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EDG
2009-Mar-01, 08:39 AM
..or within any planetary system, for that matter.

Imagine a scenario where you have a spacecraft that has limitless energy and constant mass (i.e. fuel isn't being used up to change the mass of the ship), and that can accelerate at 1g (which we'll assume is 10 m/s²). What's the quickest method by which it can leave one planet and arrive at another one?
For ease of visualisation, assume it's travelling between Earth and Mars.

Since the ship has an unlimited energy supply (so we don't need to worry about conserving fuel or energy, we just need to get there as quick as possible using whatever method is available regardless of energy requirement), I presume we don't need to bother with Hohmann transfer orbits anymore.

Ordinarily I would think that the best way would be to accelerate from Earth orbit at 1g, then turn around at the halfway point, and then decelerate into Mars orbit at 1g, but I keep thinking I'm missing something here because of course it's not accelerating from zero velocity and decelerating back to zero velocity - the ship must leave and enter planetary orbit, and the planets are moving in their orbits too. Also, the ship has to turnaround pretty quickly and start decelerating (assuming it needs to point its engines in the opposite direction) or it'll be accelerating for longer than half the trip and decelerating for less than half the trip.

So, I think I'm missing something here, but I'm not sure what - would anyone have any input?

EDG
2009-Mar-01, 05:43 PM
anybody?

2009-Mar-01, 06:07 PM
Take a look at http://home.comcast.net/~mbmcneill7/. Either the Earth Orbit Ferry or the Interstellar Vehicle could make the trip accelerating at one g, but the Interstellar Vehicle would be better suited. Since the system described requires over one mega-watt per square meter power reception and the likelihood of materials science producing such material and the implied cooling that would be required is vanishingly small, different q's and Ve's will probably be used. Also the unavoidable sail effect of radiating the IV with 10^16 or so watts will require deceleration over more than half the distance. The massive infrastructure is the strength of the system. I am currently revising the design to make the trip from Earth to the Alpha Centauri system in 40 or so years and dropping the power per square meter down into the kilowatt range---still a challenge for materials science.

EDG
2009-Mar-01, 06:16 PM
Kinda useful, but not quite what I had in mind...

I guess one thing I'm concerned about is the start and end velocities. Obviously they're not zero, but what are they if one is starting from Earth orbit, and how does one account for them mathematically in a standard accel-turnaround-decel procedure?

JohnBStone
2009-Mar-02, 12:46 AM
Earth's orbital velocity is approx 30 kms and Mars is 24 kms. The maximum difference in velocities is 54 kms when they are at opposite sides of the sun (plus I suppose up to another 16kms if you include both escape velocities).

A 1G drive can eat 54 kms in around 90 minutes (and 70 in kms under 2 hours), and Earth-Mars median straight line distance at 1G is just under 100 hours. Is there a major difference involved?

I suppose thrusting straight out means you are also fighting the sun's pull (just like thrusting from Earth's surface to orbit means subtracting Earth's gravity from your thrust until you reach orbital velocity) . So potentially you need to subtract the sun's point gravity from your thrust going towards Mars. But as this is less than one-thousandth of a G according to this thread (http://www.bautforum.com/archive/index.php/t-18792.html), not significant. There is another small amount to add as orbits are not completely aligned, again not significant.

alainprice
2009-Mar-02, 02:06 AM
You haven't set the initial condition. With that kind of acceleration possible, the orbits hardly matter.

One example is to cheat and use a Lagrangian point to start from, perhaps the earth-sun L2. This might simplify the task. Ignoring calculus and the sun's gravity, we can say that we'll just accelerate directly away from the earth to mars at the right time and perform an insertion maneuver once there. Ignoring insertion to mars orbit(say the sun-mars L1), that could make for a trip as short as 50 000 000 km. Mars is not as well behaved as earth is, being more eccentric. The difference in orbital velocities is about 5km, so we'll even ignore that. That's as quick as
75 minutes total(I got 4472 seconds for 50 million km), assuming instant turnaround midway.

The actual direction of firing won't be much different. You need to compensate for differing gravity and the decrease in the orbital velocity for mars.

EDG
2009-Mar-02, 02:09 AM
OK, so basically the accelerate-turnaround-decelerate concept can be used pretty much without modification then?

tony873004
2009-Mar-02, 02:51 AM
The velocities of Earth and Mars and even the acceleration due to the sun's gravity are rather insignificant compared to the high velocities you will achieve.

