Oil and Perpetual Steam Energy

[Solar Heretic]

First off, as I haven't fully read through the forum, I may be repeating what someone else has calculated, in which case, I apologize. However, given that the wiki "Oil" and "Energy" pages do not mention what I will do, I believe I am in the clear.

As some have noticed, sunlight is both infinite and reliable, making some people choose solar networks when they have a chance to upgrade. However, with the new "no solar" achievement, I did some calculations to see if it was possible to make a perpetual energy machine using just steam engines and solid fuel. The short answer is Yes.

Before I go to heavy into the math, I'll explain some assumptions and notations:
- All resource will instantly and magically teleport to where it needs to be.
- All pumpjacks run at 10% efficiency, to signify long term usage.
- Boilers run at half efficiency, which is accounted for at the total calculation slot
- No modules
- System will be based off of 200 pumpjacks, as the decimals are easy to deal with.

Without further ado, the Math:

Advanced Processing, Light Oil conversion only

Code: Select all

200 x Pumpjacks
- produces 20 C. Oil/s
- 18 MW used

10 x Oil Refinery (Advanced Oil Processing)
- consumes 20 C. Oil/s
- produces 2 Heavy/s, 9 Light/s, 11 Petro/s
- 4.2 MW used

27 x Chemical Plant (Advance, Light-to-Fuel only)
- consumes 9 Light/s
- produces 9 Solid Fuel/s (225 MJ/s)
- 5.67 MW used

Total Gain = 112.5 MW (post-burn), and some heavy/petrol (powers 22 1/14/10 steam config)
Total Loss = 27.87 MW
Net Gain =  84.63 MW + H & P (16.5 steam config)

Ergo, each pump has a net energy surplus of 0.42315 MW using this method.

Basic Processing, Light Oil conversion only

Code: Select all

200 x Pumpjacks
- produces 20 C. Oil/s
- 18 MW used

10 x Oil Refinery (Basic Oil Processing)
- consumes 20 C. Oil/s
- produces 8 Heavy/s, 6 Light/s, 8 Petro/s
- 4.2 MW used

18 x Chemical Plant (Advance, Light-to-Fuel only)
- consumes 6 Light/s
- produces 6 Solid Fuel/s (150 MJ/s)
- 3.78 MW used

Total Gain = 75 MW (post-burn), and some heavy/petrol (powers 14.7 1/14/10 steam config)
Total Loss =  25.98 MW
Net Gain = 49.02 WJ + H & P (9.6 steam config)

Ergo, each pump has a net energy surplus of 0.2451 MW using this method.

Advance Processing, Total Conversion

Code: Select all

200 x Pumpjacks
- produces 20 C. Oil/s
- 18 MW used

10 x Oil Refinery (Advanced Oil Processing)
- consumes 20 C. Oil/s
- produces 2 Heavy/s, 9 Light/s, 11 Petro/s
- 4.2 MW used

2.5 x Chemical Plants (Heavy to Light)
- consumes 2 Heavy/s
- produces 1.5 Light/s
- 0.525 MW uses

31.5 x Chemical Plants (Light to Fuel)
- consumes 10.5 Light/s
- produces 10.5 Solid/s (262.5 MJ/s)
- 6.615 MW used

16.5 x Chemical Plants (Petro to Fuel)
- consumes 11 Petro/s
- produces 5.5 Solid/s (137.5 MJ/s)
- 3.465 MW used

Total Gain = 200 MW (post-burn,powers 39.2 1/14/10 steam config)
Total Loss =  32.805 MW
Net Gain = 167.195 MW (32.8 steam config)

Ergo, each pump has a net energy surplus of 0.835975 MW using this method.

Now, how does this compare to solar panels. Besides the smog blasting of the setup, the only other thing to compare is the space efficiency. For the calculations, I will be using the 20 accumulator/24 solar/1 MW ratio for powering the entire day. This leads to a space usage of 296 sq.units/MW, or 49,489.72 sq.units for 167.195 MW. For the oil setup, I will only be using the production buildings above, the 1/14/10 setup, and 14 inserters for each setup. Using this, at max conversion, the setup only occupies 8,521.3 sq.units for the same amount of power. This means that the system is 5.8078 times energy dense then the solar variety, without taking account the efficiency modules that can be placed.

So what did you learn:
- 1 fluid unit of oil can be converted up-to 20 MJ, though only 10 MJ due to 50% boiler efficiency.
- The solid fuel system can produce net energy and still get material byproducts without modules.
- At max conversion, solid fuel network is about 5.8 times energy dense than the solar version.

