Re: [nbos] [AS] Travel Times
Ben Trafford
Sat Aug 29th, 2009

I'm going to give a more convoluted answer than others, because I've
found breaking it down helps me to finesse my gaming system.

1 g = 9.80665 m/s squared. Meaning, every second, the vessel would
accelerate by 9.80665 meters * .75. With the effects of cosmic dust and
light and whatnot, you can safely round a bit in whatever direction
serves your story. However, at this speed, it'll never reach its top
speed -- it'll get where it needs to go long before that.

In this case, 61 Cygni is 11.4 light years away, or
107,852,327,387,421,120 meters.

Now, what your travel time is missing is a top-end to the speed. Let's
assume, for the sake of argument, that your vessel's top end flight
is .25c (c being the speed of light), to avoid relativistic effects. The
speed of light is 299,792,458 m/s, so your top speed (in this example)
would be 74,948,114.5 m/s.

You have to figure out two things -- how long it'll take your vessel to
reach its top speed; how far it will have travelled by that point; and
then how long it'll take to travel the rest of the way at its top speed.

To figure out how long it will take to reach its top speed, we do this:

(v_f^2 - v_i^2) / (2*a) = x

where v_f equals "final velocity" and v_i equals "initial velocity" and
a equals "acceleration" and x is the "distance travelled."

In our case, it looks like this:

(74,948,114.5 ^2) - (0 ^2) / (2 * (9.80665 * .75)) =
5,617,219,867,105,110.25 meters to reach maximum velocity

Since all these figures are in meters and seconds, we know that the
ship will travel 5,617,219,867,105,110.25 meters before it reaches top
speed.

Next question is how long will it take to reach it. For that, we do
this:

(v_f - v_i) / a = t

Where t is the "time to travel" in seconds, since we're thinking in
meters per second. It looks like this:

(74,948,114.5 - 0) / (9.80665 * .75) = 10,190,107.66503682 seconds to
reach maximum velocity

Let's assume they take the same distance and time to decelerate, which
is always a safe bet. So, we double the distance and time required to
accelerate, and we have acceleration and deceleration covered. Now,
let's figure out the bits in the middle, when we're cruising along at a
happy .25c.

107,852,327,387,421,120 - (5,617,219,867,105,110.25 * 2) =
96,617,887,653,210,899.5

And how long will that take them to cover?

96,617,887,653,210,899.5 / 74,948,114.5 = 1,289,130,331

Let's round things to the nearest zero, and take this to the finish
line.

The time to reach maximum velocity: 10,190,107 seconds
The time to decelerate: 10,190,107 seconds
The time to go the rest of the trip: 1,289,130,331 seconds.
Total trip time is thus: (10,190,107 * 2) + 1,289,130,331 =
1,309,510,545 seconds or 42 years, 6 months, 11 days, 22 hours, 8
minutes and 45 seconds.

Phew. I tend to enter all this stuff into a spreadsheet and just plug
in the numbers, but you get the idea. I've put the spreadsheet I use up
here as a zip file containing Excel and ODS formats:

http://www.prodigal.ca/SublightCalcs.zip

Feel free to use and re-distribute them. The stuff in red is calculated
-- the stuff in black is what you can tweak to get different outputs.

--->Ben

On Fri, 2009-08-28 at 21:38 -0400, Doug Jessee wrote:
> Hello,
>
> I am trying to figure out some travel times at slower than light.
>
> Anyone know of any easy way to compute these times?
>
> For example, a sleeper ship that is going from earth to 61 Cygni.
> Assuming acceleration of 3/4 Gs, I was just trying to figure out how
> long it would take to travel.
>
> -Doug Jessee
> _______________________________________________
> Nbossoftware mailing list
>
>

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