Automotive Dynamics of Vehicle Motion 4 2008

Some Dynamics of Motor Vehicle Motion

Copyright z William A. Greco 2008 [email protected] page-1 of 33

The dynamics of motor vehicle motion is an important field of study as it relates tosafety, accident investigation and general ground vehicle performance. The intentof this booklet is to establish and review mathematical procedures to calculate the physicalcharacteristics of vehicle motion. Procedures and information found within these few pagesmay be of assistance to anyone interested in determining road vehicle dynamics and performancewith some emphasis relating to accident investigation and reporting.

Please note:Manufacturers published road braking performance tests provide stopping distances from 60 mph thatrange from 120 to 140 feet. These numbers are not realistic and not average, but are achievable in acontrolled test environment. The tests, are run by professional test drivers, on perfect surfaces, withbrand new tires with very high coefficients of friction. Unlike highway panic stops, the test driver isprepared for the stop to occur.

Rollover

Rollover crashes are one of the most significant safety problems for all classes of vehiclesespecially trucks (pickups, sport utility vehicles, and vans ). For 2001 through 2006, there were anaverage of approximately 210,000 rollover crashes per year. (Rollover crashes are ones in whichrollover is the first harmful event.) These rollovers resulted in an average of over 9,000 fatalities peryear and over 220,000 non-fatal injuries per year. In terms of fatalities per registered vehicle, rolloversare second only to frontal crashes in their level of severity.

Figure-1 above represents a vehicle shown from the rear turning to the left.where:m = the mass of the vehicle expressed in lbs. (pounds)v = the velocity of the vehicle expressed in ft/secr = the radius of the curve in the road to the center of gravity G of the vehicle (feet)g = acceleration due to gravity 32.173 ft/sec2 (at 40 degrees latitude)F1 and F2 represent horizontal forces affecting the left and right tires.R1 and R2 represent the vertical reaction of the left and right tires.CG = represents the center of gravity of the vehicle above ground levela = the distance from the vehicle center of gravity to the center of the tires (feet) (½ the wheel track)mg = the mass of the vehicle times the acceleration due to gravityh = the distance from the road surface to the center of gravity (feet)



Rollover (continued)

A car of mass m traveling at speed v moves on a horizontal road so thatthe center of mass describes a circle of radius r. Assume that the car is travelinginto the page (away from the reader) in Figure-1 and is turning to the left. It must besubject to a centripetal force:

the counter force is provided by frictional forces acting on the tires.

Resolving vertically = R1 + R2 = mg (equation-2)

Resolving horizontally =

Taking moments about the center of mass =

Combining equation-3 and equation-4 which both equal F1 + F2 :

And by rearrangement:

solving for R1:

From equation-2 : R1 + R2 = mgSolving for R1 from equation-2 :

Combining equation-6 and equation-7

R2 is positive on one side of the equation and negative on the other side, so R2 cancels out:



Rollover (continued from page 2)

Subtracting mass from both sides of equation-9:

By rearrangement and solving for v:

And :

It can be seen that mass cancels out when calculating rollover dynamics. The height above the roadsurface to the vehicles center of gravity, width of wheel track and radius of the curve being negotiated arevariables which determine the critical velocity that the vehicle can be traveling prior to a rollover event.The analysis is only valid if F1and F2 are being resisted by the road surface and tire coefficient of frictionor if the tires of the vehicle impact or come in contact with some solid object such as a curb while goinginto a skid. Once a vehicle goes into a skid there is a greater possibility of coming into contact with someobject (such as a curb) or surface that may cause the vehicle to overturn due to a change ofcoefficient of friction.

Skid or RolloverWhere u is used to denote the coefficient of friction, the following equation applies;

Equality implies that the tires are skidding, since by equation-2 and equation-3;

Then limiting speed (or critical skidding speed ) is given by;

or

If a vehicle is to skid rather than overturn or rollover the following condition must exist:

Both g and r cancel out leaving;

Where: u = tire coefficient of friction, a = ½ tire track to center of gravity,h = height above road surface to center of gravity




Tire Coefficient of Friction

Definition of Coefficient of Friction: The ratio of the weight of an object being moved alonga surface and the force that maintains contact between the object and the surface.

The average tire traveling 40 mph on a dry road surface (free of gravel) has a coefficientof friction between 0.7 and 0.82. On a rain soaked road the same tire at 40 mph has a coefficientof friction of 0.52 to 0.68. The upper values represent new tires with good tread and the lowervalues are for old tires with poor tread. Velocity has a significant impact on tire/road surfacecoefficient of friction. For instance at 60 mph on a dry road surface the coefficient of frictionwill be 0.68 to 0.75 and on a wet surface 0.45 to 0.52. As the speed increases the coefficientof friction drops. The coefficient of friction on all road surfaces decrease with increase of speed;the difference between the rolling and sliding coefficients increases with increase of speed.This relationship is very significant from the standpoint of safety. At low speeds below30 mph tire coefficients of friction will generally range from 0.9 to close to 1.0 (one).As the speed drops during a stop the coefficient of friction increases.

Other factors affecting the coefficient of tire and road surface friction are:

Road Surface:The straight ahead sliding coefficients of friction between nonskid tread tires and different types

of road surfaces have different ranges. Coefficients of friction are higher on Bituminous concretethan on sheet asphalt or Portland cement concrete.

Temperature:There is an optimum temperature at which certain rubber compounds are most effective.

The temperature peak at which maximum coefficient of friction is obtained is a function ofthe rubber compound. This temperature is usually between 150 degrees for a rain tire on alightweight car to 275 degrees for a NASCAR racing vehicle. A given amount of frictionwork being done on the tire generates heat which raises the temperature of the tire untila balance is achieved with the amount of heat that the tire can dissipate.

Various factors affecting tire temperature are: weight of the vehicle, amount of cornering,average speed, amount of free airflow to the tire, road temperature, air temperature,heat transmitted from brakes and engine, tire pressure, tire tread thickness and generaltire construction.

SlipSlip occurs during directional and speed changes and is the angular difference between

the direction that the tire is traveling compared to the direction it is pointed. There is arelationship between the coefficient of friction to slippage in the rubber contact area. Whether atire is accelerating, braking, or cornering, it has some small degree of slippage. In alongitudinal sense it is referred to in percent slippage. Longitudinal slip is obvious in itsextremes: in acceleration the limit is wheelspin, and in braking the limit is lockup. The extremeof cornering slip angle is 90 degrees, at which point the tire is sliding sideways. What is lessobvious is that the tire has its maximum u (coefficient of friction) at some verysmall degree of slip say 5 percent of rotation or 5 to 10 degrees slip angle. and that (u) dropsslightly beyond that point, when the tire goes into a slide. Although tire companies seldom reportthe amount of slippage and the only meaning they have to a driver is in the feel of the tire,or the warning it gives (squeal ) at the limit of breakaway. A driver automatically vies to maintainhis car at these limits by steering angle and throttle.



