here, i slightly modified it http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…1f21s1r14lflr10
If someone makes this reactor slightly better i could ad it to the list. Then we have a 1 2-pair uranium reactor a 2 2-pairs uranium reactor and a 3 2-pairs uranium reactor XD. Atm http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…1501521s1r11r10 has almost the same eu output but better eff.
Display MoreSo, each ore can become 8 near-depleted cells. Each of those, once processed and used, has a 25% chance to become a near-depleted cell. Of those 25%, 1/4 will become a near-depleted cell when used, again. It continues on like this, for a total of 100% + 25% + 6.25% + 1.5625% + 0.390625% (and so on, though I won't bother with them). With just those 5 cycles, you get a total of 1.33203125%, and this continues to approach 1 + 1/3 the longer it goes.
So, each uranium ore can potentially provide 8 + (8/3) = 10 + 2/3 cells, with processing.
Now, assuming that one uranium cell can recharge 4 near-depleted cells, it would take (10 + 2/3)/4, or 32/12, or 8/3, or 2 + 2/3 uranium cells to recharge the full output of a single uranium ore into 10 + 2/3 uranium cells. Now, if you were to use 2 + 2/3 of those in order to recharge the next ore's cells, that leaves exactly eight uranium cells that can be used elsewhere.
So, as long as you aren't stupid, and use recharged cells to recharge later near-depleted cells (rather than making a fresh uranium cell each time), each uranium ore you gather can produce a full 8 cells to be used in whatever non-breeder you choose. As mentioned by people previously, that could be as high as 4.44, giving 35.52 for each uranium ore you mine.
Oh, and that's *completely* ignoring the energy produced by the breeder reactor.
If you use a 9k temperature breeder (if I understand correctly halving the recharge time, so a single uranium cell can charge two sets of four near-depleted cells), you get an extra 1 + 1/3 cells output, since you only need half as many.
So, with the extra output, and then assuming that each uranium cell in the breeder gets at least an efficiency of 1, you get ((9 + 1/3) * 4.44) + ((1 + 1/3) * 1) = 41.44 + 1.3333... = 42.7733 per ore of input.
Now, the very first time you run the breeder, you'll need to supply a uranium cell to bootstrap the whole process, but since you can use the output of that very first cycle to provide the next uranium cells for later cycles, all other uranium you use can have that nice 8x multiplier.
I see your point. If you reuse the recharged cells you lose less uranium to breeding. But on the other hand getting a almost 10x multiplier over your reactors efficiency sounds op. Still doubting who is correct. Its so easy to overlook something.
A few simple rules and you need so much maths just to get a eff formula XD
'Simulation' of a 5 efficiency reactor
1th Cycle:
So lets take a 5 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 40 million eu. So after this cycle have ended you end up with 41 million eu. Resulting in 20.5 efficiency. This is a increase of 4.1x in efficiency.
2th Cycle
1 of those cells turns into a near depleted cell. I takes 0.125 uranium cells to recharge this. This proces makes 0.25 million eu. The recharged cell will make 10 million eu. So thats 10.25 million eu total. Combined with the 1th cycle you now made 51.25 million eu at the cost of 1.125 uranium cells. This is a efficiency of 22,777778. This is a increase of 4.55556x in efficiency. Still nowhere near your 8.5x. Also this is a increase of 1.1111x over 20.5 eff.
3th Cycle
0.25 of those cells turn into a near depleted cell. It takes 0.03125 cells to recharge this. This proces makes 0.0625 million eu. The recharged cell will make 2.5 million eu. So thats 2.5625 million eu total. Combined with the 1th and 2th cycle you now made 53.8125 million eu at the cost of 1.15625 uranium cells. This is a efficiency of 23,27027027. This is a 4.654x increase in efficiency. Also this is a increase of 1.135x over 20,5 eff
4th Cycle
0.0625 of those cells turn into a near depleted cell. It takes 0.0078125 cells to recharge this. This proces makes 0.015625 million eu. The recharged cell will make 0.625 million eu. So thats 0.640625 million eu total. Combined with the 1th and 2th cycle you now made 54,453125 million eu at the cost of 1.1640625 uranium cells. This is a efficiency of 23,38926174. This is a 4.678x increase in efficiency. Also this is a increase of 1.140939597x over 20.5 eff
Endless cycles:
After running for endless cycles you will have 54+(2/3) million eu at the cost of 1+(1/6) uranium cells this is a efficiency of 23,428. This is a increase of 1+(1/7) over 20.5 efficiency.
