tag:blogger.com,1999:blog5204863782883637837.post4975898926482046051..comments20230913T20:13:33.65604:00Comments on Accepting Ignorance: Perpetual Motion by Means of Black Holes and Dark EnergyAlrenoushttp://www.blogger.com/profile/11119846531341190283noreply@blogger.comBlogger4125tag:blogger.com,1999:blog5204863782883637837.post193598281036644525620130414T18:43:48.20704:0020130414T18:43:48.20704:00
I stand on a rock orbiting a black hole and sta...<br /><br /><br /><br />I stand on a rock orbiting a black hole and start lowering the end. The closer the end gets, the harder the hole will pull on the string. Almost immediately  especially if I pick a strong black hole  it will snap the end off. <br /><br />Let's say I find exactly where the twine snaps, and hold it there while I connect it to a turbine. If I use the twine to pull the turbine and make a spark of current, it will slow the descent of the bit that snaps off  basically, it hits my turbine with some energy instead of the surface. (So far, this is all the same as with any large, nonsingular gravity well.) If I don't hook up the turbine's magnet, the twine accelerates and then snaps against its own momentum. If I do hook up the turbine, it will be almost stationary when it snaps. <br /><br />The snapped end loses kinetic energy and the black hole won't convert it to mass. I can't extract energy because the length d that snaps off doesn't experience more acceleration than c/d until after it snaps. <br /><br />However, because I know unobtainium twine is a perpetual motion machine, I know the regular twine situation must produce a little extra as well; it is just a question of finding where. <br /><br />So why doesn't this work with normal twine in normal gravity wells? In a world idealized to within an inch of its life, it does. While the string gets pulled on less the lighter is it, I tie a weight to the end and then jack up the well's specific density, or decrease the radius and thus increase density, until the weight hardly has to move to create as much current as I want. (And thus never reaches the surface.) a=c/d, just jack up 'a' until d can be arbitrarily small. <br /><br />However, you can't jack up the density very high without making a black hole anyway, plus the twine would pile up and eventually foul the turbine  I can only do it once before I have to reset the mechanism. <br /><br />In a black hole, the twine can't pile up. Second, the mass decreases linearly with radius, but the volume decreases with the cube of radius, so the effective density increases with the square. <br /><br />Even if I decrease the string weight linearly with black hole radius, the black hole's density and thus surface gravity with increase with the square, leading to a positive infinite force for an infinitesimally small black hole, and an arbitrary force for an arbitrarily small black hole.<br /><br />I would check that the distance between the required force and excessive time dilation doesn't shrink too fast, but since I can get infinite force, it can't get small enough. <br /><br /><br /><br />I want to use the twine to pull a generator that will produce enough voltage that I can convert the resulting current's energy into enough mass to spin more twine.<br /><br />With real twine, if I make it thinner, so I need less force, I also make it weaker, so I can extract less force. It gets stronger slightly faster, but maxes out at monomolecular string, which still isn't strong enough per unit mass.Alrenoushttps://www.blogger.com/profile/11119846531341190283noreply@blogger.comtag:blogger.com,1999:blog5204863782883637837.post457892353026884861320130414T18:43:45.05304:0020130414T18:43:45.05304:00So, twine.
The summary is that black holes have ...So, twine. <br /><br />The summary is that black holes have infinite gravitational acceleration at the event horizon, so they must produce infinite energy. However, with real twine, it isn't necessarily easy to see where that energy goes. It might be that all I can do here is make sure we're on the same page. <br /><br />Still, I'll give it a shot. <br /><br />May as well start with math. Ends with a = c/d.<br /><br />Work is force times distance, W = Fd<br /><br />The breakeven point is near W = mc^2, specifically mass per unit length, W = ρc^2. (That's rho, compare lowercase p.) <br /><br />ρc^2 = Fd<br /><br />F = ma, so mad = ρc^2, a = ρc^2/dm. Converting rho back to m/d, a = c^2/d^2 = c/d. If you've got ten meters to work with, then the acceleration needs to be over 3*10^7 m/s/s in that space, then you can extract more energy than you put in. <br /><br />While there's infinite acceleration at the event horizon, it is associated with infinite time dilation, which means the acceleration is infinitely slowed from our perspective. However, <a href="https://en.wikipedia.org/wiki/Gravitational_time_dilation" rel="nofollow">the dilation decays faster with radius</a> than the gravitational pull, so to get infinite force it's just a matter of tuning the hole's mass. <br /><br />At twice the Schwarzschild radius, the dilation is about 0.7. The Schwarzschild radius decreases linearly as mass decreases. Gravity decreases linearly with mass. However, gravity also decreases with the square of radius, so a halfsize black hole is twice as strong at the event horizon, and thus also twice as strong at double the event horizon. I can put this in math too if you want.Alrenoushttps://www.blogger.com/profile/11119846531341190283noreply@blogger.comtag:blogger.com,1999:blog5204863782883637837.post698802307795675181320130413T00:01:17.38804:0020130413T00:01:17.38804:00Sure, I'm happy to try.

Work is force tim...Sure, I'm happy to try. <br /><br /><br /><br />Work is force times distance. Both these phenomena can produce arbitrary amounts of force, and thus arbitrary amounts of work over a given distance. <br /><br />This wouldn't matter if nothing could feel the force  supposedly, electrons can produce infinite repulsive electric force, being point particles, but you'd need infinite energy in the first place to get that close. There's supposedly a singularity, but like the singularity at the centre of a black hole, nothing can ever observe it. <br /><br />The string needs to be infinitely strong to transmit the force, though. <br /><br /><br /><br />The idea for my infinite turbine is to create infinite voltage. <br /><br />The force necessary to turn the magnet is proportional to how much voltage it is producing. How much voltage it produces is a proportional to how fast it turns, how strong it is, and how many windings it has. An infinitely large turbine can have infinite windings. <br /><br />If multiplying infinitesimal turn speed by infinite windings doesn't work out, then make the magnet stronger until it does. Thing is, as long as it is taking infinite force to turn, then it will produce infinite voltage. <br /><br />Or, for any given force, no matter how high, you can set up a turbine to produce a voltage proportionally high, no matter how slow the draw speed is. <br /><br />In this case, the slower the draw speed, the less twinemass will be consumed per second, until finally the twine drops out of the generator's efficiency equation entirely. But, again, the slower it is, the higher the tension has to be. <br /><br /><br /><br />Twine forthcoming.Alrenoushttps://www.blogger.com/profile/11119846531341190283noreply@blogger.comtag:blogger.com,1999:blog5204863782883637837.post649182585175447931220130412T16:42:49.98204:0020130412T16:42:49.98204:00First you posit some unobtainium that's as str...First you posit some unobtainium that's as strong as necessary, and you conclude that this gives you as much energy as necessary. I don't see how this violates conservation of anything. Later you multiply infinity by an infinitesimal, and I don't see any guarantee that the result is even welldefined. I don't see how either of these are violating conservation of energy.<br /><br />Please explain the twine case in more detail, because I don't understand how to work out the finite case from the limit cases.Eriknoreply@blogger.com