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Is there any active research on this technology for non-rocket launch systems?

clarkmoody 7y ago
I assume that the same technology could be applied for launch systems, with one major change: maximum acceleration. An electric railgun would need to be very long in order to reach high velocity without undergoing huge accelerations as in the Navy railgun. Ramping up a vehicle with payload and fuel in it without destroying it is the name of the game.
godelski 7y ago
Do you mean like going to space? There's some research but after everyone figured it that all humans inside would be instantly squished they started researching other avenues. The g forces are pretty large and would damage a lot of cargo too.
- ars 7y ago
  > would be instantly squished Not so. That's only for a very short launch. You would actually have a track a few miles long (probably up the side of a mountain) and launch using that. The g-forces would be very mild.
  - logfromblammo 7y ago
    You need about 1200km to reach LEO from the equator with max acceleration of about 3G. I did the math once, and a track from the Galapagos Islands to the peak of Chimborazo on the continent in Ecuador would do it.
    - ars 7y ago
      I was a little shocked at that number, so I ran the math again, and wow. Quite a bit longer than I expected. Maybe a circular track till you build up speed, then switch over to the mountain track? Of course the circular track would have g-forces due to that..... (Do you feel like doing more math?) I guess it's more practical to use the track just for the initial boost above the atmosphere, then a rocket to add speed. Also a track that can handle something moving at km/s is .... not easy.
      - logfromblammo 7y ago
        A circular track means you'd have to accelerate the vehicle inward proportional to the square of the velocity. If target exit velocity is 11 km/s, and the acceleration of the straight portion of the track is insignificant compared to the circular portion, in order for the combined linear and centripetal accelerations to stay under 3G, I'd make each component 20.8 m/s^2. So if 20.8 m/s^2 = (11000 m/s)^2/r, then r has to be 5800 km. Closing the ring at that radius makes the circumference larger than the distance needed for a linear track. Untrained humans could not ride it up to Earth-escape velocity. LEO only needs about 8 km/s. So that requires a radius of 20.8 m/s^2 = (8000 m/s)^2/r, 3000 km. Still too big. What if we bump the G threshold to 5G? Then centripetal portion of the acceleration should top out at 34.7 m/s^2, and required radius drops to 1850 km. Okay, cargo only then, just to LEO. 25G max acceleration, 8000 m/s exit velocity. Radius 370 km, circumference 2300 km. Still more track than needed by a linear launcher. It's always cheapest to put all the acceleration towards moving the vehicle in the direction you want it to go, instead of wasting some energy pushing it in different directions that ultimately cancel out. You could use a spiral track, but the final leg of the spiral is going to have an effective radius so large it might as well just be straight the whole way.
Robotbeat 7y ago
The closest I'm aware of for active research is SpinLaunch: https://www.nextbigfuture.com/2018/02/spinlaunch-is-using-large-centrifuges-to-accelerate-to-payloads-into-space-target-500000-per-launch.html https://www.nextbigfuture.com/2018/02/spinlaunch-is-using-la... ...but isn't using rails. Rails are kind of a pain anyway as they have a habit of eroding.
the8472 7y ago
A launch system can be stationary and larger, so can build a mass driver (big coil gun) instead of a railgun.
ars 7y ago
See here: https://en.wikipedia.org/wiki/Mass_driver https://en.wikipedia.org/wiki/Mass_driver

I assume that the same technology could be applied for launch systems, with one major change: maximum acceleration. An electric railgun would need to be very long in order to reach high velocity without undergoing huge accelerations as in the Navy railgun. Ramping up a vehicle with payload and fuel in it without destroying it is the name of the game.

godelski 7y ago

Do you mean like going to space? There's some research but after everyone figured it that all humans inside would be instantly squished they started researching other avenues. The g forces are pretty large and would damage a lot of cargo too.

ars 7y ago
> would be instantly squished Not so. That's only for a very short launch. You would actually have a track a few miles long (probably up the side of a mountain) and launch using that. The g-forces would be very mild.
- logfromblammo 7y ago
  You need about 1200km to reach LEO from the equator with max acceleration of about 3G. I did the math once, and a track from the Galapagos Islands to the peak of Chimborazo on the continent in Ecuador would do it.
  - ars 7y ago
    I was a little shocked at that number, so I ran the math again, and wow. Quite a bit longer than I expected. Maybe a circular track till you build up speed, then switch over to the mountain track? Of course the circular track would have g-forces due to that..... (Do you feel like doing more math?) I guess it's more practical to use the track just for the initial boost above the atmosphere, then a rocket to add speed. Also a track that can handle something moving at km/s is .... not easy.
    - logfromblammo 7y ago
      A circular track means you'd have to accelerate the vehicle inward proportional to the square of the velocity. If target exit velocity is 11 km/s, and the acceleration of the straight portion of the track is insignificant compared to the circular portion, in order for the combined linear and centripetal accelerations to stay under 3G, I'd make each component 20.8 m/s^2. So if 20.8 m/s^2 = (11000 m/s)^2/r, then r has to be 5800 km. Closing the ring at that radius makes the circumference larger than the distance needed for a linear track. Untrained humans could not ride it up to Earth-escape velocity. LEO only needs about 8 km/s. So that requires a radius of 20.8 m/s^2 = (8000 m/s)^2/r, 3000 km. Still too big. What if we bump the G threshold to 5G? Then centripetal portion of the acceleration should top out at 34.7 m/s^2, and required radius drops to 1850 km. Okay, cargo only then, just to LEO. 25G max acceleration, 8000 m/s exit velocity. Radius 370 km, circumference 2300 km. Still more track than needed by a linear launcher. It's always cheapest to put all the acceleration towards moving the vehicle in the direction you want it to go, instead of wasting some energy pushing it in different directions that ultimately cancel out. You could use a spiral track, but the final leg of the spiral is going to have an effective radius so large it might as well just be straight the whole way.

Robotbeat 7y ago

The closest I'm aware of for active research is SpinLaunch: https://www.nextbigfuture.com/2018/02/spinlaunch-is-using-large-centrifuges-to-accelerate-to-payloads-into-space-target-500000-per-launch.html https://www.nextbigfuture.com/2018/02/spinlaunch-is-using-la... ...but isn't using rails. Rails are kind of a pain anyway as they have a habit of eroding.

the8472 7y ago

A launch system can be stationary and larger, so can build a mass driver (big coil gun) instead of a railgun.

ars 7y ago

See here: https://en.wikipedia.org/wiki/Mass_driver https://en.wikipedia.org/wiki/Mass_driver