Calculating energy for practical applications

Simple formulae cheat sheet and examples;

Basics

metric system prefixes:
"A metric prefix is a unit prefix that precedes a basic unit of measure to indicate a multiple or fraction of the unit." [1]

Text	Symbol	Factor
giga	G	1000000000
mega	M	1000000
kilo	k	1000

This means that one kilowatt or 1 kW means 1000 watts, one megawatt or 1 MW means one million watts;

Watt (W):
"When an object's velocity is held constant at one meter per second against constant opposing force of one newton the rate at which work is done is 1 watt." [2]
In simple terms (and what we need for Factorio):
1 Watt = 1 Joule per 1 second

Joule (J):
"The work required to produce one watt of power for one second, or one "watt second" (W·s) (compare kilowatt hour - 3.6 megajoules). This relationship can be used to define the watt." [3]
1 Joule = 1 Watt for 1 second

Practical Application - Laser Turret

We have a single Laser Turret, its properties are:
Drain of 24 kW = 24 000 W (disregarding the minor drain as most problems arise when Laser Turrets are actively firing);
Energy Consumption of 2.4 MW = 2 400 000 W

We have a single accumulator, its properties are:
Energy Capacity 5.0 MJ = 5 000 000 J
Max Output 300 kW = 300 000 W

We can ask a question - how quickly a single accumulator will be drained by a single actively firing Laser Turret ?
We can calculate:
J = W * s
J / W = s
5MJ / 300kW (300kW is max output of accumulator) = 5 000 000 / 300 000 = 16.6(6)

Answer: 16 full seconds; additional observations:
Laser turret will have enough power for 2 shots
Laser turret will charge for 2 400 000 / 300 000 = 8 seconds between shots (due to limited output of a single accumulator);

Practical application: One must have 8 accumulators per Laser Turret to satisfy the drain of the power network; With 8 accumulators, a single laser turret will be able to fire for:
5MJ * 8 = 40MJ
40MJ / 2.4 MW = 16.6(6) seconds (this time with minimum charging times as total accumulator output is: 300kW * 8 = 2.4MW (matches a Laser Turret consumption);
(no brainer here, but an example on how easy it is to get from Joules to Watts and vice versa);

Sources

[1]: https://en.wikipedia.org/wiki/Metric_prefix
[2]: https://en.wikipedia.org/wiki/Watt
[3]: https://en.wikipedia.org/wiki/Joule

[anguta_morbhead]

i ask in the offical discord and also confirmed it myself the bateries don´t got a max output, so you can just slap the numbers into the j = w*s or j/w = s it works (example filling 1 mk2 baterie with 1 portable fusion reaction 100 mj / 750 kw or written out 100.000.000 joule / 750.000 wat = 133,3 secounds)

[Ragnaman]

Sure, but I don't remember if equipment batteries had limitation on how much they can output. Try doing a bit of math yourself and you'll see how easy it is to go from joules to watts and vice versa.
Personally I just crammed equipment grid with enough power generation that I always supplied more power to the equipment than it demanded, therefore never needing batteries.

[anguta_morbhead]

well in the spidertron is the same equipment grid as in the modular armor, unless i missed something you need to know the max output in some form of watt correct?

[Ragnaman]

Haven't played factorio with spidertron, can't help there.
Guide is just an example on how to make sense about Joules vs Watts.

[anguta_morbhead]

and what is the right number for the bateries max output (for spidertron + exoskelleton equiped with personal laser defence?)

[Falesz]

BSc thesis right there

[Neopolitan]

I decided to just build a stupid amount of extra accumulators, now my network has 11GJ+ capacity

[Ragnaman]

Its a fun exercise, but I eventually just took a blueprint from online repository that has a close to perfect ratio (considering vanilla day/night cycle).

also, from wiki:
"The optimal ratio is 0.84 (21:25) accumulators per solar panel, and 23.8 solar panels per megawatt required by your factory (this ratio accounts for solar panels needed to charge the accumulators). This means that you need 1.428 MW of production (of solar panels) and 100MJ of storage to provide 1 MW of power over one day-night cycle."

[Neopolitan]

In real life, it would probably be +25% that value, like you said, Factorio doesn't seem to care about resistance.

[Neopolitan]

Thanks for going in depth. I'm mainly trying to calculate how many accumulators I would need to keep my base running overnight if I swapped to 100% solar.

So if my draw is 15-18mW then I need 18MJ per second in stored power.