Movement at M4 across the galaxy...
- Thread starter Leitz
- Start date
4-5 years or so; a parsec is 3.26LY. a 2 parsec cruise, however, will only be an addition 3.5 years more
M-Drive speed is pretty irrelevant for any distance over a parsec... it will take close to 3 months to hit 0.5C and another 3 to hit .75 at 4G, and should be another 3 to 0.875...
M-Drive speed is pretty irrelevant for any distance over a parsec... it will take close to 3 months to hit 0.5C and another 3 to hit .75 at 4G, and should be another 3 to 0.875...
Icosahedron
SOC-14 1K
I never did get my head fully around the relativistic equations of motion. I wonder if Nyrath has them listed with idiot notes?
I found his notes on conversion from celestial to galactic co-ordinates to be very illuminating. When does that guy eat??!!
I found his notes on conversion from celestial to galactic co-ordinates to be very illuminating. When does that guy eat??!!
Icosahedron
SOC-14 1K
Nyrath's been at it a LONG time, Icos... His is one of the older sites on the net. And IIRC, Nyrath is also a noted game-artist... something about cybertanks...
That's what I mean. The guy must have JoT-6.
Heh, thanks! I can get a little obsessive-compulsive sometimes.
I sort of have relativistic equations with idiot notes here:
http://www.projectrho.com/rocket/rocket3aj.html#relativity
Of course, when you ask "how long will it take...", you have to specify whether it is from the standpoint of the crew, or the standpoint of the planet they left from. If you accelerated to a velocity absurdly close to the speed of light, the crew would think that the trip to the galactic core would take a few months, while for the people on your takeoff planet it would take about fifty thousand years. Everything is relative, y'know.
For the stay-at-homes on the takeoff planet, the time will be
t = sqrt[(d/c)^2 + (2*d/a)]
For the crew the time will be
T = (c/a) * ArcCosh[a*d/(c^2) + 1]
where:
d = distance in light years (3.26 light years = 1 parsec)
a = acceleration (M1 = 1.03, M2 = 2.06, M3 = 3.09, M4 = 4.12 and so on)
c = 1
t = stay at home time elapsed in years
T = crew time elapsed in years
sqrt[x] = square root of x
ArcCosh[x] = hyperbolic arc cosine of x. On Windows built-in calculator, check the "Inv" and "Hyp" checkboxes, and use the cos key
Example: how long to cross 1 parsec at M4.
d = 3.26, a = 4.12
Stay at home time:
t = sqrt[(d/c)^2 + (2*d/a)]
t = sqrt[(3.26/1)^2 + (2*3.26/4.12)]
t = sqrt[(3.26)^2 + (6.52/4.12)]
t = sqrt[10.6276 + 1.58]
t = sqrt[12.2076]
t = 3.49 years
Crew time:
T = (c/a) * ArcCosh[a*d/(c^2) + 1]
T = (1/4.12) * ArcCosh[4.12*3.26/(1^2) + 1]
T = 0.24 * ArcCosh[4.12*3.26/1 + 1]
T = 0.24 * ArcCosh[13.43/1 + 1]
T = 0.24 * ArcCosh[13.43 + 1]
T = 0.24 * ArcCosh[14.43]
T = 0.24 * 3.36
T = 0.24 * 3.36
T = 0.81 years or 9.7 months
Now, there is something else you forgot to specify. Your ship can accelerate at M4, then you can see how long it will take to travel 1 parsec. This means that at 1 parsec, your ship will be streaking past it at a speed close to lightspeed.
But what you probably want is for your ship to come to rest at a point 1 parsec away. This means you will accelerate at M4 up to the 0.5 parsec point, then decelerate to a halt at the 1.0 parsec point.
To calculate this, instead of using the full distance for d, use only half of it. Then when you get the time result, double it.
For our example, we use 1.63 for the distance.
t comes out to be 1.86 years, double that for a final result of 3.72 years
T comes out to be 0.65 years, double that for a final result of 1.3 years
The Estimable Nyrath leaves out two minor but important considerations:
1) interstellar medium drag
2) gravity.
Gravity is ugly because it slows down (very slightly, but on this scale, still significant vector effects) accelleration out, and imposes an acceleration upon arrival that mus be countered, and thus slows decelleration at far end...