[vanatteveldt]

Yes, oil is a good renewable energy source, in Factorio at least. Of course, it becomes more interesting if you can boost your pumpjacks with speed modules and your refineries/cracking with productivity and speed+efficiency in beacons.

This thread might interest you: viewtopic.php?t=7052, which also links to this poll: viewtopic.php?p=145550#p145550

[Skev]

Could have just asked

Some people (like me) have played for hundreds of hours without placing solar panels... good job working out all the numbers though!

[vanatteveldt]

Well, if I can just ask

Assuming only depleted oil sources and 1 pumpjack per beacon, and assuming level 3 modules of the appropriate type in all relevant buildings, how many pumpjacks and refinery+processing infrastructure do I need per MW of generated energy (on top of energy consumption of the infrastructure)

[Solar Heretic]

Thanks for the links. The second one was particularly useful as it help me notice the 50% boiler efficiency, which I have now taken account of in the calculations. As for not asking, it was mostly me being bored and wanting to do some hard number crunching math.

Well, if I can just ask

Assuming only depleted oil sources and 1 pumpjack per beacon, and assuming level 3 modules of the appropriate type in all relevant buildings, how many pumpjacks and refinery+processing infrastructure do I need per MW of generated energy (on top of energy consumption of the infrastructure)

As the main power draw in the system is from the pumpjacks, increasing that usage by 6 times with individual beacons alone seems rather wasteful (haven't done the math for it yet, kind of want to sleep). However, each production building has 2 module slots (making the math easy). To max the net energy, speed modules should be used, which using T3, should give a net of 321.268 MW for 200 pumpjacks, or just 1.61 MW for each pumpjack under the advance total conversion. This leads to a ratio of:

Code: Select all

1 net MW (post-burn, or hot water) = 
~ 0.6211 pumpjack
~ 0.0311 Oil Refinery
~ 0.1568 Chemical Plants

[vanatteveldt]

Your answer seems to go against the advice of viewtopic.php?p=145550#p145550, which recommends speed modules in pumpjacks (plus beacons) and a combination of efficiency and speed in processing beacons.

The reason I was asking for information is that that post falls short of giving a list of needed kit per MW, and skev offered to answer all my questions

(BTW, increasing pump electricity usage is not that strange, as long as the MW yield per extra oil extracted is high enough to compensate for the extra energy yield. OP claims that 1 oil -> 10MJ, which will be higher with modules. So, every 100% speed increase (at 0.1/s flow rate) gives an extra MW. Since pumpjack energy consumption is 0.09MW, the extra 140% energy usage is much less than the extra energy from the oil. Of course, it will be more efficient to build another pumpjack, but you have limited oil sites in most games...)

[bobucles]

So how many pumps does a speed 3 beacon have to cover to be worth it?

[DaveMcW]

So how many pumps does a speed 3 beacon have to cover to be worth it?

At 2 pumps, you pay 50% of the profit to run the beacon.
At 1 pump, you pay 100% of the profit to run the beacon.

If you use XKnight's productivity + efficiency beacon scheme farther down the line, you could even come out ahead with 1 pump per beacon.

[XKnight]

Small addition to that post:

There was discussion on reddit about rocket fuel, so...

- assembler I/II/III + beacons: number of effect receivers - 8, beacons around - 8
- chemical plant + beacons: number of effect receivers - 8, beacons around - 8
- oil refinery + beacons: number of effect receivers - 4, beacons around - 8
- pumpjack + beacons: number of effect receivers - 1, beacons around - 8

{productivity3, speed3, efficiency3, speed3 in beacons, efficiency3 in beacons}
Energy net profit per solid fuel (I) - {0,0,0,0,0} 10710. kJ
Energy net profit per solid fuel (II) - {2,0,0,8,0} 12044.7 kJ
Energy net profit per solid fuel (III) - {4,0,0,8,0} 14365.6 kJ
Energy net profit per petrolium - {2,0,0,1,6} 8136.38 kJ
Energy net profit per light oil - {2,0,0,1,6} 16272.8 kJ
Energy net profit per light oil (cracking) - {2,0,0,1,6} 5972.44 kJ
Energy net profit per heavy oil - {2,0,0,1,6} 8136.38 kJ
Energy net profit per heavy oil (cracking) - {2,0,0,1,6} 14404. kJ
Energy net profit per crude oil - {2,0,0,1,6} 14546.6 kJ
Energy net profit per crude oil (advanced processing) - {2,0,0,1,6} 15483.3 kJ
Energy net profit per depletted pump. {0,2,0,8,0} One cycle - 788.328 kJ; Power - 4729.97 kW

[dragontamer5788]

Small addition to that post:

There was discussion on reddit about rocket fuel, so...

