Forces acting on the car
Brake car
Stability car
Car handling
Patency car
The car moves at a certain rate as a result of the action on it of the driving forces and forces that have resistance to movement (Fig. 1).
To the forces that impede the movement of the car include: the strengths of resistance to rolling PF. , Resistance created by the rise of the road RA , air resistance Pw. , resistance inertia forces RJ. . To overcome these forces, the car is equipped with an energy source - engine. The torque resulting as a result of the engine is transmitted through the power transmission and the semi-axis on the drive's drive wheels. Their rotation prevents the friction force, which appears between the wheels and the surface of the road.
During rotation, the drive wheels create circumferential forces that act on the road, seeking to push it back. The road, in turn, has an equal opposition (tangent reaction) on the wheels, which causes the movement of the car.
The force that leads the car in motion is called the force of thrust and denote the pH. The relationship between these quantities or the limiting condition of the car movement in which the balance between the force of thrust and the forces of the movement is ensured, can be expressed by the formula
PK \u003d PF ± PA + PW + PJ.
This equation is called equation of traction balanceand allows you to establish how the thrust force is distributed by various types Resisters.
Resistance expensive
Resistance to tires along the road is a consequence of energy costs for hysteresis (internal) losses in the tire and on the formation of a gauge (external) loss. In addition, part of the energy is lost as a result of surface friction of the tires about the road, resistance in the bearings of the driven wheels and air resistance to the wheels. Due to the complexity of accounting for all factors, the resistance to rolling the wheels of the car is estimated by total costs, considering the power of resistance to rolling external with respect to the car. When rolling elastic wheels on a solid road, external losses are insignificant. The layers of the bottom of the tire are compressible, then stretch. A friction occurs between individual particles of the tire, the heat is allocated, which is dissipated, and the work spent on the deformation of the tire does not return completely upon subsequent restoration of the tire shape. When rolling the elastic wheel of the deformation in the front of the tire increase, and in the rear - decrease.
When the rigid wheel rolls on a soft deformable road (soil, snow), the losses for the tire deformation are practically absent and the energy is spent only for the deformation of the road. The wheel is crashed into the ground, squeezes it to the side, sprinkling individual particles, forming a rut.
When the deformable wheel rolls on a soft road, the energy is spent on overcoming both internal and external losses.
When rolling the elastic wheel on the soft road, its deformation is less than when rolling through a solid road, and the deformation of the soil is less than when rolling hard on the same soil.
The value of the resistance force rolling can be determined from the formula
PF \u003d GF COS A,
PF - the strength of resistance to rolling;
G - car weight;
a is an angle that characterizes the lifting or descent circuit;
f - The coefficient of resistance to rolling, which takes into account the effect of tire deformation forces and coating, as well as friction between them in various road conditions.
The magnitude of the rolling resistance coefficient ranges from 0.012 (asphalt concrete coating) to 0.3 (dry sand).
Fig. 1. Forces acting on a moving car
Resistance to rise. Automobile roads consist of alternating rates and descents and extremely rarely have horizontal portions of large length. The increases of lifting characterize the value of an angle A (in degrees) or the values \u200b\u200bof the liner of the road T, which is the ratio of exceeding H to the embedding in (see Fig. 1):
i \u003d H / B \u003d TG a.
The weight of the car G, moving on the rise, can be decomposed on the two-components: G Sina, directed parallel to the road, and Gcosa, perpendicular to the road. The force G sin A is called the strength of resistance to the rise and denote Ra.
On the roads with a solid coating, the angles of lifting are small and not exceed 4 - 5 °. For such small corners can be considered
i \u003d TG A ~ SIN A, then Ra - G Sin A \u003d Gi.
When moving on the descent, the power of the RA has the opposite direction and acts as a driving force. The angle A and the bias are considered positive on the rise and negative when moving on the descent.
Modern roads do not have clearly pronounced areas with a constant slope; Their longitudinal profile has smooth outlines. On such roads, the bias and strength p are constantly changing during the movement of the car.
Resistance irregularities.No road coating is absolutely even. Even new cement concrete and asphalt concrete coatings have irregularities up to 1 cm high. Under the action of dynamic loads of irregularities rapidly increase, reducing the speed of the car, reducing its service life and increasing fuel consumption. Orthodiy create additional resistance to movement.
