Bussard Collector

[Mr. B]

The field produced by a bussard collector would have to be hundreds of thousands of kilometers in diameter in order to be of any use at all, extending far beyond the confines of the warp bubble.

If that were "possible," and I don't believe it is, what would be the point of expending an obscene amount of energy to generate an incomprehensibly gigantic "net" to collect a handful of particles? I know the stated purpose of the bussard collectors is to collect interstellar hydrogen for fuel, but the concept doesn't make much sense in deep space, and certainly not at warp when you would need even more power for even more inexplicable wizardry to allow particles to come in contact with the ship. The cost/benefit could only be sensible in a nebula or skimming a gas giant... at sublight speed.

 

[buzzknox]

The configuration of the bussard collectors leaves a lot to be desired, if they are to perform as claimed. They are typically located so that the saucer is within the magnetic fields the collector would project. How much power/mass is involved in shielding the crew and equipment from these fields? It might outweigh the benefit of having the collectors at all.

Perhaps we could retcon the collectors so that they are used in close flybys of gas giants, capturing the gases by direct contact and sifting out all but the usable substances such as hydrogen.

 

[TIN_MAN]

The field produced by a bussard collector would have to be hundreds of thousands of kilometers in diameter in order to be of any use at all, extending far beyond the confines of the warp bubble.

If that were "possible," and I don't believe it is, what would be the point of expending an obscene amount of energy to generate an incomprehensibly gigantic "net" to collect a handful of particles? I know the stated purpose of the bussard collectors is to collect interstellar hydrogen for fuel, but the concept doesn't make much sense in deep space, and certainly not at warp when you would need even more power for even more inexplicable wizardry to allow particles to come in contact with the ship. The cost/benefit could only be sensible in a nebula or skimming a gas giant... at sublight speed.

Your'e forgeting that at warp speeds even the deep space interstellar particle density isn't all that rarified, even the critics of real proposals for Bussard ramjet designs concede this, and they are scientists who presumably know what they're talking about, which is why they say it would make a more effective brake (think of the parachute effect). Therefore it doesn't take "obscene amounts" of energy to collect just a "handfull of particles", the ship would actually have a surplus amount of particles to condend with. Of course this doesn't negate the possibility of scooping up hydrogen from clouds or gas giants.

 

[Longinus]

Those red orbs actually has to have some other function as well, because Phoenix already had them, and I'm pretty sure it was not meant for such voyages that would require bussard collectors...

 

[TIN_MAN]

Those red orbs actually has to have some other function as well, because Phoenix already had them, and I'm pretty sure it was not meant for such voyages that would require bussard collectors...

I always assumed the red 'orbs' were just the 'tanks' where the raw hydrogen plasma was temporarly stored as it was 'cooled' before being transfered to other places/uses? The actual collector mechanism is the ring of loovers just behind the 'orb'? Does the Pheonix have loovers?

 

[Crazy Eddie]

The field produced by a bussard collector would have to be hundreds of thousands of kilometers in diameter in order to be of any use at all, extending far beyond the confines of the warp bubble.

If that were "possible," and I don't believe it is, what would be the point of expending an obscene amount of energy to generate an incomprehensibly gigantic "net" to collect a handful of particles?

So those particles could be used as fuel for the main engines.

...

Are you, like, not aware that this exactly what a bussard collector is designed to do?

I know the stated purpose of the bussard collectors is to collect interstellar hydrogen for fuel, but the concept doesn't make much sense in deep space, and certainly not at warp when you would need even more power for even more inexplicable wizardry to allow particles to come in contact with the ship. The cost/benefit could only be sensible in a nebula or skimming a gas giant... at sublight speed.

Because, as I explained in the OP, the faster the ship moves the more quickly it can draw interstellar hydrogen. If you have an issue with feasibility (as I do) then you're barking up the wrong tree; you're better off calling up Doctor Bussard and giving him a piece of your mind.

 

[Mr. B]

The field produced by a bussard collector would have to be hundreds of thousands of kilometers in diameter in order to be of any use at all, extending far beyond the confines of the warp bubble.

If that were "possible," and I don't believe it is, what would be the point of expending an obscene amount of energy to generate an incomprehensibly gigantic "net" to collect a handful of particles?

So those particles could be used as fuel for the main engines.

Deuterium powers all ship systems. My point was you would expend exponentially more fuel powering a gargantuan collection system than you could ever hope to derive from space. It would be like sending a tanker ship across the Atlantic with only one barrel of oil in the cargo hold.

