## FANDOM

316 Pages

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

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

### CostEdit

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

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

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

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

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

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

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

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

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

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

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

### ConsumptionEdit

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