Ishtar
June 21st, 2003, 03:56 PM
I have a question about Starship maneuver drives in Traveller. Are they grav-based or do they use something like chemical thrusters to move in space, or both, or neither?
Also, what is the value of G? Is it an earth g (9.8 m/s if I'm not mistaken)?
--Courtney
Also, what is the value of G? Is it an earth g (9.8 m/s if I'm not mistaken)?
--Courtney
mad13142000
June 21st, 2003, 04:05 PM
Depends on the tech level...
At (Traveller) TL10, the standard maneuver drive is the HEPlaR - High-Energy Plasma Recombination
a variation of a fusion rocket.
At TL12, the standard becomes the thrust plate. It is more than just a Contra-Grav, the Thrust Plate pushes off of existing gravity fields to propel the ship, thus they decrease in effeciency as they travel away from a gravity well..
-MADDog
At (Traveller) TL10, the standard maneuver drive is the HEPlaR - High-Energy Plasma Recombination
a variation of a fusion rocket.
At TL12, the standard becomes the thrust plate. It is more than just a Contra-Grav, the Thrust Plate pushes off of existing gravity fields to propel the ship, thus they decrease in effeciency as they travel away from a gravity well..
-MADDog
sandman
June 22nd, 2003, 01:57 AM
AFAIK, T20 doesn't use HEPlar, (even had heard the name before subscribing here or the TML)
Also, T20 doesn't describe much thrusters, or many other technology for that matter :(
Also, T20 doesn't describe much thrusters, or many other technology for that matter :(
TKalbfus
June 22nd, 2003, 02:29 AM
The thing about the maneuver drive is that it pushes a volume in displacement tons not a mass. You could have the starship packed with solid Tungstein and it wouldn't make a difference with the acceleration so long as you have the required energy inputs you get x-Gs of acceleration. A more realistic manuever drive would work this way. Instead of producing x-Gs for a given energy input, it would produce x tons of thrust, you then assume the average density of the starship is the same as liquid hydrogen and its volume is displacement tons equals its mass and divide the ships mass by the tons of thrust produced by the manuever drive and you get the ship's acceleration. You have to remember to reduce the ship's mass as the fusion reactor consumes hydrogen. Also the ship's structural integrity will set an upper limit on acceleration so at some point the manuever drive will have to be throttled down or the ship will come apart.
daryen
June 22nd, 2003, 02:41 PM
I thought I should point out that GURPS Traveller does it *exactly* as Tom describes. GT drives use "reactionless thrusters". They are also mass based, so an empty ship travels faster than a loaded ship, and a ship gets to the gas giant for refueling faster than it leaves it.
While CT never specifically states, it is pretty obvious that they also used reactionless thrusters, though they were volume based, not mass based. T20 uses the same unidentified reactionless thrusters as CT.
In MT there were both reactionless thrusters and gravitic drives. The thrusters were bigger and more expensive, but didn't require a gravity well to operate effectively.
In TNE they introduced heplar (I forget the weird capitalization). This is a pure fusion reactor manuever drive. The also retained the reactionless thruster, though there was a TL floor for its introduction.
So, pick the system you are using, and find your answer.
While CT never specifically states, it is pretty obvious that they also used reactionless thrusters, though they were volume based, not mass based. T20 uses the same unidentified reactionless thrusters as CT.
In MT there were both reactionless thrusters and gravitic drives. The thrusters were bigger and more expensive, but didn't require a gravity well to operate effectively.
In TNE they introduced heplar (I forget the weird capitalization). This is a pure fusion reactor manuever drive. The also retained the reactionless thruster, though there was a TL floor for its introduction.
So, pick the system you are using, and find your answer.
Ishtar
June 22nd, 2003, 05:50 PM
Thanx guys.
One of the reasons I asked was that I was wondering whether or not there would be some sort of waste product from the drives- plasma or biohazardous liquid of some sort. It seems pretty clear that LHyd is the sole fuel used on most starships with a fusion power plant (by all means, correct me if I am wrong).
One more question: H2 is highly explosive. How volatile and flammable is it in liquid form (LHyd)?
