Ancient Greeks WERE smart.
It’s not just the arabic numerals. Imagine doing this stuff without algebra. They did it using geometry.
I can “prove” that the area of a circle is pi times the radius squared using limits. Greeks didn’t have the concept of limits (see my sig line). They used incredibly ingenious methods to perform their proofs. Dig up a copy of Euclid some time and flip through it.
Spanning the globe…with mathematics?. Also started by So Far So Good
CalMeacham, that method requires the angles to be measured at noon and not at any other time. Simultaneous is not enough, it has to be at the instant of both sites and the Sun being on the same meridian (therefore same plane).
It is obvious if you think the sun is vertically above one of the places. It is neither north or south nor east nor west, it is directly above. By the way, I seem to remember the place was Siena in Egypt.
As for the original question, I am sure it can be resolved quite easily but after having resolved two similar problems in this thread (as you well remember) I am staying out of this one.
Just FTR, one reason I have some ease with this type of question is that I know celestial navigation which requires some basic understanding of trigs and spherical triangles etc.
JonF–Thanks for posting the link to the previous question. This one takes up where that one left off–suppose you can’t see the whole globe, just a “chord” of it. If the answer is as you explained in the earlier thread, I don’t see it (dumb me 
). Once again, all help is appreciated.
Sailor:
Quite correct. But rad my posts again. You’ll find I said everything you did, except that it had to be at noon.
Of course CalMeacham used the height above the surface… If he had a way of finding his height from the center, he wouldn’t need to go through all of the trig, he’d just subtract height above surface from distance to center-- Done.
>> You’ll find I said everything you did, except that it had to be at noon.
Yup, and the only thing I was adding is that it had to be at noon
Of course, it is implicit in this:
>> Find a point where you know the sun is directly overhead at some exactly known time
Well, what is that “exactly known time”? Noon! it cannot be other. Your phrase gave the impression it could be any known time
To expand on this, the sun can only be directly overhead at local noon and at a place that has the same latitude as the Sun’s declination
That’s right, I believe it’s spelled Syene and it’s right near the Aswan Dam, and the Tropic of Cancer passes through. Eratosthenes had never been there, he read about the sun shining down the well on the solstice, and checked to see if it would work in Alexandria on the mediterranean coast. He found that a vertical stick cast a shadow at noon indicating that the sun was 7 degrees from the vertical. He reasoned that he must be 7 degrees north of Syene, and hired some guys to pace the distance, figuring it was about 1/51 the distance (7/360) around the Earth.
I forget the exact number he came up with, but I recall it was a few tenths of a percent off the actual value (~40000 km). I’ve always felt that more credit should have gone to the guys who had to walk ~1600 km (round trip), counting every step…
Today we think once something is discovered, it has been discovered forever but here you see this piece of knowledge was half forgotten until the 18th century when the France, England and Spain did some work to determine the circumference of the earth or as they put it: “to determine the length of the sea mile” (this is because the sea mile was defined as 1’ of the earth’s circumference)
And to add to the horror stories about teachers who don’t know sh*t, I will add my own. When I was a teenager the teacher told us about the efforts in the 18th century to determine the circumference of the Earth. I asked him exactly how this would be done and his response was that it was done “by long measuring ropes from one ship to another”.
It makes me smile now. How ignorant can you be?
Picky, picky, picky
So Far So Good, I want to help answer too, because I think people are simplifying the problem too much. (Yes, you should worry when you hear that.) Everyone so far has assumed that you know which direction is down. But what if you don’t? If you’re up there in microgravity, then you can’t get a reliable reading, particularly if you can’t see the whole Earth.
If, in fact, you don’t know which way is down, you can still calculate the radius of the Earth, but it takes a little more. You still need a measuring device. My device of choice is missiles that I can shoot out toward the Earth. These missiles travel in a straight line at a known speed, so when I see the huge explosion go off on Earth, I can figure out how far away it is, because I timed how long it took to hit. In order to perform this calculation, you need four missiles, aimed in four different directions.
You also need a sextant, so you can measure the angular distance, as seen from your position, between each of the explosions. Then, using the law of cosines, you can determine the distance between any two explosions. Here’s how that part works:
Suppose you have measured the distance to two exposions on Earth. One is 1000 km away, and one is 1016 km away. You also see that the angular distance between them is 4°. Law of cosines, as I’m sure you know, then tells you that the distance d between them is given by:
d[sup]2[/sup] = 1000[sup]2[/sup] + 1016[sup]2[/sup] - 2(1000)(1016)×cos(4°)
or in this case, d = 72.15 km. You’ve got four explosions, so you’ve got 6 distances to measure (each one to each other one).
After you’ve done that, there are a few ways of calculating where the center of the planet is. There’s a Geometric way, a Vector way, and a really pain-in-the-butt-like Algebraic way. The reason I took so long to post is that I was trying to simplify that last one. I’ll explain any of these if you like, provided CalMeacham doesn’t beat me to it. But the point is, given these four points and the distances between them, you can compute where the center of the planet is. After that, just figure the distance between any one of the four points and the center, and there’s your radius.
