Ballistics and ballistic movement

Prepared by a student of the 9th "m" class Petr Zaitsev.

Ι Introduction:

1) Goals and objectives of the work:

“I chose this topic because it was recommended to me by the class teacher-teacher of physics in my class, and I also really liked this topic myself. In this work, I want to learn a lot about ballistics and the ballistic motion of bodies.”

ΙΙ Main material:

1) Fundamentals of ballistics and ballistic movement.

a) the history of the emergence of ballistics:

In numerous wars throughout the history of mankind, the warring parties, proving their superiority, first used stones, spears, and arrows, and then cannonballs, bullets, shells, and bombs.

The success of the battle was largely determined by the accuracy of hitting the target.

At the same time, an accurate throw of a stone, hitting the enemy with a flying spear or arrow was recorded by the warrior visually. This allowed, with appropriate training, to repeat their success in the next battle.

The speed and range of projectiles and bullets, which significantly increased with the development of technology, made remote battles possible. However, the skill of a warrior, the resolving power of his eye, was not enough to accurately hit the target of an artillery duel first.

The desire to win stimulated the emergence of ballistics (from the Greek word ballo - I throw).

b) basic terms:

The emergence of ballistics dates back to the 16th century.

Ballistics is the science of the movement of projectiles, mines, bullets, unguided rockets during firing (launch). Main sections of ballistics: internal ballistics and external ballistics. The study of real processes occurring during the combustion of gunpowder, the movement of shells, rockets (or their models), etc., is the subject of the ballistics experiment. External ballistics studies the movement of projectiles, mines, bullets, unguided rockets, etc. after the termination of their force interaction with the weapon barrel (launcher), as well as factors affecting this movement. The main sections of external ballistics are: the study of forces and moments acting on a projectile in flight; study of the movement of the center of mass of the projectile to calculate the elements of the trajectory, as well as the movement of the projectile relates. The center of mass in order to determine its stability and dispersion characteristics. Sections of external ballistics are also the theory of corrections, the development of methods for obtaining data for compiling firing tables and external ballistic design. The movement of projectiles in special cases is studied by special sections of external ballistics, aviation ballistics, underwater ballistics, etc.

Internal ballistics studies the movement of projectiles, mines, bullets, etc. in the bore of a weapon under the action of powder gases, as well as other processes that occur when a shot is fired in the channel or chamber of a powder rocket. Main sections internal ballistics: pyrostatics, which studies the patterns of combustion of gunpowder and gas formation in a constant volume; pyrodynamics, which investigates the processes in the bore during firing and establishes a connection between them, the design characteristics of the bore and loading conditions; ballistic design of guns, missiles, small arms. Ballistics (studies the processes of the period of consequences) and internal ballistics of powder rockets (explores the patterns of fuel combustion in the chamber and the outflow of gases through nozzles, as well as the occurrence of forces and actions on unguided rockets).

Ballistic weapon flexibility - property firearms allowing it to be expanded combat capabilities increase the effectiveness of the action by changing the ballistic. characteristics. Achieved by changing the ballistic. coefficient (for example, by introducing brake rings) and the initial velocity of the projectile (using variable charges). In combination with a change in the elevation angle, this allows you to get large angles of incidence and less dispersion of projectiles at intermediate ranges.

A ballistic missile is a missile which, except for a relatively small area, follows the trajectory of a freely thrown body. Unlike cruise missile a ballistic missile does not have bearing surfaces to create lift when flying in the atmosphere. The aerodynamic stability of the flight of some ballistic missiles is provided by stabilizers. Ballistic missiles include missiles for various purposes, launch vehicles for spacecraft, etc. They are single- and multi-stage, guided and unguided. The first combat ballistic missiles FAU 2- were used by Nazi Germany at the end of the World War. Ballistic missiles with a flight range of over 5500 km (according to foreign classification - over 6500 km) are called intercontinental ballistic missiles. (MBR). Modern ICBMs have a flight range of up to 11,500 km (for example, the American Minuteman is 11,500 km, Titan-2 is about 11,000 km, Trider-1 is about 7,400 km). They are launched from ground (mine) launchers or submarines. (from surface or underwater position). ICBMs are carried out as multi-stage, with liquid or solid propellant propulsion systems, can be equipped with monoblock or multiply charged nuclear warheads.

Ballistic track, spec. equipped on art. polygon area for experiment, study of movement art. shells, mini etc. Appropriate ballistic devices and ballistic equipment are installed on the ballistic track. targets, with the help of which, on the basis of experimental firing, the function (law) of air resistance, aerodynamic characteristics, translational and oscillatory parameters are determined. movement, initial departure conditions and projectile dispersion characteristics.

Ballistic shooting conditions, a set of ballistic. characteristics that provide greatest influence on the flight of a projectile (bullet). Normal, or tabular, ballistic firing conditions are conditions under which the mass and initial velocity of the projectile (bullet) are equal to the calculated (table), the temperature of the charges is 15 ° C, and the shape of the projectile (bullet) corresponds to the established drawing.

Ballistic characteristics, basic data that determine the patterns of development of the firing process and the movement of a projectile (mines, grenades, bullets) in the bore (intra-ballistic) or on a trajectory (external ballistic). The main intra-ballistic characteristics: the caliber of the weapon, the volume of the charging chamber, the density of loading, the length of the path of the projectile in the bore, the relative mass of the charge (its ratio to the mass of the projectile), the strength of gunpowder, max. pressure, forcing pressure, propellant combustion progressiveness characteristics, etc. The main external ballistic characteristics include: initial speed, ballistic coefficient, throw and departure angles, median deviations, etc.

Ballistic computer, electronic device for firing (usually direct fire) from tanks, infantry fighting vehicles, small-caliber anti-aircraft guns etc. The ballistic calculator takes into account information about the coordinates and speed of the target and its object, wind, temperature and air pressure, initial velocity and angles of the projectile, etc.

Ballistic descent, uncontrolled movement of the descent spacecraft (capsule) from the moment of leaving the orbit until reaching the planet specified relative to the surface.

Ballistic similarity, a property of artillery pieces, which consists in the similarity of dependencies characterizing the process of burning a powder charge when fired in the bores of various artillery systems. The conditions of ballistic similarity are studied by the theory of similarity, which is based on the equations of internal ballistics. Based on this theory, ballistic tables are compiled that are used in ballistic. design.

Ballistic coefficient (C), one of the main externally ballistic performance projectile (rocket), reflecting the influence of its shape factor (i), caliber (d), and mass (q) on the ability to overcome air resistance in flight. It is determined by the formula C \u003d (id / q) 1000, where d is in m, and q is in kg. The less ballistic coefficient, the easier the projectile overcomes air resistance.

Ballistic camera, a special device for photographing the phenomenon of a shot and its accompanying processes inside the bore and on the trajectory in order to determine the qualitative and quantitative ballistic characteristics of the weapon. Allows to carry out instant one-time photographing to.-l. phases of the process under study or successive high-speed photography (more than 10 thousand frames) of various phases. According to the method of obtaining exposure B.F. there are spark, with gas-light lamps, with electro-optical shutters and radiographic pulsed ones.

c) speed during ballistic motion.