When the wheel hit in a long depression it hits its bottom and throws up. After a strong impact, the wheel can be separated from the coating and hit again (already with a smaller height), making decaying oscillations. Moving through short depressions and protrusions are associated with additional tire deformation under the action of force that occurs when the ledge of irregularities. Thus, the movement of the car on the irregularities of the road is accompanied by continuous blows of the wheels and the oscillations of the axes and the body. As a result, an additional dispersion of energy in the tire and details of the suspension, which is sometimes achieving significant values.
Additional resistance caused by road accidents, take into account conditionally increasing the rolling resistance coefficient.
The values \u200b\u200bof the coefficient of resistance to rolling f and slope I in the aggregate characterize the quality of the road. So often talk about the power of resistance is expensive P, equal to the amount of the forces of PF and RA:
P \u003d pf -f RA \u003d G (F cos a -f sin a) ~ G (f + i).
The expression standing in brackets is called the coefficient of resistance is expensiveand denote the letter F. Then the power of the road
P \u003d g (f cos a -f sin a) \u003d g f.
Windage.When driving a car, resistance and air environment has resistance. Power costs to overcome air resistance add up of the following quantities:
The windshield appears as a result of the pressure difference in front and behind the moving car (about 55 - 60% of the total air resistance);
Resistance created by protruding parts: steps, wings, license plate (12 - 18%);
Resistance arising from the passage of air through the radiator and podcast space (10-15%);
Friction of outer surfaces on nearby air layers (8 - 10%);
Resistance caused by the pressure difference from above and from the bottom of the car (5-8%).
With an increase in the speed of movement, air resistance increases.
Trailers cause an increase in the strength of the resistance of the air due to a significant densure of air flows between the tractor and the trailer, as well as due to the increase in the outer surface of the friction. On average, it can be assumed that the use of each trailer increases this resistance by 25% compared with a single car.
Power inertia
In addition to the strength of the road and air, the effect on the movement of the car is inertia p). Any change in the speed of movement is accompanied by overcoming the inertia strength, and its value is the greater, the larger the total car, the car:
The time of the uniform movement of the car is usually not enough compared to the overall time of its work. So, for example, when working in cities, cars are moving evenly 15 - 25% of the time. From 30% to 45% of time occupies an accelerated movement of the car and 30 - 40% - rolling and braking movement. When touching and increasing speed, the car moves with acceleration - its speed is uneven. Than faster car Increases speed, the greater the acceleration of the car. Acceleration shows how the speed of the car increases over each second. Almost the acceleration of the car reaches 1 - 2 m / s2. This means that over each second the speed will increase by 1 - 2 m / s.
The power of inertia changes in the process of moving the car in accordance with the change in acceleration. To overcome the inertia strength, part of the traction force is consumed. However, in cases where the car moves rolling after pre-overclocking or in braking, the inertia force acts in the direction of the vehicle movement, performing the role driving force. Taking this into account, some of the difficult sections of the path can be overcome with the pre-acceleration of the car.
The value of the resistance force overclocking depends on the acceleration of the movement. The faster the car accelerates, the greater the power becomes. Its value is changing even when starting from the place. If the car goes smoothly, then this power is almost absent, and with a sharp touch it can even exceed the traction force. This will lead or stop the vehicle, or to the wheels (in case of insufficient clutch coefficient).
During the operation of the car, the conditions of movement are continuously changing: the type and condition of the coating, the size and direction of the slopes, the strength and direction of the wind. This leads to a change in the velocity of the car. Even in the most favorable conditions (movement on improved motorways outside the cities and settlements) The speed of the car and the thrust force are rarely remained unchanged in, the flow of a long time. On the average. The speed of movement (defined as the attitude of the path traveled to the time spent on the passage of this path, taking into account the time of stopping time in the way) affects the effect of resistance the impact of a very large number of factors. These include: the width of the roadway, the intensity of the movement, the illumination of the road, the meteorological conditions (fog, rain), the presence of dangerous zones (railway moving, cluster pedestrians), car condition, etc.
In difficult road conditions, it can happen that the sum of all resistance forces will exceed traction, then the movement of the car will be slowed down and it can stop if the driver does not accept the necessary measures.