- assembler I/II/III + beacons: number of effect receivers - 8, beacons around - 8
- chemical plant + beacons: number of effect receivers - 8, beacons around - 8
- oil refinery + beacons: number of effect receivers - 4, beacons around - 8
- pumpjack + beacons: number of effect receivers - 1, beacons around - 8

{productivity3, speed3, efficiency3, speed3 in beacons, efficiency3 in beacons}
Energy net profit per solid fuel (I) - {0,0,0,0,0} 10710. kJ
Energy net profit per solid fuel (II) - {2,0,0,8,0} 12044.7 kJ
Energy net profit per solid fuel (III) - {4,0,0,8,0} 14365.6 kJ
Energy net profit per petrolium - {2,0,0,1,6} 8136.38 kJ
Energy net profit per light oil - {2,0,0,1,6} 16272.8 kJ
Energy net profit per light oil (cracking) - {2,0,0,1,6} 5972.44 kJ
Energy net profit per heavy oil - {2,0,0,1,6} 8136.38 kJ
Energy net profit per heavy oil (cracking) - {2,0,0,1,6} 14404. kJ
Energy net profit per crude oil - {2,0,0,1,6} 14546.6 kJ
Energy net profit per crude oil (advanced processing) - {2,0,0,1,6} 15483.3 kJ
Energy net profit per depletted pump. {0,2,0,8,0} One cycle - 788.328 kJ; Power - 4729.97 kW

Interesting. I'll have to double-check your math to fully trust it however. I'm kinda surprised that Efficiency3 actually helps with refineries, but it sorta makes sense. Each beacon saves 210kW per refinery (and with 4 refineries that's 840kW saved per beacon, with 480kW used. So net profit).

It should be noted that "compressing" to Rocket Fuel leads to a net profit with PM3 in Assembly Machines 3 (+40% productivity beats out the -25MJ loss per cycle). Do you think you can run that fancy program to figure out the best combination in AM3?

I'm betting PM3x4 + max speed beacons is the most efficient rocket fuel setup. But it can't hurt to be sure.

[XKnight]

Interesting. I'll have to double-check your math to fully trust it however. I'm kinda surprised that Efficiency3 actually helps with refineries, but it sorta makes sense. Each beacon saves 210kW per refinery (and with 4 refineries that's 840kW saved per beacon, with 480kW used. So net profit).

It should be noted that "compressing" to Rocket Fuel leads to a net profit with PM3 in Assembly Machines 3 (+40% productivity beats out the -25MJ loss per cycle). Do you think you can run that fancy program to figure out the best combination in AM3?

I'm betting PM3x4 + max speed beacons is the most efficient rocket fuel setup. But it can't hurt to be sure.

You are right, but I had to check all possible variants.
The most optimal setup is:
Energy net profit per solid fuel (III) - {4,0,0,8,0} 14365.6 kJ
Energy net profit per petrolium - {2,0,0,1,6} 8136.38 kJ
Energy net profit per light oil - {2,0,0,1,6} 16272.8 kJ
Energy net profit per heavy oil (cracking) - {2,0,0,1,6} 14404. kJ
Energy net profit per crude oil (advanced processing) - {2,0,0,1,6} 15483.3 kJ
Energy net profit per depletted pump. {0,2,0,8,0} One cycle - 788.328 kJ; Power - 4729.97 kW

[Solar Heretic]

So how many pumps does a speed 3 beacon have to cover to be worth it?

At 2 pumps, you pay 50% of the profit to run the beacon.
At 1 pump, you pay 100% of the profit to run the beacon.

If you use XKnight's productivity + efficiency beacon scheme farther down the line, you could even come out ahead with 1 pump per beacon.

It's actually slightly worse than that. If the beacon is not shared with any other pumps, then each beacon actually decreases the net gain:

Math

Code: Select all

B= beacons (2 x Speed3)
E= net power gain (MW)

      Usable Power gain     Pump power usage       Beacon power usage
E = 1*(1+0.5*B)             - 0.09*(1+0.7*B)          - 0.48*B
E = 0.91 - 0.043*B

Also, for the math with productivity modules, it ends up being a lower wattage gain. I can only assume that the setup in the other thread is due to a combination of not having enough pumps to power his setup, not liking idle converters, and wanting non-energy products more, or was tailored to a particular ratio of pump/refinery/chem plants. Just like the wiki notes, it is meant for getting the most out of each resource, which begins to fail for those in infinite supply.

