3 ms·

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

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.

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.