The calculation of the movement is called the determination of the main parameters of the movement of the car and pedestrian: speed, path, time and trajectory of movement.
When calculating the uniform movement of the vehicle use elementary ratio
where S. but , V. but and t. à - Accordingly, the path, speed and time of the car movement.
Braking at a constant clutch coefficient
If the driver slowed down during an accident, then the initial velocity of the car can be fairly accurately determined by the length of the sliding track (trace) of the tire on the road arising from the full locking of the wheels.
Experimental study of the braking process shows that due to changes in the tire clutch coefficient with expensive and oscillations caused by the presence of elastic tires and suspension elements, slowdown j. In the process of braking is complex.
Fig. 5.1. Brake diagram
To simplify the calculations, we believe that during the TN (time of the deceleration rate), the slowdown increases according to the law of the line (section AB), and during the time (time TU of the steady deceleration) remains permanent (section Sun) and at the end of the complete braking period instantly decreases to zero (point C).
The deceleration of the car is calculated based on the conditions for the complete use of the clutch by all car tires,
, m / s 2 (5.2)
whereg. = 9.81 m / s 2 ;
c. - The coefficient of longitudinal tire clutch with the road that is taken permanent.
Since the complete and simultaneous use of the clutch by all car tires is observed relatively rarely, the formula introduces the correction coefficient of braking efficiency CEand the formula acquires the following form:
, m / s 2, (5.3)
Value TO e. It takes into account the correspondence of the brake forces for the clutch forces and depends on the conditions of braking. If all the wheels were blocked when braking, then TO e. choose depending on h. .
Table 5. 1.
Meaning to in the presence of traces of souz
The most common way to determine the speed of movement vehicle Before starting the braking, it is presented by the formula in all literary sources,
where: j. but - The slowdown of the car developed during its braking, depending on the type of vehicle, the degree of loading, the state of the roadway, M / C 2;
t. n. - The growth time of the car slowing down when it is braked, which is also dependent on all the above factors, as well as slowing down, and practically changing proportionally change in the load of the car and the magnitude of the clutch coefficient, C;
S. - The length of the braking trace of the car, counting the axis of the rear wheels; If the track remained from the wheels of both axes of the car, then the car base is subtracted from the magnitude of the "Yuza" L.m.
Brake and stopping car path
Braking path, stopping path, braking trace, slowing the vehicle, etc. - to the values \u200b\u200bof these terms often have to be addicted to objectively evaluate the actions of the driver in a specific road situation.
The stopping path of the vehicle is the distance that the car overcomes from the time the driver's reaction is dangerous to its complete stop:
, m (5.5)
The braking path of the vehicle is the distance that the car overcomes from the moment the brake pedal starts up to its complete stop:
, m. (5.6)
Thus, the stopping path of the car is greater than its brake path by the distance that the car overcomes during the driver's reaction T 1.
Driver reaction time t. 1 . The value of the driver's response time (in the autotechnical examination) is a period of time from the moment the danger signal appears in the driver's field of view before starting the impact on the vehicle management bodies (brake pedal, steering wheel, accelerator pedal).
At the time of the driver's reaction, all elements of the system "Driver - a car - road - medium" (VADS) are influenced, therefore it is advisable to differentiate the reaction time values \u200b\u200bdepending on typical road traffic situations characterized by certain combinations of interrelated factors of the VADS system. The reaction time varies in large limits - from 0.3 to 1.4 or more.
Thus, when calculating the maximum allowable speed by the conditions of the visibility of the road, the minimum time of a simple sensorotor reaction should be taken equal to 0.3 s. The same time reaction should be taken when determining the minimum permissible distance between along the moving vehicles.
In the case of manifestation when moving any vehicle malfunctions affecting the safety of motion, as well as with the physical intervention of the passenger, the driver's response time can be taken as 1.2 s.
In road traffic accidents in dark time The day when the obstacle was minorly, an increase in the driver's response time is allowed to 0.6 s.
Brake Drive Relaxing Time t. 2 . During this time, the free move of the brake pedal and the brake system drive gaps are selected. The value depends on the type of brake drive and its technical condition.
The hydraulic drive of the brakes is triggered faster than pneumatic. Hydraulic drive restarting time is accepted t. 2 = 0.2 - 0.4 s. In passenger cars during emergency braking t. 2 = 0.2 C., and truly t. 2 = 0,4 from. The delay time of the triggering of a faulty hydraulic drive (if there is air in the system or malfunction of the valves in the main brake cylinder) increases. If the brakes are triggered from the second press on the pedal, it increases an average of up to 0.6 s, and at three pages - up to 1.0 s.
Time to delay the triggering of the pneumatic brake drive fluctuates within t. 2 \u003d 0.4-0.6 S., and its average value t 2 \u003d 0.4 s. In road trains having a pneumatic drive, this time increases: at one trailer T 2 \u003d 0.6 s, and at two - t. 2 = Up to 1 S..
Time of growth of deceleration t n. The time of increasing deceleration is the time from the beginning of the appearance of a slowdown or on the moment of contacting the overlays with brake drums before starting the moment of movement of the vehicle with the installed maximum slowdown or until the completion of the overlays to brake drums, and in the formation of traces of braking - prior to the formation of the latter on the carriageway.
In case of emergency braking until the wheel lock, this time almost changes in proportion to the change in the loading of the car and the magnitude of the clutch coefficient.
