 The Stars are not Far, Mars Mission 2071. |
  |
 Nov 6 2006, 05:50 PM
Post
#51
|
|||
|
Moving Target  Group: Members Posts: 394 Joined: 19-May 03 From: In your base eating your food. Member No.: 4,607  |
Not an exhaustive list but a pretty good coverage. SoTA: 2064 p.154 Module OMZ13 p.161 Yametetsu Mars mission lands SoTA: 2063 p.119-121 and p.125 Shibanokuji Freefall resort Wake of the Comet Year of the Comet Shadows of Asia p.139 Svobodniy p.31 Yang Lin Fat p.133 Sonia Mikhalina p.162 Pelawangan Space Port p.205 Tyuratam Space Port p.194 Narau Protectorate launch facilities Target Wastelands (by far the best resource) p.66-84 and p.124-129 Corporate Download p.18-20 Zurich-Orbital p.39 AresSpace You're welcome. |
||
 |
 
|
||
|
Nov 6 2006, 06:44 PM
Post
#52
|
|
|
Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
You never started checking my math.
I never claimed a linear relationship between temperature and dm/dt. You made a bad assumtion, I never claimed such a thing. Okay, my back of the envelope calculations were not the clearest and since you desire better clarity I’ll try to offer you a general formula. Average Kinetic Energy based on temperature: KE = (3/2) k T Kinetic Energy of a nonrelativistic particle: KE = (1/2) m v^2 Eliminate energy and solve for v you get: v = sqrt(3 k T / m) Thrust is the Force on the rocket. Newton’s second law: F = M a (large M for the rocket mass, small m for ejecta mass) Thrust = -v dm/dt -v dm/dt = M a dm/dt = M a / v substitute in v and you get: dm/dt = M * Sqrt (m / 3 k T) * a So where you falsely assumed a linear relationship between dm/dt and temperature there is really an Inverse square root relationship! I’m sorry Frank but you’re not even trying . . . And I’m sorry but Solid Rocket Booster (SRB) Separation occurs at 2:07 in the shuttle’s launch sequence. The Space Shuttle Main Engine then fires until 7:40, it throttles down and cuts out at 8:38. So in fact my simple calculation is much closer to correct than you pretend, if you’d like a decent undergraduate level text on classical dynamics including a nice section on rocket science I recommend this book. |
|
|
|
|
Nov 6 2006, 07:08 PM
Post
#53
|
|||||
|
Prime Runner Group: Banned Posts: 3,732 Joined: 1-September 05 From: Prague, Czech Republic Member No.: 7,665 |
Nope. I'm not. Because your basic assumption is inane. The impetus given from a man throwing a baseball off a wagon and a man dropping a baseball off a wagon are very different, and the temperatures are virtually identical. It's not even a remotely plausible claim.
Fine. So the drop from 13 million kelvin to 3600 Kelvin only produces a 60 fold drop in forward motion and not a 1000 fold drop. This still doesn't really help because that still means that you're calling for the expungement of only .43% of the vehicle's mass every second. And 510 seconds later you'd still have thrown out 89% of the vehicles mass. But a shuttle only drops 75% of its mass on the up journey. Your math gives an upward cap of efficiency which is in fact lower than the actual yields of an extremely inefficient process: the external detonation of large amounts of fuel. There's all kinds of energy lost to lateral expansion and it's still substantially more efficient than you claim it possibly could be. There is an underlying falsehood in your assumptions. I don't know where because I don't care enough. But I've got a nose. And when you start claiming that near zero kelvin rocketry can't work, I know we've been had. Equal and Opposite Reactions move things through space. They really do. Which means that your math , even if it's correct, does not represent what you are claiming that it represents. And that's why I'm not checking it. The results do not square with the real world, therefore you are wrong. -Frank |
||||
|
|
|
||||
|
Nov 6 2006, 09:31 PM
Post
#54
|
|
|
Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
I’m sorry Frank, I gave you credit for a better understanding of Physics than I should have. As a result I didn’t spell out some of the basics of how a simple rocket engine works.
