Space exploration is in the ascendant. In order to carry more things and fly farther and higher, people are beginning to use multi-engine high-thrust rockets (Fig. 25.1). How to optimize the cooperation of these engines to make the overall energy efficiency of the rocket higher and lower cost are becoming increasingly important.

Fig. 25.1
Two photographs of a large, cylindrical industrial machine with multiple nozzles and components, positioned vertically in a factory setting. An image of a sleek, metallic spacecraft stands upright on a launch pad. The vehicle has a pointed nose and fins near the base.

Multi-engine rocket

Full size image

Generally, the energy efficiency of electric heating, chemical reactions and some combustion processes is very high, because most of the energy loss in nature is ultimately lost in the form of heat dissipation. However, in the process of doing work, there is inevitably a conduction process, that is, there is heat energy loss. Therefore, the actual operating efficiency is not 100%. Rocket engines are no exception.

25.1 The Energy Efficiency of a Rocket Engine

For the energy efficiency curve of a rocket engine, for the convenience of measurement, we define that the abscissa is the mass of fuel burned per unit time of the i-th engine, mi, and the ordinate is the energy efficiency of the i-th engine, ηi, ηi is replaced by kFi/ mi, k is a constant, Fi represents the thrust of the i-th engine during unit time, the curve is as shown in Fig. 25.2.

Fig. 25.2
A line graph plots efficiency (eta) versus mass (m). The curve rises sharply to a peak at m i e, indicating maximum efficiency (eta i e), and then falls off. The graph also marks a point m i m where efficiency is lower, showing the relationship between mass and efficiency.

Energy efficiency curve of rocket engine

Full size image

Assume the energy efficiency functions η1(m) and ηi(m) of the first and the i-th engine is as shown in Fig. 25.3.

Fig. 25.3
A graph plots two efficiency curves, eta (m) and eta (n m), plotted against mass (m). The curve eta (m) rises to a peak at m i e and falls off, while eta (n m) starts lower and intersects with eta (m) at m i minus 1 before also falling off. The graph demonstrates how different masses affect efficiency differently.

The energy efficiency curves of the first and the i-th engine

Full size image

If the following equation holds:

$$ \eta_{i} \left( {m_{i} } \right) = \eta_{1} \left( {\frac{{m_{1} }}{{\beta_{i} }}} \right) $$
(25.1)

We call the first engine and the i-th engine as the energy efficiency similarity engines.

25.2 Analysis of Launch Process of Launch Vehicle

The total rocket thrust F is

$$ {\text{F}} = \sum\limits_{i = 1}^{n} {F_{i} } $$
(25.2)

The energy efficiency optimization problem of multi-engine rocket is

$$ {\text{Max S}} = {\text{Max}}\int\limits_{t = 0}^{ \propto } {Vdt} $$
(25.3)

In Eq. (25.3), S is the total distance traveled by the rocket, and V is the speed of the rocket at time t.

In the vacuum resistance-free state, the kinetic energy and pressure energy output by the engine become the potential energy of the rocket, causing the rocket to rise ΔH, with the following relationship

$$ \sum\limits_{i = 1}^{n} {\left( {\frac{1}{2}m_{i} V_{i}^{2} + \frac{{P_{i} }}{{\gamma g_{h} }}} \right)} = \left( {M - \sum\limits_{i = 1}^{n} {m_{i} } } \right)g_{h} \Delta H $$
(25.4)

gh is the acceleration of gravity at different altitudes, n is the number of working engines, Vi is the speed of fuel ejection by the i-th engine, Pi is the pressure of the gas ejected by the i-th engine, and M is the rocket total mass at the previous time t.

