How to calculate a ships Delta v

[aceassasin]

I was interested in pre planning my missions to have a higher succses rate. I have read another about calcutalting Delta V for a manuvere but i need help calculating the delta v for my rocket can anybody help ???

[Addict7]

Ok, I\'m a bit sad to use my first post this way, I am also very interested to know how to calculate the delta-v of a rocket, as far as I need it in my WIP flight manager...

[togfox]

Funny you should ask ... I\'m working on the numbers myself. If you asked next week I\'d be able to help you ...

[PakledHostage]

The Wikipedia articles an the Tsiolkovsky rocket equation and Specific Impulse both give good background to your question plus the equations you\'re looking for, but here\'re a couple of sample calculations:

Example 1:

Q. What is the total delta-V of a small single stage stack having a parachute, MK1-Pod, decoupler, Large LFT and a LV-909 LFE?

Spacecraft mass including fuel (M_fuelled) = 5.1 tonnes

Fuel mass =2.2 tonnes

Spacecraft empty mass (M_empty) = 2.9 tonnes

LV-909 effective exhaust velocity (V_e) = 5682 m/s

DeltaV = V_e*ln(M_fuelled/M_empty)

DeltaV = 5682 * ln(5.1/2.9) = 3208 m/s

Note that 'ln' is the natural logarithm.

Example 2:

Q. How much delta-V has been used up if the stack from Example 1 shows 480.0 kg fuel remaining when the tank is right-clicked with the mouse?

First, the 480.0 'kg' value shown in the popup window (KSP v0.15) is mislabelled. It should be labelled as a unit of volume. The popup window shows 500 kg for a full tank but the specs for the large LFT in the VAB show it as having a volume of 500. The actual mass of fuel burned is 2.2 tonnes * (500-480)/500 = 0.088 tonnes.

Spacecraft starting mass including fuel (M_fuelled) = 5.1 tonnes

Mass of fuel burned =0.088 tonnes

Spacecraft final mass (M_empty) = 5.012 tonnes

LV-909 effective exhaust velocity (V_e) = 5682 m/s

DeltaV = V_e*ln(M_fuelled/M_empty)

DeltaV = 5682 * ln(5.1/5.012) = 98.9 m/s

Edit: Added Example 2.

[Addict7]

Awesome! Thank you very much!

[HarvesteR]

Mind though, that this doesn\'t translate into your final velocity following a launch. When launching, you 'lose' delta-V fighting gravity and atmospheric drag, so to reach the 2300m/s required for low Kerbin orbit, you\'ll actually need somewhere around 4000m/s of delta-v (you mileage may vary with launch heading, gravity turn rate and ship configuration).

Cheers

[aceassasin]

The Wikipedia articles an the Tsiolkovsky rocket equation and Specific Impulse both give good background to your question plus the equations you\'re looking for, but here\'s a sample calculation:

Spacecraft mass including fuel (M_fueled) = 5.1 tonnes (parachute-capsule-decoupler-Large LFT-LV-909 LFE)

Fuel mass =2.2 tonnes

Spacecraft empty mass (M_empty) = 2.9 tonnes

LV-909 effective exhaust velocity (V_e) = 5682 m/s

DeltaV = V_e*ln(M_fueled/M_empty)

DeltaV = 5682 * ln(5.1/2.9) = 3208 m/s

Note that 'ln' is the natural logarithm.

Thanks but im still kinda confused about the Natural Logarithm

[PakledHostage]

Thanks but im still kinda confused about the Natural Logarithm

Most scientific calculators have a 'ln' button and a 'log' button. In the case of those calculators, 'log' is log base 10, while 'ln' is log base 'e'. e is a mathematical constant much like its more famous cousin Pi.

If you\'re doing this calculation using a scientific calculator, use the 'ln' button

If you\'re doing this calculation using Microsoft Excel, type '=LOG(5.5/2.9)' in the cell.

Note: Excel\'s log function returns the natural log. You\'d need to specify '=LOG(X,10)' if you wanted log base 10 of X.

[aceassasin]

Most scientific calculators have a 'ln' button and a 'log' button. In the case of those calculators, 'log' is log base 10, while 'ln' is log base 'e'. e is a mathematical constant much like its more famous cousin Pi.

If you\'re doing this calculation using a scientific calculator, use the 'ln' button

If you\'re doing this calculation using Microsoft Excel, type '=LOG(5.5/2.9)' in the cell.

