Solar ratios
Posted: Sat Aug 16, 2014 12:26 am
Based purely on numbers on the wiki, I mathed out some ratios:
First of all, the morning ramp-up and evening ramp-down is linear, so they can be safely approximated as half all-on and half all-off (83.33... / 2 = 41.66...). Therefore you have 208.33... + 41.66... * 2 = 250 "seconds", or 291.66... / 416.66... = 70% of a cycle, of daylight, and 416.66... - 291.66... = 125 "seconds", or 125 / 416.66... = 30% of a cycle, of night.
Solar panels are pretty straightforward; 60kW * 70% = 42kW. Since accumulators are currently lossless, we can simply subtract a further 30% from this for charging accumulators: 36kW * 30% = 10.8kW, 36kW - 10.8kW = 25.2kW.
As for accumulators, 5MJ or 5,000kJ is to 5,000 / 125 = 40kW for the entire course of one "night".
Therefore, the ratio is 40/42 = 0.952......, * 21 (I guess-and-checked this one) = 20, so 20:21 accumulators to solar panels, per 21 * 42kW = 882kW of (mean) average draw.
Remember that a key advantage of accumulators is that they can deliver an instantaneous surge of up to 300kW each, seven and a half times times what they can deliver continuously over the course of one night. This means you can safely completely ignore consumption peaks of up to seven and a half times your (mean) average (when does that EVER happen? Laser turrets maybe), and use your (mean) average draw alone to scale your power plants. Now if only the game showed average production and consumption totals in the electrical grid summary screen...
Of course, this is still theoretical; I haven't tested it just yet.
-E- I made two major goofs, one of which changes literally every other computed number. *SHOULD* be right now.
First of all, the morning ramp-up and evening ramp-down is linear, so they can be safely approximated as half all-on and half all-off (83.33... / 2 = 41.66...). Therefore you have 208.33... + 41.66... * 2 = 250 "seconds", or 291.66... / 416.66... = 70% of a cycle, of daylight, and 416.66... - 291.66... = 125 "seconds", or 125 / 416.66... = 30% of a cycle, of night.
Solar panels are pretty straightforward; 60kW * 70% = 42kW. Since accumulators are currently lossless, we can simply subtract a further 30% from this for charging accumulators: 36kW * 30% = 10.8kW, 36kW - 10.8kW = 25.2kW.
As for accumulators, 5MJ or 5,000kJ is to 5,000 / 125 = 40kW for the entire course of one "night".
Therefore, the ratio is 40/42 = 0.952......, * 21 (I guess-and-checked this one) = 20, so 20:21 accumulators to solar panels, per 21 * 42kW = 882kW of (mean) average draw.
Remember that a key advantage of accumulators is that they can deliver an instantaneous surge of up to 300kW each, seven and a half times times what they can deliver continuously over the course of one night. This means you can safely completely ignore consumption peaks of up to seven and a half times your (mean) average (when does that EVER happen? Laser turrets maybe), and use your (mean) average draw alone to scale your power plants. Now if only the game showed average production and consumption totals in the electrical grid summary screen...
Of course, this is still theoretical; I haven't tested it just yet.
-E- I made two major goofs, one of which changes literally every other computed number. *SHOULD* be right now.