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Acceleration over 200km will get you to orbit. The electricity required for a single launch is measured in US pennies. An aerodynamic vehicle could be extremel

Acceleration over 200km will get you to orbit. The electricity required for a single launch is measured in US pennies. An aerodynamic vehicle could be extremely safe; in the event of a failure, you could just glide back to the ground. The possibility of violent explosion would be practically non-existent.

ubercore 10y ago
Have a source on the electricity costing pennies?
- jsprogrammer 10y ago
  Please see my response to kragen, where I derive the cost.
- dogma1138 10y ago
  It's not going to cost pennies, it's an easy calculation treat energy efficiency as 100% efficient and calculate how much watts you would need to put say a 10,000KG payload in orbit and then calculate how much it would cost you in terms of electricity. F (d)x/(d)t = F * v = m * a * v = joules per second = watts
dogma1138 10y ago
The electricity and cost of the rail gun isn't measured in pennies. An Aerodynamic vehicle could be safe, but it would also mean it would generate considerable drag on the way up requiring more power, it also would be safe only after almost reaching orbit because it would be hypersonic out of the launch pad, if you can design a hypersonic glider NASA would like hear about it. Building a launch pad over 200KM is also not a simple feat, if you ask why people are building rockets it's because we have no clue how to build EM and by all accounts it's not sustainable for earth. EM launchers for the moon and even mars as well as large asteroid bases are considerably more sensible.
- jcoffland 10y ago
  > The electricity and cost of the rail gun isn't measured in pennies. I think you should back this up with real numbers.
bluehawk 10y ago
But where is the 200km of track? You also have to deal with the atmosphere. If the track is on the ground or low in the atmosphere, then when you come out of the "cannon" you would be torn apart. So that means the track (or at least the exit) needs to be above the atmosphere, which is quite a challenge in itself.
- jcoffland 10y ago
  You could use a 200km evacuated tube. There is of course the issue of maintaining the vacuum as the vehicle exits. This could be managed using a pair of diaphragm shutters. Then only the space between the shutters would need to be reevacuated after a launch. Shutter failure would be disastrous.
dsp1234 10y ago
According to this document from the USAF, the estimated energy needed to launch into orbit is 67GJ. Converting that to the US measurement of electricity is 18,611 kWh. According to the EIA, the US average for electricity, specifically for the transportation sector, was 10.22 pennies per kWh. So yes, a lot of pennies... edit: The same document mentions that 20 degree launch trajectory gives LEO == HEO delta-v. So that 200km track also needs to, naively, be 72km high at the end to have constant acceleration through that 200km. [0] - http://www.star-tech-inc.com/papers/lcls/low-cost_launch_2.pdf http://www.star-tech-inc.com/papers/lcls/low-cost_launch_2.p... [1] - https://www.eia.gov/electricity/monthly/epm_table_grapher.cfm?t=epmt_5_6_a https://www.eia.gov/electricity/monthly/epm_table_grapher.cf...
- dogma1138 10y ago
  That's 67GJ per KG? because otherwise either I'm very very wrong, or they are very very optimistic.
  - dsp1234 10y ago
    Looks like that's for a 300kg total weight projectile @ 20% energy efficiency. Details can be found on page 317-318.
hoorayimhelping 10y ago
>An aerodynamic vehicle could be extremely safe; in the event of a failure, you could just glide back to the ground. The possibility of violent explosion would be practically non-existent. That is a ridiculous statement, considering the deaths that have happened during reentry https://en.wikipedia.org/wiki/Space_Shuttle_Columbia_disaster https://en.wikipedia.org/wiki/Space_Shuttle_Columbia_disaste... https://en.wikipedia.org/wiki/Soyuz_11 https://en.wikipedia.org/wiki/Soyuz_11 https://en.wikipedia.org/wiki/Vladimir_Komarov https://en.wikipedia.org/wiki/Vladimir_Komarov
- extrapickles 10y ago
  Don't forget that the vehicle would be leaving the launcher at greater than reentry speeds, so it would be even worse than a normal re-entry.
- jsprogrammer 10y ago
  Not all vehicles are safe. The shuttle was >100,000kg incoming. Passenger vehicles to orbit could be on the order of 1,000kg, which would be much easier to make safe for gliding. Emergency parachutes could even be feasible.
kragen 10y ago
LEO is -29.8 MJ/kg - -62.6 MJ/kg = 32.8 MJ/kg (https://en.wikipedia.org/wiki/Orbital_speed#Tangential_velocities_at_altitude https://en.wikipedia.org/wiki/Orbital_speed#Tangential_veloc...) so one tonne is 32.8 GJ. At US$0.04/kWh (a common benchmark wholesale price for electrical energy in the US, usually stated as US$40/MWh) that works out to US$364, which is thirty-six thousand pennies. What are you launching, a sparrow? Also, doing it over only 200 km will require at least 16.7 gees. That's not survivable for people. (Or probably sparrows either.) Calculations in units(1) format in case I got something wrong (should the delta-specific-energy really be 32.8 MJ/kg rather than, say, 24 to 30?): (-29.8 - -62.6) MJ/kg * 1 tonne * US$ 0.04 / kWh (-29.8 - -62.6) MJ/kg / 200 km / gravity
- abecedarius 10y ago
  > That's not survivable for people. Got me wondering. https://en.wikipedia.org/wiki/G-force#Human_tolerance https://en.wikipedia.org/wiki/G-force#Human_tolerance reports an early experiment of a human taking 10g for 1 minute. That'd add up to 5.8 km/s, which is still under delta-v to LEO (9.4 km/s or more including air drag according to https://en.wikipedia.org/wiki/Low_Earth_orbit https://en.wikipedia.org/wiki/Low_Earth_orbit). But it's close, and the Wikipedia page doesn't say it's an upper bound. OTOH at 10g to 9.8km/s (higher delta-v to be conservative and for easier math) the track would need to be more like 490km long (and it'd take ~100sec). And maybe the deceleration on hitting the atmosphere would be worse, I don't know. It sounds more plausible for cargo.
- jsprogrammer 10y ago
  m=10,000kg a=90m/s^2 vf=7500m/s vi=0 F=ma W=Fd P=W(t^-1) t=(vf-vi)/a d=(vf-vi)^2/(2a) t=7500/90 ~83.33s d=7500^2/180 =312,500m F=10000(90) =900,000N W=900000(312,500) =281,250,000,000J P=90(281250000000)/7500 =3,375,000,000W 3,375,000kW * ((7500 / 90) / 3600 s/hr) = 78,125 kWh 78,125 kWh * $0.04/kWh = $3,125 / launch 10,000 kg/launch / 3,125 $/launch = 3.2 kg/$ = $0.3125/kg Looks very close to your numbers. I maintain my claim.
  - kragen 10y ago