Q: Why does the heat of the engine core have anything to do with the velocity of the propellant? A: Unlike what you think when you envision an “The external detonation of large amounts of fuel” what I’m assuming is that you’ve got some sort of bottle that is capable of containing a fusion reaction at a heat equivalent to that of the core of the sun. Particles in this fusion would be moving around with incredible velocities. Velocities that I have calculated for you already in the above. Then I allow you to port this plasma out into space in a linear jet that provides thrust based on a simple conservation of momentum. That’s the thrust equation I’ve shown you. Don't forget that I'm giving for free that there's some way to continue to inject fusion fuel into this reactor core to maintain the reaction at a steady state. What the space shuttle's solid rocket boosters do is they use combustion rather than fusion to create the heat that forces the propellant out the back of the rocket. Commentary: You are unwilling to accept that what I showed you actually represents the space shuttle remarkably well. The primary difference being that the shuttle is launched using the combustion of H2 and O2, but ultimately it’s a very hot core that allows a just of hot gas to be shot out the back at a velocity that is determined based on temperature. In fact the dm/dt that this very rough estimation yields is close to the actual space shuttle. You can’t accept that the space shuttle outputs 9800 kg/s at lift off because that much dm/dt could not be sustained for 510 seconds is a moot point because it is not sustained for 510 seconds. Solid rocket booster separation occurs at 2 minutes and 7 seconds in the flight. In fact I mentioned that above, but you’re stuck on that 510 seconds for some reason. After the 2:07 SRB separation the Space Shuttle Main engine continues to burn, at a much slower rate, and I haven’t attempted to represent that in any way here. And by claiming a 510 second burn to orbit you’re misrepresenting the essential two stage nature of the Space Shuttle and it’s Launch mechanism. My question for you: But I take it you don’t want to accept that this is the type of engine you would have to use. Is it that you instead want to take the fusion reaction and generate electricity with it in order to use a type of particle accelerator and boost your reaction mass to a significant fraction of the speed of light? Am I correct that that is what you are proposing? |
|
|
|
|
Nov 6 2006, 09:50 PM
Post
#55
|
|||
|
Genuine Artificial Intelligence Group: Members Posts: 4,019 Joined: 12-June 03 Member No.: 4,715 |
That's the basic principle of an ion thruster, right? You accelerate charged particles to near-relativistic velocities, so you can get away with a very low reaction mass, and get very high efficiency. The only downside, as I understand it, is that with modern power sources it's difficult to generate a lot of acceleration. A fusion reactor would solve that, I would imagine. But I'm not a rocket scientist, so I'm not claiming anything. |
||
|
|
|
||
|
Nov 6 2006, 09:58 PM
Post
#56
|
|
|
Prime Runner Group: Banned Posts: 3,732 Joined: 1-September 05 From: Prague, Czech Republic Member No.: 7,665 |
Demerzel, at its core you are proposing that the temperature of the outgas is the singular source of velocity for it. This is not true in any reference frame. Temperature motion of particles goes in every direction, and in aggregate generally sums to zero.
The temperature of the stream is, therefore largely waste energy. So the supposition "If 100% of the forward motion comes from waste energy..." is flawed to begin with. Waste energy isn't giving you any forward motion, that's why it's waste! You talk about this as if it was somehow a Maxwell's Daemon Drive (which would be awesome). It's not. It's a fusion engine that produces a very large amount of power and squirts it out the back in the form of plasma that is accelerated to high speed (probably with magnets). The acceleration is modelable in an entirely Newtonian fashion. Energy out the back equals forward acceleration. We need on the order of 10^16 Joules, but we get on the order of 10^14 Joules for every Kilogram of fuel so I'm not worried. With only a 10% efficiency we could get there with a tonne of fuel. We're probably going to want more like 20 tonnes of fuel because getting even a 10% efficiency out of high energy reactions is laughable. The standard rocket propellent is sent out the back via expansion, which doesn't require heat to be produced. If you'll recall, liquid oxygen tanks can become quite hazardous projectiles when breached, and become extremely cold when doing so. The compressed gas decompresses explosively - and consumes heat in the process of phase shifting from liquid to gas. Your fundamental assumption is without merit. Your very first sentence kills the entire argument. By comparing the motion energy of the temperature of individual particles to... really anything, you're not modelling rocketry at all. -Frank |
|
|
|
|
Nov 6 2006, 10:01 PM
Post
#57
|
|||
|
Shadow Cartographer Group: Members Posts: 3,737 Joined: 2-June 06 From: Secret Tunnels under the UK (South West) Member No.: 8,636 |
