Fuel cell powered harvester?

[lucusloc]

I tried to find an answer to this, but google failed me. Does anyone know if it possible to make a fuel cell powered harvester, and if so what ratio of drills to fuel cells I would need to make it run with no other power input? I would be willing to use a few RTGs if it gave me a positive fuel flow at nigh on the mun. Is there a percentage of ore that is needed to make this work, or is it impossible to do for any sane craft?

Edited May 20, 2015 by lucusloc

[Streetwind]

It's possible, but it depends on how efficiently you mine.

The drills always output the same amount of ore per second, but their power usage varies with the concentration of ore. You will consume far more power when you land in a poor spot, and thus you won't be able to run at a surplus with fuel cells. On the other hand, in a high concentration spot, you can likely do it no problem.

Energy efficiency on drills is also improved if you have an Engineer Kerbal on the craft. The higher level the better!

[Kourou]

Aheem, isn't it the other way round?

From what i have tested, a drill always consumes 15E/s (unless they throttle down when your source of electicity is delivering insufficient power, but i'll ignore that case for now).

Then, the basic ore rate seems to be Concentration/20 per second, so if the concentration is 8%, you'd get 8%/20 = 0.004 ore per second. You can calculate using the given rates of fuel cells, drills and the ISRU converter, that you will need a ore rate of slightly above 0.02 ore per second and drill to achieve equilibrium, so a higher rate will result in a fuel win. This means, for unmanned drill vehicles, you would need an ore rate above 40% (i have only scanned kerbin so far, on kerbin you will never find a spot close to 40%, no idea if there is a body which such a high rate, probably not or only very rare).

Now, for actually effectively farming ore cell-powered, you wanna take an engineer with you. Engineers simply increase the ore rate of drills by being in the craft. Only the highest engineer is counted, so you just need one. In my test world i only have 5* Engineers(Sandbox), so i could only test out that five-star-engineers will multiply your rate by 25, so in the 8% example, you would get 0.1 instead of 0.004 ore/s, which means that 4/5 of your mined ore will be effective fuel, while 1/5 of the ore is enough to supply the Fuel Cell Arrays.

[n0xiety]

Well if you are ok with abusing the system bugs fast forwarding at the highest level while you have positive energy flow bugs the energy and you never ran out of it even while it passes the night phase. Other than that Kourou's answer seems to be the right one for a legit way of doing it.

[Streetwind]

Aheem, isn't it the other way round?

From what i have tested, a drill always consumes 15E/s (unless they throttle down when your source of electicity is delivering insufficient power, but i'll ignore that case for now).

*checks ingame*

Huh. So it is.

So I stand corrected - I must have been going on outdated info. Thanks for pointing out my error!

[lucusloc]

Aheem, isn't it the other way round?

From what i have tested, a drill always consumes 15E/s (unless they throttle down when your source of electicity is delivering insufficient power, but i'll ignore that case for now).

Then, the basic ore rate seems to be Concentration/20 per second, so if the concentration is 8%, you'd get 8%/20 = 0.004 ore per second. You can calculate using the given rates of fuel cells, drills and the ISRU converter, that you will need a ore rate of slightly above 0.02 ore per second and drill to achieve equilibrium, so a higher rate will result in a fuel win. This means, for unmanned drill vehicles, you would need an ore rate above 40% (i have only scanned kerbin so far, on kerbin you will never find a spot close to 40%, no idea if there is a body which such a high rate, probably not or only very rare).

Now, for actually effectively farming ore cell-powered, you wanna take an engineer with you. Engineers simply increase the ore rate of drills by being in the craft. Only the highest engineer is counted, so you just need one. In my test world i only have 5* Engineers(Sandbox), so i could only test out that five-star-engineers will multiply your rate by 25, so in the 8% example, you would get 0.1 instead of 0.004 ore/s, which means that 4/5 of your mined ore will be effective fuel, while 1/5 of the ore is enough to supply the Fuel Cell Arrays.

Hey thanks, the ore collection rate was the bit i was missing. Next question: What is the highest ore rate on the mun? From my scans the bet I can find is just shy of 6%. 4% ore is way more common though so lets try that:

.04/20 = .002, *25 = .05, *10 (drills) = .5 (for optimum processing efficiency).

10 (drills) * 15 (drill power) + 30 (Converter power) = 180

Fuels Cell arrays generate 18 power, so you need 10.

Fuel output = L .45, Ox .55

Fuel consumption = L .2025, Ox .2475

Net Gain on 4% ore = L .2475, Ox .3025

Of course 10 drills is a lot of drills. If you can find 5% concentration you only need 8 drills, and at 6% you only need 7 (well, 6 and a third, so you will have extra ore with 7). With less drills you of course need less Fuel Cells, so your net gains obviously go up. Either way that *25 boost is absolutely necessary. Looks like all my fuel processing tugs are going to be manned. Thanks for the help!