You can do this in Gravity Simulator. Use the autopilot. From Earth orbit, choose "Orient Towards" and choose your spacecraft and choose Mars. Then choose "Thrust" and choose your spacecraft. Choose 10 m/s/s. At the halfway point choose "Orient Away" and choose your spacecraft and choose Mars. There will be a little bit of slop in this method as Mars has moved a little during the journey. To clean the slop, at your closest approach, choose "Orient Towards", then when you're halfway through the remaining distance, choose "Orient Away". Then at closest approach choose "Orient Retrograde" and burn until your orbit equals circular orbital velocity (use the Orbital elements box and wait for your eccentricity to drop to 0.)

JohnBStone
2009-Mar-02, 08:29 AM
That's as quick as 75 minutes total(I got 4472 seconds for 50 million km), assuming instant turnaround midway.
I get 40 hours or 2400 minutes.

see www.transhuman.talktalk.net/iw/TravTime.htm

EDG
2009-Mar-02, 08:51 AM
Yeah, me too. 50 million km at 10 m/s² acceleration with midpoint turnaround is 39.28 hours total travel time.

timb
2009-Mar-02, 11:14 AM
Yeah, me too. 50 million km at 10 m/s² acceleration with midpoint turnaround is 39.28 hours total travel time.

Don't forget to take a magazine.

Pity your spaceship defies the laws of physics.

tony873004
2009-Mar-02, 06:21 PM
It's certainly not feasable with our current technology, but what law of physics is broken by maintaining a 1 g acceleration for a few days?

alainprice
2009-Mar-02, 07:57 PM
Yeah, I screwed up between distances in Km and acceleration in m/s.

Oops. Still, 2 days is pretty darn quick. The planets don't change in their orbits a whole lot in 2 days.

timb
2009-Mar-02, 08:09 PM
It's certainly not feasable with our current technology, but what law of physics is broken by maintaining a 1 g acceleration for a few days?

If the spaceship is a closed system then it violates conservation of mass-energy.

Murphy
2009-Mar-02, 08:26 PM
I did these kind of calculations a few years ago. 1 g (9.8 m/s^2) acceleration and deceleration for the Solar system, here's my figures...

Mercury:
Max Dist. = 217,000,000 km ( 1.45 AU) - Earth to Mercury in 83 hrs at 1g (3 days, 11 hours)
Min Dist. = 82,000,000 km ( 0.55 AU) - Earth to Mercury in 51 hrs at 1g (2 days, 3 hours)
Avr Dist. = 156,000,000 km ( 1.04 AU) - Earth to Mercury in 70 hrs at 1g (2 days, 3 hours)

Venus:
Max Dist. = 260,000,000 km ( 1.74 AU) - Earth to Venus in 91 hrs at 1g (3 days, 19 hours)
Min Dist. = 40,000,000 km ( 0.27 AU) - Earth to Venus in 36 hrs at 1g (1 day, 12 hours)
Avr Dist. = 171,000,000 km ( 1.14 AU) - Earth to Venus in 73 hrs at 1g (3 days, 1 hour)

Luna:
Max Dist. = 405,696 km (0.003 AU) - Earth to Luna in 3.6 hrs at 1g (3 hours, 35 Minutes)
Min Dist. = 363,104 km (0.002 AU) - Earth to Luna in 3.4 hrs at 1g (3 hours, 23 Minutes)
Avr Dist. = 384,399 km (0.003 AU) - Earth to Luna in 3.5 hrs at 1g (3 hours, 29 Minutes)

Mars:
Max Dist. = 401,000,000 km ( 2.68 AU) - Earth to Mars in 112 hrs at 1g (4 days, 16 hours)
Min Dist. = 55,000,000 km ( 0.37 AU) - Earth to Mars in 42 hrs at 1g (1 day, 18 hours)
Avr Dist. = 254,000,000 km ( 1.70 AU) - Earth to Mars in 89 hrs at 1g (3 days, 17 hours)

Ceres:*
Max Dist. = 600,000,000 km ( 4.01 AU) - Earth to Ceres in 138 hrs at 1g (5 days, 18 hours)
Min Dist. = 229,000,000 km ( 1.53 AU) - Earth to Ceres in 85 hrs at 1g (3 days, 13 hours)
Avr Dist. = 430,000,000 km ( 2.87 AU) - Earth to Ceres in 116 hrs at 1g (4 days, 20 hours)

Jupiter:
Max Dist. = 965,000,000 km ( 6.45 AU) - Earth to Jupiter in 174 hrs at 1g (7 days, 6 hours)
Min Dist. = 591,000,000 km ( 3.95 AU) - Earth to Jupiter in 136 hrs at 1g (5 days, 16 hours)
Avr Dist. = 790,000,000 km ( 5.28 AU) - Earth to Jupiter in 158 hrs at 1g (6 days, 14 hours)