To calculate the velocity v of the projectile at an arbitrary point of the trajectory, as well as to determine the angle , which forms the velocity vector with the horizontal,

it is enough to know the velocity projections on the X and Y axes (Fig. 1).

If v and v are known, the Pythagorean theorem can be used to find the speed:

The ratio of the leg v opposite the corner to the leg v belonging to

to this corner, determines tg and, accordingly, the angle :

With uniform movement along the X axis, the projection of the speed of movement v remains unchanged and equal to the projection of the initial speed v:

Dependence v(t) is determined by the formula:

into which should be substituted:

Graphs of velocity projections versus time are shown in Fig. 2.

At any point of the trajectory, the projection of the velocity on the X axis remains constant. As the projectile rises, the velocity projection on the Y-axis decreases linearly. At t \u003d 0, it is equal to \u003d sin a. Find the time interval after which the projection of this velocity becomes equal to zero:

0 = vsing- gt , t =

The result obtained coincides with the time of lifting the projectile by maximum height. At the top of the trajectory, the vertical velocity component is equal to zero.

Therefore, the body no longer rises. For t > velocity projection

v becomes negative. This means that this velocity component is directed opposite to the Y axis, i.e. the body begins to fall down (Fig. No. 3).

Since at the top of the trajectory v = 0, the speed of the projectile is:

d) the trajectory of the body in the field of gravity.

Let's consider the main parameters of the trajectory of a projectile flying with an initial speed v from a gun directed at an angle α to the horizon (Fig. 4).

The movement of the projectile occurs in the vertical XY plane containing v.

We choose the origin at the point of departure of the projectile.

In the Euclidean physical space, the movement of the body along the coordinate

the x and y axes can be considered independently.

The gravitational acceleration g is directed vertically downward, so the movement along the X axis will be uniform.

This means that the projection of the velocity v remains constant, equal to its value at the initial time v.

The law of uniform projectile motion along the X axis is: x= x+ vt. (5)

Along the Y axis, the movement is uniform, since the gravitational acceleration vector g is constant.

The law of uniformly variable projectile motion along the Y axis can be represented as follows: y = y+vt + . (6)

The curvilinear ballistic motion of a body can be considered as the result of the addition of two rectilinear motions: uniform motion

along the X axis and equally variable movement along the Y axis.

In the selected coordinate system:

v=vcosα. v=vsinα.

The gravitational acceleration is directed opposite to the Y axis, so

Substituting x, y, v, v, av (5) and (6), we obtain the ballistic law

motion in coordinate form, in the form of a system of two equations:

The projectile trajectory equation, or y(x) dependency, can be obtained by

excluding time from the equations of the system. To do this, from the first equation of the system we find:

Substituting it into the second equation we get:

Reducing v in the first term and taking into account that = tg α, we obtain

projectile trajectory equation: y = x tg α - .(8)

e) Trajectory of ballistic movement.

Let us construct a ballistic trajectory (8).

schedule quadratic function is known to be a parabola. In the case under consideration, the parabola passes through the origin,

since it follows from (8) that y \u003d 0 for x \u003d 0. The branches of the parabola are directed downwards, since the coefficient (-) at x is less than zero. (Fig No. 5).

Let us determine the main parameters of ballistic motion: the time of ascent to the maximum height, the maximum height, the time and range of the flight. Due to the independence of movements along the coordinate axes, the vertical rise of the projectile is determined only by the projection of the initial velocity onto the Y axis.

The maximum lift height can be calculated using the formula

if substituted instead of :

Figure 5 compares vertical and curvilinear motion with the same initial velocity along the Y axis. At any moment in time, a body thrown vertically upwards and a body thrown at an angle to the horizon with the same vertical velocity projection move synchronously along the Y axis.

Since the parabola is symmetrical with respect to the top, the flight time of the projectile is 2 times longer than the time it takes to rise to the maximum height:

Substituting the flight time into the law of motion along the X axis, we obtain the maximum flight range:

Since 2 sin cos, a \u003d sin 2, then

e) the application of ballistic movement in practice.

Imagine that several shells were fired from one point, at different angles. For example, the first projectile at an angle of 30°, the second at an angle of 40°, the third at an angle of 60°, and the fourth at an angle of 75° (Fig. 6).

Figure #6 in green shows a graph of a projectile fired at 30°, white at 45°, purple at 60°, and red at 75°. And now let's look at the graphs of the flight of shells and compare them. (The initial speed is the same, and is equal to 20 km / h)

Comparing these graphs, one can deduce a certain pattern: with an increase in the angle of departure of the projectile, at the same initial speed, the flight range decreases, and the height increases.

2) Now consider another case associated with a different initial speed, with the same departure angle. In Figure 7, green color shows a graph of a projectile fired at an initial speed of 18 km/h, white at a speed of 20 km/h, purple at a speed of 22 km/h, and red at a speed of 25 km/h. And now let's look at the graphs of the flight of shells and compare them (the flight angle is the same and equal to 30°). Comparing these graphs, one can deduce a certain pattern: with an increase in the initial velocity of the projectile, at the same angle of departure, the range and height of the projectile increase.

Conclusion: with an increase in the angle of departure of the projectile, at the same initial speed, the flight range decreases, and the height increases, and with an increase in the initial velocity of the departure of the projectile, at the same angle of departure, the range and height of the projectile increase.

2) Application of theoretical calculations to the control of ballistic missiles.

a) trajectory ballistic missile.

The most significant feature that distinguishes ballistic missiles from missiles of other classes is the nature of their trajectory. The trajectory of a ballistic missile consists of two sections - active and passive. On the active site, the rocket moves with acceleration under the action of the thrust force of the engines.

In this case, the rocket stores kinetic energy. At the end of the active part of the trajectory, when the rocket acquires a speed having a given value

and direction, the propulsion system is switched off. Thereafter head part the rocket is separated from its body and flies further due to the stored kinetic energy. The second section of the trajectory (after turning off the engine) is called the section of the free flight of the rocket, or the passive section of the trajectory. Below, for brevity, we will usually talk about the free-flight trajectory of a rocket, implying the trajectory of not the entire rocket, but only its head.

Ballistic missiles are launched from launchers vertically upwards. Vertical launch allows you to build the most simple launchers and provides favorable conditions for controlling the rocket immediately after launch. In addition, vertical launch makes it possible to reduce the requirements for the rigidity of the rocket body and, consequently, reduce the weight of its structure.

The missile is controlled in such a way that a few seconds after the launch, while continuing to rise, it begins to gradually tilt towards the target, describing an arc in space. The angle between the longitudinal axis of the rocket and the horizon (pitch angle) changes in this case by 90º to the calculated final value. The required law of change (program) of the pitch angle is set by a software mechanism included in the on-board equipment of the rocket. At the final segment of the active section of the trajectory, the pitch angle is maintained, constant and the rocket flies straight, and when the speed reaches the calculated value, the propulsion system is turned off. In addition to the speed value, on the final segment of the active section of the trajectory, the trajectory is set with a high degree accuracy as well as the given direction of the rocket's flight (the direction of its velocity vector). The speed of movement at the end of the active part of the trajectory reaches significant values, but the rocket picks up this speed gradually. While the rocket is in the dense layers of the atmosphere, its speed is low, which reduces the energy loss to overcome the resistance of the environment.

