The Joy of Extreme Possibility

by Paul Gilster on October 19, 2011

Nuclear rocket designs are hardly new. In fact, it was clear as early as the 1950s that conventional chemical rocketry was inefficient, and programs like Project Rover, set up to study the use of nuclear reactors to heat liquid hydrogen for propulsion, aimed at the kind of rockets that could get us beyond the Moon and on to Mars. The NERVA rocket technology (Nuclear Engine for Rocket Vehicle Application) that grew out of all this showed great promise but ran afoul of political and economic issues even as the last Apollo missions were canceled. Nor is the public wariness of nuclear methods likely to vanish soon, yet another hurdle for future ideas.

But making people aware of what has done and what could be done is good practice, as Kenneth Chang does by example in his recent piece on the 100 Year Starship Symposium, which bears the optimistic title Not Such a Stretch to Reach for the Stars. In interstellar terms, propulsion is the biggest problem of all. Chang’s article suggests a pathway through conventional rocketry and into nuclear-thermal designs, with reference along the way to using nuclear engines to generate the electrical fields that power up an ion engine. The goal on this pathway is fusion, though Chang admits no one has yet built an energy-producing fusion reactor.

The Daedalus concept was fusion-based, and the ongoing Icarus project that followed is now examining Daedalus to note the effect of thirty years of new technology. But Chang has also talked to James Benford, whose interest in laser and microwave beaming remains strong. Leave the propellant behind and you’ve maximized payload, in addition to working with known physics and apparently achievable engineering. And there continue to be startling new concepts like those of Joseph Breeden, who finds a more extreme way to create an engineless vehicle:

From his doctoral thesis, Dr. Breeden remembered that in a chaotic gravitational dance, stars are sometimes ejected at high speeds. The same effect, he believes, could propel starships.

First, find an asteroid in an elliptical orbit that passes close to the Sun. Second, put a starship in orbit around the asteroid. If the asteroid could be captured into a new orbit that clings close to the Sun, the starship would be flung on an interstellar trajectory, perhaps up to a tenth of the speed of light.

“The chaotic dynamics of those two allow all the energy of one to be transferred to the other,” said Dr. Breeden, who came toting copies of a paper describing the technique. “It’s a unique type of gravity assist.”

What I call the ‘joy of extreme possibility’ has animated interstellar studies since the days of Robert Forward. It works like this: We know the distances between the stars are so vast as to dwarf the imagination. Indeed, most people have no notion of them, seeing an interstellar mission as merely a next step once we have explored the outer system, a kind of juiced-up Voyager. The scientists and engineers who work on these matters, knowing better, realize how far beyond our current technologies these journeys really are. So they’re not afraid to speculate even at the absolute far end of the plausible (and often beyond that). Work your way through interstellar papers like these and you pick up an infectious, jazzy brainstorming. It’s the kind of mental riffing on an idea that a John Coltrane or a McCoy Tyner does with a musical theme.

And by the way, Chang is careful to get those distances across to readers. I’m always interested in homely comparisons because you can use them to boggle audience minds when speaking about interstellar flight. This is useful, because a boggled mind often becomes a curious one, and while you can never predict these things, occasionally interstellar studies gain a new adherent. Chang cites a Richard Obousy analogy: If the Earth were Orlando and Alpha Centauri were in Los Angeles, then the Voyager spacecraft would have traveled but a single mile.

Even after all these years, that one still boggles my own mind. Chang again:

Another way of looking at the challenge is that in 10,000 years, the speed of humans has jumped by a factor of about 10,000, from a stroll (2.6 m.p.h.) to the Apollo astronauts’ return from the Moon (26,000 m.p.h.). Reaching the nearest stars in reasonable time — decades, not centuries — would require a velocity jump of another factor of 10,000.

It’s good to see the 100 Year Starship Study steadily percolating in the news. Maybe one day these concepts will not seem as esoteric as they do today. I note as I write this, for example, that my word processor flags the word ‘starship’ as a spelling error. We need to set deeper roots into the culture than that. We can start by doing what conference organizer David Neyland told Chang he wants to do, to establish a bar high enough that people “will actually go start tackling some of these really hard problems.” Of course, the real bar is set by nature, and it’s the highest bar we as a species have faced in terms of travel times and distance. But the joy of extreme possibility only ignites the spirit when everything is on the table and the challenge is immense.

tzf_img_post

{ 122 comments }

Christopher Phoenix November 16, 2011 at 19:32

@Eniac

You do? You say so, but you do not follow up with reasons. How exactly do you imagine the space drive can get around being weighed down by the fuel it needs to operate? The fuel to accelerate the payload, the fuel to accelerate that fuel, the fuel to accelerate that fuel, etc?

Okay, Eniac. I’ll explain why a space drive will not be weighed down by the fuel it needs to operate, and why the fuel does not increase just like the mass ratio of a rocket increases.

A rocket expels a part of its mass at high velocity to provide an impulse to itself. The portion of its mass that it expels compared to the portion that left is referred to as the mass ratio. The energy to accelerate that propellent has to come from somewhere else. The mass of the fuel needed to provide the energy to a rocket is relatively tiny compared to the reaction mass the rocket carries. In most cases, much of the energy provided to the craft ends up as kinetic energy of the propellent and not much of the energy ends up in the spacecraft.

You, however, are confused about the difference between fuel and propellent, mainly due to seeing rockets that expel burnt fuel as propellent. The amount of mass annihilated by the “efficient nuclear rocket” to provide all the energy the nuclear rocket uses to accelerate the spacecraft is tiny compared to the hydrogen fuel the ship expels. Propellent is not the same as fuel!! The fuel is the energy source of the rocket. The propellent is the mass the ship expels out the back to provide thrust.

