Dynamic Characteristics Of Vehicle - Uni-lj.si
1. Dynamic characteristics of vehicles1. EXERCISEDynamic characteristics ofvehiclesVehicle dynamicsPrepared by: ass. Simon Oman, Ph.D.
1. Dynamic characteristics of vehiclesContents1.Task definition. 31.1 Example of vehicle data . 42.Theoretical background of the calculation . 62.1 Basic structure of the vehicle . 62.2 Motor . 62.3 Gearbox .82.4 Differential . 82.5 Wheels .82.6 Dynamic driving conditions . 82.6.1 Kinematics of the vehicle . 82.6.2 Driving resistances . 92.6.3 Traction forces on the wheels .152.6.4 Dynamic coefficient of a vehicle .182.6.5 Limiting slopes .192.6.6 Available power on wheels and power reserves (Power balance) .202.6.7 Vehicle acceleration .222.6.8 Acceleration time .232.6.9 Path taken during the acceleration .242.7 Instructions for numerical integration by Simpson .253.References .26
1. Dynamic characteristics of vehicles1.Task definitionSelect an arbitrary road vehicle with an internal combustion engine, for which you canfind the relevant input data in accessible literature. Create an application which mustinclude three essential elements: a data entry unit, a calculation unit, and an outputunit. If you want to avoid writing the program code, you can also solve the task withany of the programs, such as EXCEL. In this case the spreadsheet must contain allthree of the abovementioned elements.The report has to include the calculations and results of the following characteristics: diagram of the external characteristics of the engine PM(n) and MM(n), diagram of the vehicle traction forces FK(v) and resistances R(v), diagram of dynamic coefficient D(v), diagram of time, path and acceleration as a function of speed a(v), t(v) in s(v), diagram of power balance PK(v) together with needed power to overcomeindividual and collective resistance P(v)resistance, diagram of vehicle speed as a function of gear ratio and engine speed v(n), a diagram of limiting slopes α (v ) diagram of power supply P(v),the following has to be determined from the diagrams: maximum allowable slope α, maximum possible acceleration of the vehicle a, needed time to accelerate from 0 to 100 km/h, maximum speed of the vehicle v.For the best grade: when determining the limit values (acceleration, slope) alsoconsider possible slip of the tires.The report must show the resolution process with all of the equations you have used.Refer to sources. Together with the report it is necessary to deliver: application in electronic version. Application has to be user friendly.To simplify the accuracy control of your results, provide a table with all inputdata used in the report.
1. Dynamic characteristics of vehicles1.1Example of vehicle dataVehicle: Renault Twingo II 1.2Tab. 1: External characteristics of the enginerpmPowerTorque[1/min](corr) 0036,5063,405999,0032,3051,40Tab. 2: Weights and loadsEmpty vehicleMax. loadLoad of front axle6060 N8100 NLoad of rear axle4040 N5400 NTotal vehicle weight10100 N13500 N
1. Dynamic characteristics of vehiclesFigure 1: Vehicle sketch with main dimensionsTab. 3: Gear rations of the gearbox and differential plus efficiencydenota Gear ratiodenotationtioniIFirst gear3,73ηIiII2,05Second gearη IIiIIIThird gear1,39η IIIiIVFourth gear1,03η IViVFifth gear0,80ηVivzReverse gear3,55η vz/Differential gear ratio idif3,56Efficiency of other //η osttransmissionsOther data:- Rolling resistance factor - f 0,01- Tire dimensions 185/55 - R15- Cross section area of the vehicle - A 2,2m2- Drag coefficient of the vehicle - c /0,90
1. Dynamic characteristics of vehicles2.Theoretical background of the calculationThe purpose of this exercise is to address the dynamic driving conditions, such asdetermining the necessary power to overcome certain driving conditions, determiningacceleration, acceleration times, maximum speed etc.2.1Basic structure of the vehicleIn this exercise, we are interested in the dynamic characteristics of the vehicle.Therefore, we will not be concerned with the structure and properties of the bodyworkof the vehicle which apart from its weight and the drag coefficient has no majorinfluence on the dynamic properties of the vehicle. Attention will be paid to the driveassembly, the diagram of which is shown in Figure 