--Courtney
One of the reasons I asked was that I was wondering whether or not there would be some sort of waste product from the drives- plasma or biohazardous liquid of some sort. It seems pretty clear that LHyd is the sole fuel used on most starships with a fusion power plant (by all means, correct me if I am wrong).
One more question: H2 is highly explosive. How volatile and flammable is it in liquid form (LHyd)?
--Courtney
sandman
June 22nd, 2003, 07:46 PM
Originally posted by Empress Nicholle:
One more question: H2 is highly explosive. How volatile and flammable is it in liquid form (LHyd)?
--Courtney Only volatile and flammable when mixed with Oxygen.
One more question: H2 is highly explosive. How volatile and flammable is it in liquid form (LHyd)?
--Courtney Only volatile and flammable when mixed with Oxygen.
mad13142000
June 23rd, 2003, 12:25 AM
Not many polutants from hydrogen - That's what makes it so attractive as an energy source. If you fuse it, you get helium. If you burn it, you get water...The energy cells making news today are using it in a slow oxidation reaction via a catalyst in a 'fuel cell', but still ending with water vapor. All in all much better than fission reactors or the horrible chemical rockets NASA uses to send stuff into orbit...
-MADDog
-MADDog
Anthony
June 23rd, 2003, 12:48 AM
Originally posted by MADDog:
Not many polutants from hydrogenThere's some recent evidence that hydrogen itself is a potential pollutant; it tends to interact in the upper atmosphere, breaking down the ozone layer. The degree of danger isn't yet clear.
Not many polutants from hydrogenThere's some recent evidence that hydrogen itself is a potential pollutant; it tends to interact in the upper atmosphere, breaking down the ozone layer. The degree of danger isn't yet clear.
BMonnery
June 23rd, 2003, 08:46 AM
Originally posted by Empress Nicholle:
I have a question about Starship maneuver drives in Traveller. Are they grav-based or do they use something like chemical thrusters to move in space, or both, or neither?
Also, what is the value of G? Is it an earth g (9.8 m/s if I'm not mistaken)?
--Courtney CT started with fusion torches and chemical thrusters (the latter only on small craft, and only in Mayday, which gave very limited fuel endurances for fighters (4G acceleration for 3 turns ISTR)). Later it was grav plates, back to rockets for TNE and then back to grav plates later.
G is 9.8 m/s2 (the squared is important). That's the acceleration due to gravity on Earth. Usually rounded to 10m/s2 for most practical purposes.
So, accelerating at 1G, after 1 second you're moving 10m/s, after 2, 20m/s, after 3, 30m/s etc.
There's no balancing force of friction like there is in an atmosphere, so with unlimited thrust you can continue accelerating for a long time, getting ever and ever closer to the speed of light.
Bryn
I have a question about Starship maneuver drives in Traveller. Are they grav-based or do they use something like chemical thrusters to move in space, or both, or neither?
Also, what is the value of G? Is it an earth g (9.8 m/s if I'm not mistaken)?
--Courtney CT started with fusion torches and chemical thrusters (the latter only on small craft, and only in Mayday, which gave very limited fuel endurances for fighters (4G acceleration for 3 turns ISTR)). Later it was grav plates, back to rockets for TNE and then back to grav plates later.
G is 9.8 m/s2 (the squared is important). That's the acceleration due to gravity on Earth. Usually rounded to 10m/s2 for most practical purposes.
So, accelerating at 1G, after 1 second you're moving 10m/s, after 2, 20m/s, after 3, 30m/s etc.
There's no balancing force of friction like there is in an atmosphere, so with unlimited thrust you can continue accelerating for a long time, getting ever and ever closer to the speed of light.
Bryn
TKalbfus
June 23rd, 2003, 10:05 AM
It seems to me that the thrusters are only reactionless, if the fuel tanks end up filled with liquid helium after the fuel is used up. If the fuel is used up and the fuel tanks are empty, then end product of the fusion reaction had to have been used as a reaction mass. I think the whole power plant/thruster combo may be considered an approximation of a fusion rocket. The volume rule is an assumption of the average density of the starship for the entire rocket burn. I know a typical fusion rocket could burn for a couple of weeks and that adding a reactionless drive doesn't add any thing except simplify the calculation of how fast the ship accelerates over time.