The increase in deceleration time depends mainly on the type of brake drive, type and state of the road surface, vehicle mass.
So, if the initial speed of the car is known V. a. Speed V. yu , appropriate start of complete braking, can be found, believing that during t. w. The car moves evenly slowly with a constant deceleration. 0,5 j..
, m / s. (5.7)
Technical ability to prevent accidents
When analyzing the circumstances of the road accident after determining the magnitude of the stopping path of the car S. about It is necessary to determine:
Removing a car ( S. a.) From the place of the departure at the moment when the danger arose for the movement;
The time required to stop the car, i.e. time at the stopping path ( t. o.);
Pedestrian time ( t. p ), which he spends on movement from the place of danger to the place of departure;
Time ( 
) during which the inhibited car moved to the departure.
The time of the pedestrian movement to the place of collision is determined by:
, C, (5.8)
where:S. n. - Path of a pedestrian from the place of occurrence of a dangerous situation to the place of departure, m;
V. n. - The speed of the pedestrian, defined either on table data, or experimentally, km / h.
If the time movement of a pedestrian to the place of collision is less than or equal to the total response time of the driver and the timing time of the brake drive ( t. n. T. 1 + T. 2 + 0,5t n. = T. ), then the pedestrian will be in the car's movement strip, while the braking has not yet come. In this case, the technical ability to prevent the hit is not, regardless of the velocity speed of the vehicle.
If a t. a. > T, That analysis is carried out in the following sequence:
Determine the distance S. a. Between the car and the place of departure at the time of danger to movement;
Compare Distance S. but With stopping by vehicle S. o. .
If the stopping path of the car (S. about ) Less distance ( S. a.), then the conclusion about the technical ability to avoid accidents, otherwise there is no such driver.
To determine the distance S. a. VNIISE recommends the following formulas:
In case of departure before the start of braking
, m, (5.9)
where L. uD - distance from the place of impact of the car to its front, m;
In the event that the inhibited car continued to move to the stop,
, m (5.10)
, m, (5.11)
where 
- The distance that overcomes the car after the departure until a complete stop.
- Evyukov S. A., Vasilyev Ya. V. Investigation and expertise of road traffic accidents / under total. ed. S. A. Evtyukova. SPB: LLC "Publishing DNA", 2004. 288
- Evyukov S. A., Vasilyev Ya. V. Examination of road traffic accidents: Handbook. SPB: LLC "Publishing DNA", 2006. 536
- Evyukov S. A., Vasilyev Ya. V. DTP: investigation, reconstruction and examination. SPb.: LLC "DNA Publishing", 2008. 390 С
- GOST R 51709-2001. Motor vehicles. Safety requirements for technical condition and verification methods. M.: Standards Publishing House, 2001. 27
- Litvinov A.S., Fourbin Ya. E. Car: Theory of operational properties. M.: Mechanical Engineering, 1986. 240 C
- Judicial autothechnical examination: allowance for experts - car vehicles, investigators and judges. Part II. Theoretical foundations and methods of experimental research in the production of autotelechnic examination / ed. V. A. Ilarionov. M.: Vnis, 1980. 492 with
- Pushkin V. A. et al. Evaluation of the road situation preceding accidents // Organization and Security road In major cities: Sat. Dokl. 8th international. conf. St. Petersburg., 2008. C. 359-363
- On approval of the Charter of the Federal budget institution Russian federal judicial examination center under the Ministry of Justice Russian Federation: Order of the Ministry of Justice of the Russian Federation of 03.03.2014 No. 49 (as amended from 01/21/2016 No. 10)
- Nadezhdin E. N., Smirnova E. E. Econometric: studies. Manual / Ed. E. N. Nadeyadin. Tula: ANO VPO "IEU", 2011. 176 with
- Grigoryan V. G. Application in the expert practice of braking parameters motor vehicles: method. Recommendations for experts. M.: VNIISE, 1995
- Decree of the Government of the Russian Federation of 06.10.1994 No. 1133 "On the forensic expert institutions of the Ministry of Justice of the Russian Federation"
- Decree of the Government of the Russian Federation on the Federal Target Program "Improving Road Safety in 2013-2020" of 10/30/2012 No. 1995-p
- Nikiforov V.V. Logistics. Transport and warehouse in supply chains: studies. benefit. M.: Grossmedia, 2008. 192 with
- Schukin M. M. Coupling devices Car and tractor: design, theory, calculation. M.; L.: Mechanical Engineering, 1961. 211 with
- Pushkin V. A. Basics of expert analysis of road accidents: database. Expert technique. Methods of solutions. Rostov N / D: IPO PI SFU, 2010. 400 С
- Shcherbakova O. V. Rationale mathematical model The process of collision in order to develop a methodology for improving the accuracy of determining the speed of movement of the road train at the beginning of the overturning on curvilinear trajectories // Bulletin of civil engineers. 2016. № 2 (55). P. 252-259
- Scherbakova O. V. Analysis of the conclusions of autothechnical expertise on road traffic accidents // Bulletin of civil engineers. 2015. № 2 (49). Pp. 160-163
The established slowdown, m / s 2, is calculated by the formula
.