The moment of turning off the propulsion system divides the trajectory of the ballistic missile into active and passive sections. Therefore, the point of the trajectory at which the engines are turned off is called the boundary point. At this point, the control of the missile usually ends and it makes the entire further path to the target in free motion. The flight range of ballistic missiles along the Earth's surface, corresponding to the active part of the trajectory, is equal to no more than 4-10% of the total range. The main part of the trajectory of ballistic missiles is the free flight section.

To significantly increase the range, it is necessary to use multi-stage missiles.

Multi-stage rockets consist of separate blocks-stages, each of which has its own engines. The rocket is launched with a working propulsion system of the first stage. When the first stage fuel is used up, the second stage engine is fired and the first stage is reset. After the first stage is dropped, the thrust force of the engine must impart acceleration to a smaller mass, which leads to a significant increase in velocity v at the end of the active part of the trajectory compared to a single-stage rocket having the same initial mass.

Calculations show that already with two stages it is possible to obtain an initial speed sufficient for the flight of the head of the rocket over intercontinental distances.

The idea of ​​using multi-stage rockets to obtain high initial speeds and, consequently, long flight ranges was put forward by K.E. Tsiolkovsky. This idea is used in the creation of intercontinental ballistic missiles and launch vehicles for launching space objects.

b) the trajectory of guided projectiles.

The trajectory of a rocket is a line that its center of gravity describes in space. A guided projectile is an unmanned aerial vehicle that has controls that can be used to influence the movement of the vehicle along the entire trajectory or in one of the flight sections. Projectile control on the trajectory was required in order to hit the target, while remaining at a safe distance from it. There are two main classes of targets: moving and stationary. In turn, a rocket projectile can be launched from a stationary launch device or from a mobile one (for example, from an airplane). With stationary targets and launching devices, the data needed to hit the target is obtained from the known relative location of the launch site and the target. In this case, the trajectory of the projectile can be calculated in advance, and the projectile is equipped with devices that ensure its movement according to a certain calculated program.

In other cases, the relative location of the launch site and the target is constantly changing. To hit the target in these cases, it is necessary to have devices that track the target and continuously determine the relative position of the projectile and the target. The information received from these devices is used to control the movement of the projectile. Control must ensure the movement of the missile to the target along the most advantageous trajectory.

In order to fully characterize the flight of a rocket, it is not enough to know only such elements of its movement as the trajectory, range, altitude, flight speed, and other quantities that characterize the movement of the center of gravity of the rocket. The rocket can occupy various positions in space relative to its center of gravity.

A rocket is a body of significant size, consisting of many components and parts, made with a certain degree of accuracy. In the process of movement, it experiences various perturbations associated with the restless state of the atmosphere, the inaccuracy of work power plant, various kinds of interference, etc. The combination of these errors, not provided for by the calculation, leads to the fact that the actual movement is very different from the ideal one. Therefore, in order to effectively control a rocket, it is necessary to eliminate the undesirable influence of random disturbing influences, or, as they say, to ensure the stability of the rocket's movement.

c) coordinates that determine the position of the rocket in space.

The study of the various and complex movements made by a rocket can be greatly simplified if the rocket's movement is represented as the sum of the translational movement of its center of gravity and rotational movement about the center of gravity. The examples given above clearly show that in order to ensure the stability of the rocket's movement, it is extremely important to have its stability relative to the center of gravity, i.e., the angular stabilization of the rocket. The rotation of the rocket about the center of gravity can be represented as the sum rotational movements about three perpendicular axes having a certain orientation in space. Fig. No. 7 shows an ideal feathered rocket flying along a calculated trajectory. The origin of the coordinate systems, relative to which we will stabilize the rocket, will be placed at the center of gravity of the rocket. Let's direct the X-axis tangentially to the trajectory in the direction of the rocket's movement. The Y axis will be drawn in the plane of the trajectory perpendicular to the X axis, and the axis

Z - perpendicular to the first two axes, as shown in Fig. No. 8.

Associate with the rocket a rectangular coordinate system XYZ, similar to the first one, and the X axis must coincide with the axis of symmetry of the rocket. In a perfectly stabilized rocket, the X, Y, Z axes coincide with the X, Y, Z axes, as shown in Fig. 8

Under the action of perturbations, the rocket can rotate around each of the oriented axes X, Y, Z. The rotation of the rocket around the X axis is called the roll of the rocket. The bank angle lies in the YOZ plane. It can be determined by measuring in this plane the angle between the Z and Z or Y and Y axes. Rotation around the axis

Y is the yaw of the rocket. The yaw angle is in the XOZ plane as the angle between the X and X or Z and Z axes. The angle of rotation around the Z-axis is called the pitch angle. It is determined by the angle between the X and X or Y and Y axes that lie in the path plane.

Automatic rocket stabilization devices should give it such a position when = 0 or . To do this, the rocket must have sensitive devices capable of changing its angular position.

The trajectory of the rocket in space is determined by the current coordinates

X, Y, Z of its center of gravity. The starting point of the rocket is taken as the starting point. For missiles long range the X-axis is taken as a straight line tangent to the arc of the great circle connecting the start with the target. In this case, the Y axis is directed upward, and the Z axis is perpendicular to the first two axes. This coordinate system is called terrestrial (Fig. 9).

The calculated trajectory of ballistic missiles lies in the XOY plane, called the firing plane, and is determined by two coordinates X and Y.

Conclusion:

“In this work, I learned a lot about ballistics, the ballistic movement of bodies, about the flight of missiles, finding their coordinates in space.”

Bibliography

Kasyanov V.A. - Physics grade 10; Petrov V.P. - Missile control; Zhakov A.M. -

Control of ballistic missiles and space objects; Umansky S.P. - Cosmonautics today and tomorrow; Ogarkov N.V. - Military Encyclopedic Dictionary.

Gorbaneva Larisa Valerievna

ballistic movement

Ballistic motion is the movement of a body in space under the action of external forces.

Consider the motion of bodies under the action of gravity. The simplest case of motion of bodies under the action of gravity is free fall with an initial velocity equal to zero. In this case, the body moves in a straight line with free fall acceleration towards the center of the Earth. If the initial velocity of the body is non-zero and the initial velocity vector is not directed along the vertical, then the body under the action of gravity moves with free fall acceleration along a curvilinear trajectory (parabola).

Let the body be thrown at an angle a to the horizon with an initial speed V 0 .

We investigate this movement, that is, we determine the trajectory of movement, flight time, flight range, the maximum height to which the body will rise, and the speed of the body.

Let us write the equations of motion for the coordinates x, y body at any moment of time and for the projections of its velocity on the axis X and Y:

,

,

Let's choose a coordinate system as shown in the figure. Wherein , .

Only the force of gravity acts on the body, which means that it moves with acceleration only along the Y axis ( .

A body moves uniformly along the X-axis (with a constant speed .

Projections of the initial velocity on the axis X and Y:

, .

