Tutorial:Solar power math

In this tutorial we will properly quantify the amount of solar panels and accumulators needed and the proper ratio that is needed between the two buildings.

Equation symbols and units

The game uses SI units and reflects how they are used in the real world. Energy $E$ and power $P$ are often used interchangeably, but power is the first time derivative of energy, i.e. how much energy is used or produced per second. Power can be used to get machines moving, as they perform work $W$. A type of work can be lifting some mass against gravitational acceleration for a set distance. Usually there is always a loss of usable energy due to thermal energy being released, but Factorio does not take such losses into consideration. Lastly there is the charge in accumulators. Charge traditionally has the unit Coulomb and the energy in a capacitor depends on the charge and the voltage across it, but as Factorio also doesn't consider equipment using different voltages, a the charge $A$ of a factorio-accumulator can be defined to use this combined value.

Therefore the important equation symbols for energy are as follows:

• $E$ ... energy, in $\mathrm {J}$ Joule, also often written in $\mathrm {kW} \,\mathrm {h}$ kilowatt-hours, where $1\,\mathrm {kW} \,\mathrm {h} =3\,600\,000\mathrm {J} =3.6\,\mathrm {MJ}$.
• $W$ ... performed work, in $\mathrm {kg} \,{\tfrac {\mathrm {m} }{\mathrm {s} ^{2}}}\,\mathrm {m}$ or $\mathrm {N} \,\mathrm {m}$ Newton-meter which is equal to $\mathrm {J}$ Joule
• $A:={\tfrac {1}{2}}Q\cdot U^{2}$ ... accumulator charge, in $\mathrm {J}$ Joule

And the important equation symbols for power are as follows, with dot-notation for the time derivative:

• $P_{\text{S}}:={\tfrac {\mathrm {d} }{\mathrm {d} t}}E={\dot {E}}$ ... supplied power, in $\mathrm {W}$ Watt, or ${\tfrac {\mathrm {J} }{\mathrm {s} }}$ Joule per second.
• ${\dot {W}}:={\tfrac {\mathrm {d} W}{\mathrm {d} t}}$ ... ongoing mechanical work, in $\mathrm {W}$ Watt
• ${\dot {A}}:={\tfrac {\mathrm {d} A}{\mathrm {d} t}}$ ... change in accumulator charge, in $\mathrm {W}$ Watt

Energy can't be generated or destroyed, only converted into a different form. This means that the balance of energy must be upheld, and therefore its rate of change must be zero. If we subtract from the supplied power both the ongoing mechanical work and the rate at which accumulators are charging (therefore taking energy from the grid), the result will be zero. As such we can write

$P_{\text{S}}-{\dot {W}}-{\dot {A}}=0$.

The player will not have to worry about overproduction, as any power generating equipment will cap its output automatically. Only too little production will no longer provide machines with the amount of power that they desire, causing them to throttle their own power demand and working slower. Therefore in almost all cases these rates will change over time, where solar panels produce energy dependant on the time of day, while machines equally draw more or less power depending on how much they are used. For example, inserters only draw their power when they swing their arms, while only using a tiny fraction of it on standby. As such we introduce notation for various statistically relevant values.

• Average value over an interval $T$ is annotated with a bar above: ${\bar {x}}={\frac {1}{T}}\int _{t}^{t+T}x(t)\mathrm {d} t$
• Peak value with an upward facing chevron: ${\hat {x}}=\mathrm {max} {\big (}x(t){\big )}$

Average output of 1 solar panel

While only the peak output of a solar panel can be easily be told at ${\hat {P}}=60\,\mathrm {kW}$, its exact power output over the day changes with available daylight. Thankfully, if we install some accumulators, they can do the integration for us.

${\begin{aligned}0&=P(t)-{\dot {A}}(t)\\{\dot {A}}(t)&=P(t)\\A(T)&=\int _{0}^{T}P(t)\,\mathrm {d} t\\\end{aligned}}$

In an experiment with one solar panel and 4 accumulators, we can see that one solar panel will produce 17.6 MJ of energy per day.

Due to rounding, this could be anywhere between 17.55 MJ and 17.65 MJ, therefore the experiment can be repeated over multiple days.

day number	accumulator charge	energy produced per day
1	17.6 MJ	17.6 MJ
2	35.3 MJ	17.65 MJ
3	52.9 MJ	17.633 MJ
4	70.6 MJ	17.65 MJ
5	88.2 MJ	17.64 MJ
6	106 MJ	17.666 MJ

Further experiments show that the true number will swing around and approach 17.64 MJ

${\begin{aligned}A(T)&=17\,640\,\mathrm {kJ} \\{\bar {P}}&={\frac {\int _{0}^{T}P(t)\,\mathrm {d} t}{T}}={\frac {A(T)}{T}}={\frac {17\,640\,\mathrm {kJ} }{420\,\mathrm {s} }}=42\,\mathrm {kW} \end{aligned}}$

A day on Nauvis lasts 7 minutes, or $T=420\,\mathrm {s}$, which means that each solar panel provides about 42 kW of power on average.