 

[Crazy Eddie]

If that were "possible," and I don't believe it is, what would be the point of expending an obscene amount of energy to generate an incomprehensibly gigantic "net" to collect a handful of particles?

So those particles could be used as fuel for the main engines.

Deuterium powers all ship systems. My point was you would expend exponentially more fuel powering a gargantuan collection system than you could ever hope to derive from space.

IS that your point? Because I'm pretty sure the whole point of this thread was to try and figure out IF that is the case, using the variables I already supplied and some rough estimate for those power requirements based on real electromagnetic fields.

In other words, the question asked by the OP: is the collection system worth it, and if so, how much energy can the bussard collectors draw from the interstellar medium? I'm not really interested in your unsubstantiated opinions here, I'm looking for someone who understands the NUMBERS.

 

[Cary L. Brown]

Guys,

I've been sorta following this, but I finally decided to try running a few numbers myself. Here's what I came up with. Please feel free to cross-check my calculations:

Basically, the ability to collect hydrogen (which is assumed to be monatomic, something I didn't initially expect) in deep space isn't nearly as restricted as it seems the original calculations would make them seem.

The question becomes "how much hydrogen do you need to collect?"

At cruising velocity (WF6, old-scale) you don't need a particularly big "scoop" at all to get a gram of hydrogen per day, in "average" conditions. In "hydrogen-heavy" regions... well, you're pretty much flush, I think.

Oh, and by the way, IF you can react 1g of total reactant mass (that would be .5g of hydrogen and .5g of "flipped" antihydrogen) you get a LOT of energy. Here are a few more calculations:

So as long as you're getting power from m/am annihilation, I think you're fine with a Bussard device.

The trick is coming up with an "inexpensive" means of flipping hydrogen into anti-hydrogen.

One last note...

In 2005, total worldwide energy consumption was 500 Exajoules (= 5 x 10^20 J).

(According to the US DOE, per this linked document: http://www.eia.doe.gov/pub/international/iealf/table18.xls )

In other words, the Enterprise has the capability to harvest enough energy to produce, in a single day, as much power as the entire planet Earth typically consumes in an entire year during our time, for all purposes.

 

[Crazy Eddie]

Fantastic, Cary, this answers alot of my questions.

With two minor nitpicks: could you possibly modify this to calcualte for grams of hydrogen PER SECOND? At least this way I'll be able to figure out for sure how big the field scoop needs to be in order to run "mostly" in gas from the ISM. While I'm still no closer to figuring out how much power that scoop will require (even after several days of intense googling) I've got half a mind to do some kind of back-of-a-napkin estimate based on industrial electromagnets and just add a couple of decimal points. Now if only I could find useful numbers for an industrial-strength electromagnet...

Inexpensive flipping of hydrogen is the easy part: I'd just [tech] some kind of fancy device that uses gravitons or subspace or some other thingamajig to do the job. Make it power intensive enough so the device, say, the energy equivalent of one gram of hydrogen in the production of .5 grams of antimatter; this way the ship would have to collect 4 grams from the ISM for every gram of fuel/energy (one part matter, one part antimatter) stored. Or something like that.

 

[Cary L. Brown]

Fantastic, Cary, this answers alot of my questions.

With two minor nitpicks: could you possibly modify this to calcualte for grams of hydrogen PER SECOND? At least this way I'll be able to figure out for sure how big the field scoop needs to be in order to run "mostly" in gas from the ISM. While I'm still no closer to figuring out how much power that scoop will require (even after several days of intense googling) I've got half a mind to do some kind of back-of-a-napkin estimate based on industrial electromagnets and just add a couple of decimal points. Now if only I could find useful numbers for an industrial-strength electromagnet...

Inexpensive flipping of hydrogen is the easy part: I'd just [tech] some kind of fancy device that uses gravitons or subspace or some other thingamajig to do the job. Make it power intensive enough so the device, say, the energy equivalent of one gram of hydrogen in the production of .5 grams of antimatter; this way the ship would have to collect 4 grams from the ISM for every gram of fuel/energy (one part matter, one part antimatter) stored. Or something like that.

You can convert the numbers from "days" to "seconds"... I've given you the conversion factor already (86400 seconds per day... that's 60 seconds/minute x 60 minutes/hour x 24 hours/day).

Just divide my numbers by 86400 and you have your "per second" value.

