Talk:Fuel Consumption

Ships by fuel consumption
I've added the "Ships by fuel consumption" table. 91.176.227.119 18:47, 15 September 2008 (UTC)

Fuel use algorithm?
Does anyone know the method used for figuring out fuel use?

Theres a number of things that contribute to the end fuel cost: Deployment speed, The ships base fuel use, and the destination.


 * It is a quiet complex algorithm as soon as you involve different ships. The consumption for each group of ships (ships of same type) is calculated through this:


 * $$ Speed = \frac{35000}{\mbox{duration} \times \mbox{speedfactor} - 10} \times \sqrt{\frac{\mbox{distance} \times 10}{\mbox{shipSpeed}}} $$


 * $$ Consumption = \frac{\mbox{basicConsumption} \times \mbox{numberOfThisShip} \times \mbox{distance}}{35000} \times \left(\frac{\mbox{speed}}{10} + 1\right)^2$$


 * To that you sill need the distance algorithm and then there is a simple addition for holding (usedd with ACS defense and expeditions). If you just want to calculate a specific consumption, I suggest using one of the tools listed under External Links. -TheDwoo 13:39, 20 December 2007 (UTC)

Weylin: Thanks for that :) I realise there are tools to calculate it, I was just curious as what equations ogame uses for fuel use. How does it figure fuel use with coordinates and multiple ships? Im still kinda lost on that


 * I'll try and do it from the top then.
 * Lets start by calculating the distance;
 * Between planet/moon/DF at the same coord: $$5$$
 * Within the same system: $$1000 + (5 \times \mbox{Number of Planets})$$
 * Between different systems: $$2700 + (95 \times \mbox{Number of Systems})$$
 * Between different galaxies: $$20000 \times \mbox{Number of Galaxies}$$
 * (The number is $$|(\mbox{start planet/system/galaxy})-(\mbox{destination planet/system/galaxy})|$$)
 * The we need the max speed of the combined fleet, that is the speed slowest ship that is participating. ($$Speed = \mbox{Base Speed} \times (1 + \mbox{Drive Level} \times \mbox{Drive Bonus Factor})$$)
 * Next is the duration of the flight (in seconds), here the speed percentage weights in; 100% = 10, 50% = 5, etc... $$Duration = \frac{35000}{\mbox{speedPercentage}} \times \sqrt{\frac{\mbox{distance} \times 10}{\mbox{maxSpeed}}} + 10$$
 * Now we finally have all we need to use the first two formulas, so lets find the sum (one way): $$\mbox{basicConsumption}_k =$$ the fuel consumption rate of the $$k$$th ship $$\mbox{number}_k =$$ amount of the $$k$$th ship $$\mbox{speed}_k = \frac{35000}{\mbox{duration} - 10} \times \sqrt{\frac{\mbox{distance} \times 10}{\mbox{shipSpeed}}}$$ $$TotalConsumption = \sum_{k=1}^n \frac{\mbox{basicConsumption}_k \times \mbox{number}_k \times \mbox{distance}}{35000} \times \left(\frac{\mbox{speed}_k}{10} + 1\right)^2$$
 * Example: 10 Large Cargo ships, between two adjacent planets. Their basic fuel usage is 50 and their basic speed is 7.500.
 * Same system, adjacent planets: $$Distance = 1000 + (5 \times 1) = 1005$$
 * Only one ship; a level 10 Combustion Drive gives $$Speed = maxSpeed = 7500 \times (1 + 10 \times 0.1) = 15000$$
 * At 100%: $$Duration = \frac{35000}{10} \times \sqrt{\frac{1005 \times 10}{15000}} + 10 \approx 2875$$
 * And it sums up to: $$\mbox{speed} = \frac{35000}{2875 - 10} \times \sqrt{\frac{1005 \times 10}{15000}} \approx 10$$ $$TotalConsumption = \frac{50 \times 10 \times 1005}{35000} \times \left(\frac{10}{10} + 1\right)^2 \approx 58$$
 * That one was kind of basic and you question was about multiple ships (I'm guessing different types) so lets do a more complex one; 16 Light Fighters (fuel: 20/speed: 12500), 8 Heavy Fighters (75/10000) and 4 Cruisers (300/15000). Start from 1:1:1 and destination 1:5:3.
 * Different systems: $$Distance = 2700 + (95 \times |5-1|) = 3080$$
 * Three different ships; level 10 Combustion Drive and level 7 Impulse Drive:
 * Light Fighter $$Speed = 12500 \times (1 + 10 \times 0.1) = 25000$$
 * Heavy Fighter $$Speed = 10000 \times (1 + 7 \times 0.2) = 24000$$
 * Cruiser $$Speed = 15000 \times (1 + 7 \times 0.2) = 36000$$
 * $$maxSpeed = min(25000, 24000, 36000) = 24000$$
 * At 80%: $$Duration = \frac{35000}{8} \times \sqrt{\frac{3080 \times 10}{24000}} + 10 \approx 4966$$
 * First the sums for the three different ships, and then the total sum:
 * Light Fighter $$\mbox{speed}_1 = \frac{35000}{4966 - 10} \times \sqrt{\frac{3080 \times 10}{25000}} \approx 7.8$$ $$consumption_1 = \frac{20 \times 16 \times 3080}{35000} \times \left(\frac{\mbox{speed}_1}{10} + 1\right)^2 \approx 89$$
 * Heavy Fighter $$\mbox{speed}_2 = \frac{35000}{4966 - 10} \times \sqrt{\frac{3080 \times 10}{24000}} \approx 8$$ $$consumption_2 = \frac{75 \times 8 \times 3080}{35000} \times \left(\frac{\mbox{speed}_2}{10} + 1\right)^2 \approx 172$$
 * Cruiser $$\mbox{speed}_2 = \frac{35000}{4966 - 10} \times \sqrt{\frac{3080 \times 10}{36000}} \approx 6.5$$ $$consumption_2 = \frac{300 \times 4 \times 3080}{35000} \times \left(\frac{\mbox{speed}_13}{10} + 1\right)^2 \approx 289$$
 * Sum $$TotalConsumption = \sum_{k=1}^3 consumption_k = 550$$
 * That is one really long answer, hope that you understand now. Please comment on how to improve so that is becomes fit for the article. -TheDwoo 00:49, 22 December 2007 (UTC)

thanks! :-)