Taking resistance such as wind resistance into account, according to Newton's second law, we have

$$ \sum\limits_{i = 1}^{n} {F_{i} } - \left( {M - \sum\limits_{i = 1}^{n} {m_{i} } } \right)g_{h} - k_{1} V^{2} = \left( {M - \sum\limits_{i = 1}^{n} {m_{i} } } \right)\frac{dV}{{dt}} $$
(25.5)

In Eq. (25.5), k1 is the wind resistance coefficient, which changes with the thinness of the air or the altitude of the rocket. The expression for F becomes

$$ F = \sum\limits_{i = 1}^{n} {F_{i} } = \left( {M - \sum\limits_{i = 1}^{n} {m_{i} } } \right)g_{h} + k_{1} V^{2} + \left( {M - \sum\limits_{i = 1}^{n} {m_{i} } } \right)\frac{dV}{{dt}} $$
(25.6)

In a short period of time t, gh and k1 are approximately constant. There are two variables, V and M1. V is the current speed of the rocket, and M1 is the current total mass of the rocket.

$$ M_{1} = \left( {M - \sum\limits_{i = 1}^{n} {m_{i} } } \right) $$
(25.7)

The total mass of the rocket has been decreasing and the wind resistance has been increasing. Derive V and M1 to get the optimal thrust F.

For each F, Fi is arranged according to the principle of maximum energy efficiency, that is, the principle of minimum fuel consumption, and the corresponding fuel combustion quality mi is calculated. Then we have

$$ F = \sum\limits_{i = 1}^{n} {F_{i} } = k_{2} \sum\limits_{i = 1}^{n} {m_{i} } \eta_{i} \left( {m_{i} } \right) $$
(25.8)
$$ {\text{m}}_{{\varvec{t}}} = \sum\limits_{i = 1}^{n} {m_{i} } $$

where mt is the total mass ejected per unit time by the rocket.

25.3 Optimal Scheduling of Multiple Engines in a Rocket

  1. 1.

    If the n engines are identical, the energy expression of the thrust becomes

    $$ F = k_{2} m_{t} \sum\limits_{i = 1}^{n} {\theta_{i} } \eta \left( {\theta_{i} m_{t} } \right) $$
    (25.9)

where θi is

$$ \theta_{i} = \frac{{m_{i} }}{{m_{{\text{t}}} }} $$
(25.10)
$$ \sum\limits_{i = 1}^{n} {\theta_{i} } = 1 $$

Consider the maximization problem of the total thrust F

$$ \max F $$
(25.11)
$$ \begin{aligned} s.t.\begin{array}{*{20}c} {} \\ \end{array} \theta_{i} > 0,i & = 1,2,...n \\ \begin{array}{*{20}c} {} & {} \\ \end{array} \mathop \sum \limits_{i = 1}^{n} \theta_{i} & = 1 \\ \begin{array}{*{20}c} {} & {} \\ \end{array} m_{t} & = cons\tan t \\ \end{aligned} $$

This problem can also be written as

$$ \max k_{2} m_{t} \sum\limits_{i = 1}^{n} {\theta_{i} } \eta \left( {\theta_{i} m_{t} } \right) $$
(25.12)
$$ \begin{aligned} s.t.\begin{array}{*{20}c} {} \\ \end{array} \theta_{i} > 0,i & = 1,2,...n \\ \begin{array}{*{20}c} {} & {} \\ \end{array} \mathop \sum \limits_{i = 1}^{n} \theta_{i} & = 1 \\ \begin{array}{*{20}c} {} & {} \\ \end{array} m_{t} & = cons\tan t \\ \end{aligned} $$

We consider three cases:

  1. (1)

    n = 2

The rocket has two variables and has

$$ \begin{aligned} & \theta_{1} + \theta_{2} = 1 \\ & \theta_{1} > 0 \\ & \theta_{2} > 0 \\ \end{aligned} $$
(25.13)

The objective function F can be expressed as

$$ F = k_{2} m_{t} \left( {\theta_{1} \eta \left( {\theta_{1} m_{t} } \right) + \theta_{2} \eta \left( {\theta_{2} m_{t} } \right)} \right) = k_{2} m_{t} \left( {\theta_{1} \eta \left( {\theta_{1} m_{t} } \right) + (1 - \theta_{1} } \right)\eta \left( {\left( {1 - \theta_{1} } \right)m_{t} } \right)) $$
(25.14)