And while a single atom per cubic meter might not seem a whole lot, it's actually quite a significant drag factor... but that also becomes form dependent. And a radiation issue.
1) interstellar medium drag
2) gravity.
Gravity is ugly because it slows down (very slightly, but on this scale, still significant vector effects) accelleration out, and imposes an acceleration upon arrival that mus be countered, and thus slows decelleration at far end...
And while a single atom per cubic meter might not seem a whole lot, it's actually quite a significant drag factor... but that also becomes form dependent. And a radiation issue.
True, I was doing a first order analysis.
1) drag of the interstellar medium
I wasn't about to open this can of worms. In reality the ship would need some kind of Bussard-esque magnetic shield to divert the medium from the hull or the ship will erode away to atoms a few years after the crew dies hideously from radiation. Such shields are not canon in the Traveller universe.
2) Gravity.
Solar escape velocity is about 42.1 kilometers per second. So the M4 ship will have to accelerate about an extra 17 minutes to escape the primary sun, and decelerate an extra 17 minutes if there is a sun at the destination.
I don't know about you but an extra 38 minutes added to an acceleration duration of 3.72 years didn't seem important. Give drives capable of 4 gs of acceleration for prolonged periods of time, gravity is not a significant factor.
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More numbers than I can think through! The question was to logically tie a scout report from 100 years (+/-) ago to a ship tumbling through space.
http://www.travellerrpg.com/CotI/Discuss/showthread.php?t=22212
Thanks!
L
http://www.travellerrpg.com/CotI/Discuss/showthread.php?t=22212
Thanks!
L
In which case, it should be going well under .1c... so anything faster than 33 years per parsec would be problematic.
Interesting scenario: the scout's J drive failed and they were limping back on M drive. Unfortunately, there was a catastrophic failure during turn-over which took out the crew, leaving the ship tumbling along at a significant fraction of c. This would give you a large distance range to start within.
Catching up with the vessel would be a major undertaking in time and energy, so perhaps there is a low power transponder beacon and just enough of an automated data dump to justify the expense to get on board and gather the rest of the data.
Catching up with the vessel would be a major undertaking in time and energy, so perhaps there is a low power transponder beacon and just enough of an automated data dump to justify the expense to get on board and gather the rest of the data.
Tash, at 0.1C, some 30,000km per sec, the interstellar medium is a radiation source sufficient to be etching the hull and causing secondary radiation sufficient to be troublesome.
Now, since we know from canon that ships can make 0.5pc per 2 years (see Imperium; it's part of CT canon) some form of shielding must be part of at least warship drives.... but if that shielding fails above 0.2C, everyone dies pretty quick.... but also pretty painfully... from radiation sickness.
Now, since we know from canon that ships can make 0.5pc per 2 years (see Imperium; it's part of CT canon) some form of shielding must be part of at least warship drives.... but if that shielding fails above 0.2C, everyone dies pretty quick.... but also pretty painfully... from radiation sickness.
Icosahedron
SOC-14 1K
With Traveller ships having typically only 4 weeks of fuel aboard, would this be the case? I imagine the ship could only accelerate for two weeks, drift at constant speed for a number of years that I'm not gonna try to calculate right now, and then decelerate for two weeks.
PS. Thanks for those figures, Nyrath, I'll file them away for future use.
Andrew Boulton
The Adminator
With no JD, there's lots of space for PP fuel.
Well, the back of my envelope says that the mission will look like this:
Accelerate at 4g for 2 weeks. Ship will then have traveled 0.0009 parsec and have a velocity of 0.157c.
Ship drifts for 20.7 years, traveling a further 0.9982 parsec
Ship decelerates at 4 g for 2 weeks. Ship will travel remaining 0.0009 parsec and have braked to a halt. All fuel has been expended.
Bottom line: 20.8 years to travel 1 parsec.
Andrew Boulton
The Adminator
You can squeeze at least 280 days fuel in using HG.
I would also consider an alternate power source for the coast/drift phase.
An old-fashioned(!) fission pile would be a more cost-and-space-effective power supply for life support and such until it was time to fire up the regular fusion powerplant and begin deceleration with the power-hungry M-drive.
I presume the OTU C-Jammer and her ilk used such a scheme.
