Your mission, should you choose to accept it, is to measure the angle between successive leaves around the stalk of different plants. There are lots of different plants to try. To clarify the mission, you’re not looking for the angle between branches or between a leaf and a branch. Instead, you should try to look straight down the stalks of various plants, and measure the angle between the directions that two successive leaves stick out radially from the axis of the stalk. I’ve marked the different kinds of angles in these photos from the Center, so you can better see what is meant.

When you’ve taken a number of measurements, what do you notice? Are any angles more common than others?

You should observe that angles around 137 or 138 degrees are the most common. Why that number? Is there anything special about that angle?

It turns out there is! What are the leaves there for? To catch sunlight for photosynthesis, which feeds the plant. Imagine that the angle between successive leaves was 90 degrees, a right angle. Then there would be just four orderly rows of leaves, stacked up right on top of each other, and they would block each other from the sun. So that would be an extremely bad angle. 90 degrees being bad suggests the question of whether there is a best angle, where the leaves on average overlap the least? And indeed there is. Using some geometry and other math, you can prove that there is an ideal, least-overlap angle, and that that angle is between 137 and 138 degrees. So the plants have naturally evolved to “discover” that ideal angle. This is a case of a basic mathematical principle dictating the structure of millions of life forms on earth. And the story gets even more interesting — it turns out that angle is closely related to the Fibonacci numbers (1,1,2,3,5,8,13, and so on, where each number is the sum of the previous two) and so can help explain why when you pick up a pineapple or sunflower or pinecone, there are usually a Fibonacci number of spirals to its seeds.

]]>How might you go about that? Well, it would be useful to know that the average lifespan of a butterfly is about two weeks. How would that help? If we knew the number of butterflies in the room, we could reason that to keep the number of butterflies at that level, that many would have to emerge every two weeks, to replace the butterflies that would all die over the course of those two weeks. So we would just divide two weeks (or 14 days or 336 hours or 20,160 minutes) by that number to get the average interval between butterflies emerging.

So that switches our question to how many butterflies are there in the room. At first, that might seem like an intractable problem. After all, the room is huge, and the butterflies are flitting everywhere. You could never count them one by one — they just won’t sit still! And how can you tell whether you’ve counted one already or not?

So we need to use other mathematical tools. Rather than trying to get an exact count, we want to *estimate*. There are lots of ways to estimate. Here are a couple you might try. (1) Take a picture of the room and carefully count how many you see in the picture. Guess what fraction of the room is visible in the picture, and then multiply to get the number visible in the whole room. Then (here’s the most inaccurate part) guess what fraction of the butterflies are visible at any time — for example, by watching a section of the foliage for a while and seeing how many you see at first versus how many seem to turn up unexpectedly). Now multiply again to get the total number of butterflies in the room, both visible and hidden. (2) First estimate the volume of the room — maybe you can find the height, width, and length of the room from the staff. Then with your classmates, pick a well-defined section of the room that you can observe carefully and try to count how many butterflies in that section. Then measure the volume of that section. Finally multiply your count by the volume of the whole room divided by the size of that section. (3) Again, start with the volume of the room. Then estimate the average distance between a butterfly and its nearest neighbor butterfly. Pick a number of closest pairs of butterflies and do your best to measure the distance between them, using a measuring tape to compare. Include both butterflies whose nearest neighbor is close, and “loner” butterflies whose nearest neighbor is far away. When you have a bunch of measurements, average them. Finally, divide the volume of the room by the volume of a sphere whose radius is what you found for the average distance between butterflies. (Imagine the entire room filled with a spherical “bubble” around each butterfly!)

When you have your estimate, do the division to guess how often a butterfly emerges from a chrysalis. When you get down to the nursery area, you can ask about your guess to see how close you’ve come!

]]>First, it might strike you that there are so many different shapes! It seems as though every kind of butterfly is different. How can different butterflies produce such different shapes? It seems as though we ought to count the number of wings on different butterflies, because one way we might explain different shapes is by different numbers of wings.

However, if you look carefully all the butterflies around you, you will discover that they all have exactly four wings. In fact, every type of butterfly in existence has four wings. So there must be some other way they achieve such shape variation.

As you look at more butterflies, though, you will start to see certain patterns that recur, similarities among those differences. Can you find some examples?

A first, and mathematically important one, is symmetry. If you imagine a line through the middle of any healthy butterfly, the left and right sides are identical, just one is flipped the other way from the first. That type of pattern is called *mirror* or *bilateral* symmetry, and you will see it over and over again in the natural world.