Then the equations of motion of the body will take the form:

,

Velocity projections on the X and Y axes at any time:

,

To find the trajectory of motion, it is necessary to find the analytical equation of the curve along which the body moves in space. To do this, you need to solve the system of equations:

Express from the second equation and substitute into the first equation. As a result, we get: . This second-order equation describes a parabola whose branches are directed downwards, the center of the parabola is displaced from the origin.

To determine the flight time of the body, we use the equation to determine y: . According to the coordinate system we have chosen, y=0 corresponds to the beginning and end of the body's movement. Then you can write: or .

This equation has two roots: . Indeed, as defined earlier, on the ground the body will be twice at the beginning and at the end of the path. Then the flight time determines the second root: .

Knowing the flight time, it is easy to determine the flight range, that is, the maximum coordinate x max:

The maximum coordinate y max determines the maximum height of the body. In order to find it, it is necessary to substitute the rise time t under into the equation, which is determined from the condition that at the highest point of the rise it is equal to 0:

Then .

In this way, .

P velocity projection on the X-axis: - remains unchanged, and the velocity projection on the Y-axis changes as follows: . To determine the speed at any height h, you need to know the time when the body will be at this height h - t h. This time can be found from the equation

Time has two meanings, since at a height h the body will be twice, the first time moving up, the second time moving down. Therefore, the speed of the body at height h is determined by the formulas:

At the first point .

At the second point

The modulus of speed at any height is determined by the formula

You can find the tangent of the slope of the velocity to the x-axis:

Most ballistic motion problems are a special case or variation of this general problem.

Example 1. At what angle to the horizon should a body be thrown so that the height of its rise is equal to the flight range?

The height of the rise of the body is determined by the formula, flight range.

According to the task H max =S, that's why

Solving this equation, we get tgα=4.

Example 2. A body is thrown at an angle α=π/6 rad to the horizon from a position with a coordinate y 0 =5m above the Earth's surface. The initial speed of the body is 10m/s. Determine the coordinate y max of the highest point of ascent of the body above the Earth's surface, the coordinate x p of the point of fall of the body on the surface of the Earth and the speed V p at this point.

R
Solution:

Selecting a coordinate system as shown in the figure.

The coordinate of the highest point of the body trajectory in the selected coordinate system is determined by the formula: or .

=6.3m

To determine the coordinates of the point of fall x p it is necessary to find the time of movement of the body to the point of landing. Time t p is determined from the condition y p =0: .

Solving this equation we get: .

Substituting the value of the quantities, we get:

\u003d 1.6 s.

The second root has no physical meaning.

Then substituting the value of t p in the formula

Let's find .

final speed of the body

Angle between OX axis and vector V P

Example 3. An artillery gun is located on a mountain with a height h. The projectile flies out of the barrel with a speed V 0 directed at an angle α to the horizon. Neglecting air resistance, determine: a) the range of the projectile in the horizontal direction, b) the speed of the projectile at the moment of fall, c) the angle of incidence, d) the initial firing angle at which the flight range is greatest.

R solution. To solve the problem, we will make a drawing, while choosing the coordinate system so that its origin coincides with the throwing point, and the axes are directed along the surface of the Earth and normal to it towards the initial displacement of the projectile.

Let's write the equations of motion and velocity of the projectile in projections on the X and Y axes:

At the time t 1, when the projectile hits the ground, its coordinates are: x=S, y= – h.

The resulting speed at the time of the fall is: .

To determine the speed of a projectile at the moment of impact V and flight range S find the time from the equation given y=-h.

By solving this equation: .

Substituting the expression for t 1 into formulas for determining the coordinates x taking into account x=S, respectively, we get:

.

To find V Need to know V x and V y .

As previously defined.

For determining V y substitute the value into the formula t 1 and we get: .

From the results obtained, the following conclusions can be drawn.

If h=0, i.e. the projectiles fall at the departure level, and, after transforming the formula , we obtain the flight range .

If, in this case, the throwing angle is 45° (sin 2α=1), then at a given initial speed V 0 maximum flight range: .

Substituting the value h=0 into the expression for determining the speed, we get that the speed of the projectile at the moment of its approach to the level from which the shot was fired is equal to its initial speed: V=V 0 .

In the absence of air resistance, the speed of falling bodies is equal in modulus to their initial throwing speed, regardless of the angle at which the body was thrown, as long as the points of throwing and falling are on the same level. Given that the projection of the velocity on the horizontal axis does not change over time, it is easy to establish that at the moment of falling the body's velocity forms the same angle with the horizon as at the moment of throwing.
Substituting the expression for S=S max into the formula for determining the throwing angle, we obtain for the angle α, at which the flight range is greatest: .

Tasks for independent solution.

F.9.1. A body is thrown horizontally with a speed of 20m/s. Determine the displacement of the body from the point of throw, ΔS, at which the speed will be directed at an angle of 45° to the horizon.

F.9.2. At what angle α should the body be thrown so that the flight range is the greatest?

F.9.3. An airplane flies horizontally at a speed of 360 km/h at an altitude of 490 m. When it flies over point A, a packet is dropped from it. At what distance from point A will the packet hit the ground?

F.9.4. A body falls freely from a height of 4m. At a height of 2 m, it elastically hits a small fixed area at an angle of 30° to the horizon. Find the total time of motion of the body and the range of its flight.

F .9.5. It is necessary to hit the target with a stone from the ground from a distance S. The target is located at a height h. At what minimum initial speed of the stone can this be done?

F.9.6. From a point with coordinates x 0 , y 0 a body is thrown at an angle α 0 to the horizon with an initial velocity V 0 (see picture). Find: the position and velocity of the body after time t, the equation of the flight path of the body, the total flight time, the maximum height of the ascent, the angle at which the body must be thrown so that its height is equal to the flight range (provided that x 0 =y 0 =0 ).

F.9.7. From a tower 20 m high, a shot was fired from a pistol at an angle of 30 ° to the horizon. Determine the take-off speed, the height of the rise and the range of the bullet if, when falling, it traveled the last 20 m of the path (tower height) in 0.5 s. Ignore air resistance.

F
.9.8. A stone is thrown on a mountainside at an angle α to its surface (see Fig.). Determine the flight range of the stone and its maximum height above the slope, if the initial speed of the stone is V 0, the angle of the mountain to the horizon β. Air resistance is ignored.

F.9.9. A body is thrown horizontally from a table. When falling to the floor, its speed is 7.8 m/s. Table height H=1.5m. What is the initial speed of the body?

F.9.10. A stone is thrown at an angle α 0 =30° to the horizon with a speed V 0 =10m/s. How long will it take for the stone to reach a height of 1m?

F.9.11. Two bodies are thrown at angles α 1 and α 2 to the horizon from one point. What is the ratio of the velocities reported by him if they fell to the ground in the same place?

F.9.12. A body is thrown horizontally with a speed of 20m/s. Determine the displacement of the body from the point of throw at which the speed will be directed at an angle of 45 ° to the horizon.

Theory

If a body is thrown at an angle to the horizon, then in flight it is affected by gravity and air resistance. If the resistance force is neglected, then the only force left is the force of gravity. Therefore, due to Newton's 2nd law, the body moves with an acceleration equal to the free fall acceleration; acceleration projections on the coordinate axes are a x = 0, and at= -g.