(7.11)
\u003d 9.81 * 0.2 \u003d 1.962 m / s 2;
\u003d 9.81 * 0.4 \u003d 3.942 m / s 2;
\u003d 9.81 * 0.6 \u003d 5.886m / s 2;
\u003d 9.81 * 0.8 \u003d 7.848 m / s 2.
The results of the calculations according to formula (7.10) are reduced to Table 7.2
Table 7.2 - Dependence of the stopping path and steady deceleration from the initial braking rate and the clutch coefficient
|
|
|
|||||||
, km / hAccording to Table 7.2, we build the dependence of the stopping path and the slowdown deceleration from the initial deception rate and the clutch coefficient (Figure 7.2).
7.9 Building a brake diagram PBX
The brake diagram (Figure 7.3) is the dependence of the slowdown and the speed of the PBX movement on time.
7.9.1 Determination of speed and deceleration on the diagram site corresponding to the timing of the drive
For this stage 
=
\u003d const 
\u003d 0 m / s 2.
In operation initial braking speed 
\u003d 40 km / h for all categories PBX.
7.9.2 Determination of the speed of the PBX on the diagram site corresponding to the time of deceleration
Speed 
, m / s, corresponding to the end of the deceleration of the deceleration time, are determined by the formula
\u003d 11.11-0.5 * 9.81 * 0.7 * 0.1 \u003d 10.76 m / s.
Intermediate velocity values \u200b\u200bin this section are determined by formula (7.12), while 
= 0
; Coefficient of clutch for category M 1 
=
0,7.
7.9.3 Determination of speed and deceleration on the section of the diagram corresponding to the time setting
Time of steady slowdown 
, C, calculated by the formula
,
(7.13)
from.
Speed 
, m / s, on the section of the diagram corresponding to the time of the steady deceleration, are determined by the formula
,
(7.14)
for 
= 0
.
The value of the steady deceleration for the working brake system of the Category M 1 is taken 
\u003d 7.0 m / s 2.
8
Definition of managing parameters PBX
Controllability PBX is its property in a specific road situation a given direction of movement or change it according to the driver's effect on the steering.
8.1 Determination of maximum angles of rotation of controlled wheels
8.1.1 Determination of the maximum angle of rotation of the external controlled wheel
Maximum angle of rotation of the outdoor controlled wheel
,
(8.1)
where R n1 min is the turning radius of the outer wheel.
The rotation radius of the outer wheel is taken equal to the corresponding prototype parameter -R H1 min \u003d 6 m.
,
\u003d 25,65.
8.1.2 Determining the maximum angle of rotation of an internal controlled wheel
The maximum angle of rotation of the internal controlled wheel can be determined by taking a king of a squash equal to the wheel track. Previously, it is necessary to determine the distance from the instantaneous center of rotation to the outer rear wheel.
Distance from instant turn center to the outer rear wheel 
, m, calculated by the formula
,
(8.2)
.
Maximum angle of rotation of an internal controlled wheel 
, hail, can be determined from expression
,
(8.3)
,
\u003d 33,34.
8.1.3 Definition of the average maximum angle of rotation of controlled wheels
The average maximum rotation angle of controlled wheels 
, hail, can be determined by the formula
,
(8.4)
.
8.2 Definition of the minimum width of the carriageway
Minimum carriage part 
, m, calculated by the formula
\u003d 5.6- (5.05-1.365) \u003d 1.915m.
8.3 Definition of critical under the conditions of traffic speed
Critical under the conditions of the traffic speed 
, m / s, calculated by the formula
,
(8.6)
where 
,
- coefficients of resistance to wheels front and rear axis Accordingly, n / hail.
Single wheel resistance coefficient 
, N / is glad, are approximately determined by empirical dependence.
where 
- internal tire diameter, m; 
- width of the tire profile, m; 
- Air pressure in the tire, kPa.
To Δ1 \u003d (780 (0.33 + 2 * 0.175) 0.175 (0.17 + 98) * 2) /57.32\u003d317.94, n / ha
To δ1 \u003d (780 (0.33 + 2 * 0,175) 0.175 (0.2 + 98) * 2) / 57.32 \u003d 318.07, n / ha
.
Turning the designed car - excessive.
To ensure traffic safety, a condition must be performed
>
.
(***)
The condition (***) is not performed, since in determining the impedance coefficients, only tire parameters were taken into account. At the same time, when determining the critical velocity, it is necessary to take into account the distribution of car mass, suspension design and other factors.
Brake power.When braking, elementary friction forces, distributed over the surface of friction linings, create a resulting torque moment, i.e. Brake moment M. Thor directed opposite to the rotation of the wheel. Breaking power arises between the wheel and expensive R TOR .
Maximum brake force R The MAX torus is equal to the tire clutch strength. Modern cars have brake mechanisms on all wheels. At a two-axis car (Fig. 2.16) Maximum brake force, n,
Projecting all the forces acting on the car when braking, on the plane of the road, we get in general Car motion equation when braking on a lift:
R Tor1 +. R Tor2 +. R K1 +. R K2 +. R P + R in + R.d. . + R g - R And \u003d R Thor +. R D + R in + R.d. . + R g - R n \u003d 0,
where R TOR \u003d. R Tor1 +. R Tor2; R d \u003d R K1 +. R K2 +. R P - the power of the resistance of the road; R etc. - friction force in the engine, shown to the leading wheels.
Consider the case of car braking only the brake system when the power R etc. = 0.