Car Wheel Clutch with Dear
In order for a fixed car to lead, one traction is not enough. More friction is needed between wheels and expensive. In other words, the car can move only with the clutch of the leading wheels with the surface of the road. In turn, the clutch force depends on the coupling weight of the car GV, i.e. vertical load on the drive wheels. The greater the vertical load, the more clutch power:
PCC \u003d fgk,
where the PCC is the force of the clutch of the wheels with the road, kgf; F - clutch coefficient; GK - Coupling, kgf. Condition of Moveless Wheels
RK< Рсц,
i.e. if the load force is less clutch, the leading wheel rolls without stopping. If the leading force is applied to the drive wheels, which is large than the power of the clutch, then the car can move only with the slip of the leading wheels.
The clutch coefficient depends on the type and condition of the coating. On the roads with a solid coating, the magnitude of the clutch coefficient is due to the friction of the slip between the tire and expensive and the interaction of the particles of the tread and the impacting of the coating. When the solid coating is wetted, the clutch coefficient decreases very noticeably, which is explained by the formation of a film from a layer of particles of soil and water. The film shares the rubbing surfaces, weakening the interaction of the tire and coating and reducing the clutch coefficient. When tire slides on the road in the contact zone, it is possible to form elementary hydrodynamic wedges, causing lifting tire elements over coating microwaves. The direct contact of the tires and roads in these places is replaced with liquid friction, in which the clutch coefficient is minimal.
On deformable roads, the clutch coefficient depends on the resistance of the soil of the slice and the internal friction magnitude in the ground. The protector protector of the driving wheel, plunging into the ground, deform and compact it, which causes an increase in the resistance of the cut. However, after a certain limit, the destruction of the soil begins, and the clutch coefficient decreases.
The magnitude of the clutch coefficient also affects the tire tread pattern. Tires of passenger cars have a protector with a small pattern, providing good adhesion on solid coatings. Tires trucks There are a large drawing of the tread with wide and high protrusions of the primer. During the movement, the soils are cut into the ground, improving the cargo of the car. Abrasion of protrusions in the process of operation worsens the tire clutch with the road.
With an increase in the internal pressure in the bus, the clutch coefficient is in charge, and then decreases. The maximum clutch coefficient corresponds to an approximately the pressure value recommended for this tire.
With full slide of the tires on the road (buxation of leading wheels or use of the braking wheels) the value of F may be 10 - 25% less than the maximum. The cross-clutch coefficient depends on the same factors, and it is usually taken equal to 0.7f. The average values \u200b\u200bof the clutch coefficient fluctuate in a wide range from 0.1 (iced coating) to 0.8 (dry asphalt- and cement concrete coating).
The tire clutch with the road is of paramount importance for the safety of motion, as it limits the possibility of intensive braking and the steady movement of the car without transverse slip.
The insufficient magnitude of the clutch coefficient is the cause of an average of 16%, and in unfavorable periods of the year - up to 70% of road accidents from the total number of them. International Commission on Combating Gliziness road coatings It has been established that the magnitude of the clutch coefficient under the safety conditions should not be less than 0.4.
Brake car
Reliable I. effective brakes Allow the driver to confidently lead the car at high speed and at the same time ensure the necessary safety of the movement.
In the process of braking, the kinetic energy of the car goes into the work of friction between the friction pads of the pads and brake drums, as well as between tires and expensive (Fig. 2).
The magnitude of the braking torque developed by the brake mechanism depends on its design and pressure in the drive. For the most common types of brake drives, hydraulic and pneumatic, pressing the plug on the block is directly proportional to the pressure developed in the drive when braking.
Torkemose modern cars Can develop a moment, significantly higher than the moment of tire clutch force with expensive. Therefore, it is quite often in practice that you are observed by the SMU when with intensive braking the wheel of the car is blocked and slide on the road without rotating. Prior to blocking the wheel between the brake linings and the drums, the grinding force is applied, and in the contact zone of the tire with the road - the force of friction of rest. After blocking, on the contrary, the brake friction will act between the driving surfaces of the brake, and in the contact zone of the tire with an expensive - the slide friction force. When blocking the wheel, the costs of friction energy in the brake and on rolling are stopped and almost all heat, equivalent to the absorbed kinetic energy of the car stands out at the point of contact of the tire with an expensive. Increased tire temperature leads to a softening of rubber and reduce the clutch coefficient. Therefore, the greatest braking efficiency is achieved in the case of rolling the wheel at the lock limit.