There are other more detailed patterns in the shapes of butterfly wings. You may notice that they tend to be divided into sections by the veins in the wings, and those sections tend to have a similar arrangement in different types of butterflies. For example, you may notice that there tends to be an oval section in the middle of most wings, surrounded by fairly straight sections that proceed out to the edges of the wings.

That basic structure allows you to notice some more things. For example, many types of butterflies have “tails,” or long protrusions on the back of the wings. And those tails generally seem to come from the third or fourth sections of the back wings being longer than on other butterflies that don’t have tails.

And that observation provides an essential clue about the overall shape variation of butterfly wings: by changing the lengths of all of the different sections, the basic structure of a butterfly wing can produce thousands of different shapes. See if you can figure out the “instructions” for producing different butterfly wings, in terms of stretching or compressing different sections of the wing.

]]>If you’re like me, you probably noticed that when a butterfly is flapping its wings, the path is usually very erratic, darting this way and that. But does a butterfly *have to* fly erratically? No, if you watch for a while, you can find times when they flay in more regular paths. So why would they fly so erratically?

That question leads us to the topic of *extrapolation*. If you knew that someone had drawn a shape and erased part of it and what was left looked like this:

how might you try to complete the shape? You would probably draw something like this:

On the other hand, if what was left looked like this:

you’d really have no idea how to complete the shape. It could turn out to be any of thousands of different paths.

How does this relate to the butterfly’s flight? Well, who might want to figure out where the butterfly’s path will take it? Who would be interested in where the butterfly would end up? Maybe something that wanted to eat it? Butterflies are not big or strong, and they are easy to spot amidst the plants. So they need other defenses, and the erratic geometry of their flight is one: they make it very difficult for predators to extrapolate their paths and intersect with them to catch them. So mathematics can be useful, even to a butterfly!

If you look at their non-erratic flight, you can also see another mathematically interesting shape. You’ll notice that sometimes they glide in straight lines, but that they also glide in curved paths. In these curved paths, they tend to go around in a circle when viewed from above, while also gradually descending. This kind of curve, that curves like a circle while descending (or ascending), has a special name in mathematics. It’s called a *helix* and it’s a shape we can see in many places around us: the handrail of a circular staircase, the thread of a screw, the rail of some roller coasters, and more. It’s also a kind of flight pattern that occurs spontaneously over and over: in nature, like the path of a spinning seed falling from a tree, or in man-made flying objects, as this illustration from the Federal Aviation Administration’s handbook for glider pilot’s shows.

What does it remind you of? A honeycomb, of course! But did it ever occur to you to wonder why bees build honeycombs in the shape they do? It turns out there is a simple mathematical reason that they do.

To understand this, try the following activity. You need twelve short straight items, like matchsticks or pins — it doesn’t really matter what, as long as they are straight and all the same length. Print out this rectangle, scaled so the long side is exactly four times the length of your sticks. Now, try to lay your sticks flat in the rectangle, to satisfy the following three rules:

- Each end of each stick is touching one of the sides of the rectangle or the end of at least one other stick.
- The rectangle is divided into seven regions by the sticks.
- No region formed is less than half the area of any other region formed.

You should work on it for a while before you look at the solution — it’s fun to try to come up with it on your own. But whether you discover it or you peek, one of the amazing facts about this puzzle is that there is only one way to do it, and in that one way, you must create a hexagon, two half-hexagons, and four three-quarter hexagons.

The reason for that is that there’s not enough total length to your twelve sticks to divide the rectangle any other way, because it turns out that the most efficient way — the way that uses the least total length of line segments — to divide an area up into many equal-sized regions is to split it up into hexagonal shapes. In other words, over the course of evolution, honeybees have figured out the way to construct their honeycombs that requires the least wax and therefore the least work. That’s what the title of this post means — honeybees have found the unique way to do the least amount of work to build their hives, which you could maybe think is a little lazy, couldn’t you?

And in fact, honeybees have done even better than I’ve mentioned so far. They build their hives in three dimensions, not two, and they make their honeycombs in double layers, so there are cells with openings on either side. If you look at how the bottoms of the cells are constructed, each consists of three small flat facets, and the facets of the cells from one side interlock perfectly with the facets from the other side. Mathematicians have determined that this configuration is also the most efficient way possible to construct a double layer of cells in three dimensions — something honeybees have “known” for millenia!

