Designing, building and launching model rockets is not easy. Especially when the designer strives to achieve the highest results in competitions.

The success of an athlete largely depends on the correct choice of engine for the model. Another step towards achieving a record is knowing the laws of motion of the model.

In this chapter, we will introduce concepts related to movement - speed, acceleration, and other factors that affect flight altitude.

The flight performance of rocket models mainly depends on the following factors:

• G CT - launch weight of the rocket model (kg);
• G T - fuel weight (kg);
• J ∑ - total impulse of the engine (motors) (kg·sec);
• P ud - specific thrust of the engine (engines) (kg sec/kg);
• V - speed of the rocket model (m/sec);
• P - thrust of the engine (engines) (kg);
• a is the acceleration of the rocket model (m/sec 2);
• t - operating time of the engine (motors) (sec);
• i is the number of stages of the rocket model.

Ideal speed of a rocket model

The flight altitude of a model rocket depends primarily on its speed achieved at the end of engine operation. First, let's look at how to find the final speed of the model without taking into account air resistance and earth gravity. We will call this speed the ideal speed of the rocket model.

To determine the speed of the rocket model, we use the following law of mechanics: the change in the momentum of any body is equal to the impulse of the force applied to the body.

The quantity of motion is the product of the mass of a body m by its speed V, and the impulse of force is the product of the force F applied to the body by the time of its action t.

In our case, this law is expressed by the formula:

where m is the mass of the rocket model;
Vк is the speed of the rocket model at the end of engine operation;
V st - speed of the rocket model at the beginning of movement (in this case Set=0);
P - engine thrust;
t - engine operating time.

Since at the moment of start V st = 0, we get:

The mass of the rocket model changes during engine operation as the fuel burns out. We will assume that fuel consumption is a constant value and that during engine operation the weight of the fuel uniformly decreases from G T to 0. To simplify the calculations, we assume that average weight fuel is equal to G T /2, then the average mass of the rocket model will be equal to:
Considering that P·t=J ∑ -Рsp·G T) and based on the average weight of the fuel, we rewrite equation (20):
where:

or

This formula is an approximate expression of the well-known formula of K. E. Tsiolkovsky. It can be written in another, more convenient form for calculation. To do this, multiply the numerator and denominator on the right side of the formula by G T /2.
Let's give a few examples of using this formula.

Problem 4. Determine the ideal speed of a single-stage rocket model if: G CT =0.1 kg; P ud =30 kg·sec/kg; G T =0.018 kg.

Solution. To solve, we apply formula (23). We get:

Formula of K. E. Tsiolkovsky

More precisely, the ideal speed of a rocket model can be determined by the well-known formula of K. E. Tsiolkovsky using logarithmic tables.
where W is the speed of gas flow from the nozzle;
m st - launch mass of the rocket model;
m k is the final mass of the rocket model;
Z - Tsiolkovsky number.

The coefficient 2.3026 appeared in the second formula when moving from the natural logarithm to the decimal one.

Problem 5. Determine the ideal speed of the rocket model using the formula of K. E. Tsiolkovsky, if: G CT =0.1 kg; G T =0.018 kg; R ud =30 kg·sec/kg.

Solution. Final weight of the rocket model:

Let's substitute the available data into the Tsiolkovsky formula:

3. Actual speed of the rocket model

The flight of a model rocket is influenced by air resistance and the presence of gravity. Therefore, it is necessary to adjust for these factors in our calculations. Only then will we obtain the actual speed of the rocket model at the end of engine operation, on the basis of which we can calculate the flight path of the model.

The actual final speed of the rocket model can be calculated using the formula:

where Vk is the ideal speed of the rocket model;
P av - average engine thrust;
g - earth acceleration;
t - time;
D - midsection diameter;
A is the coefficient.

In this formula, the expression gt takes into account the gravity of the earth, and the expression D 2 /P av ·A - the influence of air resistance. Coefficient A depends on the ideal speed and altitude of the rocket model. The values ​​of coefficient A for various ideal flight speeds and altitudes are given in table. 2.

Problem 6. Determine the actual speed of the rocket model at the end of the active part of the flight path, if P beat =30 kg·sec/kg; G T =0.018 kg; G T =0.1 kg; t=0.6 sec; P av =0.9 kg; D=3 cm.

Solution. We will determine the ideal speed of the rocket model using one of the given versions of K. E. Tsiolkovsky’s formula:

Let's calculate the actual speed of the rocket model using formula (25):
The value of coefficient A for a given flight altitude is A=0.083.
Problem 7. Determine the actual speed of the rocket model at the end of the active section, if P beat = 25 kg sec/kg; G T =0.1 kg; t=4 sec; D=3 cm; G=0.1 kg (G k is the weight of the rocket model without fuel).

Solution. Model starting weight:

Ideal speed of a rocket model:

Average engine thrust:



Based on the fact that the total impulse and operating time are the main parameters of the engine, this formula for practical use It’s more convenient to rewrite it in the form:

because

4. Flight altitude of the rocket model

Let us now consider how, knowing the speed of the rocket model, find its flight altitude. We will consider the flight of the model strictly vertically. The flight trajectory of a model rocket can be divided into two sections - active, when the engines of the rocket model are running, and passive - the flight of the model by inertia after the engines stop working. Thus, the total flight altitude of the rocket model is:
where h 1 is the flight altitude in the active section;
h 2 - flight altitude in the passive section.

The height h 1 can be calculated by assuming that the speed of the rocket model changes uniformly from 0 to V at the end of engine operation. The average speed in this section is

where t is the flight time in the active section.

In formula (27), when calculating V act, air resistance was taken into account. It's a different matter when we calculate h 2 . If there were no air resistance, then, according to the laws of mechanics, a body flying by inertia with an initial speed would gain altitude

Since in our case V start =V effective, then

To take into account air resistance, you must enter a coefficient into this formula. It has been experimentally found that it is approximately 0.8. Thus, taking into account air resistance, the formula will take the form
Then formula (26) can be written as:
Problem 8. Calculate the height of the flight path of the rocket model and its acceleration based on the data: G CT =0.08 kg; D=2.3 cm; P beat =45.5 kg sec/kg; P av =0.25 kg; f=4 sec; G T =0.022 kg; J ∑ =1.0 kg·sec (engine DB-Z-SM-10).

Solution. Ideal speed of a rocket model:

Actual speed of the rocket model:
Flight altitude of the rocket model in the active section:
Passive flight altitude:
Total flight altitude of the rocket model:

5. Changing the flight path parameters of the rocket model depending on the engine operating time

From formula (29) it is clear that the flight altitude of the rocket model mainly depends on the speed of the rocket model achieved at the end of engine operation. The higher this speed, the higher the model will fly. Let's see how we can increase this speed. Let's return to formula (25).
We see what less value gt and D 2 /P av ·A, the higher the speed of the rocket model, which means the higher the value of the model’s flight altitude.

Table 3 shows the change in the parameters of the rocket flight path depending on the engine operating time. The table is given for rocket models with a launch weight G CT = 0.08 kg and a DB-Z-SM-10 engine. Engine characteristics: J ∑ =1.0 kg·sec; P ud =45.5 kg sec/kg; G T =0.022 kg. The total impulse remains constant throughout the flight.

The table shows that with an engine operating time of 0.1 seconds, the theoretical flight altitude of the model is 813 m. It would seem that let's make engines with such an operating time - and records are guaranteed. However, with such engine operating time, the model should develop a speed from 0 to 140.6 m/sec. If there were living beings on board a rocket with such speed, then none of them would be able to withstand such an overload.

Thus, we have come to another important concept in rocket science - speed of acceleration or acceleration. G-forces associated with excessive acceleration of a model rocket can destroy the model. And to make the structure more durable, you will have to increase its weight. In addition, flying at high accelerations is dangerous for others.

6. Acceleration of the rocket model

The following forces act on the rocket model in flight: the upward thrust force of the engine, and the downward force of earth gravity (the weight of the model) and air resistance.

Let's assume that there is no air resistance. To determine the acceleration of our model, we use the second law of mechanics: the product of the mass of a body and its acceleration is equal to the force acting on the body (F=m·a).

In our case, this law will take the form:

This is an expression for acceleration at the start of flight.