ubercore 10y ago

Have a source on the electricity costing pennies?

jsprogrammer 10y ago
Please see my response to kragen, where I derive the cost.
dogma1138 10y ago
It's not going to cost pennies, it's an easy calculation treat energy efficiency as 100% efficient and calculate how much watts you would need to put say a 10,000KG payload in orbit and then calculate how much it would cost you in terms of electricity. F (d)x/(d)t = F * v = m * a * v = joules per second = watts

dogma1138 10y ago

The electricity and cost of the rail gun isn't measured in pennies. An Aerodynamic vehicle could be safe, but it would also mean it would generate considerable drag on the way up requiring more power, it also would be safe only after almost reaching orbit because it would be hypersonic out of the launch pad, if you can design a hypersonic glider NASA would like hear about it. Building a launch pad over 200KM is also not a simple feat, if you ask why people are building rockets it's because we have no clue how to build EM and by all accounts it's not sustainable for earth. EM launchers for the moon and even mars as well as large asteroid bases are considerably more sensible.

jcoffland 10y ago
> The electricity and cost of the rail gun isn't measured in pennies. I think you should back this up with real numbers.

bluehawk 10y ago

But where is the 200km of track? You also have to deal with the atmosphere. If the track is on the ground or low in the atmosphere, then when you come out of the "cannon" you would be torn apart. So that means the track (or at least the exit) needs to be above the atmosphere, which is quite a challenge in itself.

jcoffland 10y ago
You could use a 200km evacuated tube. There is of course the issue of maintaining the vacuum as the vehicle exits. This could be managed using a pair of diaphragm shutters. Then only the space between the shutters would need to be reevacuated after a launch. Shutter failure would be disastrous.

