Difference between revisions of "Specific impulse"

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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.
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The '''specific impulse''' (usually written as ''I<sub>sp</sub>'', or in-game as ''ISP'') defines the efficiency of an engine. It is thrust per the rate of fuel consumption. Or equivalently, it is change in momentum per amount of fuel consumed. It can be expressed as a duration or velocity (typically seconds and meters per second), depending whether fuel is measured by its mass, or by its weight on the surface of Earth (specifically, weight in a gravitational field exerting acceleration of exactly 9.80665 meters per second squared; note that gravity on Kerbin is slightly different at 9.81 meters per second squared). KSP uses the duration format for I<sub>sp</sub>
  
Conversely, if an engine with specific impulse <math>I_{sp,g_0}</math> (in seconds) is generating <math>F</math> newtons of thrust, the fuel it burns every second will weigh <math>\frac{F}{I_{sp,g_0}}</math> ''newtons'' on Kerbin (the planet whose <math>g_0</math> is used to convert the unit of specific impulse into “seconds”). If the specific impulse is given as <math>I_{sp}</math> (in meters/second) then it will burn <math>\frac{F}{I_{sp}}</math> ''kilograms'' of fuel every second. Note the distinction between mass (measured in kilograms) and weight (measured in newtons, depends on local gravity).
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If an engine with specific impulse <math>I_{sp,g_0}</math> (in seconds) is generating <math>F</math> newtons of thrust, the fuel it burns every second will weigh <math>\frac{F}{I_{sp,g_0}}</math> ''newtons'' on Earth (the planet whose <math>g_0</math> is used to convert the unit of specific impulse into "seconds"). If the specific impulse is given as <math>I_{sp}</math> (in meters/second) then it will burn <math>\frac{F}{I_{sp}}</math> ''kilograms'' of fuel every second.
  
== Formula ==
+
== Formulas ==
 +
=== Definition ===
 
{{Formula|math=I_{sp} = \frac{F_T}{\dot m}|where=* <math>I_{sp}</math> is the specific impulse in meters per second
 
{{Formula|math=I_{sp} = \frac{F_T}{\dot m}|where=* <math>I_{sp}</math> is the specific impulse in meters per second
 
* <math>F_T</math> the thrust in newtons
 
* <math>F_T</math> the thrust in newtons
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By multiplying this value with g<sub>0</sub> it is possible to change the unit to only ''seconds'' avoiding conversion issues between the SI and customary units. The value g<sub>0</sub> 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 I<sub>sp</sub>. The name I<sub>sp,g<sub>0</sub></sub> is used here only to clarify that both values aren't the same. This value is sometimes called weight specific impulse.
 
By multiplying this value with g<sub>0</sub> it is possible to change the unit to only ''seconds'' avoiding conversion issues between the SI and customary units. The value g<sub>0</sub> 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 I<sub>sp</sub>. The name I<sub>sp,g<sub>0</sub></sub> is used here only to clarify that both values aren't the same. This value is sometimes called weight specific impulse.
 
{{Formula|math=I_{sp,g_0} = \frac{F_T}{\dot m  \cdot g_0} = \frac{I_{sp} }{g_0}|where=* <math>I_{sp,g_0}</math> is the specific impulse in seconds
 
{{Formula|math=I_{sp,g_0} = \frac{F_T}{\dot m  \cdot g_0} = \frac{I_{sp} }{g_0}|where=* <math>I_{sp,g_0}</math> is the specific impulse in seconds
* <math>g_0</math> is the surface gravity in the required unit (usually 9.81 meters per second squared)}}
+
* <math>g_0</math> is the surface gravity in the required unit (exactly 9.80665 meters per second squared)}}
 
The formula using the it (e.g. for [[Delta-v|Δv]]) has to specify what unit it does expect and if the value is defined in the other one it has to be converted.
 
