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- Thread starter R_Kane
- Start date

S

Sophia's estimate is fair enough ballpark estimate, the maths get rather hairy once you reach an appreciable fraction of c. Relativity rears quite an ugly head indeed. Time outside the ship and inside the ship become two very different things and your passengers will have some clock adjusting to do upon arrival.

Another point to consider is fuel. While thruster plates only require electricity, you still need to generate that juice for the length of the trip.

Sincerely,

Larsen

I dug out my notes and fired up the old calculator.

Assuming 1 gee acceleration and a one parsec (3.23ly) trip distance, you're looking at ~750 days ship time and ~1480 days 'outside' time. That's a little over 2 subjective years for the crew while the people on either end of the voyage watch a little over 4 years pass.

This also assumes a constant acceleration - deceleration thrust profile, i.e. no coasting.

Relativity enters the picture at around 2/5ths of c.

Sincerely,

Larsen

How about a more reasonable scenario:

An initial burst of thrust (could be big, at least a day, if not a week long), coasting for most of the journey (since everyone has retired to low berths at this point), the appropriate deceleration at the destination.

How long would that take?

S

An initial burst of thrust (could be big, at least a day, if not a week long), coasting for most of the journey (since everyone has retired to low berths at this point), the appropriate deceleration at the destination."

Sir,

Well, as long as you stay below 2/5ths of c and thus away from any relativity problems it would be a simple time-acceleration-distance problem.

Let's see, pull out the back of that envelope...

- Figure 1 gee (that's 0.0098km/sec^2) thrust for 2 weeks (that's 1209600 seconds) to reach a velocity of 11854.08km/sec.

- During that 2 weeks you'll have travelled 7,169,347,584 km. Remember that, it's the distance we'll need to slow down too.

- A light year is roughly 624,388,608,000 km. We'll have already travelled that distance above, plus we'll need to slow down, so subtract that twice from the distance we need to go. We'll end up with 610049912832 km left. Now just divide our velocity into that for a 'coasting' trip time of ~595.64 days. Don't forget to add the 28 days spent accelerating and decelerating. We end up with a total ~623.64 days most of which was spent at ~0.04 of c. (We don't need fear relativity until we near ~0.4 of c).

Sincerely,

Larsen

Sophia,

So did I.

"You will need 1 year time at than constant 1 g thrust and you will travel about than 1/2 light year in that time of acceration."

Did you remember to account for relativity? Without accounting for relativity, one year of thrust at one gee nets you a final velocity of ~308418 km/sec. Unfortunately, that is faster than the velocity of light at 297600 km/sec.

"If you only accerate for 1 week at 1 g you will need ten's of thousand of years or more to travel at that speed."

Sorry, no. Your math is off. Look at my post above for a 1gee/2 week thrust profile, you can traverse 1ly in a little less than 2 years.

"You must also take in account outside forces and the fact there is than very thin amount of gases in space may-be 1 moleculate of h2 per cublic centimeter which can slow drown the ship."

This bit is correct. At appreciable percentages of c, collisions with even the smallest of 'moleculates' could be catastrophic.

If I may be allowed a personal question, I noticed your occupation is listed as 'collect disabilities'. May I be so bold as to ask just what your disabilities are? A very dear friend of mine, Bari Stafford, passed away from ALS last summer. He enjoyed Traveller too and even managed to participate in the TML Ship Rodeo a few weeks before his passing.

Sincerely,

Larsen

t'(1) - t'(0) = (c/a)inverse cosh(1 + (aS/c2))

t'(1) is the end time onboard ship (seconds)

t'(0) is the beginning time onboard ship (seconds)

c is the speed of light (299,792,458 m/s)

a is the ships acceleration in m/s (1-G = 9.8 m/s)

S is the distance traveled im meters

1 parsec = 3.26 light years

1 light year = 31,536,000 light seconds.

1 light year = 9.45 x 10^15 meters

1 parsec = 3.08 x 10^16 meters

There you have all the tools you need to calculate the relatavistic ship travel time to the end of its initial acceleration.

The formula for the ship's final velocity is:

V = c (1 - (1 / (1 + (aS/c^2))^2))^0.5

^ is the exponent operator.

once the cruise velocity is reached you use this formula for dertermine the travel time to cross the distance during cruise.

t' = (1 / (1 - (V^2/c^2))^0.5)) x (t - (Vx / c^2))

x is the distance crossed at constant velocity.

You then add the amount of time to accelerate as the time to decelerate and that is you total ship's time cruise for the sublight journey. To determine the travel time as perceived by the outside universe simply calculate the ship's average velocity as a percentage of the speed of light and divide the distance in light years by it. Remember to convert parsecs into light years.

For the vehicles

S

4 milies per second to reach any decent speed it will

need to have than big fuel tank which add dead weight

which limit it top velocity to about 2 time exhaust velocity unless you mult-stage. Any reaction engine will have the same problen except for photon rocket which have than exhaust velocity of the speed of light. Unless you use than interstellar ramship to bypass the mass ratio of initial mass/ final mass.

Most space chemical rocket are 80 to 90 % fuel by mass. Look at the size of the Saturn 5 rocket compare to the command modural and service modural.

that why our four space probe which are now traveling in interstellar space are going to take than long time to get anywhere. The fastest one has than velocity of 3.5 AU/year(325,500,000 miles in one year time. There is 63500 AU in one light year.

To get than faster than exhaust velocity you need than very hight mass ratio.

S

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