Due to fuel burnout, the mass of the rocket model is constantly changing. Consequently, its acceleration also changes. To find the acceleration at the end of the active section, we will assume that all the fuel in the engine has been burned, but the engine is still running at the last moment before shutting down. Then the acceleration at the end of the active section can be calculated using the formula:

If we enter into the formula the average weight of the rocket model in the active section G av = G CT -G T /2, we obtain the formula for the average acceleration:
The acceleration of the rocket model can also be determined from the approximate Tsiolkovsky formula (23), knowing that according to the well-known formula of mechanics V к =a ср ·t (t in our case is the operating time of the engine), we substitute this value for V к into formula (23)

Tsiolkovsky's approximate formula does not take into account the influence of gravity, which is directed downward and gives all bodies an acceleration equal to g. Corrected for gravity, the formula for the average acceleration during the active phase of the flight will take the form:
Once again, it should be emphasized that formulas (32) and (33) do not take into account air resistance.

Problem 9. Determine, without taking into account air resistance, the average acceleration of the rocket model if G CT =0.08/kg; G T =0.022 kg; P av =0.25 kg; t=4 sec; P ud =45.5 kg sec/kg; W=P beat g=446 m/sec.

Solution. We find the average acceleration of the rocket model using formulas (32) and (33):

As you can see, the results were the same. But since these formulas do not take into account air resistance, the actual speed calculated using the formula V act = a sr ·t will be overestimated.

Problem 10. Determine the speed of the rocket model at the end of the active section and the flight altitude without taking into account air resistance, based on the results of task 9. Compare the results with the results of task 8.

Solution. V act =a av ·t=25.7·4=102.2 m/sec.

The actual speed of the rocket model in Problem 8, solved taking into account air resistance, is 76.4 m/sec. Consequently, neglecting air resistance gives an absolute error

and relative error

Without taking into account air resistance, the flight altitude of the rocket model in the active section is:
On the passive section:

Total height: H=h 1 +h 2 =205.6+538=743.6 m.

Comparing these results with the results of problem 8, where the flight altitude of the model was calculated taking into account air resistance and was equal to 390.8 m, we obtain:

7. True acceleration of the rocket model

To determine the true acceleration of a rocket model, the formula is often used:
When deriving formula (34), two positions of the rocket model during flight are considered: at the start, when its mass is equal to G CT /g, and at the end of the active section, when the mass of the model is equal to (G CT -G T)/g. For these two positions, the acceleration of the model is calculated and its average is taken. Moreover, it is not taken into account that fuel consumption during the flight does not lead to a constant (linear) change in acceleration, but to an uneven one.

For example, consider the flight of a rocket model with a launch weight G CT = 0.08 kg and an engine DB-Z-SM-10, having data P av = 0.25 kg; t=4 sec, G T =0.022 kg; ω=0.022/4=0.0055 kg; P ud =45.5 kg sec/kg.

Using formula (30), which does not take into account air resistance, we will calculate accelerations every 0.5 seconds, assuming that the second fuel consumption is constant (ω=const).

Using formula (34), we calculate the average acceleration:
Let us determine the average acceleration using formulas (32) and (33), which also do not take into account air resistance:

Now the difference between the results obtained is clearly visible. Formula (34) for calculating the average acceleration of a rocket model is not suitable, since it is not applicable for bodies with variable mass. It is necessary to use formulas (32) and (33), which provide sufficient accuracy at any point in the flight path of the rocket model. But as shown by the results of flights of rocket models and their tests in wind tunnels, it is necessary to introduce into formulas (32) and (33) a coefficient K that takes into account air resistance, which varies within the range of 0.66÷0.8.

Thus, the formulas for the true acceleration of the rocket model are:

Let's analyze the above example to the end. Let's determine the true acceleration of the rocket model and its actual speed (take the average value of the coefficient K = 0.743)
The value of the coefficient must be selected depending on the area of ​​the midsection of the rocket model. How larger area midsection, the less you need to take the value of K from the range of its change 0.66÷0.8.

The given method for calculating the actual speed of a rocket model is the simplest and most accurate. Eliminates the need to use tables.

8. Speed ​​of multi-stage rocket models

The idea of ​​multi-stage rockets belongs to our compatriot, the wonderful scientist K. E. Tsiolkovsky. A multi-stage rocket model with the same fuel supply as a single-stage rocket achieves greater final speed, range and altitude because the engines of each stage operate sequentially, one after the other. When the engine of the lower stage runs out, it separates, the engine of the next stage begins to work, etc. With the separation of the next stage, the mass of the rocket model decreases. This is repeated until the last step. Thanks to the long acceleration and ever-decreasing weight, the model achieves a significantly higher speed than when all engines are fired simultaneously.

The weight ratios of the steps are of great importance. These relationships are even more significant than the choice of fuel for engines.

Let us assume that each stage of the rocket model uses engines with the same specific thrust, i.e., the same speed of gas flow from the engine nozzle.

The ideal speed of the last stage of the rocket model can be calculated using the Tsiolkovsky formula (24), only instead of the mass ratio m st /m to we take the value M. Formula (24) will take the form.

Chapter ten. Launching a rocket into space

At the White Sands Proving Ground, at 15:14 local time, a two-stage rocket was launched, the first stage of which was a modified V-2 rocket, and the second stage was a VAK-Corporal rocket.

Within a minute after launch, it reached an altitude of about 36 km and developed a speed of approximately 1600 m/sec. Here the V-2 separated from the VAK-Kapral, and it continued to climb, significantly increasing its speed. 40 seconds after turning on its engine, the VAK-Kapral was already flying at a speed of approximately 2.5 km/sec. The empty V-2 rocket first rose even higher (up to 161 km), and then began to fall. When, 5 minutes after launch, the V-2 rocket crashed in the desert 36 km north of the launch position, the VAK-Kapral rocket was still gaining altitude. The ascent continued for about 90 seconds. The top of the trajectory (402 km) was reached 6.5 minutes after the start.

At such an altitude, 1 km 3 of space contains fewer air molecules than in the best vacuum of any of our laboratories here, at the “bottom” of the ocean of air. At this altitude, an air molecule travels 8 km before colliding with another molecule. Thus, the VAK-Kapral missile practically reached airless space.

Naturally, after that she began to fall. The missile's impact point was in the northernmost part of the test site, 135 km from the launch site. The crash occurred 12 minutes after the start. Since the VAK-Kapral missile was small in size, the speed at which it hit the surface of the earth was very high. It took quite a long time to find her, despite the fact that radar tracking devices gave a general idea of ​​\u200b\u200bthe area where she fell. Only in January 1950 was it possible to discover and remove the remains of the heavily damaged tail section of the rocket.

The described launch was the fifth of those planned for the “Bumper Project,” which was part of the overall development program, not entirely successfully called the “Hermes Project.” “Project Bumper” involved the launch of eight V-2 missiles, three launches were successful, two were classified as “partially successful”, and three ended in failure.

The design of the VAK-Kapral missile was far from perfect. Now we can quite definitely point out two weak points this rocket. Theoretically, the second stage should have separated exactly at the moment the lower stage consumed its fuel supply. In reality, this was impossible to accomplish, since the acceleration of the V-2 rocket in last seconds the operation of its engine significantly exceeded the possible initial acceleration of the second stage, that is, the VAK-Kapral missile. These days, this problem could be solved by installing a solid fuel intermediate stage that produces higher acceleration.

The next problem, which has already been discussed a lot in the specialized literature, was the ignition of fuel in the second stage engine. Typically, in a VAK-Kapral rocket, both fuel components are mixed directly in the engine and ignite spontaneously at an altitude of several thousand meters above sea level, where the ambient air pressure is still close to normal. But at an altitude of 30 km, where the second stage separates, there is virtually no ambient air pressure. This can cause the fuel entering the combustion chamber to quickly evaporate and cause an explosion. To prevent this from happening, a sealing diaphragm is installed in the engine nozzle, which breaks when the engine starts.

The goal of Project Bumper was not only to study the problem of second stage separation in a two-stage liquid-propelled rocket, but also to achieve the highest possible altitude. According to the launch program, rockets No. 8 and 9 were intended to conduct a special experiment, which “ceremonially opened” a new test site in Florida. It had long been recognized that the White Sands site had become "cramped"; the distance from the launch position on it to the area where the shells fell did not exceed half the range of the V-2 rocket. A longer missile range could only be found on the ocean shore. In May 1949, negotiations began with the British government to establish observation and tracking stations in the Bahamas. At the same time, Cape Canaveral was chosen for the construction of launch positions. east coast Florida.

If you draw a straight line from Cape Canaveral in a southeast direction, it will pass through the Grand Bahama Islands (about 320 km from the starting positions). Great Abaco (440 km), Eleuthera (560 km), Cat (640 km), and then goes many thousands of kilometers into open ocean. Not counting the eastern end South America, the closest land in the direction of missile launch is the coast of South-West Africa (Fig. 49).