Saturn:
Max Dist. = 1,653,000,000 km (11.05 AU) - Earth to Saturn in 228 hrs at 1g (9 days, 12 hours)
Min Dist. = 1,204,000,000 km ( 8.05 AU) - Earth to Saturn in 195 hrs at 1g (8 days, 3 hours)
Avr Dist. = 1,439,000,000 km ( 9.62 AU) - Earth to Saturn in 213 hrs at 1g (8 days, 21 hours)

Uranus:
Max Dist. = 3,155,000,000 km (21.09 AU) - Earth to Uranus in 315 hrs at 1g (13 days, 3 hours)
Min Dist. = 2,817,000,000 km (18.83 AU) - Earth to Uranus in 298 hrs at 1g (12 days, 10 hours)
Avr Dist. = 2,995,000,000 km (20.02 AU) - Earth to Uranus in 307 hrs at 1g (12 days, 19 hours)

Neptune:
Max Dist. = 4,653,000,000 km (31.10 AU) - Earth to Neptune in 383 hrs at 1g (15 days, 23 hours)
Min Dist. = 4,328,000,000 km (28.93 AU) - Earth to Neptune in 369 hrs at 1g (15 days, 9 hours)
Avr Dist. = 4,489,000,000 km (30.01 AU) - Earth to Neptune in 376 hrs at 1g (15 days, 16 hours)

Pluto:
Max Dist. = 7,511,000,000 km (50.21 AU) - Earth to Pluto in 486 hrs at 1g (20 days, 6 hours)
Min Dist. = 4,273,000,000 km (28.56 AU) - Earth to Pluto in 367 hrs at 1g (15 days, 7 hours)
Avr Dist. = 5,900,000,000 km (39.44 AU) - Earth to Pluto in 431 hrs at 1g (17 days, 23 hours)

Eris:*
Max Dist. = 14,741,000,000 km (98.54 AU) - Earth to Eris in 681 hrs at 1g (28 days, 9 hours)
Min Dist. = 5,640,000,000 km (37.70 AU) - Earth to Eris in 422 hrs at 1g (17 days, 14 hours)
Avr Dist. = 10,070,000,000 km (67.31 AU) - Earth to Eris in 563 hrs at 1g (23 days, 11 hours)

*These figures for Ceres and Eris are not so certain.

This calculation doesn't account for things like gravitational influences, or the difference in orbital velocities of the planets (or their motions during the time period), but I figured that the transits are so fast that those things would make little difference and only make it a lot harder to calculate, so there're not 100% accurate, but close enough for most uses. Remember the distances quoted there are from the Earth to the other planet (not from the Sun), I got most of the figures from this website http://www.astro.uu.nl/~strous/AA/en/antwoorden/planeten.html#v335

I did this for possible Sci-Fi stories for a future in which mankind would be using Fusion and Anti-matter spacecraft. There may well be errors in the calculations so use them at your own risk. (I also did the calculations for 0.1 gs, 0.25 gs, 0.376 gs (Martian g) and 4 gs, if anybody wants those, just ask).:)

timb
2009-Mar-02, 10:13 PM
I did these kind of calculations a few years ago. 1 g (9.8 m/s^2) acceleration and deceleration for the Solar system, here's my figures......
Eris:*
Max Dist. = 14,741,000,000 km (98.54 AU) - Earth to Eris in 681 hrs at 1g (28 days, 9 hours)
Min Dist. = 5,640,000,000 km (37.70 AU) - Earth to Eris in 422 hrs at 1g (17 days, 14 hours)
Avr Dist. = 10,070,000,000 km (67.31 AU) - Earth to Eris in 563 hrs at 1g (23 days, 11 hours)

Why stop there? it would take only about a year to reach alpha Centauri.

EDG
2009-Mar-02, 10:23 PM
If the spaceship is a closed system then it violates conservation of mass-energy.

Oh, it's broken alright :). But this is a SF thing, and this is how the tech works in the setting. Still, the calculations work even if the engine doesn't ;)

speedfreek
2009-Mar-02, 10:24 PM
Why stop there? it would take only about a year to reach alpha Centauri.

If you want to stop when you get there it would take 3.6 years according to The Relativistic Rocket (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html)

timb
2009-Mar-02, 10:53 PM
If you want to stop when you get there it would take 3.6 years according to The Relativistic Rocket (http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html)

Thanks for the link. I forgot the need to stop.

Murphy
2009-Mar-02, 11:11 PM
Why stop there? it would take only about a year to reach alpha Centauri.

Yes, of course you could just keep going, but my interest in these calculations was just interplanetary trips in the Solar System. I've done plenty of interstellar calculations, but none of them involved constant 1 g acceleration, as I thought the drive types I'm thinking of wouldn't be able to do that.

