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**These are draft notes from subtitles, please help to improve them.Thank you!**###

Contents

- 1 Introduction to Physics Unit 1
- 1.1 01 l Intro to Class
- 1.2 02 l Intro to Unit 1
- 1.3 03 q What the Greeks Knew
- 1.4 03 s What the Greeks Knew
- 1.5 04 l Previous Attempts
- 1.6 05 q Talking About Error
- 1.7 05 s Talking About Error
- 1.8 06 q Archimedes Guess
- 1.9 06 s Archimedes Guess
- 1.10 07 l Who was Archimedes
- 1.11 08 q Alternatives to Guessing
- 1.12 08 s Alternatives to Guessing
- 1.13 09 l Interview with Enrico Giusti Part 1
- 1.14 10 l Eratosthenes Advantage
- 1.15 11 q Parallel Rays
- 1.16 11 s Parallel Rays
- 1.17 12 q Relevant Angles
- 1.18 12 s Relevant Angles
- 1.19 13 q Relationship Between Angle and Circumference
- 1.20 13 s Relationship Between Angle and Circumference
- 1.21 14 l Interview with Enrico Giusti Part 2
- 1.22 15 l Spotting Triangles
- 1.23 16 q Similar Triangles
- 1.24 16 s Similar Triangles
- 1.25 17 q More Similar Triangles
- 1.26 17 s More Similar Triangles
- 1.27 18 q Side Ratios for Similar Triangles
- 1.28 18 s Side Ratios for Similar Triangles
- 1.29 19 q Practice with Ratios
- 1.30 19 s Practice with Ratios
- 1.31 20 q More Practice with Ratios
- 1.32 20 s More Practice with Ratios
- 1.33 21 q Trigonometric Ratios
- 1.34 21 s Trigonometric Ratios
- 1.35 22 q Using Sin Cos and Tan
- 1.36 22 s Using Sin Cos and Tan
- 1.37 23 q Trigonometry Tables
- 1.38 23 s Trigonometry Tables
- 1.39 24 q Going from Sides to Angles
- 1.40 24 s Going from Sides to Angles
- 1.41 25 q Sides to Angles Practice
- 1.42 25 s Sides to Angles Practice
- 1.43 26 l Trigonometry Review
- 1.44 27 q Solving for Alpha
- 1.45 27 s Solving for Alpha
- 1.46 28 q Solving for Circumference of Earth
- 1.47 29 l Conclusion
- 1.48 30 l The Experiment

[Udacity Inc Presents] [Instructor Andy Brown] [Director of Photography Clay Kempf] [Second Camera and Editor Grant Yaffee] [Filmed on Location in Italy, Netherlands, England.] [Landmarks in Physics]

Welcome to Landmarks in Physics.

I'm Andy Brown, the instructor for this course, and here we are, on location in Siracusa, Italy. This course is really designed for anyone. If you have no background in physics but know a little bit of algebra, you're going to succeed in the course.

In Unit 1, we're going to begin with a question that fascinated the Greeks--how big is our planet? Today, it's easy to answer that question. You can go to Google and type "what is the circumference of the earth?" But 2,000 years ago the Greeks didn't have Google. Still, a man named Eratosthenes was able to answer this question using nothing more than some basic assumptions about the earth and the sun along with an understanding of geometry, trigonometry and shadows.

In this unit, we'll learn everything we need to answer this question ourselves, and in doing so, we'll learn some of the important mathematical tools that we'll be using throughout the course. By the end of this unit, you'll be able to go outside on a sunny day and estimate the circumference of the earth yourself.

Before we can answer the question of what the earth's circumference is, we need to start thinking about what the shape of the earth is. Now, our earth is a sphere..

When we draw this sphere we often include lines of latitude, which go from east to west, and the center line of latitude, which occurs at 0 degrees. We call it the equator.

We also draw lines on longitude, which go from north to south and are all of equal length.

Now, it's a common misconception that the ancient Greeks thought the earth was flat, but this was simply not the case. The ancient Greeks knew the earth was spherical, or at least they had good reason to believe it was. Let's start thinking about geometry by cataloging some of the evidence the Greeks had available to them and deciding which pointed towards a spherical earth.

First, the Greeks knew that the earth casts a circular shadow on the moon during a lunar eclipse. A lunar eclipse occurs when the earth gets between the sun and the moon, and the earth casts a shadow on the moon.

The Greeks also knew that when traveling at sea, you can see farther--a greater distance--from the top of the mast than from the deck of the ship. That's why when you look at old ships, you often see a crow's nest high up on the mast where a spotter could look for land.

Third, the Greeks knew that the moon and the sun appeared as disks in the sky. Finally, the Greeks also knew that the sun appears lower in the sky as you travel north. Now, this only applies for someone living in the Northern Hemisphere, but that's where the Greeks lived. Now, if we want to distinguish between a spherical earth and a disk-shaped earth, we have to use our evidence.

Now, this is the evidence that was available to the Greeks, so I want you to think through each of these four points and decide for each one whether it best supports the idea that the earth is a sphere, that the earth is a disk, or maybe it supports both or neither of these. Go ahead and select one button for each of these four points. Now, this is a tricky quiz, and some of these you could actually debate the answer. So, don't worry if you don't get it right on the first try.