Rice. 49. Florida Proving Ground

However, for the first tests carried out at Cape Canaveral under the “Bumper Project”, there was no need for observation points in the Bahamas. The missiles were launched at a relatively short range. The main purpose of these launches was to launch the VAK-Kapral missile onto the flattest possible trajectory (Fig. 50).

Rice. 50. Typical flight trajectories of missiles launched under “Project Bumper”

The new test site was so imperfect that for a long time The simplest and most common tasks at the White Sands test site, such as transporting missiles from storage to the launch site, presented real problems.

The first rocket launch from Cape Canaveral was scheduled for July 19, 1950. From the very morning, failure followed failure. While the missiles were being prepared for launch, six aircraft patrolled over the sea, warning ships and vessels of possible danger. A few minutes before launch, one of these planes suddenly made an emergency landing. As a result, the rocket launch button was not pressed in a timely manner, and since the entire schedule was disrupted, the test had to be postponed for several hours. All preparations were made again, but at the appointed time some of the electronic equipment failed. Temporary repairs caused another delay. Finally everything was ready. The pyrotechnic igniter fired right on schedule, powering the rocket's pre-stage engine. The command “Main stage, fire!” was heard. But the rocket did not rise. Then Colonel Turner, who arrived in Florida from the White Sands training ground, decided that one of the valves had failed and ordered the preliminary stage engine to be cut off. The launch did not take place on this day.

On July 24, the test was repeated with a second missile. This time everything went perfectly: the rocket rose as planned and quickly disappeared into a thin veil of cirrus clouds. Having reached an altitude of 16 km, it began to enter an inclined section of the trajectory in order to continue its flight in a horizontal plane. At the same time, the VAK-Kapral rocket separated from the first stage, which slowly descended and was blown up at an altitude of 5 km. The wreckage of the V-2 fell into the sea at a distance of approximately 80 km from the launch position. The VAK-Corporal missile, too small to carry instruments and a demolition charge, fell into the sea 320 km from Cape Canaveral.

My long experience of lecturing on missiles led me to the idea that there is one feature in missile launches under the “Project Bumper” that at first glance seems somewhat strange. Why was the VAK-Kapral rocket engine started at an altitude of only about 32 km, that is, immediately after the V-2 rocket engine stopped working? Why wasn’t this done, say, when the V-2 rocket rose to a maximum altitude of about 130 km? It turns out that the whole point was that the VAK-Kapral rocket was never launched without an accelerator, and it could not have launched itself without outside help. Therefore, if it were launched at the point of maximum lift of the first stage (V-2), it would add only 40-50 km to the maximum altitude of the V-2 rocket (130-160). The reason that the VAK-Kapral missile rose to a height of 402 km as the second stage was that it separated from the first stage not when the latter reached its maximum altitude, but when it was moving at maximum speed.

To answer this question we will have to delve a little deeper into the field of theory. Let's start with what has been known in the form of Tartaglia's law for a number of centuries. In 1540, the Italian mathematician and specialist in the field of fortification Niccolo Tartaglia, who is credited with the honor of inventing the artillery quadrant protractor, discovered a law that established a certain relationship between the firing range and the height of the gun's trajectory. He argued that the maximum range of a projectile is achieved when fired at an angle of 45° and that if the trajectory height is 1000 m, then the projectile will fly 2000 m.

This simple relationship is actually somewhat violated due to air resistance, but almost completely retains its force in two cases: with a short firing range it is very heavy projectile, similar to cast cannonballs from Tartaglia’s time, and with an extremely long firing range, when almost the entire flight of the projectile is carried out in an environment close in conditions to a vacuum. This is evidenced by the characteristics of the V-2 rocket, the maximum lift height of which was 160 km, and the longest horizontal range with a trajectory altitude of about 80 km was approximately 320 km.

Niccolò Tartaglia established this relationship experimentally; he could not explain why, in particular, the elevation angle of 45° determines the maximum firing range. Nowadays, this phenomenon can be explained very simply. The flight range of a projectile in airless space (X) is determined by the formula:

where n 0 is the initial speed of the projectile, or the speed at the end of the active part of the trajectory; Q 0 is the elevation angle, or the angle of inclination of the trajectory at the end of the active section. Obviously, sin 2Q 0 is most important when Q 0= 45. The maximum value of the trajectory height in airless space (Ym) is expressed by the formula:

and for a vertical shot:

For missiles, the trajectory height ( Y m) must be determined from the point at the end of the active part of the trajectory. Then the total height of the rocket trajectory will be:

Y=Y m +Y k

Where Y k- height at the end of the active part of the trajectory. The height of the trajectory corresponding to the maximum flight range ( Y 45°), can be calculated using the formula:

Tartaglia's law is still used today, but only for a very rough assessment of the characteristics of the system, since in essence it does not explain anything.

What determines the height reached by the projectile? For simplicity of reasoning, let us first dwell on the flight characteristics of a conventional artillery shell. As the above formulas show, the height of the trajectory of a projectile when fired at the zenith is determined by the ratio of the speed to the force of gravity. Obviously, a projectile leaving a gun barrel at a speed of 300 m/sec rises higher than a projectile having a muzzle velocity of 150 m/sec. In this case, we will be interested not so much in the height of the projectiles, but in the process of their rise and fall, as well as their speed at the moment of meeting the ground.

Let us now imagine that the projectiles do not experience air resistance; then it will be quite legal to say that a projectile that leaves the gun barrel at a speed of 300 m/sec when firing at the zenith will fall to the ground with a speed of 300 m/sec, and another, with a muzzle speed of about 150 m/sec, will have a speed of 150 m/sec when falling sec. In this case, both projectiles will reach various heights. If conventional bombs are dropped from the same heights, then their speeds when hitting the ground will be equal to 300 and 150 m/sec, respectively.

This position can be formulated as follows: the speed required to reach a certain height in airless space is equal to the speed developed by the body when falling from this height. Since it is always possible to calculate the speed of a projectile when falling from any given height, it is not difficult to determine the speed that must be imparted to it to reach that height. Here are some numbers to illustrate the above:

From these figures it is clear that heights are growing much faster than their corresponding speeds. Thus, the height indicated in the second line is four times greater than the height indicated in the first, while the speeds differ from each other only by a factor of two. Therefore, to determine the moment of separation of the VAK-Kapral rocket (second stage) from the first stage (V-2), it was not so much the height achieved that was important, but the speed obtained by the rocket.

It should be noted, however, that the above figures do not take into account air resistance, as well as the fact that the force of gravity decreases with altitude (Fig. 51). If we consider all these phenomena in relation to rockets, it turns out that for them it is not at all important at what altitude the engine stops working. Below are data showing the dependence of the lift height on the speed for rockets with an acceleration of 3g; in this case, only the change in gravity with height is taken into account, and air resistance is not taken into account.

If we compare both groups of data presented, we can draw one very interesting conclusion, namely: when a body falls from an infinite height, its speed when it hits the ground cannot be infinite. This speed is quite calculable and amounts to 11.2 km/sec.

Thus, in the absence of air resistance, a cannon whose projectile has a muzzle velocity of 11.2 km/sec could shoot to infinity. Her projectile would have escaped the sphere of gravity. Therefore, the speed of 11.2 km/sec is called the “escape velocity”, or “second escape velocity”.

Rice. 51. Earth's gravitational field.

The relative strength of the field is shown by a curve and a group of spring scales (bottom of the figure) on which identical metal weights are weighed. A weight weighing 45 kg on the Earth's surface will weigh only 11 kg at a distance of half the Earth's diameter, 5 kg at a distance of one diameter, etc. The total area limited by the curve is equal to a rectangle, that is, the actual gravitational field is equal to the field having intensity , noted at the surface of the Earth, and extending to a height of one Earth radius

As an illustration, consider the technical idea of ​​Jules Verne's novel From the Gun to the Moon. It's pretty simple: huge cannon fires a projectile at the zenith with a muzzle velocity of about 11.2 km/sec. As the projectile gains altitude, its speed continuously decreases under the influence of gravity. At first, this speed will decrease by 9.75 m/sec, then by 9.4 m/sec, by 9.14 m/sec, etc., becoming less and less every minute.

Despite the fact that the degree of reduction in speed under the influence of gravity is continuously decreasing, the Jules Verne projectile will actually use up its entire speed reserve only after 300,000 seconds of flight. But by this time he will be at a distance where the gravitational fields of the Earth and the Moon balance each other. If at this point the projectile does not have enough speed reserve of just a few cm/sec, it will fall back to Earth. But even with such a reserve of speed, it will begin to fall in the direction of the Moon. After another 50,000 seconds, it will crash onto the surface of the Moon at a falling speed of about 3.2 km/sec, spending 97 hours and 13 minutes on the entire journey.