The formula using the it (e.g. for [[Delta-v|Δv]]) has to specify what unit it does expect and if the value is defined in the other one it has to be converted.
 +
 +
=== Delta-v ===
 +
{{Formula|math=\Delta v = I_{sp,g_0} \cdot {g_0} \cdot \ln \frac{m_0}{m_f}|where=* <math>I_{sp,g_0}</math> is the specific impulse in seconds, as stated in game
 +
* <math>g_0</math> standard gravity - 9.80665
 +
* <math>m_0</math> initial, "wet" mass, including fuel
 +
* <math>m_f</math> final, "dry" mass, without fuel}}
  
 
=== Multiple engines ===
 
=== Multiple engines ===
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The result is equivalent to the [[w:Harmonic mean#Weighted harmonic mean|weighted harmonic mean]] of the engines' specific impulses, weighted by each engine's thrust.
 
The result is equivalent to the [[w:Harmonic mean#Weighted harmonic mean|weighted harmonic mean]] of the engines' specific impulses, weighted by each engine's thrust.
 +
 +
=== Relation with altitude ===
 +
In most cases, the specific impulse is nearly linearly related to the pressure (in atm units) in KSP:
 +
 +
<!--: [[File:IspvsAltEq.png]]-->
 +
<math>I_{sp}=I_{sp, vac}+(I_{sp, atm}-I_{sp, vac}) \cdot P</math>
 +
<!--Isp(P) = Isp(vac) + (Isp(atm) - Isp(vac)) * P -->
 +
 +
And if the fuel consumption is constant, it allows us to write the same thing for the thrust:
 +
 +
<!--: [[File:ThrustvsAltEq.png]]-->
 +
<math>T=T_{vac}+(T_{atm}-T_{vac}) \cdot P</math>
 +
<!--T(P) = T(vac) + (T(atm) - T(vac)) * P -->
  
 
== Physical background ==
 
== Physical background ==
In KSP the fuel consumption on most engines depend on the atmospheric pressure with the lowest consumption in vacuum. So the specific impulse is at the highest point in the 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. Because a higher thrust with the same fuel consumption it is more efficient the specific impulse also rises. The specific impulse can only be calculated using this method for [[reaction engine]]s and not [[jet engine]]s as those work on another principle.
+
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 increases 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 engine]]s and not [[jet engine]]s as those work on another principle (specifically, oxidizer is taken from the environment rather than carried with the vehicle, reducing both vehicle mass and how much the oxidizer needs to be accelerated).
  
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.
+
Although the unit of 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 ===
 
=== 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 g<sub>0</sub>. It appears that this value isn't 9.81&nbsp;m/s² which is used in real world. To determine the factor the following formula can be used:
+
To convert the specific impulse between the handy weight specific impulse and the physical usable specific impulse it had to be converted with g<sub>0</sub>.
{{Formula|math=g_{I_{sp} }=\frac{F_T}{\sum\limits_{i}(\dot m_i \rho_i)I_{sp,g_0} }|where=* <math>F_T</math> is the thrust of the engine
 
* <math>\dot m_i</math> is the mass flow of the [[fuel]] components
 
* <math>\rho_i</math> is the density of the fuel components
 
* <math>I_{sp,g_0}</math> 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&nbsp;m/s and 9.8205356&nbsp;m/s, assuming that the density of liquid fuel and oxidizer is 5&nbsp;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&nbsp;m/s and 9.82006181&nbsp;m/s by using a thrust of only 85&nbsp;kN and a weight specific impulse of 1 second for all pressures.<ref>Values determined from [[:File:Isp conversion factor.png]].</ref> 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&nbsp;m/s² making the engines a bit more efficient than expected.<ref>In [[:File:10X Xenon.png]] 10 [[PB-ION Electric Propulsion System]]s are running at full power and consuming a lower value that 10× the theoretical value calculated with 9.81 m/s².</ref>
 
  
 
=== Example ===
 
=== Example ===
The [[Kerbal X]] has six [[LV-T45 Liquid Fuel Engine]]s 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:
+
The [[Kerbal X]] has six [[LV-T45 "Swivel" Liquid Fuel Engine]]s with a specific impulse of 320s in atmosphere and one [[RE-M3 "Mainsail" Liquid Engine]] with a specific impulse of 310s. The average specific impulse of all engines is then:
 
:<math>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}</math>
 
:<math>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}</math>
To convert this value into a physical usable value. Note that the conversion factor used here is the KSP relevant value of 9.82&nbsp;m/s² and not the 9.81&nbsp;m/s² used in real world science.
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To convert this value into a physical usable value.
 