Any complex movement of a material point can be represented as an imposition of independent movements along the coordinate axes, and in the direction of different axes, the type of movement may differ. In our case, the motion of a flying body can be represented as a superposition of two independent motions: uniform motion along the horizontal axis (X-axis) and uniformly accelerated motion along the vertical axis (Y-axis) (Fig. 1).

The velocity projections of the body therefore change with time as follows:

,

where is the initial speed, α is the throwing angle.

The body coordinates therefore change like this:

With our choice of the origin of coordinates, the initial coordinates (Fig. 1) Then

The second value of the time at which the height is equal to zero is equal to zero, which corresponds to the moment of throwing, i.e. this value also has a physical meaning.

The flight range is obtained from the first formula (1). Flight range is the value of the coordinate X at the end of the flight, i.e. at a point in time equal to t0. Substituting the value (2) into the first formula (1), we obtain:

.

(3)

From this formula it can be seen that the greatest flight range is achieved at a throw angle of 45 degrees.

The highest lifting height of the thrown body can be obtained from the second formula (1). To do this, you need to substitute in this formula the value of time equal to half the flight time (2), because it is at the midpoint of the trajectory that the flight altitude is maximum. Carrying out calculations, we get

Development of the lesson "Ballistic movement"

Lesson type: learning new material.

Lesson objectives:

Educational:

By the end of the lesson, students should:

• the concept of ballistic motion;
• features of ballistic movement;
• · schedule of ballistic movement;
• the law of ballistic motion

• · describe and explain observations and fundamental experiments that have had a significant impact on the development of physics;
• · to illustrate the role of physics in the creation of the most important technical objects.

Developing:

• promote the development of speech;
• intellectual and creativity in the process of acquiring knowledge and skills in physics using modern information technologies.

Educational:

• contribute to the formation of:
• cognitive interest in the subject;
• students' outlook.

Technical equipment of the lesson:

• · Computer class;
• · Multimedia projector, screen;

Software:

· Educational electronic publication “Open Physics. Version 2.6." Part 1 - mechanics section.

Laboratory work "The movement of a body thrown at an angle to the horizon."

Setting the mood of the students

Word of the teacher: In numerous wars throughout the history of mankind, the warring parties, proving their superiority, first used stones, spears and arrows, and then nuclei, shells

The success of the battle was largely determined by the accuracy of hitting the target. At the same time, the exact throw of a stone, the defeat of the enemy by a flying spear or arrow was recorded by the warrior visually. This allowed (with appropriate training) to repeat their success in the next battle.

Significantly increased with the development of technology, the speed and, accordingly, the range of projectiles and bullets made remote battles possible. However, the resolving power of the eye was not enough to accurately hit the target.

Until the 16th century, artillerymen used tables in which, based on practical observations, angles, wind, and flight range were indicated, but the hit accuracy was very low. The problem of scientific prediction arose - how to achieve high accuracy of a projectile hit.

For the first time, the great astronomer and physicist Galileo Galilei managed to solve this problem, whose research stimulated the emergence of ballistics (from the Greek word ballo - I throw). Ballistics is a branch of mechanics that studies the motion of bodies in the Earth's gravity field.

Learning new material

So, as you probably already guessed, the topic of our lesson is “Ballistic movement”, the goal is to study ballistic movement by experimentally exploring its features.

The merit of Galileo Galilei was that he was the first to propose to consider ballistic motion as a sum of simple ones, in particular, he proposed to represent this motion as the result of the addition of two rectilinear motions: uniform motion along the Ox axis and equally variable motion along the Oy axis.

To describe the ballistic motion as a first approximation, it is most convenient to introduce an idealized computer model, in which this case model "Movement of a body thrown at an angle to the horizon" on a computer.

Under the conditions of this model, the body will be considered as material point, moving with constant acceleration of free fall, while neglecting the change in the height of the body, air resistance, the curvature of the Earth's surface, its rotation around its own axis.

This approximation greatly facilitates the calculation of the trajectory of bodies. However, such consideration has certain limits of applicability. For example, when flying an intercontinental ballistic missile, one cannot neglect the curvature of the Earth's surface. In free-falling bodies, air resistance cannot be ignored. But in order to achieve the goal in the conditions of this model, we can neglect the above values.

So let's take a closer look at the model. What parameters can we change?

Students answer: The model allows you to change:

• Firstly, the initial speed;
• secondly, the initial height;
• Thirdly, the angle of the direction of movement of the body.

Teacher's word: Right. With the help of this model, we will try to solve experimentally the first problem that Galileo Galilei set himself, i.e., we will try to find out what is the shape of the ballistic motion trajectory. To do this, we set the initial values ​​of the model parameters: speed equal to 25 m/s; an angle equal to 300. Let's choose the point of departure of the projectile at the origin, for this we set the height value equal to zero. Now let's see an experiment. What is a ballistic motion trajectory?

Students answer: The trajectory of the ballistic movement is a parabola.

Teacher's word: right! But can we definitively conclude that the shape of the ballistic trajectory is a parabola?

Student answer: No. It is necessary to check the correctness of the hypothesis expressed by Galileo by performing several experiments, each time changing the parameters of the model.

Teacher's word: Good! Let's change the angle of the projectile's direction first. To do this, we change given parameter on the model, i.e. instead of 300, set 200. And leave the rest of the values ​​unchanged. Let's consider an experiment. Has the shape of the ballistic motion trajectory changed?

Student answer: No, the shape of the trajectory has remained the same.

Teacher's word: Now let's try to increase the angle value to 400, leaving the rest of the parameters. Let's see what happens to the shape of the trajectory?

(Sets up an experiment.)

Student response: The shape of the trajectory remains the same.

Teacher's word: Let's see if its shape changes if we decrease or increase other parameters of the model. For example, let's increase the speed of the projectile to 40 m/s, leaving the angle and height the same, and observe the movement of the projectile. Has the ballistic motion trajectory changed?

Student answer: No. The shape of the trajectory does not change.

Teacher's word: And now we will reduce the value of the speed of movement to 15 m / s, leaving the value of the angle and height the same. Let's see if the shape of the trajectory changes?

Student response: The shape of the trajectory does not change.

Teacher's word: Do you think the shape of the trajectory will change if we decrease or increase the height of the body?

Student response: Probably, the shape of the trajectory will remain the same.

Teacher's word: Let's check it with the help of a computer experiment. To do this, we will change the value of the projectile lift height to 15m. Let's carefully follow the trajectory of the projectile. What is its form?

Students answer: The shape of the trajectory is still a parabola.

Teacher's word: So, based on all the experiments done, can we make a final conclusion about the change in the shape of the ballistic motion trajectory?

Students' answer: By changing all the parameters, we proved experimentally that for any values ​​​​of the angle, height, speed of the projectile, the shape of the trajectory remains unchanged.

Word of the teacher: Thus, we have solved the first task. The hypothesis of Galileo Galilei turned out to be correct - the shape of the ballistic motion trajectory is a parabola. But Galileo also proposed to consider ballistic motion as the result of the addition of two rectilinear motions: uniform along the Ox axis and equally variable along the y axis.

Therefore, our second task will be: to prove experimentally the validity of Galileo's hypothesis, that is, to make sure that the movement along the Ox axis is really uniform. If the movement is uniform, what parameter do you think should remain unchanged?

