Schrödinger's hat

While a gravitational slingshot is a good way to gain some speed, returning to earth after sling-shotting around the moon is going to be quite difficult. I don't know what you mean by irregular orbit, either. Once the object is a bit away from the moon (so earth's gravity is dominant), it will be travelling in something very close to an ellipse. Someone clever may be able to figure out a path that makes two, or even three passes -- gaining energy each time. Unfortunately at some point you're going to either circularize your orbit not able to get near earth again without inputting energy (possibly crashing into the moon), or reach escape velocity and never seeing earth again. Secondly, traveling close to the speed of light won't make you travel through time in the Back to the Future sense, all it will do is make time pass at a different rate, so that when you return to earth you will be younger.

Apologies, I made a typo (or rather copy and pasted the wrong fraction without paying attention). Also if you click on the equations it will give you the code used to generate them [math] \omega_1 = 15000 \frac{r}{min} = 15000 \frac{rev}{min} \times \frac{2\pi rad}{1 rev} \times \frac{1min}{60s} = 15000\times\frac{2\pi}{60} \frac{rads}{s} = 1571 \frac{rad}{s} [/math] Basically I'm multiplying by 1 which in this case is disguised as [math]\frac{2\pi rad}{1 rev}[/math]. Then multiplying by 1 again [math] \frac{1min}{60s}[/math] This is one way of changing units so that you don't lose track of things. Basically I can write [math] 60s=1min [/math] and then divide both sides by 60s to get the fraction. This allows you to multiply out the minutes cancelling them as you did (min/min=1) and introduces the seconds. I accidentally flipped the fraction, it's fixed now. You can also just go [math] 1min = 60s[/math] and replace min with 60s everywhere you see it (and similarly rev with 2pi rad). this sometimes leads to nested fractions though and I've found people more frequently get confused. Dividing by 60 (twice) was the correct thing to do to fix the mistake I made, the extra s unit came from that same mistake. (if you follow the revised working you see that the units work too. This is all just a formal way of doing unit conversions. Ie. your dividing by 60 to convert minutes to seconds is just compressing the multiplying and cancelling of units into a single step. I went into detail on this because you were doing many things in your conversion that were not mathematically consistent. If you follow the scheme of finding the correct '1' and multiplying by it (this does not change the value of the equation), then cancelling units you won't get strange answers. On top of this, if you make a mistake (like I did :/ ) the units will tell you there is something wrong. If you're in the habit of just randomly multiplying/dividing by 60 when you see minutes, then replacing it with seconds; you won't be any the wiser if you did the wrong operation. In the same vein, you can't randomly multiply by g to try and get the right units. This changes the equation to one that does not represent torque.

May I introduce you to the preview post button? I don't believe you have met.

Never throw away a good theory. Just have to knock 'em off their pedestal every now and again or they start looking too smug.

Huzzah. If there's a good follow up this means we've finally found a way to chip away at the edges of the standard model. It's been sittin' there lookin' all smug goin' and predicting everything for far too long. We'll show it who's boss!