2.Figure 2: Sketch of the vehicle drive assemblyThe vehicle is schematically illustrated by an engine that provides a certain torque ata given speed of the crankshaft. Torque is then transmitted over the clutch to thegearbox and then through the differential to the wheels.2.2MotorThe motor serves to convert the stored energy into mechanical rotational energy. Thestored energy can be in the form of chemical energy such as gasoline, gas oil, coal,hydrogen, etc., or in the form of electricity stored in batteries. Depending on the typeof energy stored, several types of motors are distinguished: internal combustion enginessteam enginegas turbineelectric motor
1. Dynamic characteristics of vehiclesInternal combustion engines have mostly been used as propulsion machines invehicles and electric motors have been increasing in recent years. In this exerciseswe will focus on vehicles with internal combustion engines.The engine as the propulsion machine in the vehicle should meet somerequirements: the output shaft of the engine should have a variable speed at each speed of the output shaft, it should have the same (maximum) outputpower.Today's internal combustion engines partially meet the first requirement, while theydo not meet the second one. The typical characteristic of an internal combustionengine is shown in Figure 0Močmotorjapower40P [kW]M [Nm]10020001000200030004000n [1/min]50006000700008000Figure 3: Example of the internal combustion engine external characteristicsInternal combustion engines are required to have a main usable area with as wide arange of crankshaft speeds as possible. The most up-to-date engines fulfil thisrequirement through modern technical solutions such as computer-controlled fuelinjection, changing the times of opening and closing valves, changing the lengths ofthe suction tubes, etc.The engine torque can be expressed depending on the engine power and enginespeed, which is indicated by eq. (1) [1].MM where: PM MM nMPMωM[W][Nm][min-1][ Nm] or M M 30 P Mπ nM[Nm]engine powerengine torquespeed of the crankshaft(1)
1. Dynamic characteristics of vehiclesωM 2.3[rad/s]angular velocity of the crankshaftGearboxUsually it is desirable to keep as much power on the wheels as possible. As alreadydescribed, today's engines do not allow this on a sufficiently large speed range (0 vmax). The demand for constant power on the drive wheels can be accomplished bythe use of a continuously variable transmission, but in practice, the stage gearbox isused in practice, which is a satisfactory solution. Used stage gearboxes usually havefive or more forward gears and one reverse gear. In the lower gears, this gearboxacts as a reducer, and in the higher (from 4th onwards), often as a multiplier with agear ratio ii.2.4DifferentialWhen traveling through the corner the speed of the inner drive wheel is smaller thanthe speed of the outer wheel due to different trajectory radii. Therefore the wheelsmust not be rigidly connected to one another. The torque is then transmitted from thegearbox to the wheels via the differential, which allows different revolutions of theouter and inner wheels. The differential unit also serves as a gearbox with a gearratio ikg.2.5WheelsWheels serve to transfer forces from vehicles to the road. The wheel consists of a rimand a tire. To determine the dynamics of a vehicle, it is important to know thedynamic and static radius (rd, rst) of the tire and the coefficient of friction between thetire and the road.When the data on the size of the dynamic radius of the wheel is not known (which inmost cases is a fact), eq. (2) can be used to roughly calculate it from the dimensionsof the tire.rst rd D[' '] 25,4[mm /' '] b[mm] x[%] 2100[%](2)where: 2.6rstrdDbx[mm] . static radius of the wheel[mm] . dynamic radius of the wheel[ ] . radius of the rim[mm] . tire width (section width)[%] . sidewall aspect ratioDynamic driving conditions2.6.1 Kinematics of the vehicleThe engine as a propulsion machine in the vehicle operates only in the specifiedspeed range of the crankshaft (nmin-nmax). Therefore the devices described in theprevious section are needed to overcome this issue. The speed of the vehicle istherefore dependent on the engine speed and the selected gear ratio in the gearbox.
