Iggy

No, they don't calculate precession.

There is a very commonly used trick for accurate approximations that you used here: It's quite hard to understand why you didn't use the same trick again. There really is, however, only one approximation you have to make. If I were doing this I would start with, [latex]I \omega = I \omega_2 + mr^2 \omega_2[/latex], like you did and solve it for [latex]\omega_2[/latex] without approximating giving, [latex]\omega_2 = \frac{I \omega}{I + mr^2}[/latex] The exact equation for difference in kinetic energy is then, [math]\frac{1}{2}I \omega^2 - \frac{1}{2} \left( \frac{I \omega}{I+mr^2} \right)^2 (I+mr^2) [/math] cancel terms, pull an I out of the denominator, and cancel again: [math]\frac{1}{2}I \omega^2 - \frac{1}{2} \left( \frac{I \omega^2}{1+mr^2/I} \right)[/math] then make your approximation... [math]\left( \frac{I}{1+mr^2/I} \right) \approx I(1-mr^2/I)[/math] giving, [math]\frac{1}{2}I \omega^2 - \frac{1}{2} \omega^2 I(1-mr^2/I)[/math] and there you have it, [math]\frac{1}{2} \omega^2 (I-(I-mr^2)) = \frac{1}{2} \omega^2 mr^2[/math]

An event happens one lightyear to your right at t=0. Sally is in your location moving to your right at .6c. As she passes, her clock also says t=0. According to her clock the event happens at: [latex]t' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( t - v x/c^2 \right)[/latex] [latex]= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left( 0 - 0.6 \cdot 1/1^2 \right)[/latex] [latex]= -0.75[/latex] and according to her ruler it happens at: [latex]x' = \left( \frac{1}{ \sqrt{ 1-v^2/c^2}} \right) \left( x - v t \right)[/latex] [latex]= \left( \frac{1}{ \sqrt{ 1-0.6^2/1^2}} \right) \left(1 - 0.6 \cdot 0 \right)[/latex] [latex]= 1.25[/latex]

If you could, 1) which is stronger, the ring or the disc magnet? Which would be harder to push up against the stationary magnet if you tried? 2) how many degrees, roughly, does it rotate when you first get the thing started (the most it rotates back and forth)? 3) does it act differently when you remove the stationary magnet? 4) do you have video?

Which two norths? The weakest and the stationary?

ouch Yeah, I'm sure I could simplify that to [latex]-\frac{1}{2}mr^2 \omega^2[/latex] if I approximated [latex]\omega(1-mr^2/I) \approx \omega[/latex]. But, that can't be right. That would be assuming earth doesn't change speed, and it would make all of this meaningless: I trust you know what your math means. It's like you're trying to figure it out. I don't get it. because we're subtracting the final state from the original state negative is a gain of energy. If you start with 10 apples and end up with 11, you could subtract 10-11 = -1... negative in that case means you've gained an apple. This should make sense because, if you move mass from the pole to the equator without slowing the earth, the earth will have gained kinetic energy at the end.

No, I don't suppose the bit abut the geiger counter acting as the observer matters, but the other bit, most definitely. Owl discredits spacetime in the same way he is trying to discredit superposition. He says, relativity posits a curved medium which doesn't, in fact, exist. What a headache everybody would be saved had someone been there at the start to say "spacetime is a way to organize events... not a real physical thing". The Copenhagen interpretation says that the state of the system described by the wave function means only what we can say about the system. It isn't that the cat has to be alive and dead for QM to work any more than spacetime has to be a real curved physical medium for relativity to work. It's interpretation dependent. It is enough, and it's consistent with the Copenhagen interpretation, just to say that our knowledge of the state of the cat at that time is equally consistent with both. If you deny Owl the opportunity to reify the wave function then his declaration that every cat is either alive or dead doesn't discredit anything.

Could you please demonstrate. I can't seem to get that.

When the two norths face each other there is potential energy. It's converted to kinetic energy and the magnets rotate then back to potential energy and back to kinetic and so on. It's the same way with an ordinary pendulum -- this site tries to explain. If the conversion from one type of energy to the other were 100% efficient, the process would continue indefinitely. However, you will lose energy to friction because of the air in the room and the stiffness of the string. It eventually stops when all the energy you initially gave it is lost by friction. For it to last longer you could use a thinner string, and, ideally, it would be in a vacuum.

I think Schrodinger's point was actually to illustrate the absurdity of the Copenhagen interpretation. But, at least according to wikipedia, the Geiger counter in the box would serve as an observer collapsing the wave function so that the cat is definitively dead or alive before the box is opened. While catching up on the thread I stumbled into the article Swansont mentioned -- a really good read. It brings up the wave function as an example of physics abstraction that is best not reified.

TransformerRobot, you can calculate yourself the distance needed to safely stop. If you square the velocity (in meters per second) and divide that by 19.6 and divide that by whatever g-force you might like to inflict on your hero, that will tell you how many meters it should take him to stop. For 700 kph (that's about 200 m/s), feeling 10 gees, it would come out at about 200 meters.

If you define physical that way then I think so. That's a pretty loose definition too. When the OP says "we have no access to nothingness", I think that's saying the same thing. I agree with it. It would be hard to make an argument about nothingness not look philosophical Just by formal logic, if something must be measurable to be physical and nothingness is not measurable then nothingness isn't physical.

