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-- PASS THROUGH --

-- PASS THROUGH -- 

By Vincent McConnell и Kosmo-not

Изучение азов ракетостроения очень важно для успешной постройки ракет в играх вроде Kerbal Space Program. Ракет, которые могут хорошо выполнять свои задачи. В этом руководстве мы разберём такие вещи как: расчёт полного Delta-V корабля, орбитальный манёвр перехода, получение тяговооружённости, максимальной перегрузки при ускорении двигателем, а также расчёт Delta-V для полной гомановской траектории и многое другое.

Характеристическая скорость

Характеристическая скорость (${\displaystyle \Delta v}$) имеет первостепенное значение в ракетостроении. Это, возможно, главная вещь, которую Вы хотели бы знать о вашей ракете, потому что она показывает, на что ракета способна в принципе. Среди всех вещей, описанных в этом базовом руководстве, ${\displaystyle \Delta v}$ наиболее полезна на ракетной верфи в KSP.

Для нахождения ${\displaystyle \Delta v}$ ступени вашей ракеты необходимы (на данной ступени):

• Масса полностью заправленной ракеты: ${\displaystyle m_{\text{full}}}$

• "Сухая" масса ракеты с полностью отработанным топливом: ${\displaystyle m_{\text{dry}}}$

Их можно узнать, поставив ракету на стартовый стол и заглянув в "Orbit map" в список "Info". Разумеется, под отработанным топливом имеется ввиду только то, к которому имеют доступ двигатели данной ступени.

Далее необходимо найти удельный импульс двигателей (${\displaystyle I_{\text{sp}}}$). Удельный импульс выражает эффективность двигателя: чем он больше, тем больший импульс получает корабль, выработав то же количество топлива. Например, стандартный двигатель LV-T30 в вакууме имеет удельный импульс 370 с. Дальше мы просто применяем Формулу Циолковсого, которая выглядит так

${\displaystyle \Delta v=I_{sp}\cdot \ln \left({\frac {m_{\text{full}}}{m_{\text{dry}}}}\right)}$

В игре дан весовой удельный импульс, поэтому его необходимо умножить на ${\displaystyle g=9.816{\frac {m}{s^{2}}}}$ (смотрите также Terminology about I_sp).

Рассмотрим простой пример:

Example rocket

Третяя ступень (TMI, прилунение, возвращение)
${\displaystyle m_{\text{full}}}$:	3.72 т
${\displaystyle m_{\text{dry}}}$:	1.72 т
Isp:	400 с
Δv:	3027.0 м/с
Вторая ступень (Выход на орбиту Кербина)
${\displaystyle m_{\text{full}}}$:	7.27 т
${\displaystyle m_{\text{dry}}}$:	5.27 т
Isp:	370 с
Δv:	1167.8 м/с
Первая ступень (Восхождение):
${\displaystyle m_{\text{full}}}$:	38.52 т
${\displaystyle m_{\text{dry}}}$:	14.52 т
Isp:	350 с (средний в атмосфере)
Δv:	3349.9 м/с
Всего
Δv:	7544.6 м/с

I_sp для нескольких двигателей

Чтобы подсчитать I_sp для нескольких двигателей с разными I_sp, вы должны разделить общую тягу на общий расход всех двигателей:

${\displaystyle I_{sp}={\frac {\sum _{i}^{n}(T_{i})}{\sum _{i}^{n}({\dot {m}}_{i}\cdot g_{0})}}={\frac {T_{1}+T_{2}+\dots +T_{n}}{{\frac {T_{1}}{I_{sp_{1}}}}+{\frac {T_{2}}{I_{sp_{2}}}}+\dots +{\frac {T_{n}}{I_{sp_{n}}}}}}}$

где ${\displaystyle T_{i}}$ - максимальная тяга i-того двигателя, ${\displaystyle I_{sp_{i}}}$ - удельный импульс i-того двигателя.

Это выражение определяет точный I_sp, которую можно использовать для подсчёта ${\displaystyle \Delta v}$. В случае, когда все двигатели одинаковые, общий I_sp не отличается от исходного.

Calculating transfer maneuvers

The next part of this tutorial is how to perform a transfer maneuver. This kind of action is called a Hohmann Transfer and it requires two burns at opposite points in an orbit. Adding velocity will boost our apoapsis higher. We would then simply wait until we hit our newly established apoapsis and then add more velocity to boost our periapsis to circularize. Or, we could drop our orbit by subtracting velocity by burning retro-grade.

We can also apply some ${\displaystyle \Delta v}$ calculations to find out how much thrust we will need to perform this maneuver. We will break this burn up into impulses. For example purposes, we will start at a 100 km orbit and then boost into a 200&nsbp;km orbit. Both circularized. The formula for the first burn is the following:

${\displaystyle \Delta v_{1}={\sqrt {\frac {\mu }{r_{1}}}}{\Bigg (}{\sqrt {\frac {2r_{2}}{r_{1}+r_{2}}}}-1{\Bigg )}}$

This is the formula for the final burn in the transfer:

${\displaystyle \Delta v_{2}={\sqrt {\frac {\mu }{r_{2}}}}{\Bigg (}1-{\sqrt {\frac {2r_{1}}{r_{1}+r_{2}}}}{\Bigg )}}$

Where:

• ${\displaystyle \mu }$= Gravitational Parameter of Parent Body. (3530.461 km³/s² for Kerbin).
• ${\displaystyle r_{1}}$= The Radius of our first orbit. (100 km in this case).
• ${\displaystyle r_{2}}$= The Radius of our second orbit. (200 km in this case).

