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On 24 May 2016, at 01:15, Phil Pugliese (via tml list) <xxxxxx@simplelists.com> wrote:
>
> Just wondering;
>
> How far would you have to be travelling in order to require, let's say it's the standard 'scout ship', a two-week constant 2G accell (towards the 'target'), a coasting time of one week, & then a two-week decell in order to arrive w/ only a small relative diff in velocity?
>
> Since the 'coasting' time will still require the use of the PP, take some time off both ends of the trip so that there'll still be enough power avail to survive & maneuver at the end.

1 day = 86,400 seconds
2 weeks = 1,209,600 seconds

So let's use 1 million seconds of acceleration
And balance this a bit by having G = 10 metres per second per second

After 2 weeks,

Velocity = acceleration times time
= 20 x 1,000,000 m/s
= 20,000 km/s

Distance travelled = half acceleration times time squared
= 0.5 x 20 x 1,000,000 x 1,000,000
= 10 trillion metres
= 10 billion km

Coasting for another week
= 20,000 x 1,000,000 km
= 20 billion km

Slowing down is the same time and distance covered as speeding up, so another 10 billion km, for a grand total of 40 billion km.

For reference, the speed of light is 300,000km/s so that's about 37 light hours.
It's also more than twice as far a Voyager.

Phil Kitching