Fusion for Ground Launch - Is it theoretically possible?

Euler · 2004-08-04 03:40:14

deltaV = Isp/m

Actually, deltaV=Isp*ln(mass ratio)

It also seems pretty silly to make an argument about the energy efficiency of a rocket without having energy or power be a factor in any of your equations.

I’m sorry, but you’re wrong. What matters is not momentum but change in momentum (the impulse).

The impulse imparted to the rocket is equal and opposite to the momentum of the exhaust (Newton's third law). The kinetic energy of the exhaust is equal to the energy output of the reactor, minus inefficiency. Thrust is equal to impulse/time. Lets assume that the reactor transfers power to the exhaust with 50% efficiency.

So you get:
Thrust= time derivative of exhaust mass*exhaust velocity= time derivative of exhaust mass*g*Isp.
Power= time derivative of exhaust mass*exhaust velocity^2=time derivative of exhaust mass*(g*Isp)^2
Thrust= Power/(g*Isp)

The higher the Isp, the lower the thrust.

JimM · 2004-08-04 04:36:11

deltaV = Isp/m
Actually, deltaV=Isp*ln(mass ratio)

Interesting point.

deltaV = Isp*ln(mass ratio)

...is of course the guts of the rocket equation.

However...

deltaV = Isp/m

...is something quite different. Here, 'm' is not the mass ratio (as it seems you have assumed in error) but the mass of the object, be it rocket or not. Recheck how I derived this from the equation for impulse, and I am confident you will see that I'm right.

Thrust is equal to impulse/time.

If T = thrust, Tsp = specific thrust, and t =time in seconds

T = I/t

...then if time = 1 second

Tsp = Isp/1

... thus specific thrust = specific impulse.

Therefore, the higher the Isp, the higher the thrust.

QUED (again)

BTW: We don't know what the actual power output of the reactor--or it's efficicency--is, so there's little point in speculating on 50% or whatever efficiency--although I must comment that if it's 50%, it's only a question of time, probably measured in a pretty small number of seconds, before there's a nice big mushroom cloud where the vehicle was. (All that heat has to go somewhere.)

SpaceNut · 2004-08-04 06:46:17

In the equation is not the Mass changing also as fuel is burned?

JimM · 2004-08-04 06:57:43

In the equation is not the Mass changing also as fuel is burned?

You're quite right, that's what happens with a rocket, and that's what the rocket equation handles. But in the case of the HOTL vehicle, it's not a rocket, so mass is assumed to be constant. Also in this particular case, the propellant fraction of the rocket (the VTOL) is so small (about 9%) that for a first-order approximation for comparison purposes with the HOTL, it can be ignored.

Euler · 2004-08-04 12:52:45

I = F * t
… where I = impulse, F = force, and t = time. When t = 1 second, this is Isp.

Actually, Isp is impulse/propellant mass. This is why your "deltaV = Isp/m" is only true if the mass of propellant used is 1, and even then it is only an approvimation because it does not take into account the changing mass of the vehicle.

If T = thrust, Tsp = specific thrust, and t =time in seconds
T = I/t
...then if time = 1 second
Tsp = Isp/1
... thus specific thrust = specific impulse.
Therefore, the higher the Isp, the higher the thrust.

If the engines both use up propellant at the same rate. However, by using up more propellant, the engine with lower Isp can produce the same amount of thrust while using less energy.

BTW: We don't know what the actual power output of the reactor--or it's efficicency--is, so there's little point in speculating on 50% or whatever efficiency--although I must comment that if it's 50%, it's only a question of time, probably measured in a pretty small number of seconds, before there's a nice big mushroom cloud where the vehicle was. (All that heat has to go somewhere.)

I just said 50% to get rid of the 1/2 in the equation. The point of lowering the Isp of the engine is that it will only need a relatively small reaction, which means that you have less heat and you dont have to worry as much about the heat destroying the vehicle.