Note: Excel\'s log function returns the natural log. You\'d need to specify '=LOG(X,10)' if you wanted log base 10 of X.

Thanks a bunch i found the button on my graphing calculator

[Endeavour]

How do you find the exhaust velocity or specific impulse for the engines? Is it given somewhere?

[PakledHostage]

How do you find the exhaust velocity or specific impulse for the engines? Is it given somewhere?

You calculate it.

From the VAB specs for the LV-909 engine:

Thrust = 50 kN at max throttle

Fuel flow = 2 litre per second at max throttle

Fuel density = (2.2 tonne/500 litre) = 4400 kg/m³

mass flow rate of fuel = 8.8 kg/s

I_sp = Thrust/(mass flow rate of fuel * g_kerbin) = 579.4 seconds

Effective exhaust velocity (V_e) = g_kerbin * I_sp = 5682 m/s

Note 1: Because the fuel flow and thrust both vary linearly with throttle setting, the effective exhaust velocity for the LV-909 engine is the same for all throttle settings.

Note 2: The effective exhaust velocity of the LV-T30 is the same as the LV-909 because the ratio of thrust to maximum fuel flow is the same for both engines.

Note 3: I\'m using SI units for the values given in the VAB because it makes the results of these and other calculations match the actual performance of our in-game rockets.

Note 4: The fuel density actually changes, depending on which fuel tank you use. I\'m using the fuel density for the FL-T500 fuel tank in these calculations.

[CoolMatthew]

How do you find the exhaust velocity or specific impulse for the engines? Is it given somewhere?

First find the fuel consumption of the engine in kg/s (it is given in 'fuel units'/s).

For example: A FL-T500 Fuel Tank is 2.5 kg full and .3 kg dry and holds 500 units of fuel...

So 500 units of fuel have a mass of 2.2 kg -> 0.0044 kg each

A LV-T30 LFE has a burn rate of '8' or 8 units/s or 0.0352 kg/s

Then take the thrust (which is given in Newtons) of the engine divided by the fuel consumption.

(Remember that Newtons are just m*kg/s²)

Example: A LV-T30 LFE has a burn rate of 0.0352 kg/s and gives 200 N or 200 m*kg/s² of force...

so 200/0.0352 and (m*kg/s²)/(kg/s) -> 5681.8 m/s

EDIT: okay so now you know how to do it without I_sp too...

[CoolMatthew]

Note 3: I\'m using SI units for the values given in the VAB because it makes the results of these and other calculations match the actual performance of our in-game rockets.

What?

EDIT: never mind I had International System of Units and Specific Impulse confused...

[Endeavour]

Thank you both. Now I can actually plan my rockets.

[TouhouTorpedo]

You calculate it.

From the VAB specs for the LV-909 engine:

Thrust = 50 kN at max throttle

Fuel flow = 2 litre per second at max throttle

Fuel density = (2.2 tonne/500 litre) = 4400 kg/m³

mass flow rate of fuel = 8.8 kg/s

I_sp = Thrust/(mass flow rate of fuel * g_kerbin) = 579.4 seconds

Effective exhaust velocity (V_e) = g_kerbin * I_sp = 5682 m/s

Very good, but the g value in ISP equations is always earth, 9.81. Doesn\'t matter if you\'re in space, on earth, on mars. Putting g(kerbin) is going to confuse people!

[aceassasin]

Just one last question if you have multiple engines do you add the exhaust velocity of all the engines together

[TouhouTorpedo]

No. For multiple engines -

ISP (m/s) = Total Thrust / Total mass flow rate

So for multiple engines that are all the same, the ISP does not change. For some more effecient and some less effecient engines in a stage, you\'ll get something around the middle.

[PakledHostage]

Very good, but the g value in ISP equations is always earth, 9.81. Doesn\'t matter if you\'re in space, on earth, on mars. Putting g(kerbin) is going to confuse people!

You\'re right, but you\'re kinda 'picking nits' by pointing it out... I chose to write g_kerbin instead of g₀ because I felt there could be an equal number of people who\'d be confused either way. The value used for g₀ is, after all, arbitrary and was chosen merely as a convenient convention. There\'s nothing magic about it. And ultimately, the values are the same on Earth and Kerbin.

If I made an error, it was writing the equation for V_e in terms of I_sp. I should have written the equation for V_e in terms of the more fundamental mass flow rate of fuel and thrust. I_sp is somewhat immaterial to this discussion anyway.