Sometimes units involving rotation can be a little tricky and disguise themselves as something else, but what I was mostly getting at is the whole conversion step and justifying the numbers you were using. Maybe go back a few steps and lay it out a bit more formally. So [math]\omega_1 = 15000 \frac{r}{min} = 15000 \frac{rev}{min} \times \frac{2\pi rad}{1 rev} \times \frac{1min}{60s} = 15000\times\frac{2\pi}{60} \frac{rads}{s} = 1571 \frac{rad}{s}[/math] (keeping a couple extra sig figs around for now as guard digits even though 15000 only has two.) [math] \alpha = \frac{\omega_1-\omega_0}{\Delta t} = \frac{1571 rad\, s^{-1} - 0 rad\, s^{-1}}{80 s} = 19.63 rad\,s^{-2} [/math] Multiply by moment of inertia. [math] \tau = \alpha I = 19.63 rad\,s^{-2} \times 11.5 kg m^2 = 225.8 rad\,kg\,m^2s^{-2} [/math] Our units are a bit cumbersome here, but this is correct. To find the conversion between Newtons and kg m/s^2, F=ma can be used. Put in 1 for each of the quantities and you recover: [math] 1N = 1kg\,ms^{-2} [/math] So we can switch this with the kgms^(-2) with Newtons in our equation to get [math] 225.8 rad\,kg\,m^2s^{-2} = 225.8 rad\,Nm [/math] Or, 230 rad Nm to 2 sig figs. Why are there some pesky radians hanging around, I here you ask? Well they're here to derail my point about the importance of units somewhat. There is actually a suppressed factor of rad^(-2) in the mass moment of inertia that the problem gave us, but radians are a bit odd. They're somewhat of a fundamental unit of angle, and they are the only angle-like quantity. If you use them in all your calculations you can safely ignore them -- as long as you are careful about one or two other points. For example, energy has units of Joules = Nm Torque also has units of Nm which should also equal Joules But these are very different concepts. Energy is torque multiplied by an angle. So there is a hidden rad^(-1) in the units for torque. Putting in our hidden radians gives us: [math] 230 Nm/rad [/math] Which is the units of torque (without the radians suppressed). I prefer to keep the radians in when I think about it (which isn't always), and this is good practice for someone who is learning. Unfortunately the convention in many textbooks is just to drop them, so you may be forced to just let them slide unless you know how to spot all of the missing ones. TLDR you can safely ignore units of radians as long as you know what your equation represents. Edit: I found an article backing all this radian nonsense up and going into more detail if anyone is skeptical/interested: pdfserv.aip.org/PHTEAH/vol_30/iss_3/170_1.pdf Edit: Flipped incorrect fraction, see below

Clearly the solution is to post open research problems instead of capchas, even if the spambots work it out we still win.

Pfft, when have people ever reacted that way to an impending worldwide issue that will cause a lot more harm if the response is delayed? You speak nonsense.

(what level class? University? Highschool?)

Well, now that we've established that, I have to remember how to deal with an infinite series. :/ Start off slow. The bit in the bracket is going to be less than n. The denominator of the fraction on the left is n. The sign is alternating. So it's an alternating series where the terms decrease, therefore it converges. Not terribly useful, but it's a start. We also know that it's somewhere in the vicinity of (ln(2))^2 as you correctly surmised. Because the term in the brackets is an ever-improving approximation of ln(2). But it won't be exactly. Now I have to think back to first year to make any further progress. Excuse me while I retrieve it from the long term memory archives. Loading... I'm afraid I'm coming up somewhat empty. Perhaps you could give me a little bit of context, it may provide a few clues. What is your current mathematical level? Where did you encounter this sum (was it an exercise in class? If so what type of class?) From what you've said so far I would surmise it's a Calculus 101 course or similar, but the requisite trick for this one eludes me. The n+1 vs n-1 looks like it was an intentional clue, but I see how it helps.

Somewhat. Just to be clear that would be able to be rewritten as: [math] \sum_{n=1}^{\infty}\left(\frac{(-1)^{n-1}}{n}\sum_{m=1}^n\frac{(-1)^{m+1}}{m}\right)[/math] Correct?

In the most simplistic sense: It gets cold. It gets warm again. This happens every year. People noticed that their crops, the trees around them, animals and the weather all followed a pattern. It's especially notable if you are near/on the arctic circle because it gets dark for a few days. That is as good a marker as any for the beginning of the next cycle (and the origin of many resurrection myths in winter solstice traditions). If you wanted to be a bit more accurate you could jam a rock or a stick in the ground, and trace its shadow, or mark the point its shadow reaches at noon. follow iNow's links for more details

I'm a little bit unclear on what you're trying to work out. This: [math] \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math] I interpret as: [math] \sum_{n=1}^{\infty} \left(\frac{(-1)^{n-1}}{n} (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) \right)[/math] But your inclusion of: [math] \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} [/math] Seems to imply you've bracketed it as: [math] \left(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\right) (1-\frac{1}{2}+\frac{1}{3}-...+\frac{(-1)^{n+1}}{n}) [/math] With an implicit (independant) sum on the second part. I also can't seem to find the exact link between either of these and your third statement. Either you've done some intermediate steps that I've missed because my brain isn't entirely switched on right now, I'm otherwise mis-interpreting what you've said, or you've made a mistake of some kind.