QUED

You mean Q.E.D.?

JimM · 2004-08-04 15:15:30

I = F * t
… where I = impulse, F = force, and t = time. When t = 1 second, this is Isp.
Actually, Isp is impulse/propellant mass. This is why your "deltaV = Isp/m" is only true if the mass of propellant used is 1, and even then it is only an approvimation because it does not take into account the changing mass of the vehicle.

Actually, if you take special note you will see that the 'm' in question is not the mass of propellant but the mass of the entire vehicle, propellant and all.

Further, if you care to read my most recent answer to SpaceNut, you will see that I chose to disregard the changing mass of the VTOL (changing mass does not apply to the HOTL in any case) for simplicity, and why.

Therefore, deltaV = Isp/m holds generally.
If the engines both use up propellant at the same rate. However, by using up more propellant, the engine with lower Isp can produce the same amount of thrust while using less energy.

Clearly, if the propellant (NOT the engine) with the lower Isp is used in sufficiently greater quantities, as high a deltaV (NOT thrust) can be achieved as with the higher Isp. But that would require the use of more--a lot more energy, not less. There is absolutely nothing to gain and a lot to loose by deliberately using a low-Isp propellant.

Indeed, I suspect that one of the ways to resolve the reactor efficiency problem is to use as high-Isp propellant as possible, as that will make a significant difference to efficiency in thermodynamic terms; sufficient to explain the apparent 'something-for-nothing' aspect of higher Isp delivering higher deltaV that's bothering you, I'd not be surprised. (But it's only a guess. I don't have the time or inclination to work that one out in detail.)

(Also, the heating of a highly hypersonic airflow sufficiently to deliver specified thrust, even with a fusion reactor and magnetic containment, is going to be asking a lot. At (say) 20,000 fps, it's difficult to see how the ambient air can be exposed to the fusion-generated heat for more than about, oh... say... one thousandth of a second--probably much less... and at 100,000 ft, there's not going to be a lot to use anyway. Once again, you're going to have a problem getting rid of waste heat. And think of all the global warming--not CO2, just raw heat--you'd be accused of making that doesn't apply to a VTOL. Ho-hum.)

QUED: when I was at school, this was the standard abbreviation of 'quad erat demonstrandum.'

C M Edwards · 2004-08-04 15:42:25

Not only does conservation of energy prohibit getting more power out of the thermal (inert propellant) exhaust than comes out of the fusion reaction, but, if the fusion reaction products must be vented along with the inert propellant, the first and second laws of thermodynamics prohibit getting a working thrust greater than that theoretically available from the fusion reaction alone. In other words, given a choice between using a fusion rocket engine to provide thrust and using a vented fusion reactor _with the same reaction rate_ to heat any given volume of inert propellant in a thermal rocket engine, the fusion rocket engine provides more thrust.

If thrust is all you’re after, and absolutely nothing else in the universe matters (not mass, complexity, power, or anything), then it’s better just to use a straight fusion rocket engine and forget about all that inert propellant stuff.

However…

Use of an inert propellant can simplify cooling. All of that inert propellant is potential coolant. The fusion reactor venting is cooler (having been tapped for power to heat the thermal propellant). The thermal rocket exhaust is cooler. Everything’s cooler. So, the engine mass doesn’t skyrocket as fast. As long as you were willing to settle for less thrust and specific impulses well below the 2 million Isp expected for a pure fusion rocket engine, using a reactor to heat an inert propellant is the way to go.

The engine might even ultimately weigh enough less to allow ground launch even with the reduced thrust curve, but don’t quote me on that. It all depends on whether the engine weight can be made lower than the thrust, and I have serious doubts about whether that’s doable with fusion power.

But I have to admit it’s not _impossible_. You just have to give up a lot of specific impulse.

Euler · 2004-08-04 15:53:48

Actually, if you take special note you will see that the 'm' in question is not the mass of propellant but the mass of the entire vehicle, propellant and all.

