Reactor, Large | |
---|---|

Category: | Energy Source |

Status: | Functional |

Function:
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Provides electrical power for other components. | |

Fits small ship
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Mass: | 2,776 kg |

Capacity: | 1,000 L |

Power: | 10.5 MW |

Fits large ship and station
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Mass: | 72,295 kg |

Capacity: | 8,000 L |

Power: | 300 MW |

Data Controls:
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The **Large Reactor** is an essential block in Space Engineers, a minimum of 1 reactor is required to power ships and stations.

It is important to note that this article discusses two reactor variants: Large Reactors for Large Ships (7.5m x 7.5m x 7.5m) and Large Reactors for Small Ships (1.5m x 1.5m x 1.5m). They are not to be confused with the Small Reactor variants.

## Power output and Usage

The Large Reactor is twenty-seven times the size of the Small Reactor (3x3x3 instead of 1x1x1), the large reactor being more efficient (MW/kg) by about 1.326 times (on a large ship) as much. On a small ship large reactor produces about 1.639 times more power per kg than a small one. The Large Reactor creates 300 MW on a large ship or station and 10.5 MW on a small ship. In Creative mode no ingots are required to generate power; however the Large Reactor in Survival Mode requires Uranium ingots.

One Uranium ingot (1 kg) at full power usage (300MW) last 12 seconds.

## Consumption

- Calculating uranium consumption for a given power usage and time can be done by the folliwing formula

Units :

- U : Uranium in kilograms (kg)
- t : time in seconds (h)
- P : Power in MegaWatt (MW)

$ U= t * P $ | ||

Where:
| ||

$ U $ | : Uranium (kg) | |

$ t $ | : time (h) | |

$ P $ | : Power (MW) |

The time in hours used is in decimal. eg : 1h30min = 1.5h, to transform normal hours to decimal just take the minutes and divide by 60.

eg : Convert 1h26min to decimal.

First, take just the minutes (26).

Then, do the operation : 26 / 60 = 0.433

Don't forget to add back the hour (1 + 0.433) and it gives 1.433h

For the opposite operation (converting decimal hours to normal hours), do the same as above but multiply by 60

eg : 0.433 * 60 = 26 minutes --> add the hour --> 1h26m

Given that, you can also calculate the functional time for a given amount of Uranium and the power usage doing so :

$ t=\frac {U} {P} $ | ||

Where:
| ||

$ t $ | : time (h) | |

$ U $ | : Uranium (kg) | |

$ P $ | : Power (MW) |

Finally, you can have the maximum power for the given amount of Uranium and time you want it to run :

$ P=\frac {U} {t} $ | ||

Where:
| ||

$ P $ | : Power (MW) | |

$ U $ | : Uranium (kg) | |

$ t $ | : time (h) |

## Example

Let's assume your ship, while idle, uses 4.92MW and you only have 2kg of Uranium. In how much time will your ship idle until it runs out of fuel?

For this we need to use the following formula :

$ t=\frac {U} {P} $ | ||

Where:
| ||

$ t $ | : time (s) | |

$ U $ | : Uranium (kg) | |

$ P $ | : Power (MW) |

First divide the Uranium (2kg) by he power usage (4.92MW) :

2 / 4.92 = 0.4065 h

Put this result in minutes by multiplying it by 60 (multiply by 3600 to get seconds) :

0.4065 x 60 = 24.39 min

So after about 24.39 min, the ship will run out of fuel.

## Inventory

With Realistic settings, the Large Reactor has an inventory with a capacity of 1,000L for small ships and 8,000L for large ships. The inventory only accepts the Uranium Ingots which fuel the reactor. If any of its inventory panels are attached to a conveyor system, the reactor will automatically pull ingots from valid inventories when it runs low.

## Recipe

Large Reactor | ||||
---|---|---|---|---|

Component | Large Ship/Station Required | Large Ship/Station Optional | Small Ship Required | Small Ship Optional |

Computer | 75 | — | 25 | — |

Motor | 20 | — | 5 | — |

20px Superconductor Conduits | 100 | — | — | — |

Reactor Components | 2,000 | — | 50 | — |

Large Steel Tube | 40 | — | 3 | — |

Metal Grid | 40 | — | 9 | — |

Construction Component | 70 | — | 9 | — |

Steel Plate | 800 | 200 | 40 | 20 |