One of the most important values when talking about rocket engines is the so-called specific impulse. I have read the Wikipedia article but still have trouble understanding what it really is and what its value in seconds means. What does this value really tell us?

Here's what learned so far:

  • The specific impulse the change in momentum an engine can produce with one "unit" of propellant. (What a "unit of propellant" is in this regard is still unclear to me). Momentum is the potential to change the velocity and direction of a mass, right? Momentum change divided by mass gives us the velocity change, is that correct?
  • The unit most often used is seconds, while the "correct" units would be either N·s/kg (my math is rusty, but isn't that equivalent to just m/s?) or lbf·s/lb. If the "correct" value-with-units is multiplied with standard gravity you end up with a value in seconds which is easier to compare.

But what is a useful thought experiment or example to tell me what this seconds value means? Is it equivalent to letting something fall for x seconds in a standard gravity field in vacuum?

(The question Why do we use the gravitational acceleration at the Earth's surface when calculating the specific impulse of an engine?, which this question has been marked as a duplicate of, deals with the specifics for why g0 (standard gravity) is used but IMHO doesn't give an explanation about what the meaning of specific impulse actually is in layman's terms.)


marked as duplicate by Russell Borogove, Hohmannfan Mar 25 '17 at 18:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


In simplest terms it is just the thrust produced divided by the propellant flow rate. "How much thrust am I getting for the propellant I am expending?"

So bigger is better - you are getting more thrust for the same propellant flow rate. Bigger numbers = more efficient engine = extracting more thrust from the same amount of propellant.

It is a key parameter for analysis, and is a useful figure of merit for comparing systems. Like "gas mileage" for cars. Don't fret over the units.

  • $\begingroup$ So if I pump a certain amount of fuel per second into the engine, the specific impulse is the factor to derive the thrust/force per second I get from the engine? $\endgroup$ – DarkDust Mar 25 '17 at 15:02
  • $\begingroup$ "Thrust per second" is not what you want (just thrust), but yes, for a shuttle engine with specific impulse of ~450, it uses ~1000 lbs/sec of propellant, and produces ~450,000 lbs thrust. $\endgroup$ – Organic Marble Mar 25 '17 at 15:03
  • $\begingroup$ Given various "fuel" components, do you need to qualify/constrain your statement to engine performance vs economics OR is there a "missing" that includes economics. $\endgroup$ – Enigma Maitreya Mar 25 '17 at 15:13
  • 2
    $\begingroup$ Strictly Specific impulse is "How much Impulse am I getting for the propellent I am expending". Impulse = Force x time so Specific impulse = Force x time / mass. There is a related, not identical, question here space.stackexchange.com/questions/20448/… . I still prefer my answer to that question despite the lack of votes! $\endgroup$ – Puffin Mar 25 '17 at 15:17

If you want to understand how the 'seconds' value fits the greater image, there's this rather contrived definition (which nobody uses because it's contrived and mostly useless but evocative enough.)


$I_{sp}$ in seconds is equal to the amount of time a rocket must be fired to use a quantity of propellant with weight (measured at one standard gravity) equal to its thrust.

Imagine a test setup: rocket engine plus dummy weight, such that the total mass (engine+weight) is 100kg (kilogram-force, if you want to nitpick the units).

You drive an external, flexible fuel pipe from a fuel tank which stores 100kg of fuel+oxidizer (the same as the test rig). You start the engine and keep the thrust so that it hovers, without rising or falling. You measure time from the engine start until all fuel is spent, and when it is, the time is your specific impulse. The longer the better obviously, more acceleration from the fuel.

Obviously this is not a very practical test, and this definition is not very helpful - it helps imagine why specific impulse is expressed in seconds, and what these seconds mean... but honestly, beyond "understanding" nobody cares.

What really matters are other, more practical uses of specific impulse:


$I_{sp} = { v_e \over g_0 } $

This is one trivial equation. You have the $g_0$ which is the earth gravitational acceleration, a trivial conversion factor, a constant - and you have $v_e$ - exhaust velocity, speed at which propellant is ejected from the engine. That's it. The only variable - and so you can think of specific impulse as speed of exhaust gases, only multiplied by some constant for convenience. It's not some magical property dependent on a hundred weird and obscure factors - it's just the speed of the exhaust gas. Only 'weirdified' a little by multiplying it by a constant. Really simple.


$F_\text{thrust} = g_0 \cdot I_\text{sp} \cdot \dot m$

That's the same thing as that first "useless" definition, only made more useful. That's how you can practically measure specific impulse (measuring velocity of superheated gas or building hovering test rigs with flexible pipes is not really practical). Again, "bang for the buck", how much thrust is produced - per fuel flow. The more thrust and the less fuel used the better. But you can measure how much force an engine exerts (say, how far the girders of the test cell bend when it blasts at full thrust, if it's something like Apollo's F-1, or how much the ultra-precise torsion weight turns, if it's something like a colloid thruster), and check how much fuel it uses. This way you can get the specific impulse.


$\Delta v = I_\text{sp} g_0 \ln \frac {m_0} {m_f}$

This is what this whole game is all about - where you make practical use of that painstakingly determined specific impulse.

Delta-v is the actual milleage of a rocket. Specific impulse is about the engine. But besides the engine, you have fuel, and you have the payload. This is the Tsiolkovski's Rocket Equation, and this is about "where you can go with your rocket." Good 9km/s to reach LEO. Another 4km/s or so to escape Earth and start travel over the solar system. 3km/s to capture into Mars orbit. That's the delta-v budget, which is the foundation of any mission plan. And with that you have the $m_f$, dry mass of your rocket - engine, scientific payload, telemetry, tanks, panels, everything else than fuel. And $m_0$ - launch mass. That's the above plus fuel.

That 'ln' plays a nasty trick. Usually your fuel mass will be something of order of 90-95% of the launch mass. There's very little we can do about it, because we need good thrust to overcome earth gravity before we reach orbit, and that is provided by chemical engines which have lousy specific impulse. But then, in orbit, we drop the launch stages (dry mass goes way down!) and switch to efficient engines, like ion. And so, we can produce second as much delta-V as to reach LEO, but without need to haul hundreds of tons of fuel. This is where good ISp rules.

  • $\begingroup$ Something like the "useless" experiment you give in 0) was exactly what I was looking for: it "materializes" the numbers, some picture that explain the meaning. Thanks! $\endgroup$ – DarkDust Mar 25 '17 at 17:01

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