Obviously. However, to derive it you assumed that the propellant will have a mass of 1. If the propellant does not have a mass of 1, then the formula is completely bogus.

Clearly, if the propellant (NOT the engine) with the lower Isp is used in sufficiently greater quantities, as high a deltaV (NOT thrust) can be achieved as with the higher Isp. But that would require the use of more--a lot more energy, not less. There is absolutely nothing to gain and a lot to loose by deliberately using a low-Isp propellant.

If all of the energy used to accelerate the fuel is contained in the fuel itself(as in chemical engines), then a higher Isp is always better. However, if the energy comes from another source(as in electric engines), then it is often better to have a lower Isp (which is why electric engines are often run with an Isp well below the maximum that they could produce).

In this case, if hydrogen and another fuel exit the engine with the same temperature, the hydrogen will have a higher Isp. However, the specific heat of hydrogen is much higher than that of any other material. This makes it so that the hydrogen will end up needing more energy to produce the same amount of impulse.

Euler · 2004-08-04 15:59:08

Not only does conservation of energy prohibit getting more power out of the thermal (inert propellant) exhaust than comes out of the fusion reaction, but, if the fusion reaction products must be vented along with the inert propellant, the first and second laws of thermodynamics prohibit getting a working thrust greater than that theoretically available from the fusion reaction alone.

How do they do that? You don't deed more power to get more thrust.

C M Edwards · 2004-08-05 09:33:57

Hmm...

Euler, I found a flaw in my analysis. Conditions do exist such that:

F.generator_vent + F.thermal_rocket > F.fusion_rocket

is possible. This happens where the change in temperature for the thermal exhaust is small. (I missed it because I was only looking at the efficiency of power transfer, and not paying enough attention to the amount of mass the power was being transferred to.) For example, if enough thermal propellant is used to soak up the generator power, the whole thermal propellant reaction mass might only be heated enough to vaporize it (say, to room temperature). Greater thrust can be had at a lower exhaust velocity without any change in power input.

That goes back to my earlier post about the relationship between thrust, power and exhaust velocity. Oops.

But, this may not necessarily change the outcome of the argument, because if you drop the thermal exhaust velocity to take advantage of this, the corresponding drop in specific impulse would make the rocket weight curve increase again. Obviously, that thrust curve has to get over that weight curve for the thing to fly. You can only drop the specific impulse so low before you might as well be using a chemical rocket.

Regarding engine weight, I think a further reduction in engine weight could be had by adopting a design philosophy of “It only has to run for fifteen minutes and I don’t care what it does after that.” That may be the ultimate means of addressing the cooling problem – let the heat destroy the engine.

JimM · 2004-08-07 09:47:32

Actually, if you take special note you will see that the 'm' in question is not the mass of propellant but the mass of the entire vehicle, propellant and all.
Obviously. However, to derive it you assumed that the propellant will have a mass of 1. If the propellant does not have a mass of 1, then the formula is completely bogus.

I'm sorry for the delay in answering, and also sorry for my error...

deltaV = Isp/m

...should have been written...

deltaV <is proportional to> Isp/m

This could also be written, perhaps more clearly, as...

deltaV = (Isp/m)k

... where k is a constant of unspecified valus.

You said a little while back...

...the engine with lower Isp can produce the same amount of thrust while using less energy.

I now put it to you that, by the process of reductio ad absurdum, the logic of this statement can be shown to be absurd and thus false:-

What you are telling us is that if the vehicle' propellant (not engine) had an Isp of zero that would deliver the same thrust while absorbing zero energy—and also that if the propellant had an Isp of infinity it could only deliver the same thrust if it absorbed infinite energy.

So, discarding your earlier proposal and given that deltaV is shown to be directly proportion to the Isp of the propellant and in inverse proportion to the mass of the vehicle, this means that the higher the Isp for a given ship mass the better the acceleration.

