Optimal number of fractionators to completely fractionating a given amount of hydrogen production rate

[BigMcisLove]

Let's do some math things

First, you should know the following values.
r : the input rate of hydrogen (1/s),
n : the number of fractionator in each loop,
T : the period of a loop (s).

For example, the simplest loop with a single fractionator has the values n=1 and T~4s with mk2 belt.

Now we want to express the number of loops N in terms of the above quantities.


Let's say there are total H(t) number of hydrogens on a single loop at time t. Then the change rate of hydrogen on a loop can be expressed as dH/dt.

With a proper balancer setup, one can get r/N rate of hydrogen input for each loop.

On the other side, each hydrogen will experience fractionation n times for each loop, and it will be converted to deuterium with 1% of chance for each fractionation. The total chance for one hydrogen to be converted to deuterium is 1-0.99^n, because the chance that a hydrogen WILL NOT be converted after all fractionations is 0.99^n. For simplicity, let's say p=1-0.99^n

p = 1-0.99^n : the chance that a single hydrogen is converted after one loop.

Then the expectation value of the output rate would be p*H/T.

Now we have a nice differential equation describes the number of hydrogen on a loop at time t:

dH/dt = r/N - p*H/T.

Don't be afraid. We do not have to actually solve the differential equation (easy one, though). All we want to know is the condition when dH/dt = 0, which means there is no change in hydrogen number on a loop.

After a long long time, the number H would be converge to a certain number, say H0. What we want here is that this number H0 is SMALLER than some number H1.

For example, the simplest loop with one fractionator can hold maximally ~40 hydrogens on it, and obviously we DO NOT want that the number H0 is bigger than 40, which means the system overflows.

With some simple arithmetic, one would get
r/N - p*H0/T = 0 ----> H0 = (r*T) / (p*N).

We WANT
H0<H1 ----> (r*T) / (p*H1) < N

Again, with the simplest loop example, and assume we have r = 2 (production rate of hydrogen is 2/s) and want H1=30 (# of hydrogen on a loop would not exceed 30), then we have N > (2*4) / (0.01*30) = 26.667. Therefore, we need total 27 number of fractionation loops to completely fractionate the hydrogen with a production rate 2/s !


Hope this is helpful to build your perfect interstellar factory.
If you find any typo, error, or something that I missed, please let me know with comments.

[ShuFlngPu]

Originally posted by BigMcisLove:

Let's do some math things

First, you should know the following values.
r : the input rate of hydrogen (1/s),
n : the number of fractionator in each loop,
T : the period of a loop (s).

For example, the simplest loop with a single fractionator has the values n=1 and T~4s.

Now we want to express the number of loops N in terms of the above quantities.


Let's say there are total H(t) number of hydrogens on a single loop at time t. Then the change rate of hydrogen on a loop can be expressed as dH/dt.

With a proper balancer setup, one can get r/N rate of hydrogen input for each loop.

On the other side, each hydrogen will experience fractionation n times for each loop, and it will be converted to deuterium with 1% of chance for each fractionation. The total chance for one hydrogen to be converted to deuterium is 1-0.99^n, because the chance that a hydrogen WILL NOT be converted after all fractionations is 0.99^n. For simplicity, let's say p=1-0.99^n

p = 1-0.99^n : the chance that a single hydrogen is converted after one loop.

Then the expectation value of the output rate would be p*H/T.

Now we have a nice differential equation describes the number of hydrogen on a loop at time t:

dH/dt = r/N - p*H/T.

Don't be afraid. We do not have to actually solve the differential equation (easy one, though). All we want to know is the condition when dH/dt = 0, which means there is no change in hydrogen number on a loop.

After a long long time, the number H would be converge to a certain number, say H0. What we want here is that this number H0 is SMALLER than some number H1.

For example, the simplest loop with one fractionator can hold maximally ~40 hydrogens on it, and obviously we DO NOT want that the number H0 is bigger than 40, which means the system overflows.

With some simple arithmetic, one would get
r/N - p*H0/T = 0 ----> H0 = (r*T) / (p*N).

We WANT
H0<H1 ----> (r*T) / (p*H1) < N

Again, with the simplest loop example, and assume we have r = 2 (production rate of hydrogen is 2/s) and want H1=30 (# of hydrogen on a loop would not exceed 30), then we have N > (2*4) / (0.01*30) = 26.667. Therefore, we need total 27 number of fractionation loops to completely fractionate the hydrogen with a production rate 2/s !


Hope this is helpful to build your perfect interstellar factory.
If you find any typo, error, or something that I missed, please let me know with comments.