Math

Code: Select all

E = net MW unmodded (assume non-negative)
P = MW produced/converted unmodded (non-negative)
C = MW consumed unmodded (non-negative)
Em = net MW modded w/ 1 x productivity.3 

E = P-C
E+C=P (for substitution later)

        Production   Usage     Productivity bonus
Em = 0.85*P        - 1.8*C       + 0.1*P
Em = 0.95*P - 1.8*C
Em = 0.95*(E+C) - 1.8*C
Em = 0.95*E - 0.85*C

Given that both variables are positive, you will always get a worse power per second production using productivity.

For direct efficiency modules, it only makes sense if there isn't something better (which there is in this case: speed). For beacons and efficiency3, you need to affect an infrastructure subsection totaling 0.96 MW just to recoup the beacon usage.

I haven't played around with the numbers when it comes to combining mods together, but I am under the assumption it'll fall under similar circumstances.

[XKnight]

It's actually slightly worse than that. If the beacon is not shared with any other pumps, then each beacon actually decreases the net gain:

Math, queen of all sciences...
E = base net gain (MW)
B = amount of beacons
E*(1 + 0.5*B) - 0.09*(1 + 0.7*B) - 0.48*B

Profit (B=0): E*(1 + 0.5*0) - 0.09*(1 + 0.7*0) - 0.48*0 = E - 0.09
Profit (B=1): E*(1 + 0.5*1) - 0.09*(1 + 0.7*1) - 0.48*1 = 1.5 E - 0.633
So, if 0.5*E > 0.633-0.09 = 0.5*E > 0.543 = E > 1.086 first beacon is profitable.
In our case E = 1.5483 MW, this means first beacon increases net profit.

Profit (B=2): E*(1 + 0.5*2) - 0.09*(1 + 0.7*2) - 0.48*2 = 2 E - 1.176
So, if 0.5*E > 1.176-0.633 = 0.5*E > 0.543 = E > 1.086 second beacon is profitable.
In our case E = 1.5483 MW, this means second beacon increases net profit.

...

Try to find mistake in your calculations

[Solar Heretic]

Actually, the only difference I can see in our formulas is what our E is. In your case, you denote it as net energy. For me, I used it as gross energy gain, which is why I just replaced it with 1 MW for 0.1 oil/s. And the more I look at the calculation, the more I feel like you are wrong on your end conclusion. I feel like you may have forgot your hypothesis and got confused at the end. Either that, or the oil to usable energy math I did in my original post was off.

Code: Select all

To be profitable, or break even:
B(1) >= B(0), where E=gross usable power gain (or 1 MW)

B(1) ?>= B(0)
E*(1 + 0.5*1) - 0.09*(1 + 0.7*1) - 0.48*1 ?>= E*(1 + 0.5*0) - 0.09*(1 + 0.7*0) - 0.48*0
1.5*E - 0.153 - 0.48 ?>= E - 0.09 - 0
1.5*E - 0.633 ?>= E - 0.09
1.5*E -E ?>= 0.633 - 0.09
0.5*E ?>= 0.543
0.5*1 [not] >= 0.543

Ergo, proof by contradiction, 1 beacon is less productive than no beacon.

[XKnight]

Actually, the only difference I can see in our formulas is what our E is. In your case, you denote it as net energy. For me, I used it as gross energy gain, which is why I just replaced it with 1 MW for 0.1 oil/s. And the more I look at the calculation, the more I feel like you are wrong on your end conclusion. I feel like you may have forgot your hypothesis and got confused at the end. Either that, or the oil to usable energy math I did in my original post was off.

My hypothesis? In this thread I didn't make any assumptions...
Probably we are talking about different initial conditions, in my case any modules may be used (that's why E=1.5483), and your case seems to be without modules at all.

If you don't have any modules available net profit will be E=0.9374.

