Many of the formulas used in OGame are known. Using formulas can give you a big advantage over people playing without using them.

Mines/Power plantsEdit

T = Max Temperature
L = Structure Level

Production per hourEdit

Metal Mine =  \left( 30 \times L \times {1.1}^{L} \right)

Crystal Mine =  \left( 20 \times L \times {1.1}^{L} \right)

Deuterium Synthesizer =  \left( 10 \times L \times {1.1}^{L} \times \left( -0.002 \times T + 1.28 \right) \right)

Solar Plant =  \left( 20 \times L \times {1.1}^{L} \right)

Fusion Reactor =  Energy = \left( 30 \times L \times \left( 1.05 + {\mbox{Level of Energy Technology}} \times 0.01 \right)^L \right)

Solar Satellites =  \left( \frac{T}{4} + 20 \right)

On redesigned Uni's, the Deuterium production appears to be different. Fit to the actual production over an extended range of T and L yields the relation:

Deuterium Synthesizer =  \left( 10 \times L \times{1.1}^{L} \times \left(-0.004 \times \text{Average Temp of Planet} + 1.36 \right) \right)


Metal Mine

Metal =  60 \times {1.5}^{L-1}

Crystal =  15 \times {1.5}^{L-1}

Crystal Mine

Metal =  48 \times {1.6}^{L-1}

Crystal =  24 \times {1.6}^{L-1}

Deuterium Synthesizer

Metal =  225 \times {1.5}^{L-1}

Crystal =  75 \times {1.5}^{L-1}

Solar Plant

Metal =  75 \times {1.5}^{L-1}

Crystal =  30 \times {1.5}^{L-1}

Fusion Reactor

Metal =  900 \times {1.8}^{L-1}

Crystal =  360 \times {1.8}^{L-1}

Deuterium =  180 \times {1.8}^{L-1}


Metal Mine power =  \left( 10 \times L \times {1.1}^{L} \right)

Crystal Mine power =  \left( 10 \times L \times {1.1}^{L} \right)

Deuterium Mine power =  \left( 20 \times L \times {1.1}^{L} \right)

Fusion Plant deuterium =  \left( 10 \times L \times {1.1}^{L} \right)

Production TimeEdit

Ships and Defense:  \frac { \mbox{Structural Integrity} } {2500 * (1+ \mbox{Shipyard Level})* (\mbox{Universe Speed})*{2}^{ \mbox{Nanite Factory Level}}}   hours

Buildings:  \frac { \mbox{(Metal Cost + Crystal Cost)} } {2500 * (1+ \mbox{ RoboticFactory Level})* (\mbox{Universe Speed})*{2}^{ \mbox{Nanite Factory Level}}}   hours

Research:  \frac { \mbox{(Metal Cost + Crystal Cost)} } {1000 * (1 + \mbox{ Research Lab Level}* (\mbox{Universe Speed}) ) }   hours

Planet / moon related Edit

Moon destruction Edit

The chance that the moon is destroyed is: \left ( 100 - \sqrt { Moonsize } \right ) * \sqrt { \mbox { Number of Death Stars } }

The chance that the entire attacking fleet is destroyed is: \left ( \sqrt{Moonsize} \right ) \ /\ 2

Fields Edit

The formula used for determining the maximum amount of fields on a moon: \left ( \mbox{ Moon size / 1000 } \right ) ^ 2

The formula used for determining the number of fields on which can be built on a moon: 1 + \left ( \mbox{ Level of Lunar Base } *\ 4\ \right )

However, every level of the Lunar Base uses a field. This means the effective number of fields per level of Lunar Base is 3.

Captured Resources Edit

How are resources captured when you attack a planet

Please note that you can only capture up to 50% of the overall resources on a planet.

Also note that you can only capture 50% of each resource.

For example,if you attack a planet which has 20,000 metal, 20,000 crystal and 10,000 deuterium, you will get be able to capture a maximum of 25,000 resources overall {(20k+20k+10+)/2}.

Now here is how each resources are captured and its called the "plunder algorithm":

  1. Fill up to 1/3 of cargo capacity with metal
  2. Fill up to half remaining capacity with crystal (total of 1/3 cargo capacity)
  3. The rest will be filled with deuterium
  4. If there is still capacity available fill half with metal
  5. Now fill the rest with crystal

External links Edit

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