Rollover (continued from page 4)Factors affecting the coefficient of tire and road surface friction:

Slip (continued from page-4)No matter what slip angle the peak u is at, however, the peak does not drop off sharply, or the driver

will be unable to get close to the limit without losing control. Slip accounts for wasted forces in theform of drag, and because this wasted energy contributes to heat buildup. Passenger car tires aredesigned to accept these losses for softness in ride and handling. The concept of slip angle also affectsoversteer/understeer in the consideration of an entire vehicle, but as long as the front and rear tires haveabout the same characteristics (which is why all four tires should be the same) it is primarilytheoretical and mathematical. Likewise, properly calculated longitudinal slip figures are most valuablein the design of anti-skid brake systems and Electronic Stability Control systems.

CamberThe coefficient of friction (u) is also affected a great deal by the camber angle (see Figure-2) or

inclination of a tire.

As a tire is tilted, one edge is raised off the pavement and traction is lost. A rubber tire may be expectedto flex somewhat to keep an even pressure footprint, but this flex also allows the tire carcass to roll under whensubjected to extreme cornering forces. Camber reduces the coefficient of friction (u) of a typical tire and causesa slight inward inclination of perhaps ½ a degree and actually increases the coefficient of friction. It is quitepossible for the wheel to be deflected half a degree, and the tire to be deflected over one degree with respect tothe wheel. There is an optimum camber angle for each tire on each vehicle.

Tire PressureTire pressure is another factor that can affect the coefficient of friction. The proper pressure is one at which there

is an even rubber pressure in contact with the road, all across the tire patch. Pressure must also be at a minimum, to getthe maximum rubber area on the ground and to provide some bump absorption, and it must be high enough tominimize heat-generating tire deformations. In addition, the generation of heat will raise pressures by 5 to 15 psi andcause pressures to vary from one side of the car to the other depending on which tires are working hardest at any onetime. Regardless of what pressures are recommended as ideal by engineers from their experiments an flat test surfaces,determination of actual pressures to fit a particular car and track surface are important. The most general statement thatcan be made is that in the range from 20 to 40 psi, the coefficient of friction goes up slightly with increasing pressures,except as it can be greatly decreased by road surface roughness. Enough pressure is required to keep the tire beadseated on the wheel rim, as centrifugal forces try to pull it off.



Rollover (continued from page 5)Factors affecting the coefficient of tire and road surface friction:

LoadingAnother property of the tire coefficient of friction is that it varies with the total load on the tire. As unit

pressures rise anywhere in the rubber contact patch, the rubber is less able to resist fictional shearing forces.A tire that has a friction coefficient (u) of 0.6 on a 3,000 pound car, might have a friction coefficient of 0.5 an a4,000-pound car. Where this factor gets more complicated, however, is in the lateral weight transfer under cornering,when the outside tires are more heavily loaded and the inside tires less so. At first glance, it would appear that therewould be no loss in total cornering capability, because the outer tire gains as much weight as the inner one loses. Infact, however, the change in coefficient of friction is not linear, and the outer tire loses coefficient of friction morethan the inner tire gains it.

Under no weight transfer conditions:Outside tire 750 lbs x 0.6 = 450 + Inside tire 750 lbs x 0.6 = 450 = 900 pounds of cornering force

Under large weight transfer conditions during a very fast turn on a tight radius curve =Outside tire 250 lbs x 0.3 = 75 + Inside tire 1,250 lbs x 0.61 = 762 = 837 pounds of cornering force

A loss of 63 pounds of cornering force.

These figures a loss between the two conditions, the sudden loss of 63 pounds of force in cornering powercan severely affect the handling when it is at one end of the car.

Tire DynamicsAnother factor to be considered with respect to tire coefficient of friction is the dynamics of the tire itself.

As the tire is essentially a flexible device, it has a spring rate and some internal damping. These rates are so highcompared to the suspension spring rates that they are generally ignored. However, they can become harmonic withthe chassis under certain conditions, and create a destructive chatter or wheel hop. A tire also has a self-aligningtorque, which tries to keep it traveling in a straight line, and obviously, the wider the tire, the greater the tendency.If the steering kingpin were located in the exact center of the tire, it would be very difficult to steer it. The tire wouldhave to be twisted in its own contact patch. Offsetting this pivot allows the tire to tend to roll around one edge.Another property is pneumatic trail, which means that the forces on the tire move the contact patch away from thegeometric center, upsetting static geometry. It should also be noted that tires can be made (intentionally orunintentionally) asymmetrical in cornering capabilities. This might be an advantage if the car continually had to turnin one direction (race track), or a definite disadvantage if it happens accidentally on a car intended to comer in bothdirections. Finally, most of these factors are related to steady-state conditions, whether in high-speed straight awaydriving or low-speed cornering. In practice, everything is always changing, or transient, and can't be isolated. Whenthe suspension allows the tire to oscillate vertically and move laterally over bumps, the tire coefficient of frictionalways drops.

Tire diameter also plays a part in tire coefficient of friction determination. As the diameter goes down, the contactpatch gets shorter and wider causing the car to become less stable. As section height is reduced there is less distancefor the sidewalls to deflect.

Determination of the coefficient of tire and road surface frictionTire manufacturers published coefficient of friction values are usually taken at very low speeds and on

dry surfaces.

One can easily determine the coefficient of friction of a road surface and tire by use of the work-frictionequation:

where:d = stopping distance, feetm = mass of vehicle, poundsg = acceleration due to gravity, 32.173 feet per second squaredu = coefficient of friction, dimensionless numberv = velocity of vehicle prior to braking, feet per second

rearrangement of equation-14 (mass cancels out):




Example-1Calculation of coefficient of friction by field determination:

Assume a Nissan pickup that takes 148 feet to come to a stop from60 mph or 60 x 1.46 = 87.5 feet per second, the coefficient of friction is:

Any vehicles tire – road surface coefficient of friction can be determined by field test using this method.

Example-2Calculation of initial vehicle skid or overturning.