Considering that the speed of the car during braking decreases, we can assume that the force R in ≈ 0. Due to the fact that R Mala compared to power R the torus can also be neglected, especially when emergency braking. The adopted assumptions allow you to write the car equation for braking in the following form:
R Thor +. R d - R n \u003d 0.
From this expression, after the transformation, we obtain the equation of the movement of the car during braking on the projector of the road:
φ x + ψ - Δ n a. s / g. = 0,
where φ x is the coefficient of the longitudinal clutch of tires with the road, ψ is the road resistance coefficient; δ n is the coefficient of accounting of rotating masses on the projector of the road (with a rope); a. W is the acceleration of braking (deceleration).
A slowdown is used as the vehicle braking dynamics of the vehicle but s in braking and brake path S. TOR , m. Time t. Thor, C, use as an auxiliary meter when determining the stopping path S. about.
Slow down when braking the car.Delay in braking is determined by the formula
but Z. \u003d (P Tor + R D + R in +. R d) / (Δ BP m.).
If brake forces on all wheels have reached the validity of the clutch forces, then neglecting forces R in and R G.
a. s \u003d [(φ x + ψ) / ψ BP] g. .
The coefficient φ x is usually much larger than the coefficient ψ, therefore, in the case of a complete braking of the vehicle, the value of the expression can be neglected. Then
a. s \u003d φ x g. / Δ BP ≈ φ x g. .
If during braking the coefficient φ x does not change, then slowing but It does not depend on the velocity of the car.
Brake time.Stopping time (total braking time) is the time from the moment the driver's danger is discovered until the car stops. The total braking time includes several segments:
1) driver response time t. R - Time during which the driver decides on braking and transfers the foot from the fuel supply pedal to the pedal of the working brake system (depending on its individual characteristics and qualifications is 0.4 ... 1.5 s);
2) Brake Drive Time t. PR - time from the beginning of clicking on the brake pedal before the start of the deceleration, i.e. Time to move all moving parts of the brake drive (depending on the type of brake drive and its technical condition is 0.2 ... 0.4 C for hydraulic drive, 0.6 ... 0.8 C for pneumatic acting and 1 ... 2 C for a manifold with pneumatic drive brakes);
3) Time t. y, during which the slowdown increases from zero (the beginning of the brake mechanism) to the maximum value (depends on the intensity of braking, the load on the car, type and state of the road surface and the braking mechanism);
4) braking time with maximum intensity t. torus. Determine the formula t. Tor \u003d υ / a. s Max - 0.5 t. y
For a time t. P + t. Prom car moves evenly at the speed υ , during t. y - slowly, and over time t. TOR – slowly until the complete stop.
Graphic representation of the time of braking, changing the speed, slowing down and stop the car gives a diagram (Fig. 2.17, but).
To determine the stopping time t. about , necessary to stop the car from the moment of danger, you need to summarize all the time-called time segments:
t. Oh \u003d. t. P + t. Pr + T. in +. t. TOR \u003d. t. P + t. PR + 0.5 t. y + υ / a. Z Max \u003d. t. Sum + υ / a. z Max
where t. Sumy \u003d T. P + t. PR + 0.5 t. y
If the brake forces on all wheels of the car simultaneously reach the values \u200b\u200bof the clutch forces, then accepting the coefficient δ BP \u003d 1, get
t. Oh \u003d. t. sum + υ / (φ x g.).
Braking distances - This is the distance that the car passes during the braking t. torus with maximum efficiency. This parameter is determined using the curve. t. TOR \u003d. f (υ ) and considering that in each velocity interval the car moves equifiable. Sample view of the track dependence S. torus from speed R to , R in, r T and without taking into account these forces is shown in Fig. 2.18, but.
The distance required to stop the car from the moment of danger (the length of the so-called stopping path) can be determined if we assume that the slowdown is changed as shown in Fig. 2.17, but.
Stopping path can be divided into several segments corresponding to time segments t. R, t. etc, T. y, t. Tor:
S. Oh \u003d. S. P + S. Pr + S. in +. S. torus.
Car traveled during t. P + t. Progue with a constant speed υ, define as follows:
S. P + S. pr \u003d υ ( t. P + t. etc) .
Taking that when a speed reduction from υdo υ "car moves with a constant deceleration but cf \u003d 0.5 but Z M ah, we get the way passed by the car during this time:
Δs. y \u003d [ υ 2 – (υ") 2 ] / but s m ah.
Brake path with a reduction in speed from υ "to zero during emergency braking
S. Tor \u003d (υ ") 2 / (2 but s m ah).
If the brake forces on all the wheels of the car simultaneously reached the values \u200b\u200bof the clutch forces, then R etc. \u003d. R in \u003d. R r \u003d 0 brake car path
S. TOR \u003d υ 2 / (2φ x g.).
The braking path is directly proportional to the square of the velocity of the car at the time of the start of braking, so with an increase in the initial speed, the braking path increases especially quickly (see Fig. 2.18, but).
Thus, the stopping path can be determined as follows:
S. Oh \u003d. S. P + S. Pr + S. in +. S. tor \u003d υ ( t. P + t. PR) + [υ 2 - (υ ") 2] / but z M ah + (υ ") 2 / (2 but s m ah) \u003d
= υ T. Sum + υ 2 / (2 but s m ah) \u003d υ T. sum + υ 2 / (2φ x g.).
