# Specific impulse

The specific impulse defines the efficiency of an engine. It is linked to the thrust and fuel consumption. The unit is either meters per second or only seconds.

Conversely, if an engine with specific impulse ${\displaystyle I_{sp,g_{0}}}$ (in seconds) is generating ${\displaystyle F}$ newtons of thrust, the fuel it burns every second will weigh ${\displaystyle {\frac {F}{I_{sp,g_{0}}}}}$ newtons on Kerbin (the planet whose ${\displaystyle g_{0}}$ is used to convert the unit of specific impulse into “seconds”). If the specific impulse is given as ${\displaystyle I_{sp}}$ (in meters/second) then it will burn ${\displaystyle {\frac {F}{I_{sp}}}}$ kilograms of fuel every second. Note the distinction between mass (measured in kilograms) and weight (measured in newtons, depends on local gravity).

## Formula

${\displaystyle I_{sp}={\frac {F_{T}}{\dot {m}}}}$
Where:
• ${\displaystyle I_{sp}}$ is the specific impulse in meters per second
• ${\displaystyle F_{T}}$ the thrust in newtons
• ${\displaystyle {\dot {m}}}$ the fuel consumption in kg/s

By multiplying this value with g0 it is possible to change the unit to only seconds avoiding conversion issues between the SI and customary units. The value g0 behaves like a conversion factor and doesn't change when the gravity for the craft is changing. Usually both values are called specific impulse and are abbreviated by Isp. The name Isp,g0 is used here only to clarify that both values aren't the same. This value is sometimes called weight specific impulse.

${\displaystyle I_{sp,g_{0}}={\frac {F_{T}}{{\dot {m}}\cdot g_{0}}}={\frac {I_{sp}}{g_{0}}}}$
Where:
• ${\displaystyle I_{sp,g_{0}}}$ is the specific impulse in seconds
• ${\displaystyle g_{0}}$ is the surface gravity in the required unit (usually 9.81 meters per second squared)

The formula using the it (e.g. for Δv) has to specify what unit it does expect and if the value is defined in the other one it has to be converted.

### Multiple engines

The combined specific impulse of multiple engines is calculated by the following formula:

${\displaystyle I_{sp}={\frac {\sum \limits _{i}F_{T_{i}}}{\sum \limits _{i}{\dot {m}}_{i}}}={\frac {\sum \limits _{i}F_{T_{i}}}{\sum \limits _{i}{\frac {F_{T_{i}}}{I_{sp_{i}}}}}}}$
Where:
• ${\displaystyle I_{sp}}$ is the specific impulse in meters per second
• ${\displaystyle I_{sp_{i}}}$ the specific impulse of each engine in meters per second
• ${\displaystyle F_{T_{i}}}$ the thrust of each engine in newton
• ${\displaystyle {\dot {m}}}$ the fuel consumption in kg/s

When the fuel consumption is not used in this formula, it is only important that all thrust values have the same unit (e.g. kilonewtons) and the specific impulse have all the same unit (e.g. seconds). The result is then in the same unit as the specific impulses of the engines. If all engines have the same specific impulse the resulting specific impulse will be the same.

The result is equivalent to the weighted harmonic mean of the engines' specific impulses, weighted by each engine's thrust.

## Physical background

Before 1.0 in KSP the fuel consumption on most engines depended on the atmospheric pressure with the lowest consumption (and thus highest specific impulse) in vacuum. In the real world this is usually reversed: The fuel consumption stays always the same but the thrust is increasing over time, because it is easier to have a steady fuel flow. This is how KSP behaves since 1.0, as well. Because a higher thrust with the same fuel consumption it is more efficient the specific impulse rises. The specific impulse can only be calculated using this method for reaction engines and not jet engines as those work on another principle.

Although the unit of the specific impulse is a velocity it is lower than the exhaust speed usually, because some of the fuel consumed isn't used for propelling directly, but runs the turbopumps to fuel the engine.

### Conversion factor

To convert the specific impulse between the handy weight specific impulse and the physical usable specific impulse it had to be converted with g0. It appears that this value isn't 9.81 m/s² which is used in real world. To determine the factor the following formula can be used:

${\displaystyle g_{I_{sp}}={\frac {F_{T}}{\sum \limits _{i}({\dot {m}}_{i}\rho _{i})I_{sp,g_{0}}}}}$
Where:
• ${\displaystyle F_{T}}$ is the thrust of the engine
• ${\displaystyle {\dot {m}}_{i}}$ is the mass flow of the fuel components
• ${\displaystyle \rho _{i}}$ is the density of the fuel components
• ${\displaystyle I_{sp,g_{0}}}$ is the weight specific impulse

For example liquid fuel engines have the fuel components are oxidizer and liquid fuel. When using one Rockomax "Mainsail" Liquid Engine on the launch pad at full throttle uses 48.96 units of liquid fuel and 59.84 units of oxidizer with an weight specific impulse of 280.8 seconds and a thrust of 1.5 MN. This gives a conversion factor between 9.81873052 m/s and 9.8205356 m/s, assuming that the density of liquid fuel and oxidizer is 5 kg per unit. By reducing the engine's weight specific impulse it is possible to get higher mass flow rates improving the conversion factor to a value between 9.81994836 m/s and 9.82006181 m/s by using a thrust of only 85 kN and a weight specific impulse of 1 second for all pressures.[1] The value isn't exact, because the exact mass flows aren't known. It can be assumed that the conversion factor is about 9.82 m/s² making the engines a bit more efficient than expected.[2]

### Example

The Kerbal X has six LV-T45 Liquid Fuel Engines with a specific impulse of 320 s in atmosphere and one Rockomax "Mainsail" Liquid Engine with a specific impulse of 280 s. The average specific impulse of all engines is then:

${\displaystyle I_{sp}={\frac {6\cdot 200{\text{kN}}+1500{\text{kN}}}{6\cdot {\frac {200{\text{kN}}}{320{\text{s}}}}+{\frac {1500{\text{kN}}}{280{\text{s}}}}}}=296.47{\text{s}}}$

To convert this value into a physical usable value. Note that the conversion factor used here is the KSP relevant value of 9.82 m/s² and not the 9.81 m/s² used in real world science.