Students answer: Speed, since uniform movement is movement at a constant speed.

Teacher's word: Right! This means that the velocity projection on the axis Ox Ux will remain unchanged. So, let's study the movement of a projectile fired from the origin (i.e., the height is zero) in the "Strobe" mode available on the model, since it is in this mode that the direction of the velocity vector of the fired projectile and its projection are indicated on the trajectory at regular intervals on the horizontal and vertical axes: Uх, Uу. Set the speed to 25 m/s. What parameters should we change when conducting an experimental proof?

Student response: We must change the angle and height.

Teacher's word: Good! Let's set the angle of the projectile to 450, and the value of the height to zero. Let's observe the projection of the velocity on the axis Ox - Ux. What happens to her while driving?

Student response: It will remain constant.

Teacher's word: That is, the movement along the Ox axis in this case is uniform. Decrease the value of the projectile departure angle to 150. Is now the movement along the Ox axis uniform, provided that the lift height remains the same?

Student response: Yes. The movement along the Ox axis is still uniform.

Teacher's word: Let's increase the height of the body to 20 m, and leave the angle the same. What is the movement of the body along the x-axis?

Students answer: The projectile makes a uniform movement along the Ox axis.

Teacher's word: So, we tried to change all the parameters, but at the same time we set only one speed module, equal to 25 m / s. Let's try to do the above actions by setting a different value of the velocity modulus, for example, equal to 10 m/s (the reasoning is carried out by analogy, as with the value x = 25 m/s).

What conclusion can be drawn about the nature of motion along the Ox axis after observing several experiments, each time changing the values ​​of the model parameters?

Students answer: Experimentally, we proved the correctness of Galileo's hypothesis that the movement of a body along the Ox axis is uniform.

Teacher's word: Right! Thus, we have solved the second cognitive problem. The third task is to prove the validity of the hypothesis put forward by Galileo that the motion along the Oy axis is equally variable. What parameters should we change in this case?

Student response: We will change the angle, height and speed of the projectile.

Teacher's word: Good! Then we set the initial values: the angle is equal to 150, the height is equal to 10 m and the speed is equal to 20 m/s. Let's observe what happens to the value of the velocity and the magnitude of the velocity vector of the projectile? To do this, one of the guys in the class will help me fix the values ​​of the projection of the velocity vector on the Oy - xy axis at regular intervals, for example, every 0.5 seconds.

• (The experiment is carried out, fixing the values ​​​​on the board.) t, s

Teacher's word: Let's compare these values ​​with each other, for this we will find the difference: from U2 we subtract U1, from U3 we subtract the sum of U2 + U1, etc. What do we see by comparing the values ​​of the velocity projection on the Oy axis at regular intervals?

Student response: These values ​​are equal to each other.

Teacher's word: Right. And now look at the experiment again carefully and answer the question: how does the vertical component of the velocity vector xy change up to the point showing the maximum height of the body, and after the body has passed through this point?

Students' answer: At the beginning of the movement to the point hmax, the value of the velocity projection on the Oy - Uy axis decreases to zero, then increases until the body falls to the ground.

Teacher's word: So, we have seen that as a result of ballistic movement, the value of the projection of the velocity vector on the Oy axis changes at regular intervals by the same amount. Thus, we can conclude that the motion of the body along the Oy axis is equally variable. But can we consider the conclusion we have formulated final?

Student answer: No. It is necessary to verify the correctness of the hypothesis expressed by Galileo by carrying out several studies, each time changing the parameters of the model.

Word of the teacher: Let's increase the angle of the projectile to 300, and leave the rest of the parameters the same. Let's see what will happen to the magnitude of the velocity vector?

Students answer: The value of the velocity vector changes for equal periods of time by the same amount.

Teacher's word: What can be said about the movement of the body along the Oy axis? What is it? Let's reduce the angle of the projectile to 100, will the nature of the movement change?

(Similar reasoning and calculations are carried out as above and students are invited to draw a conclusion.)

Student answer: no. The movement along the y-axis is still equally variable.

Word of the teacher: Let's try to change the value of the speed of the projectile, increase it to 30 m / s. Is the motion along the y-axis still uniformly variable?

(Similar reasoning and calculations are carried out as above and students are invited to draw a conclusion.)

Student response: Yes. The nature of the movement does not change.

Teacher's word: And if we change the height of the body, increasing it to 15 m, what will be its movement along the Oy axis now?

(Similar reasoning and calculations are carried out as above and students are invited to draw a conclusion.)

Student response: Movement along the Oy axis remains equally variable.

Teacher's word: Let's set the value of the height of the body to zero. Let's observe how the projectile will move along the Oy axis in this case?

(Similar reasoning and calculations are carried out as above and students are invited to draw a conclusion.)

Student Answer: The projectile will move uniformly.

Teacher's word: By changing all the parameters, have we become convinced of the validity of Galileo Galilei's hypothesis?

Students answer: Yes, we were convinced of the validity of the hypothesis expressed by Galileo and proved experimentally that the movement of the body along the Oy axis, under conditions of ballistic motion, is equally variable.

Teacher's word: The movement of a body thrown at an angle to the horizon is characterized by flight time, flight range and lift height. I suggest you get the formulas for calculating the basic quantities. Explanations for students:

For a kinematic description of the motion of a body, it is convenient to direct one of the axes of the coordinate system (OY axis) vertically upwards, and place the other (OX axis) horizontally. Then the motion of the body along a curvilinear trajectory, as we have already found out, can be represented as the sum of two motions that occur independently of each other - motion with free fall acceleration along the OY axis and uniform rectilinear motion along the OX axis. The figure shows the initial velocity vector of the body and its projection onto the coordinate axes.

Since the free fall acceleration does not change over time, the motion of the body, like any motion with constant acceleration, will be described by the equations:

x = x0 + x0xt + ax t2/2

y = y0 + x0yt + ay t2/2

for movement along the OX axis, we have the following conditions:

x0 = 0, x0x = x0 cos b, ax = 0

for movement along the OY axis

y0 = 0, x0y = x0 sin b, ay = - g

t flight = 2t ascent per max altitude

Next, students work in groups (4 people) to derive formulas for calculating flight time, flight range, and ascent height. The teacher is helpful.) Then the results are checked.

Teacher's word: But I want to remind you that all the results we have obtained are valid only for an idealized model, when air resistance can be neglected. The real movement of bodies in earth's atmosphere occurs along a ballistic trajectory, which differs significantly from a parabolic one due to air resistance. The greater the speed of the body, the greater the force of air resistance and the more significant the difference between the ballistic trajectory and the parabola. When projectiles and bullets move in the air maximum range flight is achieved at an angle of departure of 300 - 400. The discrepancy between the simplest theory of ballistics and experiment does not mean that it is not correct in principle. In a vacuum or on the Moon, where there is little to no atmosphere, this theory gives correct results. When describing the movement of bodies in the atmosphere, taking into account air resistance requires mathematical calculations, which we will not present due to cumbersomeness. We only note that the calculation of the ballistic trajectory of the launch and insertion into the required orbit of Earth satellites and their landing in a given area is carried out with great accuracy by powerful computer stations.

