Can someone give me the formula for burn time in seconds based on fuel units/ISP

[Silicon014]

Im trying to calculate how much burn time I have for an interplanetary ion-driven cruiser.

[Corax]

Im trying to calculate how much burn time I have for an interplanetary ion-driven cruiser.

It depends on your throttle setting; the less throttle, the longer the burn.

Keep in mind that fuel consumption calculations are "bugged" as of 0.16, i.e. the engines use disproportionately less fuel on lower throttle settings. Therefore I will assume full throttle setting throughout this experiment.

From Wikipedia:

F_thrust=I_sp*m_dot*g

F_thrust [N]

I_sp

m_dot [kg/s] (this is what you need to find: the rate at which fuel is burnt)

g=9.81 [m/s²], Earth's (or Kerbin's) standard acceleration

m_dot=F_thrust/I_sp*g

Now that you have m_dot, you divide your fuel mass (depending on your fuel tank(s)) by m_dot to get the burn duration in seconds.

For example, the LV-T30 ("liquidEngine1") has

• max thrust F_thrust=215 (kN)
  
• vacuum specific impulse I_sp=370 (s)
  

Plug that into the formula above, and you get

• m_dot=F_thrust/I_sp*g
  

• =215kN/370s*9.81m/s
  2

  
  

• =215000N/3629.7m/s
  3

  
  

• =59.23kg/s

  
  

the FL-T400 ("fuelTank") has a total mass of 2.25t, an empty mass of 0.25t, and a fuel capacity of 400 units. So the fuel mass is 2.25-0.25=2.0t

• t=m/m_dot
  

• =2000kg/59.23kg/s

  
  

• =33.77s

  
  

If you do the calculation with the I_sp at sea level, you get

• m_dot=215kN/320s*9.81m/s²
  

• =215000N/3139.2m/s
  3

  
  

• =68.49kg/s

  
  

and thus

• t=m/m_dot
  

• =2000kg/68.49kg/s

  
  

• =29.20s

  
  

which you can simply test for yourself:

Mission elapsed time at engine burn-out: 30s. Close enough, considering I_SP varies with atmospheric pressure (i.e. altitude), and the inaccuracy of the mission timer.

Oh, and as always, you should really retrace the steps for yourself and verify I didn't accidentally introduce any dimensional or other errors in my equations

Edited August 31, 2012 by Corax
higher resolution images

[Kosmo-not]

It depends on your throttle setting; the less throttle, the longer the burn.

m_dot=F_thrust/I_sp*g

You should use g₀ instead of g.

Also, keep in mind the order of operations. If you put that into your calculator, you would get basically m_dot=F_thrust*g₀/I_sp

So, rewrite it as m_dot=F_thrust/(I_sp*g₀)

[Corax]

You should use g₀ instead of g.

I give you that. But then, I should have written m_dot as á¹Â

Also, keep in mind the order of operations. If you put that into your calculator, you would get basically m_dot=F_thrust*g₀/I_sp

So, rewrite it as m_dot=F_thrust/(I_sp*g₀)

I refuse to bow to the stupidification through modern technology. If my calculator doesn't know how to interpret a formula in mathematical notation, it's so much more important that I do, and feed it to the calculator in the little bytes it can manage. I didn't mean to provide a how-to-program-your-calculator.

Regardless, commutativity says it doesn't matter whether you divide first and then multiply, or the other way around. 2*(10/4) = (2/4)*10 = (2*10)/4 = 5

Edited September 3, 2012 by Corax

[vexx32]

He's referring to how the way it was written being interpretable both ways. To avoid confusion in such a case, it is advisable to add the parenthesis to aid those who are not certain.

Haha, all these calculations make me feel like I ought to be planning my missions... Nah, I'll be alright

[Bluejayek]

I refuse to bow to the stupidification through modern technology. If my calculator doesn't know how to interpret a formula in mathematical notation, it's so much more important that I do, and feed it to the calculator in the little bytes it can manage. I didn't mean to provide a how-to-program-your-calculator.

Regardless, commutativity says it doesn't matter whether you divide first and then multiply, or the other way around. 2*(10/4) = (2/4)*10 = (2*10)/4 = 5

This isn't stupidification through modern technology; calculators use the order of operations that was established to make writing equations clear for all, and especially to make one standard that everybody uses.

As to your second point, this isn't about commutativity, since what you want to do is divide, not multiply.

A/B*C is what you wrote, which is NOT the same as A/(B*C) or A/B/C, which is what is correct, and in fact what you actually did.

And vexx, it isnt interpretable both ways! It is most clearly interpretable only in the one way, F/I*g = (F/I)*g. That is the accepted notation.