 

[Crazy Eddie]

^ Yes, but you forget to take into account the variable "NtS," which is the amount to which Newtype_alpha sucks at math.

Back of the napkin here: to collect one gram of hydrogen per second at warp six, at average hydrogen density, the collector field would need a surface area of 9.2 million km^2. This by multiplying your 106.8km^2 by seconds in a day to get the new surface area...

Did I do this right for once? Do I get a cookie?

 

[Cary L. Brown]

^ Yes, but you forget to take into account the variable "NtS," which is the amount to which Newtype_alpha sucks at math.

Back of the napkin here: to collect one gram of hydrogen per second at warp six, at average hydrogen density, the collector field would need a surface area of 9.2 million km^2. This by multiplying your 106.8km^2 by seconds in a day to get the new surface area...

Did I do this right for once? Do I get a cookie?

Well, not really...

That's because my number isn't 106.8 km^2. It's 1.0679 x 10^-4 km^2. Or, in "plain" numbering... 0.00010679 km^2.

So, you multiply THAT by 86400...

Which gives a value of 9.226656 km^2.

the area of a circle is A = pi * r^2....

So, what we're looking at here is a projected circle, of radius 1,713m, required to collect one gram of hydrogen per second at WF6.

(Naturally, this is assuming 100% efficient collection... something that is never true in reality. A good rule-of-thumb for a mature technology would probably be more along the lines of 75% efficiency, though obviously that's pure "guestimation" here)

 

[Crazy Eddie]

Are you sure? The field before converting to kilometers looks like

Max: 107 m^2
Average: 106,789 m^2
Min: 1,067,886 m^2

That's where I got the 106.8km from, because I thought that was one hundred and six thousand meters.

 

[Cary L. Brown]

Are you sure? The field before converting to kilometers looks like

Max: 107 m^2
Average: 106,789 m^2
Min: 1,067,886 m^2

That's where I got the 106.8km from, because I thought that was one hundred and six thousand meters.

That's why I suggested that anyone who wants to double-check my numbers should do so.

You did catch an error in my calculations, however... but not the error you think you did, it seems.

When I converted from m^2 to km^2, I divided by 1000^3... which was a typo... the number should have been 1000^2. (Hey, it was past midnight, after all!)

Anyway, with that correction made, the numbers actually work out a little bit differently, obviously (by a factor of 1000).

I believe that's "fixed" properly now. I doubled checked the rest of my equations, and that's the only error I find this morning...

 

[Crazy Eddie]

So if I'm reading this right, then at warp six to collect one gram a day the scoop needs to be around 106 square meters... right?

So if that's correct, the field would need to be about 9,200km^2 to collect one gram per second.

Do I get a cookie yet?

 

[Cary L. Brown]

So if I'm reading this right, then at warp six to collect one gram a day the scoop needs to be around 106 square meters... right?

So if that's correct, the field would need to be about 9,200km^2 to collect one gram per second.

Do I get a cookie yet?

Given the "maximum" expected levels of hydrogen, yeah, 106 m^2... which is basically about a 5.81 m radius circle (for 1g/day)

I would recommend using the "average" number, though... so at WF6, you'd be talking about 107,000 m^2, or a circle of radius 184.37 m, for 1g/day.

So, the amount you get per second is 1/86,400 of that, or the area you need for 1g/sec is 86,400 times that.

So, for 1g/sec at WF6, you need a circle of 9,226,569,600 m^2, or a radius of 54,193 m.

Divide that area by 1000^2, to get it in km. That's 9,227 km^2, or a radius of 54.2 km. So yeah, you pretty much have it straight.

Sorry, no cookies here, though... all out. Girlscouts skipped my place this year it seems....

My question is... why would you need 1g/sec?

Remember that a single gram of reaction mass produces a LOT of energy. It's pretty much nonsensical to assume you need that much mass for conversion.

It's also nonsensical to assume that you're going to "react" 100% of what you're collecting. A fair portion would be used to replenish lost mass (atmosphere bleed, etc) and, most likely, the majority will be diverted to be used as reaction-mass for newtonian propulsion devices (a little mass can go a long way, remember, if you can accelerate it sufficiently first, of course). Maybe... MAYBE... 20% would be converted in the m/am reaction process, I'd guess. (After all, that's still allowing you, under sustainable-operation conditions, to produce 188,324,000,000,000 joules of energy a day.)

Yeah, they could "retank" every so often, but I don't prefer that approach. I really like the advantage that m/am reaction for power generation provides... insofar as it makes the ship, potentially, self-sufficient. No need to stop at "Space Texaco" every few thousand parsecs...