But while I\'m here, I thought I\'d add another example calculation of delta-V, this time for a two stage rocket.

For a multiple stage rocket, you just add the delta-V for each individual stage, treating the upper stages as payload for the lower stages. For the rocket shown below:

Stage 1:

fuelled mass = 8.35 tonnes

mass of fuel burned = 1.05 tonnes

V_e of LV-T30 engine = 5682 m/s (ignoring the effect of the small difference in fuel density between the two LFTs)

delta-V of the lower stage is 763.6 m/s

Stage 2:

fuelled mass = 4.3 tonnes

mass of fuel burned = 2.2 tonnes

V_e of LV-909 engine = 5682 m/s

delta-V of the upper stage is 4072.0 m/s

Total:

delta-V of the entire stack = 4836 m/s

Interestingly, if you reverse the order of fuel tanks on the stack shown above (using the large tank on the lower stage), you get a total delta-V of only 4135 m/s (again, ignoring the effect of the small difference in fuel density between the two LFTs). That\'s enough of a difference that the first stack can comfortably make orbit while it may not even be possible to reach orbit with the second one!

[CaptainArbitrary]

From the VAB specs for the LV-909 engine:

Thrust = 50 kN at max throttle

Fuel flow = 2 litre per second at max throttle

Fuel density = (2.2 tonne/500 litre) = 4400 kg/m³

mass flow rate of fuel = 8.8 kg/s

For example: A FL-T500 Fuel Tank is 2.5 kg full and .3 kg dry and holds 500 units of fuel...

So 500 units of fuel have a mass of 2.2 kg -> 0.0044 kg each

A LV-T30 LFE has a burn rate of '8' or 8 units/s or 0.0352 kg/s

LV-909 effective exhaust velocity (V_e) = 5682 m/s

Bit nitpicky, I suppose, but none of these is actually anywhere near being correct, I don\'t think.

'Fuel units', in the game, are not volumetric. They\'re mass units. Specifically, they\'re kilograms. Context-click on any fuel tank in the game to see this spelled right out. I\'ll come back to this fact in a second.

Now, the relevant equation here is this one:

F = I_sp ?

where F is thrust in units of mass acceleration, Isp is specific impulse in units of velocity and ? is the mass flow rate of propellant (dm/dt, in other words).

A little sidebar here for dimensional analysis. Specific impulse is impulse per unit of propellant mass. Okay, but what\'s impulse? Impulse is the integral of mass acceleration over time. If you accelerate a one-kilogram mass by one meter-per-second-per-second for a total of one second, you\'ll have given that mass an impulse of one …in units of kilogram-meters-per-second, which is another way of saying momentum. (Some people explain it this way: impulse is force applied over time, but since 'force' is just a nickname for the time derivative of momentum, all that really means is that impulse is the finite change in momentum that results from accelerating a mass for an amount of time.)

Okay, so impulse is momentum, then. What\'s specific impulse? It\'s what we said before: Specific impulse is impulse per unit of propellant mass. Doing something with a unit of propellant mass on a rocket ship will result in a change of momentum for that rocket ship. Exactly what that change in momentum will be depends on what you do with that unit of propellant mass! If you just sit there and stare at it, or play hide-the-hydrogen with it, or whatever, then your change in momentum will be zero. But if you blast that unit out the back of the rocket somehow, then that change in momentum will be … well, something that\'s not zero, anyway. So specific impulse depends very much on what we\'re talking about. In this context, we simplify things like crazy-cuckoo-bananas and say that specific impulse is a thing rocket engines have. In real life, the actual specific impulse developed by an engine will depend on a huge number of independent factors, but in the game, we can just say that it is what it is.

But what is it? Like, specifically?

Well, it\'s this:

I_sp = F / ?

See what I did there? I just rewrote the equation from up above, moving the ? from one side to the other. See, in the game we have both thrust and mass flow rate as givens; they map to the part.cfg parameters of [tt]maxThrust[/tt] (unsurprisingly) and [tt]fuelConsumption[/tt]. At max throttle, a liquid engine in KSP will develop [tt]maxThrust[/tt] units of thrust, consuming [tt]fuelConsumption[/tt] units of propellant mass per second. So we can just read the numbers out of part.cfg and do the arithmetic to find out what exactly the Isp of a given in-game engine really is.