By equation-13

Assume a 1991 Oldsmobile Cutlass Calais that has a front track of 4.66 feet and a height above ground tocenter of gravity of 1.748 feet and it’s tire – road surface coefficient of friction is 0.6 at low speed,the Cutlass will provide the driver with a skid warning between critical skid speed and rollover speed:

Since the a/h ratio is higher than the coefficient of friction this vehicle will go into a skid prior tooverturning, allowing ample warning to the driver that an unstable condition exists, if the driver entersthe curve below critical rollover speed (equation-11b).

Example-3Assume a bus with a track width of 5.66 feet or a = ½ x 5.66 = 2.83 feet and a height above ground to

center of gravity of 4.5 feet:

This indicates that if the bus has very good tires with a coefficient of friction greater than 0.63, itwill not first go into a skid before it reaches it’s critical overturning speed. Critical skid speed is much lowerthan critical overturning speed. When going into a skid the vehicle alerts the driver of a problem priorto reaching the point of overturning. In example-3 the bus will not give any warning, as the overturncondition will occur without any skidding.

Example-4Assume a 1992 Plymouth Voyager with a front wheel track of 5 feet and a height above ground to

center of gravity of 2.08 feet that is traveling around a curve with a radius of 400 feet and has a tirecoefficient of friction of 0.7:It’s critical skid speed will be (by equation-12a):

65 miles per hour



Rollover (continued from page 7)Example-4 (continued)

The same 1992 Voyager will have a critical overturning speed of :a = ½ x 5 = 2.5 feeth = 2.08 feetr = 400 feet (curve radius)Note: One mile per hour = 1.46 feet per second and (1.46 x 60 seconds x 60 minutes)= 5,280 feet per hour

by equation-11b:

124.37 feet per second or 85.18 mph.

critical overturning speed = 85 miles per hour

Below 65 mph the voyager in the above example will make it around the curve without incident.If the driver enters the curve between 65 mph and 84 mph he will skid.If the driver enters the curve above 84 mph he will overturn.

Curve Banking

Curves are sometimes banked allowing faster speeds that vehicles can safely handle without danger of overturning.Highway curves are usually banked (inward) such that the horizontal component of the reaction force of the road ona car traveling at the design velocity equals the required centripetal force.

Dividing Fx by Fy gives:



Curve Banking (continued from page-8)

Referring to figure-3; the reaction force F of the road on the car is perpendicular to the road surface.The vertical component Fy of this force supports the weight W of the vehicle, and it’s horizontalcomponent Fx provides the centripetal force. If a car goes around a banked turn at less than the designspeed, friction will keep it from skidding outward. If the car’s speed is above the design velocitywith respect to the bank angle the car will enter into a skid. The drawback with the banking of curvesis, a given angle is only correct for one speed of vehicle, an average speed is assumed and the roadis angled to that speed.

Example-5:

A 1,000 foot banked curve is being designed to handle vehicles traveling at 50 mph. The properangle for this curve is;

50 mph x 1.46 = 73 feet per second

tangent of 0.17 = an angle of : 9.8 degrees

or 9 degrees-39 minutes

The curve must be banked at 9.8 degrees. The vertical and horizontal forces will be in equilibriumat 50 mph on this banked curve.



Forces on two wheeled Vehicles

Figure-4 shows a simplified two-wheeled vehicle. Assume that the two wheeled vehicle is comingtoward you with a velocity v, while following a curved path whose center O is to the right. To makeit travel in the curved path, the front wheel is turned so that there is a force F exerted by the groundon the tires towards O. The inertia of a body may be considered as concentrated at it’s center ofgravity. To accelerate the whole vehicle towards O, a force directed towards O is required; but thisforce must act through the center of gravity CG. The force acts at ground level through the tires.Cyclists incline their vehicles at an angle to the vertical to negotiate a curve.

Example-6: (see figure-5)A motorcyclist is turning a corner of radius 20 feet at a speed of 15 mph. The center of gravity CG

is 2 feet above the ground when the vehicle is vertical, what is the necessary angle of inclination ?

The tangent of 0.756 is 37 degrees 6 minutes.The motorcycle would have to be tilted 37 degrees 6 minutes to negotiate the 20 foot radius turn at 15 mph.



Computing the above ground Center of Gravity (h) for any vehicle

The center of gravity height above road surface, for any given vehicle can be computed by fieldexperimentation using the following procedure:

To determine the center of gravity height ( “h” variable shown on page 1 ) for any vehicle,(see figure –6)the vehicle must first be on a flat, level surface. Wheel scales are then placed under each wheel and theweight at each wheel is recorded. Next elevate the rear wheels by placing blocks of wood, concrete blocksor jacks under the rear wheels on top of the wheel scales. Elevation must be a minimum of 20 inches abovethe ground. When rear wheels are elevated, front wheel scales will show that there is now more weighton front wheels than the original recorded weight indicated while vehicle was on a flat surface.Be sure to subtract the weight of the blocks or jacks before recording the weight of the rear wheels.Subtract the weight of the combined front wheels when they are level from the combined front wheelweight when the car is angled, the answer is referred to as “D” in the example shown below.

Example-7: (refer to figure-6)A determination of center of gravity ( h ) is to be made on a vehicle that has been weighed in at 3,405 pounds

and has a wheel base of 115 inches. The combined front wheels weigh in at 1,668 pounds on a level surface.The car’s rear wheels are jacked up to 30 inches and the combined front wheels weight is recorded as 1,795 pounds.WB = 115,E= 30

Taking the sine of 0.2687 = 14.78 degrees

W = 3,405 poundsD = 1,795 – 1,668 = 127 poundsTaking the cotangent of 14.78 degrees

This vehicle has a center of gravity ( h ) of 16.26 inches or 1.36 feet above the ground.



Estimating the Initial Speed of a Vehicle from skid Marks

The initial speed which a vehicle that has been involved in a panic stop was traveling priorto brake application can be estimated by a simple equation.Expressed in English system or (IP) inch pound units:

Where:30.02 = a constant

u = coefficient of friction (dimensionless number) see pages 4 through 7Dskid = skid mark distance, feet

v = velocity prior to panic stop, miles per hour

Expressed in metric or system international (SI) units:

Where:255 = a constant

u = coefficient of friction (dimensionless number) see pages 4 through 7Dskid = skid mark distance, meters

v = velocity prior to panic stop, kilometers per hour

Note: The mass of a vehicle is not a variable to be considered in stopping distance versus skid marks.A heavy vehicle will have more kinetic energy than a light vehicle, however the increasedfriction generated by the heavier vehicle cancels out the higher kinetic energy.

