text{km}^2\). The San Gabriel Mountains cover about \(1,600\ \text{km}^2\). Suppose
that you wanted to learn to identify all the species of lupine in the San Gabriels.
Approximately how many species would you have to familiarize yourself with?\hwans{hwans:taxa-scaling}(answer check available at lightandmatter.com)

(b) What is the interpretation of the fact that the exponent, \(1/3\), is less than one?

**24**. X-ray images aren't only used with human subjects but also, for example,
on insects and flowers. In 2003, a team of researchers at Argonne National Laboratory
used x-ray imagery to find for the first time that insects, although they do not have
lungs, do not necessarily breathe completely passively, as had been believed previously; many insects rapidly compress and
expand their trachea, head, and thorax in order to force air in and out of their bodies.
One difference between x-raying a human and an insect is that if a medical x-ray machine was
used on an insect, virtually 100% of the x-rays would pass through its body, and there would
be no contrast in the image produced. Less penetrating x-rays of lower energies have to
be used. For comparison, a typical human body mass is about 70 kg, whereas
a typical ant is about 10 mg. Estimate the ratio of the thicknesses of tissue that must be
penetrated by x-rays in one case compared to the other.(answer check available at lightandmatter.com)

**25**. Radio was first commercialized around 1920, and ever since then, radio signals from our
planet have been spreading out across our galaxy. It is possible that alien civilizations
could detect these signals and learn that there is life on earth. In the 90 years that
the signals have been spreading at the speed of light, they have created a sphere with a radius of 90 light-years.
To show an idea of the size of this sphere, I've indicated it in the figure as a tiny white circle on an image
of a spiral galaxy seen edge on. (We don't have similar photos of our own Milky Way galaxy, because we can't
see it from the outside.) So far we haven't received answering signals from aliens within this sphere,
but as time goes on, the sphere will expand as suggested by the dashed outline, reaching more and more stars that might harbor extraterrestrial
life. Approximately what year will it be when the sphere has expanded to fill a volume 100 times greater than
the volume it fills today in 2010?
(answer check available at lightandmatter.com)

**26**. (solution in the pdf version of the book)
Estimate the number of jellybeans in figure r on p. 56.

**27**. At the grocery store you will see oranges packed neatly in stacks.
Suppose we want to pack spheres as densely as possible, so that the greatest
possible fraction of the space is filled by the spheres themselves, not
by empty space. Let's call this fraction \(f\).
Mathematicians have proved that the best possible result is \(f\approx 0.7405\),
which requires a systematic pattern of stacking. If you buy ball bearings or golf
balls, however, the seller is probably not going to go to the trouble of stacking
them neatly. Instead they will probably pour the balls into a box and vibrate the
box vigorously for a while to make them settle. This results in a random packing.
The closest random packing has \(f\approx 0.64\).
Suppose that golf balls, with a standard diameter of 4.27 cm, are sold in bulk
with the closest random packing. What is the diameter of the largest ball that
could be sold in boxes of the same size, packed systematically, so that there
would be the same number of balls per box?(answer check available at lightandmatter.com)

**28**. Plutonium-239 is one of a small number of important long-lived forms of high-level
radioactive nuclear waste. The world's waste stockpiles have about \(10^3\) metric tons
of plutonium. Drinking water is considered safe by U.S. government standards if it
contains less than \(2\times10^{-13}\ \text{g}/\text{cm}^3\) of plutonium.
The amount of radioactivity to which you were exposed by drinking such water
on a daily basis would be very small compared to the natural background radiation that you are
exposed to every year. Suppose that the world's inventory of plutonium-239 were ground
up into an extremely fine dust and then dispersed over the world's oceans, thereby
becoming mixed uniformly into the world's water supplies over time.
Estimate the resulting concentration of plutonium, and compare with the government standard.

\begin{handson}{}{Scaling applied to leaves}{\onecolumn}

Equipment:

leaves of three sizes, having roughly similar proportions of length, width, and thickness

balance

Each group will have one leaf, and should measure its surface area and volume, and determine its surface-to-volume ratio. For consistency, every group should use units of \(\text{cm}^2\) and \(\text{cm}^3\), and should only find the area of one side of the leaf. The area can be found by tracing the area of the leaf on graph paper and counting squares. The volume can be found by weighing the leaf and assuming that its density is 1 \(\text{g}/\text{cm}^3\) (the density of water). What implications do your results have for the plants' abilities to survive in different environments?

\includegraphics[width=170mm]{../share/intro/figs/ex-leaves}

\end{handson}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.