dsp1234 10y ago

According to this document from the USAF, the estimated energy needed to launch into orbit is 67GJ. Converting that to the US measurement of electricity is 18,611 kWh. According to the EIA, the US average for electricity, specifically for the transportation sector, was 10.22 pennies per kWh. So yes, a lot of pennies... edit: The same document mentions that 20 degree launch trajectory gives LEO == HEO delta-v. So that 200km track also needs to, naively, be 72km high at the end to have constant acceleration through that 200km. [0] - http://www.star-tech-inc.com/papers/lcls/low-cost_launch_2.pdf http://www.star-tech-inc.com/papers/lcls/low-cost_launch_2.p... [1] - https://www.eia.gov/electricity/monthly/epm_table_grapher.cfm?t=epmt_5_6_a https://www.eia.gov/electricity/monthly/epm_table_grapher.cf...

dogma1138 10y ago
That's 67GJ per KG? because otherwise either I'm very very wrong, or they are very very optimistic.
- dsp1234 10y ago
  Looks like that's for a 300kg total weight projectile @ 20% energy efficiency. Details can be found on page 317-318.

hoorayimhelping 10y ago

>An aerodynamic vehicle could be extremely safe; in the event of a failure, you could just glide back to the ground. The possibility of violent explosion would be practically non-existent. That is a ridiculous statement, considering the deaths that have happened during reentry https://en.wikipedia.org/wiki/Space_Shuttle_Columbia_disaster https://en.wikipedia.org/wiki/Space_Shuttle_Columbia_disaste... https://en.wikipedia.org/wiki/Soyuz_11 https://en.wikipedia.org/wiki/Soyuz_11 https://en.wikipedia.org/wiki/Vladimir_Komarov https://en.wikipedia.org/wiki/Vladimir_Komarov

extrapickles 10y ago
Don't forget that the vehicle would be leaving the launcher at greater than reentry speeds, so it would be even worse than a normal re-entry.
jsprogrammer 10y ago
Not all vehicles are safe. The shuttle was >100,000kg incoming. Passenger vehicles to orbit could be on the order of 1,000kg, which would be much easier to make safe for gliding. Emergency parachutes could even be feasible.

kragen 10y ago

LEO is -29.8 MJ/kg - -62.6 MJ/kg = 32.8 MJ/kg (https://en.wikipedia.org/wiki/Orbital_speed#Tangential_velocities_at_altitude https://en.wikipedia.org/wiki/Orbital_speed#Tangential_veloc...) so one tonne is 32.8 GJ. At US$0.04/kWh (a common benchmark wholesale price for electrical energy in the US, usually stated as US$40/MWh) that works out to US$364, which is thirty-six thousand pennies. What are you launching, a sparrow? Also, doing it over only 200 km will require at least 16.7 gees. That's not survivable for people. (Or probably sparrows either.) Calculations in units(1) format in case I got something wrong (should the delta-specific-energy really be 32.8 MJ/kg rather than, say, 24 to 30?): (-29.8 - -62.6) MJ/kg * 1 tonne * US$ 0.04 / kWh (-29.8 - -62.6) MJ/kg / 200 km / gravity

abecedarius 10y ago
> That's not survivable for people. Got me wondering. https://en.wikipedia.org/wiki/G-force#Human_tolerance https://en.wikipedia.org/wiki/G-force#Human_tolerance reports an early experiment of a human taking 10g for 1 minute. That'd add up to 5.8 km/s, which is still under delta-v to LEO (9.4 km/s or more including air drag according to https://en.wikipedia.org/wiki/Low_Earth_orbit https://en.wikipedia.org/wiki/Low_Earth_orbit). But it's close, and the Wikipedia page doesn't say it's an upper bound. OTOH at 10g to 9.8km/s (higher delta-v to be conservative and for easier math) the track would need to be more like 490km long (and it'd take ~100sec). And maybe the deceleration on hitting the atmosphere would be worse, I don't know. It sounds more plausible for cargo.
jsprogrammer 10y ago
m=10,000kg a=90m/s^2 vf=7500m/s vi=0 F=ma W=Fd P=W(t^-1) t=(vf-vi)/a d=(vf-vi)^2/(2a) t=7500/90 ~83.33s d=7500^2/180 =312,500m F=10000(90) =900,000N W=900000(312,500) =281,250,000,000J P=90(281250000000)/7500 =3,375,000,000W 3,375,000kW * ((7500 / 90) / 3600 s/hr) = 78,125 kWh 78,125 kWh * $0.04/kWh = $3,125 / launch 10,000 kg/launch / 3,125 $/launch = 3.2 kg/$ = $0.3125/kg Looks very close to your numbers. I maintain my claim.
- kragen 10y ago