An Optimizing Guide for Mining Bases

[Ined]

Background

I initially wanted to build a mining base on Minmus. I had a lot of questions regarding how to create a mining base to its optimal performance – I didn’t just want to haul a lot of machinery and “see what would happen”. I thought maybe I could learn from other people around the internet how they had done it, and possibly also combine my base with math given that this is what is being done by the game engine anyway.

The problem with reading about it from others was that it was very hard to find actual data as to why they had done it; it simply was not possible to determine optimal performance by looking at screenshots of other people’s bases and there were very few posts explaining exactly why the combination of machinery was used. The KSP wiki had not been updated for the parts that had changed during different versions, and there was just no guide anywhere properly covering mining bases. I found a very good one (Maximizing Mining Operational Efficiency) which unfortunately was written in 2015 prior to several changes and additions. So, I decided to start creating a summary of the various (sparse) findings, which lead to the idea of this guide.

Acknowledgements
I’d like to extend thanks to @blakemw, @Z3R0Gravitas, Atomiktoaster and @TokMor for providing a lot of data and content for which this guide relies on.

 

Intro

This guide covers mining base should contain to get the most out of it. It already assumes that the reader knows how to mine; the parts, how to find a good ore site as well as how to haul the base to the site. If this is not clear, I would recommend beginning with the Step by Step Mining Guide. Please bear in mind that KSP update v1.0.5 added new parts since the guide was last updated, so it is not fully accurate nor complete.

The only parts that are discussed in this guide are the ones that help optimize the efficiency of a mining base. That means that although ore storage is an important part of any mining base, the different containers do not change the efficiency so they have been left out.

Only stock parts are taken into account. The guide is written for v1.3 but might be viable for other versions as well.

 

Thermal Efficiency (T)

When a part is turned on, its core temperature will rise. The higher the temperature, the more efficient it will be, which leads to higher output rate. After the part surpasses its thermally efficient peak, if there is no cooling that can hold it at that value, it will keep increasing the internal temperature. At this point the output rate will lower because the part will not be as efficient anymore. When the efficiency goes down so will the amount of resource that is being processed, which in turn will lower the temperature. Eventually these equations will balance out where the part is overheated and the output rate is low, but neither values change anymore.

The parts need some time to “warm up” before they can convert the maximum amount of resource. The output rate and electricity usage are proportional to its “thermal efficiency”¹. The optimal is to have it warm up once and then constantly let it run.

The picture below shows a graph that has been plotted to show the difference between the thermal efficiency and the core temperature of a drill. For simplicity, this guide is ignoring any data regarding core temperatures and only relies on the thermal efficiency.

A graph depicting relationship of thermal efficiency and core temperature²

Source: [1, 2]

 

Engineer level multiplier (m)

Having a Kerbal engineer on board basically adds a multiplier to the effectiveness of the converters and harvesters. The engineer will increase the output rate but that also brings an increase in electricity consumption. The level of the engineer determines the multiplier. If multiple engineers are present, the game engine picks the engineer with the highest level for the calculations.

Without an engineer onboard:
m = 0.05

The formula with an engineer onboard:
m = 0.05 · 4(1 + count(★))

where count(★) represents how many stars the engineer has. This gives the following list:

--------   5%
☆☆☆☆☆  25%
★☆☆☆☆  45%
★★☆☆☆  65%
★★★☆☆  85%
★★★★☆ 105%
★★★★★ 125%

Source: [1, 2]

 

Converters

Also known as ISRUs (In-Situ Resource Utilization), these parts convert a type of resource (Ore) to another (Lf+Ox, Monoprop, LqdFuel, Ox). There are two parts that fall in this category; the Mini-ISRU (Convert-O-Tron 125) and the ISRU (Convert-O-Tron 250).

If converters are not properly cooled they will eventually overheat and turn off.

If the electricity runs out the parts will turn off (and start cooling down), and will turn back on once electricity is available again¹.

The Mini-ISRU (Convert-O-Tron 125) has a maximum cooling of 50 kW (the maximum cooling it can possibly use), but requires 100 kW. This means no matter how many cooling parts are added it will always overheat. Therefore, the Mini-ISRU can never run constantly, and should only be used for ships with limited space that need to refuel on its own (“planet hoppers”). This ship design isn’t really for mining, so Mini-ISRUs aren’t recommended for mining bases.

Furthermore, the presence of an engineer or the engineer level does not affect the Mini-ISRU in any way, despite what its description says². It is currently unknown if the description is wrong or if the implementation is.

