Timeline for Why is using a space elevator cheaper than rocket power?
Current License: CC BY-SA 4.0
14 events
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| Sep 7, 2018 at 1:32 | comment | added | Tom Spilker | @brichins I like the idea of battery nodes on the cable. That way the electrical pathlength between the source and the motor doesn't have to be tens of thousands of km. It can be much shorter, avoiding significant transmission losses. We'll have to analyze how much adding the mass of electrical cables, batteries, etc. affects the required cross-sectional area of the cable. Essentially it decreases the effective strength-to-weight ratio. But (shooting from the hip!) I don't think that's a show-stopper. | |
| Sep 6, 2018 at 23:08 | comment | added | brichins | @TomSpilker I like the idea of recharging the power in the descending car, a lot of potential energy to capture. But it might be to better store it within the car than transferring it to the ascending car. Linking opposing cars directly would require a huge connecting cable, with massive power requirements to move it. However, you could possibly have some battery nodes spaced along the elevator, which help power ascending cars and were recharged by descending ones. | |
| Sep 6, 2018 at 15:34 | comment | added | hobbs | @TomSpilker still, whatever "gravity losses" there are are only due to inefficiency in the drive system, and good engineering can reduce that (although of course never completely eliminate it). With a rocket it's inescapable. | |
| Sep 6, 2018 at 5:41 | comment | added | Tom Spilker | @immibis True, but when you try to provide power to the motor to run the car upward you can't short out the windings. However... if you're running another car downhill while one is running uphill, you can use electric power generated by the downhill one to supply most of the energy the uphill one will need. That downhill one is going to need to dissipate a lot of energy, and that transfer is a good way to do it. | |
| Sep 6, 2018 at 5:20 | comment | added | Criticizing Israel not allowed | @TomSpilker You may note that a 100% efficient motor would be made of superconductor and if intended to not move, would be shorted out. | |
| Sep 6, 2018 at 5:13 | comment | added | Tom Spilker | @immibis It depends on which theory you apply. "No energy out -> no energy in" applies to solid objects like bricks or I-beams. But given an ideal, theoretical electric motor, whether AC or DC, if you run no current through it, it generates no torque. If you then apply an external torque, the AC motor spins freely, and a DC motor with a brush commutator becomes a generator. If there's no load on the DC motor-turned-generator, it also spins freely. If either of those are hooked up to the car's drive train, that applies torque, the motors spin backward, and the car slides back downward. | |
| Sep 6, 2018 at 3:25 | comment | added | Criticizing Israel not allowed | @TomSpilker in theory, a perfectly efficient motor wouldn't need to use any energy to provide that torque. Energy would only be used to climb up, not to stay still. | |
| Sep 6, 2018 at 0:24 | comment | added | Tom Spilker | @brichins I haven't done the full analysis, but I think you're right. Yes, a car could engage a ratchet-like mechanism and support itself with no energy expended. But when it's trying to move, for any given motor and drive train there's a certain minimum torque that must be supplied to counter the weight of the car; otherwise the car spins the motor backward! To move the car upward you have to supply more torque than that. Power consumption will increase with the torque generated, so that "weight overhead" will increase consumed power. Sounds like a form of gravity loss to me! | |
| Sep 5, 2018 at 19:59 | comment | added | Nuclear Hoagie | It's sort of like the difference between jumping to the top of a tall building and taking the stairs. The leap has to be done quickly and with a huge impulse, but you can stop and take a break wherever you want on the stairs. | |
| Sep 5, 2018 at 19:57 | comment | added | brichins | Wouldn't a form of gravity loss apply to an elevator as well; possibly even a much greater net total? You have to provide a force to oppose gravity, whether for the minutes of a rocket launch or days of an elevator lift. | |
| S Sep 5, 2018 at 18:36 | history | suggested | Peter Mortensen | CC BY-SA 4.0 |
Copy edited (e.g. ref. <https://en.wikipedia.org/wiki/Delta-v>). [(its = possessive, it's = "it is" or "it has". See for example <http://www.wikihow.com/Use-Its-and-It%27s>.)
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| Sep 5, 2018 at 18:31 | comment | added | Peter Mortensen | Beamed power: subject of a TED talk - A better way to get to space: Dr. Michael Kelzenberg. | |
| Sep 5, 2018 at 18:27 | review | Suggested edits | |||
| S Sep 5, 2018 at 18:36 | |||||
| Sep 5, 2018 at 17:36 | history | answered | Evan Steinbrenner | CC BY-SA 4.0 |