I just dont use fractionators. They add alot of logistical problems.
Where are you gona feed that upgraded coal? (Forget the name off top of my head.)
Hydrogen and graphite.

So you are actually missing the last bit to your equasion- How many thermal-power stations do you need ontop of that to eat up the extra graphite?
Otherwise, your system has a logical fallacy where your graphite is gona overflow and stop your fractionators.
Also, the power cost for this system is intense so id also add needed power useage too :)

PS: If I messed up the naming and im talking about the oil seperation- ignore me. Kid has been keeping me up and I cant think very stright. :x lol

[BigMcisLove]

Originally posted by ShuFlngPu:

Originally posted by BigMcisLove:

Let's do some math things

First, you should know the following values.
r : the input rate of hydrogen (1/s),
n : the number of fractionator in each loop,
T : the period of a loop (s).

For example, the simplest loop with a single fractionator has the values n=1 and T~4s.

Now we want to express the number of loops N in terms of the above quantities.


Let's say there are total H(t) number of hydrogens on a single loop at time t. Then the change rate of hydrogen on a loop can be expressed as dH/dt.

With a proper balancer setup, one can get r/N rate of hydrogen input for each loop.

On the other side, each hydrogen will experience fractionation n times for each loop, and it will be converted to deuterium with 1% of chance for each fractionation. The total chance for one hydrogen to be converted to deuterium is 1-0.99^n, because the chance that a hydrogen WILL NOT be converted after all fractionations is 0.99^n. For simplicity, let's say p=1-0.99^n

p = 1-0.99^n : the chance that a single hydrogen is converted after one loop.

Then the expectation value of the output rate would be p*H/T.

Now we have a nice differential equation describes the number of hydrogen on a loop at time t:

dH/dt = r/N - p*H/T.

Don't be afraid. We do not have to actually solve the differential equation (easy one, though). All we want to know is the condition when dH/dt = 0, which means there is no change in hydrogen number on a loop.

After a long long time, the number H would be converge to a certain number, say H0. What we want here is that this number H0 is SMALLER than some number H1.

For example, the simplest loop with one fractionator can hold maximally ~40 hydrogens on it, and obviously we DO NOT want that the number H0 is bigger than 40, which means the system overflows.

With some simple arithmetic, one would get
r/N - p*H0/T = 0 ----> H0 = (r*T) / (p*N).

We WANT
H0<H1 ----> (r*T) / (p*H1) < N

Again, with the simplest loop example, and assume we have r = 2 (production rate of hydrogen is 2/s) and want H1=30 (# of hydrogen on a loop would not exceed 30), then we have N > (2*4) / (0.01*30) = 26.667. Therefore, we need total 27 number of fractionation loops to completely fractionate the hydrogen with a production rate 2/s !


Hope this is helpful to build your perfect interstellar factory.
If you find any typo, error, or something that I missed, please let me know with comments.

I just dont use fractionators. They add alot of logistical problems.
Where are you gona feed that upgraded coal? (Forget the name off top of my head.)
Hydrogen and graphite.

So you are actually missing the last bit to your equasion- How many thermal-power stations do you need ontop of that to eat up the extra graphite?
Otherwise, your system has a logical fallacy where your graphite is gona overflow and stop your fractionators.
Also, the power cost for this system is intense so id also add needed power useage too :)

PS: If I messed up the naming and im talking about the oil seperation- ignore me. Kid has been keeping me up and I cant think very stright. :x lol

AFAIK, there are hydrogen sources other than plasma refining right?
This calculation is just for a pure mathematical curiosity. It only considers the hydrogen, not others.
For the power consumption, it actually cost less than other deuterium sources if you use mk3 belt!

[Coops]

I read somewhere that 10 was optimal. more than that it becomes inefficient.

[BigMcisLove]

Originally posted by Coops:

I read somewhere that 10 was optimal. more than that it becomes inefficient.

We need more calculation to get 10 fractionators as a conclusion. This one is assuming very generous situation, i.e. you can set the number of fractionator per loop without any consideration.

For sketch, T, p, H1 are all functions of n (sort of). Not sure we can solve it analytically, but anyway we get very strict equation for n and in principle can get a VERY optimal solution for that. I here showed some guide for close enough value of optimal number of fractionators.

[BigMcisLove]

Originally posted by BigMcisLove:

Originally posted by Coops:

I read somewhere that 10 was optimal. more than that it becomes inefficient.

We need more calculation to get 10 fractionators as a conclusion. This one is assuming very generous situation, i.e. you can set the number of fractionator per loop without any consideration.