The ISRU (Convert-O-Tron 250) is purposely³ designed to run multiple converter processes simultaneously⁴, each requiring 200 kW of cooling⁵. Because it has a maximum cooling limitation of 500 kW⁵, it means it can only run two processes simultaneously without overheating.

ISRU Convert-O-Tron 250 running multiple processes⁴

Source: [1, 2, 3, 4, 5]

Electric charge consumed
These parts have an Electric charge input, which unfortunately is not the same as the electricity consumed. This “complicates” calculations a bit but is not very difficult to do the math. The Electric charge input is the same for all types of processes; 30 Electric charge/s.

The consumption is calculated using the following formula:

x · E · T · m Electric charges/s

Where the factors are affecting the consumption are:

• x: the number of processes that are running (always 1 for the Mini-ISRU)
• E: the electric charge input (30)
• T: the thermal efficiency of the part
• m: a multiplier for how skilled an onboard engineer is (see chapter above)

Ore consumed
The consumption of ore is always proportional to the consumption of electric charges, meaning if only half the electricity is consumed for a conversion process then only half the ore input will be consumed for that process. Because the electricity consumption is determined by the thermal efficiency and the engineer multiplier, then so is also the ore consumption:

O · T · m Ore/s

Because different processes use different amounts of ore as input it cannot simply be multiplied by x for more processes; they have to be calculated individually and added.

The factors used in the calculation are:

• O: the ore input, which is different for each process
• T: the thermal efficiency of the part
• m: a multiplier for how skilled an onboard engineer is (see chapter above)

The ore input O is taken from the Wiki for each part:

Convert-O-Tron	125	250
Lf+Ox	2.5 ore/s	0.5 ore/s
Monoprop	2.5 ore/s	0.5 ore/s
LqdFuel	2.25 ore/s	0.45 ore/s
Ox	2.75 ore/s	0.55 ore/s

A note to the table above: The smaller converter can process more ore per second, but the outputted fuel is lower than for the bigger converter. In other words, it “wastes” a lot of ore. The produced values are covered in the next subchapter.

Resource produced
Similarly to the consumption, the production is calculated using the formula:

R · T · m Units/s

Where R is a factor representing the expected resource output of a specific process, taken again from the Wiki:

Convert-O-Tron	125	250
Lf+Ox	0.225 Lf/s + 0.275 Ox/s	0.45 Lf/s + 0.55 Ox/s
Monoprop	0.5 Monoprop/s	1 Monoprop/s
LqdFuel	0.45 Lf/s	0.9 Lf/s
Ox	0.55 Ox/s	1.1 Ox/s

Example 1: A thermally efficient (T = 1.0) ISRU running Lf+Ox without an engineer onboard will:

• consume (1 · 30 · 1.0 · 0.05) = 1.5 Electric charges/s
• consume (0.5 · 1.0 · 0.05) = 0.025 ore/s
• produce (0.45 · 1.0 · 0.05) = 0.0225 units of Lf/s
• produce (0.55 · 1.0 · 0.05) = 0.0275 units of Ox/s
• need 200 kW of cooling.

Example 2: A thermally efficient (T = 1.0) ISRU running Lf+Ox with a fully leveled engineer onboard will:

• consume (1 · 30 · 1.0 · 1.25) = 37.5 Electric charges/s
• consume (0.5 · 1.0 · 1.25) = 0.625 ore/s
• produce (0.9 · 1.0 · 1.25) = 1.125 units of Lf/s
• produce (1.1 · 1.0 · 1.25) = 1.375 units of Ox/s
• need 200 kW of cooling.

Example 3: A thermally efficient (T = 1.0) ISRU running LqdFuel and Ox without an engineer onboard will:

• consume (2 · 30 · 1.0 · 0.05) = 3 Electric charges/s
• consume ((0.45 + 0.55) · 1.0 · 0.05) = (1.0 · 1.0 · 0.05) = 0.05 ore/s
• produce (0.9 · 1.0 · 0.05) = 0.045 units of Lf/s
• produce (1.1 · 1.0 · 0.05) = 0.055 units of Ox/s
• need (2 · 200) = 400 kW of cooling.

