How the Dyson Sphere Program balances the performance of miners, smelters and plants

 

How the Dyson Sphere Program balances the performance of miners, smelters and plants

Gameplay Dyson Sphere Program is built on the processing of some components into others. However, how to balance the mining, processing and processing cycle so that the conveyor does not stand idle and overloaded? Exactly how many collectors and factories should be built per miner? Explained in this guide.


So, it all starts with the miner ore (Mining Machine) or a miner oil (Oil Extractor). After placement, click on the building and examine the block menu in the upper left corner of the window that appears:

balances the performance of miners, smelters and plants

The performance of a miner is determined by how many nodes it spans. With a successful location of the miner, the amount of output (K (output)) will be 200-250 units of raw materials per minute (units / min.), With an unsuccessful one, covering 2-3 nodes, 60-90 units / min.

performance of miners, smelters and plants

Try to arrange the earners so that their performance is generally the same - there are no overly powerful and overly weak. This makes it easier to control production from the entire mine as a whole.

performance of miners, smelters and plants

Remember that over time, the nodes will deplete and performance will begin to drop. In this regard, do not get carried away with "growth" construction. By the time you need to increase production capacity, raw materials may already be depleted.


But the paragraph above does not apply to oil or hydrogen, which can be downloaded without restrictions (as, at least, it was in the version of the game that was released in Early Access on Steam). If only there were enough wells or hydrogen vacuum cleaners.


Example # 1

Let's take an example of the productivity of one field. In the screenshot below, there are five stone miners, each with its own productivity, but if you add up everything, then the deposit produces about 640 units / min. All the mined stone I send for processing into silicon ore.

Example # 1

Smelters and Assemblers have two main characteristics:

  • the amount of raw materials / products that must be supplied to the input (K (input)) to start production;
  • the rate of processing of products at the entrance to products at the exit (K (inlet) => K (outlet)).

Both characteristics are best viewed before choosing a product for recycling:

performance of miners, smelters and plants

The screenshot shows that for the release of 1 unit. silicon ore you need:
  • submit 10 units to the entrance. stone (stone);
  • manufacturing time - 10 seconds.

That is, the productivity of one smelter is 6 units / min, and for uninterrupted operation it needs to supply 100 units. stone per minute.
In the same minute from the mine, we definitely get 600 units. stone. Dividing 600 by 100, we get 6. That is how many smelters need to be placed in order to fully use the stone supplied from the mine.

If from one smelter we get 6 units / min. silicon ore, then with six smelters we get 36 units. products per minute.

performance of miners, smelters and plants

Now let's calculate how many units of pure silicon (high-purity silicon) we get from 36 units of silicon ore. This operating mode has the following parameters:
  • K (entrance) = 2 units. ores;
  • K (release) = 1 unit. silicon;
  • manufacturing time - 2 seconds.

Ideally, 30 processing cycles will happen in 1 minute, 30 units will be made. silicon and spent 60 units. ore. But we only produce 36! That is, even one smelter will be idle for 24 seconds. Unless, of course, you bring silicon ore from another source.
How much silicon ore needs to be fed to the input so that one pure silicon smelter is not idle? Counting in reverse order, we get: 10 smelters of stone into silicon ore, to which you need to supply 1000 units. stone.


Example No. 2

Now let's take a look at refined oil and hydrogen obtained from crude oil using plasma refining technology in an Oil Refinery.

Example No. 2

As follows from the description of the plasma cleaning technology:
  • K (entrance) = 2 units. crude oil;
  • K (release) = 2 units. petroleum products and 1 unit. hydrogen;
  • manufacturing time - 4 seconds.

Ideally, 15 processing cycles will happen in 1 minute, 30 units will be spent. crude oil and received 30 units. petroleum products and 15 units. hydrogen.
Now let's look at practice. One of my Oil Extractors produces 2.24 units. crude oil per second (the value varies from place to place), which is equivalent to 134.4 units / min. Dividing 134.4 by 30, we get 4 - this is the number of oil refineries I need for my well.

120 units will be received from 4 factories in 1 minute. petroleum products and 60 units. hydrogen. At the beginning of the game, they can be used, for example, to produce sulfuric acid and red cubes, respectively.

performance of miners, smelters and plants

In the first case, everything is simple:
  • K (entrance) = 6 units. petroleum products, 8 units stone and 4 units. water (water);
  • K (release) = 4 units. acid
  • manufacturing time - 6 seconds.