Without scrutinizing your calculations in detail (including units makes this a lot easier). You didn't put units on your answer (put units on your answers). If it was supposed to be SI, I can say 93 kN is highly unlikely. Did you read my section about unit conversion? Just in case the message wasn't clear. Put units on all your quantities as you define them. Put units in your calculations for unit conversion. Keep track of units in your calculations. Yes, write them in even if it's a bit tedious. Units UNITS UNITS UNITS UNITS

The tug on the string would travel at the speed of sound (of the string). Much slower than light. No matter how rigid the object, the way its components are interacting with each other is by exchanging photons (for the most part, nuclear forces and gravity can be involved as well, but gravity travels at the speed of light as far as we know and so do the strong/weak forces), so the signal can only travel at or less than the speed of light.

We've told you before, go and seek professional help. Doctors should be able to help you with your gastric problems, and if you are being threatened, that's what the police are there for. You should also seek some kind of counselling.

Phantom limb syndrome is where an amputee seems to feel a missing limb. Alien hand syndrome a different phenomenon. The afflicted person's limb will act seemingly of its own accord, often taking actions which distress or contradict the owner's intentions. Wiki for more info.

Even without a microscope there are plenty of ways you can prove the existence of something microbe like. Experiments can (and were) done to show that diseases were transmitted by some contamination in pus and mucus from infected patients. That contaminated water would make one ill. That certain types of decay and growths would only occur if you exposed the relevant substance to some source of contaminant. If you listen carefully to your body/brain you'll notice that you do have a number of different reactions to various things. Why do you think we have phrases like 'he was conflicted'? Why is it that people with brain damage can have completely different reactions to things they see on their right side, to things they see on their left? What about alien hand syndrome? I suppose that these people must be possessed by demons and have two different souls. So you can replace (or augment, I suppose as you said nothing about shape) a primitive, but useful theory with an egotistical one that serves no purpose? So you'll let me monkey with your brain? Maybe insert a dopamine delivery system, some electrodes to inhibit various areas. Reckon you wouldn't want what I decide you should want? Want to try it?

I still don't understand why you want unicode. To a computer, unicode still looks like 0F8A or 0110111000100101. It only comes out as a character after you put it through some kind of renderer. Fundamentally no different to an integer, other than data types/methods for strings in high level languages usually carry around a lot of baggage that you won't need for a mathematical algorithm. A sequence alignment algorithm can just as easily be programmed to use atoms consisting of integers as atoms consisting of characters. Unless you need to render it for some reason and get human input, the best way of doing this is with a lower level data type. To elaborate on your dna example (or a 4-bit per character representation of a number or code), you wouldn't have gttcaag as a sequence, it would be an element of the sequence. In this case the sequence would be (using [] to denote different elements in an array): [gttcaag] [gaagtgc][gttcaaa][gtagatc] Which would be just as similar to [gttcaaa] [gaagtgc][gttcaaa][gtagatc] as it would be to [aaaaaaa] [gaagtgc][gttcaaa][gtagatc] (unless you had some kind of weighting on your algorithm)

Hmm, I'd wager that it's going to be more complicated than all that Probably involving lift ratios and stall speed and a number of other things I haven't really thought about enough to give you a good answer. These factors will probably depend heavily on geometry, as well. One way of getting an answer simply might be to use a top down approach. Start with energy. Find some efficiency figures for basic windmills (probably around 10% as a good ballpark unless it's a high performance design). Then find the kinetic energy contained in the air. That'll give you the max rotation rate for a given torque and wind-speed (the combination of friction and energy going to your generator). Maybe you could describe what you are trying to achieve and a bit more about the situation, then we could possibly be of more use.