Example 4: A thermally efficient (T = 1.0) ISRU running LqdFuel and Ox with a fully leveled engineer onboard will:

• consume (2 · 30 · 1.0 · 1.25) = 75 Electric charges/s
• consume ((0.45 + 0.55) · 1.0 · 1.25) = (1.0 · 1.0 · 1.25) = 1.25 ore/s
• produce (0.9 · 1.0 · 1.25) = 1.125 units of Lf/s
• produce (1.1 · 1.0 · 1.25) = 1.375 units of Ox/s
• need (2 · 200) = 400 kW of cooling.

 

Resource Harvesters

The resource harvesting parts are the 'Drill-O-Matic' Mining Excavator (Senior) and the 'Drill-O-Matic Junior' Mining Excavator (Junior). They are both radially mounted harvesters that mine for ore. The Junior requires at least 2.5% ore concentration to function, while the Senior has no restrictions.

The Senior harvests 5x more ore than the Junior (assuming equal environmental properties, see the formula below). It weights 5x more, has 5x more energy consumption but costs 6x more. Technically you can purchase 6 Juniors for the price of 1 Senior, but the price difference is so small that this should not affect the decision on which drill to choose. Unlike the converters, the smaller version does not suffer any serious penalty (beside the ore concentration restriction). For bigger mining operations, I personally would just pick the Senior. Keeping the part count down allows the game engine to run at more stable framerates.

If the electricity runs out the drill(s) will turn off. However, once the electricity is available again they will not automatically turn back on like the converters do¹.

Source: [1]

Operational modes
The resource harvesters have two operational modes; Surface Harvester and Asteroid Harvester. The mode is automatically determined when the part comes in contact with either a surface (planet/moon) or an asteroid. Different modes give it different bonuses.

Electric charge consumed
Harvesters have a very similar electricity consumption calculation as converters, with the exception that the harvesters have a secondary mode where the consumption is determined differently.

The electric charges that a harvester consumes in Surface Harvester mode depends on the type of drill and if there is a Kerbal engineer onboard (as well as its level). It also takes into account the thermal efficiency, i.e. you only pay electricity for actual results. The electric consumption calculation is done using the formula:

E · T · m Electric charges/s

The factors are the following:

• E: the electricity required (3 for Junior, 15 for Senior)
• T: the thermal efficiency of the part
• m: a multiplier for how skilled an onboard engineer is (see chapter above)

For Asteroid Harvester mode, the electric charge consumption is fixed. A Junior uses 0.3 Electric charges/s while a Senior uses 1.5 Electric charges/s.

The different modes and drills can be summarized in the following table:

Drill-O-Matic	Surface Harvester	Asteroid Harvester
Senior	15 · T · m Electric charges/s	1.5 Electric charges/s
Junior	3 · T · m Electric charges/s	0.3 Electric charges/s

A side note: a harvester’s electric charge consumption is independent of the site’s ore concentration.

Harvest rate
The harvest rate is how much ore/s a drill actually harvests after all the factors are taken into account. Given the factors mentioned above we denote a few more factors:

• B: the base harvest rate, depending on celestial body and drill type
• c: the site’s ore concentration

The base harvest rate B is given as:

Drill-O-Matic	Surface Harvester	Asteroid Harvester
Senior	1.5 ore/s	5.0 ore/s
Junior	0.3 ore/s	1.0 ore/s

Applying the factors to a formula gives, for one drill¹:

B · c · T · m ore/s

Example 1: A thermally efficient (T = 1.0) Senior on a site with 10% ore concentration with a level 2 engineer onboard will:

• consume (15 · 1.0 · 0.65) = 9.75 Electric charges/s
• harvest (1.5 · 0.1 · 1.0 · 0.65) = 0.0975 ore/s

Example 2: A thermally efficient (T = 1.0) Junior on a site with 10% ore concentration with a level 2 engineer onboard will:

• consume (3 · 1.0 · 0.65) = 1.95 Electric charges/s
• harvest (0.3 · 0.1 · 1.0 · 0.65) = 0.0195 ore/s

Example 3: A thermally efficient (T = 1.0) Senior on a site with 10% ore concentration with a fully leveled engineer onboard will:

• consume (15 · 1.0 · 1.25) = 18.75 Electric charges/s
• harvest (1.5 · 0.1 · 1.0 · 1.25) = 0.1875 ore/s.

Example 4: A thermally efficient (T = 1.0) Senior on an asteroid site (c = 1.0) without an engineer onboard will:

• consume 1.5 Electric charges/s
• harvest (5.0 · 1.0 · 1.0 · 0.05) = 0.25 ore/s.