Now, for this first option, I said it could support both of these theories. I could imagine a disk casting a circular shadow on the moon, and I could also imagine a sphere. Now, if you chose sphere, you have a good reason to argue with me, because if the earth were disk-shaped, it would be a pretty huge coincidence of the alignment of that disk for it to actually cast a circular shadow. Think about this one a little more if that wasn't clear, and we can talk about it in the forums. For the second point, when traveling at see you can see farther from the top of the mast, I said this definitely, definitely points to the earth being a sphere. Let's think about why. Here's my zoomed drawing of a portion of the earth, and here's a boat--and exaggerated boat. It's much larger than any actual boat on the earth would be, but that's okay. In fact, this is a tool we're going to use a lot in physics. I exaggerate this drawing because I want to make a point clear. In my head when I'm thinking how a larger ship may allow a person to see farther, I'm going to exaggerate my test case in my head. Let's think of how far this guy could see-- the guy standing on the deck. He could clearly see something down here, no problem. He could see something over here. That's fine. But then this seems to be about as far as he can see. He can never see something over here, because, well, the earth is in the way. Now, this sailor--she's standing up on the top of the boat. This gives her a better angle and actually an unobstructed view towards this farther place on the earth. This clearly points towards a spherical earth, because on a flat disk-shaped earth, this wouldn't happen. Actually, both sailors would be able to see anything. The third option is about the sun and the moon, but we're not talking about the sun and the moon. We're talking about the earth, so I don't think this is really relevant. Again, you can make some arguments about this pointing towards sphere or disk and you should do that. Go to the forums and talk about this point, but I thought the best answer was neither. This fourth point--that the sun appears lower in the sky as you travel north-- I say best supports that idea that the earth is a sphere. And let's talk about why. We'll, here's our spherical earth. Well, it's supposed to be spherical. My drawing may not be the best you've ever seen. And here's the sun. In reality, the sun is much, much, much larger than the earth, and it's much farther away. For our purposes, this drawing will be fine. First, let's imagine this guy. He's standing pretty close to the equator. He's not very far north at all. For him, if he wants to see the sun, he can't be looking in this direction. he's got to look all the way up here. He's going to have to really crane his neck if he wants to look at the sun. We call this "high in the sky." Now, let's imagine his friend--his green-haired northern-dwelling friend. Well, for her to look at the sun it's not very difficult at all. She doesn't have to crane her neck. In fact, just from the geometry of where she's standing and the sun's position, the sun appears very low in the sky to her. This would be really close to the horizon, and she doesn't have to crane her neck at all. This also points toward a spherical earth. You wouldn't see any effect like this on a flat earth. From the data we had or from the data the Greeks had, some of it doesn't support either claim--the spherical or the disk hypothesis. But two of these points really seem to point towards a spherical earth. None point towards a disk-shaped earth, which means we can say with some certainty that even with the knowledge that the Greeks had, we can know that the earth is a sphere. Now, it's this last point--that the sun appears lower in the sky as we travel north that Eratosthenes will exploit to determine the circumference of the earth.

Now, before we get to Eratosthenes' solution, let's talk about some previous attempts made to calculate the circumference of the earth. One famous think, Plato, estimated the earth's circumference at 400,000 stadia. What's a stadia? Well, a stadia is a unit that the Greeks used, and it's a unit of length. In physics, we'll often be using different units. It's not something we should be intimidated by. Actually, we don't even know what the Greeks meant when they say 1 stadium or 2 stadia, but our best guess is that one stadium is about equal to 185 m, which is a more familiar unit and one that we'll be using frequently in this class. Let's convert this number--400,000 stadia-- into a better unit, maybe kilometers. Now, one way to do that is to start by writing out what we were given. We know that we have a number, 400,000 stadia, and I'm being careful to writing the units, because the units are pretty relevant here. I'm going to write this as over 1, because I can always do that without changing the number. I'm then going to multiply this by a conversion factor. The conversion factor that I chose is the one I know-- that 185 m is equal to 1 stadia. The interesting thing about this is that stadia appears here, and it appears here--numerator and denominator--so we can cancel. We now have a number that's represented as meters-400,000 * 185 m. But that's not exactly what I want either. So, we'll need to use another conversion factor, and since I know that there are 1000 m in 1 km, I can write this conversion factor. Notice that I put meters in the denominator so that it will cancel with the meters in the numerator over here. So, we can see all of the units have canceled. All I'm left with are kilometers, and to carry out this calculation, I do 400,000 * 185 * 1/1000. That gives me a result of 74,000 km. Now, the true circumference--and this is the number you get if you type into a search engine the question what is the circumference of the earth--it's 40,000 km. Plato was wrong, but how wrong was Plato? How far off was he? Error is something we often want to talk about in physics, but how do we do it?

Talking about error is something that we do all the time in physics. It's a very important skill to have, but it's not as obvious or straightforward as it may seem. Here, Plato's guess was 74,000 km. The true circumference is 40,000. Well, how are we going to state this error? There are a few options. We could say Plato overestimated the circumference by 34,000 km. We could say the true circumference is about half of what Plato calculated. Or we could say that Plato's guess was 85% larger than the true circumference. Now, all of these statements are true, and they all talk about Plato's error in making his guess about the circumference of the earth. Now, my question is which of these do you think is most informative? If we're trying to gauge the accuracy of Plato's guess, which of these three statements is the best quantification of Plato's error? I realize this is somewhat subjective.

I think that this is the best answer. The problem with this answer, even though it may seem appealing initially, is that without knowing the true circumference, this number--34,000 km-- is sort of meaningless. What if the earth had a circumference of 10 bajillion km? Well, then an error of 34,000 km would be nothing in comparison to that. So, I really have to reference point for what this number means. This is actually a pretty good description of Plato's error-- that the true circumference is about half. But this one's a little better, because it gives me a little more precision. I have an actual number. Plato's guess was 85% larger than the true circumference. This is the best statement of error.