Example-8:If a vehicle’s skid marks are measured as 130 feet from initial brake application to a complete stop andthe tire coefficient of friction with the road surface is 0.75, what was the vehicles initial speed priorto brake application ?

by equation-17:

by equation-17a:

130 feet x 0.3048 feet per meter = 39.6 meters

87 kph x 0.6213712 kph per mph = 54.1 mph

This vehicle had an approximate velocity prior to brake application of just over 54 mph.



Minimum Stopping Distance

The minimum stopping of a vehicle is determined from the modified form of the work friction equation:

Where:Vo = initial velocity prior to brake application, ft per second

d = minimum stopping distance, feetu = coefficient of friction (dimensionless number) see pages 4 through 7g = acceleration due to gravity, 32.173 ft/ sec squared

Example-9:What is the minimum stopping distance of a vehicle traveling at 45 mph on a rain slick road with poor

tires, if the tire coefficient of friction is 0.58 ?

45 miles per hour x 1.46 feet per second/mph = 65.7 feet per second

Increasing the speed of this same vehicle will cause a drop in coefficient of friction, assume that the driverincreases his speed to 90 mph causing the coefficient of friction to drop to 0.53 and then applies the brakes,what is the minimum stopping distance at the new conditions ?

90 miles per hour x 1.46 feet per second/mph = 131.4 feet per second

Under most conditions, doubling the speed of a vehicle causes the minimum stopping distance to quadruple,however the change in coefficient of friction will cause the stopping distance to more than quadruple.Velocity has a significant impact on tire/road surface coefficient of friction.



Acceleration, Deceleration, G Forces and Collisions

AccelerationThe acceleration of a body is the rate at which it’s velocity increases. When a car speeds up

it’s acceleration is not necessarily uniform. Even if the vehicles engine exerted a constant forcethe acceleration would be roughly uniform, but engine power and frictional forces varywith speed. Various factors contribute to a non-uniform acceleration. This article will concentrateon situations where acceleration is considered to be uniform.

DecelerationWhen a vehicle slows down, it is said to undergo a deceleration or retardation, the value of

which is equal to the rate at which the velocity decreases. A deceleration is regarded as anegative acceleration.

The Description of MotionA body is in motion if, as time elapses, it’s position changes. If the body is restricted to

motion in a straight line--linear motion--then as time elapses, its distance from some fixed pointon the line changes. If the distance moved in a given time is noted, the average velocity can beobserved. More detailed observations of the distance traveled at intervals of a second or less willshow how the velocity varies from time to time. If the body is 17 in, from a starting point after43 seconds, and 20 in, from the starting point 1 second later, its velocity between the 43rd and44th second was 3 in./second. If the velocity is known from second to second in this way,its rate of change--the acceleration--can be calculated, and at the end of any interval an averageacceleration can be worked out.

Four quantities used to describe the linear motion of a vehicle are, time, distance, velocity andacceleration. The relationship between these four quantities allows many diverse computations tobe completed.

Equations of MotionConsider a vehicle moving in a straight line path with uniform acceleration ( a ). The initial

velocity ( Vi ) after a time ( t ) when it has traveled a distance ( d ) will have a final velocity (Vf).The five quantities a, Vi, t, d, and Vf are related by four basic equations.

d = distance, feetVi = initial velocity, feet per secondVf = final velocity, feet per seconda = acceleration, feet per second squaredt = time, seconds

The distance traveled by the vehicle is the product of average velocity and elapsed time,d = average velocity x t.

The average velocity is the average of the initial velocity Vi and final velocities,

And distance =This is the first of the four equations.

The second equation is, in effect, a definition of the acceleration.Acceleration a = rate of change of velocity



Acceleration, Deceleration, G Forces and Collisions (continued)

The remaining two equations are derived from equations 19 and 20.Expressing equation-20 in the form,

And substituting in equation 19;

Solving for Vf =

And substituting the values of Vf:

Time can also be computed by;

Dividing acceleration by the acceleration due to gravity will yield G force.Example-10:

A vehicle traveling at 60 mph is involved in a panic stop, the vehicle had a tire coefficient of 0.7 ona dry road surface, the drivers weight is 180 pounds,a. What is the distance required to come to a full stop ?b. How much force will the driver have to resist during the panic stop ?

60 mph x 1.46 = 87.6 feet per second

From equation 18:

10.a (answer)= 170 feetby equation-24 the time to come to a stop is calculated:

Average deceleration by equation 20=

The stop will cause 0.7 G’s of negative force for 3.9 seconds. The driver will be thrust forward in his seat.There will be (180 x 0.7) = -126 pounds of negative horizontal force countered by the drivers 180 pounds.The resultant (see page-16) is 219.7 pounds at an angle of 35 degrees. The driver will feel 219.7 – 180 = 39.7 poundsFor 3.9 seconds during this event. He must maintain steering control of the vehicle while resisting this force.The 180 pound driver will weigh 219.7 pounds during the panic stop (see page-16 and figure-8).



Acceleration, Deceleration, G Forces and Collisions (continued)Example-10: (continued)

Example-10: (continued) In the vector diagram above (Figure-8) 126 pounds of horizontal braking force is exertedforward, the drivers weight (180 pounds) is being acted upon by gravity which points straight down toward thecenter of the earth, during braking the resultant of the vector which can be worked out graphically (as shown above)or mathematically (as shown below) is 219.7 pounds at an angle with respect to gravitational pull of 35.17 degrees.therefore the driver during this panic stop will weigh almost 220 pounds and have a forward rotational pull of35 degrees on his body while resisting 219.7 – 180 = 39.7 pounds.

Example-11:A vehicle comes to a leisurely stop from 40 mph which requires 8.2 seconds, if the driver weighs 180 pounds,

how far did the vehicle travel during the stop and how much negative force will the 180 pound driver experience ?by equation-19:

40 mph x 1.46 feet per second per mph = 58.4 feet per secondby equation 19;

The vehicle traveled 239.4 feet during braking.

The driver would resist = (-0.22 x 180) = -39.6 pounds of horizontal force during the braking event.

Tangent of 0.22 = approximately 12 and one half degrees

The driver in this case would weigh 184.3 pounds and have a rotational pull of 12 and ½ degreesWhile feeling 184.3-180 = 4.3 pounds of force on his body.