So far:
The 25% of uranium cells turn into near depleted mechanic (@9000 heat) increase the efficiency by:
1+(1/7)x for a 3 efficiency reactor
1+(1/7)x for a 4 efficiency reactor
1+(1/7)x for a 5 efficiency reactor
I think its pretty safe to asume that this mechanic increases the output by 1/7 for any eff reactor. So now the formula for a 9000 heat breeder is:
Total eff@9000 heat = (8*Reactor eff+1)*(0,5+1/14)
I will update the 6000 heat formula later. Currently marked it red since it isnt completely correct (through its only about 10% off).
'Simulation' of a 3 efficiency reactor
1th Cycle:
So lets take a 3 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 24 million eu. So after this cycle have ended you end up with 25 million eu. Resulting in 12,5 efficiency. This is a increase of 4.16x in efficiency.
2th Cycle
1 of those cells turns into a near depleted cell. I takes 0.125 uranium cells to recharge this. This proces makes 0.25 million eu. The recharged cell will make 6 million eu. So thats 6.25 million eu total. Combined with the 1th cycle you now made 30.25 million eu at the cost of 1.125 uranium cells. This is a efficiency of 13.444444 (ok i realise that the near depleted cell part of my formula isnt working correctly). This is a increase of 4.48x in efficiency. Still nowhere near your 8.5x. Also this is a increase of 1.075555556x over 12.5 eff.
3th Cycle
0.25 of those cells turn into a near depleted cell. It takes 0.03125 cells to recharge this. This proces makes 0.0625 million eu. The recharged cell will make 1.5 million eu. So thats 1.5625 million eu total. Combined with the 1th and 2th cycle you now made 31.8125 million eu at the cost of 1.15625 uranium cells. This is a efficiency of 13.75675676. This is a 4.59x increase in efficiency. Also this is a increase of 1.100540541x over 12.5 eff
4th Cycle
0.0625 of those cells turn into a near depleted cell. It takes 0.0078125 cells to recharge this. This proces makes 0.015625 million eu. The recharged cell will make 0.375 million eu. So thats 0.390625 million eu total. Combined with the 1th and 2th cycle you now made 32.203125 million eu at the cost of 1.1640625 uranium cells. This is a efficiency of 13.83221477. This is a 4.61x increase in efficiency. Also this is a increase of 1.106577181x over 12.5 eff
Endless cycles:
After running for endless cycles you will have 33+(1/3) million eu at the cost of 1+(1/6) uranium cells this is a efficiency of 14+(2/7). This is a increase of 4+(16/21) efficiency.
Also is a increase of 1+(1/7)x over 12,5 efficiency.
It takes 2 uranium to make and breed 8 near depleted cells to uranium cells. The ratio is 1:4. Its impossible to get 8 uranium cells from just 1 ore. You always need 1 uranium cell to breed. Like i told you 8 eff is the highest you can get with free breeding. You had 8.5x eff so it can never be correct. If that would be true breeding would be op 
So lets take a 5 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 40 million eu. So after this cycle have ended you end up with 41 million eu. Resulting in 20,5 efficiency. This is a increase of 4.1x in efficiency
So lets take a 4 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 32 million eu. So after this cycle have ended you end up with 33 million eu. Resulting in 16,5 efficiency. This is a increase of 4.125x in efficiency.
So lets take a 3 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 24 million eu. So after this cycle have ended you end up with 25 million eu. Resulting in 12,5 efficiency. This is a increase of 4.16x in efficiency.
So lets take a 2 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 16 million eu. So after this cycle have ended you end up with 17 million eu. Resulting in 8,5 efficiency. This is a increase of 4.25x in efficiency.
So lets take a 1 eff reactor with a max eff breeder. You get 4 uranium cells for every ore you find. The proces of breeding those 4 uranium cells makes 1 million eu. Now we put those 4 cells in the reactor. They produce 8 million eu. So after this cycle have ended you end up with 9 million eu. Resulting in 4,5 efficiency. This is a increase of 4.5x in efficiency.
There is no way that the fact that every uranium cell has 25% chance to turn into a near depleted cell will raise this efficiency to anywhere near your results. That would mean it has to double the efficiency i just calculated. Its just impossible. I dont know what you exactly did wrong but the answer cannot be correct.
Wait.. Isn't that 33% actually 25% ?
For 1 cycle its 25% but for endless cycles it 33+(1/3)%. Its because there is a 25% chance that a uranium cell turns into a near depleted cell but there is also a chance that this happens again with the same cell.