Let's do some classical mechanics then: You may have seen a bunch of algebraic equations if you did a highschool mechanics course. [math]x = x_0 + v_0 t + \frac{1}{2}at^2[/math] [math]v = v_0 + at[/math] And so on. Really they are just special cases and/or algebraic rearrangements of this equation: [math] F = m\ddot{x} [/math] Where the dots represent time derivatives. Which you might recognise a bit better as: [math] F = ma[/math] The most basic example (and where all those algebraic equations come from) is if the force is a constant. Then we can just take a as a constant (or F/m, it makes no difference) and integrate to get x as a function of time. So the first step is: [math] \ddot{x} = a[/math] We get [math] \dot{x} = \int \ddot{x} dt = at + C [/math] Then we identify the constant of integration by realizing that [math]\dot{x}[/math] or the first derivative of position represents velocity ([math]v = \dot{x}[/math]). We put in 0 for time and get [math] v(0) = C[/math] or C is initial velocity. This is the second equation above. Then we can integrate again: [math] x = \int dot{x} dt = \int at + v_0 dt = \frac{1}{2}at^2 + v_0 t + D[/math] The new constant of integration we recognise by the same logic as the initial position so: [math] x(t) = \frac{1}{2}at^2 + v_0 t + x_0[/math] But this is only one case (constant acceleration/force). If you have some force that varies with time (but not position) you can use the same concept. Just integrate it twice and you know where the particle is. There are other cases though. If you have an object attached to a spring, then the force depends on position (but not time), so we have: By Hooke's law the force on the object will be proportional to how far it moves from the rest position of the spring ([math]F=k(x-x_0)[/math]. So if we set x_0 to 0 (rest position is x=0) we get: [math] F = m\ddot{x} = -kx [/math] Or [math] \ddot{x} = -\frac{k}{m}x [/math] The minus sign comes from the force acting to push the object back where it came from. This is something called a differential equation. We can't just solve it by integrating because we don't know how x (the thing we need to find) depends on t yet. If we did we wouldn't need to solve it. I'm going to wave my hands here a bit and magic up the fact that the solution to this equation is a combination of cos and sin functions1. So let's try [math]y=A\sin(\omega t)[/math] The A is just there as an arbitrary scaling factor, and the omega represents the speed at which the sin function oscillates. Basically its a way of writing down the broadest, most general type of sin function I can think of2. Differentiate it once, we get [math]A\omega\cos(\omega t)[/math] Differentiate it again, we get [math]-A\omega^2\sin(\omegat)[/math] So for our trial function we can see that [math]\ddot{y} = -\omega^2y[/math] This looks remarkably like the differential equation, all we have to do is notice that [math]\omega^2 = \frac{k}{m} \rightarrow \omega = \sqrt{\frac{k}{m}}[/math] or [math]x = A\sin{\sqrt{\frac{k}{m}}t}[/math] So it turns out that arbitrary constant is the magnitude of the oscillations, and [math]\sqrt{\frac{k}{m}}[/math] is the angular velocity (or the frequency, up to a constant of 2pi). I've also left out the case where the objects starts off-set (you'd need to include a cos function or phase offset in your answer with another arbitrary constant). This is actually quite important in more general cases. A second order (with two derivatives) differential equation requires two arbitrary constants to cover all solutions. All this is just from two little equations (F=ma and F=k(x-x_0) ) and a function (position) of one variable (time). It just gets more useful (and more complicated) from here. Well the post above was just about functions of one variable (time) which only involved one type of derivative (also time). In one little niche (one dimensional motion). Once you start adding more, you need more maths to deal with it all. There's plenty in geometry too, if that didn't satisfy you. 1There is a rather nifty property of this type of differential equation. It's actually one way of defining the exponential function ([math]y=e^t[/math] is the solution to [math]\dot{y} = y[/math]). We can use tricks like the chain rule and substitution of variables for when it's e^(ax) rather than e^(x). It all gets rather hairy because of the minus sign and imaginary numbers come into the picture and then we eventually wind up with sin and cos anyway (e^(ix) = cos(x) + i*sin(x)). e i (the square root of negative 1) and pi do tend to show up together a lot, but I digress. 2Phase offset is important, too, but let's not worry about that right now.