For sketch, T, p, H1 are all functions of n (sort of). Not sure we can solve it analytically, but anyway we get very strict equation for n and in principle can get a VERY optimal solution for that. I here showed some guide for close enough value of optimal number of fractionators.

With some simple model for T, p, and H1, it seems like the most optimal solution is about 11 fractionators per a loop per 1 hydrogen/s, which is fair close to 10.

[singularity314]

Seems a bit simpler to automate the production of orbital collectors. Once you produce 40 of them you can completely fill a gas giant and probably not worry about deuterium again.

[josmith7]

Originally posted by singularity314:

Seems a bit simpler to automate the production of orbital collectors. Once you produce 40 of them you can completely fill a gas giant and probably not worry about deuterium again.

They're normally the way I go too -- but you either need a lot of vein utilization or you need to cover a lot of gas giants in order to never worry about deuterium again.


Take the gas giant in my starting system, it's a little lower than average with 0.82/s hydrogen and 0.04/ deuterium. With 40 orbital collectors and a vein utilization of 0 they collectively (once accounting for the energy burned to keep them going) provide a "whopping" 6.6/s deuterium. That's only 22% of a single Mk.3 belt.
And for comparison a single fractionator with a quad stacked Mk.3 belt (120/s hydrogen flowing through) outputs 1.2/s deuterium (2.4 if you max proliferate it as well) -- so with the right belts and build just 6 fractionators could outproduce my entire gas giant with 40 orbital collectors!

Though, admittedly, vein utilization scales up orbital collectors very quickly - if I did the math right by VU 2 it is producing nearly 2 belts; 51.4/s; after VU 5 (the last before white science) it'd be making 134.1/s. That's in part because their 30 MW power consumption doesn't go up with VU; while the production rate does -- so while initially it was eating 48.4% of my output by VU 5 it's down to only 22.4%; so that fraction drops rapidly while output also scales up.
(And since VU has no effect on fractionators the balance quickly tilts in the favor of gas giants)

[mark.the.fiddler]

Is there an optimal number? I place about 700 fractionators on every planet orbiting a giant. It takes 40 deterium per rocket. Most people, including experienced players, hit the wall immediately after starting mass rocket production, because even your best storage will be drained before you know it. I watched half a million full tanks of deuterium go dry long before

I place the deuterium plants in complete double sided rings around the poles. I believe, they the first two rings where you can actually fit a double row of fractionators. I also set the blueprints to help build the rings.

Ill share a few lessons I learned the hard way. :P make sure when your input hydrogen joins the hydrogen loop that it joins the loop like the leg of a capital T. The leg is input the top of the T is part of the loop. Otherwise you'll find your factory at a dead stop because the hydrogen can't circulate.

Once the ability to stack resources is acheived, restack the hydrogen loop just before the hydrogen input feed. "Not after" the input feed ;). Your deuterium output will be higher with higher stacks.

If your output line is getting full before the end of the loop. Branch it off and send it to your logistics Station. Start a new output line where you cut the other off.

I'll guesstimate that my deuterium factories produce in excess 18000 deuterium per minute. Since I produce 5400 rockets per minute on 4 different planets, 18000 deuterium is a drop in the bucket. I need lots of planets with lots of polar deuterium rings.

[matrimvidz]

The equation shouldn't have to be that complicated.

Assuming proliferation of the hydrogen, One fractionator gives 2% deuterium.
100%/2% = 50 Fractionators will give back the same amount of deuterium as you put new hydrogen into the loop.
If no proliferation, then 100 Fractionators.
After the initial massive dump of hydrogen to fill the loop, you only put in what you get out.

[Schalimah]

This is a neat little thougt experiment.
But in the game your goal should not be how many fractionators does it take to empty a belt, but how do I keep the belt full so my fractionators keep pushing out the maximum amount of deuterium.
Given a fully proliferated 4 stacked tier 3 belt, you can expect a fractionator to produce on average 2.4 deuterium per second.
With every hydrogen converted, the efficency of the next fractionator is reduced due to less hydrogen on the belt.
Ideally you would restack and replenish after each fractionator pass.

https://steamcommunity.com/sharedfiles/filedetails/?id=2804389034

[josmith7]

Originally posted by matrimvidz:

The equation shouldn't have to be that complicated.

Assuming proliferation of the hydrogen, One fractionator gives 2% deuterium.
100%/2% = 50 Fractionators will give back the same amount of deuterium as you put new hydrogen into the loop.
If no proliferation, then 100 Fractionators.
After the initial massive dump of hydrogen to fill the loop, you only put in what you get out.

As Schalimah said, you want to keep the loop as full as possible, ideally always seeing 120/s hydrogen flowing through each fractionator (4-stacked hydrogen).

