Maybe these swords would be more popular if they could produce hundreds of tiny metal objects per minute traveling at several hundred meters per second in an operator-chosen direction.

* Rayleigh scattering (cities in North Carolina do not scatter light in any unusual way)

But that's not very first world, is it?

The Wikipedia article you linked says that one cell produces 0.9 V. I wouldn't want to attach a 9 V battery to a 3.7 V battery.

You're referring to ammunition, not bullets, aren't you?

@Separatrix The 400 yard number refers to where the arrows fell on the ground. Plus, it's made up. English longbows shot far but not that far. 70 m is the distance to archery targets today. Archers today aren't even remotely strong enough to pull one of the super strong English longbows with great range. Obviously, being able to deliver a headshot at 70 m and hoping to hit anyone in a large group of people a few hundred yards away, are quite different goals. No one who knows anything about archery would argue that an English longbow is more precise than a modern compound bow.

"Some more weight"? You probably almost need to double the dry mass of the stages to get twice the thrust. The first stage of the Saturn V had a dry mass of 131 t which already is very close to its payload to LEO. Doubling the dry mass of the second stage already exceeds the original payload in extra mass. If we neglect the extra mass of the third stage, this means, you're now about half way to LEO, nowhere close to escaping, and used all of your fuel up.

@Frostfyre Done.

@NathanielFord I added it to the question.

@NathanielFord No ballpark involved. I don't like those.

It means 8 times the volume and 8 times the planet's mass but not 8 times its surface gravity and not 8 times the gravitational potential from the surface to infinite distance. Since I can't put formulas in here, I'll have to write it down. I can do that later and post a link here.

This would be true if you could burn the fuel twice as fast as the Saturn V burns it because you have to account for the temporally longer pull of gravity on what remains of the rocket which costs you energy. I'll take a more detailed look onto this later. PS: You're defeating the purpose of the setting (no (purely) chemical launches possible). :D ;-)