Problem of math

[Harmon]

Sorry for not checking in with this more.

I thank you all a dozen times.

Still not sure exactly how I will handle this campaign, its been on my mind for a looonnnnggg time and I want to do it, hopefully this will give me the kick I need to get past my distance and speed hang ups.

Many thanks for the thoughts and assistance. I knew EN World was the place to ask.

 

[Arc]

As far as the long vs short travel goes, consider using a semi-Newtonian acceleration system, but with a higher order polynomial instead of quadratic equation. For example, if your ship accelerates at 1.25 au/day^3 under the formula x = at^3, .5 au will take 1.2 days (accelerate for .25 au, decelerate for .25 au), but 60k au will take 58 days. Use some derivation of the formula x = k(at^n + at^(n-1) + ... + x0), for 0 < k, a > 1, n > 2, and you'll be able to find something that'll provide the necessary times. If you have a rough estimate of how long you want various distances to take, it shouldn't be too hard to come up with a set of constants that solve for those times and distances.

The nice thing about this is you can nice separate space travel (semi-Newtonian physics) from atmospheric travel (fully Newtonian), since they operate under different systems of equations altogether. Make up some technobabble, and you're set.

 

[Cheiromancer]

....but 60k au will take 58 days

How long would it take to get to Alpha Centauri? If your formula is cubic then doubling travel time will octuple the distance traveled. I think it opens up interstellar travel, which was what Harmon wanted to avoid.

 

[Pbartender]

How long would it take to get to Alpha Centauri? If your formula is cubic then doubling travel time will octuple the distance traveled. I think it opens up interstellar travel, which was what Harmon wanted to avoid.

Not necessarily. Remember, you've got the speed of light as a hard speed limit, so you very quickly reach a practical limit of diminishing returns... Speeding up from .9C to .99C is a lot harder than speeding up from .1C to .9C, for example. What's more, from the outside observer's point of view the difference between you traveling at .90C and .91C is negligible, for all intents and purposes. And that's not even talking into account logistics problems of accelerating for that long (which is why bussard ramjets are so handy).

The only real benefit you get is from the apparent relativistic effects -- the trip takes a little less time for you, inside the ship, but still takes approximately the same time for everyone else outside the ship.

At one point, I had an automated Excel spreadsheet that, if you entered a distance, would calculate travel times (and apparent travel times for the passengers), based on different rates of constant acceleration and deceleration.

If I remember right (Correct me if you care to, Umbran or anyone else... it was LONG time ago and I don't feel like re-doing the math ), someone traveling in a spaceship to Alpha Centauri at a contant acceleration of 1G would get there in roughly 5 years (actual time), but only 2 years and a handful of months would have passed on board the ship.

 

[Cheiromancer]

Well, 60,000 AU is almost a light year (over 300 light days, anyway) so if you are getting there in 58 days something funny is going on. I don't know what kind of natural phenomena a cubic distance formula would correspond to.