But what if I told you that the top sides of its wings are the same dull color, and only *look* bright blue because of the mathematics of addition and subtraction of functions? It’s true!

To see how this can be, you first need to know that rays of light act like waves. They are not waves in the air (those are sound waves), they are waves in the “electromagnetic field,” which is like an energy level that permeates everywhere in the universe. The ray of light makes that energy level fluctuate up and down (extremely rapidly!) as it travels through space. And every wave has a characteristic “wavelength,” which is just the distance from one peak of the wave to the next. A ray of light of a specific wavelength will cause us to see a specific color when it enters our eye. For example, we would see a ray of light with a wavelength of 480 nanometers (that’s just 48 *millionths* of a centimeter!) as blue, or a ray with a wavelength of 576 nanometers as yellowish-green. The white light from the sun is a mixture of rays with a whole lot of different wavelengths.

What does this have to do with the color of a butterfly’s wings? The scales on the top side of the wings have microscopic little structures that reflect light from different surfaces that are just tiny distances apart, distances near the minute size of the wavelength of light. Those tiny separations interact with the mathematics of the wave nature of light to ensure that only blue light gets back to our eyes, making the top surface of the wings look blue.

To see how this works, consider a simple model of what’s occurring on the surface of the butterfly’s wings. Imagine two mirrors a tiny separation apart, 720 nanometers, just one and a half times the wavelength of blue light. Let’s suppose that the top mirror is only a half-mirror: it lets half the light through and reflects half the light. Here’s a diagram that gives an idea of what happens:

Half of the light reflects off of the top mirror (the one to the left in the diagram), and half of the light reflects off the bottom mirror. But the waves of the two reflections align perfectly with each other and strengthen each other, so (just about) all of the blue light reflects back. But now look what happens if we shine that yellowish-green light with a wavelength of 576 nanometers onto the same arrangement of mirrors:

This time, the waves of the two reflections are perfectly *out* of alignment and they quench each other, which means that no yellowish-green light gets reflected back.

This way, the butterfly’s wings ensure that lots of blue light gets reflected back to your eyes and very little of any other color of light does, making the wing appear a bright blue — even though the tiny “mirrors” on its scales are made from exactly the same dull brownish material that you see on the underside of the wing! In reality, as you can see in the microscopic picture below, the structures on the butterfly’s wings are much more complicated than just a pair of mirrors (which turns out to help the butterfly look similar colors from many different directions) but the basic principle remains the same — the butterfly is using the addition and subtraction of electromagnetic waves to control what colors of light reach your eyes.

]]>First, you will need straight beams with regularly spaced holes. Ideally the center one will be labeled with a “0”, the two adjacent to the center labeled with “1”s, the next two out labeled with “2”, and so on. You can print out the following PDF, cut out the rectangular beam, and then use a hole punch to make the holes. Be careful to position the holes accurately! Positioning is key to balance.

If you need or want more exactly-constructed beams, you can use the girders from many different construction sets, like Meccano (see for example item #2). Just be sure to use a beam with equally-spaced holes, and an odd number of them, so that there will be a single center hole.

Second, you will need the ornaments. They can really be any objects that you want to hang, with one **major** proviso: they need to come in a limited number of types, and the weight of the ornaments (including the hooks!) of each type must be identical, and there must be a lightest type, a type that weighs twice as much, a type that weighs three times as much, and so on. Again, for emphasis, those weights should include the weights of the hooks.

Here’s one way to achieve this. Print out five of the caterpillar ornaments below on the heaviest cardstock that will work with your printer. You must use the same weight of cardstock for all ornaments of a given kind. On a very sensitive scale (with 1-gram or better resolution), weigh the five ornaments together. Separately weigh five identical paper clips to use as hooks (you must use all identical paper clips as hooks for all ornaments for this to work!). Divide the measured weights by five to get the weights *C* of a single caterpillar and *P* of a single paper clip.

Now make two test copies of the chrysalis (you will be throwing these two out), on the card stock that you want to use for this ornament. Cut them out and weigh them together, dividing the weight by two to get the test weight *T* of the chrysalis. Then the scaling factor for the final copies of the chrysalis is *s* = √((2*C*+*P*)/*T*). Calculate the scaling factor, rounding it to the nearest percent. Then print out several copies of the chrysalis, scaling by the computed scaling factor, on the desired card stock. Cut them out, and verify with your scale that a chrysalis with paper clip weighs the same as two caterpillars each with its paper clip.