The rocket equation relates the delta-v of a rocket with the exhaust velocity and the initial and final mass of the rocket. The mass the rocket ejects is not fuel!! In fact, the rocket equation ignores the energy source of the rocket entirely. Do you get that? The mass ratio of the rocket has to do with propellent, not the energy source of the rocket!!

A space drive will convert stored energy directly into motion. It does not expel mass. Thus, there is no mass ratio, as the space drive carries no mass to throw the other way. The energy the space drive needs to fly to the Moon can easily be provided by a handful of uranium-235, but the nuclear rocket needs to carry propellent to expel and extra fuel to make up for the energy lost in accelerating the propellent. That is exactly the point of a space drive- it does not have to carry propellent.

Why isn’t the space drive weighed down by the fuel it needs to operate? The fuel doesn’t weight that much!! Consider the nuclear rocket. The fuel was just the U-235 rods. The reaction mass, which the rocket equation deals with, weighed a lot, but contributed no energy. Remember, the fuel is the energy source and the propellent is the mass the rocket expels at high velocity. If I can convert energy directly to motion without expelling propellent, I can forget about mass ratios. I just need energy to operate, and the source of that energy will not weight any more than the reactor on a nuclear rocket does.

Here, I will prove it. According to the NASA website Warp Drive When, sending a Shuttle-sized vehicle on a 50 year one-way trip to visit our nearest neighboring star (subrelativistic speed) would take over 7 x 10^19 Joules of energy. Let’s assume I have a drive core that converts matter to energy with 100% efficiency.

E=MC^2 (M is the mass of the fuel I will utterly annihilate for energy and C is the speed of light)

7×10^19 joules= M (in kg)* 299,792,458 meters per second ^2

M= 7×10^19 joules/299,792,458 meters per second^2

M=778.855 kilograms

A space shuttle orbiter weighs about 70,000 kilograms. I’d hardly call carrying about 779 kilograms of fuel to propel a shuttle sized ship to Proxima Centauri “the same exponential law known as the rocket equation”. Generating this much power is a problem, but the mass of fuel does not increase by the “same exponential law known as the rocket equation”. Compare the space drive that consumes 779 kg of fuel to the best rocket we could design based on today’s knowledge- a matter/antimatter rocket- which will require ten railway tanker sized propellent tanks!!

I’m guessing your counter-arguement is that the power generator for my spaceship will weigh me down. Maybe, but that doesn’t have much to do with our argument. You argued that the fuel for my space drive will increase just as fast as the propellent for a rocket does, but that is utterly absurd. The weight of fuel and a power generator, neither of which are ejected from my craft at high velocity, has nothing to do with the mass ratio of a rocket. A rocket carries fuel and mass to eject. The space drive craft does not.

Frankly, I don’t understand how you can be so thick that you don’t understand that the entire reason anyone wants a “space drive” in the first place is to avoid the bane of all rocket designers- mass ratios. Nor do I understand how you could possibly confuse the reaction mass a rocket carries with the energy source of the rocket. The rocket equation does not deal with fuel, it deals with propellent- the mass the rocket ejects at high velocity. Find a way to propel a craft without needing propellent, and you avoid the rocket equation, even though you still have to find a way to power this drive.

Eniac November 16, 2011 at 23:32

Not just some, but most. This is referred to as propulsive efficiency.

Wrong. Some, not most. See http://en.wikipedia.org/wiki/Propulsive_efficiency, or see your own link http://en.wikipedia.org/wiki/Spacecraft_propulsion#Power_use_and_propulsive_efficiency .
The propulsive efficiency is greater than 50% except at very low speed.

Sigh… not this again… FUEL IS NOT THE SAME AS PROPELLENT!!!!!

I don’t like to repeat myself, but here it is again: The most efficient rocket is one where all spent fuel is used as reaction mass, and nothing else. Those are the ones I am talking about. I am not interested in inefficient rockets. Forget about inefficient rockets for this purpose. It is obvious that an efficient space drive is better than an inefficient rocket. We want a fair comparison.

Why would fuel for a space drive behave anything like this? You aren’t throwing mass overboard and moving due to the reaction with a space drive- thus there is no mass ratio.

Simple, because fuel has mass and needs to be carried, just like the propellant. Not throwing spent fuel overboard just makes it worse. Even with a space drive, you do better throwing the spent fuel overboard, otherwise it will be dead mass.

Let’s look at your “efficient nuclear rocket”. Let’s assume it is a classic “atomic rocket” like the Project NERVA engines.

NERVA happens to be a very inefficient rocket, because it carries propellant that is not spent fuel. An efficient nuclear rocket expels fission fragments at 0.05-0.1 c. It is just as good as a nuclear driven space drive, and the mass ratio starts getting bad around delta-v = 0.1-0.2 c.

Strange, there doesn’t seem to be several kilograms of U-235 rods for every kilogram of ships mass- I thought fuel followed the rocket equation- oops I meant the “powered flight equation”.

It does indeed. Nuclear fuel contains much more energy per mass, so the nuclear rocket equation sets in at much higher delta-v. You have to carry a heck of a lot of U-235 rods to get to 0.5 c. For both types of engine. No mystery here.

Now, why don’t you explain to me why the ship with the space drive will have to carry as much U-235 fuel as the rocket ship had to carry propellent in order to achieve the same velocity? The rocket didn’t need to carry that much fuel- even though it carried a lot of propellent. My craft doesn’t need propellent. Why would it need more fuel? It might even need less, since it doesn’t lose energy accelerating the exhaust.

