Convert-o-tron 65% load?

[rino2727]

i have a mining rig complete with sufficient cooling for 4 big drills and 4 convert-o-tron 250s two are working just fine with all 4 temps optimal but the other 2 are saying 65% load with all tanks filled with 7200 ore total with mines still running. What does this mean, is it running at 65% capacity, if so, what is causing this. i have plenty of power, a lvl 2 engineer, plenty of cooling, could it be the docking port can only transfer so much fuel at once?

[king of nowhere]

14 hours ago, rino2727 said:

 could it be the docking port can only transfer so much fuel at once?

 

i don't know what's the issue there, but i can at least confirm it is not a problem with transfer. the game does not model it.

 

however, I have to ask, is having one convert-o-tron at 65% such a big deal? refueling a ship with a big convert-o-tron takes a few weeks, a few months even for a very big one. in the time scale of transfer windows, that's short. if one of four chemical plants is working at two/thirds capacity, you'll take a few extra days to refuel, nothing more. this game is full of those little glitches that you just have to live with

[rino2727]

well its 2 and i just want to run at max efficiency but i think it might be docking ports or just some kind of bug cuz i can reset them and they seem to work fine, sometimes it does it sometimes it doesn't

[Lt_Duckweed]

On 10/28/2023 at 7:25 AM, rino2727 said:

but the other 2 are saying 65% load

The actual bug is that they aren't all saying 65% load.  In a situation like this where you have enough ore to feed all of them, they will all run at the same speed

On 10/28/2023 at 7:25 AM, rino2727 said:

is it running at 65% capacity

Yes, 65% is the capacity it should be running at.

On 10/28/2023 at 7:25 AM, rino2727 said:

a lvl 2 engineer

And this is why.  With no engineer, they will run at 5% load, and you get +20% load for having an engineer on board, and another +20% load for each level the engineer is. 5 (base) + 20 (engineer) + 40 (lvl 2) = 65%