The question is where did this number come from? How did we get at 85%? To calculate percent error is a pretty straightforward calculation. All we do is look at the absolute error. In this case, the absolute error was Plato's guess, 74,000, minus the actual circumference of the earth, 40,000. We'd look at this absolute error, and we'd divide by the true circumference. In this case, 40,000 km. When I do this, I get 74,000 - 40,000, so 34,000/40,000. The absolute error over the known circumference. We can see we can do some cancelling. We're left with 34/40, which reduces to 17/20 or 85%. Now, this is something we'll be doing quite frequently in this class when we want to talk about error. Now, Plato wasn't the only person to make a guess about the circumference of the earth. After Plato, another famous and brilliant think made a guess at the circumference of the earth. That was Archimedes. Now, Archimedes guessed that the earth had a circumference of 300,000 stadia. I want you to tell me how far off was Archimedes. What was his percent error? You don't need to include the percent sign. Some additional information you may need to use to make this calculation is that the true circumference of the earth is 40,000 km. One stadium is 185 m, and 1 km is 1,000 meters. Enter your answer here.

Well, let's see. Before I can make any error calculation, I'm going to need to get 300,000 stadia into kilometers. So, let's go through that process. To do this conversion, we write our original number, 300,000 stadia, multiply it by our first conversion factor in which the stadia cancel out, multiply by our second conversion factor in which the meters cancel out, and then we're left with 55,500 km. Now, we can calculate the error by looking at the absolute error, which is how far off Archimedes was. That's 55,000 - 40,000 and dividing that by the true circumference, which we know is 40,000 km. When we do this, we get 15,000/40,000, which is 15/40, which is 3/8, which is 37.5%. He was still closer than Plato, but quite a ways off. Then since both Plato and Archimedes were just guessing, I don't know how either of them had any confidence that the numbers they made were correct. Now, Plato and Archimedes were both really smart guys, but both of their guesses were quite far off from the true circumference of the earth. Let's go back to Syracuse, the birthplace of Archimedes, to learn more about some of the accomplishments he made over the course of his life.

Archimedes was born about 2,300 years ago, and he made important contributions in math, physics, engineering, and astronomy. In physics, Archimedes was the first to explain the principle of the lever, which we will be talking about quite a bit in the third unit. He invented an ingenious device known as the Archimedes screw, which can be used to move water from low places to high. Many of his inventions were actually used to defend Syracuse, which in his day was frequently attacked. He created catapults and ballistas, which could fire projectiles huge distances when enemy ships attacked. He's even rumored to have constructed a heat ray, which used mirrors on the bluffs of Syracuse to focus the sun's rays on an attacking ship. In math, Archimedes was able to calculate the area under a parabola. This is a shape that we'll be using extensively in the next unit. He defined the Archimedes spiral, which you can learn more about in one of the challenge homework problems in this unit, but the discovery that he was most proud of was his proof that a sphere inscribed in a cylinder has exactly 2/3 the volume and the surface area of the cylinder.

Now, despite being a really brilliant thinker, Archimedes estimate of the earth's circumference seems to have been a guess. And that's not the best way to do physics. Let's look at alternatives to guessing. One thing we could do to maybe improve on just a blind guess is we could have many people make guesses and then take the average. Or we could look at the geometry of a sphere and somehow find a clever way to calculate the circumference. We could also wrap a really long string around the earth and measure the length. Or we could determine the circumference of the moon and assume the earth is the same size. Of these four alternatives, which do you think is the best? Just choose your favorite answer.

Now, I imagine you might have suspected that this was the best answer. Look at the geometry of a sphere and find a clever way to calculate this circumference. Now, you may have been tempted by this one, but, well, yes in science it is generally good to take a lot of data. That's a great habit to be in. But if every single piece of data you're taking is just a wild guess with no real reason to believe it, this isn't going to help. This one's a little impractical--wrapping a string around the earth-- and there's no reason to believe the moon and the earth have the same circumference, not to mention, how do we calculate the circumference of the moon? This leaves geometry, which means this is a job for math. Let's go to Florence to learn a little more about Eratosthenes' approach to to measuring the circumference of the earth.

[Enrico Giusti, Curator, Garden of Archimedes] Eratosthenes was the first to measure the diameter of the earth. He used astronomical techniques to perform this task. The idea was that, in Summer Solstice, the sun was vertical on Syene, but it's more angled in Alexandria. And Eratosthenes measured the distance between Syene and Alexandria, and, from this and from the measure of the angles, was able to deduce the angle between the two cities and, therefore, the radius of the earth. The sun is vertical on the Summer Solstice, because Syene is on the Tropic, so you have no shadow. Actually, Eratosthenes noticed that, from the bottom of a pit, you could see the sun exactly vertical on your head. So, for this calculation, Eratosthenes used only a little trigonometry, and this was all.

So even though Eratosthenes and Archimedes were both spending a lot of time thinking about the earth and, in fact, the earth's circumference, Erathothenes seemed to have an advantage. So because he was so well-traveled, Erathosthenes knew that, here in Syene, the sun's rays struck vertically on the Summer Solstice, which happens on about June 21st or the 22nd. He knew this because these rays of the sun actually made it all the way to the bottom of the well, which wouldn't happen if they were coming in at some other angle. He also knew that in a city a little bit north of Syene called Alexandria, the sun's rays did not strike exactly vertically. He knew this because he could hold a metal bar or any other straight object vertically so that it pointed directly into the ground at Alexandria. And if the sun were overhead, this shouldn't cast any shadow, but it did. It turned out this actually made a small shadow. Okay. So now we're getting somewhere. It looks like to solve this problem, we're going to need to really explore the geometry involved, and since we use geometry all the time in physics, now's a good time to learn or if you already know, review some of the basics, so let's back up a bit.

