I'm still a science major but had to take and extended leave for various reasons (don't ask) to be better suited in the long run in multiple facets. Soon I will be going back and which makes things, well, makes me feel there is more to life than negative reinforcement. There is a sample problem in the text book "Halliday/Resnick/(Walker) Fundamentals of Physics" 7th, 8th and 10th editions at least, where it appears a quantity was simply pulled out of someones ass. Mulling over the problem, as elevator rides provide great opportunity for reflection on things, I wondered: "Where the F* did they get that from?" In bold. --------------------------------------------------------------------------------------------------------------------- Sample Problem 1.01 Estimating order of magnitude, ball of string The world’s largest ball of string is about 2 m in radius. To the nearest order of magnitude, what is the total length L of the string in the ball? KEY IDEA We could, of course, take the ball apart and measure the total length L, but that would take great effort and make the ball’s builder most unhappy. Instead, because we want only the nearest order of magnitude, we can estimate any quantities required in the calculation. Calculations: Let us assume the ball is spherical with radius R = 2 m. The string in the ball is not closely packed (there are uncountable gaps between adjacent sections of string). To allow for these gaps, let us somewhat overestimate the cross-sectional area of the string by assuming the cross section is square, with an edge length d = 4 mm. Then, with a cross-sectional area of d^2 and a length L, the string occupies a total volume of V = (cross-sectional area)(length) = Please Register or Log in to view the hidden image! This is approximately equal to the volume of the ball, given by Please Register or Log in to view the hidden image! which is about Please Register or Log in to view the hidden image! because Pi is about 3. Thus, we have the following Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! (Note that you do not need a calculator for such a simplified calculation.) To the nearest order of magnitude, the ball contains about 1000 km of string! --------------------------------------------------------------------------------------------------------------------- Now, not about the methodology, but say I chose an edge length, not of 4 mm, but 4 km or even light years? How often did one have to guess, and or, deliberately or not screw up in assignments and exams? I would appreciate to hear any personal experiences. Thanks. Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image!
Wish I could still do those powers of ten exponents in my head. Can't say I ever got very good at math though. Somehow I come up with 2*10^18m instead of 2*10^6m.
If you measure the diameter of a string, it is probably around 3 mm. However, in a real-life ball of string, there are gaps of air between the string overlaps. So the text book is just using 4 mm so that value will be large enough to account for the gaps. They also use a square as the cross-section shape of the piece of string, (instead of a circle), just because the area of a square is an easier formula than the area of a circle.
Would you think the answer/sloppiness appropriate if "nearest order of magnitude" wasn't called for? :EDIT: Maybe I could see what replacing 4 mm with 3 mm would yield in complexity and naswer, busy ATM...
For those who are unaware, ×10^ signifies how many zeros follow. For example, 2×10^3 Means 3 zeros follow the 2. 2×10×10×10=2000. Please Register or Log in to view the hidden image!
Use your browser. ---------------------------------------------------------------------------------------------------------------- Also, this has a lot to do with playing around with Maple. Like having a template ready whereby changing one number and see it execute a series of commands; showing all steps to a calculus question and so on...
There are a lot of assumptions here, and they are not the only possible ones. Just to compare, let's make some different assumptions and see where we get to. Again, assume the ball of string is approximately spherical, which seems reasonable if it's all wound round and round into a big ball. Now, let's assume the string has a circular cross-section (more realistic than the square), with diameter $d$. We aren't told what the diameter is, and it's going to have a fairly big effect on the final answer. To get a fair comparison with the previous answer, let's assume the diameter is 4 mm, although that number actually strikes me as a little larger than it should be. There's still the problem of how to account for the "gaps" between different parts of the windings of the string. Being conservative, I'm going to assume, without brilliant justification, that the gaps can be accounted for by pretending that the diameter of the string is half as large again as its assumed diameter - i.e. I'll use a diameter of 6 mm instead of 4 mm. Now, the total volume the string of length $L$ occupies is $\pi (d/2)^2 L$, which is the "effective" circular cross-sectional area of the string multiplied by its length. Then we have $\pi \left(\frac{d}{2}\right)^2 L = \frac{4}{3} \pi R^3$ or $L=\frac{16R^3}{3d^2}$ Plugging in $R=2$ m and $d=6$ mm gives $L=1185$ km. To the nearest order of magnitude, the string is, again, 1000 km long. The previous calculation, by the way, works out at $L=2000$ km. The important point is that we can be fairly sure that the length of the string isn't 100 km or 10,000 km. It's likely to be somewhere in the thousands-of-kilometers range, even if the amount of "gap" between windings was several times larger than what I assumed. The point is: we're only making a (reasonably good) back-of-the-envelope estimate of the length here. We could make a better estimate if we knew the actual diameter of the string, obviously. We could make an even better estimate if we investigated just how much "gap" you get when you wind string into a ball. We could work that out by experiment with a large-enough ball in lots of different ways. (For example, one less-obvious way would be to weigh a small known length of string and to weigh a larger ball of string of measured diameter, then use the density of the string.) Scientists often want a "ball-park" answer like this before they try to work out an accurate answer (possibly with an experiment). The fact that we can get a reasonably accurate answer with not too much effort shows that this kind of educated-guess approximate thinking is a useful scientific tool to have in your arsenal.
