Splitology

baturinsky · Aug 10th 2012

What happens when current splits to different consumers?

It's really quite interesting. I was experimenting with it for a while, to make automatic timers for reactors without extra mods.

You can find how it works yourself, btw - it could be really fun. Was for me. Here is what I know at the moment.

When there is more than one possible target for EU packet at the moment, it splits. Not alternating between target, not choosing at random, just physically splits.
It splits evenly between all possible targets. Or so i thought. and it is true - as long as you use only optic fiber with length up to four meters. I was using fiber for test stands, so has not noticed this detail until I started to replacing it with copper and gold to lower costs of production model.

When I replaced some parts of fiber with copper, currents was suddenly completely jumbled. So I started simple experiments - splitting current from one batbox to two others through different wires. It was quite apparent that split ratio is very close to total resistance of cable, i.e. if we have 1m copper and 2m copper (or 1m gold) then ratio should be 2. But it was is 2.1 - close, but not exactly. Fiber (anywhere between 0 and 4 meters) and 1m copper is 4.16666. Fiber and 1m gold is 9.33333
Suspicious numbers. I was very excited when I solved this thing - all these ratios (21/10, 25/6, 28/3) adds up to 31. So, packet is divided to integer sized packets, round down. Not sure what happens to remainder, but it seems to not be lost completely.

So, complete (?) rule of splitting is this: when there are many possible takers for a packet, then it is split to smaller integer packets. Ratio is inversely proportional to resistances (i.e. supposed eu/t loss) of pathes to targets, with bottom cap of 0.1 EU/t. So, fiber and tin up to 4 meters, as well as connection without wires (side against side) are completely equal.

This can be used, beside other things, to downgrade current all the way to 1 eu packets. 1eu can be made without losing energy, by merely a three-ways split, two of which are 1-4m fiber and last (1eu one) is 4m copper.

Luminators can be charged from BatBox by copper tin, but only if there is always at least six (or exactly zero) luminators and machines (or storages or transformers) consuming energy from same source each tick. This can be easily done by adding five transformers to the wire, that transport energy back to the source BatBox.

Any number of 100 Eu/t reactors can be fed directly to two MFSU - unless MFSU fill up. Useful if you run reactors in microcycles.

And if there is a very long cable, it's better if there is only one storage taking energy from it most of time, because otherwise losses are doubled or more.

Gaxx · Aug 10th 2012

Quote from baturinsky

Suspicious numbers. I was very excited when I solved this thing - all these ratios (21/10, 25/6, 28/3) adds up to 31. So, packet is divided to integer sized packets, round down. Not sure what happens to remainder, but it seems to not be lost completely.

my intuition says that's the 0,5EU loss per block of the slitter cable.

GregoriusT · Aug 10th 2012

Hmm, i thought it would "split" like in RP's Pneumatic Tubes. It just fills the closest Target up and then it searches for the next Target to fill (my Factories are confirming this). But when every Target is equivalent close to the Splitterpoint, then it completely depends on how the E-net is coded. So it either alternates the Packets, or it splits one Packet into two halfsized ones and sends them to both Locations.

Player, if you read this, could you explain that to us more precicely? Because you wrote the new E-net.

eculc · Sep 10th 2012

Quote from Gaxx

my intuition says that's the 0,5EU loss per block of the slitter cable.

as far as I can tell, He isn't using a splitter cable. he's just splitting directly.

MagusUnion · Sep 10th 2012

Quote

So it either alternates the Packets, or it splits one Packet into two halfsized ones and sends them to both Locations.

It's the latter. An easy test to see this in effect is to make 3 Batboxes, fill one full of energy, and have three cables sitting on the output of the filled Batbox

Then, add ONE Batbox to the left, and wait till it has 1000 EU

___

Next, add the last Batbox to the other side of the cable

___

The rate at which each box will gain energy is equal to one other. BUT, the level of charge is not only equal to what was in the first box, but the rate at which the EU is transfered is always constant to each machine attached.

So the Enet is smart enough to distribute EU according to each machine attached, as well as account for losses. To see this in action, have one side of 4 block distance of insulated copper going to the left, and the other side having uninsulated gold cable to the right. Take measurements of the three centermost cables closest to the output, and record results...

You'll see that the Enet can be quite 'smart' in the distribution of its energy...

Player · Sep 10th 2012

Energy is usually being split according to the conductance of the available paths, not taking those into account which have a higher loss than the total amount of eu emitted from the energy source. Paths leading to targets which currently don't accept energy (usually because their internal buffer is full) will be ignored as well.

The previous calculation yields the minimum amount of eu sent to the various destinations, however there are some targets which don't get their full share because they are either almost filled up and the energy gets partially rejected or the share is not larger than the path loss. Those lost shares are being collected and sent in further distribution rounds, just like the original method, but with randomly eliminating targets in stalemate situations. This situation occurs when all shares were unable to be distributed as in everyone would get less than 1 eu = effectively 0. Reducing targets increases the share of the others, ultimately to one which gets everything or even zero if the last possible path is too lossy.

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