So here we have our circular earth, which I'm trying to represent in 3 dimensions by showing the equator, and somewhere here, we have Syene, and somewhere here-- and I'm going to exaggerate this distance to make the drawing easier--we have Alexandria. Now to begin solving this problem, Eratosthenes did what physicists often do when they're tackling a really difficult problem. He made an assumption. And now the assumption that he made was that the sun was so far away, way, way, way off the screen, that when the sun's rays arrive at the earth, they're coming more or less parallel no matter where on the earth you are. Now these lines are parallel-- we assume they'll never touch-- and why was this a safe assumption to make? Well, let's take a look. Let's draw a little play version of the sun over here, and now this is totally separate from what's going on over here, and let's look at the sun's rays. Here they are, radiating outwards from the center from the sun, and you can tell that they're certainly not parallel. What happens is we get really, really, really far away from the sun like way out here, where we've got more rays like this. We can start to see that if we're really far away from the sun--maybe if our earth is over here-- well, in this neighborhood, the rays do look more or less parallel. They're not exactly parallel--they seem to be diverging a little bit--but maybe it's good enough for our purposes. We'll examine this more in the homework. Okay. So we have our rays arriving parallel at Syene and Alexandria. Now what? Now how am I supposed to figure out the circumference of the earth based on some parallel rays and the observation that they're striking vertically here at Syene. Well, before we jump into the calculation of the circumference, let's try and understand what's going on here first. Well, we've already talked about a person standing here in Syene--their shadow would be directly below them, because the sun's rays are striking vertically. Here in Alexandria, a person would have a tiny little shadow, and you can see that the angle from the head of the shadow to the head of the person-- this line actually follows the sun's ray. We'll be using that more later, but, for now, let's think. If this person started walking north, closer to the pole up here, what would happen to his shadow? Would his shadow lengthen as he walked north, or would it get smaller?

And I said it would lengthen. We can see that as we walk toward Syene, it must get smaller, because, in the limiting case where we actually arrive at Syene, we have no shadow, so it seems reasonable that as we get further away, the shadow gets bigger. Now it's good to do these sort of qualitative thought exercises, but, okay, it gets bigger. How much bigger? Can we describe that mathematically? And this is a perfect example of encountering a problem in physics where we don't have the tools just yet to solve it, so let's build up some of those tools.

Here we've zoomed in on the relevant portion of the globe. We can see Syene and Alexandria. Now, here the sun's rays strike vertically, which by vertical we mean perpendicular to the earth. Here at Alexandria, it doesn't strike vertically. If I drew a line perpendicular to the earth, I can see there's some angle here, which we don't know just yet. Now, we're trying to find the circumference of the earth. We found out that as we move further north from Alexandria this angle actually gets bigger and the shadow gets bigger as well. It seems like there's some relationship--hopefully a mathematical one-- between this angle and the circumference of the earth. Let's see if we can figure it out. I'm actually going to start by making a wish list. This wish list is going to contain everything that I wish I knew in order to solve this problem. Hopefully, we'll be able to come back to it later and address everything on it. I wish I knew this angle, and I'm going to call it the Greek letter a. We're dealing with the Greeks. Why not use the Greek letter? It's pretty common to call angles a in physics. Okay, so on my wish list, I wish I had that angle--the angle a. Now, what are we going to do with this angle a? Let's not even worry about how to calculate it yet. What are we going to do with it? Well, whenever I get stuck on a geometry problem, I start drawing lines, and they usually help me. Since a circle is involved here, or a sphere-- but we're only looking at a circular cross section of it-- drawing lines from the center is usually a good idea. So, I'm going to draw a line from the center to Syene and from the center to Alexandria. Okay, now there's still now a ton jumping out to me. Maybe I'll extend this line--that sometimes helps me see what's going on. Ah! Okay. Something just jumped out to me. This angle a shows up in more than one place. I'll tell you there are other angles on this drawing that are equal to a--maybe one, maybe more. I'm going to mark some contenders and why don't you tell me which ones are correct. What about this angle? Is that equal to a? Check that box if you think it is. Or what about this angle? Is that equal to a? This one? Or what about this one? For this quiz, just go ahead and check all of the angles that you think are equal to this one.

Let's see. This one seems to be clearly not equal to a. This is a huge angle--greater than 90 degrees. But I do know that this angle is identical to a. The reason I know that--well, they're opposite each other, so that's always going to be the case that opposite angles are identical. But let's talk about why. If I examine this entire arc here, I know a straight line--well, I've made half of a circle here, so this is 180 degrees. This angle here is some a degrees-- I don't know what that is--which means that this angle marked in red must be 180 - a. Now, since this line here is also equal to 180 degrees, this segment of it is 180 - a. This segment must also be equal to a. But that's not so interesting. What I care more about is this angle here, which also happen to be equal to a. Again, we can talk about why. Let's extend this line and this line a little bit. Now we have these two parallel lines. The black line here is parallel with this purple line up here. We can go through this same sort of thinking to prove that this angle is a, this one here is 180 - a, this one has to be a, and then this one over here would be 180 - a. But this is what we care about. This angle here is identical to this angle here. Now we're making some real progress. Let me clean this up a bit so we can see what's going on.