Calculations

Energy net profit per solid fuel (I) - {} 10710. kJ
Energy net profit per solid fuel (II) - {} 10650. kJ
Energy net profit per solid fuel (III) - {} 10746. kJ <- all assemblers decrease net profit, so rocket fuel is useless
Energy net profit per petrolium - {} 5998. kJ
Energy net profit per light oil - {} 11996. kJ
Energy net profit per light oil (cracking) - {} 3718.67 kJ
Energy net profit per heavy oil - {} 5998. kJ
Energy net profit per heavy oil (cracking) - {} 8871. kJ
Energy net profit per crude oil - {} 8449.3 kJ
Energy net profit per crude oil (advanced processing) - {} 9374.2 kJ
Energy net profit per depletted pump. {} One cycle - 847.42 kJ; Power - 847.42 kW

If you want to use only speed3 modules + beacons

Calculations

modules = {speed3, speed3 in beacons};
Energy net profit per solid fuel (I) - {0,0} 10710. kJ
Energy net profit per solid fuel (II) - {0,0} 10650. kJ
Energy net profit per solid fuel (III) - {0,0} 10746. kJ <- still useless
Energy net profit per petrolium - {0,0} 5998. kJ
Energy net profit per light oil - {0,0} 11996. kJ
Energy net profit per light oil (cracking) - {0,0} 3718.67 kJ
Energy net profit per heavy oil - {0,0} 5998. kJ
Energy net profit per heavy oil (cracking) - {0,0} 8871. kJ
Energy net profit per crude oil - {0,0} 8449.3 kJ
Energy net profit per crude oil (advanced processing) - {0,0} 9374.2 kJ
Energy net profit per depletted pump. {2,0} One cycle - 829.42 kJ; Power - 1658.84 kW <- beacons are still useless, speed3 modules are used only directly in pumpjack

For speed3 + beacons for pumpjack with "number of effect receivers - 2, beacons around - 8":

Calculations

modules = {speed3, speed3 in beacons};
Energy net profit per solid fuel (I) - {0,0} 10710. kJ
Energy net profit per solid fuel (II) - {0,0} 10650. kJ
Energy net profit per solid fuel (III) - {0,0} 10746. kJ <- still useless
Energy net profit per petrolium - {0,0} 5998. kJ
Energy net profit per light oil - {0,0} 11996. kJ
Energy net profit per light oil (cracking) - {0,0} 3718.67 kJ
Energy net profit per heavy oil - {0,0} 5998. kJ
Energy net profit per heavy oil (cracking) - {0,0} 8871. kJ
Energy net profit per crude oil - {0,0} 8449.3 kJ
Energy net profit per crude oil (advanced processing) - {0,0} 9374.2 kJ
Energy net profit per depletted pump. {2,8} One cycle - 497.42 kJ; Power - 2984.52 kW <- now beacons are profitable

If you want to use only speed3 and efficiency3 modules + beacons (pumpjack: number of effect receivers - 1, beacons around - 8)

Calculations

modules = { speed3, efficiency3, speed3 in beacons, efficiency3 in beacons};
Energy net profit per solid fuel (I) - {0,0,0,0} 10710. kJ
Energy net profit per solid fuel (II) - {0,2,0,0} 11130. kJ
Energy net profit per solid fuel (III) - {1,3,0,0} 11182.8 kJ <- still useless
Energy net profit per petrolium - {0,2,0,0} 6199.6 kJ
Energy net profit per light oil - {0,2,0,0} 12399.2 kJ
Energy net profit per light oil (cracking) - {0,2,0,0} 4077.07 kJ
Energy net profit per heavy oil - {0,2,0,0} 6199.6 kJ
Energy net profit per heavy oil (cracking) - {0,2,0,0} 9274.2 kJ
Energy net profit per crude oil - {0,2,0,0} 8939.86 kJ
Energy net profit per crude oil (advanced processing) - {0,2,0,0} 9874.84 kJ <- not enough to start using speed beacons
Energy net profit per depletted pump. {2,0,0,0} One cycle - 879.484 kJ; Power - 1758.97 kW

But actually, even all tier-1 modules significantly increase your profit (pumpjack: number of effect receivers - 1, beacons around - 8).

Calculations

modules = { productivity1, speed1, efficiency1, speed1 in beacons, efficiency1 in beacons};
Energy net profit per solid fuel (I) - {0,0,0,0,0} 10710. kJ
Energy net profit per solid fuel (II) - {0,0,2,0,0} 11010. kJ
Energy net profit per solid fuel (III) - {1,0,3,0,1} 11412. kJ <- still useless
Energy net profit per petrolium - {1,0,1,0,3} 6186.59 kJ
Energy net profit per light oil - {1,0,1,0,3} 12373.2 kJ
Energy net profit per light oil (cracking) - {0,0,2,0,0} 4012.39 kJ
Energy net profit per heavy oil - {1,0,1,0,3} 6186.59 kJ
Energy net profit per heavy oil (cracking) - {2,0,0,0,5} 9711.13 kJ
Energy net profit per crude oil - {2,0,0,0,5} 9309.35 kJ
Energy net profit per crude oil (advanced processing) - {2,0,0,0,5} 10218.4 kJ
Energy net profit per depletted pump. {0,2,0,0,0} One cycle - 893.271 kJ; Power - 1250.58 kW <- without beacons

And the last one for all tier-2 (pumpjack: number of effect receivers - 1, beacons around - 8).