Acceleration, Deceleration, G Forces and Collisions (continued)Example-12:

A vehicle traveling at 70 mph impacts a large tree, the vehicle’s front section is totally caved in 4 feet fromwhere the bumper was prior to the crash, if the occupant weighed 180 pounds, what force did he experience ?

70 mph x 1.46 = 102.2 feet per secondby equation 24 the time required to crush the vehicles front section =

the occupant would experience a deceleration of:

the occupant would have to resist = ( -40.7 x 180 ) = (7,326-180) = 7,146 pounds negative of horizontal force.

The occupant would weigh: 7,328 pounds for 0.08 seconds, assuming 2 feet of seat belt across the mid sectionAnd 2 feet across the chest with a 2 inch wide belt 48 x 2=96 square inches of contact area, the occupant’s seatbelt would absorb 76 pounds per square inch.

Conservation of Energy during a CollisionSometimes vehicles impact and collide with one another and simply bounce off of each other.Consider the case of a large vehicle with mass m1, with velocity v1 making a perfectly elasticimpact with a smaller vehicle of mass m2 with a velocity of v2.Conservation of energy by the first law of thermodynamics would conclude that:

(m1)(v1)+(m2)(v2) = (m1)(V1)+(m2)(V2) ( equation-25 )Where: v1 and v2 are velocity before impact

V1 and V2 are velocity after impactA loss of energy is given off to distortion and deformation of vehicle body parts (as will be shown on.

Page 18) Neglecting any losses due to distortion and deformation and taking into account the coefficient ofelasticity’s (e), the equation becomes:

(m1)(v1)(e1)+(m2)(v2)(e2) = (m1)(V1)(e1)+(m2)(V2)(e2) ( equation-26 )

And to calculate the velocity of the vehicles immediately after impact

Example-13:Vehicle A and Vehicle B are moving in the same straight line direction, Vehicle A with

a mass of 4,500 pounds and is traveling at 50 mph strikes vehicle B in the rear,Vehicle B weighs 2,750 pounds and is moving at a velocity of 35 mph prior to impact.Vehicle A is slowed down to 40 mph by the impact, assuming an elasticity coefficientof 0.9 for vehicle A and 0.75 for vehicle B, what is the velocity of vehicle B immediately after impact ?




Conservation of Energy during a Collision (Continued)The law of the conservation of energy is fundamental, but, in a collision, the total energy of the vehicles

after collision is less than it was before the collision. The deficit occurs because some of the kinetic energyhas been transformed into other kinds of energy.

When two vehicles collide, there is always some distortion to both of them. They become eithertemporarily or permanently dented. Energy is given off by one vehicle to deform the other, energyis transferred and stored in the compressed region for either a short time or if the material isinelastic it will remain deformed and the stored energy is lost as heat. If the deformed material iselastic, it will decompress itself and return to it’s original shape. Thus stored energy once againbecomes kinetic energy.

Example-14:A 4,500 pound SUV moving at 8 mph hits a stationary 3,000 pound car in the rear that is stopped at

a traffic light, if the drivers foot in the 3,000 pound car slips off of the brake:a What will the common speed of both vehicles be after the collision ?b If they both apply their brakes after the collision, what distance will they travel

assuming a 0.8 coefficient of friction (very slow speed, good tires and dry road) ?c How long will it take to stop the vehicles after the collision ?d. How much of the original (pre-collision) kinetic energy is lost and absorbed by the vehicles ?

Note: The term slug is a gravitational system unit of mass that when subjected to1 pound of force accelerates at 1 foot per second squared where weight in poundsis divided by 32.173 feet per second squared at 40 degrees latitude.

Initial velocity of the SUV = 8 x 1.46 = 11.68 feet per secondTotal momentum of the two vehicles before collision = (139.9 x 11.68) + 0 = 1,634 slug feet per second

Total Mass of the two vehicles =

Common speed (V) of both vehicles immediately after collision =

What distance will they travel before coming to a stop, if both drivers apply the brakes immediately after thecollision ?

4.8 mph x 1.46 = 7 feet per second

How long will it take to stop the vehicles if both drivers apply the brakes immediately after thecollision ?

How much of the original (pre-collision) kinetic energy is lost and absorbed by the vehicles ?Total kinetic energy before collision = Total kinetic energy after collision =

Or about 40% of the original force was absorbed by the vehicles. G Force involved in this accident:




Static Friction ForceIt has been experimentally established that coefficients of friction (and therefore the forces of friction)

are nearly independent of the size of contact area. Friction is related to pressure and gravitational acceleration.

Example-15:On a dry road, a 6,000 pound vehicle A traveling at 45 mph strikes a 2,500 pound vehicle B in the rear that is

stopped at a traffic light. The driver of the 2,500 pound vehicle keeps his foot pressed on the brake pedalthroughout the entire event until coming to a stop.

If the 2,500 pound vehicle B has a tire friction coefficient of 0.75, what common speed will the two vehicles betraveling immediately after impact ?

Maximum force of static friction of 2,500 pound vehicle B;(0.75) x (2,500) x (32.173) = 60,324 pounds of static friction

Initial velocity of vehicle A = (45 x 1.46) = 65.7 feet per secondMass of vehicle A =

Total mass of the two vehicles:

Total momentum of the two vehicles before the collision = (186 x 65.7) + 0 = 12,220 slug feet per second

Common speed (V) of both vehicles immediately after collision =

The G force on this impacted vehicle would work out to –0.75 G. even though it was struck by aconsiderably heavier vehicle than the vehicle in example-14 which is traveling at a muchlower rate of speed. Application of the brakes during a collision will cut down on the G forceexperienced by the impacted vehicle, however, driver’s that are rear ended often have their footslip off of the brake peddle.

Velocity as a function of Acceleration

On page 14 (equation 20) states that: and on page 15 (equation 23) states that:

If a vehicle is starting out from a zero velocity then: so that:

Solving for (a) from equation 28 = combined with equation 20: solving for Vf from: becomes;

Where: d = distance, feet Vi = initial velocity, feet per second Vf = final velocity, feet per seconda = acceleration, feet per second squared t = time, seconds

Example-16: Note: The actual velocity will be much lower, about 88 MPH, the answer is based on an averageacceleration. Acceleration will be affected by other factors and will not be constant (see appendix).A 1965 Mustang with 210 hp, 289 cumin. engine, 4-speed manual transmission and 4 barrel carburetor could travelthe quarter mile from a standing start in 16.5 seconds, what was it’s final velocity at the end of the quarter mile run ?From equation 29 = (which is incorrect---see appendix for actual answer)

Some Dynamics of Motor Vehicle MotionCopyright z William A. Greco 2008 [email protected] page-20 of 33


Distance as a Function of AccelerationExample-17:A 1965 Mustang with 210 hp, 289 cuin. engine and 4 barrel carburetor could reach 60 mph in 8.5 secondsfrom a standing start, how far did the Mustang travel upon reaching 60 mph ?

