Timeline for Why is using a space elevator cheaper than rocket power?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 16 at 5:53 | comment | added | Peter - Reinstate Monica | @KnudsenNumber Yes, a float is the proposed solution, partly in order to evade storms. | |
| Jun 20, 2019 at 0:47 | comment | added | Knudsen Number | @TomSpilker Could you just anchor the elevator to a large floating platform in the ocean? You could use the water to damp out the vibrations and actually stick some motors on the platform and use active control while your at it. | |
| Sep 11, 2018 at 13:44 | comment | added | Davis Broda | While you might need some sort of engine to stabilize the elevator, why does it have to be a conventional chemical rocket engine, with all of its inefficiencies. Use an Ion drive, or some similar low-thrust high-efficiency engine to maintain position much more cheaply than sending up conventional rocket fuel | |
| Sep 9, 2018 at 19:00 | comment | added | TonyK | Thank you for this excellent answer, Tom! I have long wondered about how a space elevator can supply the transverse delta-V. | |
| Sep 7, 2018 at 10:18 | comment | added | Ruadhan2300 | "overstressing the cable is distinctly suboptimal." - Glorious levels of understatement for what would absolutely be an apocalyptic level of destruction across the entire equatorial region of the planet. | |
| Sep 6, 2018 at 18:11 | comment | added | Peter Cordes | re: you'd need lots of electric power. You'd need a lot of current to produce a strong magnetic field. But with superconducting coils, the amount of power should only be proportional to the actual work done, right? Maybe some losses in the electronics that modulate the current over time, but not resistive losses from just keeping the current flowing. | |
| Sep 6, 2018 at 18:00 | comment | added | Peter Cordes | Oh, I missed that the first "hardware" was "elevator hardware", not "rocket hardware". Makes sense now. The overall point about expending energy on the launch vehicle was already clear I think. (I already knew that, but I think it would have been clear even if I didn't. The earlier sentences explained it nicely.) I may have been skimming slightly because I already understood where you were going overall in that paragraph. Neat answer about how exactly the horizontal forces play out. | |
| Sep 6, 2018 at 17:54 | comment | added | Tom Spilker | @PeterCordes Thanks for pointing out this source of confusion. What I meant is that 1) the mass of the hardware (on the ground) needed to generate the electric power that propels the elevator car to the GEO radius is much less than the mass of a rocket that would do the same task, and 2) the ground-based generator isn't destroyed when the task is complete; it can power another car, and another, and another... I'll edit my answer to clarify that—again, thanks! Of course, a ground-based power plant is much heavier than a rocket, but the car's fraction of that power and mass is fairly small. | |
| Sep 6, 2018 at 17:44 | comment | added | Peter Cordes | Phrasing: None of the hardware involved, which for a given payload is much lighter than the rocket hardware necessary, becomes high-energy junk! I think you're pointing out that rockets are heavy compared to the payload. So maybe "which for a given payload is much heavier than the payload itself"? Anyway, I don't think your current sentence makes your point clearly, because either I'm missing it or it's backwards. | |
| Sep 6, 2018 at 0:50 | comment | added | Tom Spilker | SInce the bottom end is essentially fixed in position it provides an effectively infinite-impedance reflection interface. Downward traveling waves would grow in amplitude, be essentially totally reflected at the bottom (unless a damping mechanism is acting), lose amplitude as they ascend, all the while interfering with later downward-travelling waves. | |
| Sep 6, 2018 at 0:40 | comment | added | Tom Spilker | @BlackThorn Whoa, you're taking the guitar-string analogy much too far! I meant only to indicate that launching traveling waves at a given point can result in interference patterns when reflections from the two ends meet. But here's a very important point: The cable is not thickest at the base! It's thickest at GEO radius, where the upward force due to the cable above that point and the counterweight balance (with a bit of bias tension added) the downward force due to the suspended cable below that point. From GEO, it gets thinner in both directions. | |
| Sep 6, 2018 at 0:26 | comment | added | BlackThorn | Also, the oscillations would not be like guitar strings at all as the cable would be tapered... much much thicker at the base than the end. That means that the end will behave like a whip, exaggerating any upward propagating disturbances, but downward propagating disturbances will quickly die on their own, so no harmonics can last on the cable. | |
| Sep 6, 2018 at 0:22 | comment | added | BlackThorn | Good observations, but I want to point out a few things you didn't mention. An operating elevator would likely have numerous cars and sensors all up and down that would allow it to quickly respond to oscillations with highly tuned control systems. The cars would throw their weight around to neutralize unwanted disturbances. To maintain the proper angular momentum/velocity, simply send a car down with equivalent weight to the ascending car. It can contain shuttles, rock samples, ore, or junk. I highly doubt you would need rockets on the cable to preserve its attitude. | |
| Sep 6, 2018 at 0:00 | history | edited | Tom Spilker | CC BY-SA 4.0 |
Added analogy to upside-down pendulum.
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| Sep 5, 2018 at 23:15 | history | answered | Tom Spilker | CC BY-SA 4.0 |