Accumulators in MJ

[Schmendrick]

Day has 25000 ticks.
This is the cycle (realtively speaking)

0.75 -> 0.25 (half a day), 100% bright
0.25->0.45 getting dark (evening)
0.45->0.55 night
0.55->0.75 get light (morning)

Interesting. I suppose earlier when I came up with 25200 I was assuming you used a nice round number of minutes for a day length.

However, I ran some tests (every tick, check game.daytime and watch for values > .9999) and a day seems to be 24966 ticks, though one interval (of the 16 I watched) was 24967.

What is going on here?

And how long is a 'tick'?

A tick is 1/60th of a second, though a tick really isn't a relevant time unit unless you're modding. For gameplay, seconds are quite adequate, and the game is consistent with them as far as I've seen.

[therapist]

Then you're heading into artificial time unit territory. I doubt it would be less confusing. Also, Megawatt/day would result in a base unit of J/s². It should be Megawattday.
I already explained why I think kWh and MWh is not feasible. I was responding to kWs and MWs, which is a higly unusual unit to use, even more because it's exactly the same as kJ/MJ, but with a name nobody uses.

Edit: wolfram alpha doesn't even know the (mega)Wattsecond

Also Schmendrick: Days aren't 50/50 in division between light and dark. It's about 1 minute of total dark and the rest of the time at least some light.

I still think that knowing the amount of MW produced per day, and the amount of MW used per day, and the amount of stored MW per day gives a useful frame of reference for most typical homeowners who are used to managing their kilowatthours. Tying it to in-game days is just the only logical unit to pass time with.

[Gammro]

If the time a day and a night lasts are clear(like now) and easily available(wiki and/or ingame), I still think J is the easiest to work with. Adding a non-standard unit will only confuse people(wth is a kWd?). I think factorio players on average are intelligent enough to know what a joule is.

[Sander Buruma]

If I understand correctly then if we count that a solar panel produces 42kW on average and if we assume for the sake of simplicity that a night is 30% of the day (as 10% is actual night +20%/2 from dawn and +20%/2 from dusk) then we need about 180 seconds of electricity from accumulators then we need 0.30 * 7 * 60 * 42 kW = 5292kWs or 5.3MWs (or MJ) of capacity). So in effect, every solar panel then needs the support of 1.0584x accumulators... (or lets just say, 1.06x, for simplicity sake) or about 17 accumulators for every 16 solar panels... this is not looking pretty (pretty being nice round numbers, not long winded fractional numbers)
This is all assuming that solar panels are the only source of electricity.

[Schmendrick]

When you take into account the ramping, the math works out to about .842 accumulators per solar panel actually, so 1:1 is still a good rule of thumb (before accounting for laser fire of course, which makes all the clean-ish approximations kind of superfluous anyway).