Now of course in the case of a rocket the ship mass will decrease as propellant is used up, but as I have earlier explained, in the case of the VTOL ship with a presumed Isp of 10000, this would be a relatively small effect and has been ignored to ease comparison with the HOTL sc/ram/jet-powered alternative. (This also works to the advantage of the HOTL, BTW)

A HOTL with an Isp of (say) 1000 from ambient air would thus be at a very considerable disadvantage; it's deltaV (and hence acceleration and hence impulse) would be very much poorer than the VTOL. On the other hand it has the apparent advantage of what might be called an 'infinite propellant tank' but to be set against that are the following, largely overlooked here, disadvangages:

(1) It has to fly through that 'tank', creating considerable drag (mostly in the supersonic and hypersonic speed ranges) which it has to fight against. This is probably equivalent to at least a halfing of the effective Isp.

(2) At 'cruising altitude', said to be 100,000 ft, there is going to be very little ambient air to use as propellant, which will have the effect of curtailing thrust considerably.

(3) It has to apply several thousands of degrees of heating to air that at least in the case of the scramjet, will only be exposed to the heater for a minute fraction of a second--and will have already entered the champer at a very high temperature due to collision with the hypersonic vehicle, but what matters is what I might call the deltaHeat.

(4) It is still going to need to be able to fly as a rocket anyway, to regularise its orbit and to re-enter at the end of its mission.

(5) It will basically be confined to shuttle-style missions between earth and LEO, while the VTOL, with additional fuel (easily accommodated) could fly around the solar system more or less at will, possibly re"fuel"ing at Titan, Mars, or wherever else is appropriate.

(6) It can't start from rest at the end of the runway powered by sc/ram/jets. They don't work until considerable speed is already achieved. We still have not been told how this is to be done.

C M Edwards · 2004-08-07 12:41:37

...the engine with lower Isp can produce the same amount of thrust while using less energy.
I now put it to you that, by the process of reductio ad absurdum, the logic of this statement can be shown to be absurd and thus false:-
What you are telling us is that if the vehicle' propellant (not engine) had an Isp of zero that would deliver the same thrust while absorbing zero energy—and also that if the propellant had an Isp of infinity it could only deliver the same thrust if it absorbed infinite energy.
So, discarding your earlier proposal and given that deltaV is shown to be directly proportion to the Isp of the propellant and in inverse proportion to the mass of the vehicle, this means that the higher the Isp for a given ship mass the better the acceleration.

Sorry, no.

The objection is about these engines requiring too much power, which is correctly inferred (but admittedly not shown anywhere in this discussion) to mean requiring too much heat. It's not about them having too much specific impulse.

Engine power is proportional to some power of specific impulse. Your "reduction to absurdity" is a trivial observation regarding one data point. (A mathematically trivial case, that is - meaning it's true but conveys no useful information.)

You would do better to consider the behavior over a range of specific impulses that will actually give a power output, rather than at the one value that you know will not. For example, consider rocket engines with exhaust velocities of 10 m/s (about what you'd expect from a child's squirt gun), and 1000 m/s (about what you'd expect from a rocket powered dragster).

If you're like me, your first impulse is that the rocket powered dragster produces the most thrust. But if the one with 10 m/s were the Hoover Dam instead? It has an exhaust velocity similar to a squirt gun. Adjust the gates just right, and it can put out the same power as the rocket powered dragster. (You would have to stopper the Hoover Dam down quite a bit to reduce it to the output of a measly dragster rocket, I might add.)

This equation gives the power of each:

P = 1/2 * F * v

So, this equation gives the thrust of each:

F = 2 * P / v

Given the same power, the engine with the smaller value of v gives more thrust. Alternately, given a smaller exhaust velocity, a rocket engine MUST have more thrust to develop the same power.

