Archived from groups: alt.games.vgaplanets4 (More info?)
Like it is with guesses: Sometimes they are wrong. I assumed more complexity for the fuel comsumption as there actual is. So
excuse me and here it is how it should work:
First calculate the fuel consumption per kT and lj at given speed V as
q(v) = B(i) * ( 2^i * v / MaxV)^2 * WarpDrag/100 * (100%-PMBonuses)
MaxV is the max speed of the used engines
PMBonuses is the sum of the percentual plasma manifolds exotic tech bonuses
WarpDrag ist the warp drag factor of the used hull
i is either 2,1 or 0 and is dependent of in which interval of the three
[0, MaxV/4) [MaxV/4, MaxV/2) [MaxV/2, MaxV] the vlaues of v is.
B(i) is the either vC/4-K/100000 or vC/2-K/100000 or vC-K/100000
This are the burn rates per kTlj at the speeds MaxV/4, MaxV/2 and MaxV
Then the fuel consumption DF(d,v) as a function of the total distance d, which can be traveled within the next turn is at travel
speed v is given by
DF(d,v) = M*q(v)*d where M is the total mass at the beginning of the movement.
If one now sends a ship over several turn to a distant target and one wants to know how mach fuel is
needed then one have to do the following:
Assume the speed stays constant and only the fuel load is allowed to change due to fuel burning by the movement.
Then one can write abbreviated
Fn+1 = Fn-(Fn+H) q d = Fn(1-q d) -H q d
which leads to
Fn=F0 (1-qd)^n - H (1- (1-qd)^n)
d is here the distance which can be traveled within a whole turn.
Then one gets the fuel consumption over a distance of an integral multiple of d as
DF = M*( 1 - [1- q(v)*d]^n )
_______________________
n is the number of turns.
If one wants to know how many turns a fuel load lasts then one have resolve
for n:
n = LN[ H/(F+H)] / LN[1-q(v)*d]
___________________________
Here H is the total mass without the fuel mass.
If one wants to know how many fuel one have to take for the journey then one have to resolve
F=(H+F) (1 - [1- q(v)*d]^n ) for F:
F= H ( [1- q(v)*d]^(-n) -1 )
______________________
I hope this time it fits.
As mentioned changing orders for the plasma manifold exotechs are not displayed and also the client seems to display the mass of
the last turn. So one have to calculate the the mass manually.
The mass depends on
- nacked hull mass
- SupWs
- LWs
- SW, each 1kT
- PD, each 1kT
- Engines, each 1KT (?)
- personnel, 100 personnel=1kT
- Cargo load, don't forget pods
- docked Wings (?)
- Ord&Rep load, (?)
- Fuel load
- mc, 1000mc=1kT (?)
- Armor (?)
GFM GToeroe
Like it is with guesses: Sometimes they are wrong. I assumed more complexity for the fuel comsumption as there actual is. So
excuse me and here it is how it should work:
First calculate the fuel consumption per kT and lj at given speed V as
q(v) = B(i) * ( 2^i * v / MaxV)^2 * WarpDrag/100 * (100%-PMBonuses)
MaxV is the max speed of the used engines
PMBonuses is the sum of the percentual plasma manifolds exotic tech bonuses
WarpDrag ist the warp drag factor of the used hull
i is either 2,1 or 0 and is dependent of in which interval of the three
[0, MaxV/4) [MaxV/4, MaxV/2) [MaxV/2, MaxV] the vlaues of v is.
B(i) is the either vC/4-K/100000 or vC/2-K/100000 or vC-K/100000
This are the burn rates per kTlj at the speeds MaxV/4, MaxV/2 and MaxV
Then the fuel consumption DF(d,v) as a function of the total distance d, which can be traveled within the next turn is at travel
speed v is given by
DF(d,v) = M*q(v)*d where M is the total mass at the beginning of the movement.
If one now sends a ship over several turn to a distant target and one wants to know how mach fuel is
needed then one have to do the following:
Assume the speed stays constant and only the fuel load is allowed to change due to fuel burning by the movement.
Then one can write abbreviated
Fn+1 = Fn-(Fn+H) q d = Fn(1-q d) -H q d
which leads to
Fn=F0 (1-qd)^n - H (1- (1-qd)^n)
d is here the distance which can be traveled within a whole turn.
Then one gets the fuel consumption over a distance of an integral multiple of d as
DF = M*( 1 - [1- q(v)*d]^n )
_______________________
n is the number of turns.
If one wants to know how many turns a fuel load lasts then one have resolve
for n:
n = LN[ H/(F+H)] / LN[1-q(v)*d]
___________________________
Here H is the total mass without the fuel mass.
If one wants to know how many fuel one have to take for the journey then one have to resolve
F=(H+F) (1 - [1- q(v)*d]^n ) for F:
F= H ( [1- q(v)*d]^(-n) -1 )
______________________
I hope this time it fits.
As mentioned changing orders for the plasma manifold exotechs are not displayed and also the client seems to display the mass of
the last turn. So one have to calculate the the mass manually.
The mass depends on
- nacked hull mass
- SupWs
- LWs
- SW, each 1kT
- PD, each 1kT
- Engines, each 1KT (?)
- personnel, 100 personnel=1kT
- Cargo load, don't forget pods
- docked Wings (?)
- Ord&Rep load, (?)
- Fuel load
- mc, 1000mc=1kT (?)
- Armor (?)
GFM GToeroe
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