Having calculated in advance the duration of this flight, Jules Verne aimed his cannon at the calculated meeting point, that is, where the Moon was supposed to appear four days after the command “Fire!”

Despite the fact that the initial data in the novel is very close to the truth, the technical details of the implementation grandiose project either incomplete or very vague. Thus, an arbitrary amount of pyroxylin (181,000 kg) is placed in the barrel of a giant “gun” cast directly in the ground, and the author believes that this amount of pyroxylin will be sufficient to provide the projectile with a muzzle velocity of 16 km/sec. Elsewhere in the novel it is stated that for a projectile with such a high initial velocity, air resistance will not matter, because, supposedly, it will take only a few seconds to overcome the atmosphere.

The last remark is similar to the statement that an armor plate 1 m thick will not be able to stop a 16-inch projectile, since it covers a distance of 1 m in 0.001 seconds.

If the experiment with Jules Verne’s “gun” had been carried out in practice, the researchers would probably have been greatly surprised, since the projectile would have fallen 30m from the muzzle of the “gun”, rising to approximately the same height. In this case, the projectile would be flattened, and part of it could even evaporate. The fact is that Jules Berne forgot about the air resistance encountered by the projectile in the 210th gun barrel. After the shot, the projectile would find itself between two very hot and extremely powerful pistons, that is, between the wildly expanding gases of pyroxylin from below and a column of air heated by compression from above. Of course, all passengers of such a projectile would be crushed by the enormous force of the projectile's acceleration.

In addition, it is doubtful that such a “gun” could fire at all. Somehow, in their spare time, Aubert and Vallier calculated more accurately the estimated characteristics of Jules Verne’s “gun”. They came to amazing results. It turns out that the projectile had to be made of high-quality steel, such as tungsten, and be a solid solid body. The caliber of the projectile was determined to be 1200mm, and its length was 6 calibers. The cannon barrel had to be up to 900m long and dug into a mountain near the equator so that the muzzle was at least 4900m above sea level. Before firing, it would be necessary to pump the air out of the barrel and close the muzzle hole with a fairly strong metal membrane. When fired, the projectile would compress the remaining air and the latter would tear off the membrane the moment the projectile reached the muzzle.

A few years after Oberth, von Pirquet again looked at this problem and came to the conclusion that even such a “moon gun” could not accomplish the task of sending a projectile to the Moon. Von Pirke “increased” the height of the mountain by: 1000m and “installed” additional charges in the barrel, but even after that it was impossible to say with certainty whether the construction of such a weapon would be feasible and whether the funds that the country could allocate in the budget for the implementation would be enough for it. conventional war.

In short, it is impossible to fire a cannon into space through an atmosphere like the Earth’s and through a gravitational field like ours. The Moon is another matter: it would really be possible to use such a “gun” there, and its projectile, experiencing less gravity and without overcoming the atmosphere, of course, could fly to the Earth.

On Earth, the laws of nature favor rockets more than projectiles. Large rockets tend to rise slowly until they reach high altitudes, and only then begin to pick up speed. And although a rocket overcomes the same force of gravity as a projectile, and perhaps even greater, since it has to withstand the struggle with this force during a longer ascent, air resistance for it, with sufficiently large dimensions, is not such a serious obstacle .

Jules Verne's technical idea was that of using "brute force." Later, to overcome the force of the earth's gravity, another theory was put forward, based on an “easier” method. It was first outlined by H.G. Wells in his novel “The First Men in the Moon”; here a substance called “cavorite” is used, which supposedly not only resists the influence of gravity, but also creates a “gravitational shadow,” that is, a space where this force is absent.

Currently, we know very little about the laws of gravity. It is known, for example, that the force of gravity decreases in proportion to the square of the distance from the body creating “gravitational attraction”. In Fig. 51 graphically shows how the gravitational force changes depending on the distance. Mathematicians, for their part, tell us that this decrease is due to the law of geometry, according to which the area of ​​a sphere is proportional to the square of its radius. Of course, this characteristic of the gravitational force is not exclusive and it must have many other features. In this regard, we know much more about what qualities gravity does not possess. For example, it has been established that the force of gravity does not depend on the type of matter present; it is not affected by light and shadow, electricity and magnetism, ultraviolet and X-rays, as well as radio waves; it cannot be screened.

Therefore, it is quite understandable that all attempts to explain the nature of the force of gravity have so far been unsuccessful. However, one can call the explanation “classical”, which was proposed back in 1750 by a certain Le Sage from Geneva. According to this explanation, the entire universe is filled with "ultraterrestrial corpuscles" moving at high speed and creating constant pressure on the surface of all bodies. This pressure, according to Le Sage, presses a person to the surface of the Earth. If in our time someone put forward such a hypothesis, he would have to answer the question of where the heat that occurs when corpuscles hit bodies disappears, but in 1750 the law of conservation of energy had not yet been discovered.

Le Sage's hypothesis was accepted for many decades, but later it was found that corpuscles must penetrate any solid body, losing speed. For this reason, the shielding effect can be measured at least from the satellites of Jupiter. But all the studies said that such an effect does not exist.

When Albert Einstein became interested in this problem, he decided to look around him for some similar, difficult to explain natural phenomenon and soon found it. It was inertia and mainly centrifugal force. Einstein argued that a person in a rotating circular room would find himself in a certain “inertial field” that would cause him to move from the center of the room to the periphery. In this case, the force of inertia becomes greater the further a person is from the center of rotation. Einstein further stated that the “gravitational field” is equivalent to the “inertial field” due to a certain change in coordinates, but he did not explain anything else.

The implication of Einstein's suggestion is that gravity is probably not a "force" in its own right, as it is commonly understood. But then there cannot be any screens from gravity. If, nevertheless, gravity is associated with the general concept of “force,” then it is legitimate to put forward a hypothesis about the screening of this force, as G. Wells did in his novel. But then we come to an even stranger paradox.

The curve points in Fig. 51 are points of gravitational potential. It has a certain value on the surface of the Earth and decreases with distance from it. At some “infinite” distance from the Earth, the gravitational potential is zero. In order to move a body from a point with a higher potential to a point with a lower potential, it is necessary to do some work. For example, to lift a body weighing 1 kg to a height of 1 m, an effort equal to 1 kgm is required - a kilogram meter (a unit of work adopted in the metric system of measures). To lift a body weighing 1 kg to a height where the gravitational potential is zero, it is necessary to do work of the order of 6378. 10 3 kgm, and this work is equivalent to the release of all the kinetic energy of a body weighing 1 kg, accelerated to the second escape velocity.

Now suppose that Wells' Cavorite creates zero potential. Consequently, a person who steps on a cavorite leaf will have to overcome the full gravitational potential of the Earth. Let's say that a person weighs 75 kg. Then the muscles of his legs will have to produce work equal to only... 6378. 10 3. 75=47835- 10 4 kgm! And this is in just one step, for distance has no meaning; All that matters is the difference in potential. Thus, the brave traveler finds himself in a very difficult situation: either his muscles will not withstand such an exorbitant load and he will not be able to enter spaceship, or his muscles will somehow miraculously endure this test, but then he will not need the ship itself, since with such muscles he will be able to jump straight to the Moon.

It is said that there is a laboratory in the United States working on the problem of anti-gravity, but nothing is known about the details of its work. Of course, it would be interesting to know what theories and principles underlie these studies and whether it is already possible to talk about some kind of general starting point in this field of science. After all, all the explanations of the force of gravity that have been put forward so far should obviously be considered incorrect, because if Einstein’s thought is correct, then it closes all avenues for research.

Therefore, let us agree for now to focus on rockets as the most realistic means of overcoming earth's gravity. To understand the essence of a rocket flight into space, let’s solve this hypothetical example. Let's say that we set out to lift some payload weighing X kg to an altitude of 1300 km above sea level. From the table on page 244 it is clear that to rise to this height the rocket must reach a speed of more than 4 km/sec.

If it were necessary to create a rocket specifically to reach this height, then the decision on its likely dimensions would have to be postponed until all other problems were solved. The size of a rocket is not in itself an indication of its capabilities, except that a larger rocket will likely be more powerful. The central question here will be the determination of the rational relative mass of the rocket, that is, the relationship between the mass of the rocket in the launch position and the mass of the rocket after it has used up all the fuel. The initial mass of the rocket at the moment of launch (m 0) is the sum of the mass of the rocket itself (m p), the mass of the payload (m p) and the mass of fuel (m t). The final mass of the rocket at the moment of fuel consumption (m 1) is formed by the mass of the rocket itself (m p) and the mass of the payload (m p), and the ratio m 0 / m 1 is precisely the relative mass of the rocket.