JimM · 2004-08-07 14:16:18

You would do better to consider the behavior over a range of specific impulses that will actually give a power output, rather than at the one value that you know will not. For example, consider rocket engines with exhaust velocities of 10 m/s (about what you'd expect from a child's squirt gun), and 1000 m/s (about what you'd expect from a rocket powered dragster).
If you're like me, your first impulse is that the rocket powered dragster produces the most thrust. But if the one with 10 m/s were the Hoover Dam instead? It has an exhaust velocity similar to a squirt gun. Adjust the gates just right, and it can put out the same power as the rocket powered dragster. (You would have to stopper the Hoover Dam down quite a bit to reduce it to the output of a measly dragster rocket, I might add.)

Sorry, you're wrong too.

The example you give is misleading. Suppose, instead, we compared one squirt gun with an exhaust velocity of 1 m/s and another squirt gun, otherwise identical, with an exhaust velocity of 100 m/s. The volume and mass of water squirted is identical; only the velocity differs. So which delivers the most thrust?

Clearly, the answer is the gun with the most powerful spring (or however squirt guns do it) In other words, exhaust velocity is directly proportional to thrust, when mass of water (or, in our case, propellant) is identical.

And so it becomes clear that exhaust velocity and hence thrust, and hence deltaV, is directly proportional to Isp, everything else being equal.

Thus, a VTOL machine using LH2 propellant (let's say, Isp 10000) at the rate of (let's say) 1 ton/sec, will deliver greater—significantly greater—thrust (and deltaV) than a HOTL using ambient air (let's say Isp 1000) also at the rate of 1 ton/sec.

So it is clear (I'd say, obvious) that the only putative advantage a low-Isp HOTL has over a high-Isp VTOL is that it can expell a greater mass/sec, or the same mass for longer, or perhaps both.

But this is a purposeless advantage, given the low mass of LH2 propellant needed to get the VTOL to LEO, already demonstrated to be less than the mass-penalty of adding on wings, ramjets, and whatever else is needed to make a HOTL work...

...plus all the other reasons I listed earlier.

Euler · 2004-08-07 18:30:32

Say you have a your VTOL with isp=10,000, and an HTOL with isp=1000. They both have identical reactors, producing identical amounts of power. The HTOL uses propellant at a rate 100 times greater than the VTOL does, and therefore is able to produce 10 times more thrust. That is its advantage.

JimM · 2004-08-08 04:20:39

Say you have a your VTOL with isp=10,000, and an HTOL with isp=1000. They both have identical reactors, producing identical amounts of power.

... except that HOTL couldn't, where it mattered.

Firstly, at take off: without yet another (expensive and heavy) means of propulsion, its thrust at the proposed start of its run down the runway would be zero.

Secondly, at 'cruise' at 100,000 ft, where it could not find enough atmosphere to turn into propellant—except perhaps by having enormous intakes that would (a) multiply the drag problem manyfold and (b) add significantly to the mass and complexity of the vehicle.

The HTOL uses propellant at a rate 100 times greater than the VTOL does, and therefore is able to produce 10 times more thrust. That is its advantage.

Except that it would not; it could only only produce the same thrust, at best. And for various practical reasons already referred to, it certainly could not use propellant at a higher rate than the VTOL during 'launch' (take off) or probably, at 'cruise'.

And in any case, even if you were right (which I do not accept) the power reactor in the VTOL can be made bigger with much greater ease and almost no weight penalty—not possible for HOTL

All-in-all, HOTL is an over-complex, impractical and self-limiting way to get to orbit, if it could ever be made to work.

Furthermore VTOL is amost endlessly scaleable; HOTL is not.
------------------
VTOL is a true spaceship; HOTL is not, but no more than an aircraft that pops momentarily out of the atmosphere before dropping back down again (how?) to where it's at home.