The optimization condition is

$$ F^{\prime } \left( {\theta_{1} } \right) = 0 $$
(25.15)

It is easy to see that

$$ \theta_{1} = \frac{1}{2} $$
(25.16)

for the optimization point. Then we have

$$ \theta_{2} = 1 - \theta_{1} = \theta_{1} = \frac{1}{2} = \frac{1}{n} $$
(25.17)

That is, the optimal control method is to keep

$$ m_{1} = m_{2} = \frac{{m_{t} }}{2} = \frac{{m_{t} }}{n} $$
(25.18)

The total F is

$$ F = k_{2} m_{t} \eta \left( {\frac{{m_{t} }}{2}} \right) = k_{2} m_{t} \eta \left( {\frac{{m_{t} }}{n}} \right) $$
(25.19)

Since the shape of the overall efficiency curve of the rocket is the same as that of a single engine, so the second derivative of F is also less than zero

$$ F^{\prime \prime } \left( {\theta_{1} } \right) < 0 $$
(25.20)

F is the unique maximum value.

$$ {\text{maxF}} = k_{2} m_{t} \eta \left( {\frac{{m_{t} }}{2}} \right) $$
(25.21)

The overall energy efficiency ηt of the rocket is the only maximum value.

$$ max\eta_{t} = \eta \left( {\frac{{m_{{\text{t}}} }}{n}} \right) $$
(25.22)
  1. (2)

    n = 3

The rocket has three variables, based on known conditions, we have

$$ \theta_{1} + \theta_{2} + \theta_{3} = 1 $$
(25.23)
$$ \begin{aligned} \theta_{1} & > 0 \\ \theta_{2} & > 0 \\ \theta_{3} & > 0 \\ \end{aligned} $$

The F expression becomes

$$ F = k_{2} m_{t} \left( {\theta_{1} \eta \left( {\theta_{1} m_{t} } \right) + \theta_{2} \eta \left( {\left( {\theta_{2} } \right)m_{t} } \right) + \theta_{3} \eta \left( {\left( {\theta_{3} } \right)m_{t} } \right)} \right) $$
(25.24)

Assuming that θ3 is fixed and an optimization point, only θ1 and θ2 are variables, we have

$$ \theta_{1} + \theta_{2} = 1 - \theta_{3} = constant $$
(25.25)

Based on the conclusion of n = 2 above, there are

$$ \theta_{1} = \theta_{2} $$
(25.26)

is the optimal point.

Assuming that θ2 is fixed and is an optimization point, only θ1 and θ3 are variables, we have

$$ \theta_{1} + \theta_{3} = 1 - \theta_{2} = constant $$
(25.27)

According to the conclusion of n = 2 above, there are

$$ \theta_{1} = \theta_{3} $$
(25.28)

is the optimal point.

Similarly, assuming that θ1 is fixed and is an optimization point, only θ2 and θ3 are variables, we have

$$ \theta_{2} + \theta_{3} = 1 - \theta_{1} = constant $$
(25.29)

According to the conclusion of n = 2 above, we have

$$ \theta_{2} = \theta_{3} $$
(25.30)

to be the optimal point.