Example 5: A thermally efficient (T = 1.0) Senior on an asteroid site (c = 1.0) with a fully leveled engineer onboard will:

• consume 1.5 Electric charges/s
• harvest (5.0 · 1.0 · 1.0 · 1.25) = 6.25 ore/s. This is the max ore rate a drill can harvest!

As a comment to the examples; an exactly equal mining base on an asteroid with a fully leveled engineer will harvest about 640% ore/s compared to having the base on an “okay” site on a moon with an “okay” engineer. The drawback is that asteroids get depleted.

Source: [1]

Interpreting the in-game menus
A typical menu showing information of the drill will look like the picture below:

It is important to understand the values that are being presented here:

Thermal Efficiency – Percentage for how efficient the drill is based on heat. This value will start at 0%, reach 100%, and lower again if the drill is unable to cool down. This value represents T in formulas.

Core Temp – The internal temperature of the part. If the temperature reaches 500K for the Junior it becomes thermally efficient (100%).

Ore rate – This is an extremely misleading name/value! It does NOT mean how much ore the drill is currently harvesting per second! It only considers the base harvest rate (B) and the site’s ore concentration (c), i.e. this value is (B · c). It is also wrong to suggest that this value is some form of “highest possible” ore harvest rate that can be achieved on this site, because the engineer multiplier can be more than 100%. Frankly I don’t understand why the developers thought putting half the equation as a data field because it doesn’t really help the player in any way. Furthermore, calling it “ore rate” is technically correct, but is not the final ore rate for the entire drill.

Surface Harvester load – Also a poorly chosen name. This value is only available in Surface Harvester mode and not Asteroid Harvester mode, which might be the reason for the name. The value represents the remaining factors of the equation, which takes into account the thermal efficiency (T) and the engineer level onboard (m), i.e. the value shown here is (T · m). This means that as the drill’s Core Temp will rise, so will the Thermal Efficiency, which in turn will make the Surface Harvester load to increase.

Multiplying the Ore rate with the Surface Harvester will give the actual ore per second that is being harvested and filled into the ore tanks by this drill. This value can only be seen in the Resource menu for how much ore is being “added” to the ship per second. The value however is rounded in the view, and if multiple drills are running it will show the sum of them all.

Example 1: In the picture above, B · c = 0.034250 and T · m = 0.25. The drill is currently harvesting B · c · T · m = 0.034250 · 0.25 = 0.0085625 ore/s.

 

Number of Harvesters per Converter

To match the number of drills to a converter, we need to do some calculations. Ideally if you have brought up a converter you want to let it run at 100% or close to that. That means we just need to satisfy the input requirements (electric charge and ore). In this chapter, we will consider the ore.

The ore input requirements for the Convert-O-Tron 250 running LqdFuel and Ox is:

(0.45 + 0.55) · T · m = 1 · T · m = T · m

The ore output of a ‘Drill-O-Matic’ Mining Excavator in Surface Harvester mode is:

1.5 · c · T · m

Thus, we need to find the value x, that here will represent how many of drills are needed. We will denote T with subscripts to separate them; the thermal efficiency of a harvester is not the same as of the converter:

x · (drill ore output) = (converter ore input)

x · (1.5 · c · T_H · m) = T_C · m

If we assume the parts are on the same vehicle, then the modifier m affects both parts equally. We can remove it from the equation:

x · (1.5 · c · T_H) = T_C

Similarly, we can assume the parts are running thermally efficient, T_H = T_C = 1.0:

x · (1.5 · c) = 1

Writing the equation on explicit forms:

x = 2 / (3c)

c = 2 / (3x)

This means that the number of drills we will need is only depending on the surface ore concentration! Let’s make a table with some important concentration points. We also recall that a Junior is harvesting 1/5 of the rate of the Senior, so we need 5x more Junior drills to satisfy the Convert-O-Tron 250 ore input:

Surface ore concentration	Number of Seniors needed	Number of Juniors needed
16.66%	4	20
13.33%	5	25
11.11%	6	30
9.52%	7	35
8.33%	8	40
7.41%	9	45
6.66%	10	50
6.06%	11	55
5.56%	12	60
5.13%	13	65
4.76%	14	70
4.44%	15	75

You can of course bring fewer harvesters than what is listed in the table, but that means that the ISRU will not run at its maximum – you’ve just hauled a part to another celestial body that is not running at its full potential. If you bring more drills they will not be running at maximum because the ISRU will not keep up using all the ore they are producing. That again means that you’ve hauled drills that aren’t running at its full potential.