In 1 minute there will be 10 processing cycles, 1 Chemical Plant will consume 60 units. petroleum products, 80 stone and 40 water. The result will be 40 units. acid. Since we get 120 units per minute. oil, you can build 2 chemical factories.

performance of miners, smelters and plants

In the case of red cubes, the calculation is slightly different:
  • K (entrance) = 2 units. graphene, 2 units. hydrogen;
  • K (release) = 1 red cube
  • manufacturing time - 6 seconds.
In 1 minute, there will be 10 processing cycles and the laboratory will spend 20 units. hydrogen and 20 graphene. The result will be 10 red cubes. Considering that we have 60 units from one oil rig. hydrogen, then you can build 3 matrix laboratories (Matrix lab).

performance of miners, smelters and plants

Mining graphene is not particularly difficult:
  • K (entrance) = 2 units. coal;
  • K (release) = 1 unit. graphene;
  • manufacturing time - 2 seconds.
Here in 1 minute there will be 30 processing cycles, we will spend 60 coal and get 30 graphene. And we need 60 graphene per minute, which means we need two smelters. The miner must have a productivity of 120 units. coal per second.


Example No. 3

But what are we all about raw materials, but about raw materials. Let us estimate how many resources are needed for the production of conveyor belt (conveyor belt) the second level (Mk.2). From the help we know what is needed:

performance of miners, smelters and plants

  • K (entrance) = 3 units. conveyor belt of the first level (Mk.1), 1 electromagnetic turbine;
  • K (release) = 3 units. conveyor belt of the second level;
  • manufacturing time - 1 second.
Let's count for 300 units. conveyor of the second level. In total there will be 100 processing cycles, 300 units will be required. first level belts and 100 turbines. Let's expand these components to the basic ones.

performance of miners, smelters and plants

a) For the production of a first level conveyor belt, you will need:
  • K (entrance) = 2 units. iron (iron ingot), 1 unit. gears;
  • K (release) = 3 units. conveyor belt of the first level;
  • manufacturing time - 1 second.
Since we want to get 300 units. tape, you have to take 200 iron and 100 gears.

performance of miners, smelters and plants

b) The second component is an electromagnetic turbine, made from:
  • К (input) = 2 electric motors and 2 electromagnets (Magnetic coil);
  • K (outlet) = 1 turbine;
  • manufacturing time - 2 seconds.

We need 100 turbines, which is equal to 200 electric motors and 200 electromagnets.
Summing up points a) and b), we get: 200 iron, 100 gears, 200 electric motors and 200 electromagnets.

performance of miners, smelters and plants

The electric motor is made from:
  • K (entrance) = 2 units. iron, 1 gear and 1 electromagnet;
  • K (release) = 1 electric motor;
  • manufacturing time - 2 seconds.
This means that 200 electric motors turn into 400 iron, 200 gears and 200 electromagnets. Our application is simplified: 200 iron, 100 gears and 200 electromagnets + new 400 iron, 200 gears and 200 electromagnets = 600 iron, 300 gears and 400 electromagnets.

performance of miners, smelters and plants

The electromagnet is made from:
  • K (input) = 2 magnets (magnet), 1 unit. copper (copper ingot);
  • K (release) = 2 electromagnets;
  • manufacturing time - 1 second.
Since we get 2 units at once. products, 400 electromagnets will be made from 400 magnets and 200 copper. In turn, 1 magnet and 1 gear are made of iron, 1 unit each. for each.

performance of miners, smelters and plants

performance of miners, smelters and plants

That is, 600 iron, 300 gears and 400 electromagnets are converted into 600 iron, + 300 iron, + 400 magnets and 200 copper = 900 iron, + 400 iron and 200 copper, which is finally equal to 1300 iron and 200 copper.

This is exactly how much raw materials are needed to produce 300 units. conveyor of the second level. Note that the production of individual elements differs in time. The same electric motors and turbines are produced twice as long as all other components.

***

That's all. Keep in mind that the Dyson Sphere Program is in Early Access on Steam, so certain aspects will change in the release version.

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