My craft does not carry propellant. It uses the spent fuel. Comes for free, and the propulsive efficiency is close to 100% at for speeds around the exhaust velocity. The fact that your craft does not make use of the spent fuel will not help it, and will not reduce its burden. In fact, you’ll have to throw the spent fuel overboard or else you’ll do much worse than the rocket equation.

If you can give me a convincing argument of why the space drive ship needs to carry as much U-235 fuel as the nuclear rocket has to carry propellent, even though the rocket’s fuel is not nearly that massive as you say the space drive’s fuel has to be and the space drive is likely to be much more efficient than a rocket, I promise I will accept your view and repeat the mantra “the space drive is subject to the same exponential law, generally known as the rocket equation” to everyone else I converse with on Centauri Dreams.

Good. The space drive needs to carry as much fuel as the rocket because it requires the same amount of energy to achieve the same delta-v. Neither the space drive nor the rocket need to carry extra propellant, because the rocket uses its spent fuel. The rocket is quite efficient (see http://en.wikipedia.org/wiki/Spacecraft_propulsion#Power_use_and_propulsive_efficiency), the space drive would be hard pressed to exceed that, and not by much.

You are confused by thinking about rockets that carry extra propellant. These are not good rockets, and should not be used as straw-men. True, we do not have a “good” nuclear rocket, yet, but then again, we do not have a space drive, either.

The mass of the fuel needed to provide the energy to a rocket is relatively tiny compared to the reaction mass the rocket carries.

Wrong. This is not the case for “good” rockets, such as regular rockets when we are talking conventional energy, and fission fragment rockets when we are talking nuclear. Or the photon rocket, when we are talking complete mass conversion. In all cases, the the propellant is the spent fuel, and it comes for free.

The amount of mass annihilated by the “efficient nuclear rocket” to provide all the energy the nuclear rocket uses to accelerate the spacecraft is tiny compared to the hydrogen fuel the ship expels.

As I have patiently tried to explain many times, the “efficient nuclear rocket” carries NO extra propellant whatsoever. It has NO hydrogen propellant. It ONLY uses the spent fuel, which is available at no extra cost.

The mass ratio of the rocket has to do with propellent, not the energy source of the rocket!!

It has to do with whatever you have to carry to provide propulsion. It applies to fuel the same as propellant.

Compare the space drive that consumes 779 kg of fuel to the best rocket we could design based on today’s knowledge- a matter/antimatter rocket- which will require ten railway tanker sized propellent tanks!!

Actually, if you are talking about complete mass-energy conversion, the best rocket is the photon rocket. And, guess what, it would require not a whole lot more than 779 kg of fuel for the same 50 year journey. No railway tanker sized propellent tanks. And if you wanted to go in 10 years instead of 50, the rocket equation would still get you, for both engines.

Christopher Phoenix November 20, 2011 at 0:28

My craft does not carry propellant. It uses the spent fuel. Comes for free, and the propulsive efficiency is close to 100% at for speeds around the exhaust velocity. The fact that your craft does not make use of the spent fuel will not help it, and will not reduce its burden. In fact, you’ll have to throw the spent fuel overboard or else you’ll do much worse than the rocket equation.

I’m not convinced. Now it is my turn to explain why a space drive is superior to a rocket.

There is no such thing as “free propellent”. Propellent is propellent. Call it fuel (even though that is incorrect), call it propellent, call it reaction mass, call it the stuff that shoots out the back end of your rocket- any rocket, whether it is a design like the fission fragment or not, has to carry mass to shoot out the back end.

Your rocket is hurling mass out the back end, mass that the space drive would avoid carrying in the first place. Do you dispute this? I’d like to see you prove that a rocket does not have to carry the weight of the exhaust before it is expelled. All engines need a working mass, the mass against which a system operates to produce acceleration. In a rocket, that working mass is carried along with it.

http://en.wikipedia.org/wiki/Working_mass

So, your rocket uses its spent fuel as reaction mass. What do we gain? Well, we don’t eliminate the need to carry reaction mass. Your rocket has exhaust, does it not? All the weight of that exhaust has to be carried before your eject it at high speed. Your rocket is as limited as any other in the need to carry propellent. No propellent=no exhaust. No exhaust=no ejected mass=no action=no reaction=NO MOVEMENT!!! Your fission-fragment rocket has to carry reaction mass in order to propel itself, just like any other rocket. That reaction mass is not “free”, it is weight the rocket has to carry with it.

Good. The space drive needs to carry as much fuel as the rocket because it requires the same amount of energy to achieve the same delta-v. Neither the space drive nor the rocket need to carry extra propellant, because the rocket uses its spent fuel.

You seem to be under the impression that since a fission fragment rocket uses the spent fuel as working mass, it does not suffer the mass-penalty of carrying working mass to begin with. That is not true- all the mass the ship expels as working mass is mass the “space drive” would avoid carrying in the first place.

Want proof? A “space drive” doesn’t carry any working mass to push against. Your fission fragment rocket does. My “space drive” has to carry all of the energy it needs to reach a certain delta-v, but it does not have to carry working mass to push against, like your rocket- which was my argument in the first place. Space drives don’t carry any propellent at all, while all rockets carry mass that is expelled in the exhaust.

My argument is that a space drive is far superior to a rocket, since it does not have to carry working mass to push against along with it. All rockets- including fission fragment rockets- have to carry all the working mass they push against along with them.

It is totally unfair to compare space drives with rockets because the space drive does not need to carry working mass with it to produce movement. Feed energy in, and the ship magically accelerates- try beating that with a fission fragment rocket!!