So, we have the optimal point

$$ \theta_{1} = \theta_{2} = \theta_{3} = \frac{1}{3} = \frac{1}{n} $$
(25.31)

That is, the optimal control method is to keep

$$ m_{1} = m_{2} = m_{3} = \frac{{m_{t} }}{3} = \frac{{m_{t} }}{n} $$
(25.32)

The maximum value of the total F is

$$ max F = k_{2} m_{t} \eta \left( {\frac{{m_{t} }}{3}} \right) = k_{2} m_{t} \eta \left( {\frac{{m_{t} }}{n}} \right) $$
(25.33)

The maximum value of the overall energy efficiency ηt is

$$ \max { }\eta_{t} =\upeta \left( {\frac{{m_{t} }}{3}} \right) = \eta \left( {\frac{{m_{t} }}{n}} \right) $$
(25.34)
  1. (3)

    n = k

The rocket has k variables, the above conclusion can be extended to the case of n = k, the optimal point is

$$ \theta_{1} = \theta_{2} = ... = \theta_{k} = \frac{1}{k} $$
(25.35)

That is, the optimal control method is to keep

$$ m_{1} = m_{2} = ... = m_{k} = \frac{{m_{t} }}{k} $$
(25.36)

The maximum value of F is

$$ max F = k_{2} m_{t} \eta \left( {\frac{{m_{t} }}{k}} \right) $$
(25.37)

The maximum value of the overall energy efficiency ηt is

$$ \max { }\eta_{t} = \left( {\frac{{m_{t} }}{k}} \right) $$
(25.38)
  1. 2.

    If the n engines are the energy efficiency similarity engines, the energy expression of the total thrust becomes

    $$ F = F = k_{2} m_{t} \sum\limits_{i = 1}^{n} {\theta_{i} } \eta_{i} \left( {\theta_{i} m_{t} } \right) $$
    (25.39)

where

$$ \begin{array}{*{20}c} {\theta_{i} = \frac{{m_{i} }}{{m_{{\text{t}}} }}} \\ {\sum\limits_{i = 1}^{n} {\theta_{i} } = 1} \\ \end{array} $$
(25.40)

Consider the maximization problem of the total thrust F

$$ \max F $$
(25.41)
$$ \begin{aligned} s.t.\begin{array}{*{20}c} {} \\ \end{array} \theta_{i} > 0,i & = 1,2,...n \\ \begin{array}{*{20}c} {} & {} \\ \end{array} \mathop \sum \limits_{i = 1}^{n} \theta_{i} & = 1 \\ \begin{array}{*{20}c} {} & {} \\ \end{array} m_{t} & = cons\tan t \\ \end{aligned} $$

This problem can also be written as

$$ \max k_{2} m_{t} \sum\limits_{i = 1}^{n} {\theta_{i} } \eta_{i} \left( {\theta_{i} m_{t} } \right) $$
(25.42)
$$ \begin{aligned} s.t.\begin{array}{*{20}c} {} \\ \end{array} \theta_{i} > 0,i & = 1,2,...n \\ \begin{array}{*{20}c} {} & {} \\ \end{array} \mathop \sum \limits_{i = 1}^{n} \theta_{i} & = 1 \\ \begin{array}{*{20}c} {} & {} \\ \end{array} m_{t} & = cons\tan t \\ \end{aligned} $$

We consider three cases:

  1. (1)

    n = 2

The rocket has two variables and has

$$ \theta_{1} + \theta_{2} = 1 $$
(25.43)
$$ \begin{aligned} \theta_{1} & > 0 \\ \theta_{2} & > 0 \\ \end{aligned} $$

The objective function F can be expressed as

$$ F = k_{2} m_{t} \left( {\theta_{1} \eta_{1} \left( {\theta_{1} m_{t} } \right) + \theta_{2} \eta_{2} \left( {\theta_{2} m_{t} } \right)} \right) $$
(25.44)

The optimization condition is

$$ F^{\prime } \left( {\theta_{1} } \right) = 0 $$
(25.45)

It is easy to see that

$$ \theta_{1} = \frac{1}{{1 + \beta_{2} }} = \frac{{\beta_{1} }}{{\beta_{1} + \beta_{2} }} $$
(25.46)

for the optimization point. Then we have

$$ \theta_{2} = 1 - \theta_{1} = \frac{{\beta_{2} }}{{1 + \beta_{2} }} = \frac{{\beta_{2} }}{{\beta_{1} + \beta_{2} }} $$
(25.47)