For instance the maximum velocity needed to do those 1 g trips, is about 4% of the speed of light (for the most distant trip to Eris). That's well within the probable limits of Fusion drives, so it's got a level of realism (at least for Sci-Fi). When you try to accelerate at 1 g for too long, you start getting up to the speed of light pretty quickly, which brings major problems for the drive system (i.e. some drives cannot go beyond a certain speed, and others just require more and more fuel mass, which makes them a lot less useful).

Fusion ships are unlikely to be able to get anywhere near the speed of light (unless they use some kind of Bussard Ram Scoop, and thus have infinite propellant) 50% to 25% C sounds more likely. So all the interstellar trip calculations I've done involve accelerating up to less than half the speed of light and then costing for a long time and then decelerating.

One of the main motivations for a 1 g trip through the Solar System (besides being the quickest way) is to have constant artificial gravity onboard for the passengers. But on a very long interstellar trip lasting decades to centuries that is of less concern, as I assume the passengers will be in some kind of cryogenic stasis (or at least that the ship will be big enough to have large rotating environments, etc).

2009-Mar-03, 03:53 AM
That's well within the probable limits of Fusion drives, so it's got a level of realism...Since we don't know how to design a fusion system capable of supplying the thrust needed to achieve one g acceleration for a ship of sufficient mass to transport a few humans over interstellar distances, how do we know? Has anyone shown that a fusion powered ship designed with comfortable safety factors can be built. My fear is that the mass of the ship including its pre-fusion elements, propellant, shielding, and fusion reactor can never be proportioned to function near one g. What will function, if anything will, is collimated photon beaming at power levels near 10^17 watts to photovoltaic receivers supplying power to ion engines with propellants supplied by paser.

Murphy
2009-Mar-03, 06:40 AM
Ok, it's all theoretical at this point, but aren't Ion Engines one of least powerful types of possible drive?
Unless you mean it in a different way to the Ion drives that exist today...?

As for Fusion, I don't see why it can't be done, the exact capabilities of the theoretical engine I couldn't predict, but the idea of laser detonating hundreds of tiny deuterium pellets every second and using magnetic fields to contain and blast the resulting Plasma out of a nozzle, is all quite theoretically sound. The actual engineering of it is another thing of course, that might take 100 years, but it’s by no means impossible. It's essentially the same principle as the Orion type nuclear bomb propulsion, and we know that could have worked.

It might turn out that a Fusion engine can only get a useful payload to 0.1 g, but then all you have to do is strap 10 of them together and you're on your way a 1 g ride. Anyway the idea of calculating 1 g travel times for me was to see just how fast Humans could travel around the Solar system with theoretical tech (1 g because it's just about the highest acceleration a normal Human body can take for long periods of time), not necessarily because I think that is what will happen.

timb
2009-Mar-03, 08:48 AM
Anyway the idea of calculating 1 g travel times for me was to see just how fast Humans could travel around the Solar system with theoretical tech (1 g because it's just about the highest acceleration a normal Human body can take for long periods of time)

Are you sure about that? Most g-force experiments seem to concentrate on brief exposures (the sponsors are mostly interested in aircraft turns) rather than endurance, but users of the Libelle suit (http://findarticles.com/p/articles/mi_qa3731/is_200012/ai_n8907444) reportedly find 9g "comfortable". That would cut your travel times to a third.

Peter B
2009-Mar-03, 10:52 AM
It might turn out that a Fusion engine can only get a useful payload to 0.1 g, but then all you have to do is strap 10 of them together and you're on your way a 1 g ride.

Er, is that right? If a given spacecraft can accelerate at 1 G, and ten of them separately will each accelerate at 1 G, strapping them all together won't suddenly make them accelerate at 10 G, any more than 10 cars strapped together will accelerate 10 times faster than an individual car.

JMV
2009-Mar-03, 02:07 PM
Er, is that right? If a given spacecraft can accelerate at 1 G, and ten of them separately will each accelerate at 1 G, strapping them all together won't suddenly make them accelerate at 10 G, any more than 10 cars strapped together will accelerate 10 times faster than an individual car.
I guess he assumes that the mass of the engine is insignificant compared to the mass of the payload.

2009-Mar-03, 03:30 PM
actual engineering of it is another thing of course, that might take 100 years, but it’s by no means impossible. Sadly, I don't know enough about the design of fusion systems to reasonably predict how likely the engineering is considering the current state of materials science. If someone here knows enough to define the design envelope or links to how it might be done, that would be nice information to make available. We need sufficient data to size the mass of the ship including each element of the fusion system and the necessary infrastructure and to project how the ratio of mass-to-power varies over the ranges required for interplanetary and interstellar travel. At the most recent Joint Propulsion Conference (2008) a paper was presented taking the position that interstellar travel is near impossible. I was not there and have not read the paper, but I would like to know how solidly their position was defended since I presented a paper at the 2003 version of that conference showing how it could be done even though I glossed over the detail design of the subsystems required.