This formula will give us our velocity for the burn in km/s (multiply by 1000 to convert it into m/s). It's important to make sure that you will have the ${\displaystyle \Delta v}$ in the stage to make this burn. Again, you can do that by using the ${\displaystyle \Delta v}$ calculations above.

Calculating fuel flow

Next, we will explain how to calculate fuel flow in mass to see how much fuel a burn uses up in a specific amount of time.

If we know the ${\displaystyle \Delta v}$ needed for the burn and the total mass of the rocket before the burn, we can calculate how much fuel is required to complete the burn.

First, we calculate the mass of the rocket after the burn is complete. To do this, we use the Tsiolkovsky Rocket Equation, inputting the initial mass and ${\displaystyle \Delta v}$ of the burn. We can then solve the equation for the final mass (“dry mass”) after the burn. The difference between these two masses will be used to determine the length of time that is needed to complete the burn.

The equation for mass flow rate of fuel, given I_sp and thrust, is:

${\displaystyle {\dot {m}}={\frac {thrust}{I_{sp}}}}$

where ${\displaystyle {\dot {m}}}$ is the mass flow rate of fuel consumed. Again if the specific impulse is given in seconds it needed to multiplied by 9.81 m·s⁻² (see also Terminology about I_sp).

Dividing the difference between initial mass and final mass for the burn by the mass flow rate of fuel, we can determine how many seconds are required.

Usually, when the thrust is in kN and the specific impulse is in m/s the result is in Mg/s (= t/s). As the density of the liquid fuel/oxidizer mixture is 5 Mg/m³ this gives 1/5 m³/s = 2 dm³/s (= l/s).

Orbital velocity

Rather easy is the formula to calculate the orbital velocity of an orbit. This assumes circular orbit or the velocity of a specific point in an orbit. For this, we simply do this calculation:

${\displaystyle {\sqrt {\frac {\mu }{r}}}}$

Where:
${\displaystyle \mu }$ = Gravitational Parameter of parent body. (km³/s²)
${\displaystyle r}$ = radius of orbit. (km)

If we input the radius of the orbit in Kilometers, our orbital velocity will come out in Kilometers per second. In a 100 km orbit, our radius will be 700 km. Meaning our velocity will be ~2.2458 kilometers per second (km/s), or 2245.8 m/s.

Delta-v map

A ${\displaystyle \Delta v}$ map consists of approximate amounts of ${\displaystyle \Delta v}$ needed to get from one place (whether it is on the ground or in space) to another. The ${\displaystyle \Delta v}$ values we have for our ${\displaystyle \Delta v}$ map are approximate and include a fudge factor (in case we slip up on our piloting). Our map is as follows:

Launch to 100 km Kerbin orbit:	4700 m/s
Trans-Munar Injection:	900 m/s
Landing on the Mun:	1000 m/s
Launch from Mun and return to Kerbin:	1000 m/s
Total ${\displaystyle \Delta v}$:	7600 m/s

If we design our rockets to have 7600 total ${\displaystyle \Delta v}$, and the acceleration of the launch stages are adequate, we can have confidence that our rocket is able to land on the Mun and return to Kerbin. A rocket with a little less ${\displaystyle \Delta v}$ can accomplish this goal, but it is less forgiving of less efficient piloting.

Thrust to Weight Ratio

Calculating Thrust to Weight Ratio is only very simple steps.

It is important to know the thrust to weight ratio of your rocket to ensure your rocket will actually liftoff. If your TWR is less than 1, you can bet that you won't make an inch in altitude when starting from the launch pad. The minimum optimal TWR to have for your rocket at launch is 2.2.

To lift off the rocket's thrust need to exceed the gravitational force. The formula for this is simply the thrust of all of your current stage engines divided by the weight of your ship, fully fuelled.

${\displaystyle F_{T}>F_{G}=m\cdot g\implies TWR={\frac {F_{T}}{F_{G}}}={\frac {F_{T}}{m\cdot g}}>1}$

To calculate the acceleration simply use Newton's second law:

${\displaystyle F=m\cdot a=F_{T}-F_{G}=F_{T}-m\cdot g=m\cdot a\implies a={\frac {F_{T}}{m}}-g}$

These calculations only work when counteracting gravity. While coasting on an orbit the gravitational acceleration isn't important and thus the TWR may be below one and still work. The acceleration is at minimum directly after launch when the craft is heavy and at maximum immediately before running out of fuel, when the tanks are dry:

${\displaystyle a_{min}\approx {\frac {F_{T}}{m_{total}}}-g}$ and ${\displaystyle a_{max}\approx {\frac {F_{T}}{m_{dry}}}-g}$

The dry mass also includes the fully fuelled upper stages of the craft. To determine the g-force simply divide achieved acceleration by ${\displaystyle g_{0}=9.81{\frac {m}{s^{2}}}}$. As the craft is in free fall, the gravitational acceleration isn't felt by the crew so the accelerations appear to be higher for the crew leading to cancelling out the factor g:

${\displaystyle {\text{g-force}}_{min}\approx {\frac {F_{T}}{m_{total}\cdot g_{0}}}}$ and ${\displaystyle {\text{g-force}}_{max}\approx {\frac {F_{T}}{m_{dry}\cdot g_{0}}}}$

As the weight of the ship depends on the current gravitation (${\displaystyle g}$) the TWR differs between the celestial bodies.

In conclusion: This guide will hopefully have helped with designing your rockets to allow you to get the job done -- whatever it may be -- with no test flights first. We hope this guide has been helpful to new and continuing KSP pilots alike.