Calculations

modules = { productivity2, speed2, efficiency2, speed2 in beacons, efficiency2 in beacons};
Energy net profit per solid fuel (I) - {0,0,0,0,0} 10710. kJ
Energy net profit per solid fuel (II) - {0,0,2,0,0} 11130. kJ
Energy net profit per solid fuel (III) - {4,0,0,8,0} 12062.6 kJ <- still useless
Energy net profit per petrolium - {2,0,0,0,5} 6413.71 kJ
Energy net profit per light oil - {2,0,0,0,5} 12827.4 kJ
Energy net profit per light oil (cracking) - {0,0,2,0,0} 4219.81 kJ
Energy net profit per heavy oil - {2,0,0,0,5} 6413.71 kJ
Energy net profit per heavy oil (cracking) - {2,0,0,0,5} 10481.9 kJ
Energy net profit per crude oil - {2,0,0,0,5} 10216.7 kJ
Energy net profit per crude oil (advanced processing) - {2,0,0,0,5} 11101.3 kJ
Energy net profit per depletted pump. {0,2,0,0,0} One cycle - 986.377 kJ; Power - 1578.2 kW

Short summary:
without modules: 847.42 kW
with speed3: 1658.84 kW (speed in pumpjacks, beacons are useless everywhere)
with speed3 + beacon-2 (beacon per 2 pumpjacks): 2984.52 kW (these beacons should be used everywhere where 2 pumpjacks can be affected)
with speed3 + efficiency3 + beacon-1: 1758.97 kW (speed in pumpjacks, efficiency in other assemblers, beacons are useless everywhere)
with all tier-1: 1250.58 kW (speed in pumpjacks, pumpjacks only with speed beacon-4+, assemblers with productivity sometimes with efficiency, efficiency beacons near other assemblers, without rocket fuel)
with all tier-2: 1578.2 kW (speed in pumpjacks, pumpjacks only with speed beacon-2+, all assemblers with productivity, efficiency beacons near other assemblers, without rocket fuel)
with all tier-3: 4729.97 kW (speed in pumpjacks, pumpjacks with all possible speed beacons, all assemblers with productivity, efficiency sometimes speed beacons near other assemblers, with rocket fuel)

[Solar Heretic]

My hypothesis? In this thread I didn't make any assumptions...
Probably we are talking about different initial conditions, in my case any modules may be used (that's why E=1.5483), and your case seems to be without modules at all.

The term hypothesis was me mixing terminology from both statistics (null hypothesis) and proof-base math (statement). I'll try to avoid mixing them up for the remainder of my post.

If E denotes the modded net energy gain from pumpjacks w/ 2 x speed.3, then you're using the wrong formula for several reasons:
- Modules effects are completely additive regardless of the source. Your usage of the formula partly implies that the direct modules take effect first, then the beacons modifies off the new numbers, resulting in a partly multiplicative effect. The same applies to any combination of mods. (I tested in game to make sure of this)
- Using modded net energy for E is incorrect, as the growth of speed and "de-efficiency" is not equal (+50%/+70% respectively). They need to be split up to their base components (production and consumption), and evaluated from there.
- As a moot curiosity because it doesn't affect the math below due to the bullet directly above, how did you got E=1.5483 for the modded max speed pumps, as my net gain calculation is 1*2-0.09*2.4=1.784 MW?

If you put in the math into the new formula designed for the initially modded pumpjack, you still get the same result: the power gain drops at a constant rate for each additional beacon (0.043 MW/beacon for your initially modded max speed pumps).

Below is the math(s):

Code: Select all

E = base usable energy produced (0.1 oil/s, or 1 usable MW)
Em = modded usable energy produced, using speed3 only
B = number of beacons filled w/ Speed3

Em = E*(100% + direct mod boost + becaon mod boost * B) - 0.09*(100% + direct mod boost + beacon mod boost * B) - 0.48*B
Em = E*(1+1+0.5B) - 0.09*(1+1.4+0.7B) - 0.48B
Em = 2E + 0.5EB - 0.216 - 0.543B
Substitute E=1
Em = 2 + 0.5B - 0.216 - 0.543B
Em = 1.784 - 0.043B

B(0) = 1.784 MW
B(1) = 1.741 MW
B(2) = 1.698 MW
...