Bob B.
2009-Mar-03, 08:37 PM
Imagine a scenario where you have a spacecraft that has limitless energy and constant mass (i.e. fuel isn't being used up to change the mass of the ship), and that can accelerate at 1g (which we'll assume is 10 m/s²). What's the quickest method by which it can leave one planet and arrive at another one?
For ease of visualisation, assume it's travelling between Earth and Mars.

I just completed a simple computer simulation of this scenario. I think that the best way to get there is to always thrust along the velocity vector (except as noted later in this post). This means we start out traveling along a tangent to Earth's orbit, since we're already moving in that direction. Because we are accelerating so quickly to such high velocity, the trajectory is practically a straight shot until we reach Mars' orbit, with only a very small amount of curvature. The route taken looks like this:

http://www.braeunig.us/pics/E-M-1g.gif

Of course this trajectory has a small launch window occurring once every 26 months when Earth and Mars are properly aligned.

In this scenario, the spacecraft thrusts continuously in the direction of travel for 36 hours. The craft then turns around and thrusts in the opposite direction for about 35.75 hours. For the last 45-50 minutes, the spacecraft points its nose toward the Sun and thrusts in that direction to cancel out the remaining radial velocity. This gets us to Mars in about 72.5 hours. The total distance traveled is 172 million kilometers, the average velocity is 660 km/s, and the maximum velocity is 1,300 km/s at the midway point.

For this scenario I assumed Earth and Mars at their mean distances from the Sun. For most launch windows, the travel time and distance will vary.

Murphy
2009-Mar-03, 11:41 PM
Originally Posted by Peter B
Er, is that right? If a given spacecraft can accelerate at 1 G, and ten of them separately will each accelerate at 1 G, strapping them all together won't suddenly make them accelerate at 10 G, any more than 10 cars strapped together will accelerate 10 times faster than an individual car.

Well here's the way I think about it. No, 10 cars strapped together wont go 10 times faster, but if you removed the 10 engines and placed them in a single car (with the same mass as the original) then they would go about 10 times faster (theoretically...), though they'll burn through their fuel 10 times faster as well. Obviously the car analogy doesn't work very well in relation to rockets out in space, as there’re no obstacles like road friction and wind resistance there.

Originally Posted by JMV
I guess he assumes that the mass of the engine is insignificant compared to the mass of the payload.

That would certainly help, but I don't think it has to be that way. It all depends on how powerful the thrust is.

If we look at something like Project Daedalus for an example (http://en.wikipedia.org/wiki/Project_Daedalus), let's say our futuristic Fusion drive has an exhaust velocity of 10,000,000 m/s and a thrust of 10,000,000 N, and the engine system itself weighs 1,000 metric tonnes. (I've no idea how realistic these numbers are, but for this hypothetical example let's just assume they are).

So, basic physics tells us that F = m*a.
Therefore m = F/a, so the amount of mass the rocket can accelerate up to a certain speed is; the force divided by the acceleration. If we go for 0.1 gs, we get: 10,000,000 / 0.98 = 10,204,082 kg or 10,204 tonnes of mass. Suppose half of the mass is propellant (5,102 tonnes), that leaves 4,102 tonnes of structure and payload (subtracting the 1,000 tonne engine).

So, if we want to accelerate the same payload to 0.2 gs, or 0.5 gs, or 1 g, you just have to add another engine (or five, or ten, etc). At least I think that's how it's done...

Originally Posted by timb
Are you sure about that? Most g-force experiments seem to concentrate on brief exposures (the sponsors are mostly interested in aircraft turns) rather than endurance, but users of the Libelle suit reportedly find 9g "comfortable". That would cut your travel times to a third.

Eh, well I'd be pretty sceptical as to the idea that you could comfortably survive 9 gs for any extended length of time. Sure you can do it for a few minutes if you happen to be a trained fighter pilot in a g-suit, but I'm talking about ordinary people travelling in space for weeks. I'm pretty sure experiments have been done in centrifuges to see the effects of high-g on the Human body, but no one (not even fighter pilots or astronauts) can hold out for very long under more than a few gs.

I think you start to go unconscious after a few minutes of really high-g, whether or not you're wearing a g-suit. I've hear from the world of Sci-Fi that it might be possible to continuously accelerate at about 4 gs if you used some kind of advance protective suits and were laying on big Gel cushions, but it would still be an immense strain (and painful), so it would only be used by the "Space Military" or some other emergency situation. I can't imagine having normal civilian passengers travelling like that.

