We don't use multistage rockets to get to orbit because we lack vision. We use them because they're the most effective way to do so.
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Multistage Rockets are the most efficient way to get to orbit.
I hate to be too rude, but that particular statement is BS(See my math further down). Even a cursory understanding or jet propulsion will tell you that
rockets are the least efficient type of jet engine.. This is because in an air-breathing jet engine there is an additional source of mass: the incoming air; oxidizer doesn't need to be carried with the craft. Because of this, the total mass is less, a given expenditure of energy will cause a grater change in momentum, since the energy that must be expended to achieve a given exhaust velocity increases by the square of velocity, where change in momentum is linear with respect to velocity. Just because it's the easy way doesn't mean it's the safest or most efficient.
Further, the space shuttle is an SSTO craft; it does not implement staging; this is unlike rockets such as the Saturn V. Are you implying that the Saturn V is an efficient design?
I suggest you run some ideal rocket equations for fuel use.
Nowhere in my design is there a conventional rocket. Ramjets are air breathing engines, meaning that the conventional equation for rocket propulsion is inaccurate; regardless, we can still come up with some equations:
To make an extremely conservative fuel estimate, assume 2% efficiency in conversion of fuel energy to altitude. Further ignore the fact that we don't have to carry the weight of spent fuel.
e = efficiency
Energy requirement E=m
Tg\Deltah/e
E=(m
Ship+m
Payload+m
Fuel)g\Deltah/e
k
h2 is a constant representing the energy released when 1kg of fuel combines with oxygen. Dimensions J/kg
E=M
FuelK
H2
system of linear eqs
( m
Ship + m
Fuel )( g \Deltah / e + V
2/(2e) - V
02 / 2e) = M
Fuel K
H2
M
Fuel=M
Ship(g \Deltah + V
2/2 - V
02 / 2)/( K
h2 e - V
2/2 + V
02 / 2 - g\Deltah)
Plugging in absurdly conservative numbers:
e = 2%
M
Ship (mass of everything except fuel) = 500000kg
V
0=mach 1.5
V=mach 6
\Deltah = 100000m
K
H2=247000000J/kg
M
Ship=500000kg
M
Fuel=1305000kg of LH
2 fuel
Giving a fuel to payload ratio of .383
The actual ratio using a less conservative equation should give a ratio closer to 1
Compare to 109000kg/(2*590000kg+760000kg)=.0561 for the space shuttle orbiter
Needing three different motors to get to orbit... will mean that the craft would have to be absolutely gargantuan
Jet engines weigh next to nothing in comparison to the fuel they need. 4 ramjet engines would actually be fairly light, figure 20000kg for all four.http://www.grc.nasa.gov/WWW/BGH/ramjet.html" (A Boeing 747 jet engine weighs about 5000kg, ramjets are lighter than turbofans). The type of ion drive I'm picturing would come in at less than 1000kg. Most of the weight will be in the enormous reactor powering the thing. Further, the fans are part of a carrier craft, which doesn't have to be carried into orbit.
...even if you could get enough thrust out of the ion engine (which you can't)...
I'll do the math on this as well, but I believe the kind of ion drive I'm thinking of would generate sufficient thrust. It isn't a conventional ion drive, but rather an atmospheric one; because air ionizes to some extent when contacting a hypersonic body, a simple electrostatic drive outside the spacecraft could accelerate the protons towards the back of the ship, while pushing electrons forward. The same work would be done on both sets of particles, but the protons would have a greater change in momentum, thus the ship accelerates forward. MythBusters built one out of a kit in an afternoon, and it levitated; can't be that hard to do, Ja?
I'm starting to think I should just move bits and pieces of my design to their own threads. I'll start a new one for the ion drive, because that's a discussion of it's own.
I'd like to hear some other possible designs other people have found/come up with; I didn't want this thread to be just for my own design.