1. Dynamic characteristics of vehiclesThe speed of the vehicle is determined by the Eq. (3) and the speed of the wheelswith the Eq. (4).v n k π rd30(3)where: v [m/s]. vehicle speednk [min-1] . rotational speed of the wheelsnk nmii i kg(4)where: ii [/] . current gear ratioikg [/] . gear ratios of the differential350300v onpath0010002000300040005000n [obr/min]Figure 4: Diagram of vehicle speed as a function of gear ratio and engine speedFigure 4 shows the vehicle speed as a function of used gear and speed of thecrankshaft. The "Acceleration path" curve shows the acceleration path where shiftbetween gears is performed at maximum crank shaft speed. In the case of properlycalculated gear ratios the acceleration path is always within the range determined bythe speed of the maximum torque and the maximum engine power.2.6.2 Driving resistancesEvery movement on the Earth's surface is subjected with the energy loss due todifferent resistances. In the case of linear movement of the vehicle, the followingdriving resistances shall appear:: Rf [N] . rolling resistance, Rs [N] . slope (hill) resistance, Rz [N] . air resistance (drag),
1. Dynamic characteristics of vehicles Ri [N] . resistance of mass inertia andRp [N] . resistance of trailer.2.6.2.1Rolling resistance RfThe rolling resistance acts in the contact area between the tires and the road surface.It is the result of the deformation of the tires and the road surface. In case of levelroad driving, Rf is calculated by equationRf f Zi f G(5)in the case of driving the vehicle in the slope, the rolling resistance is reducedR f f G cos(α )(6)where: G [N] . vehicle weightf [/] . rolling resistance coefficientα [o] . angle of the slopea.) Driven wheelb.) Driver wheelFigure 5: Rolling resistance on driver and driven wheelTab. 4: Indicated values of the rolling resistance coefficient fRoad typefasphalt concrete (tarmac), smooth 0,010concrete, smooth0,010concrete, rough0,014curbstone, very good0,015curbstone, good0,020curbstone, poor0,033macadam, poor0,035field road, very good0,045field road, good0,080field road, poor0,160sand, non-compacted, dry0,150 – 0,300
1. Dynamic characteristics of vehiclesAnother cause for the formation of a rolling resistance is the tangential displacementsin the tire's supporting surface, which cause slipping. These tangential displacementsdepend on the design of the tire running layer, which can be radial or diagonal. Thereare practically no tangential movements in the radial tires, therefore the rollingresistance is lower. The rolling resistance coefficient is determined by experimentsand is a function of:f f (Qtp , p p , Qc , v )(7)where: Qtp . [/]quality of tire running layer Qc [/] . quality of the road pp [Pa] . tire pressure v [m/s]. vehicle speedWhen calculating, we assume that the rolling resistance does not change with vehiclespeed. The rolling resistance of the vehicle at straight driving is composed of: basic rolling resistance of the tires, toe-in toe-out effect and resistance due to driving on uneven road.The proportions of resistance due to the toe-in toe-out effect and due to uneven roadare usually very small and can therefore be neglected.2.6.2.2Air resistance (drag) RzThe air resistance consists of the following components: pressure resistance resulting from all normal pressure forces acting on thesurface of the vehicle, or resistor of the shape,, friction resistance, which is the result of all tangential forces acting on thesurface of the vehicle, or the resistance of the surface,, resistance that occur as a result of essential parts of the vehicle (locks,mirrors, .) which in any way deviate from the basic vehicle profile and resistance resulting from the flow of air through the engine cooler andthrough the interior of the vehicle.Rz in case the speed of the air w 0 is calculated using equation:Rz where: 1 ρ A c v22(8)ρ [kg/m3] .air densityc [/] .drag coefficient that includes all above mentioned influences(c c1 c2 c3 c4)A [m2] .the surface obtained as a vehicle's projection to a planeperpendicular to the direction of motion; this is the so-called front surfacev [m/s].vehicle speed
1. Dynamic characteristics of vehicles w [m/s].absolute value of air speedIn case when the speed of the air is not zero, the relative vehicle speed is calculatedtaking into account the speed of the air w (v v w).The front surface of the vehicle is calculated using equation: for personal vehiclesA 0.9 B H (9)for commercial vehiclesA 0.78 B H(10)where: B [m] . vehicle widthH [m] . vehicle heightFigure 6: Vehicle's projection to a plane perpendicular to the direction of motion
1. Dynamic characteristics of vehiclesFigure 7: Drag coefficient for different vehicle types2.6.2.3Slope resistance RS (resistance of hill)The weight component parallel to the slope Rs is called a slope resistance or anascent resistance. From the parallelogram of forces it is clear that it is defined asR S G sin(α )(11)Usually the angle α is defined in %. This corresponds to the tangent of the anglebetween the slope and the horizontal planetan α α %(12)100The resistance of the slope can be positive or negative, depending on the direction oftravel. The slope resistance brakes the vehicle when driving uphill (Rs 0), andaccelerate the vehicle when driving downhill (Rs 0).
