Calculating burn time from twr

[GrihanOs]

How do I calculate engine burn time from the ship's twr (for given d/v value)?

Edited December 27, 2013 by GrihanOs

[Smidge204]

Time = Delta-V / (TWR * 9.81)

9.81 is assuming TWR is relative to Kerbin (substitute surface gravity for any other body).

=Smidge=

[Dave Kerbin]

I'm pulling this from memory and sort of figuring it out as I write, anyone feel free to point out if I've incorrectly inverted the delta-v equation or screwed up how to determine the correct average TWR.

For a fast estimate which is useful for burns of a minute or two using efficient engines, but less accurate for longer burns or big fuel guzzlers you can just divide the Delta-v by TWR * 9.8 to get the rough number of seconds for the burn. You can divide that by 2 to figure out how many seconds before the node to start burning for best results.

To calculate the exact time you'll need to know more then just the TWR, you'll need enough information to know what ratio of the ship's mass (fuel) will be burned since as mass is reduced the TWR will increase (you'll be accelerating faster near the end of the burn then at the beginning).

Start with a couple of figures,

The ISP and total thrust of the engine(s).

Let's assume we have a LV-N atomic rocket so those figures are 800 ISP and 60 kN thrust

The current mass of the ship

Let's assume 10 tons

The delta-v you need

Let's assume 2000m/s

From here we can calculate what the final mass of the ship will be. We determine this with Mass / EXP(deltaV / ISP / 9.8). In the example it's 10 / EXP(2000 / 800 / 9.8) which gives us a final mass of 7.75 tons.

From there we can find the average TWR for the burn, which is Thrust / ((Initial_Mass + Final_Mass) / 2). In our example that's 60 / ((10 + 7.75) / 2) or 6.76.

Plug the average TWR into our original equation and we get 2000 / (6.76 * 9.8) and we get 30.19 seconds.

[GrihanOs]

Thank you!

[numerobis]

...30.19 seconds...

I'm getting 294s, which I think is because you calculated your average TWR incorrectly -- there's no 9.8 in that line. I'm not 100% convinced the average TWR is valid, but I'm not 100% convinced it's wrong either. You're certainly getting good-enough figures, once you fix the bug.

Smidge's approximation is generally pretty good. In Dave Kerbin's scenario, Smidge's approximation tells you you're burning for 2000 / (60 / 10) = 333s. That's only 10% longer than the truth. The approximation gets worse if you expel a greater fraction of the spacecraft in the burn.

How I calculate the burn time: first, calculate the amount of propellent we need to burn, then the amount of propellent we burn in a second, and divide.

First off, let's use exhaust velocity to simplify the equations:

Ve = Isp g0

That's 7856 m/s for an LV-N in the KSP world, where g0 = 9.82.

Propellent mass: the ideal rocket equation states:

m1 = m0 e^{dV / Ve}

The difference in mass is the propellent we're burning; write it in terms of values we know:

mprop = m1 - m0 = (1 - e^{-dV/Ve}) m1

With Dave Kerbin's stats (dV = 2000 m/s, Isp = 800s, m1 = 10t) we need 22.5% of the mass to be propellent, which is 2.25 tonnes.

The mass flow of an engine is

mdot = thrust / Ve

For a LV-N, that's 60 kN / 7856 = 7.64 kg/s.

Divide the mass of propellent we need to burn by the amount we burn per second, and we get the number of seconds we need to burn:

t = mprop / mdot = (1 - e^{-dV/Ve}) m1 Ve / thrust

For Dave Kerbin's scenario, that's

t = 0.225 * 10t * 7856 m/s / 60 kN = [b]294.2856[/b]... s

Edited December 27, 2013 by numerobis