astroshak wrote: ↑Wed Aug 19, 2020 2:44 amThen you have to find a common point - how many reactors in a 2xN setup produce the same MW as a given number of Boilers+2 Steam Engines, and how many Solar Panels/Accumulators provide that much power?
[... everything in between ...]
It is very expensive compared to boilers.
I think this is exactly what Koub meant. This topic isn't about comparing solar power to other means of power production but only about the ratio between solar panels and accumulators.
This is a really interesting thought! The math would be quite complicated. Therefore I'll elaborate it with logic only:mrvn wrote: ↑Tue Aug 11, 2020 3:22 pmIf you use more solar cells then they will provide more power earlier in the day and last longer in the evening. That means less power will be drawn from the accumulators over night and you can use less of them. They will also be recharged faster during the day and then there will be excess power that will simply go to waste (assuming our constant consumption of power). But to some accumulators are expensive and solar cells are cheap. Space is also often not a problem or you probably wouldn't be using solar. So wouldn't a 30:18 (or whatever ratio makes it work again) be better if you assign accumulators a higher price than solar cells?
So what we really should have is a table with solar cells at one axis and accumulators at the other axis and the garanteed constant power as values in each cell. There would be a region in the table with sensible combinations of solar cells and accumulators. But also regions that don't make sense. Any cell that doesn't improve on the one with one less accumulator is out too. That's just wasted. On the other side though, asides from having 0 accumulators there is nothing to obviously exclude. Every added solar cell will increase the power even one accumulator will alow as constant draw. But it quickly becomes near 0.
There would be a line starting at 1,1 and go through 25:21 that reflects the "optimal" 0.84 ratio. The line of maximum accumulators. How many additional W/solar cell does that line give? I think it's something around 40W/cell.
The question I would want answered is: Given a cost ration X:Y for solar cells to accumulators what ratio gives you the most W per cost and how much is it?
The ratio 100 solar panels to 84 accumulators is optimized for solar panels first, than for accumulators. We could do that the other way around. We would need the fewest accumulators if we would just draw power from them if solar panels would produce no power at all. For that we would need an infinite amount of solar panels however, since the power output of each solar panel is just tiny at the end of dawn and the beginning of dusk. That this would be worse than 100:84 is obvious but we don't have to go that far.
Would it make any sense at all to reduce the number of accumulators in favour of more solar panels? No! Because of the following reasons:
- Space: Solar panels use more than twice the amount of space than accumulators. There is no way we will save more than two accumulators with each additional solar panel.
- Material: Accumulators are way cheaper than solar panels. Even if you consider 10 Oil being worth 1 copper or iron ore (personally I consider oil free like water since it is infinite)
solar panels cost 67.5 resources https://kirkmcdonald.github.io/calc.htm ... dnkvP+tgM= (~40 with productivity modules https://kirkmcdonald.github.io/calc.htm ... 77Gcz6bS4=)
and accumulators only ~29.4 resouces https://kirkmcdonald.github.io/calc.htm ... ulator:r:1 (~13.6 with productivity https://kirkmcdonald.github.io/calc.htm ... vB95nURV4=)
Again accumulators are 2 to 3 times better in terms of material costs.