Formulas

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

Production per hour
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 = $$ \left( 10 \times L \times{1.1}^{L} \times \left(-0.004 \times \text{Average Temp of Planet} + 1.36 \right) \right) $$(1.44 not 1.36)

Cost
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 = $$ 180 \times {1.8}^{L-1} $$

Consumption
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 Time
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

Moon destruction
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
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
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