C M Edwards · 2004-08-09 08:00:20

Suppose, instead, we compared one squirt gun with an exhaust velocity of 1 m/s and another squirt gun, otherwise identical, with an exhaust velocity of 100 m/s. The volume and mass of water squirted is identical; only the velocity differs. So which delivers the most thrust?
Clearly, the answer is the gun with the most powerful spring (or however squirt guns do it) In other words, exhaust velocity is directly proportional to thrust, when mass of water (or, in our case, propellant) is identical.
And so it becomes clear that exhaust velocity and hence thrust, and hence deltaV, is directly proportional to Isp, everything else being equal.

Ah. We may be coming down to the difference in our arguments. The difference appears to be what we're holding equal.

A little earlier in this discussion, I posted that entropy and the mechanics of power transfer was a prime reason why a thermal rocket powered by a vented reactor was less thrust efficient than a rocket based on a direct reaction. Euler called me on it, and two posts later I had to back down from that position because I hadn't allowed for differences in reaction mass between the two types of engine.

If you insist that the two types of engine must each have the exact same reaction mass, then the entropy objection is again valid, and I can now go back and stand by my earlier assertion.

I'm not going to take that option, because I'm ultimately more interested in thrust and the power required to produce it. Reaction mass is incidental to me, so I'm not bothering with the assumption that reaction mass is constant.

Your comparison of two water guns with equal reaction mass neglects the power required to produce the thrust in each case. Thus, it neglects to address my objection.

What about the engine power in each case?

C M Edwards · 2004-08-09 15:07:28

Post Script for Jim:

Another difference is that we might be arguing two different threads under the same topic. Can't really say I care about VTOL vs. HTOL.

SpaceNut · 2004-08-11 09:17:17

Info on ongoing research by nasa on propulsion design.

Redesigning Rockets: NASA Space Propulsion Finds a New Home
http://www.space.com/businesstechnology … 40811.html

SpaceNut · 2004-08-27 09:00:37

I was not quite sure of which topic to post this under but her goes.
We all know that developement of Nuclear power for space use is a must but in the same vein we know how dangerous it can be as well. Though a slow process of learning and or understanding we can build safer and more robust reactors. We must over come or fears of it for the eventual use of it in space to shorten the length of time to the planets and to give the much needed power resource for use for extended flights.

Analysis: Nuclear Power Gaining Popularity
http://www.spacedaily.com/news/nuclear-civil-04l.html

SpaceNut · 2004-09-02 11:45:58

Sealed nuclear reactor sure would give the activist little to worry about when it comers to nuclear energy use. I wonder if it would be adaptable for space use?

Public Release: 2-Sep-2004
US plans take-away nuclear power plants

http://www.eurekalert.org/pub_releases/ … 090204.php

A small, sealed, transportable nuclear reactor could meet the energy needs of developing countries without the risk of by-products getting in the wrong hands for weapons programmes. The sealed units, being developed by the US Department of Energy, would be delivered to a site, and collected when the fuel runs out after about 30 years. Its tamper-proof cask would be monitored and heavily alarmed.

Contact: Claire Bowles
claire.bowles@rbi.co.uk
44-207-331-2751
New Scientist

GCNRevenger · 2004-09-02 12:05:51

Another option is to use a Pebble-Bed reactor... in this type of reactor, the fuel is held tightly inside ceramic beads that are only a millimeter or so across. Then, they are wrapped inside another ceramic and graphite shell before being placed into the reactor instead of usual fuel rod assemblies. The packages would be so hard to open and retrieve useful fuel that they are essentially un-recyclable.

The idea of a small power reactor that would be maintenance free for thirty years would be a momentous boon to world energy needs.

SpaceNut · 2004-09-02 12:31:50

So a little research later into the subject I find that it is Helium cooled rather than water as the typical reactor is and that the pebbles are recycled until they are no longer of any value with in the reaction chamber. Also none are in operation but some are in the construction stages. Technology is still on the cutting edge of developement.