Of course, nobody would ever measure the size of the Earth this way, because it’s not a perfect sphere. Oh yeah, that, and you usually can see the whole Earth. Oh yeah, that, and nobody has missiles like that. Oh yeah, that, and nobody wants to shower the Earth with missiles just to get a measurement. But other that that, it’s a sound method.
By allmeans, Achernar, go to it.
But for my money, you like blowing things up to much. I’m into non-desructive testing, myself.
Fact of the matter is, we don’t know how accurate Eratosthenes’ measurement was, we can only estimate how accurate it should have been. You see, he reported his answer in stadia. The problem is, nobody today is sure just how long a stadium, as a unit of length, is. In fact, the usual method is to assume that his measurement was accurate, and then to use that and the known radius of the Earth to determine the length of a stadium.
And Achernar, if you can see a segment of the Earth, you can extrapolate it to a circle. The center of that circle is down.
Achernar, I will study your post & see what I can find a way to use. I would rather not make war against the Earth to answer my question, but desperate times call for desperate measures !!
The kind of answer I am looking for is one that requires observation and measurement only–no missles, no lasers, no radars, no nothing except a view of a “chord” of the earth that I can see out the wondow/porthole.
Whether the Earth is a perfect sphere or not isn’t as important as the measurements and math I must have and use to get an answer.
CalMeacham, I tried to contact you through the site you mentioned earlier–did I succeed? There was nothing there that seemed to point directly to you.
The definition of the problem is so muddled by now nobody knows what the heck the problem is. I mean, every other post adds or takes away some condition. I think it would be good if you would restate it with clear conditions. I am not understanding the “chord” part (but that’s ok as I do not intend to resolve it) because I cannot see how music would have anything to do with it.
Let’s see. If you can measure the diameter of the circle (as an angle, with a sextant or whatever) and you can measure the distance to the visible horizon, then the problem is pretty easy.
If you can measure the diameter of the circle and you can measure the distance to the center of the circle (nearest point), then the problem is just a tad more complicated.
If you do not have these two givens then WTF you got?
So Far So Good:
You did indeed reach me. I replied using the “reply” feature, but it bounced back. Sorry. I’ll contact you through the e-mail address you gave.
Sounds like other folks want to help, too. Nothing like a good math problem to get the juices flowing.
Sailor–(and whoever else wants it)
I am a specific distance above the surface of the earth, say 1000 miles. I can see a portion of the earth out my “window.” The shape of the part I can see is thus: flat like a ruler on one side, and the “roundness” of the earth on the other side-- like you took scissors and cut a straight-line slice off of a paper circle. The part I can see does not include a “diameter” of the earth. I will try to use keyboard characters to approximate what I can see, & hope I don’t create some obscure smilie.
|)
looks like a reasonable approximation.
I am limited to observing only–I can measure angles (from my point of view), but I can’t measure the distance to the horizon. The only “given” number I have is my altitude above the survace of the earth.
Hope this helps make things clear . . . if not please let me know where I need to explain/describe more.
Cal-- I will be watching my eamil. Thanks.
And once again thanks to all who are participating here.
Hmm… Well, that’s not at all the question I was trying to answer. I still want to help, though, so I think I’ll point out that it’s necessary to state a couple more things. Why is your view of the Earth shaped like that? Is that the shape of your window, and it’s filled with Earth? Is the flat side of the shape due to the Earth’s terminator (the line dividing night and day) or is it because that’s the edge of your window? Is the round side the horizon, or is that the terminator? I think what you have in mind is this: You have a square window, and the Earth (which is not doused in shadow) is peeking up from the bottom, kind of like a sunrise, before half the Sun is visible. If that’s the case, I can answer this. What you’ll need is a compass and a straightedge, and again, a sextant.
Put a transparency sheet up over your window, and mark off three points on the Earth’s horizon. Then take it down, and get ready to do a little Geometry. Suppose your points are called P, G, and R. Draw line segment PG, and then draw its perpendicular bisector. If you don’t know how to draw a perpendicular bisector of a segment, then I’m sure I or someone else here can explain it. Then, draw the perpendicular bisector of line segment GR. Where these two bisectors cross is the center of the circle that P, G, and R are on, and thus it’s the center of the Earth, C. Measure the angular distance of segment CP, or CG, and CR, and you have the angular radius of the Earth. From there, it’s just applying the knowledge You already have to convert that angular radius into a real radius.
Achernar–
The flat part is the edge of the window, the round part is the horizon . . . it looks like your reply is going in the direction I want to go, so let me play with it for a bit. Thanks a bunch.
I think you can calculate mass only, not circumference, from the orbit. To check the size, just wait for the shadow to cross the moon, and then measure the moon.