4 ms·

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^

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
Your calculations are correct, with the caveat that that's the average power assuming constant acceleration, not the maximum power. But you can't launch a one-kilogram thing and get it through Earth's atmosphere; you need an aerodynamic fairing and ablative tiles and whatnot, not to mention whatever kind of inductive apparatus you're using to grab hold of it in the first place. Thus my remark about the sparrow! Until your objects get long and thin enough that drag matters, Newton's impact depth approximation applies to the atmosphere: if you're going straight up, you have roughly 10 g of air per cross-sectional mm² that you will run into on the way, assuming you can keep the hypersonic aerodynamics sufficiently under control to keep from just totally tumbling end over end, which is harder than it sounds. At the shallow angles available running up mountainsides, the situation is several times worse. It's unfortunate that people incapable of doing calculations themselves are downvoting you. Calculations like these, plus experiments to validate them, are how we got rockets in the first place.
dsp1234 10y ago
The electricity required for a single launch is measured in US pennies. 78,125 kWh x $0.04/kWh = $3,125 / launch I maintain my claim. The two claims above are incompatible.
- jsprogrammer 10y ago
  $ is a unit of pennies $=100p The calculation above was for a 10Mg payload. Payloads up to 3.2kg would cost under 100p under the same assumptions.
  - dogma1138 10y ago
    Your calculation is very very incorrect, not the math, but the concept you are missing too many variables. The US navy actually is building railguns, their efficiency is very very low due to the resistance from inductance it seems that if we go the the equations for railguns your 200KM EM gun cannot be built. Overall the US is designing a 64MJ railgun, this gun can't put anything into orbit, it will have a range of about 20 miles, the ship that is going to be equipped with it is going to have a 78MW power plant and while it can power a single rail gun it will not be able to power multiple ones. By the US Navy's own calculations it would require 28MW to launch a projectile at 32MJ which which means yeah.... these figures are all off by orders of magnitude. It seems there is much more to railguns than classical mechanics.
    - jsprogrammer 10y ago
      I believe the barreled design to be fatally flawed. An earthquake or other disruption during launch could be catastrophic to the vehicle. Much better for the vehicle to ride next to the track; in failure scenarios, the vehicle can simply detach and glide back to earth. Yes, it means you have to go through the atmosphere, but it doesn't take far to clear. That will effect the calculations some, but not much. What will really effect the calculations though, is the cost of electricity. Generation costs will surely drop below $0.04. Further, transmission loss can be almost entirely mitigated by generating and supplying the required power on-track. Launches in favorable conditions (moon and/or planet alignments) would probably make some launches even cheaper. Of course, if the launcher were operating continuously, the savings would be used up during unfavorable conditions.
      - kragen 10y ago
        You're suggesting that we should magnetically levitate the launch vehicle in free air above a track running along the ground? That seems like it makes the problem a lot harder — instead of having to push it through just the air between here and space, which is about ten tonnes per cross-sectional square meter, you're pushing it through another two or three hundred kilometers of air, which is about another 200 tonnes of air per square meter. You know that shroud of plasma surrounding a re-entering spacecraft? That's the power required to push an orbital-speed object through air — but in that case without even maintaining velocity, let alone rapidly accelerating, and in that case it's the rarefied air of the stratosphere. You're proposing to do that for the majority of the track. That seems like a bad idea. Yes, generation costs will likely drop significantly below US$0.04/kWh eventually. But that's a Kardashev-Type-1 kind of event. Generating the power on-track may not turn out to be less expensive than long-distance transmission, because it depends on things like sunlight availability. Of the few suitable sites, most are pretty cloudy on one side. No moon or planet alignments significantly reduce the energy barrier to get to orbit.

kragen 10y ago

Your calculations are correct, with the caveat that that's the average power assuming constant acceleration, not the maximum power. But you can't launch a one-kilogram thing and get it through Earth's atmosphere; you need an aerodynamic fairing and ablative tiles and whatnot, not to mention whatever kind of inductive apparatus you're using to grab hold of it in the first place. Thus my remark about the sparrow! Until your objects get long and thin enough that drag matters, Newton's impact depth approximation applies to the atmosphere: if you're going straight up, you have roughly 10 g of air per cross-sectional mm² that you will run into on the way, assuming you can keep the hypersonic aerodynamics sufficiently under control to keep from just totally tumbling end over end, which is harder than it sounds. At the shallow angles available running up mountainsides, the situation is several times worse. It's unfortunate that people incapable of doing calculations themselves are downvoting you. Calculations like these, plus experiments to validate them, are how we got rockets in the first place.

dsp1234 10y ago

The electricity required for a single launch is measured in US pennies. 78,125 kWh x $0.04/kWh = $3,125 / launch I maintain my claim. The two claims above are incompatible.

jsprogrammer 10y ago
$ is a unit of pennies $=100p The calculation above was for a 10Mg payload. Payloads up to 3.2kg would cost under 100p under the same assumptions.
- dogma1138 10y ago
  Your calculation is very very incorrect, not the math, but the concept you are missing too many variables. The US navy actually is building railguns, their efficiency is very very low due to the resistance from inductance it seems that if we go the the equations for railguns your 200KM EM gun cannot be built. Overall the US is designing a 64MJ railgun, this gun can't put anything into orbit, it will have a range of about 20 miles, the ship that is going to be equipped with it is going to have a 78MW power plant and while it can power a single rail gun it will not be able to power multiple ones. By the US Navy's own calculations it would require 28MW to launch a projectile at 32MJ which which means yeah.... these figures are all off by orders of magnitude. It seems there is much more to railguns than classical mechanics.
  - jsprogrammer 10y ago
    I believe the barreled design to be fatally flawed. An earthquake or other disruption during launch could be catastrophic to the vehicle. Much better for the vehicle to ride next to the track; in failure scenarios, the vehicle can simply detach and glide back to earth. Yes, it means you have to go through the atmosphere, but it doesn't take far to clear. That will effect the calculations some, but not much. What will really effect the calculations though, is the cost of electricity. Generation costs will surely drop below $0.04. Further, transmission loss can be almost entirely mitigated by generating and supplying the required power on-track. Launches in favorable conditions (moon and/or planet alignments) would probably make some launches even cheaper. Of course, if the launcher were operating continuously, the savings would be used up during unfavorable conditions.
    - kragen 10y ago
      You're suggesting that we should magnetically levitate the launch vehicle in free air above a track running along the ground? That seems like it makes the problem a lot harder — instead of having to push it through just the air between here and space, which is about ten tonnes per cross-sectional square meter, you're pushing it through another two or three hundred kilometers of air, which is about another 200 tonnes of air per square meter. You know that shroud of plasma surrounding a re-entering spacecraft? That's the power required to push an orbital-speed object through air — but in that case without even maintaining velocity, let alone rapidly accelerating, and in that case it's the rarefied air of the stratosphere. You're proposing to do that for the majority of the track. That seems like a bad idea. Yes, generation costs will likely drop significantly below US$0.04/kWh eventually. But that's a Kardashev-Type-1 kind of event. Generating the power on-track may not turn out to be less expensive than long-distance transmission, because it depends on things like sunlight availability. Of the few suitable sites, most are pretty cloudy on one side. No moon or planet alignments significantly reduce the energy barrier to get to orbit.