:<math>I_{sp} = g_{I_{sp}} \cdot 296.47 \text{s} = 2911.34 \frac{\text{m}}{\text{s}}</math>
 
:<math>I_{sp} = g_{I_{sp}} \cdot 296.47 \text{s} = 2911.34 \frac{\text{m}}{\text{s}}</math>
 
This value can then be used to calculate the fuel consumption:
 
This value can then be used to calculate the fuel consumption:
 
:<math>\begin{align}
 
:<math>\begin{align}
   I_{sp} &= \frac{6 \cdot 200 \text{kN} + 1500 \text{kN}}{\dot m} = 2911.34 \frac{\text{m}}{\text{s}} \\
+
   I_{sp} &= \frac{6 \cdot 200 \text{kN} + 1500 \text{kN}}{\dot m} = 2907.38 \frac{\text{m}}{\text{s}} \\
   \dot m &= \frac{6 \cdot 200000 \text{N} + 1500000 \text{N}}{2911.34 \frac{\text{m}}{\text{s}}} \\
+
   \dot m &= \frac{6 \cdot 200000 \text{N} + 1500000 \text{N}}{2907.38 \frac{\text{m}}{\text{s}}} \\
         &= 927.4 \frac{\text{kg}}{\text{s}}
+
         &= 928.7 \frac{\text{kg}}{\text{s}}
 
\end{align}</math>
 
\end{align}</math>
 
Because the engines use [[liquid fuel]] and [[oxidizer]] with a density of 5000&nbsp;kg/m³ it is possible to calculate the volume consumed.
 
Because the engines use [[liquid fuel]] and [[oxidizer]] with a density of 5000&nbsp;kg/m³ it is possible to calculate the volume consumed.
 
:<math>{\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}}</math>
 
:<math>{\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}}</math>
Of course this 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.
+
Of course these values of the craft are valid for [[Kerbin]]'s atmosphere. 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 ==
 
== Comparing engines ==
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* Third newton's law: <math>\frac{dv}{dt}m = F</math>, where the mass has now three components: payload mass + engine mass + fuel mass (<math>m_p + m_e + m_f</math>), which we can integrate to obtain:
 
* Third newton's law: <math>\frac{dv}{dt}m = F</math>, where the mass has now three components: payload mass + engine mass + fuel mass (<math>m_p + m_e + m_f</math>), which we can integrate to obtain:
  
 
+
<math>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<t\leq m_{f}\frac{I_{sp}}{F}</math> (notice that we are considering no gravitational field here) (this is a version of the so called ideal rocket equation).
<math>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<t\leq m_{f}\frac{I_{sp}}{F}</math> (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
 
With the above, we can compute the total work done as
 
  
 
<math>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</math>
 
<math>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</math>
 
  
 
or
 
or
  
 
+
[[ Image:Work_done_by_engine.png ]]
<math>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)</math>
+
<!-- W = I_{sp}^{2}\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).
 
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 <math>m_p</math>, we can compare a single Poodle engine (<math>I_{sp} = 390</math>, <math>m_e = 2</math>, <math>F=220</math>) against four nuclear engines (<math>I_{sp} = 800</math>, <math>m_e = 2.25*4 = 9</math>, <math>F=4*60 = 240</math>). For a small vehicle (say <math>m_p = 1, m_f = 10</math>), the nuclear engine is better (i.e. the ratio of the works W) by a factor of 1.9. With more fuel <math>m_p = 1, m_f = 20</math>, 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).
 