The stopping path, as well as stopping time, depends on a large number of factors, the main of which are:
vehicle speed at the time of the start of braking;
qualifications and physical condition of the driver;
type I. technical condition working brake system of the car;
pavement state;
car load;
condition of car tires;
method of braking, etc.
Intensity intensity indicators.To test the effectiveness of the brake system, the largest allowable braking path is used as indicators and the smallest allowable slowdown in accordance with GOST R 41.13.96 (for new cars) and GOST R 51709-2001 (for auto-operating cars). The intensity of braking cars and buses under traffic safety conditions is checked without passengers.
The greatest permissible brake path S. Tor, M, when driving with an initial speed of 40 km / h on a horizontal section of the road with a smooth, dry, pure cement, or asphalt concrete coating, has the following values:
passenger cars and their modifications for the carriage of goods ..........14,5
buses S. full mass:
up to 5 tons inclusive ............................................ 18.7
more than 5 tons ....................................... ... .................. 19.9
full weight trucks
up to 3.5 tons inclusive ................ ...........................19
3.5 ... 12 T inclusive .................................... .. ... 18,4
more than 12 t ...................................................... .. ... 17.7
motor tractor with trucks with full weight:
up to 3.5 T inclusive ......................... .................. 22.7
3.5 ... 12 T inclusive ..................................... ... .22,1
more than 12 t .......................................................... 21.9
Distribution of brake force between car bridges.When braking the car inertia R and, (see Fig. 2.16), acting on the shoulder h. C causes the redistribution of normal loads between the front and rear bridges; The load on the front wheels is increased, and the rear is reduced. Therefore, normal reactions R. Z 1 I. R. z 2. , acting respectively on the front and rear axle bridges during braking, significantly different from loads G. 1 I. G. 2 , which perceive bridges in static condition. These changes are evaluated by the coefficients of changing normal reactions. M. P1, I. m. P2, which for the case of car braking on the horizontal road is determined by formulas
m. p1 \u003d 1 + φ H. H. C / l. 1 ; m. P2 \u003d 1 - φ H. H. C / l. 2 .
Consequently, normal road reactions
R. z 1 \u003d. m. P1 G. 1 ; R. z 2 \u003d. m. P2. G. 2 .
During the car inhibition, the largest values \u200b\u200bof the reaction change coefficients are within the following limits:
m. p1 \u003d 1.5 ... 2; m. P2 \u003d 0.5 ... 0.7.
The maximum intensity of braking can be provided with the complete use of the clutch by all the wheels of the car. However, the braking force between the bridges can be distributed unevenly. Such unevenness characterizes Brake power distribution coefficientbetween the front and rear bridges:
β O \u003d. R TOR1 / R TOR \u003d 1 - R Tor2 / R torus.
This coefficient depends on various factors from which the mains are: the distribution of the car weight between its axes; intensity of braking; reaction change coefficients; Types of wheeled brake mechanisms and their technical condition, etc.
With the optimal distribution of brake force front and rear wheels The car can be brought to blocking simultaneously. Ad hoc
β o \u003d ( l. 1 + φ about H. c) / L.
Most brake systems provide a constant ratio between the brake forces of the front and rear axle (R Tor1 I. R Tor2. ), therefore, total strength R The torus can reach the maximum value only on the road with the optimal coefficient φ about. On other roads full use Coupling weight without blocking at least one of the bridges (front or rear) is impossible. However recently appeared brake systems With the regulation of the distribution of brake forces.
The distribution of the total brake force between the bridges does not correspond to the normal reactions varying during braking, therefore the actual deceleration of the car is less, and the time of braking and the braking path is more theoretical values \u200b\u200bof these indicators.
To approximate the results of the calculation to experimental data in the formula, the coefficient of braking efficiency is introduced TO E. , which takes into account the degree of use of theoretically possible efficiency of the brake system. On average for passenger cars TO E. = 1,1 ... 1.2; For trucks and buses TO E. = 1.4 ... 1.6. In this case, the calculated formulas have the following form:
a. s \u003d φ x g / K. e;
t. Oh \u003d. t. Sum +. TO e υ / (φ x g.);
S. TOR \u003d. TO e υ 2 / (2φ x g.);
S. O \u003d υ. T. Sum +. TO e υ 2 / (2φ x g.).
Methods of car braking. Cooperating brake system and engine.This method of braking is used to avoid overheating brake mechanisms and accelerated tire wear. Brake moment on wheels is created at the same time brake mechanisms and engine. Since in this case, the brake pedal is preceded by the release of the fuel supply pedal, the angular velocity of the engine of the engine should decrease to the angular velocity idle move. However, in fact, the drive wheels through the transmission are forcibly rotated crankshaft. As a result, an additional force of r TD resistance to movement appears proportional to the friction force in the engine and the deceleration of the car.
The inertia of the flywheel counteracts the inhibitory action of the engine. Sometimes the opposition of the flywheel turns out to be more inhibited engine action, as a result of which the intensity of the braking is somewhat reduced.
Joint braking of the working brake system and the engine more efficiently than braking only the brake system if slowing down when braking a. Z. from More than a slowdown in braking with a disconnected engine a. s, i.e. a. Z. from > a. s.
On the roads with a small clutch coefficient, joint braking increases transverse stability Car under the conditions of drift. When braking in emergency situations, the clutch is useful to turn off.