${\displaystyle I_{sp}=g_{I_{sp}}\cdot 296.47{\text{s}}=2911.34{\frac {\text{m}}{\text{s}}}}$

This value can then be used to calculate the fuel consumption:

{\displaystyle {\begin{aligned}I_{sp}&={\frac {6\cdot 200{\text{kN}}+1500{\text{kN}}}{\dot {m}}}=2911.34{\frac {\text{m}}{\text{s}}}\\{\dot {m}}&={\frac {6\cdot 200000{\text{N}}+1500000{\text{N}}}{2911.34{\frac {\text{m}}{\text{s}}}}}\\&=927.4{\frac {\text{kg}}{\text{s}}}\end{aligned}}}

Because the engines use liquid fuel and oxidizer with a density of 5000 kg/m³ it is possible to calculate the volume consumed.

${\displaystyle {\dot {V}}={\frac {\dot {m}}{\rho }}={\frac {927.4{\frac {\text{kg}}{\text{s}}}}{5000{\frac {\text{kg}}{{\text{m}}^{3}}}}}=0.185{\frac {{\text{m}}^{3}}{\text{s}}}=185{\frac {\text{l}}{\text{s}}}}$

Of course these values of the craft are valid for Kerbin's atmosphere. Because the air gets thinner with altitude the efficiency is rising and the fuel consumption is falling. Also because of the staging pattern engines will be dropped until reaching vacuum and thus changing the efficiency again. This time the efficiency lowers, because the higher efficient engines drop first leaving only the Mainsail engine with the lowest efficiency of all engines on the craft.

## Comparing engines

Because different engines have different masses, simply comparing their specific impulse is sometimes not enough.

However, we can improve the comparison by using the total energy (work) that the engine produces.

• Definition: ${\displaystyle F=I_{sp}{\dot {m_{f}}}}$
• total burn time (${\displaystyle m_{f}}$ is the total fuel mass): ${\displaystyle t_{max}=m_{f}{\frac {I_{sp}}{F}}}$
• Third newton's law: ${\displaystyle {\frac {dv}{dt}}m=F}$, where the mass has now three components: payload mass + engine mass + fuel mass (${\displaystyle m_{p}+m_{e}+m_{f}}$), which we can integrate to obtain:

${\displaystyle v(t)=I_{sp}\log \left({\frac {\left(m_{p}+m_{e}+m_{f}\right)}{\left(m_{p}+m_{e}+m_{f}\right)-{\frac {F}{I_{sp}}}t}}\right)\,\,0 (notice that we are considering no gravitational field here) (this is a version of the so called ideal rocket equation)

With the above, we can compute the total work done as

${\displaystyle W=\int _{x_{0}}^{x_{1}}F\cdot dr=\int _{0}^{t_{max}}F{\frac {dr\left(t\right)}{dt}}dt=F\int _{0}^{t_{max}}v\left(t\right)dt}$

or

${\displaystyle W={\frac {I_{sp}^{2}}{F}}\left(m_{f}+(m_{p}+m_{e})\log \left({\frac {m_{p}+m_{e}}{m_{p}+m_{e}+m_{f}}}\right)\right)}$

With this formula we can compare engines. The higher the W, the better since the work can be converted to either potential energy (i.e. distance from the gravitational body) or kinetic energy (i.e. speed).

For example, for the same vehicle with a given mass ${\displaystyle m_{p}}$, we can compare a single Poodle engine (${\displaystyle I_{sp}=390}$, ${\displaystyle m_{e}=2}$, ${\displaystyle F=220}$) against four nuclear engines (${\displaystyle I_{sp}=800}$, ${\displaystyle m_{e}=2.25*4=9}$, ${\displaystyle F=4*60=240}$). For a small vehicle (say ${\displaystyle m_{p}=1,m_{f}=10}$), the nuclear engine is better (i.e. the ratio of the works W) by a factor of 1.9. With more fuel ${\displaystyle m_{p}=1,m_{f}=20}$, it increases to 2.2. Increasing the payload (e.g. m_p = 10) increases this ratio to 2.8 (i.e. almost 3 times better). A more interesting case is the 48-7S (${\displaystyle I_{sp}=350}$, ${\displaystyle m_{e}=0.1}$, ${\displaystyle F=30}$) vs the LV-909 (${\displaystyle I_{sp}=390}$, ${\displaystyle m_{e}=0.5}$, ${\displaystyle F=50}$), two small engines. Even with a higher ${\displaystyle I_{sp}}$, the LV-909 is less efficient than the 48-7S. This is because its mass is 5 times higher than the 48-7S, which is not compensated by the increase of only ~10% of ${\displaystyle I_{sp}}$. Of course, when leaving a gravitational body, the thrust F must be able to compensate the gravity pull. This naturally adds a constraint to the choice of engines (i.e. the PB-ION Electric Propulsion System is the most efficient thruster but it is not able to lift itself, let alone fuel and a payload, off of Kerbin).