The example for thinking in terms of magnitude that I've heard most often is to answer the question "How many piano tuners are there in London"? This is a example of a question where you may think you know very little about the subject and yet you can probably get close enough to the answer. Make up your own reason for needing to know the answer. You are thinking of moving to London and starting a piano tuning business, etc. Sure, you could Google this one. That isn't the point. It's just another way of pointing out how close an educated guess can be if you think in terms of magnitudes. If you asked a large room full of people the piano turner question some might guess 50,000, some 1,000, some 2, some 300,000. You need a better method that just guessing without the "educated" part. Please Register or Log in to view the hidden image! To figure out the problem you might guess at the population of London even if you weren't from that part of the world. You know that at one time London was the biggest city in the world, you know that today there are cities larger than 10 million so you guess 8 million. Having a piano isn't as popular today as it was at one time. How many families are there in London? Maybe you define a family to be 4 people. Some will be larger and some smaller but let's just divide 8 million by 4 to get 2 million families. You might guess that only the top 25 percent (500,000) could afford a piano and that only half of those actually do have one (250,000). How many pianos at day can a tuner tune? I'll guess 4 and maybe they work 20 days a month for 11 months so 220 days a year. That means each tuner can tune 880 pianos a year. That would mean 284 piano tuners. Lets assume that many work for a small company so maybe two tuners are in business together. That's 142 tuner companies listed in the yellow pages. When I answered that scenario, those were my guesses and as I remember the actual answer was less than 200 piano tuners listed in the London yellow pages. An even quicker example would just be to guess the diameter of the Earth at the equator. You could guess based on your own experience as to how many miles across a time zone typically is. Guessing 1,000 miles would be a reasonable guess. There are 24 times zones across the Earth so a guess of 24,000 miles would be a good one and since the equator is further around than the latitudes of the U.S. then you might guess 25,000 miles around the equator. I think that's very close to the actual diameter. That is the point your textbook is trying to make.
My father could tune pianos but I doubt that he ever did more than 2 or 3 in a year and he did it for free. I live in a city of 250,000 people and there probably isn't one professional piano tuner here who does it as his/her main work. Just sayin'. Please Register or Log in to view the hidden image!
I might have fudged as I was multitasking... At first thought, I would have been fine with a square cross section for gaps and executing a whole series of equations by changing the quantity of one thing via math software that has become ever more prevalent in education. The 4 mm edge is what... well... made me go: "Holy Haagen-Dazs BATMAN!" But soon I should talk about free body diagrams for such questions as: "At what height should a rock be dropped on a stupid persons head, who's brain is as dense as a singularity, to knock some sense into him/her?" Please Register or Log in to view the hidden image!
I'm sayin' that you couldn't have picked a worse example if you tried. Please Register or Log in to view the hidden image!
That's the point, dummy. Please Register or Log in to view the hidden image! And yet, there are about 200 piano tuners in London.
When I first read your question, my wild guess was "a few hundred" - just as accurate as your calculated estimate. Please Register or Log in to view the hidden image!
Good for you. That's not the point but since the post was about getting the magnitude right, you got the magnitude right and then later ruined your point by implying that there were no commercial piano tuners anymore. Are you implying that every time you guess you get the magnitude right or do you acknowledge that applying a little thought is helpful when it's a subject you know nothing about?
So is the assumption that strings have a square cross-section. But even a less-than-perfect assumption can lead to a ballpark-correct answer, as we have seen here.