You have to carry the mass of uranium to expel from your fission fragment rocket as well as the mass of uranium that is converted into energy, and you need a lot more uranium to run your fission-fragment than I need to run my space drive, because the uranium is the working mass. Take the exhaust velocity of a fission fragment rocket, the delta-v your want, the final mass of your space vehicle (including the mass of the engine), and figure out the initial mass and mass ratio. See? Even though you expel the spent fuel as reaction mass, you are still carry a lot of working mass with you to reach a speed at which interstellar travel is practical.

A space drive won’t have to carry any mass to accelerate out the back and it won’t lose any energy accelerating that exhaust. It is obvious that a reactionless drive would be superior to a rocket. You are the only person that so vehemently argues otherwise. You are utterly alone in your arguments- scientists and engineers who work at NASA don’t make the arguments you make. They simply point out that such a drive would be very nice, but violates Conservation of Momentum. The problem with a “space drive” is that they don’t exist, not that they don’t have any advantage over rocket ships!!

Christopher Phoenix November 20, 2011 at 0:53

Wrong. Some, not most. See http://en.wikipedia.org/wiki/Propulsive_efficiency, or see your own link http://en.wikipedia.org/wiki/Spacecraft_propulsion#Power_use_and_propulsive_efficiency .
The propulsive efficiency is greater than 50% except at very low speed.

I’m afraid you mixed up cycle efficiency and propulsive efficiency. Cycle efficiency is how efficient the engine is at transforming the energy from the rocket’s energy source into mechanical energy. Propulsive energy is the efficiency, in percent, with which the energy contained in a vehicles propellent is converted to useful energy- energy of the craft. In a rocket, a lot of energy goes into accelerating the exhaust.

Due to your statement that the propulsive efficiency is greater than 50% except at low speed, I am afraid you might not understand the scenario where the propulsive efficiency is 100%.

Imagine an accelerating rocket. Now, imagine that the rocket dynamically alters its exhaust velocity, so that as seen by a non-accelerating observer at rest alongside the flight path, the exhaust velocity is always equal and opposite the vehicle velocity. The exhaust will stop in space and ALL the kinetic energy ends up in the craft. No rocket actually does this, since it would take an impractical amount of propellent, but it is an interesting hypothetical scenario.

Another fact I think you missed is that the more mass efficient a rocket is, the less energy efficient it is. If I have a higher exhaust velocity, which I need to get the same impulse from a smaller mass of propellent, I need to output more energy- after all, KE=.5M*V^2, while momentum is MV=MV. The more mass efficient my rocket is, the more energy I lose in the exhaust.

Actually, if you are talking about complete mass-energy conversion, the best rocket is the photon rocket. And, guess what, it would require not a whole lot more than 779 kg of fuel for the same 50 year journey. No railway tanker sized propellent tanks. And if you wanted to go in 10 years instead of 50, the rocket equation would still get you, for both engines.

I think not!! As I patiently explained above, the more mass efficient a rocket is, the less energy efficient it is. There is not better example of this than the photon rocket. Think about how much energy is lost in the energy of the beam of light- the propulsive efficiency is terrible!! The rocket is mass efficient, all right, but not energy efficient. By the way, a photon rocket still suffers from mass ratios and will need a lot more than 779 kg of fuel to reach the same speed as my space drive craft.

I’ll prove it. My calculations assumed that the space drive converted the energy of the fuel directly into the motion of the craft, but that won’t be true about a photon drive. A lot of energy will be lost as a titanic beam of light shining out the back of your photon rocket. Try standing behind a photon rocket. After your recover from the sunburn- or don’t recover from the vaporization of your body- think about how much energy is being lost as that light beam, instead of going into the motion of the craft.

Eniac November 21, 2011 at 2:59

Your rocket is hurling mass out the back end, mass that the space drive would avoid carrying in the first place. Do you dispute this?

Yes, I do. The second part. The space drive will have to carry the same amount of fuel, whether it hurls it out the back or not. Unless it can produce energy out of nothing.

My “space drive” has to carry all of the energy it needs to reach a certain delta-v, but it does not have to carry working mass to push against, like your rocket

But that’s just it, my rocket uses not a gram more than that fuel that your space drive has to carry. It puts the fuel to double use: First, burn it for energy, then use it as working mass. It is not a revolutionary concept, it is what all chemical rockets do.

You have to carry the mass of uranium to expel from your fission fragment rocket as well as the mass of uranium that is converted into energy, and you need a lot more uranium to run your fission-fragment than I need to run my space drive, because the uranium is the working mass.

You are forgetting that the reaction mass is the spent uranium, the same that was used as fuel. There is no “uranium to expel from your fission fragment rocket”. Fission fragments are not uranium. They need not be carried, they come with the fuel.

You obviously have some sort of a block in understanding this simple principle. Reflect on it a little.

A lot of energy will be lost as a titanic beam of light shining out the back of your photon rocket.

This is true. However, if you calculate the energy that goes into the beam and that which goes into the ship if it is accelerated to near light speed, you will find the latter is greater than the former. This is always true when the delta-v approaches or exceeds exhaust velocity, which also happens to be when the rocket equation sets in. At delta-v much lower than exhaust velocity, the space drive does much better. But that is not the regime we are interested in, it is a regime of generous mass ratios where the rocket equation is not a problem, for either drive.

You are utterly alone in your arguments- scientists and engineers who work at NASA don’t make the arguments you make. They simply point out that such a drive would be very nice, but violates Conservation of Momentum.

Ordinarily, that is enough. Not many real scientists and engineers will go any further than that. I would expect the few that do to at least discuss the weight of the fuel, and not try to make it disappear by calling it “stored potential energy”, with a tacit assumption that it is massless. If you can find someone who actually considers the mass of the fuel needed to power a space drive and comes to your conclusions, please give me a reference. I do not think you will.