Ben W Bell
June 23rd, 2003, 10:12 AM
It only uses reactionless against volume for simplification so people wouldn't have to worry about it (which seems at odds against some of the other incredibly detailed sections and calculations.)
Zutroi
June 23rd, 2003, 08:43 PM
Originally posted by Tom Kalbfus:
It seems to me that the thrusters are only reactionless, if the fuel tanks end up filled with liquid helium after the fuel is used up. If the fuel is used up and the fuel tanks are empty, then end product of the fusion reaction had to have been used as a reaction mass. I think the whole power plant/thruster combo may be considered an approximation of a fusion rocket. The volume rule is an assumption of the average density of the starship for the entire rocket burn. I know a typical fusion rocket could burn for a couple of weeks and that adding a reactionless drive doesn't add any thing except simplify the calculation of how fast the ship accelerates over time. Or not. Technically, your car works on a reactionless principle, I've never yet had to empty the CO out of my fuel tank to refill. It's quite possible (although it would be stupid to waste 'free' thrust IMHO) to just dump the helium out your 'exhaust pipe'
It seems to me that the thrusters are only reactionless, if the fuel tanks end up filled with liquid helium after the fuel is used up. If the fuel is used up and the fuel tanks are empty, then end product of the fusion reaction had to have been used as a reaction mass. I think the whole power plant/thruster combo may be considered an approximation of a fusion rocket. The volume rule is an assumption of the average density of the starship for the entire rocket burn. I know a typical fusion rocket could burn for a couple of weeks and that adding a reactionless drive doesn't add any thing except simplify the calculation of how fast the ship accelerates over time. Or not. Technically, your car works on a reactionless principle, I've never yet had to empty the CO out of my fuel tank to refill. It's quite possible (although it would be stupid to waste 'free' thrust IMHO) to just dump the helium out your 'exhaust pipe'
TKalbfus
June 23rd, 2003, 11:54 PM
The Traveller Starships look like they should have trails of flame behind them, that's why I think the whole Power plant/thruster combo is a game mechanic to approximate a fusion rocket. A fusion rocket is a fusion power plant with a leak. The plasma squirts out the back and pushes the starship forward, but the fusion rocket is also a power plant that can power laser weapons and life support, the trade off is, the more plasma that leaks out the back, the less power there is left over to power the laser weapons. That is why, I think the game separates the thrusters from the power plants. No modern rocket launch system has a power source and a propulsion system that requires a separate power source to operate.
bromdenlong
June 26th, 2003, 01:37 PM
Why is it that canon ships with full streamlining, clearly meant to land on any terrestrial planet (like the free/far traders)only have maneuver-1, i.e. 1-g acceleration? A ship with 1-g acceleration could not take off from a planet with Earth's surface gravity. It would just float neutrally, like a toy balloon that has lost too much helium to float up to the ceiling, but still retains too much helium to sink to the floor.
Granted, as someone pointed out in another post somewhere, one could posit the value of 1-g as 10 meters/second/second, rather than 9.8, but liftoff would still be *awfully* slow.
Is it just me, or did someone really foul up on this one? Shouldn't any fully streamlined ship intended to land on a variety of worlds in a variety of systems have 2-g maneuver drives?
I believe 2-g should suffice - I believe I have read in an astronomy text or two that the Earth is about as large as terrestrial planets can get. If a planet forms with much more mass, it becomes a gas giant.
Granted, as someone pointed out in another post somewhere, one could posit the value of 1-g as 10 meters/second/second, rather than 9.8, but liftoff would still be *awfully* slow.
Is it just me, or did someone really foul up on this one? Shouldn't any fully streamlined ship intended to land on a variety of worlds in a variety of systems have 2-g maneuver drives?
I believe 2-g should suffice - I believe I have read in an astronomy text or two that the Earth is about as large as terrestrial planets can get. If a planet forms with much more mass, it becomes a gas giant.
dougmedic
June 26th, 2003, 01:52 PM
Originally posted by Isaac_1963:
Why is it that canon ships with full streamlining, clearly meant to land on any terrestrial planet (like the free/far traders)only have maneuver-1, i.e. 1-g acceleration? A ship with 1-g acceleration could not take off from a planet with Earth's surface gravity. It would just float neutrally, like a toy balloon that has lost too much helium to float up to the ceiling, but still retains too much helium to sink to the floor.