60 mph x 1.46 = 87.6 feet per secondfrom equation-20:

From equation-23:

The Mustang would have traveled 372 feet to reach 60 mph.

Example- 18: (see figure-9)A vehicle travels at 50 mph over an old hump back bridge, if the car’s weight is 4,000 pounds and the radius (R)

of the bridge is 180 feet, find the force exerted by the car on the road at the bridges highest point. Also find thespeed at which the car will lose contact with the road at zero G . Note: Suspension will keep the wheels on the road.

Forces acting on the car are at the highest point on the bridge, Car Weight = (m)mass x (g)acceleration due togravity downward, and the normal force N exerted in the upward direction,

Where:W = weight of the car, and v = 50 mph = 73 feet per second

The force exerted by the car on the road at the highest point of the bridge while traveling at 50 mph is 319 pounds.The car will lose contact with the road when N = 0, when v squared = R x g.

The vehicle body will lose road contact if it travels over the bridge at a velocity above 52 mph.In the above example the car will weigh 319 pounds at 50 MPH, a 180 pound driver will weigh 14 pounds.




Example-19:A vehicle that is traveling at 45 mph speeds up to 55 mph in 3 seconds, how far did the vehicle travel

during the time that it took to accelerate to the higher velocity ?45 mph = 65.7 feet per second and 55 mph = 80.3 feet per second

The vehicle traveled 219 feet during the acceleration.

Example-20: (See Figure-10)A car weighing 2,400 pounds fails to stop at an intersection and crashes into a 6,000 pound truck.

Witnesses and the driver of the truck agree that the truck was traveling at the posted speed limit,approximately 45 mph when struck by the car. The wreckage became locked and traveled 42 feetat an angle of 63.41 degrees before coming to rest. The posted speed limit in the car’s direction was 35 mph.Determine how fast the car was traveling prior to making contact with the truck.Determine the magnitude of the collision and deceleration value.

Let the car be traveling in the x direction and the truck be traveling in the y direction.m1 x v1 = (m1 + m2) v x sin (63.41_degrees)

And m2 x v2 = (m1 + m2) v x cos (63.41_degrees)

Solving for v2 =

The tangent of 63.41 degrees is 1.998

The car was traveling at 56.3 mph or 21.3 mph above the posted speed limit prior to the collision.



Acceleration, Deceleration, G Forces and Collisions (continued)Example-20: (continued)

Velocity of the combined mass immediately after the collision:

The combined mass initially traveled 32.14 feet per second or 22 mph.

a = Deceleration = -12.3 feet per second squared

Deceleration force =

The combined frictional decelerating force to bring the wreckage to a stop was 3,210 pounds.

Vehicle Performance as Related to EngineTorque, Gearing and Horsepower

Torque versus Horsepower = Torque is the twisting effort that an engine supplies through the crankshaft.Horsepower is a rate of performing work. Horsepower requires movement, torque does not require movement.Torque can be applied to an object without the object actually moving.

In automotive calculations torque relates to acceleration. Horsepower relates to maintaining a speedonce the vehicle has been accelerated.

Horsepower:Relating bhp, ihp and fhp

Bhp (brake horsepower) is the power delivered by the engine; ihp (indicated horsepower) is the power developedagainst the moving pistons and fhp (friction horsepower) is the power used to overcome friction in the engine.So that bhp = ihp – fhp

SAE net horsepower is an indication of how much power is actually available on the production vehicle as itis tested with all auxiliary equipment and a full exhaust system installed.

The equation for calculating indicated horsepower (ihp) is:

Where:P = mean effective pressure (mep), (cylinder expansion pressure), psiL = Length of stroke, feetA = area of piston in square inchesN = number of working (expansion) strokes per minute of the engine

Example-21:What is the indicated horsepower of a Ford 8 cylinder, 289 cuin engine with a four barrel carburetor and a

mep of 125 psi at 4,400 rpm ?(the length of stroke on the 289 cuin. Ford engine is 2.87 inches or 0.239 feet, the bore is 4 inches or 12.56 sqin.)

Subtracting the fhp or friction horsepower which also includes all of the auxiliary equipment such as the fan,alternator, power steering ect. plus friction from the pistons the bhp will be about 210.


Copyright z William A. Greco 2008 [email protected] page-23 of 33Vehicle Performance as Related to EngineTorque, Gearing and Horsepower (continued)

The relationships between mep (mean effective pressure), torque and horsepower are:

An engine will develop more torque at intermediate RPM’s than it will at high RPM’s, because at thelower RPM’s volumetric efficiency is higher, the combustion chamber scavenges better at lower speeds.

At higher RPM’s volumetric efficiency is lower. Maximum horsepower is dependent on engine RPM,maximum torque is not. Therefore maximum horsepower is attained at higher RPM’s and maximumtorque is obtained at lower than maximum engine RPM.

Example-22:In Example-21 it was shown that the indicated horsepower of a ford 8 cylinder, 289 cuin engine with a four barrel

carburetor and a mep of 125 psi at 4,400 rpm is 216 hp and we assumed 210 Brake horsepower after frictionvalues were accounted for. What is the torque value at 210 BHP and 4,400 RPM ? Using equation-31:

Ford published a maximum torque value for this engine as 300 pound_ feet @ 2,400 RPM. Using this particularengine, shifts for maximum acceleration using a manual transmission should be executed at 2,400 RPM, with run upto maximum speed in high gear to 4,400 RPM.