Christopher Phoenix November 22, 2011 at 3:14

@Eniac

Yes, I do. The second part. The space drive will have to carry the same amount of fuel, whether it hurls it out the back or not. Unless it can produce energy out of nothing.

You seem to think that because you expel the spent fuel as the exhaust of your rocket, you don’t suffer the penalty of carrying reaction mass. That is not true.

You are forgetting that the reaction mass is the spent uranium, the same that was used as fuel. There is no “uranium to expel from your fission fragment rocket”. Fission fragments are not uranium. They need not be carried, they come with the fuel.

You obviously have some sort of a block in understanding this simple principle. Reflect on it a little.

Reflect on this, Eniac- it doesn’t matter that your rocket uses spent fuel as reaction mass!! It gains you nothing against the rocket equation.

The rocket equation tells us how much delta-v a rocket will get from expelling a certain amount of its mass at a certain exhaust velocity. It is that MASS, the mass of the exhaust, that the space drive seeks to eliminate, not the energy source. You can’t improve your rocket’s performance beyond what the rocket equation predicts just because your rocket uses its spent fuel as reaction mass. All rockets have to carry a large amount of fuel/propellent to expel. You seem to have a very, very large mental block in understanding this basic fact of rocketry.

The mass of the fuel a space drive has to carry will be a lot less then the working mass the rocket drags along with it. Your rocket ejects a large portion of its mass as reaction mass. Rockets for which the fuel and propellent are the same still eject a large portion of their mass.

Think about it this way- a rocket ejects a portion of its mass at high velocity in order to propel itself by the principle of action-reaction. The energy to eject this mass must come from somewhere. In some rockets, like NERVA, the energy source is separate from the reaction mass. In other rockets, like chemical rockets, the fuel and propellent are the same.

In both examples, the goal is to expel mass from the engine- the exhaust. As the rocket equation shows us, the rocket must eject a certain amount of mass at a certain exhaust velocity to reach a certain delta-v. It does not matter that the chemical rocket’s energy source is in the propellent- it still has to eject a large amount of reaction mass to achieve the desired delta-v. Bundling the fuel and the propellent is a more elegant design, but it doesn’t save you from ejecting a large fraction of your ship’s mass as exhaust!!

In fact, we can ignore the details of any one propulsion system and base all our comparisons on theoretical rockets with a certain exhaust velocity and mass ratio, instead of arguing about the merits of having a the fuel bundled with the propellent.

My point is that your rocket still has to eject a certain amount of mass to reach a certain delta-v, an amount that depends on the exhaust velocity of your rocket. It doesn’t matter that your bundled the energy source with the reaction mass.

Eniac November 22, 2011 at 22:25

The rocket equation tells us how much delta-v a rocket will get from expelling a certain amount of its mass at a certain exhaust velocity.
Actually, it is not the expelling that does the damage, it is having to carry it with you. Whether you burn it to power a space drive or a rocket does not make a difference, as long as you do not take extra reaction mass.

All rockets have to carry a large amount of fuel/propellent to expel. You seem to have a very, very large mental block in understanding this basic fact of rocketry.

This should be “… to provide power and expel”. The space drive eliminates the second reason, but not the first. No matter how many reasons, the amount stays the same, whether you use it to power (space drive) or to power and expel (rocket). Reflect!

The mass of the fuel a space drive has to carry will be a lot less then the working mass the rocket drags along with it.

No, not really. Only if you carry extra propellant in the rocket, which you wouldn’t.

In some rockets, like NERVA, the energy source is separate from the reaction mass. In other rockets, like chemical rockets, the fuel and propellent are the same.

The former are inefficient. It is the latter you have to compete against. When using nuclear power, that would be the fission fragment rocket, with an exhaust velocity of around 0.1 c. Can you calculate how much U235 you would need for fuel to reach 0.1 c with a space drive? Surprise! it is just as much as the fission fragment rocket needs for fuel/propellant at the same delta-v.

Bundling the fuel and the propellent is a more elegant design, but it doesn’t save you from ejecting a large fraction of your ship’s mass as exhaust!!

That is correct. The point you keep missing (terminally, I suspect) is that the space drive needs to carry the same mass of fuel, because it requires the same power. It will even have to eject the spent fuel, because keeping it on board would be very bad. Beyond rocket equation bad.

My point is that your rocket still has to eject a certain amount of mass to reach a certain delta-v, an amount that depends on the exhaust velocity of your rocket.

And my point is that your space drive still has to carry a certain amount of mass to reach a certain delta-v, an amount that depends on the energy density of the fuel. Because that amount is determined by the kinetic energy that needs to be provided, it will be the same for both drives. Except for an efficiency factor, which for rockets is easily above 50%, given a delta-v around exhaust velocity.

Actually, the efficiency condition is that all the energy the fuel provides is converted to kinetic energy of the exhaust. With this relationship, you can substitute “energy density” in my statement above with “exhaust velocity”. Then it becomes identical to yours, except mine is about space drives and yours about rockets.

And, if I may make a request, please don’t keep aping my friendly put-downs. Make up your own.

Eniac November 22, 2011 at 22:29

You seem to think that because you expel the spent fuel as the exhaust of your rocket, you don’t suffer the penalty of carrying reaction mass. That is not true.

You seem to think that because you don’t expel the fuel as the exhaust of your drive, you don’t suffer the penalty of carrying it. That is not true.

I cannot make it any clearer than this. I will give up trying.

Christopher Phoenix November 23, 2011 at 18:22

@Eniac

That is correct. The point you keep missing (terminally, I suspect) is that the space drive needs to carry the same mass of fuel, because it requires the same power. It will even have to eject the spent fuel, because keeping it on board would be very bad. Beyond rocket equation bad.