Granted, as someone pointed out in another post somewhere, one could posit the value of 1-g as 10 meters/second/second, rather than 9.8, but liftoff would still be *awfully* slow.
Is it just me, or did someone really foul up on this one? Shouldn't any fully streamlined ship intended to land on a variety of worlds in a variety of systems have 2-g maneuver drives?
I believe 2-g should suffice - I believe I have read in an astronomy text or two that the Earth is about as large as terrestrial planets can get. If a planet forms with much more mass, it becomes a gas giant. I believe that in CT/MT there has always been an unwritten assumption that the maneuver drives (whether the generic CT thrusters, or MT contra-grav or thrusters) have always had a contra-grav component suffient to "negate" planetary gravity, and provide the rated acceleration on top of that. :eek: For game playability, CT/MT never addressed the issue of varying planetary gravities. Remember, SDB are supposed to lurk (hovering) in GG atmospheres on their CG/thrusters graemlins/file_22.gif
Why is it that canon ships with full streamlining, clearly meant to land on any terrestrial planet (like the free/far traders)only have maneuver-1, i.e. 1-g acceleration? A ship with 1-g acceleration could not take off from a planet with Earth's surface gravity. It would just float neutrally, like a toy balloon that has lost too much helium to float up to the ceiling, but still retains too much helium to sink to the floor.
Granted, as someone pointed out in another post somewhere, one could posit the value of 1-g as 10 meters/second/second, rather than 9.8, but liftoff would still be *awfully* slow.
Is it just me, or did someone really foul up on this one? Shouldn't any fully streamlined ship intended to land on a variety of worlds in a variety of systems have 2-g maneuver drives?
I believe 2-g should suffice - I believe I have read in an astronomy text or two that the Earth is about as large as terrestrial planets can get. If a planet forms with much more mass, it becomes a gas giant. I believe that in CT/MT there has always been an unwritten assumption that the maneuver drives (whether the generic CT thrusters, or MT contra-grav or thrusters) have always had a contra-grav component suffient to "negate" planetary gravity, and provide the rated acceleration on top of that. :eek: For game playability, CT/MT never addressed the issue of varying planetary gravities. Remember, SDB are supposed to lurk (hovering) in GG atmospheres on their CG/thrusters graemlins/file_22.gif
TheEngineer
June 26th, 2003, 04:49 PM
Hi there,
I just took a look at the good old Megatraveller Starship Operators Manual....:
...plates may be overdriven for up to 40%...
...overdriving the plates by up to 400 % (as in the case of a 1g ship trying to do a lateral hover at takeoff or landing) takes the upmost care, and can only be done for brief periods of time (under 5 minutes)....
If there is interest I could provide an extract of
the paragraph...
Best regards,
Mert
I just took a look at the good old Megatraveller Starship Operators Manual....:
...plates may be overdriven for up to 40%...
...overdriving the plates by up to 400 % (as in the case of a 1g ship trying to do a lateral hover at takeoff or landing) takes the upmost care, and can only be done for brief periods of time (under 5 minutes)....
If there is interest I could provide an extract of
the paragraph...
Best regards,
Mert
Jame
June 27th, 2003, 11:29 AM
I tried to ask this in a previous topic, but never got a quite satisfactory answer (or maybe did, but don't remember).
How long does it take a 1-g drive, 2-g drive, and etcetera to accelerate to 10 percent speed of light? Does a 6-g drive accelerate 6 times faster, or is each increase an exponential increase?
How long does it take a 1-g drive, 2-g drive, and etcetera to accelerate to 10 percent speed of light? Does a 6-g drive accelerate 6 times faster, or is each increase an exponential increase?
The Oz
June 27th, 2003, 11:55 AM
MATH WARNING!
The equation is:
Velocity = (Acceleration)x(time)
so, Time (to reach a certain velocity) = (that velocity)/(acceleration)
1G = 9.8 meters/second squared, let's round that to 10 meters/second squared just to make the math easier. 2 G is twice that or 20 meters/second squared, and 6 G is then 120 meters/second squared.