Why did you multiply by g? Also you're going to want everything in the same units. What is one revolution in radians? What is one minute in seconds? Given that, what's the conversion between revs/minute and rads/second? The equation you want is [math] \tau = I \alpha [/math] where I is the moment of inertia in SI units, tau is the torque and alpha is the angular acceleration. You're on roughly the right track for finding the angular acceleration. Formally (given constant torque) the equation is: [math]\alpha = \frac{\Delta \omega}{\Delta t} [/math] But your general logic is sound. You should be careful with unit conversions though. A good way of getting them right is along these lines: Let's say I want to convert 150 km/h to m/s [math] 1km = 1000m\; \rightarrow \; 1 = \frac{1000m}{1 km} = 1000 {m}{km} [/math] [math]60 s = 1min[/math] [math]60 min = 1h[/math] [math]\rightarrow 60\times60 s = 1h \rightarrow 1 = \frac{1h}{3600s} = \frac{1}{3600}\frac{h}{s}[/math] This means I can multiply or divide anything by [math] \frac{1}{3600}\frac{h}{s}[/math] without changing it, because this quantity is equal to 1. So I get my original figure 150km/h and multiply/divide to cancel out the units (treating them as if they were algebraic quantities. [math]150 km/h \left(\frac{1}{3600}\frac{h}{s}\right) \left(1000\frac{m}{km}\right)[/math] Rearranging: [math]150 km/h\frac{m}{km}\frac{h}{s} \frac{1000}{3600}[/math] Then you can cancel things out to get: [math] 150 \frac{1}{3.6} m/s = 41.67 m/s[/math] This is a bit laborious and abstract, but guarantees you won't mix up dividing/multiplying the ratios between your units. Here's a good source for explanations and equations related to rotational motion http://hyperphysics....hbase/circ.html Here's another thread with some questions for testing/improving your understanding of torque: http://www.sciencefo...__1#entry637202 They're fairly conceptual rather than rote application of the formulae and some of them require calculus, so don't be disappointed if you don't understand them all. Perhaps you could tell us what level you are at mathematically (high school, college, some calculus/no calculus etc -- I'm guessing high school from that question but cannot be sure) and we could come up with some more appropriate questions. Also have a read of that hyperphysics site and perhaps ask questions based on more specific concepts.

We dismiss that which does not have any measurable effect. Until there is a strict definition of this mind/soul/whatever and we can come up with some measurable effect that cannot be explained without it, it is completely irrelevant to my model of reality and can thus be dismissed. Or that the single identity is a mirage. I can -- if i'm not hungry at the time -- decide I don't want to eat. If enough time passes, the hormones and electrical signals my digestive system produces will override this feeling and produce a desire to eat. Even if we assume that I am a single entity, why can multiple pieces not make a single system? Your argument of 1+1=1 is also completely invalid, because I am made of cells, not numbers. By this type of logic, I am the same thing as two of me because of banach-tarski. Neuroscience is making great strides in understanding how the brain works (even if it is still missing the larger piece of the puzzle), and none of it involves some metaphysical mind.

The only thing I can think of is muscle wire (I think that's a brand name, can't remember the generic term, sorry). It doesn't so much expand/contract as bend or assume a different shape when heated by electricity. I think there's also some exotic stuff involving nanotubes, but I don't think it is mass market yet.

Why does it need to be a character? If you're doing it in a computer surely storing it as an integer would be fasters/more efficient. If you need a visual representation for some reason (maybe aligning things by eye somehow?) then wouldn't representing them as colours or simple patterns of colours be more efficient? Perhaps you could explain a bit more about the nature of this data and what you are doing?