It is known, for example, that in the V-2 rocket m p was 3 tons, m p was equal to 1 t, and m t reached 8 tons. Consequently, the initial mass of the V-2 was 3 + 1 + 8 = 12 tons. The final mass was 3 +1 = 4 tons, and the relative mass was 3: 1.

Our next step should probably be to determine the relative mass required for the rocket to reach a speed of 4 km/sec. However, here we encounter a rather interesting problem. It turns out there are a lot of answers to this question. Theoretically, the relative mass required to impart a speed of 4 km/sec to a rocket can be arbitrary, since it depends on the rate of exhaust of fuel combustion products. It is enough to change the value of this speed, and we will get a different value of the relative mass. Therefore, until we determine the rate of exhaustion of combustion products, we will not be able to find the most rational relative mass of the rocket. It must be remembered that any specific value of the outflow velocity will give only an unambiguous answer corresponding to the accepted condition. We need to obtain a solution in general form.

The solution to this dilemma is extremely simple. It is based on the use of measurement of any rate of combustion products as a standard. To do this, we need to know only one thing - the relative mass at which the rocket can be imparted a speed equal to the speed of the outflow of combustion products. With a higher exhaust speed we will get a higher speed, and with a small one we will get a correspondingly lower speed of the rocket. But whatever these speeds may be, the relative mass of the rocket, which is necessary to impart to it a speed equal to the exhaust speed, must be constant.

The speed of a rocket is usually denoted by v, and the speed of exhaustion of combustion products by c. In our example, what should the relative mass be equal to at v = c? It turns out that it is equal to 2.72:1, in other words, a rocket with a launch weight of 272 conventional units should have a weight of 100 units upon reaching a speed equal to the rate of exhaustion of its combustion products. This number has already been mentioned by us and represents the constant known to every mathematician e = 2.71828183.., or rounded 2.72.

This is exactly the general solution we were looking for. Written in the form of a formula, this dependence of the maximum speed of the rocket on the rate of exhaustion of combustion products and the relative mass of the rocket looks like this:

v = c ln(m 0 /m 1)

Using this formula, one can easily determine what relative mass would have to be had if the speed of the rocket were to be increased twice as much as the exhaust speed. Substituting the value v = 2c into the formula, we obtain a relative mass equal to the square of e, that is, approximately 7.4:1. Accordingly, a rocket with such a relative mass can be accelerated to a speed of 3s.

In our example, to lift a rocket to a height of 1300 km, it is necessary to develop a speed of only 4 km/sec, and this is approximately twice the speed of the combustion products of the V-2 rocket. Therefore, a rocket with a gas exhaust speed similar to that of the V-2 rocket and a relative mass of 7.4: 1 should rise to a height of about 1300 km.

The dependence we have shown is theoretically correct, but requires some clarification in practice. It is completely valid only for airless space and in the absence gravitational field. But when taking off from Earth, the rocket must overcome both air resistance and the force of gravity, which has a variable value. A V-2 rocket with a relative mass of 3:1 should therefore have a higher speed than the exhaust speed of its engine (2 km/sec). However, its actual maximum speed was only 1.6 km/sec. This difference arises from air resistance and gravity and varies from rocket to rocket.

For example, a small pyrotechnic rocket develops a speed equal to 2-3% of the theoretical maximum speed. The V-2 rocket accelerated to a speed of 70% of its maximum design speed. The larger the rocket, the smaller the difference between these two values; a rocket capable of escaping Earth's gravity would likely have up to 95% of its maximum design speed.

All this suggests that high values The rocket's flight speed can be achieved either by increasing the rate of exhaustion of combustion products, or by choosing a higher relative mass, but it is preferable to use both of these factors. The increase in the relative mass of missiles depends entirely on the level of development rocket technology, while increasing the flow rate of combustion products is mainly a problem of chemistry. To give a general idea of ​​what can be expected in this regard from some of the fuel mixtures currently in use, their main experimental characteristics are given below.

Of these fuels, nitromethane has been studied most thoroughly, which is a so-called monofuel because it contains both a fuel and an oxidizer. This fuel has not found widespread use, since experts consider it to be explosive due to shocks and impacts. The latter mixture - oxygen with hydrogen - has been tested case by case and requires further research, but it can already be said that it is not an ideal rocket fuel, despite the supposedly high rates of combustion products provided by it. Thus, the temperature of liquid oxygen exceeds the boiling point of liquid hydrogen by as much as 70°C, which makes handling and maintaining liquid hydrogen in the mixture very difficult. Another disadvantage is that hydrogen, even in a liquid state, is very light and therefore must take up a large volume, which leads to larger tanks and total weight rockets.

Currently, alcohol, aniline and hydrazine are widely used as rocket fuels. In parallel, work is underway with other chemical compounds, however, the general impression that emerges from analyzing the formulas of these substances is that, from the point of view of energy content and combustion characteristics, the greatest progress seems to have been achieved in the field of improving the oxidative part of fuel mixtures.

One of the very promising ideas In this direction, one can name a proposal to replace liquid oxygen with liquid ozone, which is oxygen that has three atoms in each molecule, unlike ordinary, diatomic oxygen. He has a higher specific gravity; a cylinder that usually contains 2.7 kg of liquid oxygen can hold almost 4.5 kg of liquid ozone. The boiling point of liquid oxygen is -183°C, and that of liquid ozone is -119°C. In addition to its higher density and boiling point, ozone has another advantage, which is that the decomposition of liquid ozone produces very large quantity heat. The fact is that atoms of ordinary oxygen can group into ozone molecules only when absorbing energy of the order of 719 g/cal, which is observed during lightning discharges and irradiation with ultraviolet rays. If ozone is used as an oxidizer, then during the combustion of fuel it again turns into molecular oxygen, releasing the energy it absorbed. Calculations show that fuel oxidized with ozone would provide a gas flow rate approximately 10% higher than when the same fuel was oxidized with oxygen.

However, all these advantages are currently losing their significance due to the fact that liquid ozone is very unstable and, with slight overheating, can turn into oxygen with an explosion. The presence of any impurities in it, as well as contact with certain metals and organic substances, only accelerates this process. It is possible, of course, that there is a substance in nature that would make ozone safe, but the search for such an anticatalyst has not yet been successful.

All of the fuel components we have listed (hydrogen peroxide, nitric acid, ozone and some unmentioned nitrogen compounds, for example NO 4) are oxygen carriers and ensure combustion by oxidizing the fuel with oxygen. However, chemists know another type of combustion, in which the active element is not oxygen, but fluorine. Due to its extremely high activity, fluorine remained little known to science for a long time. It was impossible to store this substance even in laboratory conditions; it “burned through” the walls of containers and easily destroyed everything it came into contact with. Nowadays, in the study of the properties of fluorine, great success. It has been discovered, for example, that compounds of uranium and fluorine are very stable and do not react even with pure fluorine. Thanks to new substances obtained by chemists, it is now possible to preserve pure fluorine for a long period of time.

Bench testing by Rokitdyne of a large liquid rocket engine in the Santa Suzanna Mountains near Los Angeles

Liquid fluorine is a yellow liquid that boils at -187°C, that is, 4°C below the boiling point of oxygen; its specific gravity is slightly higher than the specific gravity of liquid oxygen and is equal to 1.265 (specific gravity of oxygen 1.15). While pure liquid fluorine reacts actively with liquid hydrogen, its oxide (F 2 O) is not so active and therefore can be useful and quite acceptable as an oxidizing agent in rocket engines.

Thus, since the dimensions of fuel tanks depend on the density and energy parameters of the fuel components, the relative mass of the rocket to a certain extent depends on the fuel mixture used. The main task of the designer is to select a fuel at which the launch weight of the rocket would be minimal. The possibilities for reducing the weight of tanks and the engine are quite limited. The only promising rocket component in this regard is the turbopump unit. Currently, the fuel supply system for the turbopump and steam gas generation includes tanks for hydrogen peroxide and permanganate, as well as a steam gas generator and a system of valves and pipelines. All this could be eliminated if it were possible to use the main rocket fuel to operate the unit. This issue is now being addressed by creating turbines that can operate at significantly higher temperatures than what was considered the limit 10 years ago. If necessary, such a turbine could operate on a re-enriched fuel mixture so that the combustion temperature remains within acceptable limits. In this case, some of the fuel would inevitably be lost, but these losses would still be less than the weight of the turbopump unit.