Primary test of mastery of knowledge

Frontal survey

What does ballistics study?

What idealized model is used to describe ballistic motion?

What is the nature of the motion of the body during ballistic horizontal motion?

What is the nature of the motion of the body during ballistic vertical motion?

What is a ballistic trajectory?

Development of practical skills to solve problems

(work in pairs at the computer)

Teacher's word: Guys, I suggest you solve problems, the correctness of which you will check with the help of a virtual experiment.

Group I. An arrow fired from a bow vertically upwards fell to the ground after 6 s. What is the initial boom speed and maximum lift height?

Group II. The boy threw a ball horizontally from a window at a height of 20 m. How long did the ball fly to the ground and with what speed was it thrown if it fell at a distance of 6 m from the base of the house?

Group III. By how many times must the initial velocity of the body thrown up be increased in order for the height of the lift to increase by 4 times?

Group IV. How will the time and distance of a body thrown horizontally from a certain height change if the throwing speed is doubled?

Group V. The goalkeeper, knocking the ball out of the goal (from the ground), informs him of a speed of 20 m/s, directed at an angle of 500 to the horizon. Find the time of flight of the ball, the maximum height of the rise and the horizontal range of the flight.

Group VI. From a balcony located at a height of 20 m, a ball is thrown at an angle of 300 upwards from the horizon with a speed of 10 m/s. Find: a) the coordinate of the ball in 2 s; b) how long will it take for the ball to hit the ground? c) horizontal flight range.

Homework Information

FOR ALL 63 - 70 textbook V.A. Kasyanov "Physics -10" - answer the questions p. 71.

Get the equation of the trajectory y = y (x) of the movement of a body thrown at an angle to the horizon.

OPTIONAL Set at what angle of throw the flight range is maximum.

OR Plot the time dependences of the horizontal xx and vertical xy projections of the velocity of a body thrown at an angle to the horizon.

Reflection

Today in class we learned new topic using the capabilities of a computer.

Your opinion about the lesson: ...

Today I found out...understood...surprised...

This topic is for understanding...

Information from external ballistics

External ballistics - this is a science that studies the movement of a bullet (grenade) after the cessation of the action of powder gases on it.

Having flown out of the channel a of the barrel under the action of powder gases, the bullet (grenade) moves by inertia. A grenade with a jet engine moves by inertia after the expiration of gases from the jet engine.

Trajectory and its elements

trajectorycalled the curved line described by the center of gravity of the bullet in flight.

A bullet flying through the air is subjected to two forces: gravity and air resistance.

The force of gravity causes the bullet to gradually descend, and the force of air resistance continuously slows down the movement of the bullet and tends to knock it over.

As a result of the action of these forces, the bullet's flight speed gradually decreases, and its trajectory is an unevenly curved curved line in shape.

Options trajectories	Parameter characteristic	Note
1. Departure point	Center of muzzle	The departure point is the start of the trajectory
2. Horizon weapons	Horizontal plane passing through the departure point	The horizon of the weapon looks like a horizontal line. The trajectory crosses the horizon of the weapon twice: at the point of departure and at the point of impact
3. Elevation line	A straight line that is a continuation of the axis of the bore of the aimed weapon
4. Elevation angle	The angle enclosed between the line of elevation and the horizon of the weapon	If this angle is negative, then it is called the angle of declination (decrease)
5. Throw line	Straight line, a line that is a continuation of the axis of the bore at the time of the bullet's departure
6. Throwing angle	The angle enclosed between the line of throw and the horizon of the weapon
7. Departure angle	The angle enclosed between the line of elevation and the line of throw
8. Drop point	Point of intersection of the trajectory with the horizon of the weapon
9. Angle of incidence	The angle enclosed between the tangent to the trajectory at the point of impact and the horizon of the weapon
10. Full horizontal range	Distance from departure point to drop point
11. Top of the trajectory	The highest point of the trajectory
12. Trajectory height	The shortest distance from the top of the trajectory to the horizon of the weapon
13. Exceeding the trajectory above the aiming line	The shortest distance from any point of the trajectory to the line of sight
14. Target elevation angle	The angle enclosed between the line of sight and the horizon of the weapon	The target's elevation angle is considered positive (+) when the target is above the weapon's horizon, and negative (-) when the target is below the weapon's horizon.
16. Meeting point	Intersection point of the trajectory with the target surface (ground, obstacles)
17. Point of aim (aiming)	The point on or off the target at which the weapon is aimed
18. Meeting angle	The angle enclosed between the tangent to the trajectory and the tangent to the target surface (ground, obstacles) at the meeting point	The smaller of adjacent corners, measured from 0 to 90°
19. Line of sight	A straight line passing from the shooter's eye through the middle of the sight slot (level with its edges) and the top of the front sight to the aiming point
20. Sighting range	Distance from the point of departure to the intersection of the trajectory with the line of sight
21. Aiming angle	The angle enclosed between the line of elevation and the line of sight
Elevation	Giving the axis of the bore the desired position in the vertical plane
Ascending branch	Part of the trajectory from the departure point to the summit
Horizontal aiming	Giving the axis of the bore the desired position in the horizontal plane
target line	A straight line connecting the departure point with the target	When firing direct fire, the target line practically coincides with the aiming line
Slant Range	Distance from point of origin to target along target line	When firing direct fire, the slant range practically coincides with the aiming range.
descending branch	Part of the trajectory from the top to the point of impact
final speed	Bullet speed at point of impact
Shooting plane	The vertical plane passing through the line of elevation
Total flight time	The time it takes for a bullet to travel from point of departure to point of impact
Aiming (pointing)	Giving the axis of the bore of the weapon the position in space necessary for firing	In order for the bullet to reach the target and hit it or the desired point on it
aiming line	A straight line connecting the middle of the sight slot to the top of the front sight

direct shot

Straight shot called a shot in which the trajectory of the bullet does not rise above the line of sight above the target throughout its entire length. The range of a direct shot depends on the height of the target and the flatness of the trajectory. The higher the target and the flatter the trajectory, the greater the range of a direct shot and, therefore, the distance at which the target can be hit with one sight setting.

The practical value of a direct shot lies in the fact that in tense moments of the battle, shooting can be carried out without rearranging the sight, while the aiming point in height will be selected along the lower edge of the target.

Each shooter must know the value of the range of a direct shot at various targets from his weapon and skillfully determine the range of a direct shot when shooting.

The range of a direct shot can be determined from the tables by comparing the height of the target with the values ​​\u200b\u200bof the greatest excess above the line of sight or with the height of the trajectory.

Direct shot and rounded direct shot ranges

from small arms caliber 5.45 mm

When firing, you need to know that the distance on the ground, during which the descending branch of the trajectory does not exceed the height of the target, is called affected space (the depth of the affected space Ppr.).

Depth (Ppr.) depends:

on the height of the target (it will be the greater, the higher the target);

from the flatness of the trajectory (it will be the greater, the flatter the trajectory);

from the angle of inclination of the terrain (on the front slope it decreases, on the reverse slope it increases).