Now here we have the earth, Syene where the sun's rays strike vertically, Alexandria where they strike at an angle a, and this angle ais the same as this angle a, and we're hoping we can use that to calculate the circumference. Well, let's see. We know there's 360 degrees in a full circle. This alpha is going to represent some fraction of that 360 degrees-- maybe a half, maybe a quarter--we don't know yet. What we do know is that the ratio of this length-- I'm going to call this length d--to the full circumference of the earth must be equal to the ratio of this angle over 360 degrees. Now, I'll talk about that more in a second, but let's just write that down. The ratio of this distance d to the full circumference of the earth is also equal to the ratio of a to 360 degrees. A good way to confirm that this equation makes sense is let's think of some limiting cases. What if a were 0? That would be a really, really tiny angle covering absolutely no distance. That would mean this length would have to be 0 as well. Good. Now, what if a were a full 360 degrees? What would d have to be for this formula to make sense? Check the appropriate box. Would d have to be 0, 1, or C--the circumference of the earth?

That's right. It would have to be C. If a is 360 degrees, that means we're looking at the entire circumference. This d would be identical to C. Now, we're starting to see an equation emerge. This circumference somehow depends on this distance and this angle a. Now, I guess we'll have to add the distance to our wishlist, because we don't know that yet either. Before we go any further, let's just rewrite this equation. Let's solve it for C. I'm going to bring this C up here, and I'm going to bring these--a and 360--over to the other side. So, this is it. This is our answer for the circumference of the earth. We just need to know a and d. Now, let me just talk about this equation a little bit before me move on. First of all, there is nothing wrong with the way we wrote it before. This was a totally true expression that you could use to calculate the circumference. The reason I like this a little better is because it explicitly tells us this is the circumference. Then the way we've written it over here is nice as well. I see I have here 360/a. Now, these are both in degrees. When I do degrees divided by degrees, I get some unit-less number. This is just some multiplier that multiplies a number with a unit of distance. I can see just by looking at this equation that what I will get is a unit of distance. And I can see how it depends on this distance from Alexandria to Syene and the angle of the shadow. I think that's a nice way to show this equation. Now, let's just worry about this angle and the distance and keep track of our units as we do. This may require a new mathematical tool as well. Now, let's go to Florence Italy to talk to an expert in the field.

The angles he measured by degrees, of course. [Enrico Giusti, Curator, Garden of Archimedes ] For the unit of length, he used the stadium, but he didn't measure with a rope or something, but he estimated the distance between Syene and Alexandria by the time that a camel would need to go from one city to another. Eratosthenes measure was rather accurate for the time. Even for modern standards, it was say about 10%. The source of error was essentially the estimation of the distance, because the angles he could measure with a good accuracy.

All right. So, we timed a camel and got the distance d. Maybe not the most accurate of odometers, but, hey, we'll talk about error later, and we'll go with this for now. The distance he measured was 5,000 stadia--we see that strange unit popping up again. Now we just have to get a, which will require trigonometry-- the study of triangles, specifically right triangles. Now, before we get jumping into the study of right triangles, let's just quickly show where they're going to show up in this problem. I mentioned before that if someone were standing here-- I'm going to represent someone standing by vertical black line-- he would cast a shadow like this. You see--straight line, right angle--90-degree angle--straight line, and we can imagine the sun's ray completing the triangle. Now, hidden in this right triangle are the mysteries of the circumference of the earth. Let's figure out what's going on inside a right triangle.

Let's learn some trigonometry. Now, if you've already mastered trig, you don't have to watch this next part. But you may be entertained, and if you haven't mastered trig, you're in a for a treat, because this is one of the branches of mathematics that we use all the time in physics, and it's really fascinating. Let's start by looking at a right triangle. Isn't that nice with it's right angle marked by a square and it's three sides? Okay, what can we learn from this triangle? Let's first make a copy of it. Here is a copy, identical in every way-- same angles and same sides. We call that "congruent." That just means we have the same sides and the same angles. If this has a length of 7, this has a length of 7. If this has a length of 12, this has a length of 12. Likewise for the angles. What if this triangle were a little bigger? What if I were to expand it proportionally so as it grows it grows the same way in each direction the same amount in each direction? Well, clearly this is not identical to this triangle. These two are not congruent. But they are similar in some ways, so let's come up with a name to describe two triangles that are similar in such a way. In fact, let's call them similar. These are similar triangles, and they're so vital to doing physics. Now, each of these triangles has three angles. This one I've labeled 1, 2, and 3. This one 4, 5, and 6. Do they have anything in common? Are they equal? Is this statement true, that angle one equals 4, angle 2 equals 5, angle three equals angle 6? Check which every of these statements are true about these two triangles.

They're all true. Similar triangles have exactly identical angles. In fact, we could have rotated this triangle. We could have moved it around. We could make it bigger or smaller. As long as these three angles are identical to these three angles, it doesn't matter. These triangles are still similar. Let's do a quick little quiz.

The quiz is which of the flying triangles are similar to this triangle here? Just check all that apply.

I said that this was a similar triangle and this was a similar triangle. Let's take a look why. If I rotate this one just right, make it a little bit smaller, I've got the exact same triangle I had before. Likewise for this one. If I move it, rotate it, and make it a bit bigger, I have the same triangle I had before. Since I can't do that with any of these three triangles, they are not similar.