Brake with periodic termination of the brake system.The inhibited non-slip wheel perceives a large brake force than when moving with partial slippage. In case of free rolling, the angular velocity of the wheel ω k, radius r. to and progressive speed υ to the movement of the wheel of the wheel are associated with addiction υ to = ω K. R. to . The wheel moving with partial slippage (υ * ≠ ω K. R. K), this equality is not respected. The difference of velocities υ K and υ * determines the speed of sliding υ , i.e. υ с = υ -ω K. R. to.
The degree of slippage wheeldefined as λ = υ SC. / υ K. . The slave wheel is loaded only by the forces of resistance to movement, so the tangent reaction is small. The application to the braking torque wheel causes an increase in the tangent reaction, as well as an increase in deformation and elastic tire slipping. The clutch coefficient of the tire with a road surface increases in proportion to slipping and reaches a maximum when slipping around 20 ... 25% (Fig. 2.19, but -point IN).
Workflow maintenance of maximum tire clutch with road coating illustrates a graph (Fig. 2.19, b.). With an increase in the braking torque (section OA)the angular velocity of the wheel decreases. In order not to give the wheel to stop (blocked), the braking moment is reduced (plot CD).The inertia of the pressure control mechanism in the brake drive leads to the fact that the pressure reduction process occurs with some delay (section Aq). Location on EF. Pressure is stabilized for a while. The growth of the angular velocity of the wheel requires a new increase in the braking torque (section GA)to the value corresponding to 20 ... 25% slip values.
At the beginning of the sliding, the slowdown of the wheel increases and the linear proportionality of the dependence is disturbed: ω \u003d f (M. TOR ). Plots DE.and FG. characterized in inertia executive mechanisms. The brake system in which the pulsating pressure control mode is implemented in working cylinders (cameras) is called anti-lock.The depth of the pressure modulation in the brake drive reaches 30 ... 37% (Fig. 2.19, in).
The wheels of the car due to the cyclic loading of the braking torque rolling with partial slippage, approximately equal to the optimal, and the clutch coefficient remains high during the braking period. The introduction of anti-lock devices reduces tire wear and allows you to increase the transverse stability of the car. Despite the complexity and high cost, anti-lock brake systems are already legalized by the standards of many foreign countries, they are installed on the passenger cars of secondary and higher classes, as well as buses and cargo cars for long-distance transport.
Example number 1.
Install the slowdown and speed of the car before starting braking on a dry asphalt coating, if the length of the braking tracks of all wheels is 10 m, the slowdown time of 0.35 ° C, which is set to slowing 6.8 m / s 2, the car base is 2.5 m, the clutch coefficient - 0.7.
DECISION:
In the current road transport, in accordance with the recorded track, the vehicle speed before braking was approximately 40.7 km / h:
j \u003d g * φ \u003d 9,81 * 0,70 \u003d 6.8 m / s 2
The formula is indicated:
t 3 \u003d 0.35 s is the rise of the deceleration.
j \u003d 6.8 m / s 2 - installed slowdown.
SJ \u003d 10 m - the length of the fixed trace of the braking.
L \u003d 2.5 m - the car base.
Example number 2.
Install the stopping pathway of the car VAZ-2115 on a dry asphalt concrete coating, if: the driver reaction time is 0.8 s; Time to delay the triggering of the brake drive 0.1 s; Time of growth of deceleration 0.35 s; established slowdown 6.8 m / s 2; The speed of movement of the car VAZ-2115 - 60 km / h, the clutch coefficient is 0.7.
DECISION:
In the current traffic situation, the stopping path of the VAZ-2115 car is approximately 38 m:
The formula is indicated:
T 1 \u003d 0.8 s is the driver's response time;
T 3 \u003d 0.35 s - the time of deceleration of deceleration;
J \u003d 6.8 m / s 2 - the established slowdown;
V \u003d 60 km / h - VAZ-2115 car speed.
Example number 3.
Determine the stopping time of the VAZ-2114 car on the wet asphalt concrete, if: the driver's response time is 1.2 s; Time to delay the triggering of the brake drive 0.1 s; Time of growth of deceleration 0.25 s; established slowdown 4.9 m / s 2; Car speed VAZ-2114 50 km / h.
DECISION:
In the current traffic situation, the stopping time of the VAZ-2115 car is 4.26 s:
The formula is indicated:
T 1 \u003d 1.2 s is the driver's response time.
T 3 \u003d 0.25 C is the rise of the deceleration.
V \u003d 50 km / h - vehicle speed VAZ-2114.
J \u003d 4.9 m / s 2 - slowdown in VAZ-2114 car.
Example number 4.
Determine the safe distance between the VAZ-2106 vehicle moving in front and the KAMAZ car moving at the same speed. To calculate the following conditions: the inclusion of the stop signal from the brake pedal; The driver's response time when choosing a safe distance - 1.2 s; KAMAZ car brake drive triggering time - 0.2 s; The increase in the deceleration of the car KAMAZ - 0.6 s; Slowing car KAMAZ - 6.2 m / s 2; Slowing car VAZ - 6.8 m / s 2; Time to delay the triggering of the brake drive of the car VAZ - 0.1 s; The growth time of the car VAZ is 0.35 s.
DECISION:
In the current traffic situation, the safe distance between cars is 26 m:
The formula is indicated:
T 1 \u003d 1.2 s is the driver's response time when choosing a safe distance.