It actually looks right to me. I'm not saying I can't be proved wrong, but I think if you're saying his maths isn't right, the onus is on you to point out where. It seems accurate to me. As I understand the way a plasma engine would work however, it doesn't work on high temperature forcing the ejecta out, but on the fact that the particles in a plasma are highly charged and are accellerated out using a powerful elextromagnetic field. This means that it works in quite a substantially different way to the rockets that you both are talking about. Now this field can be powered by the same reaction that is producing the plasma, but it's not free energy. You have to take this into account as well. That's more complicated and perhaps not necessary to work out for this argument which is why I stuck to basic physics in calculating the energy costs of accelerating and decellerating a given mass at the beginning of this thread. But I am pretty sure that a plasma drive is more of a high-efficiency, low power drive suitable to sustained travelling rather than the less efficient, but super-punching conventional rocket power. That may be why Demerzal has rightly interpreted Frank to be talking about "temperature" based propulsion, i.e. conventional rocketry. |
||
|
|
|
||
|
Nov 6 2006, 10:03 PM
Post
#58
|
|||
|
Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
The unfortunate thing is that it wouldn't really, while fusion is a great way to generate a ton of energy it's energy in the form of heat. If you want to do anything with that energy, other than vent it out the back as in the plasma rocket I've been trying to describe, you've got to convert it into another form of energy. Now you start having to talk about efficiency losses in heat engines. Converting to electricity to create a immense particle accelerator would require a way to turn that heat into energy. The problem is a heat engine requires not just a heat source but a heat differential. And while space is cold, it's also a vacuum so it's also a great insulator, all you get is radiative heat loss there is no conduction to the vacuum and no convection currents to carry away waste heat. You could allow water to boil away into the vacuum of space and carry away a lot of heat, but there you go wasting a ton of mass again... |
||
|
|
|
||
|
Nov 6 2006, 10:04 PM
Post
#59
|
|
|
Midnight Toker Group: Members Posts: 7,686 Joined: 4-July 04 From: Zombie Drop Bear Santa's Workshop Member No.: 6,456 |
A quick fact check shows that fuel is, in fact, 92% of the launch system's mass (give or take; I compared gross weights to empty weights so the estimate could be slightly high).
After the SRBs are jettisoned the shuttle and the external fuel tank actually weigh so much that the thrust from the Shuttle Main Engines provides less force than gravity and the shuttle begins to slow down until it has burned through a certain amount of fuel. By the last 10 seconds it has depleted more than 75% of its mass in fuel and actually has to throttle down to keep the acceleration under 3gs. Math that shows the shuttle losing more than 75% of its mass while leaving the atmosphere can't be rejected on that fact alone because the shuttle does lose more than 75% of its mass while leaving the atmosphere |
|
|
|
|
Nov 6 2006, 10:13 PM
Post
#60
|
|||
|
Shadow Cartographer Group: Members Posts: 3,737 Joined: 2-June 06 From: Secret Tunnels under the UK (South West) Member No.: 8,636 |
Just a word on these efficiency figures that are starting to be repeated. I can't comment on them as far as conversion of heat energy to electricity goes, but if you're talking about accelerating a plasma out of the ship using electromagnetic force, which I think you've confirmed that you are, then only about 20% of the plasma can provide thrust because the other 80% resulting from Deuterium-Tritium is neutrons which have no charge and can't be accelerated. You're back to depending on Demerzel's heat-energy equations for that. So the efficiency might be a bit less than you would hope. |
||
|
|
|
||
|
Nov 6 2006, 10:15 PM
Post
#61
|
|||||
|
Prime Runner Group: Banned Posts: 3,732 Joined: 1-September 05 From: Prague, Czech Republic Member No.: 7,665 |
Granted. But the launch mass is 2.04 million kilograms, and it burns off only 1.59 million kilograms during the 510 second jump into orbit. There is additional fuel onboard, but that's because those selfish astronauts want to get back. ;)
Unless we're doing He3 + D reactions, in which case all products would carry a positive charge. But regardless, Demerzel's Heat equations depend upon something which is not true: the concept that all motion comes from the temperature derived motion of the particles themselves. That's not true. It's not even a little bit true. 1 Kilogram of fuel gets 10^14 Joules. I don;t know what form all that energy takes, because for health and safety reasons I don;t have a fusion engine in my yard. But I don''t actually care. If you can focus even 10% more of the resulting energy to go in a direction of your choice than radially throught the universe (and you'd better, you're standing right next to this thing) - you can get there on a tonne of fuel. That's much less than Demerzel is reporting. Maybe he's under guessing the temperature? Maybe he isn;t calculating the forward velocity obtainable by magnetically reflecting the newly liberated protons? I don't know. I don't even care, because I'm not actually building one of these things. What I do know is that the actual energy being produced by the actual reaction is many orders of magnitude larger than the total amount of energy needed to get there and back in the allotted time frame, so I'm done. I honestly don't even care what hang-up is causing Demerzal to come up with the wrong answer. But conservation of mass and energy proves beyond doubt that his answer is incorrect. And that's good enough for me. -Frank |
||||
|
|
|
||||
|
Nov 6 2006, 10:29 PM
Post
#62
|
|
|
Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
I think this argument may have stemmed from the basic premise that I tried to give a simple example of how 1g acceleration for 4 days (2 days out 2 days back) is untenable. Unfortunately it degenerated to a defense of a short and oversimplified counter example.
