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Comment by jsprogrammer | Hacker News Reader

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.

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LEO is -29.8 MJ/kg - -62.6 MJ/kg = 32.8 MJ/kg (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

> That's not survivable for people.

Got me wondering. 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). 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.

> That's not survivable for people. (Or probably sparrows either.)

A peregrine falcon can take 20Gs [1], at least over short periods. Maybe we need to start an astrobird training program.

[1]: https://www.quora.com/How-much-G-force-can-a-bird-take

    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.

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.

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.

$ 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.

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.

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.

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.

The track would not run along the ground. You'd want to get height as quickly as possible to minimize atmosphere and track length.

Originating somewhere around Mojave and launching towards Las Vegas could work.

Alternatively, you may want to launch over the ocean for safety reasons, but it seems like you may be subject to more weather concerns.

You don't want to run the track along the ground? Where then, on a dirigible at the mesopause as suggested in http://postbiota.org/pipermail/tt/2011-August/009613.html?

Bouyant structures could be used, as well as wings, to suspend the track.

You can't run along the ground because you don't want to go too fast through atmosphere.

jsprogrammer has the merit of having actually done some calculation, which you evidently have not — although if you think resistance is a thing that inductance can have, maybe you are not in a position to do any calculations. Either way, you should rethink your attitude.

Naval railguns don't have the luxury of accelerating over hundreds of kilometers; they are optimized for muzzle velocity, not efficiency. The things I've seen videos of launch projectiles at about Mach 7.5 (2500 m/s) over about 7 meters. That implies an average acceleration of at least 45000 gees, which means that you're going to have to accept significant inefficiencies that you can avoid in a design that accelerates three thousand times more gently.

In practice, both rotary and linear electric motors typically have efficiencies of over 80%, often over 95%. The proposal in question is a 200-km-long linear electric motor running up the side of a mountain. It's clearly feasible, but it won't cost pennies per launch.

You do need to partially evacuate the launch tube to shove your launch vehicle through hundreds of kilometers of it.

It is not plausible that a naval railgun uses only 28 megawatts. Traveling 7 meters at an average of Mach 3.75 takes 5.6 ms; if the total energy output is 64 MJ, that's an average of 11.3 gigawatts†, which is a lot more than 28 megawatts or for that matter 78 megawatts. So what you do is you charge a big low-ESR capacitor bank (at less than 78 megawatts) before the shot, then discharge it during the shot (at tens of gigawatts). If you're doing that, though, you have no limit on how many railguns you can run from your 78 MW power plant, just a limit on how many total shots per second you can fire among all of them. Your entire paragraph on the topic is incoherent nonsense.

For comparison, a .22 LR 30-grain (1.94 g) copper-plated hollowpoint bullet traveling at 500 m/s out of a 510 mm AR-15 barrel only has at most 2 ms to accelerate to its final 240 J energy and therefore requires over 120 kW of power.

Classical mechanics are perfectly adequate for all of this. No relativistic or quantum effects are relevant. Your suggestion otherwise is absurd.

Calculations in units(1) format for those who want to check them:

    (mach 7.5)^2/2 / 7 meters / gravity
    7 meters / (mach 7.5/2)
    64 MJ / (7 meters / (mach 7.5/2))
    30 grains
    510 mm / (.5 500 m/s)
    30 grains (500 m/s)^2/2
    30 grains (500 m/s)^2/2 / (510 mm / (.5 500 m/s))

† With constant acceleration the power output ramps up linearly and ends up at twice the average. With constant power the acceleration ramps down instead, which means that you have to start out at even higher accelerations to get the same average acceleration, and your total time in the barrel is shorter, so your average power is higher, although I don't feel like doing the simple calculus to quantify this at the moment. In either case you have at least a point where the power is a few times higher than this average.

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.

> The electricity and cost of the rail gun isn't measured in pennies.

I think you should back this up with real numbers.

>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_disaste...

https://en.wikipedia.org/wiki/Soyuz_11

https://en.wikipedia.org/wiki/Vladimir_Komarov

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.

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.

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.

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.

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.p...

[1] - https://www.eia.gov/electricity/monthly/epm_table_grapher.cf...

That's 67GJ per KG? because otherwise either I'm very very wrong, or they are very very optimistic.

Looks like that's for a 300kg total weight projectile @ 20% energy efficiency. Details can be found on page 317-318.

Have a source on the electricity costing pennies?

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

Please see my response to kragen, where I derive the cost.

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