The thermal energy from the turbine exhaust gases, consisting of water and alcohol vapor, as well as carbon dioxide, could be used in a heat exchanger to evaporate some oxygen to create a boost in the oxidizer tank. After cooling in the heat exchanger, the gases would be diverted back into the fuel tank to create pressurization there. As a result, condensed alcohol vapor would flow back into its tank. A small amount of water condensed from vapor would practically not reduce the calorific value of the fuel, but carbon dioxide could be used to increase boost.

The measures considered can only slightly improve the performance of the rocket; the most important thing is that to rise to a height of 1300 km, the rocket must have a relative mass of about 7.5:1. And this requires a fundamentally new solution to many engineering issues. This solution is the creation of multi-stage rockets, the first examples of which were the German Reinbote rocket and the American Bumper rocket.

When implementing the “Bumper Project,” the principle was based on the principle of combining existing missiles.

This solution offers a number of significant practical advantages; in particular, there is no need to wait for the development of each stage of the system; The performance characteristics of missiles, as a rule, are already known, and besides, such a system costs much less. But in this case, the result is a rocket in which the stages have different relative masses. And since these stages operate on different fuels, they show different rates of exhaust combustion products. Calculating the performance of a multi-stage rocket is quite complex, but we will simplify it somewhat by using a two-stage rocket as a basis, in which both stages run on the same fuel and have the same relative masses (each 2.72:1). Let us also assume that the experiment is carried out in airless space and in the absence of any gravitational field. The first stage will give our rocket a speed equal to the exhaust speed (1s), and the second will double it (2s), since the final speed of the second stage will be equal to twice the exhaust speed. With a single-stage design, this would require creating a rocket with a relative mass of 7.4: 1, which is nothing more than 3, or 2.72 X 2.72. It follows from this that in a multi-stage rocket the final speed corresponds to the maximum acceleration speed of a single-stage rocket with a relative mass equal to the product of the relative masses of all stages.

Knowing this, it is quite easy to calculate that a launch to an altitude of 1300 km should be carried out by a two-stage rocket, in which each stage has a relative mass of 3:1. Both stages must operate on ethyl alcohol and liquid oxygen at an exhaust velocity of about 2 km/sec, at sea level. In this case, the first stage would practically not be able to develop a speed equal to the exhaust speed, since in real conditions it would have to overcome gravity and air resistance, but the second stage, which does not deal with these negative aspects, would be able to develop a speed close to double flow rate of combustion products. To get an idea of ​​how big such a rocket would have to be, let's assume that the second stage payload weighs 9 kg. Then all weight characteristics will take the following form (in kg):

This weight is almost equal to the weight of the Viking rocket No. 11, which reached an altitude of 254 km with a payload of 374 kg, which is significantly greater than the weight of the second stage in our example.

Twenty years ago, scientists discussed two problems with great fervor; will the rocket be able to go beyond the earth's atmosphere and will it be able to overcome the force of gravity. At the same time, concerns were expressed that the rocket would develop too high a speed in a very short period of time and would spend the overwhelming majority of its energy overcoming air resistance. Today, most of these fears can be considered unfounded; rockets have left the Earth's atmosphere more than once. Practice has shown that as soon as a rocket reaches the tropopause in optimal mode, almost all obstacles to its further upward movement will be eliminated. This is explained by atmospheric layer, lying below the tropopause, contains 79% of the total mass of air; The stratosphere covers 20% of the mass, and less than 1% of the total air mass is scattered in the ionosphere.

The degree of rarefaction of air in the upper layers of the atmosphere is even better illustrated by the average free path of air molecules. It is known that at sea level, 1cm 3 of air at +15°C contains 2.568 X 10 19 molecules, which are constantly in rapid motion. Since there are so many molecules, they often collide with each other. The average distance in a straight line that a molecule travels from one collision to another is called the mean free path. This parameter does not depend on the speed of movement of the molecule, and therefore on the temperature of the medium. At sea level, the average free path of air molecules is 9.744 X 10 -6 cm, at an altitude of 18 km it already reaches 0.001 mm, at an altitude of 50 km it is 0.1 mm, and at 400 km from the Earth it approaches 8 km.

At even higher altitudes, the concept of the mean free path of molecules loses all meaning, since the air here ceases to be a continuous medium and turns into a cluster of molecules moving around the Earth in independent astronomical orbits. Instead of a continuous atmosphere, at these altitudes there is a region of “molecular satellites”, which astrophysicists call the “exosphere”.

In the upper layers of the atmosphere there are zones of high temperatures. So, at an altitude of 80 km the temperature is 350 ° C. But this value, which is quite impressive at first glance, essentially only expresses the fact that air molecules here move at a very high speed. A body that gets here cannot heat up to such a temperature while remaining here for a short time, just as people who are in a spacious barn, in one corner of which hangs a light bulb with a filament heated to several thousand degrees, cannot die from the heat.

In the specialized literature, the question of finding such an “optimal speed” of a rocket that would be sufficient to overcome air resistance and gravity, but not so high as to cause overheating of the rocket, has been raised more than once. Practice shows that this question practical significance does not, since large liquid rockets moving quite slowly in lower layers atmosphere, cannot have accelerations that would ensure their acceleration even to the “optimal speed” in this section of the trajectory. By the time the rockets reach this speed, they are usually beyond the lower layers atmosphere and are not exposed more danger overheating

Several years ago, the first large solid-fuel rockets appeared, which necessitated changes in many already established rocket design standards during their development. The National Aviation Advisory Committee (NACA) conducted a series of studies for this purpose in order to select the most appropriate shapes for the hull, tail, and wings of rockets intended for flight at high speeds. Experimental models were built and launched with solid fuel engines, the payloads of which were so large and the operating time of the engines so short that there was almost no danger of exceeding the “optimal speed”. Subsequently, solid fuel rockets, especially the Deacon rocket, began to be used for scientific research, and above all for cosmic ray research.

Cosmic rays are fast moving elementary particles(mainly protons). When such a particle approaches the Earth, the Earth's magnetic field deflects it, and it may happen that it does not enter the atmosphere at all. In the uppermost layers of the atmosphere, protons collide with oxygen or hydrogen atoms, resulting in qualitatively new cosmic rays, which in technology are called “secondary” in contrast to those coming from space, that is, “primary”. The maximum density of cosmic rays is observed at an altitude of about 40 km, where secondary rays have not yet had time to be absorbed by the atmosphere.

The source of origin of primary cosmic rays is still unknown, since the Earth's magnetic field deflects them so strongly that it is impossible to determine the initial direction of their movement in space.

The intensity of cosmic radiation near the Earth's surface is practically independent of the time of year and day, but it varies at different magnetic latitudes. It has minimum values ​​at the magnetic equator, and maximum values ​​above the magnetic poles at an altitude of 22.5 km.

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For further calculations, let's take the R-9 / R-9A (8K75)SS-8/(Sasin) intercontinental ballistic missile. For which the basic parameters are defined in the directory:

Initial mass

Rocket diameter

Velocity of separated particles

Let us further define the parameters of the atmosphere:

Density of air on the surface of the Earth

Height above sea level

Radius of the Earth

Earth mass

Earth's rotation speed at the equator

Earth's gravitational constant

Using the initial conditions and a system of equations, you can determine the trajectory of the ICBM using the differentiation method described in paragraph 1.3.

Since we differentiate the equations discretely with a certain step, this means that the ICBM will stop further movement only in the case when the height at which the ICBM is located becomes less than zero. To eliminate this shortcoming, we will use the method described in paragraph 1.4, but we will apply it to our case:

We will look for the coefficients a and b of the variables And , Where – height of the ICBM above ground level, – deflection angle. As a result, we get the equations:

In our case
, as a result we get

By determining the deflection angle at which the height of the ICBM will be equal to the level of the Earth. Let's find the flight range of an ICBM:

The engine operating time is determined by the formula:

Where
– warhead mass. For a more realistic flight, we will take into account the mass of the stage shell; for this we will add the coefficient to this formula
, which shows the ratio of the stage mass to the fuel mass.

We are now able to determine the trajectory of the ICBM under given initial conditions.

Chapter 2. Results

2.1. Parametric curves of single-stage MBR

The initial parameters used in the construction of Fig. 1.

Instantaneous fuel combustion rate Mu = 400 kg/s;

Graph of ICBM flight range versus angle of attack

In Fig. 1. it can be seen that the maximum flight range is at the angle of attack =38 degrees, but this is the value of the optimal angle of attack with constant parameters of the instantaneous fuel combustion rate and final mass. For other values ​​of Mu and Mk, the optimal angle of attack may be different.