So for me, I presume that the ship is able to collect significantly more matter than is required to power the ship's systems, under "normal" loading, indefinately.

Given the sheer power output available, and the limits of materials and energy as we understand them today... I really think that the ratio of "collected" to "consumed for power generation" is pretty large.

If I were creating the ship, myself, with as much "magic" technology as I wanted, I'd give it a 5 km^2 "projected circle" for the primary collection "cone, and I'd convert 20% in the m/am reaction (using about 10% of the power output from that, or 2% of the energy potential in the available mass supply, to drive the antimatter "flip" system). The rest, I'd divert into cryogenic slush tanks, to serve as thrust mass, fusion "auxiliary power" fuel, and even replenishment of lost mass (atmosphere bleed, etc).

Even "only" 20% of the output from that 5km^2 "scoop" would be a LOT of energy.

 

[Crazy Eddie]

So yeah, you pretty much have it straight.

YAY!

My question is... why would you need 1g/sec?

Remember that a single gram of reaction mass produces a LOT of energy. It's pretty much nonsensical to assume you need that much mass for conversion.

Three reasons:
1) The antimatter generator probably consumes something like 30% of the reactor power.

2) The warp drive probably uses up the rest of that power.

3) In order for a M/AM reaction to produce "plasma," you'd need more matter than antimatter.

So if your reactor is 70% efficient ordinarily, and your antimatter generator uses up 30% for itself, then 1g of intake gives you 386mg of hydrogen and 16mg of antimatter. At 70% efficiency, the reactor has a maximum power output of about 10 terawatts, of which 7 terawatts would be used by the warp engines.

I don't think 7 terawatts is unreasonable for a ship that is traveling several hundred times faster than light, especially if some of that power is used to generate the scoop field. Of course, the point of this thread is partially to figure out how much of that energy is allocated to the ramscoop and how much is pure space warp.

It's also nonsensical to assume that you're going to "react" 100% of what you're collecting. A fair portion would be used to replenish lost mass (atmosphere bleed, etc) and, most likely, the majority will be diverted to be used as reaction-mass for newtonian propulsion devices

Okay... well, if you only need 16mg of matter to react with the antimatter (and let some undefined but totally recyclable "warp plasma" absorb that energy for power conversion) then you're taking on 370mg of hydrogen per second. That means that a twelve-day flight at warp six would allow for an intake of about 380kg of hydrogen (given average density) which can then be used by the impulse engines. How long this will last kind of depends on how quickly the impulse engines and/or fusion reactors consume fuel when they are active.

I kind of like this, because it reminds me of the diesel/electric analogy I once thought of for starships, the way submarines have to run on diesels for a while to recharge their batteries, which becomes extremely important for modern diesel boats that do their most important work while submerged under battery power.

I'm still not sold on whether or not this would be more efficient than occasionally skimming needed materials from the atmospheres of various planets the ship may come across. Oxygen and nitrogen are hard to come by in the ISM, and water harder still; it's got to be easier and less complicated to get these things from natural sources like planets and comets than it is to try and make them from scratch, especially in the case of hydrogen where the ship can refuel in minutes instead of in days.

 

[Myasishchev]

Cary, that's great stuff, but if you don't assume the magic quark-flipping device, the only use for protium (as opposed to deuterium) would be P-P fusion, less energetic by a factor of about 2000. The electron-positron pair produced during the P-P chain annihilates for a pair of gamma rays adding up to 1.02MeV, compared to the pions which decay into a pair of gamma rays adding up to 1.88 GeV in a pure proton-antiproton annihilation reaction.

I'll add that the initial fusion of two protons releases neutrinos on the order of .42 MeV, but this is of course wasted, unless Trek science permits the harvesting of neutrinos (definitely possible, as they appear perfectly able to detect them, and detection requires stopping them somehow).

However, the P-P chain would permit them to replenish deuterium stocks to mix with their antihydrogen-2. It is likely the impulse reactors are able to utilize both P-P and the deuterium + protium-->tritium + tritium-->helium chain. A fair amount of its energy is likely going into sustaining the fusion reaction, however. I wonder what kind of battery they need to start the motor to produce a self-sustaining reaction. Antimatter sparkplugs, or simply powerful chemical batteries?

Still, as I definitely prefer not to assume a quark-flipper, the actual energy obtained from a run--through a route that also must never have been previously cleared--would be far less under that stricture.