SpaceNut · 2004-10-07 10:29:29

So we are now getting some interest in fusion in the form of a $12.5M In Subcontracts Awarded For Fusion Experiment At Princeton. The awarding of two subcontracts for the fabrication of major components for the National Compact Stellarator Experiment (NCSX), now under construction at the Laboratory.

http://www.spacedaily.com/news/energy-tech-04zzq.html

National Compact Stellarator Experiment design

Martian Republic · 2004-10-07 12:52:26

So a little research later into the subject I find that it is Helium cooled rather than water as the typical reactor is and that the pebbles are recycled until they are no longer of any value with in the reaction chamber. Also none are in operation but some are in the construction stages. Technology is still on the cutting edge of developement.

Well, that good news! If we are able to solve the fusion problem, it will solve a lot of problems both down here and in space. It might be possible to build aerospace plane out of that technology. Even if you happen to have crashed it into a mountain, it would cause less pollution than either chemical fission air planes or rockets. You would have less fuel on board and it would be just hydrogen, helium and the nuclear process would stop and there would be no nuclear waist to deal with. It also generates more power too. It would be the idea power source, if we can develop it.

Larry,

SpaceNut · 2004-11-11 08:03:52

Sandia Imagists Overcome Maelstrom Obscuring Z Machine's Drive Force

Last year, its central mechanism, called a Z-pinch, fused isotopes of hydrogen to create nuclear fusion.

New Mars Forums

Announcement

#51 2004-08-04 03:40:14

Re: Fusion for Ground Launch - Is it theoretically possible?

#52 2004-08-04 04:36:11

Re: Fusion for Ground Launch - Is it theoretically possible?

#53 2004-08-04 06:46:17

Re: Fusion for Ground Launch - Is it theoretically possible?

#54 2004-08-04 06:57:43

Re: Fusion for Ground Launch - Is it theoretically possible?

#55 2004-08-04 12:52:45

Re: Fusion for Ground Launch - Is it theoretically possible?

#56 2004-08-04 15:15:30

Re: Fusion for Ground Launch - Is it theoretically possible?

#57 2004-08-04 15:42:25

Re: Fusion for Ground Launch - Is it theoretically possible?

#58 2004-08-04 15:53:48

Re: Fusion for Ground Launch - Is it theoretically possible?

#59 2004-08-04 15:59:08

Re: Fusion for Ground Launch - Is it theoretically possible?

#60 2004-08-05 09:33:57

Re: Fusion for Ground Launch - Is it theoretically possible?

#61 2004-08-07 09:47:32

Re: Fusion for Ground Launch - Is it theoretically possible?

#62 2004-08-07 12:41:37

Re: Fusion for Ground Launch - Is it theoretically possible?

#63 2004-08-07 14:16:18

Re: Fusion for Ground Launch - Is it theoretically possible?

#64 2004-08-07 18:30:32

Re: Fusion for Ground Launch - Is it theoretically possible?

#65 2004-08-08 04:20:39

Re: Fusion for Ground Launch - Is it theoretically possible?

#66 2004-08-09 08:00:20

Re: Fusion for Ground Launch - Is it theoretically possible?

#67 2004-08-09 15:07:28

Re: Fusion for Ground Launch - Is it theoretically possible?

#68 2004-08-11 09:17:17

Re: Fusion for Ground Launch - Is it theoretically possible?

#69 2004-08-27 09:00:37

Re: Fusion for Ground Launch - Is it theoretically possible?

#70 2004-09-02 11:45:58

Re: Fusion for Ground Launch - Is it theoretically possible?

#71 2004-09-02 12:05:51

Re: Fusion for Ground Launch - Is it theoretically possible?

#72 2004-09-02 12:31:50

Re: Fusion for Ground Launch - Is it theoretically possible?

#73 2004-10-07 10:29:29

Re: Fusion for Ground Launch - Is it theoretically possible?

#74 2004-10-07 12:52:26

Re: Fusion for Ground Launch - Is it theoretically possible?

#75 2004-11-11 08:03:52

Re: Fusion for Ground Launch - Is it theoretically possible?

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