TokMor

So I found one more oversight on my part, if one is only running during the day, then you only need half the daily power output of running all day. This actually puts solar power as more efficient to run on planets that are Kerbin distance from the sun or closer. Also assuming that your location gets sun for at least 50% of the day. Just remember to restart your drills often. Time to update the top post once again. Sharpy, to your question: assuming you are on minimus with 10% ore concentration and a 3 star engineer, then you would need 5.88 drills per ISRU, and either 5.57 power cell arrays for 32.3 m/1fps or 4.79 gigantor panels for 25.6m/1fps if you only run during the day time.

Correct, those are from the drill wiki. I did some experiments on the launch pad and they look pretty accurate. For Terwin and FancyMouse, I suppose if the expectation is that the devs will correct the Ec consumption/generation during warp/inactive periods then we should be optimizing to the "proper" numbers, but since it is a game one could argue that we should optimize to the game rules rather than reality. Any ideas if there are plans to right that?

Thank you for pointing that out. I guess I have to take the less lazy method to test this then. I found that if I warped at exactly 1,000X, it increased the production of my drills by a factor of 1,000 but it didn't increase the power consumption of them. I thought it must have just been one of my mods bugging something, but perhaps it's not just me after all. Edit: So I put a ship in orbit with a period of 2,063.4s, and the amount of energy that was generated by the solar panel was 35,145, for an average output of 17.033, which puts us in much better shape for looking at solar as part of our power solution. The fact that batteries are needed to get through the night still makes trying to run through the night with solar not an option (new mass per 1f/s is 223.97 when there are enough batteries), but running half the day on just solar panels bring us down to 39.56 mass for 1 fuel/s. I'll update the top post with these new findings and see if perhaps using both fuel cells and solar panels gets us a better result for our efficiency.

Hello! I'm creating this thread mostly to benefit the types of people who want to optimize their mining operation as much as possible. There are already some great threads that cover the basics of how to set up a mining operation. I want this thread to help people who are planning a large scale mining operation to figure out exactly how many of each component that they need to get on site to have the optimal performance with no part being over-utilized nor being the bottleneck. I plan to start with surface operations, and expand from there assuming that there is interest. Base rules and assumptions Use engineers. You already know this if this thread interests you, but I'll include numbers to show just how much they help. Balance the drill production to ISRU consumption Create the most efficient electrical grid Maximize the fuel generated each second based on the mass of the equipment placed in situ On engineers So we all know that high level engineers make mining better, but I for one was surprised to see just how much of a difference that it makes. Lets assume we find our mining site, and it has 10% ore concentration. If we don't bring an engineer with us, we will need 100 drills all running at the same time just to supply a single ISRU converter, and the setup will draw 1530 power each second Just adding a single greenhorn engineer reduces the drill count to 20 and power consumption to 330 If we level that engineer all the way up to the max of 5, we need just 4 drills and 90 power The spreadsheet linked below shows how many drills and power you need for each IRSU based on the ore concentration and the level of engineer running the rig https://docs.google.com/spreadsheets/d/1wPRf8Pw42jjs7-f-ypjNiFG8oXsOhQ7j8CzbLIc7FeU/edit?usp=sharing Yes, there are fractions of drills listed, but chances are you will try to build a base with more than one IRSU, so you can pick a number that gives you a roughly even number of drills. Balancing Drill and ISRU production Placing the drills and ISRU in the same "craft" and balancing their outputs means that we don't need to spend mass on lots of ore storage tanks. Combining these steps also lets us eliminate the need for a ore transfer step in our supply chain, making logistics easier. All of our mass dedicated to storage (with the exception of one small ore tank) can therefore go to storing refined fuels. Power production This seems to be the most glossed over topic in most of the posts I've seen on the matter. I've seen some recommend solar, some recommend fuel cells, but few go into exactly how many are needed. For right now, if you want your operation to run both day and night, fuel cells are the way to go. If you only want to run during day time, solar power can be more efficient as long as you are no further away from the sun than Kerbin. I will be expanding this section in the near future however. Here is an example in the mean time though. (Also, the thermoelectric generator also ended up giving a worse efficiency than fuel cells in case you were wondering) To compare fuel cell power to solar power, I needed to collect some data to give myself a baseline. I ask you the reader, please review and let me know if you feel I am missing things and/or doing something wrong. My new experiment was performed by putting a craft in LKO, 125km circular at 0 degree inclination. I drained almost all the power in the craft then opened one gigantic panel on the craft angled so it would be orthogonal to the sun. The period of the orbit was 2063.4s, and the net power gain during one rotation was 35,145 which gives an average output of 17.033 energy each second. But, being higher up allowed the craft to view the sun for longer than it really should have. Correcting for that amount, the average comes out to be 12.35 per second, or 24.7 during the day time. (I'll be using fractions of parts in these calculations since I'll assume our actual bases will really be much bigger. Also, I will measure efficiency by calculating the amount of liquid fuel generated when creating a fuel/oxidizer mix) Let's assume we are setting up a mining site at minimus and we luck out with 10% concentration right on the equator. From referencing the table, we need 4 drills to supply 1 IRSU. This draws a total of 90 power and weighs 7.25. If we supply this with fuel cell arrays, we need 5.05 of them for a total mass of 8.512. The fuel cells will draw 0.102 fuel per second, for a net of 0.348 fuel/second. Given this setup, the required mass to produce 1 fuel per second is 24.5 Now at this point I thought "Well, since this method consumes fuel, solar must give a better output" Assuming I calculated that average power per second correctly, we need 7.366 solar panels to supply the power. Minimus has a rotational period 40,400s, so we need a daily power output of 3,675,415. This requires 1,837,707 worth of battery storage to work through the night, or 459.43 of the 4k batteries. With all those batteries, our mass required to generate one fuel per second is a staggering 225, over 9 times less efficient than using fuel cells. If we removed the batteries and just relied on solar power for half the day then we get a much better rate of 25.63, since only half the daily energy production is needed if we are only running half the day. This does put us a bit above the efficiency of fuel cell arrays, as long as you can turn on your drills at the right time each day. Math equations (mostly taken from the wiki) http://wiki.kerbalspaceprogram.com/wiki/%27Drill-O-Matic%27_Mining_Excavator Drill output: 0.05 * concentration (between 0 and 1) * engineer skill Engineer skill: 5x for level 0, 9x for level 1, 13x for level 2, 17x at level 3, 21x at level 4 and 25x at level 5 ISRU output: 0.45 fuel per second when making fuel/oxidizer mix