What about the REACTION MASS the rocket has to carry? Rockets don’t just carry fuel. They carry a lot of reaction mass, and bundling the energy source of the rocket with the propellent doesn’t make it go away. If your rocket has an exhaust velocity of 0.1 C, guess how much propellent it needs to carry? 1.7 times the mass of the ship!! That is not fuel. That is the mass of the exhaust that goes squirting out the back. A space drive would not have to carry that mass, since a space drive only needs energy to operate.

It does not matter that you bundled the fuel with the propellent- the immutable laws of physics demand that you squirt a certain mass of propellent away from your ship at a certain exhaust velocity to reach a certain delta-v, and the more propellent your squirt or the faster the propellent is squirting out, the faster your rocket can go. You just keep missing this simple fact- intentionally, I guess. You can continue repeating “the space drive requires the same mass of fuel”, but that doesn’t change that the space drive doesn’t need to carry the same mass of propellent!! In fact, a space drive does not carry any propellent at all.

So it doesn’t even matter for this argument that my space drive ship has to carry the fuel needed to power the space drive. I don’t have to drag along a huge mass of propellent to squirt out the back, and this is the main advantage of a hypothetical space drive. This is my point- eliminating the propellent constraints of rockets is the goal of space drives. If you are so clever, prove me wrong.

This should be “… to provide power and expel”. The space drive eliminates the second reason, but not the first. No matter how many reasons, the amount stays the same, whether you use it to power (space drive) or to power and expel (rocket). Reflect!

I reflect- on the fact that you are both wrong and stubborn. The amount does not “stay the same”. Once you eliminate the need to carry propellent to expel, you don’t have to worry about mass ratios. Got that? A mass ratio tells how much of the mass of the ship is expelled as exhaust. No expelling=no mass ratio.

Christopher Phoenix November 23, 2011 at 18:23

@Eniac

That is correct. The point you keep missing (terminally, I suspect) is that the space drive needs to carry the same mass of fuel, because it requires the same power. It will even have to eject the spent fuel, because keeping it on board would be very bad. Beyond rocket equation bad.

What about the REACTION MASS the rocket has to carry? Rockets don’t just carry fuel. They carry a lot of reaction mass, and bundling the energy source of the rocket with the propellent doesn’t make it go away. If your rocket is trying to reach the speed of 10% the speed of light and has an exhaust velocity of 0.1 C, guess how much propellent it needs to carry? 1.7 times the mass of the ship!! That is not fuel. That is the mass of the exhaust that goes squirting out the back. A space drive would not have to carry that mass, since a space drive only needs energy to operate.

It does not matter that you bundled the fuel with the propellent- the immutable laws of physics demand that you squirt a certain mass of propellent away from your ship at a certain exhaust velocity to reach a certain delta-v, and the more propellent your squirt or the faster the propellent is squirting out, the faster your rocket can go. You just keep missing this simple fact- intentionally, I guess. You can continue repeating “the space drive requires the same mass of fuel”, but that doesn’t change that the space drive doesn’t need to carry the same mass of propellent!! In fact, a space drive does not carry any propellent at all.

So it doesn’t even matter for this argument that my space drive ship has to carry the fuel needed to power the space drive. I don’t have to drag along a huge mass of propellent to squirt out the back, and this is the main advantage of a hypothetical space drive. This is my point- eliminating the propellent constraints of rockets is the goal of space drives. If you are so clever, prove me wrong.

This should be “… to provide power and expel”. The space drive eliminates the second reason, but not the first. No matter how many reasons, the amount stays the same, whether you use it to power (space drive) or to power and expel (rocket). Reflect!

I reflect- on the fact that you are both wrong and stubborn. The amount does not “stay the same”. Once you eliminate the need to carry propellent to expel, you don’t have to worry about mass ratios. Got that? A mass ratio tells how much of the mass of the ship is expelled as exhaust. No expelling=no mass ratio.

Christopher Phoenix November 23, 2011 at 19:51

@Eniac

And, if I may make a request, please don’t keep aping my friendly put-downs. Make up your own.

Put-down: A remark intended to humiliate or criticize someone. That is not very friendly. It seems you have plenty of time to make up plenty of put-downs- not surprising really, as you don’t make any good arguments in support of your statements. Repeating your statements and sprinkling some “friendly put-downs” is not a very good debating style.

By the way, I have a question- do you really find it so difficult to understand the principle that rockets carry mass to squirt out the back, and that the mass of this propellent doesn’t go away just because you bundled it with the fuel, or are you so stubborn because you don’t want to admit you’re wrong? I think it is the latter.

As for this “efficient rocket” and “inefficient rocket” thing- NERVA is not capable of an interstellar precursor mission because its exhaust velocity is too low. A fission fragment does not thermalize the output of the reactor, and thus achieves a much greater exhaust velocity. It is not carrying propellent separate from the fuel that is the problem. It is the temperature limitations of a solid rocket engine that limits NERVA.

An Orion-style nuclear pulse drive, on the other hand, carries separate propellent. The fuel was the uranium or plutonium in an atomic pulse unit, but the propellent could be any cheap inert material that is placed between the bomb and the pusher plate. It could be as light as polyethylene or as heavy as tungsten. It might be shipboard waste as well as ice, frozen methane, or other material obtained from the surface of Mars, among the rings of Saturn, or anywhere else the ship decided to stop. Go read this excerpt of “Project Orion: The Story of the Atomic Spaceship” by George Dyson.

http://books.google.com/books?id=r_Gu4f0QxrkC&pg=PA3&lpg=PA3&dq=project+orion+propellent&source=bl&ots=F47WYDU4b5&sig=VfAz_XT6HpWahSXXIGg1aU4-uk8&hl=en&ei=dnXNTr6ME8SliQLhlfm8Cw&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCQQ6AEwAQ#v=onepage&q=project%20orion%20propellent&f=false

Nuclear pulse drives would have great performance. Obviously, carrying separate propellent or even obtaining more propellent from the ship’s destination is not a problem. Next you’ll try to tell me that if I roll up my luggage really small and shove it in my fuel tank, it will weigh less.