Now, I'm being a little cryptic here. What do I mean by "ratios?" Let's go back to our similar triangles to figure out what I mean. Now, I have two triangles here. They're similar because the angles match, but the sides are different. I'm going to label these sides a, b, c and these sides A, B, and C. You can see that even though these sides are all longer, there's some correspondence between A and side a. Likewise with C and c and B and b. But what exactly is that correspondence? What do these things all have in common? The fact is they have a ratio in common. What I mean by that is they're linked through ratios. If I look at the ratio side A to side B, I'm going to find that's exactly the same as the ratio of side a to side b. In fact, it doesn't matter how big or small I make this triangle, this equality will always hold true. And that's a really powerful mathematical tool that we can use. For example, let's say I didn't give you variable names here, but I gave you some actual numbers. So, 3 m, 4 m and 5 m are replacing our sides from before, and these are still unknown. But let's say I told you that--I don't know--side B was equal to 6 m. I can use this to calculate side C, because I know that C/B has to equal, well, the corresponding sides--4/3. More precisely 4 m over 3 m. Notice that when we do this, the units cancel out, so it doesn't even matter that we were using meters. We could have been using inches or stadia or furlongs or any unit we'd like. Now, I know this ratio--C/B--is equal to 4/3. But notice that I have one other piece of information. B is actually equal to 6 m, so let's write that in. Now if I want to solve for C, I multiply both sides by 6 m, and that gives me the following: C = 4/3 * 6 m, also known as 8 m. Okay, that wasn't so bad. We used the power of ratios to calculate an unknown side of a triangle, and this is a huge tool to have at our disposal. Can you do the same to solve for side A? How long is side A in meters. And you can just enter the number. You don't have to type in the m.

Side A is 10 m. There's a variety of ways you could have done this. You could have compared A to B or A to C. In fact, you could have used some other tricks. You could have used the Pythagorean Theorem or compared across these triangles, but we're not going to worry about that right now.

Now, let's do some practice. In this question we're using this triangle--5 m, 12 m, and 13 m on the side, which makes this angle a 22.6 degree angle, and we don't know how to calculate that yet, but we will shortly. My question, before we begin calculating ratios is whether this triangle-- and all I've told you about this triangle is that it has a right angle here and a 22.6-degree angle here-- I want to know are these two definitely similar? What's the best answer? Yes, they are definitely similar, they could be similar, but maybe their not, or no they are definitely not similar triangles.

The answer is that they must be similar. They absolutely have to be. The reason is that even though I only explicitly gave you two angles-- the fact that this is a right angle and that this is 22.6 degrees-- we also know that the angles of a triangle must sum up to 180 degrees. Well, what number adds to 90 and 22.6 to give you 180? That's 57.4 degrees. So, both this angle and this angle are 57.4 degrees. Since that means that these triangles both have the exact same three angles, they must be similar.

Now that we've shown that 2 angles is really enough to prove similarity, let's answer some questions. If I told you that this side had a length of 18 m, what would the length of this side be in meters? Again, you don't need to put the m. I'll put that there for you.

To solve this, I looked at the ratio--the ratio of this side to that side-- and compared it to this and that. Let's see--over here we have 5 m over 12 m, and I know that must, must, must be equal to whatever this side is--I'll call it x/18 m. Again, these units cancel out. This ratio 5/12 seems to be pretty important. I'm going to multiply both sides by 18, and when I carry out this multiplication, I find 7.5 m. Now, this ratio, 5/12--I'm going to put it in parentheses to show that it's important-- every single time we see a 22.6 degree angle in a right triangle I know that look at this side over that side will give me this ration 5/12. That may seem trivial, but it's actually deeply, deeply important. In fact, since we're going to be using these ratios so often, let's agree on a way to refer to the sides of a right triangle.

If we have a triangle like this one here and we're looking at this angle, this is what we call our reference angle for now. We have three sides we can talk about. Now this one, the largest side, will always be called the hypotenuse-- so I'm going to abbreviate the hypotenuse with Hyp. This side here at the bottom is opposite the angle we're referring to so let's call that the opposite side. This side being next to the angle is adjacent. We're going to label that AdJ. Adjacent just means "next to," and one thing to be careful of is not to call this side the adjacent side. Even though, yes, it is next to this angle, we give this one the special name hypotenuse because it is the longest side. Now if we think back to our sides from before or side lengths. We have 5 meters, 12 meters, and a hypotenuse of 13 meters. What we now know is that anytime we have a similar triangle to this one-- a right triangle with an angle of 22.6 degrees here-- the ratio of opposite to adjacent will always, always be 5/12. This ratio is so important that we're going to give it it's own name. The first trigonometric ratio you're going to learn is the tangent, and the tangent of an angle is always equal to the opposite over the adjacent, which in this case would have been 5/12. I would write that by saying the tangent of 22.6 degrees equals 5/12, but that's not the only ratio we can talk about. We can also compare the opposite to the hypotenuse. When we do that, we're using what's called a sine, so I should note that I am making abbreviations here--tan for tangent sin for sine. Sine is just the opposite over the hypotenuse. Can you tell me what would the sine of 22.6 degrees be? Remember the tangent of 22.6 was 5 over 12, what's the sine? Enter in your answer here--you can enter it as a fraction with the numerator up here and the denominator down here.