With simultaneous braking by the engine and brakes, the achievement of the magnitude of the clutch force on the driving wheels occurs with a smaller power of pressing the pedal than when braking only brakes. Long-term braking (for example, while driving on started descents) As a result of the heating of the brake drums, sharply reduces the friction coefficient of friction linings, and consequently, the braking moment. Thus, braking with an indispensable engine, used as an additional way to reduce speed, allows you to increase the life of the brakes. In addition, when braking with an incompanied engine increases transverse stability car.
Fig. 2. Forces acting on the car wheel when braking
There are emergency and service braking.
Serviceit is called braking to stop the car or reducing the speed of movement in a pre-assigned driver. The reduction in the speed in this case is carried out smoothly, more often by combined braking.
Emergencyit is called braking, which is made in order to prevent the departure to an unexpected or noticed obstacle (subject, car, pedestrian, etc.). This braking can be characterized by the stopping path and the braking car.
Under stopping pathunderstand the distance that conducted a car From the moment the driver's danger detects until the car is stopped.
Brake paththey call a part of the stopping path, which will pass the car from the moment of starting the braking of the wheels until the car stops.
The total time T0 required to stop the car from the time of the obstacle ("stopping time") can be represented as a sum of several components:
t0 \u003d \u200b\u200bTR + TPR + TU + TT,
where Tp is the driver's response time, C;
tPR - time between the beginning of clicking on the brake pedal and the beginning of the brakes, C;
tU - time to increase the deceleration, C;
tT - Time of complete braking, p.
Amount tNP + TY. It is often called the timing time of the brake drive.
The car during each of the components of the time intervals passes a certain path, and their sum is the stopping path (Fig. 3):
S0 \u003d s1 + s2 + s3, m,
where S1, S2, S3 is respectively the journey through the car during TR, TPR + TU, TT.
During the TR, the driver is aware of the need to brake and transfers his leg with the fuel supply pedal to the brake pedal. Tier time depends on the qualification of the driver, its-truck, fatigue and other subjective factors. It varies from 0.2 to 1.5 s or more. When calculating, TR \u003d 0.8 s is usually taken.
TNP time is necessary for choosing gaps and moving all drive parts (pedals, brake cylinder pistons or brake chamber diaphragms, brake pads). This time depends on the design of the brake drive and its technical condition.
Fig. 3. Brake path and car safety distance
On average for a good hydraulic drive You can take TPP \u003d 0.2 C, and for pneumatic - 0.6 s, in road trails with a pneumatic drive brake, the TPR can reach 2 s. The segment Tu characterizes the time to gradually increase the deceleration from zero (the beginning of the brakes) to the maximum value. This time is an average of 0.5 s.
During the TP + TPP time, the car moves evenly with the initial velocity Va. During TU, the speed is somewhat reduced. During the time TT, the slowdown is saved by approximately constant. At the time of stopping the car, the slowdown decreases to zero almost instantly.
Stopping the path of the car without taking into account the resistance force of the road can be determined by the formula
S \u003d (T * V0 / 3.6) + KE (VA2 / 254FX)
where S0 is a stopping path, m;
Va - vehicle speed at the initial moment of braking, km / h;
kE is a braking efficiency coefficient that shows how many times the actual slowing down the car is less theoretical, as possible on this road. For passenger cars Ke ~ 1.2, for trucks and buses KE ~ 1.3 - 1.4;
FX - tire clutch coefficient with expensive,
t \u003d TR + TPR + 0.5TU.
The expression is Ke \u003d V2 / (254 Wow) - represents the braking path, the value of which, as can be seen from the formula, is proportional to the square of the velocity with which the car moved before the start of braking. Therefore, with an increase in the speed of movement twice, for example, from 20 to 40 km / h, the brake path will increase by 4 times.
The efficiency standards of the foot brake of cars under operating conditions are shown in Table. 1 (initial braking speed of 30 km / h).
When braking on snow and slippery roads, the brake forces of all vehicles reach the clutch force values \u200b\u200balmost simultaneously. Therefore, at FC<0,4 следует принимать кэ= 1 для всех автомобилей.
It is known that to ensure the movement, the traction effort should be greater than the total resistance to the movement of the car.
The horizontal power of the Republic of Kazakhstan (traction force), which arises from the action on the rotational moment of the MVR in the zone of its contact with the coating, is directed towards, inverse movement (see Fig. 5.1).