The speed of light (in a vacuum) is 300,000,000 meters/second (again, that's a slight rounding).
10% of this is 30,000,000 meters/second.
So....
Time (at 1 G)=30,000,000 meters per second/10 meters per second squared = 3,000,000 seconds = 833.3 hours at 1 G to get to 10% light speed
Time (at 2 G)=30,000,000 m/s divided by 20 m/sec squared = 1,500,000 seconds = 416.7 hours at 2 G to get to 10% light speed.
Time (at 6 G)=30,000,000m/s divided by 120 m/second squared = 250,000 seconds = 69.4 hours at 6 G to reach 10% light speed.
So the time to reach a certain speed (at constant acceleration) is inversely proportional to the acceleration. In plain English that means that it takes 1/6 the time at 6G to reach a certain velocity as it would take at 1G.
Given that the time it takes to travel to a far gas giant is only 166.7 hours even at 1G (numbers from LBB2) it's not very likely that even a 6G ship would ever approach 10% of lightspeed in a Traveller game.
The equation is:
Velocity = (Acceleration)x(time)
so, Time (to reach a certain velocity) = (that velocity)/(acceleration)
1G = 9.8 meters/second squared, let's round that to 10 meters/second squared just to make the math easier. 2 G is twice that or 20 meters/second squared, and 6 G is then 120 meters/second squared.
The speed of light (in a vacuum) is 300,000,000 meters/second (again, that's a slight rounding).
10% of this is 30,000,000 meters/second.
So....
Time (at 1 G)=30,000,000 meters per second/10 meters per second squared = 3,000,000 seconds = 833.3 hours at 1 G to get to 10% light speed
Time (at 2 G)=30,000,000 m/s divided by 20 m/sec squared = 1,500,000 seconds = 416.7 hours at 2 G to get to 10% light speed.
Time (at 6 G)=30,000,000m/s divided by 120 m/second squared = 250,000 seconds = 69.4 hours at 6 G to reach 10% light speed.
So the time to reach a certain speed (at constant acceleration) is inversely proportional to the acceleration. In plain English that means that it takes 1/6 the time at 6G to reach a certain velocity as it would take at 1G.
Given that the time it takes to travel to a far gas giant is only 166.7 hours even at 1G (numbers from LBB2) it's not very likely that even a 6G ship would ever approach 10% of lightspeed in a Traveller game.
daryen
June 27th, 2003, 12:56 PM
Originally posted by The Oz:
1G = 9.8 meters/second squared, let's round that to 10 meters/second squared just to make the math easier. 2 G is twice that or 20 meters/second squared, and 6 G is then 120 meters/second squared.
...
Time (at 6 G)=30,000,000m/s divided by 120 m/second squared = 250,000 seconds = 69.4 hours at 6 G to reach 10% light speed.Uh, not to be completely annoying, but shouldn't 6G be 60 m/s/s instead of 120 m/s/s?
That would make the result for 6G 138.8 instead of 69.4. (Which fits the 1/6 of 833.3 you were looking for.)
1G = 9.8 meters/second squared, let's round that to 10 meters/second squared just to make the math easier. 2 G is twice that or 20 meters/second squared, and 6 G is then 120 meters/second squared.
...
Time (at 6 G)=30,000,000m/s divided by 120 m/second squared = 250,000 seconds = 69.4 hours at 6 G to reach 10% light speed.Uh, not to be completely annoying, but shouldn't 6G be 60 m/s/s instead of 120 m/s/s?
That would make the result for 6G 138.8 instead of 69.4. (Which fits the 1/6 of 833.3 you were looking for.)
Ishtar
June 28th, 2003, 04:25 PM
Another question, then
Why is acceleration more important than velocity when determining relative ship positions? It's been a long time since I've used the old vector starship movement system in CT, so forgive me if I seem a little ditzy (no, I'm a redhead, not blonde) ;) .
And don't vessels reach a terminal velocity? How do you figure what that is?