The initial parameters used in the construction of Fig. 2.

Attack angle = 30 deg.

Final mass (warhead) Mk = 2.2 tons.

Graph of ICBM flight range versus instantaneous fuel combustion rate

Figure 2 shows that the optimal value of the instantaneous fuel combustion rate = 1000 kg/s. It is clearly visible that this value is not possible. This contradiction occurs due to the fact that the R9 ICBM under consideration is heavy (missile mass = 80.4 tons) and the use of one stage for it is not possible.

To find optimal parameters we will use the gradient descent method. For a single-stage rocket, assuming that the angle of attack is constant, the optimal parameters are:

Instantaneous fuel combustion rate Mu = 945 kg/s;

Attack angle = 44.1 deg.

Before this, our research was carried out under the assumption that the angle of attack is equal to a constant, let's try to introduce another dependence, let the angle of attack depend on the height as
.

The optimal parameters in this case are:

Instantaneous fuel combustion rate Mu = 1095 kg/s;

Constant C = 0.0047.

Graph of flight range at optimal parameters

Rice. 3. 1 – if dependent
, 2 – if dependent

In Fig. 3. It can be seen that when the angle of attack is not equal to a constant, the missile range is greater. This is due to the fact that in the second case the rocket leaves the earth’s atmosphere faster, that is, it is slowed down less by the atmosphere. In further research we will take the dependence
.

March 24, 2014 at 07:05 pm

Educational/game program for calculating the payload of a rocket, taking into account several stages and gravitational losses

• Cosmonautics,
• Physics,
• Games and game consoles

Parameters not taken into account

• To simplify the problem, the following are not taken into account:
• Air friction losses.
• Change in thrust depending on atmospheric pressure.
• Climb.
• Loss of time for the separation of steps.
• Changes in engine thrust in the area of ​​maximum speed pressure.
• Only one layout is taken into account - with a sequential arrangement of steps.

A little physics and mathematics

Speed ​​calculation

The rocket acceleration in the model goes like this:


The flight altitude is assumed to be constant. Then the rocket thrust can be divided into two projections: Fx And Fy. Fy must be equal mg, these are our gravitational losses, and Fx- this is the force that will accelerate the rocket. F is constant, this is the thrust of the engines, m changes due to fuel consumption.
Initially, there was an attempt to analytically solve the equation of rocket motion. However, it was not successful, since gravitational losses depend on the speed of the rocket. Let's do a thought experiment:

1. At the beginning of the flight, the rocket simply will not lift off from the launch pad if the thrust of the engines is less than the weight of the rocket.
2. At the end of the acceleration, the rocket is still attracted to the Earth with force mg, but this does not matter, since its speed is such that it does not have time to fall, and when it enters a circular orbit, it will constantly fall to the Earth, “missing” it due to its speed.

It turns out that the actual gravitational losses are a function of the mass and speed of the rocket. As a simplified approximation, I decided to calculate gravitational losses as:

V1- this is the first cosmic speed.
To calculate the final speed we had to use numerical modeling. The following calculations are performed in one second increments:

The superscript t is the current second, t-1 is the previous one.

Or in a programming language

for (int time = 0; time< iBurnTime; time++) { int m1 = m0 - iEngineFuelUsage * iEngineQuantity; double ms = ((m0 + m1) / 2); double Fy = (1-Math.pow(result/7900,2))*9.81*ms; if (Fy < 0) { Fy = 0; } double Fx = Math.sqrt(Math.pow(iEngineThrust * iEngineQuantity * 1000, 2)-Math.pow(Fy, 2)); if (Fx < 0) { Fx = 0; } result = (result + Fx / ms); m0 = m1; }

Maximum payload calculation

Knowing the resulting speed for each allowable payload, the payload maximization problem can be solved as a problem of finding the root of a nonlinear equation.

It seemed to me most convenient to solve this equation using the half division method:

The code is completely standard

public static int calculateMaxPN(int stages) ( deltaV = new double; int result = 0; int PNLeft = 50; while (calculateVelocity(PNLeft, stages, false) > 7900) ( PNLeft = PNLeft + 1000; ) System.out.println (calculateVelocity(PNLeft, stages, false)); int PNRight = PNLeft - 1000; double error = Math.abs(calculateVelocity(PNLeft, stages, false) - 7900); System.out.println("Left " + Double.toString (PNLeft) + "; Right " + Double.toString(PNRight) + "; Error " + Double.toString(error)); boolean calcError = false; while ((error / 7900 > 0.001) && !calcError) ( double olderror = error; if (calculateVelocity((PNLeft + PNRight) / 2, stages, false) > 7900) ( PNRight = (PNLeft + PNRight) / 2; ) else ( PNLeft = (PNLeft + PNRight) / 2; ) error = Math .abs(calculateVelocity((PNLeft + PNRight) / 2, stages, false) - 7900); System.out.println("Left " + Double.toString(PNLeft) + "; Right " + Double.toString(PNRight) + "; Error " + Double.toString(error)); if (Math.abs(olderror - error)< 0.0001) { //аварийный выход если алгоритм уйдет не туда PNLeft = 0; PNRight = 0; calcError = true; } } result = (PNLeft + PNRight) / 2; calculateVelocity(result, stages, true); return result; }

How about playing?

Now, after the theoretical part, you can play.
The project is located on GitHub. MIT License, feel free to use and modify, and redistribution is encouraged.

The main and only window of the program:

You can calculate the final speed of the rocket for a specified PN by filling out the parameter text fields, entering the PN at the top and clicking the "Calculate Velocity" button.
You can also calculate the maximum payload for given rocket parameters; in this case, the “PN” field is not taken into account.
There is a real rocket with five stages "Minotaur V". The "Minotaur V" button loads parameters similar to this rocket in order to show an example of how the program works.
This is essentially a sandbox mode in which you can create rockets with arbitrary parameters, studying how different parameters affect the rocket's payload.

Competition

Competition mode is activated by pressing the Competition button. In this mode, the number of controllable parameters is greatly limited to ensure the same competition conditions. All stages have the same type of engines (this is necessary to illustrate the need for several stages). You can control the number of motors. You can also control the fuel distribution by stages and the number of stages. Maximum fuel weight is 300 tons. You can add less fuel.
Task: using minimal amount engines to achieve maximum PN. If there are a lot of people willing to play, then each number of engines will have its own classification.
Those interested can leave their results with the parameters used in the comments. Good luck!

In which there is no thrust or control force and moment, it is called a ballistic trajectory. If the mechanism that powers the object remains operational throughout the entire period of movement, it belongs to the category of aviation or dynamic. The trajectory of an aircraft during flight with the engines turned off at high altitude can also be called ballistic.

An object that moves along given coordinates is affected only by the mechanism that drives the body, the forces of resistance and gravity. A set of such factors excludes the possibility of rectilinear movement. This rule works even in space.

The body describes a trajectory that is similar to an ellipse, hyperbola, parabola or circle. The last two options are achieved at the second and first cosmic velocities. Calculations for movement along a parabola or circle are carried out to determine the trajectory ballistic missile.

Taking into account all the parameters during launch and flight (weight, speed, temperature, etc.), the following trajectory features are distinguished:

• In order to launch the rocket as far as possible, you need to choose the right angle. The best is sharp, about 45º.
• The object has the same initial and final speed.
• The body lands at the same angle as it launches.
• The time it takes for an object to move from the start to the middle, as well as from the middle to the finishing point, is the same.

Trajectory properties and practical implications

The movement of a body after the influence of the driving force on it ceases is studied by external ballistics. This science provides calculations, tables, scales, sights and produces optimal options for shooting. The ballistic trajectory of a bullet is the curved line described by the center of gravity of an object in flight.

Since the body is affected by gravity and resistance, the path that the bullet (projectile) describes forms the shape of a curved line. Under the influence of these forces, the speed and height of the object gradually decreases. There are several trajectories: flat, mounted and conjugate.

The first is achieved by using an elevation angle that is less than the angle of greatest range. If the flight range remains the same for different trajectories, such a trajectory can be called conjugate. In the case where the elevation angle is greater than the angle of greatest range, the path becomes called a suspended path.

The trajectory of the ballistic movement of an object (bullet, projectile) consists of points and sections:

• Departure(for example, the muzzle of a barrel) - this point is the beginning of the path, and, accordingly, the reference.
• Weapons horizon- this section passes through the departure point. The trajectory crosses it twice: during release and during fall.
• Elevation area- this is a line that is a continuation of the horizon and forms a vertical plane. This area is called the firing plane.
• Trajectory vertices- this is the point that is located in the middle between the starting and ending points (shot and fall), has the highest angle along the entire path.
• Tips- the target or sighting location and the beginning of the object’s movement form the aiming line. An aiming angle is formed between the horizon of the weapon and the final target.