Ok... 2130... If I fell off a bike going ridiculous speeds, I would want an airbag system to deploy like spirit and opportunity had when they hit the surface of mars... [skip to 0:50] Or you could put some maneuvering rockets on the suit that act to quickly decelerate the rider before he hits something. Bullets and missiles could be handled with a futuristic version of the IDF video you found. Darpa is working on something similar for RPG's that you could google. As far as being hit by a fighter jet going some high mach number... the only realistic solution is to get out of the way. The radar system that detects incoming missiles could also serve to say decide if something really big is on its way and fire maneuvering thrusters to move out of its way.

It sounds like a story Marvel already told The problem, as I'm sure people have already explained, is that a suite no matter how strong, doesn't protect you from inertial forces. If you jump off the empire state building wearing a strong suite of armor, the armor could survive the impact intact, but that wouldn't do the person inside any good. You'd still need a mop to get them out. Falling off a bike going 700 km/h and hitting a wall is the end of the rider. Unless the suit breaks the laws of physics, it's the end of the story. Fade to black. The IDF tank system shoots scatter-shot at an incoming missile. It detonates the missile prematurely. That would work great for a personal anti-missile system, but would serve no other purpose. It wouldn't help with bullets or impacts or falling off a bike, as examples. If you have to break the laws of physics to accomplish what you want, and it sounds like you do, then you might as well give the wearer inertial dampeners. EDIT -- > inertial negation would have the added bonus of making the bike faster (the rider would effectively weigh nothing) and allow the rider to undertake unrealistic maneuvers without feeling any g-force <--EDIT

What would physically impossible mean? I could, for example, define "physical" as anything that can be measured. It would be easy to show that nothingness can't be measured, because if you measure it, it is something. It's then trivial to show that nothingness is not physically possible.

My understanding is that Popper distinguished between universal and singular statements. You cannot prove true, but could prove false, a purely universal statement (eg all swans are white). Purely universal statements are the negation of purely existential statements and they likewise cannot be proven true but can be proven false (eg non-white swans don't exist). owl's statements on the matter, such as: are either purely universal and cannot be proven true, or... are equivalently the negation of a purely existential statement and likewise can't be proven true. In other words, the quote above could be rephrased "there exists no frame in which the diameter varies". owl is correct, you cant prove them true. Problem is... he is the one saying they are true. In other words, to prove "there exists a variation" true, you need only one experiment. No amount of experiments would ever prove "no variation exists" true.

but, you do agree that reducing, [latex]\frac{1}{2}I \omega^2 - \frac{1}{2} \left(\omega(1-mr^2/I) \right)^2 (I+mr^2) [/latex] to [latex]\frac{1}{2}mr^2 \omega^2[/latex] via algebra is what you did, and the correct way to go? //edited latex

good thing we have university professors of psychology, logic, and experimental design all present. one of them should be able to diagnose the problem here

that makes very good sense did you use some force of algebra to get you there?

Would you say I includes the drop?

I don't know what the world would look like if the US were not a part of it, but I would answer your question and the Captain's comment about not being able to spend money on every what-if the same. I believe the best approach is similar to what JPL calls "faster, better, cheaper" JPL used to spend lots of money on each spacecraft, trying to bring the risk of failure as low as possible to each mission. Better, they later thought, not to put all their eggs in each mission's basket. Because, even the expensive space baskets fail. They started launching lots of cheap craft with a higher individual chance of failure but accomplishing better overall results. While the US stations a few thousand troops along the DMZ trying to deter resumed hostilities, it's good to have something like the six party talks, and good that the US, China, and other countries give North Korea food and fuel aid. The more low-cost approaches trying to prevent the worst outcome, and the more nations involved, the better. It's the same with Iran. I saw in the news a few days ago that UN inspectors are seeking evidence, they called a "smoking gun" that would force Iran to capitulate. It's not just US spy drones looking for that evidence. It would help -- just today, for example, diplomats were meeting in Rome from the EU and US looking for ways to put pressure on Iran to resume talks over its nuclear program. Having a "don't wait for the translation" moment would be very helpful right now, and it's not just the US thinking that or trying to make it happen.

Thank God North Korea's president will remain unchanged. That should provide some stability.

Interesting. I think Michel hit the nail right on the head. The correct relationship would be: [latex]\frac{4}{3} \pi R^3 < \frac{4}{3} \pi R_P (R_E)^2[/latex] Where [latex]R_P[/latex] is the polar radius and [latex]R_E[/latex] is the equatorial radius. Since the equatorial radius is 301/300 times the polar radius all we have to do is solve for C: [latex]C \frac{4}{3} \pi R^3 = \frac{4}{3} \pi R \left( R \frac{301}{300} \right)^2[/latex] which gives, [latex]C = \left( \frac{301}{300} \right)^2[/latex] Your approximation is quite good. No, I agree. More explanation is better. I'd still love to see the wiki page Let read the next bit... Does the "I" on the left side of the equation have the same value as the "I" on the right?

Oh my! Could that be right... How would that work out, Newts?

you might find a problem here, Adding 1/300 to earth's radius adds what to its volume? You guessed "2/300". Perhaps that is the error...? How much bigger is the right side of the equation: [latex]\frac{4}{3} \pi R^3 < \frac{4}{3} \pi R^2 \left( \frac{301R}{300} \right)[/latex] The left side is the volume of a sphere of radius R and the right side is the volume of the sphere with a 1/300 bulge [volume of a spheroid]. Can you agree the volume isn't 2/300, but 1/300 more? .29 * 6371 * 1/300 = 6.16 Would you agree? I was, in fact, going to suggest some editing. We're 1500 words in, and so far nothing has appeared relevant to your conclusions. By the way, when you said that your equation was on wikipedia, can you point me to that, because I'm having a hard time reading the way the equations are rendered on the site you gave. Thank you.