Alternatively, using calculus and proof by contradiction:
Em = 2E + 0.5EB - 0.216 - 0.543B
d.Em/d.B = 0.5E - 0.543
Second deriv(Em,B) = 0

Let's assume that my statement regarding that beacon were wasteful is wrong, then that means beacons either break even or produce a net of power...
d.Em/d.B >= 0
0.5E - 0.543 >= 0
Substitute E=1
0.5*1 - 0.543 >= 0
-0.043 >= 0
Because the last part is not true, this disproves that beacons either break even or produce a net power gain.

QED: Speed beacons are not efficient at any number.

[XKnight]

- Modules effects are completely additive regardless of the source. You're usage of the formula partly implies that the direct modules take affect first, then the beacons, resultuing in a partly multiplicative effect. The same applies to any combination of mods. (I tested in game to make sure of this)
- Using modded net energy is incorrect, as the growth of speed and "de-efficiency" is not equal (+50%/+70% respectively). They need to be split up to their base components (production and consumption), and evaluated from there.

- Nope, there is no multiplicative effect.
- They are splitted.

I don't want to manualy perform all calculations, but you have forced me to do this at least for two assemblers. So, please be very very attentive.

First part

modules = { productivity3, speed3, efficiency3, speed3 in beacons, efficiency3 in beacons};
The most optimal setup is:
Energy net profit per solid fuel (III) - {4,0,0,8,0} 14365.6 kJ

Modules: 4 x productivity3, 8 x speed3:
Productivity bonus: 4 x (-15% speed, +80% energy consumption, +10% productivity)
Speed bonus: 8 x (+50% speed, +70% energy consumption)

Sum of all modules:
Speed: 100% - 4 x 15% + 8 x 50% = 440%
Productivity: 140%
Energy: 100% + 4 x 80% + 8 x 70% = 980%

Base energy consumption for assembler III: 210kW
Modded energy consumption: 210kW * 980% = 2058kW
Beacons (8 beacons are used, each beacon is shared between 8 assemblers): 480kW * 8 / 8 = 480kW
Total energy consumption: 2058kW + 480kW = 2538kW
Base crafting speed for assembler III: 1.25
Modded crafting speed: 1.25 * 440% = 5.5
Base recipe production time (rocket fuel): 30 seconds
Real production time (according to crafting speed): 30 sec / 5.5 = 5.4545 sec
Energy required to complete one recipe: 2538kW * 5.4545 sec = 13843.63kJ
Produced items: 1.4 rocket fuel
Burning rocket fuel will give us 1.4 * 225000kJ * 0.5 (lost in boilers) = 157500kJ
Profit per cycle: 157500kJ - 13843kJ = 143657kJ
Used resources (10 solid fuel), 143657kJ/10 = 14365.7kJ per solid fuel.
Energy net profit per solid fuel (III) - {4,0,0,8,0} 14365.6 kJ. As predicted.

Second part (you are really the first man who convinced me to manually calculate so far):

Second part

modules = { productivity3, speed3, efficiency3, speed3 in beacons, efficiency3 in beacons};
The most optimal setup is:
Energy net profit per petrolium - {2,0,0,1,6} 8136.38 kJ

Modules: 2 x productivity3, 1 x speed3, 6 x efficiency3
Productivity bonus: 2 x (-15% speed, +80% energy consumption, +10% productivity)
Speed bonus: 1 x (+50% speed, +70% energy consumption)
Efficiency bonus: 6 x (-50% energy consumption)

Sum of all modules:
Speed: 100% - 2 x 15% + 1 x 50% = 120%
Productivity: 120%
Energy: 100% + 2 x 80% + 1 x 70% - 6 x 50% = 30% (min cap is 20%)

Base energy consumption for chemical plant: 90kW
Modded energy consumption: 210kW * 30% = 63kW
Beacons (7 beacons are used, each beacon is shared between 8 assemblers): 480kW * 7 / 8 = 420kW
Total energy consumption: 63kW + 420kW = 483kW
Base crafting speed for chemical plant: 1.25
Modded crafting speed: 1.25 * 120% = 1.5
Base recipe production time (petroleum to solid fuel): 3 seconds
Real production time (according to crafting speed): 3 sec / 1.5 = 2 sec
Energy required to complete one recipe: 483kW * 2 sec = 966kJ
Produced items: 1.2 solid fuel
1.2 solid fuel will give us 1.2 * 14365.7kJ = 17238.76kJ (14365.6363kJ is calculated in first section)
Profit per cycle: 17238.76kJ - 966kJ = 16272.76kJ
Used resources (2 petroleum), 16272.76kJ / 2 = 8136.38kJ per petroleum
Energy net profit per petrolium - {2,0,0,1,6} 8136.38 kJ. As predicted.