Solarpanel output as a function

For further calculations we assume the power output function of a solar panel looks as follows:

${\begin{aligned}P(t)&={\begin{cases}{\tfrac {60}{84}}t&0\,\mathrm {s} \leq t<{\frac {5\,040}{60}}\,\mathrm {s} \\60&{\frac {5\,040}{60}}\,\mathrm {s} <t\leq {\frac {17\,640}{60}}\,\mathrm {s} \\60-{\tfrac {60}{84}}\cdot (t-{\frac {17\,640}{60}}\,\mathrm {s} )&{\frac {17\,640}{60}}\,\mathrm {s} <t\leq {\frac {22\,680}{60}}\,\mathrm {s} \\0&{\frac {22\,680}{60}}\,\mathrm {s} <t\leq {\frac {25\,200}{60}}\,\mathrm {s} \end{cases}}\\\int _{0}^{T}P(t)\,\mathrm {d} t&=17\,640\,\mathrm {kJ} \end{aligned}}$

Knowing how much power a solar panel provides on average we can construct a fitting load to test how much max accumulator charge ${\hat {A}}$ is needed.

In a second experiment we take two solar panels on separate grids and put a load of ${\tfrac {17\,600\,\mathrm {kJ} }{420\,\mathrm {s} }}\approx 41.9\,\mathrm {kW}$ on one, and one with ${\tfrac {17\,640\,\mathrm {kJ} }{420\,\mathrm {s} }}=42\,\mathrm {kW}$, which can be achieved with...

• 8 Assembly machines 2 (5 kW per unit), and
• 3/4 fast inserters (500 W per unit), and
• 1/0 regular inserters (400 W per unit).

To that we add a generous 100 accumulators and observe the energy in the system.

Experimental values for the 41.9 kW grid show the first peak has a height of about 4.24 MJ, with each consecutive peak growing higher as the true output of a solar panel is indeed more than 41.9 kW.
Experimental values for the 42 kW grid show each peak reaching about 4.24 MJ, but falling short for 5 seconds if it has to start from no charge at all. This may be due to the fact that machines start to behave differently when power is low. However, if we start the experiment with some slight pre-existing charge, the system will be stable.

Accumulators per solar panel

4.24 MJ is enough to fill 0.848 accumulators as they can hold a charge of 5 MJ each. Conservatively we can therefore set 0.85 accumulators per solar panel, or a ratio of 20 solar panels to 17 accumulators. A less conservative approach is to use 0.84 accumulators per solar panel, or a ratio of 25 solar panels to 21 accumulators.

Analytic solution: 2646/3125

We can set the previously defined power function as the rate of change of the accumulator and subtract it's average output as the load which the setup should be able to continuously supply. First we check when the power reaches 42 kW

${\begin{aligned}42&={\frac {60}{84}}\cdot t_{\text{start}}\\t_{\text{start}}&={\frac {3528}{60}}\end{aligned}}$

Then we shift the function for power in time and reduce it by the 42 kW so that the function will start at zero.

${\begin{aligned}{\dot {A}}(t)&=P(t-{\tfrac {3528}{60}})-{\hat {P}}={\begin{cases}{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}\cdot t&0\,\mathrm {s} \leq t<{\frac {1\,512}{60}}\,\mathrm {s} \\60\,\mathrm {kW} -42\,\mathrm {kW} &{\frac {1\,512}{60}}\,\mathrm {s} <t\leq {\frac {14\,112}{60}}\,\mathrm {s} \\60\,\mathrm {kW} -{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}\cdot \left(t-{\frac {14\,112}{60}}\,\mathrm {s} \right)-42\,\mathrm {kW} &{\frac {14\,112}{60}}\,\mathrm {s} <t\leq {\frac {19\,152}{60}}\,\mathrm {s} \\-42\,\mathrm {kW} &{\frac {19\,152}{60}}\,\mathrm {s} <t\leq {\frac {21\,680}{60}}\,\mathrm {s} \\{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}\cdot (t-{\frac {21\,672}{60}}\,\mathrm {s} )-42\,\mathrm {kW} &{\frac {21\,672}{60}}\,\mathrm {s} <t\leq {\frac {25\,200}{60}}\,\mathrm {s} \end{cases}}\\A(T=420\,\mathrm {s} )&=\int _{0}^{T}{\dot {A}}(t)\,\mathrm {d} t=0\,\mathrm {J} \end{aligned}}$

To find the critical point in time $t_{\text{crit}}$ at which the accumulator charge reaches it's peak ${\hat {A}}=A(t_{\text{crit}})$, we need to see when it's derivative passes zero. This happens in it's 3rd section.

