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Old 01-19-2011, 05:17 AM
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Quote:
Originally Posted by txen View Post
Since the ships are accelerated and decelerated by lasers on the earth side, that constrains the time at earth some. If I remember correctly the ships accelerate at 1.5 g so that will take just under half a year to reach the 0.7c crusing speed. Guessing that there is just one earth side laser system it can either be speeding up or slowing down one ship. That means one ship incoming and one leaving per year. Each needs almost six months of laser time. So a ship comes back and as soon as it arrives another is sent to Pandora. The incoming ship has one year for refitment and can be sent after the next incoming ship arrives.
...THANK YOU for pointing this out. I totally knew about the laser / photon-sail system on the Earth end, but somehow forgot to check to make sure my version satisfies the constraint that only one ship can be accelerated or decelerated at a time. (I agree with your guess that there's only one laser system). As a matter of fact, this changes things. If you look at the diagram I drew, the ship labeled "t = 0" is about to leave earth... except it actually CAN'T, because the ship labeled "t = 14.21083" still has 3.5 months of deceleration left to go. So something has to be different... that incoming ship has to be either more than 5.5 months away when t=0 leaves, or it has to be already in orbit.

So far, I see 3 possible alternatives that would satisfy the laser acceleration constraints while still satisfying the canon (non-asterisk) assumptions:

Quote:
1) The loiter around Earth is less than a year, short enough that a ship can arrive at Earth and leave again before the next one comes in range (Ship A arrives, Ship A leaves; Ship B arrives, Ship B leaves; Ship C arrives, Ship C leaves; etc).

2) The scenario txen described - i.e. There is a loiter of about a year around Earth, and a ship leaves shortly after the one behind it arrives (Ship A arrives, loiters for a year; Ship B arrives, Ship A leaves; Ship C arrives, Ship B leaves; etc) In this scenario, each ship would be about 1 year ahead of the one behind it instead of 1.29 years. However, because the total cycle would still be 15.5 years and there are 12 ships, there would be a long stretch at some point where ship #12 leaves Earth and then ship #1 doesn't arrive for 4.5 years. This would be discarding my previous assumption that the ISVs are spaced equally throughout the 15.5-year cycle.

3) We preserve the assumption that ISVs are spaced equally throughout the cycle, but make it a longer cycle such that the timing of arrivals and departures work out. This would mean that the loiter around Earth is longer than a year, with more than one ISV in orbit at a time.
I'm going to have to play around with the numbers and brainstorm logistics to figure out which scenario seems most reasonable, and what the revised parameters would be. Hopefully, I'll figure something out and post a revised version that satisfies all constraints.
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