If the surface ore concentration is between two values (say at 15%), I would round down the number of drills needed (meaning 5 Seniors). The ISRU will not be at full capacity but I consider it better than bringing too many drills; they are quite big and heavy. You could also try adding Juniors to fill the gap between two numbers of Senior, but that requires a bit more calculations and it will not be covered in this guide.

 

Daylight energy

Let’s assume the ore concentration on the surface is 10%. That means an efficient mining base would consist of 7 Seniors and 1 ISRU. We will consider them running at full thermal efficiency with a 5-star engineer onboard (always plan for the “worst” case!).

The electric charge consumption of one drill is:
B · c · T · m = (15 · 1.0 · 1.25) = 18.75 Electric charges/s

The electric charge consumption of one ISRU running LqdFuel and Ox is:
(2 · 30 · 1.0 · 1.25) = 75 Electric charges/s

The total electric consumption would be:
7 · 18.75 + 75 = 206.25 Electric charges/s

Given that a Gigantor XL Solar Array generates 24.4 Electric charge/s, we would need:
206.25 / 24.4 = 8.45 --> 9 panels

 

Cooling

I initially wanted to write a chapter about how to properly perform cooling, but after some consideration and the sheer length of this guide I decided not to. BlakeMW’s reply in a Reddit thread is excellent, so I am just going to link to it: How do I calculate how many radiators I need for efficient processes?

The computational problem the game engine has to face is - for anyone who has taken a discrete mathematics class - the maximum bipartite matching problem. KSP trivially fails to solve this.

There are still a lot of unanswered questions regarding cooling, especially how the engineer (level) helps. Z3R0Gravitas begun an investigation in his thread How are thermal mechanics modified by engineer level?, but did not arrive at any answers it seems. This guide will be updated in the future if any results can be arrived at.

 

Example of an optimized Mining Base

• ISRU SuperCrawler

I am making a note here: huge success.

However, due to the extent of this guide I might have made an error or missed something. If this is the case, please highlight it!

I wouldn't mind having @RoverDude dropping in and looking through it!

[TacoMuffin]

This is a really weird instruction. Needs more tacos and less noodles 

 

- TacoMuffin

[tocken]

Thank you for the guide, this helped a lot. I am not sure your numbers are correct for asteroid harvesting. For example, with plenty of power, a level 2 engineer (0.65), a "Drill-o-Matic Mining Excavator" (not the JR) running at 100% thermal efficiency on an asteroid that reports 94.917% Resources, I am getting 0.46 ore/sec. No ISRUs are running or anything like that. Your formula predicts around 3.1 ore/sec in this scenario (5 * .65 * 1 * .95). Also, electric draw of the miner is only about 0.03 charge / second (testing w/ it on and off and comparing draw).

[Atario]

On 9/10/2017 at 4:36 AM, Ined said:

The ISRU (Convert-O-Tron 250) is purposely³ designed to run multiple converter processes simultaneously⁴, each requiring 200 kW of cooling⁵. Because it has a maximum cooling limitation of 500 kW⁵, it means it can only run two processes simultaneously without overheating.

This is incorrect.  The actual number is 750kW, per the in-game specs and the .cfg file setting, so you can run three processes at once for 600KW total without overheating, given sufficient cooling.

[Wijbrandus]

There is a lot of great information in this topic, especially for a returning player who's been out of the business of Kerbals for over a year.

How do you get the fuel out of the ISRU and into a spaceship?  If I land nearby, is there now a part that lets me run a hose line and gas up?  Or do I need to bring this into orbit and fill up a gas station?

[Egdod]

@Wijbrandus: KIS/KAS mod for connecting ships with fuel hoses, or simply fly up and dock with the ship you want to fuel in vanilla.  I'm assuming you know this, but since you phrased the question as such I will point out that the ISRU does not hold any fuel itself, the fuel is stored in fuel tanks attached to the ISRU.  Alternatively, you can store the harvested ore in ore tanks, fly the ore up to an orbiting station with an ISRU there and convert in space.

 

[Egdod]

...and one more thing, so you don't make the same mistake I did with my first drill setup.  You must have at least one ore container in the rig.   The ISRU cannot convert ore directly from the drill apparently, the ore has to go into a container, however briefly, before the ISRU can convert it.  Best practice is to test the thing at the runway or launch pad first.  Saves a lot of tears later.

[PrvDancer85]

Awesome guide, thanks

[MongooseNX]

Thank you for making this guide...excellent information.