If we're talking about a passenger shuttle that has all types of people on board (like old people and babies, etc.), then I don't see how we can go beyond 1 g.

Edit: BTW try out this calculator program, which deals with constant acceleration thrusters, etc. http://www.5596.org/cgi-bin/thrusters.php (That site also has some other interesting calculators http://www.5596.org/equations.html).

timb
2009-Mar-04, 12:14 AM
Eh, well I'd be pretty sceptical as to the idea that you could comfortably survive 9 gs for any extended length of time. Sure you can do it for a few minutes if you happen to be a trained fighter pilot in a g-suit, but I'm talking about ordinary people travelling in space for weeks. I'm pretty sure experiments have been done in centrifuges to see the effects of high-g on the Human body, but no one (not even fighter pilots or astronauts) can hold out for very long under more than a few gs.

I think you start to go unconscious after a few minutes of really high-g, whether or not you're wearing a g-suit. I've hear from the world of Sci-Fi that it might be possible to continuously accelerate at about 4 gs if you used some kind of advance protective suits and were laying on big Gel cushions, but it would still be an immense strain (and painful), so it would only be used by the "Space Military" or some other emergency situation. I can't imagine having normal civilian passengers travelling like that.

For taking maximum benefit from the Oberth effect (http://en.wikipedia.org/wiki/Oberth_effect) around compact objects such as BDs, WDs, and neutron stars (if you're lucky enough to have one in your system) it might be desirable for the pilots to be able to pull 40g or more around periapsis. This is conceivable if the pilots are immersed in liquid of the same density as the body because the acceleration manifests itself mainly as an increase in pressure, which the body is relatively tolerant of. Above about 15g the air in the lungs becomes a problem, so researchers have looked at breathable liquids (http://en.wikipedia.org/wiki/Breathing_liquid). So far none has been found with a specific gravity close enough to the body's to be useful in this regard. While a marine habit is usually considered a disadvantage for the prospective space traveller, creatures that naturally breathe water may be at an advantage where extreme acceleration is required.

grant hutchison
2009-Mar-04, 12:59 AM
Liquid breathing is going to be hard work because of the mass and viscosity involved, and it's not very good at eliminating CO2. Perhaps a more realistic prospect might be liquid ventilation, in which external pressure is used to pump liquid in and out of the lungs through, for instance, a tracheostomy tube.
Once you have the body filled with, and surrounded by, liquid that matches average tissue density, a residual problem at very high acceleration is the difference in density between our various body tissues: your skeleton tends to drop to the bottom of your immersion tank, while your fat tends to float to the top. Barring serious re-engineering of the human frame, that will presumably place an upper limit on tolerable acceleration, but it would be quite a high limit.

Grant Hutchison

JMV
2009-Mar-04, 03:19 AM
If we look at something like Project Daedalus for an example (http://en.wikipedia.org/wiki/Project_Daedalus), let's say our futuristic Fusion drive has an exhaust velocity of 10,000,000 m/s and a thrust of 10,000,000 N, and the engine system itself weighs 1,000 metric tonnes. (I've no idea how realistic these numbers are, but for this hypothetical example let's just assume they are).

So, basic physics tells us that F = m*a.
Therefore m = F/a, so the amount of mass the rocket can accelerate up to a certain speed is; the force divided by the acceleration. If we go for 0.1 gs, we get: 10,000,000 / 0.98 = 10,204,082 kg or 10,204 tonnes of mass. Suppose half of the mass is propellant (5,102 tonnes), that leaves 4,102 tonnes of structure and payload (subtracting the 1,000 tonne engine).

So, if we want to accelerate the same payload to 0.2 gs, or 0.5 gs, or 1 g, you just have to add another engine (or five, or ten, etc). At least I think that's how it's done...

Let's say you have a single engine rocket weighting 10,000 tonnes and that one engine weights 1,000 tonnes. You try to increase the acceleration by adding nine more similar engines. This way you increase the thrust by a factor of ten, but here's the catch: by adding more engines you have increased the mass of the vehicle as a whole to 19,000 tonnes. Using the a = F/m equation you'll see that the acceleration has increased only by a factor of 5.26.

Well then, how many engines would you have to add to increase the acceleration by a factor of ten? Here's where it gets interesting, the answer is: infinite number of engines. You see, as the number of engines increases, the acceleration multiplier approaches asymptotically the ratio between the single engine vehicle mass and the mass of one engine, but never reaches it. In my example that ratio is 10, so it is not physically possible to get ten fold increase in acceleration by adding more engines.
I see in your example you used 10,204 tonnes as the mass of the single engine vehicle and 1,000 tonnes as the mass of one engine. In that case the ratio is 10.204 and the number of engines needed is 451 or so.