Eniac November 24, 2011 at 14:21

What about the REACTION MASS the rocket has to carry?

We use the spent fuel, so we do not have to carry extra propellant. Not a gram, beyond what we need to carry for energy.

A space drive would not have to carry that mass, since a space drive only needs energy to operate.

Ah yes? Really? And where do you get the energy if you do not take fuel, which has mass?

If your rocket is trying to reach the speed of 10% the speed of light and has an exhaust velocity of 0.1 C, guess how much propellent it needs to carry? 1.7 times the mass of the ship!! That is not fuel.

It is too. A fission fragment rocket uses no more propellant than what was the fuel. Not a gram extra. If your space drive is trying to reach the speed of 10% the speed of light and uses fission power, guess how much fuel it needs to carry? 1.7 times the mass of the ship!!

Christopher Phoenix November 28, 2011 at 2:40

@Eniac

You seem to think that because you don’t expel the fuel as the exhaust of your drive, you don’t suffer the penalty of carrying it. That is not true.

I cannot make it any clearer than this. I will give up trying.

Stop using straw man attacks when you lose an argument. It is highly unattractive.

I stated that a rocket has to carry a large amount of propellent whether it bundles the fuel with the propellent or not, and that a method of propulsion that converts energy directly into motion without expelling propellent would have a major advantage over a rocket since it does not have propellent constraints. You replied by claiming that I think that a space drive doesn’t suffer the mass penalty of carrying fuel and proceeded to attack that highly twisted misrepresentation of my actual position. Does the word STRAW MAN mean anything to you?

http://en.wikipedia.org/wiki/Straw_man

koyagi November 28, 2011 at 10:44

Such a spaceship would warp space-time thus gaining exponential advantage over the rocket equation. Thus a space-drive should have the capacity to warp space-time.

Eniac November 28, 2011 at 22:39

I stated that a rocket has to carry a large amount of propellent whether it bundles the fuel with the propellent or not, and that a method of propulsion that converts energy directly into motion without expelling propellent would have a major advantage over a rocket since it does not have propellent constraints.

So you did, many times. What you fail to acknowledge is that this statement rests on an unsupported assumption: The fuel that is required to power the space drive masses much less than the fuel/propellant needed by an efficient rocket to reach the same final velocity.

This assumption is false.

For a fair comparison, both drives get to use fuel with the same energy density, and the rocket gets to use its spent fuel as the only reaction mass.

Eniac November 29, 2011 at 21:44

@koyagi: You are talking of a warp-drive. We are talking about a space drive, aka reactionless drive, which is a mythical device that can accelerate without ejecting reaction mass. It is easy to calculate the minimum energy requirement for such a drive: it is equal to the kinetic energy imparted to the craft. Unfortunately, if the energy has to be obtained from fuel, this requirement makes the drive only marginally superior to a good rocket under even the most optimistic assumptions of efficiency. We have no clue what the energy requirements for the equally mythical warp drive are. My guess is: astronomical.

Christopher Phoenix December 1, 2011 at 0:51

@Eniac

So you did, many times. What you fail to acknowledge is that this statement rests on an unsupported assumption: The fuel that is required to power the space drive masses much less than the fuel/propellant needed by an efficient rocket to reach the same final velocity.

Which should be obvious, unless you have an educational impediment. Fuel+Working Mass weighs more than Fuel alone. Your argument rests on the unproven assumption that a rocket can carry the same amount of fuel as a space drive and use that fuel as reaction mass to achieve the same delta-v as a space drive. That is not true- the need to carry reaction mass limits the performance of any rocket far below the predicted performance of a hypothetical space drive.

I’ll prove it. Your precious “efficient rocket” doesn’t gain any leverage on the rocket equation, no matter where it keeps its energy source. It could even have energy beamed to it, like a laser thermal rocket, and it will still suffer from the rocket equation. Why? Because the rocket equation deals with the amount of mass the rocket has too eject at high velocity, not the energy source!! In fact, the energy required to eject the mass is a secondary consideration to the mass the rocket ship needs to carry to eject. Chemical rockets have their propellent serve double duty as their fuel source, but that doesn’t help them any against the rocket equation.

Rocket designers often refer to the propellent as “fuel” since the mass of propellent is a primary consideration and chemical rockets use their propellent as fuel anyway. That is what lead to this “fuel is the same as propellent” misconception in the first place.

For a fair comparison, both drives get to use fuel with the same energy density, and the rocket gets to use its spent fuel as the only reaction mass.

In which case the rocket will lose. Unless you limit the space drive to using fuels with such a low energy density that the mass of fuel the ship has to carry is more than the mass of propellent+high energy density fuel the rocket has to carry (an extremely unfair comparison), the rocket will lose in this comparison. The whole point of a space drive is to eliminate the need to carry a large amount of working mass along. If you don’t have to eject a portion of the mass of your ship at high velocity in order to move, you don’t have to worry about mass ratios.

As for comparing the drives on the basis of energy- a hypothetical space drive will be more efficient than a rocket, for the simple reason that no energy is lost accelerating the propellent. All the energy that is not loss as waste heat ends up as the motion of the vehicle.