T 22 \u003d 0.2 s is the time of delaying the brake drive of the car KAMAZ.
T 32 \u003d 0.6 s is the increase in the deceleration of the car KAMAZ.
V \u003d 60 km / h - vehicle speed.
J 2 \u003d 6.2 m / s 2 - deceleration of the car KAMAZ.
J 1 \u003d 6.8 m / s 2 - Slowing car VAZ.
T 21 \u003d 0.1 s is the time of delaying the brake drive of the car VAZ.
T 31 \u003d 0.35 s is the increase in the vase vehicle slowing down.
Example number 5.
Determine the safe interval between moving in the passing direction by cars VAZ-2115 and KAMAZ. Car speed VAZ-2115 - 60 km / h, Car speed KAMAZ - 90 km / h.
DECISION:
In the current traffic situation with the passing movement of vehicles, a safe side interval is 1.5 m:
The formula is indicated:
V 1 \u003d 60 km / h - VAZ-2115 car speed.
V 2 \u003d 90 km / h - the speed of movement of the car KAMAZ.
Example number 6.
Determine the safe velocity of the VAZ-2110 car under visibility conditions, if visibility in the direction of movement is 30 meters, the driver's reaction time when oriented in the direction of movement - 1.2 s; Time to delay the triggering of the brake drive - 0.1 s; Slowness increase time - 0.25 s; The established slowdown is 4.9 m / s 2.
DECISION:
In the current traffic situation, the safe velocity of the VAZ-2110 car under visibility conditions in the direction of movement is 41.5 km / h:
The formulas indicate:
t 1 \u003d 1,2 s is the driver's reaction time when oriented towards the movement;
t 2 \u003d 0.1 s - the time of delaying the triggering of the brake drive;
t 3 \u003d 0.25 s - time of deceleration increases;
jA \u003d 4.9 m / s 2 - established deceleration;
SV \u003d 30 M is the distance of visibility in the direction of movement.
Example number 7.
Install the critical velocity of the VAZ-2110 car on the turn by the transverse slip condition, if the rotation radius is 50 m, the transverse clutch coefficient is 0.60; Cross-ending angle - 10 °
DECISION:
In the current traffic situation, the critical velocity of the VAZ-2110 car on turn on the transverse slip condition is 74.3 km / h:
The formula is indicated:
R \u003d 50 m - rotation radius.
F y \u003d 0,60 is a cross-clutch coefficient.
B \u003d 10 ° - the angle of the crosslock of the road.
Example number 8.
Determine the critical speed of the vehicle VAZ-2121 car on the rotation of a 50 m radius under the overturning condition if the height of the center of gravity of the car is 0.59 m, the car's track of the VAZ-2121 - 1.43 m, coefficient transverse roll Pressure mass - 0.85 .
DECISION:
In the current traffic situation, the critical speed of the vehicle VAZ-2121 car on turning under the overturning condition is 74.6 km / h:
The formula is indicated:
R \u003d 50 m - rotation radius.
Hz \u003d 0.59 m - Height of the center of gravity.
B \u003d 1.43 m - Car KAZ-2121 car.
q \u003d 0.85 is the coefficient of the transverse roll of the undercorns.
Example number 9.
Determine the brake route of the car GAZ-3102 in the conditions of ice at the speed of 60 km / h. Loading a car 50%, the time of delaying the brake drive is 0.1 s; Slowness increase time - 0.05 s; The clutch coefficient is 0.3.
DECISION:
In the current traffic situation, the brake route of the car GAZ-3102 is approximately 50 m:
The formula is indicated:
t 2 \u003d 0.1 s - the time of delaying the triggering of the brake drive;
t 3 \u003d 0.05 s - time of deceleration of deceleration;
j \u003d 2.9 m / s 2 - established slowdown;
V \u003d 60 km / h - Gas-3102 car speed.
Example number 10.
Determine the time of braking car VAZ-2107 at a speed of 60 km / h. Road and Technical Conditions: Screen Snow, the time of delaying the triggering of the brake drive - 0.1 s, slowing down the rise time is 0.15 ° C, the clutch coefficient is 0.3.
DECISION:
In the current road transport situation, the braking time of the VAZ-2107 car is 5.92 s:
The formula is indicated:
t 2 \u003d 0.1 s is the retreating time of the brake drive.
t 3 \u003d 0.15 s is the rise of the deceleration.
V \u003d 60 km / h - vehicle speed VAZ-2107.
j \u003d 2.9 m / s 2 - Destination of the car VAZ-2107.
Example number 11.
Determine the movement of the KAMAZ-5410 car in the inverted state at a speed of 60 km / h. Road and Specifications: Loading - 50%, wet asphalt concrete, clutch coefficient - 0.5.
DECISION:
In the current traffic situation, the movement of the KAMAZ-5410 car in the inverted state is approximately 28 m:
j \u003d g * φ \u003d 9.81 * 0.50 \u003d 4.9 m / s 2
The formula is indicated:
j \u003d 4.9 m / s 2 - established slowdown;
V \u003d 60 km / h - the speed of movement of the car KAMAZ-5410.
Example number 12.
On the road, 4.5 m wide occurred a counter collision of two cars - freight ZIL130-76 and a passenger gas-3110 "Volga", as established by the consequence, the speed of the truck was about 15 m / s, a passenger - 25 m / s.