If we look in particular at the LV-909 engine used elsewhere in thread as an example, we find that [tt]maxThrust[/tt] is 50 and [tt]fuelConsumption[/tt] is 2. Okay, but in what units? Long story short, the units are kilonewtons and kilograms per second. I\'ll leave demonstrating that as an exercise for the reader, cause I\'m getting tired of typing and want to start wrapping this up.

If you do the arithmetic, dividing 50 kilonewtons by 2 kilograms per second, you get 25,000 meters per second. That\'s the specific impulse — in meters per second —of the LV-909 engine in KSP. I say 'in meters per second' specifically there because there are two ways to quantify specific impulse: there\'s as I\'ve done here, in terms of thrust and mass flow of propellant, and then there\'s another equivalent (but different) way that uses thrust and weight of propellant. This was very much the way it was done back in the early days of rocketry, because weights (in pounds) were customarily much preferred to masses (in slugs, and I\'ll be a bit surprised if you\'ve even heard of that unit, which sorta demonstrates my point). If you quantify specific impulse by using pounds of thrust and pounds per second of mass flow, you end up canceling out the pounds and being left with just seconds. The difference between Isp-as-seconds and Isp-as-a-velocity is just a constant numerical factor; it\'s the numerical factor that relates our units of weight and mass. That\'s g, the acceleration due to gravity of a freely falling body at sea level on the Earth. Except since we\'re talking about Kerbin and not Earth here, isn\'t it the acceleration at Kerbin sea level? No! The definition of our unit of weight has nothing to do with Kerbin\'s surface gravity; it\'s Earth\'s surface gravity that provides the constant conversion factor we need to go from kilograms to units of weight. The fact that that constant conversion factor was based on an empirical quantity related to our planet is irrelevant …which is why, frankly, it\'s best by far to just skip the whole notion of Isp-as-seconds and stick to Isp-as-thrust-over-fuel-consumption. Makes the math easier.

(That said, we can do the arithmetic and see that an Isp-by-mass of 25,000 meters per second is equivalent of an Isp-by-weight of about 2,550 seconds …which is nearly six times the very high Isp of LH2/LOX rocket propellant, but whatever, that\'s just one of the myriad ways in which the 'stock' parts that come with KSP are just temporary placeholders.)

Okay, so now that we know the Isp of the LV-909, we can apply the rocket equation. You\'ve seen others do it here, but I\'ll run through it just show you again how it\'s done. The rocket equation is:

?v = I_sp log m? / m?

Here Isp is your specific impulse in meters per second, just like we figured out earlier, m? is the total mass of your rocket ship before you start the engine, and m? is the total mass of your rocket ship after you\'ve stopped the engine. The difference between m? and m? is the amount of propellant, in units of mass, you used during the burn.

So say your rocket has a fully fueled mass of 30 tons, of which 5 tons is the propellant you plan to burn. That makes m? equal to 30,000 kilograms and m? equal to 25,000 kilograms —your starting 30 tons minus the 5 tons of fuel you plan to get rid of during your burn. If you\'re using the LV-909 engine, with an Isp of 25,000 meters per second, your ?v is:

?v = 25,000 m/s log (30,000 kg / 25,000 kg)

The units of mass (as well as three of those zeros) just cancel out on the right-hand side, making it:

?v = 25,000 m/s log (30/25)

You can find the logarithm of 30/25 by looking it up in a table or using a calculator; it\'s about 0.182. And 25,000 meters per second times 0.182 is about 4,558 meters per second.

Just doing a quick sanity check, we see that the answer we got has units of velocity —meters per second —which is exactly what we were looking for. So it seems we have a good answer here. It turns out that using the LV-909 on a 30-ton rocket to burn 5 tons of propellant will give you about 4,500 meters per second of ?v — which incidentally is pretty close to what you\'d need to get from the launch pad to LKO, so …you know. That\'s good.

Now, I typed all that out in kind of a rush, so while I\'m fairly sure all the math is right, I make no actual promises. Anybody who finds a mistake in it will …um …receive a prize.

[Endeavour]

So if you want to use exhaust velocity in your equation, you use the natural log, but to use specific impulse you use log base 10?

[UmbralRaptor]

'Fuel units', in the game, are not volumetric. They\'re mass units. Specifically, they\'re kilograms. Context-click on any fuel tank in the game to see this spelled right out. I\'ll come back to this fact in a second.

I\'m fairly sure that the context menu is wrong. In particular, note that the ratio fuel mass from the CFG file to fuel units varies from tank to tank.

