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A tribe of nomadic aliens is located in a planetary system that can supply them with sufficient resources to get to the next resource rich system. The next such system is 8 ly away, and there is only empty space in between. They can carry enough resources, a "load", to go 5 ly. They can make storages of any size on the way.

What is the minimum amount of loads they will need to get to the next system?

Is there a limit to the distance between systems that they can cross?

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If they travel at around .8 c, they should make the trip in under five years.  In their frame of reference, due to Minkowski contraction of their path, their apparent journey will be under 5 LY.  So, zero loads are needed along their path.  If they can go faster, closer to c, then they can travel to much farther systems without reloading.  The limits would depend on engineering:  ratio of reaction mass to cargo, drive efficiency, passenger tolerance for acceleration and deceleration, etc.

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3 minutes ago, TheVat said:
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If they travel at around .8 c, they should make the trip in under five years.  In their frame of reference, due to Minkowski contraction of their path, their apparent journey will be under 5 LY.  So, zero loads are needed along their path.  If they can go faster, closer to c, then they can travel to much farther systems without reloading.  The limits would depend on engineering:  ratio of reaction mass to cargo, drive efficiency, passenger tolerance for acceleration and deceleration, etc.

What they can do is given. 5 ly in the planetary systems FoR.

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What is the minimum amount of loads they will need to get to the next system?

Spoiler

If any of this is right, could you only say so in a spoiler message and not give any upvotes? It seems a lot of people like to try to solve it without reading spoilers, and if others are like me we're tempted to cheat if we see it's already been solved.

They have to travel a distance of 1.6 1-way 1-"loadtrip"-lengths. They'll make an odd number of 1-way legs between storage points. Optimal seems to be making the most use of the last 1-way.

Working backward, the last trip will be a distance of 1 loadtrip, ending at the new system with 0 resources left. So they need 1 load at a storage point there, and they need 3 legs to bring it there. The max distance they can carry 1 load is 1/3 loadtrips, leaving 1/3 and then 2/3 loads at the storage. So they need 2 loads stored at 1+1/3 loadtrips from the final destination.

They need 5 trips min to carry 2 loads a max distance if 1/5 loadtrips (carrying .6+.6+.8 to the storage) again using 1 load to do so.

Finally they need 7 trips to carry 3 loads to the storage at 1+1/3+1/5 from the destination, but they don't have to make full 1/7-length trips, because they only have to go 1/15 left to add up to 1.6. So that costs 7/15th to do so, and the total cost is 3 7/15 or 3.4666 loads.

I'm guessing that shortening only the first legs, and then maximizing the leg length the rest of the times, is probably best because you're making the most trips between the earlier storage points.

Is there a limit to the distance between systems that they can cross?

Spoiler

I cheated and wolfram alpha says that the sum of 1/(2n-1) for n=1 to infinity, diverges to infinity, so there should be no limit to how far they can go if the system has unlimited resources. (But they'd need eg. over a million storage points and millions of trips between them, to go 40 LY.)

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@md65536, you are right. Well done! Instead of one upvote, here are 5 *****.

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On 6/12/2023 at 6:44 PM, md65536 said:
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If any of this is right, could you only say so in a spoiler message and not give any upvotes? It seems a lot of people like to try to solve it without reading spoilers, and if others are like me we're tempted to cheat if we see it's already been solved.

Spoiler

I don't see others trying to solve it anymore. OTOH, your answer is too good not to be marked. If anyone is curious but gave up, they can go directly to your answer. I think it's time to upvote it. +1

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