--Courtney
Why is acceleration more important than velocity when determining relative ship positions? It's been a long time since I've used the old vector starship movement system in CT, so forgive me if I seem a little ditzy (no, I'm a redhead, not blonde) ;) .
And don't vessels reach a terminal velocity? How do you figure what that is?
--Courtney
Hecateus
June 28th, 2003, 08:11 PM
OZ,
correct me if I am wrong, but I believe your calculations don't take into account the effects of relativity in terms of time and fuel needed. Though I realise you are just keeping things simple. smile.gif
Also anyone approaching relativistic speeds within a star system risks getting torn to little bits by the concentration of dust and gas therein...as compared with interstellar space.
On another note, many ships are streamlined,and possibly make adequate lifting-bodies...they might not need to make relatively heavy use of thrusters and anti-grav plates.
correct me if I am wrong, but I believe your calculations don't take into account the effects of relativity in terms of time and fuel needed. Though I realise you are just keeping things simple. smile.gif
Also anyone approaching relativistic speeds within a star system risks getting torn to little bits by the concentration of dust and gas therein...as compared with interstellar space.
On another note, many ships are streamlined,and possibly make adequate lifting-bodies...they might not need to make relatively heavy use of thrusters and anti-grav plates.
The Oz
June 28th, 2003, 09:28 PM
Daryen said,
Uh, not to be completely annoying, but shouldn't 6G be 60 m/s/s instead of 120 m/s/s?
That would make the result for 6G 138.8 instead of 69.4. (Which fits the 1/6 of 833.3 you were looking for.) Yes, it does. I must have been thinking of the acceleration at 2-Gs when I multiplied by 6.
Thanks for catching my error. :rolleyes:
Hecateus said,
correct me if I am wrong, but I believe your calculations don't take into account the effects of relativity in terms of time and fuel needed. Though I realise you are just keeping things simple. You're right that I'm ignoring relativistic effects. I'm doing that because at the low velocities of most Traveller ships, relativistic effects are minimal. You're got to really get up into large fractions of lightspeed to have relativistic effects become apparent.
Empress Nicholle said,
Why is acceleration more important than velocity when determining relative ship positions? It's been a long time since I've used the old vector starship movement system in CT, so forgive me if I seem a little ditzy (no, I'm a redhead, not blonde) .
And don't vessels reach a terminal velocity? How do you figure what that is? Taking the second question first....
Yes, technically there is a maximum of just under lightspeed, but Traveller ships will never even get close to that. However, we can take a SWAG (Scientific Wild-Ass Guess) at a maximum safe speed for Traveller ships.
The maximum speed an unarmored Traveller spaceship can safely reach is about 0.5% of lightspeed, which is about 1500 kilometers/second (km/s). This figure is based on the Striker armor rating of an unarmored Traveller ship hull and the impact energy (in TNT-equivalent) of a 0.1 gram piece of spacedust at that speed. If you want the math I can show it to you.
A Traveller ship accelerating at 1-G will reach this maximum safe speed in about 41 hours of continuous acceleration. Since safe jump distance from a large gas giant is only 17.6 hours at 1-G (and that includes a turnover at the halfway point and deceleration the rest of the way) it's unlikely that a Traveller ship would ever need to reach this theoretical maximum safe speed. Armored Traveller ships could reach higher safe speeds.
Remember that objects in a vacuum do not experience friction and so they keep whatever speed they had when their drives are shut down, unless something else (like a planet's gravity) acts to change their speed.
To answer the question about why acceleration is often more important than velocity, it's necessary to understand that most Traveller ships will never be moving at all that high a speed. Usually Traveller ships are just going to or from a safe jump distance from an inhabited world. This doesn't take very long, even for a 1-G Free Trader (7 hours for a size-A world at 1-G) and the max velocity the Free Trader reaches is only 126 km/s. A 6-G pirate can match that velocity in about 40 minutes or less. So the acceleration is more important since the velocities are so low that a high-G ship can easily gain the speed needed to intercept.
Uh, not to be completely annoying, but shouldn't 6G be 60 m/s/s instead of 120 m/s/s?
That would make the result for 6G 138.8 instead of 69.4. (Which fits the 1/6 of 833.3 you were looking for.) Yes, it does. I must have been thinking of the acceleration at 2-Gs when I multiplied by 6.