Follow the same procedure with the flower ornament, except use the formula *s* = √((3*C*+2*P*)/*T*) for the scaling factor. Verify that one of the scaled flowers with a paper clip weighs the same as three caterpillars with three paper clips. Finally, repeat with the butterfly using the formula *s* = √((4*C*+3*P*)/*T*) and check that a butterfly with paper clip weighs the same as four caterpillar-paper clip combinations.

Now you have a supply of ornaments for making balanced mobiles. You can twist the paperclips however is helpful for making them into hooks, as long as you don’t break the paperclips or add any tape or anything else that would add weight. You can write down the scaling factor for each ornament, and as long as you use the same paper clips and the same cardstock, you can print and cut out as many ornaments as you like using those scaling factors.

Finally, here are the PDFS for the four types of semicircular beams referred to in the post.

]]>Notice that the branch on which the glass models are hanging is supported only at one point, so it is free to pivot, and yet it hangs perfectly horizontally instead of tilting to the right or left. (You can find a clearer picture on the designers’ website.) In other words, it is in balance. What can you learn from this state of balance?

You may know that we often use something called a “balance” for weighing things, so here’s a question you can ponder as you gaze at *Chrysalis*: which is heavier, two model chrysalises or one model chrysalis and two model leaves?

It’s tempting to suppose that since the sculpture is in balance, those weights must be the same. Yet there’s more to it than that, making the situation even more interesting. To explore this, try the following activity: Start by hanging a perforated beam by its center hole (which may be labeled “0”). The beam should roughly balance horizontally, although it may be fairly wobbly. Next, get a supply of mobile ornaments; there should be different types, and they may also be labeled with numbers like 1, 2, 3, 4.

Your goal is to hang an ornament or ornaments on each side of the perforated beam so that it will balance. (Don’t use the center hole, it is just for suspending the beam.) Experiment with different configurations. What kinds of arrangements will balance?

Likely you will discover that hanging two identical ornaments the same distance away from the center of the beam will cause it to balance. And that makes sense: if the two sides are the same, then they should balance each other out.

But do the sides *have to* be the same in order for the beam to balance? Not necessarily. Try this: hang a “3” ornament in the hole immediately to the left of the center of the beam. Is there anywhere you can hang a “1” ornament to the right of the center to achieve balance?

With some experimentation, you should find that a “1” ornament in the “3” hole just balances a “3” ornament in the “1” hole. And there are other combinations that will balance as well. Try a “2” ornament in the “2” hole against a “1” ornament in the “4” hole. Or put the same ornaments in the “2” and “3” holes on one side and in the “5” hole on the other side. Again, it should balance.

Can you find a rule that will tell you when an arrangement will balance? Again, with some trial and error, you may be able to come up with this rule: take the number on each ornament and multiply it by the number of the hole it is in. Add up those products on each side. If you get the same answer on each side, then the beam should balance. For example, a “2” ornament in the “3” hole and a “3” ornament in the “2” hole on one side should balance a “4” ornament in the “2” hole and a “1” ornament in the “4” hole on the other side, because 2×3 + 3×2 = 12 = 4×2 + 1×4.

Another piece in *Imago* suggests a more advanced activity: Suppose instead of being straight, our beam were semicircular. It’s clear that if we hang the beam from its midpoint, it still balances by itself. But where should the other holes be so that the same rule as before will determine when ornaments hung on the semicircular beam will balance?

There are two ways of going about this. The more challenging approach gives participants beams with just the “1” holes already punched, setting the goal of punching additional holes labeled “2”, “3”, etc. so that the rule will still hold. (Note that each participant or group will likely need multiple semicircular beams to work with, as it may take several tries to find the correct hole locations.) Alternatively, each participant or group can be supplied with three different possible hole-punched beams, and they can experiment to find which one works.

In either case, participants should be able to discover that the holes punched at equal intervals in terms of *horizontal* distance will follow the same balancing rule as the holes on the straight beam did. This may be surprising, because then the holes are *not* distributed equally around the semicircle. For further mathematical challenge, at the high school level, it’s possible to use a little bit of trigonometry and similar triangles to demonstrate why it is only the horizontal spacing of the holes that matters.

Because of the allure and attraction of gold, our first math walk stop in Fair Park will be to ask, “How much is all of that gold on the Tower Building worth?” Before we go through the calculations, stop and take a moment to guess. Is it 1,000 dollars worth of gold if we crumpled it up into a solid nugget of gold? $10,000? A million dollars? Let’s figure it out.

