Day, night, solar, and accumulators in pictures

[Permittivity]

If you want a denser, mathematical approach to this topic, you should check out this post.

Factorio has a day/night cycle that controls how much power your solar panels produce. For 50% of the day, your solar panels produce full power. For the next 20% of the day, the power produced linearly decreases to zero. You produce no power for 10% of the day, after which the power produced linearly increases back to full power over the remaining 20% of the day. That is represented by the black line in this plot:

Solar Output.png (9.18 KiB) Viewed 2583 times

I have divided the plot here into 100 smaller squares. The vertical divisions each represent one tenth of the solar panel's maximum power, while horizontal divisions represent one tenth of the game day. Energy is power multiplied by time, so each square represents a fixed amount of energy being produced. To find out how much energy a solar panel gives you over the course of one day, we just need to count the number of squares underneath the black line. Since it's much easier to count whole squares, we are going to cut out the orange triangles and then flip them over and slide them down to the blue locations. Now we count the grey and blue area. That is 70 squares, so a solar panel produces 70 blocks of energy each day.

Our factories tend to use about the same amount of power during the night as they do during the day. That means that a factory consuming 70 blocks of energy has usage that looks a lot more like this grey line:

Steady State.png (10.51 KiB) Viewed 2583 times

The tooltip for a solar panel tells us that a solar panel produces 60 kW of power. That means each of the ten vertical divisions represents 6 kW. Our grey line showing our factory's consumption is seven blocks high, so one solar panel will supply a factory that draws 7 * 6 kW = 42 kW of power. We need one solar panel for every 42 kW that our factory consumes, or 23.8 solar panels per 1 MW.

That's basically it for the solar panels. When you first research solar panels, you'll still be using steam engines to power your base through the night. Once you start oil production, you get access to accumulators. These are just big batteries that store energy when the solar panel is producing more power than you need, and then provide that energy back when it gets dark. That is shown in our next graph:

Accumulator.png (10.73 KiB) Viewed 2583 times

Where the black line is above the grey line, the solar panels are producing excess power. The orange regions mark the energy that we are going to store in the accumulator, and the blue regions represent energy that is going to be taken from the accumulator to power your base during the night. Remember that both the black and grey lines represent the same amount of daily energy. Grey plus orange is the same 70 blocks area as grey plus blue, so we know that the areas of the orange and blue regions are the same. Conveniently that is exactly what we wanted. The amount of energy we store during the day should be the same as the energy we withdraw during the night.

So, how much energy is that? It's time to count blocks again. We'll measure the area that was orange in our previous graph:

Area Under Curve.png (10.66 KiB) Viewed 2583 times

Triangular areas haven't gotten any easier to count, so once more we flip the orange part of this graph over and slide it into the blue area to make everything rectangular. The grey and blue area that we want to count is not a whole number of blocks this time. I've divided several of the blocks into fifths to help us count accurately. There are 15 whole blocks, plus nine fifths of blocks. That's 16.8 blocks of energy that our accumulator needs to store.

Now, we need to know exactly how much energy is in a block. We already found out that one vertical division is 6 kW of power. To calculate how much time one horizontal division represents we will go back to the wiki and find out that a day is 25000 ticks. One tenth of that is 2500 ticks. Divide by 60 ticks per second, and we come up with 41 and two thirds seconds. Multiple that by the 6 kW from earlier and we find out that the area of one block is 2500 ticks / 60 ticks per second * 6 kW which equals 250 kJ.

• The 70 blocks of energy that a solar panel produces per day is 70 * 250 = 17500 kJ.

• The 16.8 blocks of energy that the accumulators need to store is 16.8 * 250 = 4200 kJ.

• The tooltip for an accumulator tells us that it holds 5 MJ of energy. This is 20 blocks or 5000 kJ, which means that one solar panel uses 4200 kJ / 5000 kJ = 0.84 of an accumulator.

Now, we've found out everything there is to know. We'll need one solar panel for every 42 kW of power that our factory consumes. That's nearly 24 solar panels per MW. We also need 0.84 accumulators for every solar panel. That's the 25 solar panel to 21 accumulators ratio that you will often here cited.

Keep in mind that this analysis is based on our base using the same amount to power at all times. Because of lights that come on at dark, our base actually uses slightly more power at night than it does during the day. That means we need very slightly more accumulator storage than this. Similarly, when the biters attack we may end up briefly using a lot of extra power for laser turrets. Since laser turrets don't tend to fire for very long their average power usage is not that high, but we still need to have extra accumulator storage to account for their ability to use a lot of energy in a short period of time.

So, while you need at least 0.84 accumulators per solar panel for any base that is powered only by solar, a base that uses a lot of lights may need slightly more, and a base that uses laser turrets heavily may need moderately more. It would require an exceptional situation to require more than a 1 to 1 ratio.