Thanks for catching my error. :rolleyes:
Hecateus said,
correct me if I am wrong, but I believe your calculations don't take into account the effects of relativity in terms of time and fuel needed. Though I realise you are just keeping things simple. You're right that I'm ignoring relativistic effects. I'm doing that because at the low velocities of most Traveller ships, relativistic effects are minimal. You're got to really get up into large fractions of lightspeed to have relativistic effects become apparent.
Empress Nicholle said,
Why is acceleration more important than velocity when determining relative ship positions? It's been a long time since I've used the old vector starship movement system in CT, so forgive me if I seem a little ditzy (no, I'm a redhead, not blonde) .
And don't vessels reach a terminal velocity? How do you figure what that is? Taking the second question first....
Yes, technically there is a maximum of just under lightspeed, but Traveller ships will never even get close to that. However, we can take a SWAG (Scientific Wild-Ass Guess) at a maximum safe speed for Traveller ships.
The maximum speed an unarmored Traveller spaceship can safely reach is about 0.5% of lightspeed, which is about 1500 kilometers/second (km/s). This figure is based on the Striker armor rating of an unarmored Traveller ship hull and the impact energy (in TNT-equivalent) of a 0.1 gram piece of spacedust at that speed. If you want the math I can show it to you.
A Traveller ship accelerating at 1-G will reach this maximum safe speed in about 41 hours of continuous acceleration. Since safe jump distance from a large gas giant is only 17.6 hours at 1-G (and that includes a turnover at the halfway point and deceleration the rest of the way) it's unlikely that a Traveller ship would ever need to reach this theoretical maximum safe speed. Armored Traveller ships could reach higher safe speeds.
Remember that objects in a vacuum do not experience friction and so they keep whatever speed they had when their drives are shut down, unless something else (like a planet's gravity) acts to change their speed.
To answer the question about why acceleration is often more important than velocity, it's necessary to understand that most Traveller ships will never be moving at all that high a speed. Usually Traveller ships are just going to or from a safe jump distance from an inhabited world. This doesn't take very long, even for a 1-G Free Trader (7 hours for a size-A world at 1-G) and the max velocity the Free Trader reaches is only 126 km/s. A 6-G pirate can match that velocity in about 40 minutes or less. So the acceleration is more important since the velocities are so low that a high-G ship can easily gain the speed needed to intercept.
Ishtar
June 29th, 2003, 06:35 PM
Originally posted by The Oz:
Hecateus said,
[QUOTE]correct me if I am wrong, but I believe your calculations don't take into account the effects of relativity in terms of time and fuel needed. Though I realise you are just keeping things simple. You're right that I'm ignoring relativistic effects. I'm doing that because at the low velocities of most Traveller ships, relativistic effects are minimal. You're got to really get up into large fractions of lightspeed to have relativistic effects become apparent.
Absolutely!
To be purposefully annoying tongue.gif
The equation (for relativistic effects for a frame of motion compared to a 'rest frame')is something like (correct me, please, if I'm wrong) 1/sqr root(1-v2/c2) v being the velocity of the object and c being the speed of light. The closer v gets to c then, the smaller the denominator gets and the more pronounced the relativistic effects, right? v has to be a significant fraction of c to even get close to giving a big number.
Courtney
Hecateus said,
[QUOTE]correct me if I am wrong, but I believe your calculations don't take into account the effects of relativity in terms of time and fuel needed. Though I realise you are just keeping things simple. You're right that I'm ignoring relativistic effects. I'm doing that because at the low velocities of most Traveller ships, relativistic effects are minimal. You're got to really get up into large fractions of lightspeed to have relativistic effects become apparent.
Absolutely!
To be purposefully annoying tongue.gif
The equation (for relativistic effects for a frame of motion compared to a 'rest frame')is something like (correct me, please, if I'm wrong) 1/sqr root(1-v2/c2) v being the velocity of the object and c being the speed of light. The closer v gets to c then, the smaller the denominator gets and the more pronounced the relativistic effects, right? v has to be a significant fraction of c to even get close to giving a big number.
Courtney
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