Re: [nbos] [AS] Travel TimesDoug Jessee Sat Aug 29th, 2009
Thanks for all your help!
I appreciate it.
On Sat, Aug 29, 2009 at 2:56 AM, Ben Trafford <ben-at-prodigal.ca> wrote:
> 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.
> 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
> 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
> (v_f^2 - v_i^2) / (2*a) = x
> where v_f equals "final velocity" and v_i equals "initial velocity"
> 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
> Next question is how long will it take to reach it. For that, we do
> (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,
> 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) =
> 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
> 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
> 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:
> Feel free to use and re-distribute them. The stuff in red is
> -- the stuff in black is what you can tweak to get different outputs.
> 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
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