Also since you are only getting 8 near depleted cells per ore the absolute maximum efficiency boost (meaning with no uranium wasted during the breeding proces or using a 1 eff reactor) is 8x. It is not possible to go higher than this simply because you only get 8 cells per uranium ore. immibis had like 8,5x efficiency boost.
huh? 38 eff? Thats seems extremely high. I think you maths could be a bit off. Why the x4/3? How you get over 10 near depleted cells from 1 ore? You get only 8 then that ore is gone.
EDIT: I see what you done now. You use the fact that 33% of your uranium get turned back into near depleted cells and you apply that to the 8 near depleted cells you get from 1 ore making 10+( 1/3). Then later you apply this 33% again resulting in a double dip which also explains the extremely high eff you had.
Well with overclocker upgrade i see a basic system for more modules. Am i right? XD. Would be pretty cool i guess.
EDIT: i see there already 3 modules
Iam suprised there so few comments XD
Apparently
'Lastly, a word of warning: Whilst performing cropxperiments, you may encounter sorts of weed, capable of spreading across your crops whilst destroying and assimilating previous plants.
Instant removal of weed and continous observation of (exspecially empty) crops is advised.
As well, be cautious of new breeds. Currently unknown, there may be rare plants capable of inflicting serious or even lasting damage to your body or mental state. Unless our storage of FAP is low, we recommend all employees to exercise caution when breeding new crops.'
Other than that i dont see much explaination about 'unknown crops'.
http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…1501521s1r11r10
According to the Reactor planner its a
Mark-1-0 EC
First core made btw
http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner.html?3u0e50e5c0=1p10101001501521s1r11r10</a>
Is cheaper and does exactly the same. Its already in the list.
Further if you really want max eu output cheap then you can use this design:
http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…1501521s1r11r10
The formula can easily be changed to work with any temperature and any breeding setup. You just need to know how many uranium cells and eu you get per uranium ore from the breeding. Havent really tested the formula so iam not 100% i didnt made some error somewhere in the calculations. I do not have much time to do this since iam rather busy with school lately.
If you just want to know the formula scroll down if you want to see whats behind it read it all.
So now some maths on breeder efficiency. Let say we take this design: http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…114010101001019 (note: this is the max eff you can get with a breeder)
Whats the max efficiency boost in percents @6000 and @9000 heat taking into account breeders also produce eu.
1 uranium ore = 8 Near depleted cells
@6000 heat
To get the reactor started we need 1 uranium cell and 4 depleted cells. This cost 1,5 uranium in total. In return you get 4 uranium cells and 2million eu. Do note the fact breeder always nets a 2 million eu for each cycle means lower efficiency reactors will get a higher boost in %.
Lets say we use this reactor to make eu:
http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…7ps011111101110
Specs:
3 eff = 6 million eu/uranium cell
4 uranium cells/cycle (ok this does not change the maths but easier example)
With the breeder you make 4 uranium with 1,5 uranium ore. This means the 1,5 uranium ore made 4*6+2= 26 million eu. This means for each uranium ore you find you get 17,33 million eu. This is a efficiency of 8,67.
BUT as most here will know uranium cells have 25% chance to turn into a near depleted uranium cell. So the actual efficiency is even higher. Whats the worth of 1 near depleted uranium cell? Its 1/8 of a uranium ore. In this example its 17,33/8 = 2,16625 million eu. Its easy to figure out what the gain is after 1 cycle. But whats the gain after 10 cycles? or 100 cycles, or endless cycles? The chance that a uranium cell turns into a near depleted cell is 25%. The chance this happens again with the same uranium cell is (0.25^2)*100
= 6,25%. The formula to calculate the chance is 0.25^cycle. After 2 cycles the gain is (0.25^1 + 0.25^2)*100 = 31,25%. Now we want to know what the gain is after x cycles. As you repeat this with more cycles the gain will come closer and closer to 33,33333333%. Thus the gain is 33,33333333%. So this gives us a extra 2,16625/3 = 0,722 million eu or 0.36 more eff making the total eff 9.03.
In the end the formula for total eff @6000 heat is this: Total eff = ((8*Reactor eff+2)/1,5)*(0.5+(1/48 ))
If you want eff boost in percents do this (Total eff/Reactor eff)*100
@9000 heat
This formula is almost the same but you only need 1 uranium ore to make 4 uranium cells thus the 1,5 in the formula changes to 1. Also the uranium cell used to breed only produces half the eu per breeding cycle (9000 heat means half the time). So the formula is now: Total eff = (8*Reactor eff+1)*(0.5+(1/48 )). In this example it would mean total eff is 13,02.
If you want eff boost in percents do this (Total eff/Reactor eff)*100
I hope i didnt do any miscalculations but the answers seem to be realistic.