I just treat them as arbitrary fuel units to be cancelled out as soon as possible:

LV-909 == 50 kN/(2 fuel units / second) == 25 (kN * second)/(fuel unit)

FL-T500 == 500 fuel units / (2.5-0.3 tonnes) ~= 227.27 fuel units/tonne

25 (kN * second)/(fuel unit) * 227.27 fuel unit/tonne ~= 5681.818 (kN * second) / tonne) == 5681.818 (tonne*meter*second)/(second^2*tonne) == 5681.818 m/s

As an example, here\'s a rocket out in flatish space before and after burning 1 tank of fuel:

?V == Ve*ln(Mi/Mf)

Solving for Ve: Ve == ?V/ln(Mi/Mf)

?V == 2440.8 m/s

Mi == 1 + 0.8 + 2.5 + 2 == 6.3

Mf == 1 + 0.8 + 2.5 + 2 == 4.1

Substituting: Ve ~= 5682 m/s (Which is what I would expect for an LV-T30, T45, or 909 attached to an FL-T500)

I\'ve included the SFS file in an attachment if anyone wants to duplicate it.

edit:

So if you want to use exhaust velocity in your equation, you use the natural log, but to use specific impulse you use log base 10?

You always use a natural log with this. For converting from exhaust velocity to specific impulse, you divide by a factor of 9.81, and for converting from specific impulse to exhaust velocity you multiply by a factor of 9.81.

[PakledHostage]

Thanks, UmbralRaptor, for confirming my and coolMathew\'s results through your own analysis and experimentation!

CaptainArbitrary, does UmbralRaptor get a prize?

[Absolution]

Would someone be kind enough to indulge me and double check my math? These are the relevant values for a rocket I am modding from the ground up. You will notice the values are a lot higher than a 'stock' KSP part. This is due to my effort to design a more realistic rocket. The reason why I ask is that my delta-V appears to be notably higher than what I\'ve seen considered to be sufficient for obtaining orbit (~6300 vs ~4500). I can achieve an orbit of 150km reasonably well with this rocket and find it to be fairly easy to fly and thus assume my delta-V is not being wasted too badly.

-----------------------------------------------------------------

Tank and Engine Properties:

LFE-7E3-A1 (Stage 1 Engine) = 2000kN thrust, 62 units/s burn rate

LFE-7E1-A1 (Stage 2 Engine) = 373kN thrust, 14.17 units/s burn rate

LFE-705-A1 (Stage 1 Tank) = 82.51 tonnes (mass of fuel in tank not including tank mass), 7500 units Fuel

LFE-702-A1 (Stage 2 Tank) = 33 tonnes (mass of fuel in tank not including tank mass), 3000 units Fuel

*I determined a standard fuel density then calculated how much fuel could fit into the tank I modeled accounting for its internal volume. So fuel density should be identical for each tank.

-----------------------------------------------------------------

delta V Calculations:

Stage 1:

fueled mass = 132.79 tonnes

mass of fuel burned = 82.15 tonnes

Ve of LFE-7E1-A1 = 2932.55 m/s

delta-V of the lower stage is 2848 m/s

Stage 2:

fueled mass = 42.96 tonnes

mass of fuel burned = 33 tonnes

Ve of LFE-7E1-A1 engine = 2392.34 m/s

delta-V of the upper stage is 3497 m/s

Total:

delta-V of the entire stack = 6345 m/s

-----------------------------------------------------------------

Does that add up?

[Endeavour]

A question for the math savvy people here. With the specific impulse now changing with altitude, does it require some more complicated math to get an accurate measurement?

[UmbralRaptor]

@PakledHostage: Yay!

@Endeavour: In principle, yes. In practice, I\'d be tempted to ignore it and just pretend there are additional atmospheric ?V losses. Especially since how much it impacts your available ?V depends on your ascent profile, so unless you feel like doing line integrals...

@Absolution: I get:

Isp1 == (2000*7500) / (62*82.51) ~= 2932.1959 m/s

Isp2 == (373*3000) / (14.17*33) ~= 2393.0198 m/s

?V1 == 2932.1959 * ln((132.79/(132.79-82.15)) ~= 2826.716 m/s

?V2 == 2393.0198 * ln(42.96/(42.96-9.96)) ~= 631.187 m/s

? ~= 4129 m/s

For the 3497 m/s, I\'m guessing you did 2392.34*ln(42.96/9.96)?