On the other hand, a rocket uses its energy to accelerate the propellent. Much of the energy expended in accelerating the propellent does not end up in the craft and is lost as the kinetic energy the exhaust. Not only that, but the more mass-efficient the rocket is, the less energy efficient it is. Why? Momentum is Mass times Velocity. To get the same amount of momentum out of a smaller amount of propellent, I need to accelerate it faster- to obtain a higher exhaust velocity. However, KE=.5M*V^2. Accelerating the propellent to a higher speed means expending more and more energy.

Thus, a space drive doesn’t need to carry propellent and is also much more energy efficient. Why else would Mark Millis and others try to figure out how a space drive could be built? You obviously think you are smarter than everyone at NASA and the Tau Zero Foundation, but you’re not.

I’m not claiming a space drive is possible- in fact, in absence of new physics, I don’t see how one can be made to work. However, to argue that a working space drive has no advantage over a rocket is absurd. So don’t go and use a straw man against me again by attacking the underlying physics of a hypothetical space drive rather than the question of its advantages over a rocket.

Christopher Phoenix December 1, 2011 at 3:33

@koyagi

Such a spaceship would warp space-time thus gaining exponential advantage over the rocket equation. Thus a space-drive should have the capacity to warp space-time.

That’s a warp drive, which could be used as a reactionless drive of sorts, but the craft doesn’t move, space moves around it…

I was referring to a space drive, an idealized form of propulsion that creates propulsive forces anywhere in space without expelling propellent using only the interactions between the spacecraft and the surrounding space. To find out more, go to the website Warp Drive When.

http://www.nasa.gov/centers/glenn/technology/warp/warp.html
http://en.wikipedia.org/wiki/Breakthrough_Propulsion_Physics_Program
http://www.grc.nasa.gov/WWW/bpp/
http://www.centauri-dreams.org/?p=16202

It is easy to calculate the minimum energy requirement for such a drive: it is equal to the kinetic energy imparted to the craft. Unfortunately, if the energy has to be obtained from fuel, this requirement makes the drive only marginally superior to a good rocket under even the most optimistic assumptions of efficiency.

Eniac is wrong. Don’t fall for it- a space drive does not have to carry large amounts of reaction mass, unlike a rocket, and current analysis suggest a major gain in capability should a propulsion breakthrough that allows us to create propulsive forces without expelling propellent materializes. Just read the “Getting There” page at the Tau Zero Foundation- scroll down to “Enough Fuel For the Journey”- and the “Why is Interstellar Travel So Tough” section at Warp Drive When- read the part describing the limitations of rockets. I’ve patiently explained the reasons why Eniac is wrong, provided links to peer-reviewed papers, and still Eniac stubbornly repeats the same statements without once backing them up.

Christopher Phoenix December 1, 2011 at 3:42

@Eniac

We have no clue what the energy requirements for the equally mythical warp drive are. My guess is: astronomical.

You could go research the topic, instead of guessing, you know.

The problem with the Alcubierre drive isn’t just energy. You need exotic matter. According to one set of calculations, you’d need negative one Jupiter mass to create a warp bubble one meter across. Manipulating positive one Jupiter mass is already pretty freaky. As for the energy requirements- some warp drives would require more energy than exists in the universe. There are other problems as well- instabilities, lethal radiation, and tidal forces. Its a neat idea, but it seems unlikely we could create a warp bubble.

http://en.wikipedia.org/wiki/Alcubierre_drive

Eniac December 1, 2011 at 21:41

I’ll prove it.

You did no such thing. Again you went and ignored the mass of the fuel a space drive has to carry to feed its power plant.

Fuel+Working Mass weighs more than Fuel alone

Not in the case of “my precious” efficient rocket. By definition. Not a gram more.

A hypothetical space drive will be more efficient than a rocket, for the simple reason that no energy is lost accelerating the propellent.

This is what I mean when I say the perfect space drive with a perfect power plant has a marginal advantage. A real and imperfect rocket loses less than 50% of its energy to accelerating propellant out the back, if accelerating to a delta-v beyond its exhaust velocity. Very few power plants can do better than 50%, and who knows about the space drive.

And don’t even get me started on power density. Chemical rockets have the largest power to mass ratio of any known power plant, by an order of magnitude. Even a massless space drive would not be capable of rocket-like performance simply because there are no power plants that are light and powerful enough.

Or heat rejection. At an efficiency of 50%, the power plant/space drive will generate enormous amounts of heat. In a rocket, that heat goes out the back. What do you do about it in your space drive?

Eniac December 2, 2011 at 8:06

We are not getting anywhere. Perhaps, instead of trading insults (a very one-sided trade, recently), we should look at the physics of the matter. You seem to have at least some superficial exposure to elementary physics, so perhaps you can figure yourself the answer to this challenge question:

Given a massless power plant and a massless reactionless drive, both of which are 100% efficient, how much fission fuel would be required to accelerate 1 ton of payload to 0.1c? You may assume that 0.7% of the mass of the fuel is converted to energy according to E = mc^2.

Eniac December 10, 2011 at 12:21

Here is the solution: Converting all the fuel’s energy into kinetic energy, we obtain:

0.007 mf c^2 = 1/2 ms v^2 = 0.005 ms c^2 (assuming v = 0.1 c)

This gives us the mass of the fuel (mf) as 5/7 of the ship mass (ms). You could indeed build a ship like that, but 5/7 th of its mass would have to be fuel, leaving 2/7 for payload. The situation is quite similar to that of a fission fragment rocket with an exhaust velocity of 0.1c.

If you wanted to reach 0.2 c, the fuel needed would be four times as much, or 20/7 ms. That is, the fuel does not have enough energy to propel itself, much less any payload. The only way out is to drop used fuel overboard as you go. The resulting equation and exponential mass ratio would look eerily familiar: They would be equivalent to the rocket equation.

Comments on this entry are closed.