That is, the optimal control method is to keep

$$ \begin{array}{*{20}c} {m_{1} = \frac{1}{{1 + \beta_{2} }}m_{{\text{t}}} } \\ {m_{2} = \frac{{\beta_{2} }}{{1 + \beta_{2} }}m_{{\text{t}}} } \\ \end{array} $$
(25.48)

The total thrust F is

$$ F = k_{2} m_{t} \eta_{1} \left( {\frac{1}{{1 + \beta_{2} }}m_{t} } \right) $$
(25.49)

Since the shape of the overall efficiency curve of the rocket is the same as that of a single engine, so the second derivative of the F is also less than zero

$$ F^{\prime \prime } \left( {\theta_{1} } \right) < 0 $$
(25.50)

F is the only maximum value.

$$ {\text{max}}F = k_{2} m_{t} \eta_{1} \left( {\frac{1}{{1 + \beta_{2} }}m_{t} } \right) $$
(25.51)

The overall energy efficiency ηt of the rocket is the only maximum value.

$$ max\eta_{t} = \eta_{1} \left( {\frac{1}{{1 + \beta_{2} }}m_{t} } \right) $$
(25.52)
  1. (2)

    n = 3

The rocket has three variables, based on known conditions, we have

$$ \theta_{1} + \theta_{2} + \theta_{3} = 1 $$
(25.53)
$$ \begin{aligned} \theta_{1} & > 0 \\ \theta_{2} & > 0 \\ \theta_{3} & > 0 \\ \end{aligned} $$

The F expression becomes

$$ F = F = k_{2} m_{t} \left( {\theta_{1} \eta_{1} \left( {\theta_{1} m_{t} } \right) + \theta_{2} \eta_{2} \left( {\theta_{2} m_{t} } \right) + \theta_{3} \eta_{3} \left( {\theta_{3} m_{t} } \right)} \right) $$
(25.54)

Assuming that θ3 is fixed and is an optimization point, only θ1 and θ2 are variables, we have

$$ \theta_{1} + \theta_{2} = 1 - \theta_{3} = constant $$
(25.55)

Based on the conclusion of n = 2 above, we have

$$ \begin{array}{*{20}c} {\theta_{1} = \frac{1}{{1 + \beta_{2} }}} \\ {\theta_{2} = \frac{{\beta_{2} }}{{1 + \beta_{2} }}} \\ \end{array} $$
(25.56)

is the optimal point.

Assuming that θ2 is fixed and is an optimization point, only θ1 and θ3 are variables, there are

$$ \theta_{1} + \theta_{3} = 1 - \theta_{2} = constant $$
(25.57)

According to the conclusion of n = 2 above, there are

$$ \begin{array}{*{20}c} {\theta_{1} = \frac{1}{{1 + \beta_{3} }}} \\ {\theta_{3} = \frac{{\beta_{3} }}{{1 + \beta_{3} }}} \\ \end{array} $$
(25.58)

is the optimal point.

Similarly, assuming that θ1 is fixed and is an optimization point, only θ2 and θ3 are variables, we have

$$ \theta_{2} + \theta_{3} = 1 - \theta_{3} = constant $$
(25.59)

According to the conclusion of n = 2 above, there are

$$ \begin{array}{*{20}c} {\theta_{2} = \frac{{\beta_{2} }}{{\beta_{2} + \beta_{3} }}} \\ {\theta_{3} = \frac{{\beta_{3} }}{{\beta_{2} + \beta_{3} }}} \\ \end{array} $$
(25.60)

is the optimal point.