Maximum Speed, Road and Aerodynamic ResistanceAerodynamic forces created by the motion of a vehicle affect it’s performance. The component of the resultant

force which opposes the forward motion is called aerodynamic drag. Beside aerodynamic drag, road resistance isanother factor to be considered when calculating the maximum speed of a vehicle. The power consumption ofroad resistance increases as the square of the vehicle velocity, aerodynamic drag increases as the cube of velocity,therefore road resistance becomes a small percentage of total power use compared to aerodynamic drag.Road resistance at 100 mph will use only about 7 horsepower.Equations-34 and 35 will yield the amount of force and brake horsepower required to overcome aerodynamic drag:

Where:Force = pounds of force

A = frontal area of the vehicle (width x height), square feetV = vehicle velocity, feet per minute

CD = Coefficient of DragBHP = Brake Horsepower required to overcome aerodynamic drag force

Example-23:A 1965 Fastback Mustang has a width of 68.2 inches and a height of 51.5 inches, making it’s frontal area

24.39 square feet. This Mustang has a coefficient of drag of 0.308 and sports a 210 BHP engine, ignoring road dragwhat is the calculated aerodynamic drag force developed and horsepower that this vehicle will use toreach 110 mph (160.6 feet per second) ?

By equation-34:

By equation-35:

The maximum calculated speed for this vehicle is approximately 110 mph.



Vehicle Performance as Related to EngineTorque, Gearing and Horsepower (continued)

GearingTorque multiplication between the engine and the road wheels is required for acceleration to

take place. Acceleration or initial movement takes place when the torque transmitted to thedriving wheels exceeds the torque necessary to overcome the tractive resistance caused bythe combined rolling and air resistance. Axle ratios and transmission ratios are combined belowfinal drive or high gear (final gear). In high gear the transmission ratio is usually 1:1.Equation-37 gives the calculation for axle ratio with respect to maximum speed.

Where Tire_Radius is in inches.

Example-24:What is the required final gear axle ratio for a vehicle with 15 inch radius tires that is designed to travel at

110 mph with the engine turning at 4,400 RPM ?

If the vehicle above in example-24 had a 3:1 axle ratio, what maximum speed would the axle allow ?

Gearing as related to AccelerationThe combined axle ratio and transmission gear ratio are required to calculate force and speed during

acceleration. The equations required to predict force and speed during acceleration are shown below.

Where: Tire_Radius is in feet.

Example-25:A vehicle has the following transmission ratios; first gear 2.20 to 1, second gear 1.66 to 1, third gear 1.31 to 1

and 1:1 in fourth gear. The axle ratio is 3.45 to 1. The maximum engine torque value for this vehicle is282 foot pounds at 2,400 rpm and the tire radius is 15 inches, the vehicle weighs 2,606 pounds.What is the maximum available force provided by the drive train, speed and G force in second gear ?

Tire radius = 1.25 feet

In second gear the vehicle will have 1,098 pounds of available force from the drive train, and optimum shiftpoint should be made at 38 mph at 2,400 rpm and will be accelerating at 0.42g x 32.173 = 13.5 feet per secondsquared. Missing the optimum shift point, acceleration will drop off and the engine will begin to load up.In theory this car in second gear could achieve a maximum of 69 mph at 4,400 RPM.With a 4.11 differential this same vehicle in second gear would have a maximum of 0.502 g’s of acceleration.


Copyright z William A. Greco 2008 [email protected] page-25 of 33Appendix I

Example-26:Example-16 indicated that a 1965 Mustang with 210 hp, 289 cuin. engine, 4-speed manual transmission and

4 barrel carburetor could travel the quarter mile from a standing start in 16.5 seconds, and had a final velocity at theend of the quarter mile run of 109 MPH. This speed at the end of the quarter mile would be impossible for this car,the calculation was only good for the assumption of constant acceleration. Three shifts were required and all fourgears varied in acceleration values, unfortunately the last gear always consumes the longest amount of time and hasthe lowest acceleration value.

The ’65 Mustang owners manual indicates that the transmission gearbox had the following ratio’s:1st gear = 2.78 2nd gear = 1.93 3rd gear = 1.36 and 4th gear 1:1

The rear axle ratio was 3 : 1, tire radius = 15 inches (1.25 feet); the car weighed 2,955 poundsThe Mustang has a coefficient of drag of 0.308 and a frontal area of 24.39 square feet. (see example-23)If the 210 BHP engine was rated at 300 Pound feet of torque at 2400 rpm then:In first gear the car developed the following maximum force by equation 38:

And the optimum speed in first gear before shifting into second by equation 39 would be:

The 1,418 pounds of force generated by the car must overcome the drag aerodynamic drag force =By Equation- 34:

1,418 pounds generated – 27 pounds of drag at 26 MPH = 1,390 pounds available

Solving for time by Equation-20:

Where: t = time in seconds, Vf = final velocity, feet per second, Vi = initial velocity, feet per secondA = acceleration, feet per second squared

Optimum time spent in first gear = (26 mph x 1.46 = 37.96 feet per second)

Calculated Values for 1st ,2nd ,3rd ,4th Gears at Optimum Shift PointsGear Pounds Force

AvailableAcceleration

g’sAcceleration

Ft sec squaredSpeed MPHShift Point

Time in gearSeconds

1st gear 1,390 0.47 15.12 26 2.52nd gear 929 0.31 10.11 37.3 1.203rd gear 581 0.20 6.33 53 2.644th gear Average 263 0.11 3.65 -------- 8.81

The total run took = 2.5 + 1.20 + 2.64 + 8.81 = 16.56 seconds.In third gear the Mustang had travelled = 237.6 feet and the final shift from third to fourth was at a speed of 53 mph,the remainder of the run at an average acceleration of 3.65 ft per second squared, covered the remaining 1,082.4 feetin 8.81 seconds and finished the quarter mile at a speed of 84.6 mph.



Appendix I (continued)

A simpler method of calculating elapsed time for the quarter mile is shown below:This method is not very accurate, it does not take gearing or coefficient of drag into consideration,

however it yields a quick although inaccurate answer.Many individuals use rule of thumb equations to get simple, quick answers. Garbage in, garbage out.

Equation –40 should only be used as a comparison between two vehicle performances.

Elapsed time =

Where:ET = elapsed time in secondsHp = horsepower

Example-27:The 1962 corvette weighed 3,065 pounds and used a 327 cubic inch engine that came with either

a 300 hp, 340 hp or the fuel injected 360 hp engine. Using the 340 horsepower version what would theelapsed time for the quarter mile be ?

The 1965 Mustang with 210 hp weighing 2,955 pounds would have done the quarter mile in:

The Corvette would probably have completed the ¼ mile run about 2 seconds before the Mustang.

A simple method of calculating speed in mph for the quarter mile is shown below:

Equations 40 and 41 are very conservative and should be considered as quick and simple, in many cases they areinaccurate. For more exact answers the method shown in example-26 should be used.