But each time a fractionator converts a hydrogen to deuterium there's one less hydrogen on the belt (either a stack gets lower or there's an empty spot) -- and so you want to restack the belt and inject fresh hydrogen to keep it full.

If you just have a proliferated loop with 1 injector point that's 50 fractionators long (which I believe is what you were suggesting) then the first fractionator would be seeing 120/s flow in (and thus outputting 2.4/s deuterium) but the 50th would only be seeing 44.6/s hydrogen (and thus only outputting 0.9/s deuterium)
Overall, those 50 fractionators are only operating at an average of 63.5% of their full potential (76.3/s output instead of 120/s) -- because the belt isn't being kept full.

[Dirkels]

When restocking the hydrogen loop I use a 3 step process:
- Replenish the line.
- An automatic piler.
- Replenish again any gaps created by the piler.
- second automatic piler.
- Final replenishment of gaps created by the second piler.
When fed with a 4 high stacked mk3 proliferated input line this keeps my loop at maximum capacity, and with only 8 fractionaters I produce about 17 Duet/s (1020/min)

[Kyrros]

The problem with all of this - at least on an intellectual level - is that at some point in the last year, the way fractionators 'buffer' internal hydrogen has changed so that small gaps in the feeder lines have much less impact. As long as there's an adequate buffer in an individual fractionator, it will still function at near 100% rate, even if the feed belt is less than 100% itself.

In a purely 'numbers' approach - the most efficient number of fractionators in a loop is 1. Every number above that, comes with a sliding scale of inefficiency - the scale is not linear, though, and starts to rise more noticeably after 12-15 fractionators.

So the 'optimal' number of fractionators is purely subjective based on the player. In most cases, the ideal number in a loop is between a number needed to keep a buffer intact in all units in the loop - which rises as you increase belt speed and capacity - and a number that allows the player to 'feel' like it's not too inefficient.

[Ohm is Futile]

Originally posted by Kyrros:

The problem with all of this - at least on an intellectual level - is that at some point in the last year, the way fractionators 'buffer' internal hydrogen has changed so that small gaps in the feeder lines have much less impact. As long as there's an adequate buffer in an individual fractionator, it will still function at near 100% rate, even if the feed belt is less than 100% itself.

In a purely 'numbers' approach - the most efficient number of fractionators in a loop is 1. Every number above that, comes with a sliding scale of inefficiency - the scale is not linear, though, and starts to rise more noticeably after 12-15 fractionators.

So the 'optimal' number of fractionators is purely subjective based on the player. In most cases, the ideal number in a loop is between a number needed to keep a buffer intact in all units in the loop - which rises as you increase belt speed and capacity - and a number that allows the player to 'feel' like it's not too inefficient.

:sphere:

Hmm, used to be that any missing piece of hydrogen would slow the loop down... which would explain why you'd say the most efficient number is 1, but then you mention the buffer which would make this accurate, at least on paper:

Originally posted by matrimvidz:

The equation shouldn't have to be that complicated.

Assuming proliferation of the hydrogen, One fractionator gives 2% deuterium.
100%/2% = 50 Fractionators will give back the same amount of deuterium as you put new hydrogen into the loop.
If no proliferation, then 100 Fractionators.
After the initial massive dump of hydrogen to fill the loop, you only put in what you get out.

In practice you probably don't want such a long loop but putting in a single sorter to refill the line every 25 (example) fractionator should be more than sufficient to keep a buffered loop going at max capacity non-stop.

My question I suppose is how fast do fractionators "tick" since they're buffered? Do they roll for deuterium 30 times per second regardless of the belt type? Is it tied to the belt type? Intuitively, I'd think it's still tied to belt type. The fractionator would swap a unit of hydrogen as fast as the belt can handle it and either put it back as hydrogen or as deuterium. No matter what, it can't go faster than its outputs can handle I could think, unless it can also buffer on the other side. So, in theory, the most "efficient" use of space is however many fractionators as your belt can process before you put in a sorter. Granted, the odds of 30 fractionators all pulling hydrogen at the same time are astronomically low...

[Kyrros]

Originally posted by Ohm is Futile:

My question I suppose is how fast do fractionators "tick" since they're buffered?

...

No matter what, it can't go faster than its outputs can handle I could think

You are correct, and also inadervtently answered your own question. It ticks as either the rejected hydrogen exits, continuing on its way, or the newly converted Deut is output'ed... which is then replaced by a new one from the buffer (or belt if the buffer is empty)

Output belt determines overall speed, input belt determines overall efficiency.