When inspection of the accident site, brake trails are fixed. Rear tires truck Left Track Uza 16 m long, rear tires a passenger car - 22 m. As a result of the investigative experiment, it was established that at the moment when each drivers had technical opportunity Detect a counter car and evaluate the road atmosphere as a dangerous, the distance between cars was about 200 m. In this case, the cargo car was from the collision site at a distance of about 80 m, and the passenger-120 m.
Set the presence of a technical ability to prevent car clashes from each drivers.
For the study adopted:
For car ZIL-130-76:
For GAZ-3110 car:
DECISION:
1. Stopping car path:
freight
Passenger
2. The condition for the possibility of preventing the collision awarded driver response to the obstacle:
We check this condition:
The condition is performed, therefore, if both drivers correctly appreciated the created road situation and at the same time took the right decision, then the collision would have been avoided. After stopping cars between them, there would be a distance s \u003d 200 - 142 \u003d 58 m.
3. The speed of cars at the time of the start of complete braking:
freight
passenger
4. The path traveled by cars by the NTZ (Pattolation):
freight
passenger
5. Moving cars from the collision site in the inverted state in the absence of a collision:
freight
passenger
6. The ability to prevent collisions from drivers in the created setting: for a truck
Condition is not performed. Consequently, the driver of the car ZIL-130-76, even with a timely response to the emergence of the GAZ-3110 car, did not have the technical ability to prevent a collision.
for a passenger car
The condition is performed. Consequently, the GAZ-3110 car driver with a timely response to the appearance of the ZIL-130-76 car had a technical opportunity to prevent collision.
Output. Both drivers were inexplicitly reacted to the appearance of danger and both slowed down with some delay. (S "Y d \u003d 80 m\u003e s" o \u003d 49.5 m: s "y d \u003d 120 m\u003e s" o \u003d 92.5 m). However, only the car-3110 passenger car in the created setting has an opportunity to prevent collision.
Example 13.
The LAZ-697N bus, which was moving at a speed of 15 m / s, was shot down a pedestrian, which went with a speed of 1.5 m / s. Pedestrian hit is applied to the front of the bus. Pedestrian managed to pass through the length of the bus movement of 1.5 m. Full movement of a pedestrian 7.0 m. The width of the roadway in the accident area is 9.0 m. Determine the ability to prevent the pedestrian on a pedestrian by tracing a pedestrian or emergency braking.
For the study adopted:
DECISION:
We will check the possibility of preventing a pedestrian by a pedestrian in front and rear, as well as emergency braking.
1. The minimum safe interval during a pedestrian
2. Width of the dynamic corridor
3. The coefficient of maneuver
4. The condition of the performance of the maneuver, taking into account the road situation during a pedestrian:
rear
in front
Traveling a pedestrian is possible only from behind (from the back of the back).
5. Cross offset of the bus required for a pedestrian side of the back:
6. In fact, the required longitudinal movement of the bus for its displacement to the side by 2.0 m
7. Removing the car from the location of the pedestrian at the time of the dangerous situation
6. Condition of a safe pedestrian:
The condition is performed, therefore, the bus driver had a technical opportunity to prevent a pedestrian from hitting his backside.
7. Length of the bus stop
As S. UD \u003d 70 m\u003e S o \u003d 37, b m, the safety of pedestrian transition could also be provided by emergency braking of the bus.
Conclusion. The lifeline of the bus had a technical opportunity to prevent hitting on a pedestrian:
a) by tracing a pedestrian from the back of the back (with an unchanged speed of the bus);
b) by emergency braking from the moment the pedestrian movement on the carriageway.
Example 14.
The car brand ZIL-4331 as a result of damage to the front left wheel tire suddenly drove on the left side of the roadway of the road, where the frontal collision was happening with the GAZ-3110 counter car. Drivers of both cars in order to avoid collisions were used inhibition.
The question of the expert was raised by the question: whether they had a technical opportunity to prevent collision by braking.
Initial data:
- driving part - asphalt, wet, horizontal profile;
- the distance from the place of collision to the beginning of the rotation of the car ZIL-164 left - S \u003d 56 m;
- Length of the braking trace from the rear wheels of the car GAZ-3110 - \u003d 22.5 m;
- the length of the braking trace of the car ZIL-4331 to the blow - \u003d 10.8 m;
- The length of the braking trace of the car ZIL-4331 after impact until the complete stop - \u003d 3 m;
- The speed of movement of the car ZIL-4331 in front of the incident -v 2 \u003d 50 km / h, the vehicle speed of the gas-3110 is not installed.
The expert adopted the following values \u200b\u200bof the technical values \u200b\u200brequired for the calculations:
- slowing down cars in emergency braking - j \u003d 4m / s 2;
- the time of the driver's reaction - T 1 \u003d 0.8 s;
- the time of delaying the operation of the brake drive of the car GAZ-3110 - T 2-1 \u003d 0.1 C, the car ZIL-4331 - T 2-2 \u003d 0.3 C;
- the increase in the growth of the car GAZ-3110 - T 3-1 \u003d 0.2 C, the car ZIL-4331 T 3-2 \u003d 0.6 s;
- weight of the car GAZ-3110 - G 1 \u003d 1.9 t, the weight of the car ZIL-4331 - G 2 \u003d 8.5 tons.