Well, sine is opposite just 5 meters over hypotenuse which is 13. Notice I didn't actually include the m when I was writing because it's a ratio and the unit will cancel out. The last ratio we have is the cosine abbreviated cos. This is equal to the adjacent side over the hypotenuse. The cosine of 22.6 degrees is 12 over 13, so here you have your trigonometric ratios. A common way to remember these three ratios is with the following mnemonic-- SOH CAH TOA--this means that the sine is equal to the opposite of the hypotenuse, cosine adjacent over hypotenuse, and the tangent is opposite over adjacent. If this helps you remember these three functions, that's great.

Let's get some practice using these ratios. Here we have a right triangle. This angle is 37 degrees and the sides are 3, 4, and 5. I want to know what's the sine of 37 degrees, and it's the opposite over the hypotenuse so in this case 3/5. What about the cosine of 37 and the tangent of 37?

The cosine is adjacent over hypotenuse. This is the adjacent side. This is the hypotenuse 4 and 5. And the tangent is opposite over adjacent just 3/4.

Okay, you're doing great. You've really learned a lot of powerful mathematics in the last 20 minutes, so let's think how we're going to use this. It doesn't seem to make sense to keep calculating the sine of 37 degrees every single time we encounter a triangle with a 37 degree angle. This number is never changing. This sine will always be 3/5. What they did in Eratosthenes' day was calculate large tables of trigonometric ratios. They would find the sine, cosine, and tangent of a variety of angles and save it away for future reference--here's an example table of trigonometric ratios. In the columns, we have sine, cosine, and tangent and we have these values computed for, well just a few degrees here--40, 41, 42 degrees. In a better table, we'd have a larger range of angles and higher resolutions so we'd have 40.1 degrees, 40.2 degrees, and so on, but this will work for now. What this table allows us to do is something really amazing, so let's say we have a triangle and it's a right triangle with an angle here of 42 degrees and let's say we know this length its 1000 meters. How can I find this length? This is something we haven't done yet. What I can do is say, "Ah, I know that the ratio of this side to this side is totally independent of how long these sides actually are. " In fact, this side being adjacent to 42 degrees and this side being hypotenuse looks like we're going to want to use a cosine which compares adjacent to hypotenuse. In fact what I can say is that the cosine of 42 degrees which I can look up in my table is 0.743 must be equal to the adjacent side, I'll call that x for now, over the hypotenuse and the hypotenuse is the length of 1000 meters. This is a huge, huge tool to have because now all we have to do is multiply by 1000 to solve for x and we find that x is equal to 743 meters. Now I'm going to give you a shot. Let's call this unknown length y. Can you go through a similar set of steps may be you won't be using cosine this time to figure out how long is this side y.

Let's see. How do we want to do this? Now, I'm looking at the side opposite this 42 degree angle. And I'm going to be comparing it to my known side of the 1000 meters. You also could've compared it to the x, but I'm going to do it this way. So, opposite and hypotenuse--it looks like I'm going to need sine. Okay. So sine of 42 degrees I know that equals 0.669. But I also know that it must equal the opposite over the hypotenuse. That's what the sine is and that equals y over 1000. Now, I can multiply both sides by 1000 just like I did before to get that y = 669 meters. Pretty cool. Now we're really getting to the point where we can start to calculate the circumference of the area. Before we move any further, there's one more tool I think we should learn.

What we've shown already is that if I know a single one of these angles and one of the sides as well, I can learn everything there is to know. I can learn the other angle by using the fact that they must add up to 180 degrees. I can find the other sides by using these Trigonometric Ratios. Great! But there's one thing we can't do quite yet, and that's to go from sides back to angles. So, let's say I gave you a right triangle and I told you this side was 231 meters and this side was 266 meters. All right, let's say now you want to know something about one of the angles. This angle for example. Now, initially this seems a bit tricky and it is. Well, what we can do is even though I don't know what this angle is yet, maybe i know something about this angle. For this angle, would it be easier for me to calculate the sine, the cosine, or the tangent of this angle? Check up here which should be easiest. Tangent, sine, or cosine?

Since I have the side that is opposite and the side that is adjacent, remember this is adjacent and this is hypotenuse, that looks like tangent would be easiest. Why don't you go ahead and calculate the tangent of this angle? And I'm going to start calling it angle--I don't know--a. It seems like a good letter. And I know the tangent of a is equal to the opposite over the adjacent. And when I do this, I get about 0.8684 and that's really basically 0.869. We're very close to this. I can actually figure out, hey, this must be a 41-degree angle more or less. We're not at exactly 0.869 but close enough. Using this fact, I can say with confidence that alpha is a 41-degree angle. And now maybe you are seeing how we can use trigonometric ratios to calculate the angle at which the sun's rays are striking the earth in a particular location. We're making some serious progress. Before we move on, let's do one more practice problem.

Here I've given you a right triangle and I told you two of the sides - 401.0 meters and 523.5 meters. Now, I want to know, using this table of trig values and what you know about Trigonometric Ratios, can you tell me what is a and enter your answer in degrees. You don't need to enter the degrees. We'll assume they're there.

And the best way to solve this is to say well,"What sides do I know?" Relative to a, this is the adjacent side. This is the hypotenuse. If I look over here, I see that cosine uses the adjacent and the hypotenuse. Okay, so cos a is equal to 401.0 over 523.5. When I plugged that in, I get an answer of 0.766. Now I can go back to my table of cosine values and see--oh hey, 0.766 that's the cosine of 40 degrees. Alpha must have been 40 degrees. Great work. This is a lot of material we've just covered. We've learned the first two weeks of any normal trigonometry class, and you've done it all just now--I'm very impressed.