1. Dynamic characteristics of vehiclesFigure 8: Slope resistanceWhen designing vehicles, we take into account the largest allowable road inclinationin Europe, which is 26%. For vehicles intended for special use, however, themaximum ascent should be further defined.2.6.2.4Resistance of mass inertia RiAt accelerated movement a part of the power is used to accelerate translatorymasses Ri′ , and the other part to accelerate the vehicle's rotational masses R ' 'i [2].Ri Ri' Ri"(13)im2 ikg2JG a ; Ri" ( J m 2 η z 2k ) agrdrdTotal resistance of mass inertia is therefore [2]Ri' Ri i2i2GJa (1 J m m2 kg η z 2 k )grd mrd mRi Ri' δ where: G a δg(14)(15)(16)δ [/] . coefficient of rotational massesCoefficient of rotational masses δ cannot be calculated due to unknown mass inertiaof all rotating parts but can be estimated using equation (17) [2].δ 1.03 k im2where:(17)
1. Dynamic characteristics of vehicles k 0.04 (personal vehicles) 0.07 (commercial vehicles)im gear ratioCoefficient of rotational masses can also be estimated using experiential equation(Eq. (18)) [2].δ 1 k1 k 2 ii2(18)where: k1 [/] . coefficient of rotational masses of wheelsk2 [/] . coefficient of rotational masses of enginezk I k grst rdk1 (19)where: zkIkrstrd[/] . number of wheels[kgmm2] mass inertia of the wheel[mm] . static radius of the wheel[mm] . dynamic radius of the wheelk2 I m ikg2 gη t rst rd(20)where: Im [kgmm2] average mass inertia of rotating parts of the engineηt [/] . transmission efficiencyValues of coefficients k1 and k2 can also be estimated based on experience:k1 0,076k2 0,007.2.6.3 Traction forces on the wheelsThe torque, which is available on the engine shaft at a certain engine speed, istransmitted through the transmission to the wheels. Value of the torque on wheelscan be calculated by Eq. (21). To overcome the forces of driving resistance, atraction force is required on the wheels, which is calculated by Eq. (22) [2]. 30 v ii ikg η i η kg η oM K ,i (v) M M (n m (v )) ii ikg η i η kg η o M M ii ikgπ rd where: MK,i(v) . [Nm] torque on the wheels in ith gear(21)
1. Dynamic characteristics of vehicles MM(v) . [Nm] torque of the engineη i [/] . efficiency of transmission in ith gear ηkg [/] . efficiency of differential η o [/] . efficiency of other transmissionsFk ,i (v) M K ,i ( v )(22)rdwhere: FK,i(v). [N]traction force on the wheels in ith gearIn case of ideal motor (constant power at each speed) or continuous variabletransmission with efficiency 1, the traction force on the wheels would be ideal. Idealtraction force can be calculated by Eq. (23).Fid (v) Pkonst (max)v(23)where: Fid(v) . [N] ideal traction force on the wheelsPkonst. [W] ideal constant power of the engineThe traction forces FK on the wheels are opposed by the driving resistancesFK R R f R z Rs Ri(24)The equation is called a motion equation, or a balance of forces. Using this equation,we can calculate the total traction force Fk needed to overcome the sum of drivingresistances or the amount of traction force used to overcome a given force ofresistance. This equation is used in assessing vehicle driving characteristics.