So, we have the optimal point, and the optimal control method is to keep

$$ \begin{array}{*{20}c} {\theta_{1} = \frac{{\beta_{1} }}{{\beta_{1} + \beta_{2} + \beta_{3} }}} \\ {\theta_{2} = \frac{{\beta_{2} }}{{\beta_{1} + \beta_{2} + \beta_{3} }}} \\ {\theta_{3} = \frac{{\beta_{3} }}{{\beta_{1} + \beta_{2} + \beta_{3} }}} \\ \end{array} $$
(25.61)

The maximum total trust F is

$$ \max F = k_{2} m_{t} \eta_{1} \left( {\frac{1}{{1 + \beta_{2} + \beta_{3} }}m_{t} } \right) $$
(25.62)

The maximum overall energy efficiency ηt is

$$ \max { }\eta_{t} = \eta_{1} \left( {\frac{{P_{0} }}{{1 + \beta_{2} + \beta_{3} }}} \right) $$
(25.63)
  1. (3)

    n = k

The rocket has k variables, the above conclusion can be extended to the case of n = k, the optimal point and the optimal control method is to keep

$$ \theta_{i} = \frac{{\beta_{i} }}{{\mathop \sum \limits_{l = 1}^{k} \beta_{l} }} $$
(25.64)

The maximum total trust F is

$$ \max F = k_{2} m_{t} \eta_{1} \left( {\frac{1}{{\mathop \sum \limits_{l = 1}^{k} \beta_{l} }}m_{t} } \right) $$
(25.65)

The maximum overall energy efficiency ηt is

$$ \max { }\eta_{t} = \eta_{1} \left( {\frac{1}{{\mathop \sum \limits_{l = 1}^{k} \beta_{l} }}m_{t} } \right) $$
(25.66)

25.4 Optimal Number of Operational Engines in a Rocket

  1. 1.

    A rocket has m identical engines, if the following equations are satisfied

    $$ \eta \left( {\frac{{m_{{\text{t}}} }}{n}} \right) \ge \eta \left( {\frac{{m_{{\text{t}}} }}{n - 1}} \right) $$
    (25.67)
    $$ \begin{aligned} & \eta \left( {\frac{{m_{{\text{t}}} }}{n}} \right) \ge \eta \left( {\frac{{m_{{\text{t}}} }}{n + 1}} \right) \\ & n \le m \\ \end{aligned} $$

Then the number n is the number of engines with optimal operation.

  1. 2.

    A rocket has m energy efficiency similarity engines, if the following equations are satisfied

    $$ \eta_{1} \left( {\frac{{P_{0} }}{{\mathop \sum \limits_{l = 1}^{n} \beta_{l} }}} \right) \ge \eta_{1} \left( {\frac{{P_{0} }}{{\mathop \sum \limits_{l = 1}^{n1} \beta_{l} }}} \right) $$
    (25.68)
    $$ \begin{aligned} n & \le m \\ n1 & \le m \\ \end{aligned} $$

n1 is any combination other than the optimal combination of n units this time, and also include other combinations of n units. The number n is the number of engines with optimal operation.

25.5 Optimal Switching Rule for Multiple Engines in a Rocket

  1. 1.

    A rocket has m identical engines, and the number n is the number of engines currently in optimal operation. If mt increases to mt1, the following relationship holds:

    $$ \eta \left( {\frac{{m_{{{\text{t}}1}} }}{n}} \right) = \eta \left( {\frac{{m_{{{\text{t}}1}} }}{n + 1}} \right) $$
    (25.69)
    $$ n \le m $$

Then mt1 is the optimal switching point between the operation of n engines and the operation of n + 1 engine. When the total required mt is greater than mt1, the optimal number of engines in operation is switched from n to n + 1. If mt increases until mt /n = m1m, there isn’t the point mt1, then mt/n = m1m is the switching point from n to n + 1.

If mt is reduced to mt2, the following relationship is established

$$ \eta \left( {\frac{{m_{t2} }}{n}} \right) = \eta \left( {\frac{{m_{t2} }}{n - 1}} \right) $$
(25.70)
$$ n \le m $$

Then mt2 is the optimal switching point between the operation of n engines and the operation of n-1 engines. When the total required power mt is less than mt2, the optimal number of engines in operation is switched from n to n-1. If mt reduces until mt /(n-1) = m1m, there isn’t the point mt2, then mt /(n-1) = m1m is the switching point from n to n-1.