Rockets: features of launch and movement

There are guided and unguided ballistic missiles. The formation of the trajectory is also influenced by external and external factors (resistance forces, friction, weight, temperature, required flight range, etc.).

The general path of a launched body can be described by the following stages:

• Launch. In this case, the rocket enters the first stage and begins its movement. From this moment, the measurement of the height of the ballistic missile’s flight path begins.
• After about a minute, the second engine starts.
• 60 seconds after the second stage, the third engine starts.
• Then the body enters the atmosphere.
• Lastly, the warheads explode.

Launching a rocket and forming a movement curve

The rocket's travel curve consists of three parts: the launch period, free flight and re-entry into the earth's atmosphere.

Live projectiles are launched from a fixed point on portable installations, as well as Vehicle(ships, submarines). The flight initiation lasts from tenths of a thousandths of a second to several minutes. Free fall constitutes the largest portion of a ballistic missile's flight path.

The advantages of running such a device are:

• Long free flight time. Thanks to this property, fuel consumption is significantly reduced in comparison with other rockets. For prototype flight ( cruise missiles) more efficient engines are used (for example, jet engines).
• At the speed at which the intercontinental weapon moves (approximately 5 thousand m/s), interception is very difficult.
• The ballistic missile is capable of hitting a target at a distance of up to 10 thousand km.

In theory, the path of movement of a projectile is a phenomenon from general theory physics, section of the dynamics of rigid bodies in motion. With respect to these objects, the movement of the center of mass and the movement around it are considered. The first relates to the characteristics of the object in flight, the second to stability and control.

Since the body has programmed trajectories for flight, the calculation of the ballistic trajectory of the missile is determined by physical and dynamic calculations.

Modern developments in ballistics

Because the combat missiles of any kind are dangerous to life, the main task of defense is to improve the points for launching destructive systems. The latter must ensure the complete neutralization of intercontinental and ballistic weapons at any point in the movement. A multi-tier system is proposed for consideration:

• This invention consists of separate tiers, each of which has its own purpose: the first two will be equipped with laser-type weapons (homing missiles, electromagnetic guns).
• The next two sections are equipped with the same weapons, but designed to destroy the head parts of enemy weapons.

Developments in defense missile technology do not stand still. Scientists are modernizing a quasi-ballistic missile. The latter is presented as an object that has a low path in the atmosphere, but at the same time sharply changes direction and range.

The ballistic trajectory of such a missile does not affect its speed: even at an extremely low altitude, the object moves faster than a normal one. For example, the Russian-developed Iskander flies at supersonic speeds - from 2100 to 2600 m/s with a mass of 4 kg 615 g; missile cruises move a warhead weighing up to 800 kg. During flight, it maneuvers and evades missile defenses.

Intercontinental weapons: control theory and components

Multistage ballistic missiles are called intercontinental missiles. This name appeared for a reason: due to the long flight range, it becomes possible to transfer cargo to the other end of the Earth. The main combat substance (charge) is mainly an atomic or thermonuclear substance. The latter is located in the front of the projectile.

Next, a control system, engines and fuel tanks are installed in the design. Dimensions and weight depend on the required flight range: the greater the distance, the higher the launch weight and dimensions of the structure.

The ballistic flight trajectory of an ICBM is distinguished from the trajectory of other missiles by altitude. The multi-stage rocket goes through the launch process, then moves upward at a right angle for several seconds. The control system ensures that the gun is directed towards the target. The first stage of the rocket drive separates independently after complete burnout, and at the same moment the next one is launched. Upon reaching a given speed and flight altitude, the rocket begins to rapidly move down towards the target. The flight speed to the destination reaches 25 thousand km/h.

World developments of special purpose missiles

About 20 years ago, during the modernization of one of the medium-range missile systems, a project for anti-ship ballistic missiles was adopted. This design is placed on an autonomous launch platform. The weight of the projectile is 15 tons, and the launch range is almost 1.5 km.

The trajectory of a ballistic missile for destroying ships is not amenable to quick calculations, so predict the enemy’s actions and eliminate this weapon impossible.

This development has the following advantages:

• Launch range. This value is 2-3 times greater than that of the prototypes.
• Flight speed and altitude make military weapon invulnerable to missile defense.

World experts are confident that weapons of mass destruction can still be detected and neutralized. For such purposes, special reconnaissance out-of-orbit stations, aviation, submarines, ships, etc. The most important “counteraction” is space reconnaissance, which is presented in the form of radar stations.

The ballistic trajectory is determined by the reconnaissance system. The received data is transmitted to its destination. The main problem is the rapid obsolescence of information - for short period Over time, the data loses its relevance and may diverge from the actual location of the weapon at a distance of up to 50 km.

Characteristics of combat systems of the domestic defense industry

Most powerful weapon Currently, an intercontinental ballistic missile is considered to be stationary. Domestic missile system"R-36M2" is one of the best. It houses the heavy-duty 15A18M combat weapon, which is capable of carrying up to 36 individual precision-guided nuclear projectiles.

The ballistic flight path of such a weapon is almost impossible to predict; accordingly, neutralizing a missile also poses difficulties. The combat power of the projectile is 20 Mt. If this ammunition explodes at a low altitude, the communication, control, and missile defense systems will fail.

Modifications of the above missile launcher can also be used for peaceful purposes.

Among solid fuel missiles, the RT-23 UTTH is considered especially powerful. Such a device is based autonomously (mobile). In the stationary prototype station (“15Zh60”), the starting thrust is 0.3 higher compared to the mobile version.

Missile launches carried out directly from stations are difficult to neutralize, because the number of projectiles can reach 92 units.

Missile systems and installations of the foreign defense industry

The height of the ballistic trajectory of the American Minuteman-3 missile is not very different from the flight characteristics of domestic inventions.

The complex, which was developed in the USA, is the only “defender” North America among weapons of this type up to today. Despite the age of the invention, the stability indicators of the gun are quite good in present time, because the complex’s missiles could withstand missile defense, as well as hit a target with high level protection. The active part of the flight is short and lasts 160 seconds.

Another American invention is the Peakkeeper. It could also ensure an accurate hit on the target thanks to the most favorable trajectory of ballistic movement. Experts say that combat capabilities the given complex is almost 8 times higher than that of the Minuteman. The Peacekeeper's combat duty was 30 seconds.

Projectile flight and movement in the atmosphere

From the dynamics section we know the influence of air density on the speed of movement of any body in various layers of the atmosphere. The function of the last parameter takes into account the dependence of density directly on flight altitude and is expressed as a function of:

N (y) = 20000-y/20000+y;

where y is the height of the projectile (m).

The parameters and trajectory of an intercontinental ballistic missile can be calculated using special computer programs. The latter will provide statements, as well as data on flight altitude, speed and acceleration, and the duration of each stage.

The experimental part confirms the calculated characteristics and proves that the speed is influenced by the shape of the projectile (the better the streamlining, the higher the speed).

Guided weapons of mass destruction of the last century

All weapons of this type can be divided into two groups: ground and airborne. Ground-based devices are those that are launched from stationary stations (for example, mines). Aviation, accordingly, is launched from a carrier ship (aircraft).

The ground-based group includes ballistic, cruise and anti-aircraft missiles. Aviation - projectile aircraft, ADB and guided air combat missiles.

The main characteristic of calculating the ballistic trajectory is the altitude (several thousand kilometers above the atmospheric layer). At a given level above the ground, projectiles reach high speeds and create enormous difficulties for their detection and neutralization of missile defense.

Well-known ballistic missiles that are designed for average range flights are: “Titan”, “Thor”, “Jupiter”, “Atlas”, etc.

The ballistic trajectory of a missile, which is launched from a point and hits specified coordinates, has the shape of an ellipse. The size and length of the arc depends on the initial parameters: speed, launch angle, mass. If the projectile speed is equal to the first cosmic speed (8 km/s), a military weapon, which is launched parallel to the horizon, will turn into a satellite of the planet with a circular orbit.

Despite constant improvements in the field of defense, the flight path of a military projectile remains virtually unchanged. At the moment, technology is not able to violate the laws of physics that all bodies obey. A small exception are homing missiles - they can change direction depending on the movement of the target.

Inventors anti-missile systems they are also modernizing and developing a weapon to destroy funds mass destruction new generation.