For example, for the same vehicle with a given mass <math>m_p</math>, we can compare a single Poodle engine (<math>I_{sp} = 390</math>, <math>m_e = 2</math>, <math>F=220</math>) against four nuclear engines (<math>I_{sp} = 800</math>, <math>m_e = 2.25*4 = 9</math>, <math>F=4*60 = 240</math>). For a small vehicle (say <math>m_p = 1, m_f = 10</math>), the nuclear engine is better (i.e. the ratio of the works W) by a factor of 1.9. With more fuel <math>m_p = 1, m_f = 20</math>, 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 (<math>I_{sp} = 350</math>, <math>m_e = 0.1</math>, <math>F=30</math>) vs the LV-909 (<math>I_{sp} = 390</math>, <math>m_e = 0.5</math>, <math>F=50</math>), two small engines. Even with a higher <math>I_{sp}</math>, 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 <math>I_{sp}</math>.
 
A more interesting case is the 48-7S (<math>I_{sp} = 350</math>, <math>m_e = 0.1</math>, <math>F=30</math>) vs the LV-909 (<math>I_{sp} = 390</math>, <math>m_e = 0.5</math>, <math>F=50</math>), two small engines. Even with a higher <math>I_{sp}</math>, 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 <math>I_{sp}</math>.
Of course, when leaving a gravitational body, the trust F must be able to compensate the gravity pull. This naturally adds a constraint to the choice of engines (i.e. the ion trust is the most efficient but it is not able to lift anything out of Kerbin).
+
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 [[IX-6315 "Dawn" 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).
  
 
== See also ==
 
== See also ==

Latest revision as of 02:29, 3 January 2023

The specific impulse (usually written as Isp, or in-game as ISP) defines the efficiency of an engine. It is thrust per the rate of fuel consumption. Or equivalently, it is change in momentum per amount of fuel consumed. It can be expressed as a duration or velocity (typically seconds and meters per second), depending whether fuel is measured by its mass, or by its weight on the surface of Earth (specifically, weight in a gravitational field exerting acceleration of exactly 9.80665 meters per second squared; note that gravity on Kerbin is slightly different at 9.81 meters per second squared). KSP uses the duration format for Isp

If an engine with specific impulse (in seconds) is generating newtons of thrust, the fuel it burns every second will weigh newtons on Earth (the planet whose is used to convert the unit of specific impulse into "seconds"). If the specific impulse is given as (in meters/second) then it will burn kilograms of fuel every second.

Formulas

Definition

Where:
  • is the specific impulse in meters per second
  • the thrust in newtons
  • 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.

Where:
  • is the specific impulse in seconds
  • is the surface gravity in the required unit (exactly 9.80665 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.

Delta-v

Where:
  • is the specific impulse in seconds, as stated in game
  • standard gravity - 9.80665
  • initial, "wet" mass, including fuel
  • final, "dry" mass, without fuel

Multiple engines

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

Where:
  • is the specific impulse in meters per second
  • the specific impulse of each engine in meters per second
  • the thrust of each engine in newton
  • 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.

Relation with altitude

In most cases, the specific impulse is nearly linearly related to the pressure (in atm units) in KSP:

And if the fuel consumption is constant, it allows us to write the same thing for the 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 increases 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 (specifically, oxidizer is taken from the environment rather than carried with the vehicle, reducing both vehicle mass and how much the oxidizer needs to be accelerated).

Although the unit of 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.

Example

The Kerbal X has six LV-T45 "Swivel" Liquid Fuel Engines with a specific impulse of 320s in atmosphere and one RE-M3 "Mainsail" Liquid Engine with a specific impulse of 310s. The average specific impulse of all engines is then:

To convert this value into a physical usable value.

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

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

Of course these values of the craft are valid for Kerbin's atmosphere. 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:
  • total burn time ( is the total fuel mass):
  • Third newton's law: , where the mass has now three components: payload mass + engine mass + fuel mass (), which we can integrate to obtain:

(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

or

Work done by engine.png

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 , we can compare a single Poodle engine (, , ) against four nuclear engines (, , ). For a small vehicle (say ), the nuclear engine is better (i.e. the ratio of the works W) by a factor of 1.9. With more fuel , 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 (, , ) vs the LV-909 (, , ), two small engines. Even with a higher , 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 . 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 IX-6315 "Dawn" 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).

See also

Notes