1. Dynamic characteristics of vehicles12000.001st gear1.prestava10000.002nd gear2.prestava3rd gear3.prestava8000.00F [N]4th gear4.prestava6000.005th gear5.prestava6th gear6.prestava4000.00Idela tractionIdealnavlečnaforcesilaAir alniSum uporovof .00300.00v [km/h]Figure 9:: Diagram of traction forces and driving resistances for vehicle VW Golf GTI30000Ideal tr. forceIdealnasila1st gearPrestava12nd gearPrestava23rd gearPrestava34th gearPrestava45th gearPrestava56th gearPrestava67th gearPrestava7Rolling resist.KotalniuporAir resistanceZračniuporSum anja25000F [N]20000150001000050000050100150200250300350v [km/h]Figure 10: Diagram sil of traction forces and driving resistances for vehicle MercedezSLS
1. Dynamic characteristics of vehiclesFigure 9 and Figure 10Error! Reference source not found. show the size ofindividual forces depending on vehicle speed. From this chart maximum vehiclespeed can be determined. Maximum vehicle speed is located at the intersection ofthe sum of resistances curve and the curve of traction force (the one that intersecstthe sum of resistances curve at the highest speed F(vmax)).2.6.4 Dynamic coefficient of a vehicleAt a certain vehicle speed, the maximum traction force on the wheels that is opposedby the sum of resistances can be determined. The difference in the maximum tractionforce and the driving resistance forces is the reserve of the traction force Frez, whichcan be used for acceleration. In order to facilitate comparison between differentvehicles, a dynamic coefficient (D) is calculated according to Eq. (26) [2], which takesinto account the traction force on the wheel and the air resistance. The dynamiccoefficient is essentially the reserve of the traction force on the wheel reduced by theweight of the vehicle.Frez ,i (v) FK ,i (v) Rcel (v)where: Frez,i(v) . reserve of the traction force in ith gearDi ( v ) where: (25)FK ,i (v) R z (v)GDi(v) . [/]G . [N] R f R j Ri(26)Gdynamic coefficient of the vehicleweight of the vehicle0.900.800.70D []0.601st gear1.prestava0.502nd gear2.prestava0.403rd gear3.prestava4th gear4.prestava0.305.prestava5th gear0.206.prestava6th 300.00v [km/h]Figure 11: Diagram of the vehicle dynamic coefficient
1. Dynamic characteristics of vehiclesFigure 11 shows the values of the dynamic coefficient depending on the vehicle(data: vehicle VW Golf GTI).2.6.5 Limiting slopesA relatively high driving resistance also represents the force required to overcome theslope. The power to overcome this force is greatly increased with the speed andbecomes the main limitation of the maximum vehicle speed.The maximum climb that the vehicle can handle at a certain speed is derived fromEq. (27) [2].D(v ) f cos(α (v )) sin(α (v )) f 1 sin 2 (α (v )) sin(α (v ))(27)It is assumed that the speed during the driving through the slope is constant. Bysquaring the equation we obtain(1 f 2 ) sin 2 (α (v )) 2 D(v )sin(α (v )) ( D 2 (v ) f 2 ) 0(28)From there followssin(α (v )) D (v ) f 1 f 2 D 2 (v ) 1 f 2(29) D (v ) f 1 D 2 (v ) f 2 i α i (v) arcsin i(30)2 1 f The condition must be fulfilled 1 D(v) f in order for the squareroot to be realnumber.The limiting slope represents the slope angle which the vehicle can still overcome ata certain constant speed. αi(v) . [ o]limiting slopeThe size of the slope in practice is given in percentages (j [%]) and not in degrees(α [ o]). The mathematical link between the two methods of giving a slope is givenby the Eq. ( 31).j tan(α [ o ])100[%] j. [%] slope intensity(31)
1. Dynamic characteristics of vehicles160.00140.00120.001st gear1.prestavaj [%]100.002.prestava2nd gear80.003.prestava3rd gear60.004.prestava4th gear40.005.prestava5th gear6.prestava6th 00300.00v [km/h]Figure 12: Diagram of limiting slopesFigure 12 shows the limiting slopes depending on the vehicle speed for each gear(data: vehicle VW Golf GTI). It is obvious that the vehicle in 1st gear can overcomethe slope of 140 %. This means that the vehicle has engine powerful enough toovercome that kind of slope but in practice this would not be possible since the tireswould sleep or the vehicle would tip over the rear wheels.2.6.6 Available power on wheels and power reserves (Power balance)Instead of the balance of forces, we can carry out the balance of power inputs on thedrive wheels PK.P K P M η P f P s P z Pi(32)The power brought to the driving wheels is equal to the total power required toovercome the driving resistance. This term is called the power balance. The powerbalance is very suitable for solving fuel economy problems as well as for analysingindividual parameters of the motor vehicle system. Links between forces and powerPf R f v [W] ;Pz R z v [W] ;Ps Rs v [W]Pi Ri v [W](33)Curves 1-5 in Figure 13 represent the power on the wheels in a certain gear, whilethe thicker curve represents the sum of resistance powers at a certain speed. Theintersection of this curve with the power curves determines the maximum speed ofthe vehicle.