The power of the Republic of Kazakhstan causes the horizontal force of the reaction T, which represents the force of friction (clutch) of the wheel coated in the zone of their interaction, while T \u003d RK.
Fig.5.1. The condition of the possible movement of the car
But the wheel has to overcome even rolling resistance. The strength of resistance to rolling PF is determined by a certain dependence:
where GK is an effort transmitted to leading wheel, Gk \u003d (0.65: 0.7) G - for trucks and (0.5: 0.55) G - for passenger, where G is the weight of the car; - Round resistance coefficient.
where but - distance from the vertical axis of the wheel to the location of the reaction R on the weight of the GK transmitted by the wheel; - rolling radius of a pneumatic wheel; \u003d λ * R, where R is the radius of an undeformed wheel, λ is the coefficient of reducing the radius of the wheel depending on the cruelty of the tires (λ \u003d 0.93 - 0.96).
It has been established that the almost value remains constant to the speed V \u003d 50 km / h and is depending on the type of coating in the range \u003d (0.01-0.06). With increasing speed increases, because When the wheel is on the irregularity of the kinetic energy, directly proportional to V², is spent in much greater staining to overcome these obstacles.
At v\u003e 50 km / h F is determined by the dependence
V-,
where is the coefficient of rolling resistance at V to 50 km / h.
Using the positions of theoretical mechanics and rice. 5.1, you can write: T \u003d RK -
T \u003d RK - T \u003d RK - (5.4)
Obviously, the movement of the car is possible at T\u003e RK.
The greatest value of the friction force, and therefore the traction force is determined by the dependence of the TMAH \u003d φ ∙ GSz, where φ is the clutch coefficient; GCC Coupling car weight transmitted to the drive wheel.
Naturally, the friction force (clutch) reaches the greatest value (with the same coupling weight transmitted by the wheel) at the maximum value of the clutch coefficient φ.
The clutch coefficient is a variable value and depends on many factors (the state of the carriage part, braking mode, the presence of lateral forces, tire pressure, tread pattern, speed, etc.). φ varies in wide limits (φ \u003d 0.1-0.7) and therefore it can only be considered to be considered as a parameter uniquely characterizing the coating.
The maximum possible value of φmax leading wheels with a coating under these conditions corresponds to the time preceding the beginning of their buxation, and braking wheels - the transition from braking brake pads to brake on the drum to slip on the coating of blocked wheels by the USE.
The coefficient of longitudinal clutch φ1 is distinguished, corresponding to the beginning of slipping or driving the wheel when rolling or braking without lateral power yk; and the transverse clutch coefficient φ2 is the transverse component of the clutch coefficient, which occurs when the rolling wheel of the driving wheel occurs at an angle to the motion plane under the influence of the sideways YK, when the wheel, rotating, slides.
The transverse clutch coefficient of φ2 is used to assess the resistance of cars against driving when moving horizontal curves when a transverse centrifugal force acts onto the car; φ2≈ (0.85-0.9) φ1.
The clutch coefficient is the most important characteristic Transport and operational qualities car expensive. From φ, it depends not only the possibility of selling the car's traction force, but also the resistance of the car against the drift on the curves, the possibility of timely stop of the car in front of the obstacle or pedestrian. Insufficient tire clutch with coated wheel is often the root cause of road accidents (accidents). It has been established that increasing the clutch coefficient by 2 times allows to reduce the number of accidents by 1.5 times.
Many factors affect the values \u200b\u200bof clutch coefficients. It has been established that the condition of the road surface has a larger effect on the clutch coefficient value than its type. It's related
so that in ideal conditions, under any coatings, solid protrusions of mineral particles are pressed into the tire and therefore the wheel can slip mainly as a result of the deformation of the tread rubber.
As coat wear wear, their roughness decreases, and therefore, their clutch with the wheel decreases. The clutch coefficient is most resistant to cement concrete coatings in a dry state with the duration of their service to 10-12 years, asphalt concrete - 5-8 years. With wear (erasing) of coatings on a 50-60% clutch coefficient decreases by 30-40%. In other words, over time, the clutch coefficient decreases.