A good source of info on propulsion is this article on wiki... I continue to contest that it'll be highly unlikely that you could get 1g of accelleration for something on the order of days... If you want to switch the argument to ion engines we can, but you've got enormous efficiency issues, plus you're talking about scaling a technology to broken proportions. |
|
|
|
|
Nov 6 2006, 10:36 PM
Post
#63
|
|||
|
Prime Runner Group: Banned Posts: 3,732 Joined: 1-September 05 From: Prague, Czech Republic Member No.: 7,665 |
From that site you linked:
But with Shadowrun tech, there is a contained fusion drive. Carrying the major metropolitan power generator on board isn't clear impractical in Shadowrun's 2071. That's the deal. The energy involved is very large, but in Shadowrun that's a solved problem. -Frank |
||
|
|
|
||
|
Nov 6 2006, 10:47 PM
Post
#64
|
|
|
Shadow Cartographer Group: Members Posts: 3,737 Joined: 2-June 06 From: Secret Tunnels under the UK (South West) Member No.: 8,636 |
edit: Deleted post - the argument is dead. Just objected to Demerzel's posts that seem correct to me being dismissed as "hangups."
|
|
|
|
|
Nov 6 2006, 10:51 PM
Post
#65
|
|||||||
|
Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
Where do you get this number?
I do, it makes heat. What is heat? Well that’s actually a pretty deep physics question and one I can’t really answer without putting you through an actual thermodynamics class. The simplest answer and the on you’ll probably have to accept is that it relates to the kinetic energy of the constituent particles.
I think this statement alone best exemplifies your basic misunderstanding of the physics involved. Energy is a scalar; you cannot direct it any way at all. |
||||||
|
|
|
||||||
|
Nov 6 2006, 11:02 PM
Post
#66
|
|
|
Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
Okay Frank, lets use the example there of a Ion Drive. It’s got an efficiency of 25kW per N of thrust.
So if I have a 100,000 kg (Space Shuttle sized) craft. 1g nearly 10m/s^2 would require 1,000,000 N of force. 25kW/N means 25,000,000 kW, or 25GW OUTPUT from a power generator. Now if I give you a better than possible efficiency of 50% that means that 25GW of power output is waste heat. Where do you propose that goes? In SR the generators you’re talking about put it in a river or into the ocean. Where is your enormous heat sink? |
|
|
|
|
Nov 6 2006, 11:11 PM
Post
#67
|
|||||
|
Prime Runner Group: Banned Posts: 3,732 Joined: 1-September 05 From: Prague, Czech Republic Member No.: 7,665 |
A glowing contrail of plasma?
I'm seriously not concerned. That's 6.9 x 10^6 kW days out of 5/3 of a kilogram. If you need only 25 kW per Newton, you're going to be able to - theoretically - keep our craft going the whole time on 12 kilograms of fuel. Obviously you're never going to get that kind of efficiency. But if the absolute maximum is 12 kilograms, you'll understand why I'm willing to handwave the possibility with dozens of tonnes to work with. -Frank |
||||
|
|
|
||||
|
Nov 6 2006, 11:30 PM
Post
#68
|
|||||
|
Midnight Toker Group: Members Posts: 7,686 Joined: 4-July 04 From: Zombie Drop Bear Santa's Workshop Member No.: 6,456 |
You seem the be forgetting to factor in the mass of the jettisoned fuel tank and SRBs. The maximum amount of fuel the shuttle can retain is 40,414.4kg (gross weight of orbiter-empty weight of orbiter), and that is being generous. It will be less depending on the shuttle's cargo. The orbiter itself is only 109,000 kg when full. It takes a lot more fuel to reach orbit than it does to land. When landing, gravity does most of the work. That's the great thing about conservation of energy and the relationship between height and potential energy. |
||||
|
|
|
||||
|
Nov 6 2006, 11:49 PM
Post
#69
|
|
|
Moving Target Group: Members Posts: 170 Joined: 18-September 06 Member No.: 9,412 |
Wow, who says roleplayers are a bunch of nerds...? :)
|
|
|
|
|
Nov 6 2006, 11:51 PM
Post
#70
|
|||||
|
Shadow Cartographer Group: Members Posts: 3,737 Joined: 2-June 06 From: Secret Tunnels under the UK (South West) Member No.: 8,636 |
Well you gave yourself a 20% efficiency earlier. So with enough energy to run New York for an afternoon, you've still got 80% of it you're trying to radiate off from your spaceship somehow. Going to get hot in there. |
||||
|
|
|
||||
|
Nov 6 2006, 11:51 PM
Post
#71
|
|
|
Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
The glowing contrail of plasma would be the output energy from the reaction the waste energy would have to go into some sort of heat reservoir. Basically it would heat your ship. The waste heat is just that waste and you have to do something with it, it is all the energy that you cannot convert to useful work due to the fact that no power generator is 100% efficient.