Furthermore simply adding more engines could be counterproductive for your cause. You increase the dry mass of the vehicle without increasing the total impulse provided by the propellants, therefore decreasing the delta-v the vehicle is capable of. See Tsiolkovsky's rocket equation (http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation) for more info on that.

timb
2009-Mar-04, 05:56 AM
Liquid breathing is going to be hard work because of the mass and viscosity involved, and it's not very good at eliminating CO2. Perhaps a more realistic prospect might be liquid ventilation, in which external pressure is used to pump liquid in and out of the lungs through, for instance, a tracheostomy tube.

Yes, mechanical ventilation would almost certainly be required. Even deep sea divers breathing trimix notice the greater effort required to breath. The tracheostomy tube would have the advantage of making the first breath less, uh, challenging.

Once you have the body filled with, and surrounded by, liquid that matches average tissue density, a residual problem at very high acceleration is the difference in density between our various body tissues: your skeleton tends to drop to the bottom of your immersion tank, while your fat tends to float to the top. Barring serious re-engineering of the human frame, that will presumably place an upper limit on tolerable acceleration, but it would be quite a high limit.

There's always the old brain in a vat trick, but nothing living is going to beat solid state electronics so there is no advantage in going past the point where the pilot can still get out and do other work after the flight.

grant hutchison
2009-Mar-04, 06:04 PM
Another possibility would be non-invasive ventilation, either positive- or negative-pressure (or both).
You isolate the head from the rest of the body with some sort of rigid structure sealed around the neck, and generate a head-thorax pressure gradient to facilitate flow in/out of the lungs. That's the principle of the old "iron lung", for instance. You can synchronize with the subject's own ventilatory efforts by building a pressure sensor into the system somewhere, to trigger each ventilation cycle.
So you might imagine our pilot wearing a rigid helmet with a neck seal. When he tries to breathe in, he triggers a ventilator cycle that provides liquid flow into his helmet at a regulated overpressure, while removing liquid from his evironment at a similarly regulated underpressure. The pilot's lungs fill as a result.
An expiratory effort would trigger the reverse movement.

I can report that abdicating your ventilatory cycle to an external pressure source is an interesting sensation, but not particularly unpleasant so long as adequate flows are delivered promptly.

Grant Hutchison

Murphy
2009-Mar-04, 11:36 PM
JMV, yeah, you are right of course, I kind of knew there was something not right about what I was saying, but I couldn't quite pin it down.

That was just a hurried example, though the Drive I picked obviously just isn't suitable for 1 g travel, you'd have to use something much more powerful.

Looking though some websites, Atomic Rockets has an Engine list page detailing possible drives (http://www.projectrho.com/rocket/rocket3c2.html), for sustained 1 g travel you'd have to go for something like IC-Fusion MAX, which he thinks could deliver 100,000,000 N of thrust. That would certainly solve the problem.:)

As for alterative breathing systems, how about just injecting Oxygen directly into the blood stream? Not in bubbles of air obviously (that would just kill the person), but perhaps some kind of system where the blood is temporarily taken out of the body into some machine which oxygenates it and removes CO2 and is then re-injected right back, bypassing the lungs entirely (kind of like how dialysis machines clean the blood today, replacing the job done by the kidneys).

grant hutchison
2009-Mar-05, 12:02 AM
... but perhaps some kind of system where the blood is temporarily taken out of the body into some machine which oxygenates it and removes CO2 and is then re-injected right back ...Sounds like you're describing the technique of ECMO/ECCO2R (http://ccforum.com/content/4/3/156) (extracorporeal membrane oxygenation / extracorporeal CO2 removal).

Grant Hutchison

EDG
2013-Apr-02, 12:27 AM
Sorry for the bump, but I'm just pondering this again and happened to come across this thread again in my search for an answer (I didn't notice asking this specifically before) :)

So... let's say you have a ship that constantly accelerates at 1g up to the midway point to its destination, then it flips around and decelerates at 1g to reach its destination with zero velocity (I know, it's not realistic. Let's just assume for now that the planets aren't moving relative to eachother).

I've been assuming that one would use s = ut + 0.5at² to figure out the travel time to the destination, by:

s = ut + 0.5at²

u = 0 so ut = 0 so

s = 0.5at²

so

t = SQRT(2s/a)

where t is in seconds, s = distance to midpoint in metres, a = acceleration in m/s².

OK, that works from the start to the midpoint. Usually what I've done is just calculate that travel time to midpoint and then double it (since it's decelerating in exactly the same way from the midpoint).