The analysis process is shown in Fig. 25.4.

Fig. 25.4
A graph plots efficiency (eta) versus mass (m), with multiple efficiency points marked. The curve eta (m) rises to a peak and falls off. Vertical lines mark points such as m t slash (n plus 1), m t slash n, and m t slash (n minus 1), illustrating different mass ratios and their corresponding efficiencies.

Energy efficiency comparison curve of the rocket with the identical engines

Full size image
  1. 2.

    A rocket has m energy efficiency similarity engines, and the number n is the number of engines currently in optimal operation. If mt increases mt1, the following relation holds:

    $$ \begin{array}{*{20}c} {\eta_{1} \left( {\frac{{{\text{m}}_{t1} }}{{\mathop \sum \limits_{l = 1}^{n} \beta_{l} }}} \right) = \eta_{1} \left( {\frac{{{\text{m}}_{t1} }}{{\mathop \sum \limits_{l = 1}^{k1} \beta_{l} }}} \right)} \\ {n \le m} \\ \end{array} $$
    (25.71)

Then mt1 is the optimal switching point between the operation of n engines and the operation of k1 engines. When the total required power mt is greater than mt1, the optimal number of engines in operation is switched from n to k1. If mt increases until mt /\({\sum }_{l=1}^{n}{\beta }_{l}\)=m1m, there isn’t the point mt1, then mt /\({\sum }_{l=1}^{n}{\beta }_{l}\)=m1m is the switching point from n to k1.

If mt is reduced to mt2, the following relation is established

$$ \begin{array}{*{20}c} {\eta_{1} \left( {\frac{{m_{t2} }}{{\mathop \sum \limits_{l = 1}^{n} \beta_{l} }}} \right) = \eta_{1} \left( {\frac{{m_{t2} }}{{\mathop \sum \limits_{l = 1}^{k2} \beta_{l} }}} \right)} \\ {n \le m} \\ \end{array} $$
(25.72)

Then mt2 is the optimal switching point between the operation of n engines and the operation of k2 engines. When the total required power mt is less than mt2, the optimal number of engines in operation is switched from n to k2. If mt reduces until mt /\({\sum }_{l=1}^{k2}{\beta }_{l}\) =m1m, there isn’t the point mt2, then mt /\({\sum }_{l=1}^{k2}{\beta }_{l}\) =m1m is the switching point from n to k2.

The analysis process is shown in Fig. 25.5.

Fig. 25.5
A graph plots efficiency (eta) versus mass (m), with a single efficiency curve eta (m) rising to a peak and then gradually declining. The graph marks points m l slash summation n l beta l, m l slash summation n l beta l, m l slash summation n l beta l, and m l slash summation n l beta l, emphasizing the influence of different mass parameters on efficiency.

Energy efficiency comparison curve of the rocket with the similar efficiency engines

Full size image

k1 and k2 are any combination other than the optimal combination of n units this time, and also include other combinations of n units. Point \({\text{m}}_{t}/{\sum }_{l=1}^{k1}{\beta }_{l}\) is the point closest to \({\text{m}}_{t}/{\sum }_{l=1}^{n}{\beta }_{l}\) to the left of point \({\text{m}}_{t}/{\sum }_{l=1}^{n}{\beta }_{l}\). Point \({\text{m}}_{t}/{\sum }_{l=1}^{k2}{\beta }_{l}\) is the point closest to \({\text{m}}_{t}/{\sum }_{l=1}^{n}{\beta }_{l}\) to the right of point \({\text{m}}_{t}/{\sum }_{l=1}^{n}{\beta }_{l}\). . If all engines are identical, we have k1 = n + 1 and k2 = n-1.