Matter of taste, I guess.

Edit: something occurs to me about reaction mass. As a matter of storage, is it better or worse to use a far more massive element or compound than hydrogen? It strikes me that one could store more mass in the space space for less energy with liquid oxygen, for example, than compressed hydrogen, since the compressibility of a substance has to do with its electrons. Does the single shell of hydrogen take up substantially less space than the two of oxygen, enough to outweigh the sixteen times (or eight if deuterium) greater mass of the oxygen?

 

[Cary L. Brown]

Cary, that's great stuff, but if you don't assume the magic quark-flipping device, the only use for protium (as opposed to deuterium) would be P-P fusion, less energetic by a factor of about 2000.

True, but the Trekkian technology is principally based around matter/antimatter annihilation, with fusion used solely as "supplementary" (or "auxiliary") power.

Given that m/am annihilation is central to the topic, you really have two choices... carry your own antimatter along with you (refueling either via "Space Texaco" or by some natural deposits found along the way), or generating it (with the "magic flipping device")

The only way that Treknology really seems to work, and be practical, is to have that "magic flipper."

And, of course, what this means is "you have to spend significantly less energy to create antimatter than you get from a m/am reaction."

So this is a conceit which I think is CRUCIAL to anything in "Treknology" being remotely practical.

If we want to talk about REAL technology... your point is entirely valid, of course.

The electron-positron pair produced during the P-P chain annihilates for a pair of gamma rays adding up to 1.02MeV, compared to the pions which decay into a pair of gamma rays adding up to 1.88 GeV in a pure proton-antiproton annihilation reaction.

I'll add that the initial fusion of two protons releases neutrinos on the order of .42 MeV, but this is of course wasted, unless Trek science permits the harvesting of neutrinos (definitely possible, as they appear perfectly able to detect them, and detection requires stopping them somehow).

However, the P-P chain would permit them to replenish deuterium stocks to mix with their antihydrogen-2. It is likely the impulse reactors are able to utilize both P-P and the deuterium + protium-->tritium + tritium-->helium chain. A fair amount of its energy is likely going into sustaining the fusion reaction, however. I wonder what kind of battery they need to start the motor to produce a self-sustaining reaction. Antimatter sparkplugs, or simply powerful chemical batteries?

Well, we have a clue to this given in "Where No Man Has Gone Before."

Remember, Mitchell notices something that Kelso missed, even though he saw it... that the "points in the impulse packs are decayed to lead."

What might this mean? I suspect that you're looking at some sort of fission-based initiator for the fusion system. Remember, hydrogen bombs are initiated by the use of an atomic bomb as a "detonator." I suspect that they use something along the same lines (albeit far better controlled) to initiate fusion... and that you need the fusion system running at full output in order to get the m/am system running (because combining it "cold" can do ugly things, it seems).

Still, as I definitely prefer not to assume a quark-flipper, the actual energy obtained from a run--through a route that also must never have been previously cleared--would be far less under that stricture.

Matter of taste, I guess.

Or rather, "real science" versus "Trek science." Trek science, at its best, is an extension of "real science" using a few mystical, magical "not yet discovered" bits of pseudo-science.

The "efficient flipper" is one of those. But then again, so is "warp drive." And "subspace." And "transporters." And on and on...

Now, about "routes not having been cleared." This makes sense, it seems... but only if you have very tightly held, controlled "traffic lanes" and a LOT of traffic along them. Remember, space is big... really, really big. I wonder how many trillions of starships would it take to "clear" any area of space to any significant extent?

Edit: something occurs to me about reaction mass. As a matter of storage, is it better or worse to use a far more massive element or compound than hydrogen? It strikes me that one could store more mass in the space space for less energy with liquid oxygen, for example, than compressed hydrogen, since the compressibility of a substance has to do with its electrons. Does the single shell of hydrogen take up substantially less space than the two of oxygen, enough to outweigh the sixteen times (or eight if deuterium) greater mass of the oxygen?

Good question. If everything else was equal, you're better off with the highest-density fluid you can get (which is why some ion engine designs are based around mercury as a propellant).

But remember, we're talking about BUSSARD COLLECTORS as a fueling method... so you don't need to carry fuel with you (except as a buffer). I doubt, very much, that you'll collect much in the way of anything but monoatomic hydrogen with such a system.

So, even if some heavier element is theoretically better for "low-energy storage," the whole point of the bussard system makes that sort of a moot point, doesn't it?