Let's quickly review what we learned. Well first, we learned about similar triangles. Two triangles are similar if their angles are identical even though their sides don't have to be the same length. And we've learned that these triangles have many special properties. One really important property is that the ratio of side lengths are the same for two similar triangles. What this means is that if I have a right triangle with a 40-degree angle here, I can immediately tell you the ratio of this side to that side or this one to that one. And that's a very powerful tool. In fact, these ratios are so powerful that we gave names to them - sine, cosine, and tangent. And we can remember what those names mean by SOH, CAH, TOA. Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. Finally, we learned that we can go back and forth between sides and angles and side lengths and angles or maybe even degrees. And this is something that we'll be doing frequently throughout the course. Let me say good work. This is a huge accomplishment learning this much already. Now, let's put it to use and figure out what the circumference of the earth is.

Now, we can go back to the original problem--the thing we've been trying to solve this whole unit. As you remember we have the sun's rays striking perpendicular to the earth here at Syene. We've seen that these rays all came in parallel like what's around here and here in Alexandria which is north of Syene, the rays no longer strike the earth perpendicular to the surface of the earth. They strike at this angle a. We showed before that must be equal to this angle a as well. And by comparing the full angular measure of the circle 360 degrees to this portion a, we will define that if we knew d and we knew a, we can calculate the circumference of the earth. We got d by timing a camels distance from Syene to Alexandria. Now we just need to find a. To determine this angle a, let's imagine what it would be like to be standing here in Alexandria. Well, if I were actually standing there, I can now say the earth is flat because I don't notice the curvature of the earth on a daily basis and I can imagine sticking some sort of pole into the ground. And maybe I know the length of this pole. Now I know the sun's rays are coming in at a certain angle. I'm going to draw the rays in red just to make it a little easier to see. So here comes the sun's rays, you can tell they're being blocked by the pole in some places. So when the pole blocks the sun's rays, we get a shadow--here's the pole shadow. And now you can see we're almost there. You can see the right triangle that's emerged. We have the sun's ray that just barely missed the edge of the pole, we have the shadow of the pole, and we have the pole itself forming a right triangle. Now from this drawing, you can actually see that the angle we called alpha. If I imagine making a perpendicular line here perpendicular to the earth, this is the angle we called alpha. There's no problem. That's also equal to this angle a and now we're almost there. So here's our triangle with three sides--one, two, three, opposite, adjacent to a, and hypotenuse and we just need to know what this angle is. Now that we have a right triangle, we just need to consult the trigonometric table and we should be able to figure out what alpha is and what's the circumference of the earth is. So we set up the experiment. We have our vertical bar with its shadow and what Eratosthenes data may have looked like with the length of the bar would have been something around 1 meter and the length of the shadow about 0.126 meters which is 12.6 cm. He, of course, has access to his trigonometric tables and here's a portion of one such table. Now, can you put yourselves in Eratosthenes' shoes and tell me what is the value of alpha?

Well, we have a right triangle--we know something about the side opposite a-- that's the length of the shadow--and the side adjacent to a--that's the length of the bar. Opposite and adjacent are the tangent, so that tells us the tangent of a is equal to the opposite Now, I know that the tangent of this angle is 0.126, I can go into my table of trig values, and I see--hey, that corresponds to an angle of 7.2 degrees.

Now for the moment of truth--calculating the circumference of the earth in the same way that Eratosthenes did. Now we have this data. The important stuff seems to be this angle alpha. This distance d which was the distance from Syene to Alexandria and that's 5000 stadia and remember 1 stadium was equal to 185 meters. We have this equation which we found out from the geometry of the earth, and now I want to know can you tell me in kilometers what is the circumference of the earth.

Well, to solve this, we can plug in C equals 360 over alpha, which was 7.2 degrees times D, which is 5000 stadia. When we did this out with the degree is cancelled--we get a value for the circumference of 250,000 stadia. Now we're still not done because we need to convert--we're going to take our number so we'll multiply it by 1 kilometer for every 1000 meters. When we do this, we get a final answer of 46,250 kilometers. Amazing! We did this all just with some shadows, some basic knowledge of geometry, and a little bit of trigonometry that we have to learn as well Not only is it amazing that we could calculate the circumference using such basic knowledge, but the skills that we gained in doing so are skills we used throughout this course--congratulations! Now, let's see the experiment in action.

I'm in Syracuse with my assistant Roberta and apparently most of Syracuse as well. We were going to do the experiment we've been learning about--Eratosthenes' experiment where he determines the circumference of the earth measuring shadows. All you're going to need to do to perform this experiment is get some help. Thank you Roberta. >>You're welcome. You're going need a string or some sort of straight stick, a weight on the end of that string to make sure that it's truly vertical and some sort of measuring device. Now my measuring device is a ruler and it's marked in inches. You might have one marked in centimeters, but like we mentioned before the only thing relevant in this experiment is the ratio and the units don't matter. It's really not important. Now let's go ahead and make our first measurement. Can you just put it this way--perfect--and down so your thumb touches it--perfect, perfect. And the length here is 42 inches. Now without changing anything, without moving anything. Rebecca is holding perfectly still. Gracia! I'm going to make the next measurement which is that of the shadow and I get a result of 8-1/2 inches. Now you'll notice over here, it's not really clear where the shadow begins and ends. This is uncertainty in my measurement and that's something we're going to have to deal with. When you do the experiment at home, you'll have uncertainty as well. In fact, uncertainty is inherit to any measurement you make in physics. We'll talk about that later but thank you very much Roberta. >>You're welcome.