Appendix II

Drunk Driving

Although not directly related to Motor Vehicle Dynamics, drunk driving is one of the most frequentcauses of death on the highway. An alcohol related death occurs every 22 minutes in the US or 66 per day.Drunk driving remains the number one killer of automobile operators and passengers, the number twokiller is falling asleep at the wheel or distraction at a very in-oportune moment.

When a person has been consuming an alcoholic beverage, the measure of how much alcohol they haveconsumed is known as the BAC (Blood Alcohol Concentration). In most of the United States the BACis considered to be 0.08, in Pennsylvannia the BAC is 0.08.

where:BAC = Blood alcohol Concentration

Ounces = Ounces of Alcoholic Beveragepercent_alcohol = percent by volume of Alcoholic Beverage

weight = body weight of individual in poundshours = number of hours that it took to consume the alcohol

Example 28:

A 180 pound man who consumes a 6 pack of 12 ounce beers containing4.0 percent alcoholic content in 2 hours would have a blood alcohol concentration of:

0.09 BAC = drunk, should not drive

Equation 42 is sobering mathematics.

Appendix III.

Looking at example 26 the following information was made available:Calculated Values for 1st ,2nd ,3rd ,4th Gears at Optimum Shift Points

Gear Pounds ForceAvailable

Accelerationg’s

AccelerationFt sec squared

Speed MPHShift Point

Time in gearSeconds

1st gear 1,390 0.47 15.12 26 2.52nd gear 929 0.31 10.11 37.3 1.203rd gear 581 0.20 6.33 53 2.644th gear Average 263 0.11 3.65 -------- 8.81

Curve fitting time in gear vs speeds gives us the following (special case) equation from the above chart:

Where X = time in seconds. Equation 43 has a 0.999 correlation coefficient.



Please keep in mind that equation-43 is used only as a special case for example 26 and is not to be used forgeneral purposes.To find the rate of change in speed vs time, equation-43 is placed in the differential form:

Pulling the constant out of the derivative:

Product rule: The derivative of a function that is the product of two functions.

(f g)’(x) = f(x) g’(x) + g(x)f’(x)

Differentiating a power with a constant base:D/dx Cu = (ln a) au du/dx

(d/dx) (1/x) = -1/x2

Simplifying and collecting powers:



Appendix III.

Differentiation of equation 43 – Continued:

Power rule: d/dx xn = n x(n-1) 1-0.296 = 0.704

a-n = 1/an

Put terms, factors in order and compute ln(0.1632):

(-) and (-) = +

The derivative is :

Substituting t (time) in seconds for x:



Appendix III.

Differentiation of equation 43 – Continued:Performance graphs of example-26, ’65 Mustang.Graph-1 below indicates the rate of acceleration increase in Miles per Hour at time (seconds).Graph-2 indicates the speed at time (seconds).

By graph-1 it can be seen that acceleration diminishes in relation to time.



As graph-1 on page 30 indicates that as time from initial start and speed increase, the rate ofacceleration decreases. There are many reasons for the decrease in the rate of acceleration.1. As speed increases, the vehicle must overcome greater air resistance force.2. Less force is available for acceleration because of the lower gear ratios in higher gears.

Although the calculations shown here are strictly newtonian in nature and do not support thenumber 3 reasoning:3 . It is suggested that inertia is a fundamental property that has not been properly

addressed by quantum field theory. The acquisition of a mass-energy field may stillrequire a mechanism (scientific explanation) to generate an inertial reaction force uponacceleration. The contention here is that acceleration is acted upon by inertia, in shortacceleration must overcome inertia, the higher the inertia the harder it is to accelerate.

Available horsepower used for acceleration for example 26

HPAccel = HPDrive_Wheels – ( Dragair + HPspeed ) (equation-44)Where:HPAccel = available horsepower for acceleration

HPDrive_Wheels = Horsepower at the drive wheels (considerably less than the manufacturersrated horsepower at the flywheel, most of the brake horsepower is used to driveaccessories, overcome mechanical friction, overcome tire slipageand maintain speed).

Dragair = Air resistance.HPspeed = Horsepower being used to maintain the speed that has been attained.

Equation 44 indicates that available horsepower for acceleration at the drive wheels is beingshared by air resistance plus the amount of horsepower to maintain speed.

Force versus distance from example 26Gear Total Time

SecondsTotal Distance

(feet)Force availableFor acceleration

(pounds)1 2.5 47 13902 3.7 76 9283 6.34 238 5814 16.56 1320 263

Using single constant and single exponent form curve fitting, distance (x) and force available (y),the following Polynomial was produced with a correlation value of 0.984.

by ax 0.47978113y x



Integrating using the indefinite integral:

To find the integral of x-0.4797, find the anti-derivative. The formula for the anti-derivative of abasic monomial is:

1 – 0.4797 = 0.5203

The indefinite integral also has some unknown constant. This can be proven by completing theopposite operation (derivative) in which C would go to 0. The value of C can be found in caseswhen an initial condition of the functions is given, however C in this case goes to zero since novalue for C was required.

Graph 3 indicates the force available for acceleration during the Mustangs ¼ mile run.

0.4797 0.47978113 constant 8113x cdt ct c x

0.5203

81130.5203x



Equation 45 below gives the horsepower available at the drive wheels at total time taken toarrive at the total distance covered from start of the ¼ mile run for the ’65 Mustang inexample-26.

Where:hp = horsepower at the drive wheelst = time from the start of the ¼ mile run

550 = 550 foot pounds per second per horsepowerx = distance from starting point of ¼ mile run

At the start of the run the mustang is spending 2.5 seconds in first gear and would have reached47 feet to attain the 300 Pound feet of torque at 2400 rpm prior to the next shift point.

At the end of the run:

Some basic acceleration (a), velocity (v), distance equations (x):

Note:The author drove a 1964 ½ Mustang used in example-26 from May 1964 to June 1976.

0.5203

0.4797 81138113 0.5203

550 550

xx

hpt t

( 45)equation

0.52038113 470.5203 84 @2.5 550

hp drive wheels

0.52038113 13200.5203 72 @

16.56 550hp drive wheels

2 1(distance_difference)x x x

2 1

2 1

(average_velocity)x x xvt t t

2 1 2 1x x v t t 0x x v t

x v t0

limt

x dxvt dt

2

2( )dv d dx dadt dt dt dt

dv dvadt dx

dx dvvdt dx

1 2( )va a a t

t

Automotive Dynamics of Vehicle Motion 4 2008

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Transcript of Automotive Dynamics of Vehicle Motion 4 2008