The depth of the affected space (Ppr.) can be determined from the tables of excess trajectories above the line of sight by comparing the excess of the descending branch of the trajectory by the corresponding firing range with the height of the target, and if the target height is less than 1/3 of the trajectory height, by the thousandth formula:

where Ppr- depth of the affected space in m; Vts- target height in m; β is the angle of incidence in thousandths.

The space behind a cover that is not penetrated by a bullet, from its crest to the meeting point is called covered space . The covered space will be the greater, the greater the height of the shelter and the flatter the trajectory.

The part of the covered space in which the target cannot be hit with a given trajectory is called dead (unaffected) space. Dead space will be the greater, the greater the height of the shelter, the lower the height of the target and the flatter the trajectory. The other part of the covered space (Pp), on which the target can be hit, is the hit space.

The depth of dead space (Mpr.) Is equal to the difference between the covered and affected space:

Mpr \u003d Pp - Ppr

Knowledge of the value of Pp. and Mpr. allows you to correctly use cover to protect against enemy fire, as well as take measures to reduce dead spaces by choosing the right firing positions and firing at targets with weapons with a more hinged trajectory.

Normal (table) firing conditions

The tabular trajectory data corresponds to normal firing conditions.

The following are accepted as normal (table) conditions:

Weather conditions:

· atmospheric (barometric) pressure on the horizon of the weapon 750 mm Hg. Art.;

· air temperature at the weapon horizon +15° С;

· relative air humidity 50% ( relative humidity called the ratio of the amount of water vapor contained in the air to the largest amount of water vapor that can be contained in the air at a given temperature);

· there is no wind (the atmosphere is still).

Ballistic conditions:

· bullet weight, muzzle velocity and departure angle are equal to the values ​​indicated in the shooting tables;

· charge temperature +15°С;

· the shape of the bullet corresponds to the established drawing;

· the height of the front sight is set according to the data of bringing the weapon to normal combat;

· heights (divisions) of the sight correspond to the tabular aiming angles.

Topographic conditions:

· the target is on the horizon of the weapon;

· there is no side slope of the weapon.

If the firing conditions deviate from normal, it may be necessary to determine and take into account corrections for the range and direction of fire.

The influence of external factors on the flight of a bullet

With the increase atmospheric pressure the air density increases, and as a result, the air resistance force increases and the range of the bullet decreases. On the contrary, with a decrease in atmospheric pressure, the density and force of air resistance decrease, and the range of the bullet increases.

As the temperature rises, the air density decreases, and as a result, the air resistance force decreases and the range of the bullet increases. Conversely, as the temperature decreases, the density and air resistance force increase, and the range of the bullet decreases.

With a tailwind, the speed of the bullet relative to the air decreases. As the speed of the bullet relative to the air decreases, the force of air resistance decreases. Therefore, with a fair wind, the bullet will fly further than with no wind.

With a headwind, the speed of the bullet relative to the air will be greater than with no wind, therefore, the air resistance force will increase, and the range of the bullet will decrease.

The longitudinal (tail, head) wind has little effect on the flight of a bullet, and in the practice of shooting from small arms, corrections for such a wind are not introduced.

The side wind exerts pressure on the side surface of the bullet and deflects it away from the firing plane depending on its direction: the wind from the right deflects the bullet to the left side, the wind from the left - to the right side.

The wind speed is determined with sufficient accuracy by simple signs: with a weak wind (2-3 m / s), a handkerchief and a flag sway and flutter slightly; with a moderate wind (4-6 m / s), the flag is kept unfolded, and the scarf flutters; with a strong wind (8-12 m / s), the flag flutters with noise, the handkerchief is torn from the hands, etc.

Changes in air humidity have little effect on air density and therefore bullet range, so it is not taken into account when shooting.

Penetrating (lethal) action of a bullet

For firing from a machine gun, cartridges with ordinary (with a steel core) and tracer bullets are used. The lethality of a bullet and its penetrating effect mainly depends on the distance to the target and the speed that the bullet will have at the moment of meeting the target.

№ p.p.	Name of the obstacle (protective equipment)	Firing range, m	Penetration % or Bullet Penetration Depth
	Steel sheets (at a meeting angle of 90°) with a thickness of:
	2 mm.
	3 mm.
	5 mm.
	Steel helmet (helmet)		80-90%
	Bulletproof vest		75-100%
	A breastwork made of hard-packed snow		50-60 cm.
	Earthen barrier from compacted loamy soil		20-25 cm.
	Wall made of dry pine beams 20 cm thick.
	brickwork	If the circle is divided into 6000 equal parts, then each such part will be equal to: The length of the arc corresponding to this angle is 1/955 (rounded 1/1000) of the length of the radius of this circle. Therefore, the division of the goniometer is usually called a thousandth. The relative error that results from this rounding is 4.5%, or rounded 5%, that is, the thousandth is 5% less than the goniometer division. In practice, this error is neglected. The division of the protractor (thousandth) makes it easy to switch from angular to linear units and vice versa, since the length of the arc corresponding to the division of the goniometer at all distances is equal to one thousandth of the length of the radius equal to the firing range. An angle of one thousandth corresponds to an arc equal at a distance of 1000 m - 1 m (1000 m: 1000), at a distance of 500 m - 0.5 m (500: 1000), at a distance of 250 m - 0.25 m (250: 1000), etc. d. An angle in a few thousandths corresponds to an arc length AT, equal to one thousandth of the range (D/1000) multiplied by the angle containing At thousandths, i.e. The resulting formulas are called thousandth formulas and have wide application in shooting practice. In these formulas D- distance to the object in meters. At- the angle at which the object is seen in thousandths. AT- the height (width) of the object in meters, i.e. the length of the chord, not the arc. At small angles (up to 15°), the difference between the length of the arc and the chord does not exceed one thousandth, therefore, at practical work they are considered equal. Measurement of angles in goniometer divisions (thousandths) can be made:the goniometric circle of the compass, the binocular and periscope reticle, the artillery circle (on the map), the entire scope, the sniper scope's side adjustment mechanism, and handy items. The accuracy of angular measurement with a particular instrument depends on the accuracy of the scale on it. When using improvised objects to measure angles, it is necessary to determine their angular value in advance. To do this, you need to stretch out your hand with an improvised object at eye level and notice any points on the ground near the edges of the object, then use a goniometer (binoculars, compass, etc.) to accurately measure the angular value between these points. The angular value of an improvised object can also be determined using a millimeter ruler. To do this, the width (thickness) of the object in millimeters must be multiplied by 2 thousandths, since one millimeter of the ruler, when it is 50 cm away from the eye, corresponds to an angular value of 2 thousandths according to the thousandth formula. Angles expressed in thousandths are written through a dash and read separately: first hundreds, then tens and ones; in the absence of hundreds or tens, zero is written and read. For example: 1705 thousandths are written 17-05, read - seventeen zero five; 130 thousandths are written 1-30, read - one thirty; 100 thousandths are written 1-00, read - one zero; one thousandth is written 0-01, read - zero zero one. such a firing range at which the height of the trajectory is equal to the height of the target, it can also be defined as the greatest range to the target, at which it is no longer possible to obtain a direct shot. a complex thermodynamic process of a very rapid, almost instantaneous conversion of the chemical energy of gunpowder into heat, and then into the kinetic energy of the powder gases that set the bullet in motion.