The clutch coefficient depends: from the material from which the tire is made (the largest clutch coefficient is provided with tires made of high-hydraese rubber); type of tire tread pattern (on a wet tire coating with a tread pattern having greater dismemberment, provide a higher clutch coefficient); The degree of wear of the tire tread (with the complete abrasion of the tread pattern, the clutch coefficient is reduced by 35-45%, and about 20-25% on wet and dirty coatings).
The clutch coefficient is reduced due to the presence of dirt, dust, tire wear products, etc., for them are filled with the surfaces of the surfaces of the tire protectors, which reduces their roughness.
Studies have shown that the clutch coefficient decreases with increasing speed. This is due to the fact that at high speeds of movement of the tire does not have time to fully deform, since the duration of contact with the coating is not sufficient for this, and therefore, the irregularities of the coating are pressed into the bus to a smaller depth. On dry coatings, a decrease in the clutch coefficient with an increase in speed is less noticeable.
The moisture, wetting the contact zone between the bus and the coating, acts as a lubricant separating rough surfaces (coatings and wheels), reducing the clutch coefficient. With a layer of water on a coating of a thickness of several millimeters and strong wear of tires and speeds close to 100 km / h, an aquaplaning phenomenon may occur when the water wedge generated between the tire and coating, creating hydrodynamic lifting force, dramatically reduces the wheel pressure on the road, as a result This contact front wheels with a coated can be completely stopped with a loss of car manageability.
If there is dirt on the coating, etc. φ changes greatly during the rain. In the first period of rain, a relatively thick dirt film is formed, which plays the role of lubricant that reduces the clutch coefficient. Gradually, the lubricant is dodged, partially washed off with the rain and the clutch coefficient begins to increase, nevertheless, without reaching the value φ on a dry coating.
In general, the clutch coefficient varies widely within a year due to the change in climatic conditions. Naturally, φ is the highest summer and decreases in winter. Therefore B. winter We carry out various events that increase the clutch coefficient (cleaning of road surfaces from snow, ice, elimination of ice and sliding coatings by sprinkling sand, slags, antifungal mixtures, etc.).
With a uniform movement, there is no acceleration, therefore, the dynamic factor in the type D is equal to the coefficient of the total resistance of the road ψ, that is, d \u003d ψ \u003d f to + i.
That is, using the dynamic characteristic with a well-known coefficient of resistance to rolling wheels F K, you can find the amount of overcome lifting i.with uniform movement of a car with full load.
According to the task ψ \u003d 0.082, when moving along the road V category, we accept f to \u003d 0.03.
Then for a uniform movement, the magnitude of the limit angle of lifting:
α max \u003d arctg (D MAX - F K), hail.
The calculations for this formula are carried out without taking into account the action on the car forces of the aerodynamic resistance, since when overcoming the maximum possible lifting, the vehicle speed is not large.
| Kamaz | Mercedes. | |
| Dmax | 0,489 | 0,435 |
| FK. | 0,03 | 0,03 |
| α |
WITHOUT DRAWING MOVEMENT POSSIBLE PERFORMING Conditions:
D C \u003d a ∙ φ x ∙ cos α max / (L-HD ∙ (φ x + f k)) ≥ D max.
D C - dynamic clutch factor
a- Distance from the center of the masses to the rear axle of the car
α MAX - extreme angle of overcome lifting
L- Wheel base of the car (because Kamaz 6 * 4 wheel formula, then for L take the distance from the front axle to the balance axis)
HD- height of the center of gravity
f K - rolling resistance coefficient
HD \u003d 1/3 * HD, where a HD-overall height
a \u003d m 2 / m a * l, where m 2 -t car coming on rear axle (rear trolley), M a - Full weight car.
According to the task of the clutch coefficient with the road φ x \u003d 0.2. For the car KAMAZ:
a \u003d 125000/19350 * 3.85 \u003d 2.48m
HD \u003d 1/3 * 2,960 \u003d 0.99
D C \u003d 2.48 * 0.2 * COS 25 ° / (3.85-0.99 * (0.2 + 0.03)) \u003d 0.124< D max = 0,489.
