[Spaceships] Voyage Time Calculation with Different Drives

Ulzgoroth · 07-20-2011, 11:19 AM

Basically? Newtonian rules of motion. Brush up, or get somebody good at high-school level physics to solve it for you.

In this case, it's relatively simple. In the first and last 30 hours combined the ship only covers a little more than 1 percent of an AU (1/2 A t^2 each), which I'd discount as negligible, and just treat as starting and ending the flight at 10 MPS rather than 0. Then again, that's not much simplification and I'm using a (google) calculator anyway, so I'll go ahead full detail.

So, not counting the first and last 30 hours, we're starting and ending at ~10.33 mps, and covering ~2.988 AU. Thus the equation for either (post burn) half-flight is 1/2 * 0.0005 G * t^2 + 10.33 mps * t = 1.494 AU.

The quadratic formula tells us that Ax^2+Bx+C = 0 has solutions x = (-B +- sqrt(B^2-4AC))/2A. In this case, A = .0005 G/2 = .0005/2 * 9.8 m/s^2, B = 10.33 mps, and C= -1.494 AU. Plugging that right into my search-bar, units and all (with + for the +-, since we're only interested in positive time), I get t ~= 78.04 days for the half-trip. Double to 156.08 days for the full trip, then add back the 60 hours of thruster burn to bring it up to about 158.6 days.

07-20-2011, 11:19 AM	#2
Ulzgoroth Join Date: Jul 2008	Re: [Spaceships] Voyage Time Calculation with Different Drives Basically? Newtonian rules of motion. Brush up, or get somebody good at high-school level physics to solve it for you. In this case, it's relatively simple. In the first and last 30 hours combined the ship only covers a little more than 1 percent of an AU (1/2 A t^2 each), which I'd discount as negligible, and just treat as starting and ending the flight at 10 MPS rather than 0. Then again, that's not much simplification and I'm using a (google) calculator anyway, so I'll go ahead full detail. So, not counting the first and last 30 hours, we're starting and ending at ~10.33 mps, and covering ~2.988 AU. Thus the equation for either (post burn) half-flight is 1/2 * 0.0005 G * t^2 + 10.33 mps * t = 1.494 AU. The quadratic formula tells us that Ax^2+Bx+C = 0 has solutions x = (-B +- sqrt(B^2-4AC))/2A. In this case, A = .0005 G/2 = .0005/2 * 9.8 m/s^2, B = 10.33 mps, and C= -1.494 AU. Plugging that right into my search-bar, units and all (with + for the +-, since we're only interested in positive time), I get t ~= 78.04 days for the half-trip. Double to 156.08 days for the full trip, then add back the 60 hours of thruster burn to bring it up to about 158.6 days. __________________ I don't know any 3e, so there is no chance that I am talking about 3e rules by accident.