1. Dynamic characteristics of vehiclesIn case of a numerical calculation of the characteristics, it is best to calculate thepower balance with the following equations:Pi ,( cel ) (v) Fi ,( Rcel ) (v)v(34)3,6where: Pi,cel(v) . [kW] power on wheels depending on vehicle speed in ith gear Fi , ( R cel ) (v) . [N] traction force on the wheels in ith gearDifference between the power on wheels and the total power of resistances ΔP orPrez represents the power reserve which can be used for vehicle acceleration (seeFigure 13).Prez Pi ,( cel ) PRcel Prez . [N]PRcel. [N](35)power reservetotal power of resistancesChart of power balance180.00160.001.prestava1st gear140.002.prestava2nd gearP [km/h]120.003.prestava3rd gear100.004.prestava4th gear80.005.prestava5th gear60.006.prestava6th gear40.00MočkotalnegauporaRollingres. power20.00Močzračnega poweruporaAir .00MočuporovTotalpower of resis.v [km/h]Figure 13: Power chart (data: vehicle VW Golf GTI)
1. Dynamic characteristics of vehiclesChart of power reserves160.00140.00120.00100.001st gear1.prestavaP [kW]80.002nd gear2.prestava60.003rd gear3.prestava40.004th gear4.prestava20.005th gear5.prestava0.00-20.00 0.0050.00100.00150.00200.00250.00300.006th gear6.prestava-40.00-60.00v [km/h]Figure 14: Diagram of power reserves (data: vehicle VW Golf GTI)2.6.7 Vehicle accelerationThe most important result of the vehicle dynamics calculation is the acceleration ofthe vehicle, which we usually want to be as large as possible. The acceleration iscalculated using the equation on the basis of already calculated values (36) [2].ai (v) (Di (v) f )where: gδai(v) [m/s2] vehicle accelerationg [m/s2] . gravity accelerationδ [/] . coefficient of rotational masses(36)
1. Dynamic characteristics of vehiclesAcceleration chart8.007.00a [m/s2]6.001.prestava1st gear5.002.prestava2nd gear4.003.prestava3rd gear3.004.prestava4th gear5.prestava5th gear2.006.prestava6th 00.0050.00100.00150.00200.00250.00300.00v [km/h]Figure 15: Diagram of the vehicle accelerationFigure 15 shows the value of acceleration in different gears depending on the vehiclespeed. The curve »Acceleration path« shows maximum possible acceleration fromthe start to the maximum vehicle speed (data: vehicle VW Golf GTI).2.6.8 Acceleration timeWhen calculating the acceleration time, we proceed from the expression (37) fromwhich we obtain the expression (38) for calculating the acceleration time between twospeeds (v1 and v2)a dvdv dt dtat (v ) v2v11dva (v )(37)(38)where: t(v) . [s] acceleration time from v1 to v2v1 . [m/s] vehicle start speedv2 . [m/s] vehicle end speedDuring acceleration it is necessary to shift from lower gear to a higher one, whichalso requires a certain amount of time, so the actual accelerating times are greaterthan the times so calculated.
1. Dynamic characteristics of vehicles2.6.9 Path taken during the accelerationWhen calculating the path taken during the acceleration, we proceed from theexpression (39) from which we obtain the expression (40) for calculating the pathtaken during the acceleration between two speeds (v1 and v2).v dS dS vdtdts (v ) t (v 2 )t ( v1 )(39)(40)v(t ) dtwhere s(v) . [m] path taken during the acceleration from v1 to v2t(v1) . [s] time at vehicle start speedt(v2) . [s] time at vehicle end speed60800700506004040030Path [m]Time 0Vehicle speed [km/h]Figure 16: Diagram of acceleration time and pathFigure 16 shows time needed to accelerate to a certain speed path taken during theacceleration. This diagram also takes into account the time needed for gear shifting.Shifting time is tpres 0,5 s.
1. Dynamic characteristics of vehicles2.7Instructions for numerical integration by SimpsonSimpson's rule h a b f (x )dx 3 f (a ) 4 f 2 f (b) Rbab a2Generalized Simpson's ruleh h f (x )dx 3 ( fbah b an0 4 f1 2 f 2 . 2 f n 2 4 f n 1 f n )(41)(42)(43)(44)
1. Dynamic characteristics of vehicles3.References[1]Kraut, B.: Krautov strojniški priročnik, Tehniška založba Slovenije, Ljubljana, 1993[2]Simić, D.: Motorna vozila, Naučna knjiga, Beograd, 1988
The purpose of this exercise is to address the dynamic driving conditions, such as determining the necessary power to overcome certain driving conditions, determining acceleration, acceleration times, maximum speed etc. 2.1 Basic structure of the vehicle In this exercise, we are interested in the dynamic characteristics of the vehicle.
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