But... at the midpoint u (the initial velocity) isn't zero any more, it's a pretty large number (if accelerating constantly up to that point). That means ut has to come back into the equation doesn't it? And if it does, then the equation gets kinda messy because you'd end up with t = SQRT([2(s-ut)]/a), which I'm not sure how to solve (I guess i need to substitute something for one of the t?).

Is this even the right way to figure this out though?

neilzero
2013-Apr-02, 02:36 AM
The 30 kilometer per second of Earth's orbit hardly matters as we can chose a different frame of reference = speed is relative. Since the travel time is days rather than months, a straight line is close to the shortest and quickest distance. The energy requirment however is huge and generally dynamic breaking is not possible so the energy is worse than wasted = we have to dispose of the waste heat.
Worse: Nearly all matter in our solar system is traveling less than 500 kilometers per second, so if we exceed that speed, the potential for a colision, and the damage from even tiny collisions increases rapidly above that speed. The vector also matters so flying toward the sun at 600 kilometers per second is far worse than flying away from the Sun at 600 kilometers per second. Neil

a1call
2013-Apr-02, 03:35 AM
If you accelerate at 1G off the Earth, you will experience 2Gs at start, never quite reducing to 1G (experience) until you stop accelerating at midpoint. then you would experience 0G until you start decelerating.

You can travel experiencing just above 1G by slowly accelerating and gradually increasing your acceleration to complement the reduction of the gravitational force due to Inverse square law. Still. you will always start at greater than 1G (experience), or else you would have to stay put.

Grey
2013-Apr-02, 04:48 PM
OK, that works from the start to the midpoint. Usually what I've done is just calculate that travel time to midpoint and then double it (since it's decelerating in exactly the same way from the midpoint).

But... at the midpoint u (the initial velocity) isn't zero any more, it's a pretty large number (if accelerating constantly up to that point). That means ut has to come back into the equation doesn't it? And if it does, then the equation gets kinda messy because you'd end up with t = SQRT([2(s-ut)]/a), which I'm not sure how to solve (I guess i need to substitute something for one of the t?).

Is this even the right way to figure this out though?If you really want to, you can work out the velocity at midpoint, and re-solve the equation (taking care to make sure that you now use a negative acceleration, since you're accelerating in the direction opposite your velocity and motion). However, you're intuition is correct here, that the deceleration phase is symmetrical to the acceleration phase. So if you just work out half the journey and then double it as you did originally, you'll find that you get the same answer as working it out the hard way (but with much less work ;) ).

EDG
2013-Apr-03, 04:56 AM
If you really want to, you can work out the velocity at midpoint, and re-solve the equation (taking care to make sure that you now use a negative acceleration, since you're accelerating in the direction opposite your velocity and motion). However, you're intuition is correct here, that the deceleration phase is symmetrical to the acceleration phase. So if you just work out half the journey and then double it as you did originally, you'll find that you get the same answer as working it out the hard way (but with much less work ;) ).

That's good enough for me :). Thanks!

swampyankee
2013-Apr-03, 09:35 AM
I'd be surprised if 9 "g" could be tolerated for more than a few seconds. There are some scholarly articles about human acceleration tolerance, such as http://www.dtic.mil/dtic/tr/fulltext/u2/a138520.pdf and http://www.spacemedicineassociation.org/timeline/1956/27003.pdf ; these may provide hours of enjoyment.

One point to remember is that the acceleration fighter pilots feel is essentially vertical, from their head to their backsides, as a pilot's position is basically sitting upright (some aircraft have reclined seats, but too much reclining makes it difficult for a pilot to perform a visual scan of the sky or see the runway on landing). Regardless, a supine (lying on his/her back) rocket pilot would probably require technological support, possibly like a ventilator, to be able to survive 9 "g" for more than a few dozen seconds.

publiusr
2013-Apr-06, 07:50 PM
Nuclear salt water rockets look to be a good way to keep 1g for hours if you can make it work.

Either that or rotate this:
http://cosmiclog.nbcnews.com/_news/2013/04/05/17606782-scientists-develop-fusion-rocket-technology-in-lab-and-aim-for-mars

fertilizerspike
2013-Apr-07, 05:14 AM
If the spaceship is a closed system then it violates conservation of mass-energy.

There are no known examples of closed systems, except imaginary ones.

neilzero
2013-Apr-07, 04:53 PM
Let's consider a Mars to Earth flight. The passengers experience 1.38 g at takeoff, decreasing to 1.00 g at the L1 point between Mars and the Sun, unless Earth is near the Mars/Sun L1 point. Likely the best route is not very close to the Mars/Sun L1. Since Earth has more gravity, it will cancel somewhat more than of the deceleration than the reverse flight direction, so the travelers will experience somewhat less than one g during a deceleration to Earh's somewhat higher orbital speed. Neil