Since gold leaf is gold spread out thinly over an area, we need to figure out the surface area of the building that is covered by gold. We will model the front of the tower as a giant rectangle, the front of the eagle as another rectangle (which is of course a rough approximation, but it’s a small percentage of the entire gold area), and each wing as another rectangle. So we just need the dimensions of each rectangle.

To estimate these dimensions, we can start with the height of the tower from the top of the frieze depicting the history of Texas to the feet of the eagle. We see that it is divided into nine vertical sections that appear to be equal in height. So we will focus on measuring the height of one of these sections, in particular, the bottommost section.

If you stand dead center in front of the building, you will notice as you walk toward the building or away from it that when you are very close to the building, the frieze section appears taller than the first gold section above it, but when you are very far from the building the gold section appears taller. Move back and forth until you get to a spot where the two sections appear to be exactly the same height (lining a pencil up at about arm’s length so that it appears to match one of the sections in height and moving it up and down to compare to the other section can help).

Once you’ve found such a spot, measure the distance to the center of the front wall of the building, and measure the height of the frieze directly with a measuring tape. (I get that the frieze is eleven feet high and that if I stand 22 feet away, the sections appear to be the same height.)

Now take a look at the following diagram — it’s a vertical cross-section of the situation where your eye is at the point marked “I.” The fact that the two sections appear to be the same height means exactly that the two angles labeled theta are the same. Now with just the two dimensions and a little trigonometry, we can figure out the unknown height of the gold section. If you’ve learned some trigonometry, see if you can work it out.

Here’s one way, in which we’ve cut the lower triangle in half with a horizontal line:

The angle theta/2 has to have tangent 5.5/22 feet, so it is about 14 degrees. Therefore, angle MIA is about 42 degrees, and so leg MA is 22 feet times the tangent of 42 degrees, or about 19.8 feet. Therefore, the gold section BA is 19.8 – 5.5 = 14.3 feet, which we will round to 15 feet for ease of calculation (and because our eyeballing and measuring was very approximate, and the architect was more likely to use round numbers of feet for key measurements of the building).

So the front section of the tower is roughly 9×15=135 feet tall, and we can directly measure how wide it is at the base. I got that it’s about 7 feet wide, but you should make your own measurement.

Now, what about that eagle? We’ll just guess that it’s the same height as one vertical section of the tower; that shouldn’t be too far off. So we’re modeling the front of the eagle as another 15 foot by 7 foot rectangle, and the wings as 15 feet high. So all that remains is the length of the wings, which is the horizontal dimension of the two sides of the tower.

But look closely at the sides of the tower. You will see a faint rectangular grid, with that horizontal dimension three and a half boxes wide. Moreover, two of the boxes side by side appear to form a square! It’s like the builders wrapped the building in graph paper to help us. Therefore, the horizontal dimension we want is 3.5/2 times the height of one vertical section, or about 27 feet.

Adding up all of the sections of gold, there are 135×7 + 15×7 + 15×27 + 15×27 = 1,860 square feet of gold leaf on the building.

How much gold does it take to make 1,860 square feet of gold leaf? Searching the internet, we can find one resource that tells us that 1000 “leaves” of gold cover 79 square feet and weigh up to 23 grams. Dividing, the 1,860 square feet we need to cover would need 1,860/79 or about 23.5 groups of 1,000 leaves, which would therefore weigh 23.5×23, or about 542 grams.

How much is that much gold worth, if it were crumpled up into one big nugget? To get that answer, we have to convert 542 grams into the unit that gold is bought and sold in: the troy ounce. A troy ounce consists of 31.1 grams, so the gold we’d need to cover the Tower Building would weigh 542/31.1, or approximately 17.43 troy ounces.

To finish off, as of the time of this writing gold has been trading in a range around $1,300 per troy ounce for the past year. Therefore, the raw value of the gold on the Tower building is about 17.43×1,300, or $23,000. Is that more or less than you guessed? If it’s less, then it’s probably because gold leaf is so incredibly thin. By comparison, enough household aluminum foil to cover the Tower Building would weigh about 6,700 grams, or over twelve times as much. Combining that with the fact that gold is over seven times more dense than aluminum tells us that gold leaf is almost 100 times thinner than aluminum foil — yet it’s strong enough to last centuries on the outside of buildings!

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