For car Mercedes.:
A \u003d 115000/200000 * 4.2 \u003d 2.42m
HD \u003d 1/3 * 2,938 \u003d 0.98m
D C \u003d 2.42 * 0.2 * COS 22 ° / (4.2-0.98 (0.2 + 0.03)) \u003d 0.113 Turning to the dynamic passport of the car, we will see that because D SC Conclusion: at a given value φ x \u003d 0.2 on the road with limit angles of lifting and full load, cars move with slipping leading wheels. The calculation in this course work of the limit angles of the car's overcome lifts allows us to conclude that the magnitude of these angles depends, first of all, from three factors: the mass of the car, the values \u200b\u200bof the traction force and the magnitude of the coefficient of resistance to the rolling of the wheels. 10. Determination of the limit force of thrust on the hook on all transmissions and checking the possibility of motion subject to the bounce on the road ψ \u003d 0.11and φ x \u003d 0.6The definition of the lowest transmission on the kitter will move without stopping on the specified road. The force of thrust on the hook characterizes the ability of the car to towing trailed links. The magnitude of the limit force of the thrust on the hook of the car is determined by the formula: where - the limit force of thrust on the hook, n; - maximum load force on the transfer, H; - air resistance force corresponding to the movement mode with maximum traction force, H; - power of general road resistance, N. To check the possibility of moving the car by condition, it is necessary to determine the force of the clutch of the leading wheels with the road and compare the value obtained with the limit value of the thrust force on the hook for each transmission. P.cc \u003d m 2 ∙ l ∙ φ x / (A-HD ∙ (φ x + F K)) - the force of the grip. Example of calculation for the car KAMAZ: 1 Transfer: 84,147kn; \u003d 0.007kn; \u003d 28.5kn. 84,147-0.007-28.5 \u003d 55.64kn 2 Transmission: 43,365kn; \u003d 0.0254kn; \u003d 28.5kn. 43,365-0.0254-28.5 \u003d 14.84kn 3 Transfer: 35.402kn; \u003d 0.0382kn; \u003d 28.5kn. 35.402-0.0382-28.5 \u003d 6,86kn P.cc \u003d 125000 * 3.85 * 0.6 / (2.48-0.98 * (0.6 + 0.02)) \u003d 151.1kn Example of calculating for Mercedes car: 1 Transfer: 97,823kn; \u003d 0.005kn; \u003d 29.43kn. 97,823-0.005-29,43 \u003d 68,388kn 2 Transmission: 55,59kn; \u003d 0.0169kn; \u003d 29.43kn. 55,59kn -0.0169-29,43 \u003d 26,14kn 3 Transfer: 33,491kn; \u003d 0.0464kn; \u003d 29.43kn. 33,491-0.0464-29,43 \u003d 4.01kn P.sc \u003d 115000 * 4.2 * 0.6 / (2.42-0.98 * (0.6 + 0.02)) \u003d 159.9kn Based on the fact that on any gears, it can be said that when the car is moving, there is no slipping of the leading wheels. Comparative table of obtained estimated parameters of traction and high-speed properties, imprisonment. Conclusion: In this section, a study of the traction and speed properties of two cars was almost the same power. Despite the fact that the Mercedes engine has the same power, and the Mercedes car itself, as a whole, is heavier, a high moment on medium-sized turns and an increased gear ratio of transmissions allow it to surpass the car KAMAZ in traction properties and developed effort on the hook. KAMAZ car has more maximum speed, deal. In turn, the car, Mercedes is able to overcome sharp rise, which makes it indispensable in difficult areas. Physical processes in the spack of contact of the leading tractor and car wheel with the same way. However, in contrast to the car, the tractor is a traction machine. The tractor wheel is loaded with a large leading moment than the car, and works on agricultural backgrounds that differ significantly from road conditions. Therefore, the process of bucking of the tractor wheel is the norm, and not an exception. During the rotation of the wheel to the angle of βk in the absence of deformations of the crumpled and soil shift, the path traveled by the wheel should be equal to the distance LP between the soils. However, due to the deformation of the soil, the real path of SP is less theoretical on ΔSmax. The wheel axis along with the movement forward as it should move back (to the side opposite to its movement) by value equal to the deformation of the soil shift ΔSmax under the last soil. This is the physical essence of the bucking: δ \u003d (ln-Sn) / Ln \u003d ΔSmax / ln .. The bucking (as a kinematic factor) is estimated by the bucking coefficient, which is determined as the ratio of the rate of reduction of speed to its possible theoretical value in% or fractions: Δ \u003d (VT - VK) / VT or VK \u003d VT (1-Δ), where VT, Vc- Theoretical and valid speed of the transit movement of the wheel. Efficiency of the bucks ηδ: ηδ \u003d VK / VT; Δ \u003d (Vt- VK) / VT \u003d 1- ηδ. Theoretically, the stopping occurs with the beginning of the tractor movement, when the leading moment appears on the wheel and the tangent force PK. The experimental definition of the bucking of the tractor proposals is that on the measuring field of the field to compare the total number of revolutions of the leading wheels when the tractor moves at idle the NK and under the load of the NK. The load on the hook should be set step from the minimum value to the value at which there is an intensive bucking of the wheels. Since the path in all cases is the same, then the bucking can be found from the ratio of the total numbers of revolutions of the leading wheels when the tractor is moved without load and with the load on the hook, i.e.δ \u003d (1- NK / NK) 100%. The number of revolutions of the leading wheels is measured in the process of traction tests regulated by GOST 7057-81. Since the path passed in each experiment can be different, the formula for determining the bucking has the form δ \u003d 100%, where NK, N˝k.x - the total number of revolutions of the left and right leading wheels of the tractor when driving without load on SK SK; NK, N˝k is a total number of revolutions, respectively, the left and right leading wheels on the SV path when the tractor is moved under load. It should be noted that this method of determining the bucking is universally used as standard, incorrect. It adopted such assumptions: when driving without load, the driving wheel is missing; The radius of the leading wheels does not depend on the load on the tractor hook and other test conditions. However, the error of the adopted assumptions is small, therefore, with the operational assessment of the tractor, it is neglected.
Kamaz Mercedes.
External speed characteristic N e max \u003d 183kW (2100) M E max \u003d 989nm (1300) N E Max \u003d 180kW (2100) M E max \u003d 972nm (1100)
Conclusion: KAMAZ car is more powerful than Mercedes, which can be seen from an external high-speed characteristic, as well as he has a greater torque.
Traction and Power Balance Maximum load force in the car KAMAZ P T T MAX \u003d 84,147N. At the point where the schedule PT and (RD + RV intersects), i.e. RT \u003d RD + RV, the speed is maximum under these conditions of motion V Max MAZ \u003d 5.22m / s (on the third gear). Maximum load force at the MERCEDES P T T T Max \u003d 97,823n. At the point where the schedule PT and (RD + RV intersects), i.e. RT \u003d RD + RV, the speed is maximal under these motion conditions, V maxmerc \u003d 5.2 m / s (on the third gear).
Conclusion: based on graphs of traction and power balances, it can be noted that on the same gears when moving on the same speeds, the Mercedes car has a greater maximum traction and traction power, and a greater supply of traction force and power that can be used on Acceleration of the car, overcoming the forces of resistance to movement, towing the trailer, etc. .. Consequently, the car Mercedes has the best traction properties. This is also connected with the fact that the transmission efficiency has more from the Mercedes car, because this car has one leading bridge.
Dynamic passport D max \u003d 0.435 corresponding to it speed v \u003d 1,149m / s D max \u003d 0.489 corresponding speed V \u003d 1,029m / s
Conclusion: Dynamic factor in Mercedes a / m is larger than KAMAZ, because The load force is straightforward to him. The traction properties of the Mercedes car better than that of KamAZ because the Mercedes Mercedes is most overcome than that of KamAZ
Acceleration, time and dispersion path Maximum acceleration j a \u003d 0.638 m / s 2. Maximum acceleration j a \u003d 0.533 m / s 2
Time and way of overclocking on the way: 400m 1000m t \u003d 90 seconds t \u003d 205sek t \u003d 121sek t \u003d 226sek
Conclusion: Mercedes spent on overclocking longer than KAMAZ, because It accelerates slowly. The distance traveled during acceleration, the Mercedes is as bigger. So The pickup of the car KAMAZ is better than Mercedes. However, it is impossible to just judge what car has better richness, because Methods for determining parameters are approximate and can differ significantly from real data.
An extreme angle of lifting and checking the possibility of movement by condition Limit angle \u003d 25º Extreme angle of lifting \u003d 22º
Conclusion: Lifts overcome by cars in specified conditions are different. The maximum angle of overcome lifting from the car KAMAZ is more than that of Mercedes. When checking on the bar, we see that cars will move without slipping. Cars can move without stopping on this road at all speeds (which are used on the road of this category)