If you continue these calculations you will come to "Energy net profit per crude oil (advanced processing) - {2,0,0,1,6} 15483.3 kJ", but for me this is obvious.

[Solar Heretic]

Ok, I found the point of difference in all our math: the beacon sharing.

Your formula is based on sharing the beacon effect among as many machines as possible, effectively distributing the beacon burden across all machines. My math was based on if beacons were not shared, meaning that each pump had to pay the beacon cost on their own. This was primarily to answer the 1 pump-all beacon scenario listed below (to simulate scenarios where none of the oil spots were within beacon reach of each other). I do concur that as soon as the beacon gets shared, the system actually start producing more net energy.

Also, I prefer the full work being shown all the time as the math sometime holds details that is sometimes missed in summaries. That's how I roll I guess.

Quote

[quote="Solar Heretic"][quote="DaveMcW"][quote="bobucles"]So how many pumps does a speed 3 beacon have to cover to be worth it?[/quote]

At 2 pumps, you pay 50% of the profit to run the beacon.
At 1 pump, you pay 100% of the profit to run the beacon.

If you use XKnight's productivity + efficiency beacon scheme farther down the line, you could even come out ahead with 1 pump per beacon.[/quote]

It's actually slightly worse than that. If the beacon is not shared with any other pumps, then each beacon actually decreases the net gain:

...
Math

Code: Select all

B= beacons (2 x Speed3)
E= net power gain (MW)

      Usable Power gain     Pump power usage       Beacon power usage
E = 1*(1+0.5*B)             - 0.09*(1+0.7*B)          - 0.48*B
E = 0.91 - 0.043*B

Also, for the math with productivity modules, it ends up being a lower wattage gain. I can only assume that the setup in the other thread is due to a combination of not having enough pumps to power his setup, not liking idle converters, and wanting non-energy products more, or was tailored to a particular ratio of pump/refinery/chem plants. Just like the wiki notes, it is meant for getting the most out of each resource, which begins to fail for those in infinite supply.

Math

Code: Select all

E = net MW unmodded (assume non-negative)
P = MW produced/converted unmodded (non-negative)
C = MW consumed unmodded (non-negative)
Em = net MW modded w/ 1 x productivity.3 

E = P-C
E+C=P (for substitution later)

        Production   Usage     Productivity bonus
Em = 0.85*P        - 1.8*C       + 0.1*P
Em = 0.95*P - 1.8*C
Em = 0.95*(E+C) - 1.8*C
Em = 0.95*E - 0.85*C

Given that both variables are positive, you will always get a worse power per second production using productivity.

For direct efficiency modules, it only makes sense if there isn't something better (which there is in this case: speed). For beacons and efficiency3, you need to affect an infrastructure subsection totaling 0.96 MW just to recoup the beacon usage.

I haven't played around with the numbers when it comes to combining mods together, but I am under the assumption it'll fall under similar circumstances.[/quote]

[XKnight]

Ok, I found the point of difference in all our math: the beacon sharing.

Your formula is based on sharing the beacon effect among as many machines as possible, effectively distributing the beacon burden across all machines. My math was based on if beacons were not shared, meaning that each pump had to pay the beacon cost on their own. This was primarily to answer the 1 pump-all beacon scenario listed below (to simulate scenarios where none of the oil spots were within beacon reach of each other). I do concur that as soon as the beacon gets shared, the system actually start producing more net energy.

And again, beacon sharing is not used for pumpjack (number of effect receivers - 1), but it is still worth it.

- assembler I/II/III + beacons: number of effect receivers - 8, beacons around - 8
- chemical plant + beacons: number of effect receivers - 8, beacons around - 8
- oil refinery + beacons: number of effect receivers - 4, beacons around - 8
- pumpjack + beacons: number of effect receivers - 1, beacons around - 8

[Koub]

Don't forget to have a look at this old thread :
viewtopic.php?f=8&t=7052 : It's basically on the same topic.