Conclusion:
Total eff@6000 heat = ((8*Reactor eff+2)/1,5)*(0.5+(1/48 ))
Total eff@9000 heat = (8*Reactor eff+1)*(0,5+1/14)
Eff boost in percent = (Total eff/Reactor eff)*100
So now some theory on breeder efficiency. Let say we take this design: http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…114010101001019 (note: this is the max eff you can get with a breeder)
Whats the max efficiency boost in percents @6000 and @9000 heat taking into account breeders also produce eu.
1 uranium ore = 8 Near depleted cells
@6000 heat
To get the reactor started we need 1 uranium cell and 4 depleted cells. This cost 1,5 uranium in total. In return you get 4 uranium cells and 2million eu. Do note the fact breeder always nets a 2 million eu for each cycle means lower efficiency reactors will get a higher boost in %.
Lets say we use this reactor to make eu:
http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…7ps011111101110
Specs:
3 eff = 6 million eu/uranium cell
4 uranium cells/cycle (ok this does not change the maths but easier example)
With the breeder you make 4 uranium with 1,5 uranium ore. This means the 1,5 uranium ore made 4*6+2= 26 million eu. This means for each uranium ore you find you get 17,33 million eu. This is a efficiency of 8,67.
BUT as most here will know uranium cells have 25% chance to turn into a near depleted uranium cell. So the actual efficiency is even higher. Whats the worth of 1 near depleted uranium cell? Its 1/8 of a uranium ore. In this example its 17,33/8 = 2,16625 million eu. Its easy to figure out what the gain is after 1 cycle. But whats the gain after 10 cycles? or 100 cycles, or endless cycles? The chance that a uranium cell turns into a near depleted cell is 25%. The chance this happens again with the same uranium cell is (0.25^2)*100
= 6,25%. The formula to calculate the chance is 0.25^cycle. After 2 cycles the gain is (0.25^1 + 0.25^2)*100 = 31,25%. Now we want to know what the gain is after x cycles. As you repeat this with more cycles the gain will come closer and closer to 33,33333333%. Thus the gain is 33,33333333%. So this gives us a extra 2,16625/3 = 0,722 million eu or 0.36 more eff making the total eff 9.03.
In the end the formula for total eff @6000 heat is this: Total eff = ((8*Reactor eff+2)/1,5)*(0.5+(1/48 ))
If you want eff boost in percents do this (Total eff/Reactor eff)*100
@9000 heat
This formula is almost the same but you only need 1 uranium ore to make 4 uranium cells thus the 1,5 in the formula changes to 1. So the formula is now: Total eff = (8*Reactor eff+2)*(0.5+(1/48 )). In this example it would mean total eff is 13,54.
If you want eff boost in percents do this (Total eff/Reactor eff)*100
I hope i didnt do any miscalculations but the answers seem to be realistic.
I don't get it - how can a max number of cycles be less,than 1 and not have a critical state?
http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…1501521s1r11r10
(not suggesting this setup,just to show this strange..thing)
It takes melting parts in account too. There is a part that melts at 0.62 of the cycle thus it can run only 0.62 cycles.
Can you show breeder-version?
Well,one HD,okay..now-better.
as you can see-my setup goes with 1HD/2CS,in other reactors its ~1HD/3CS,I don't remember which setup it was-guess,have to look upWell,yeah - it builds up heat pretty quick,that's supposed to be good,right?
It should build up heat in a reasonable time without melting any parts. At 908 seconds parts start to melt long before reactor heat reached 9k heat.
http://www.talonfiremage.pwp.blueyonder.co.uk/reactorplanner…7ps011111101110
This one got 3 eff can run alot longer and has over twice the effective eu/tick output. Its also cheaper and has a possibility to run with a breeder making the eff way higher (i think 8 eff or so)
Maybe if you improve yours so that it can be a good alternative to this one i may ad it.
Your breeder heat configuration melts parts after 908 seconds
It could be because it rounds numbers or something i dunno. But the diference between 24,94 and 25 is noticable. Its the diference between a crater and a intact reactor after 25 cycles. Not that anyone will actualy run 25 cycles in a row with instant refuel through. Takes almost 70 hours to run 25 cycles. Before this reactor blows up some update prolly fucked up your world.
Hybrid reactors will never even get near the efficiency of a separate reactor and breeder. We tried that with CASUC long time ago.
Also i found the error in my breeder. Starting heat was set to 9000 which explains the melting after 472 secs. Fixed it it should now work properly.
Not really its just that the planner fails at calculating when components melt. According to the planner first components start to melt after 472 secs which is impossible since there is only 25000 heat by then and its evenly distributed across all hd's and cooling cells.
