• Halved the fuel cell consumption rate of the Fulgoran reactor.

This also halved the output from 6GW to 3GW as well. Which is a breaking change for any existing reactor designs. Keeping the consumption at 6WG and doubling the effectivity from 2 to 4 compensates for this so the output is still 6GW, but the input is still only 1.5GW.

I do not know if your intent was to also reduce the output in addition to the input, but the higher outputs greatly helps heating the surface pre-Aquilo (before you can make your own radiative towers).

Ah yes, I recall this property catching me out before. Thanks.

Intent was just to halve the input rate, leaving the heat produced by one fuel cell unaffected, and leaving the constructible reactors unaffected. Effectivity alone would not do this.

Feel free to play with the numbers before I next get a chance to investigate and fix.

I figured it might have been an accident based on the patch notes. And I already mentioned the numbers to hopefully get your intended effect. Leave consumption at 6GW and set effectivity to 4 (up from 2). That makes the input 1.5GW (down from 3GW) and the output still 6GW. Unless I am missing something about that. I assume the heat produced by a single was solely based on the output. And then the "consumption" of the cell itself was the input.

Effectivity refers to the input source and doubling it would double the burn length of fuel cells while leaving the input rate unchanged, giving twice as much energy per cell.

Consumption, consulting the docs, refers to the total energy flow for both input and output. This means I can't change input without changing output. Due to the positive effect on the reactor gameplay, and without wanting to affect the constructible reactor, I'll treat this change as intentional for now and see if anybody complains. I'll also clarify the 2.13.10 changelog in the next published version.

As far as I can tell, consumption is actually the output of the burner (maybe it called consumption because that is how much power actual burners consume, whereas reactors actually output it as heat). effectivity is how "efficient" the burner is. So, the actual input consumption is calculated by consumption / effectivity. effectivity has no actual effect on the burn length of the fuel.

The input (constant consumption) is the burn rate of the fuel. Each fuel has an energy value. Each burner has a constant consumption. The burn length is going to be the energy value of the fuel / burner's constant consumption rate. consumption of 6GW and effectivity of 4 gives it an input consumption rate of 1.5GW and an output of 6GW. 1 W = 1 J/s. So, a Mixed Oxide Fuel Cell has 96 GJ, which is 96GW. 96J / 1.5GW = 64s burn time.

consumption = 6GW | effectivity = 4 gives

• constant consumption (input) of 1.5GW (consumption / effectivity)
• efficiency of 400% (effectivity)
• output of 6GW (consumption)
• burn time of 64s (96J / 1.5GW)

https://i.imgur.com/c7PZhRt.png

consumption = 6GW | effectivity = 2 gives

• constant consumption (input) of 3GW (consumption / effectivity)
• efficiency of 200% (effectivity)
• output of 6GW (consumption)
• burn time of 32s (96J / 3GW)

consumption = 3GW | effectivity = 2 gives

• constant consumption (input) of 1.5GW (consumption / effectivity)
• efficiency of 200% (effectivity)
• output of 3GW (consumption)
• burn time of 64s (96J / 1.5GW)

https://i.imgur.com/vDJq3xh.png

I timed and verified the burn times of 1 and 3 (the ones with imgur links). They are in fact both exactly 64s. And neither consumption or effectivity will change the buildable Mixed Oxide reactor because they are both values on the Fulgoran Reactor. It still has a base constant consumption/output of 160MW (100% efficiency): https://i.imgur.com/IZbwKTi.png