So basically if I give you a 50% efficient engine, and you need 25GW of power to keep it accelerating at 1g, then you’ve got 25GW of waste heat to deal with. Water absorbs a huge amount of heat per unit mass compared to the things your ship is likely made of like metal. If your ship absorbs heat just as well as water, at 25 Celsius, 25 GW of waste energy would heat your ship 62.5 Kelvin per second. How long will your crew survive? Okay I know what you’re saying that heat will go somewhere, it’s not confined to the ship. But there is no conduction of heat from the ship into space (a vacuum), only radiative heat loss will occur. In order for the radiative heat loss to emit 25GW of energy it would have to be glowing hot. So you want some sort of huge radiators that will reduce that heat, then you’ve got to add that weight, and increase your power output and in the end you’re going to be talking about an immense amount of heat sink. Also the hotter your heat sink the worse your efficiency. Basically, your 2 day there and 2 day back at 1g trip is just ridiculous. Take two months to get there and two to get back and there you go, but why are you in this ridiculous hurry? |
|
|
|
|
Nov 7 2006, 12:16 AM
Post
#72
|
|
|
Midnight Toker Group: Members Posts: 7,686 Joined: 4-July 04 From: Zombie Drop Bear Santa's Workshop Member No.: 6,456 |
Well, if you want to get into pure speculative theoretical limits then 2kg of carbon could hold 5.594MJ of heat before melting and if flattened into a 100 square meter sheet could emit (eT)22.041GW at this temperature, where (eT) is the emissivity correction factor of the carbon at this temperature.
The actual mass and specific heat of the radiator doesn't factor into the equation except as they effect temperature. Since a high-temperature radiator is good a low-mass and low specific-heat radiator actually benefits you in some ways. |
|
|
|
|
Nov 7 2006, 12:19 AM
Post
#73
|
|
|
Running Target Group: Members Posts: 1,206 Joined: 9-July 06 From: Fresno, CA Member No.: 8,856 |
Carbon is a poor conductor however so as a heat sink it would be unreliable...
|
|
|
|
|
Nov 7 2006, 12:31 AM
Post
#74
|
|
|
Shadow Cartographer Group: Members Posts: 3,737 Joined: 2-June 06 From: Secret Tunnels under the UK (South West) Member No.: 8,636 |
SR2070 technology - how about this: You create a nano-material with strictly controlled bonding forces between the molecules. At specific temperature threshold, they "evaporate" away from the ship. You're in Space so you don't care about the super fine cloud of particles that your ship leaves behind it like an all-consuming fart. It's inherantly degrading, but it's a neat way to get rid of heat and you can make it eight-foot thick if you want. I'm sure SR2070 technology can also come up with a neat little capilliary system to get the heat out to the heat sink "shell".
If we're talking science-fiction (and we are), you could even vary the strength of the bonds in the material as needed with varying current. My idea is so brilliant that I should patent it, but instead I shall donate it to the public domain, for the greater glory of Dumpshock. :D |
|
|
|
|
Nov 7 2006, 12:38 AM
Post
#75
|
|||
|
Midnight Toker Group: Members Posts: 7,686 Joined: 4-July 04 From: Zombie Drop Bear Santa's Workshop Member No.: 6,456 |
Except when you fly back through the particles on your return trip and they punch holes in your oxygen tanks like the superheated birdshot that they are. Far better to use a lattice of carbon nanotubes as a giant radiator sail as it is possible to create nanotubes with conductive properties. You're still screwed if you run into a cloud micrometeorites, of course. They'll probably swiss-cheese the radiator. The important thing is to have a high melting point, good conductivity, and enough strength to stay together when flattened into an extremely large and thin sheet and put under the stress of acceleration. |
||
|
|
|
||
|
|
Lo-Fi Version | Time is now: 26th January 2025 - 11:30 AM |
Topps, Inc has sole ownership of the names, logo, artwork, marks, photographs, sounds, audio, video and/or any proprietary material used in connection with the game Shadowrun. Topps, Inc has granted permission to the Dumpshock Forums to use such names, logos, artwork, marks and/or any proprietary materials for promotional and informational purposes on its website but does not endorse, and is not affiliated with the Dumpshock Forums in any official capacity whatsoever.