${\begin{aligned}{\dot {A}}(t_{\text{crit}})=0&=60\,\mathrm {kW} -{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}\cdot \left(t_{\text{crit}}-{\frac {14\,112}{60}}\,\mathrm {s} \right)-42\,\mathrm {kW} \\t_{\text{crit}}&={\frac {(60\,\mathrm {kW} -42\,\mathrm {kW} )}{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}}+{\frac {14\,112}{60}}\,\mathrm {s} ={\frac {15\,624}{60}}\,\mathrm {s} =260.4\,\mathrm {s} \\\end{aligned}}$

${\begin{array}{rcccccc}A(t_{\text{crit}})=\displaystyle \int _{0}^{t_{\text{crit}}}{\dot {A}}(t)\,\mathrm {d} t&=&\displaystyle \int _{0\,\mathrm {s} }^{{\frac {1\,512}{60}}\,\mathrm {s} }{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}\cdot t\,\mathrm {d} t&+&\displaystyle \int _{{\frac {1\,512}{60}}\,\mathrm {s} }^{{\frac {14\,112}{60}}\,\mathrm {s} }60\,\mathrm {kW} -42\,\mathrm {kW} \,\mathrm {d} t&+&\displaystyle \int _{{\frac {14\,112}{60}}\,\mathrm {s} }^{{\frac {15\,624}{60}}\,\mathrm {s} }60\,\mathrm {kW} -{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}\cdot \left(t-{\frac {14\,112}{60}}\,\mathrm {s} \right)-42\,\mathrm {kW} \,\mathrm {d} t\\&=&{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}\cdot {\frac {1}{2}}\cdot \left({\frac {1\,512}{60}}\,\mathrm {s} \right)^{2}&+&\left(60\,\mathrm {kW} -42\,\mathrm {kW} \right)\cdot \left({\frac {14\,112}{60}}\,\mathrm {s} -{\frac {1\,512}{60}}\,\mathrm {s} \right)&+&\left(60\,\mathrm {kW} -42\,\mathrm {kW} \right)\cdot \left({\frac {15\,624}{60}}\,\mathrm {s} -{\frac {14\,112}{60}}\,\mathrm {s} \right)-{\frac {60\,\mathrm {kW} }{84\,\mathrm {s} }}\cdot {\frac {1}{2}}\cdot \left({\frac {15\,624}{60}}\,\mathrm {s} -{\frac {14\,112}{60}}\,\mathrm {s} \right)^{2}\\&=&{\frac {1\,134}{5}}\,\mathrm {kJ} &+&3\,780\,\mathrm {kJ} &+&{\frac {1\,134}{5}}\,\mathrm {kJ} \\&=&{\frac {21\,168}{5}}\,\mathrm {kJ} \\&=&4\,233.6\,\mathrm {kJ} \end{array}}$

The true perfect ratio for solar panels to accumulators therefore turns out to be...

${\frac {{\tfrac {21\,168}{5}}\mathrm {kJ} }{5\,000\,\mathrm {kJ} }}={\frac {2\,646}{3\,125}}=0.846\,72$

It takes 0.84672 accumulators per solar panel, or a ratio of 2646 accumulators to 3125 solar panels. Speaking conservatively we can take a higher ratio of 0.85 with 17 accumulators to 20 solar panels.

Best ratios

These ratios approach the one we got from the analytic solution, but to get more accurate, they need a larger number of individual equipment to do so.

more accumulators than necessary

accumulators	solar panels	ratio
1	1	1
6	7	0.857142857
17	20	0.85
28	33	0.848484848
39	46	0.847826087
50	59	0.847457627
61	72	0.847222222
72	85	0.847058824
83	98	0.846938776
94	111	0.846846847
105	124	0.846774194
221	261	0.846743295
337	398	0.846733668
453	535	0.846728972
569	672	0.84672619
685	809	0.846724351
801	946	0.846723044
917	1083	0.846722068
1033	1220	0.846721311
1149	1357	0.846720707
1265	1494	0.846720214
2646	3125	0.84672

fewer accumulators than necessary

accumulators	solar panels	ratio
1	2	0.5
2	3	0.666666667
3	4	0.75
4	5	0.8
5	6	0.833333333
11	13	0.846153846
116	137	0.846715328
1381	1631	0.846719804
2646	3125	0.84672

Testing ratios

The most commonly recommended ratio is 21/25 or 0.84, let's see how it fairs against 17/20 or 0.84, and then check out ratios for 0.8467 and 0.8469. And for fairness, these tests start off with fully charged accumulators before a load is applied.

fell 0 kJ accumulator charge for 18 seconds

lowest recorded accumulator charge: 959 kJ (11.282 kJ per accumulator)

lowest recorded accumulator charge: 107 kJ (1.289 kJ per accumulator)

fell 0 kJ accumulator charge for half a second

The tests support the analytic solution. The ratio for 0.8467 barely falls short of being able to sustain 42 kW of power, while 0.8469 almost fully discharges to 0.025%.