Automotive Dynamics of Vehicle Motion 4 2008

Some Dynamics of Motor Vehicle Motion Copyright z

William A. Greco 2008

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The dynamics of motor vehicle motion is an important field of study as it relates to safety, accident investigation and general ground vehicle performance. The intent of this booklet is to establish and review mathematical procedures to calculate the physical characteristics of vehicle motion. Procedures and information found within these few pages may be of assistance to anyone interested in determining road vehicle dynamics and performance with some emphasis relating to accident investigation and reporting. Please note: Manufacturers published road braking performance tests provide stopping distances from 60 mph that range from 120 to 140 feet. These numbers are not realistic and not average, but are achievable in a controlled test environment. The tests, are run by professional test drivers, on perfect surfaces, with brand new tires with very high coefficients of friction. Unlike highway panic stops, the test driver is prepared for the stop to occur.

Rollover Rollover crashes are one of the most significant safety problems for all classes of vehicles especially trucks (pickups, sport utility vehicles, and vans ). For 2001 through 2006, there were an average of approximately 210,000 rollover crashes per year. (Rollover crashes are ones in which rollover is the first harmful event.) These rollovers resulted in an average of over 9,000 fatalities per year and over 220,000 non-fatal injuries per year. In terms of fatalities per registered vehicle, rollovers are 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/sec r = 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/sec 2 (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 level a = 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 gravity h = the distance from the road surface to the center of gravity (feet)



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Rollover (continued)

A car of mass m traveling at speed v moves on a horizontal road so that the center of mass describes a circle of radius r. Assume that the car is traveling into the page (away from the reader) in Figure-1 and is turning to the left. It must be subject 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 = mg Solving 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:

Some Dynamics of Motor Vehicle Motion

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Copyright z


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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 road surface to the vehicles center of gravity, width of wheel track and radius of the curve being negotiated are variables 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 friction or if the tires of the vehicle impact or come in contact with some solid object such as a curb while going into a skid. Once a vehicle goes into a skid there is a greater possibility of coming into contact with some object (such as a curb) or surface that may cause the vehicle to overturn due to a change of coefficient of friction. Skid or Rollover Where 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



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Tire Coefficient of Friction Definition of Coefficient of Friction: The ratio of the weight of an object being moved along a 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 coefficient of friction between 0.7 and 0.82. On a rain soaked road the same tire at 40 mph has a coefficient of friction of 0.52 to 0.68. The upper values represent new tires with good tread and the lower values are for old tires with poor tread. Velocity has a significant impact on tire/road surface coefficient of friction. For instance at 60 mph on a dry road surface the coefficient of friction will be 0.68 to 0.75 and on a wet surface 0.45 to 0.52. As the speed increases the coefficient of 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 below 30 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 concrete than 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 of the rubber compound. This temperature is usually between 150 degrees for a rain tire on a lightweight car to 275 degrees for a NASCAR racing vehicle. A given amount of friction work being done on the tire generates heat which raises the temperature of the tire until a 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 general tire construction. Slip Slip 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 a relationship between the coefficient of friction to slippage in the rubber contact area. Whether a tire is accelerating, braking, or cornering, it has some small degree of slippage. In a longitudinal sense it is referred to in percent slippage. Longitudinal slip is obvious in its extremes: in acceleration the limit is wheelspin, and in braking the limit is lockup. The extreme of cornering slip angle is 90 degrees, at which point the tire is sliding sideways. What is less obvious is that the tire has its maximum u (coefficient of friction) at some very small degree of slip say 5 percent of rotation or 5 to 10 degrees slip angle. and that (u) drops slightly beyond that point, when the tire goes into a slide. Although tire companies seldom report the 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 maintain his car at these limits by steering angle and throttle.



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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 the form of drag, and because this wasted energy contributes to heat buildup. Passenger car tires are designed to accept these losses for softness in ride and handling. The concept of slip angle also affects oversteer/understeer in the consideration of an entire vehicle, but as long as the front and rear tires have about the same characteristics characteri stics (which is why why all four tires should be the same) it is primarily theoretical and mathematical. mathematical . Likewise, properly calculated calculat ed longitudinal slip figures are most valuable in the design of anti-skid brake systems and Electronic Stability Control systems.

Camber The 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 expected to flex somewhat to keep an even pressure footprint, but this flex also allows the tire carcass to roll under when subjected to extreme cornering forces. Camber reduces the coefficie nt of friction (u) of of a typical tire and causes a slight inward inclination of perhaps ½ a degree and actually increases the coefficient of friction. It is quite possible for the wheel to be deflected half a degree, and the tire to be deflected over one degree with respect to the wheel. There is an optimum camber angle for each tire on each vehicle. Tire Pressure Tire 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 get the maximum rubber area on the ground and to provide provide some bump absorption, and it must be high enough to minimize heat-generating tire deformations. In addition, the generation of heat will raise pressures by 5 to 15 psi and cause pressures to vary from one side of the car to the other depending on which tires are working hardest at any one time. 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 that can 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 bead seated on the wheel rim, as centrifugal forces try to pull it off.



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Rollover (continued from page 5) Factors affecting the coefficient of tire and road surface friction: Loading Another 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 a 4,000-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 there would be no loss in total cornering capability, because the outer tire gains as much weight as the inner one loses. In fact, however, the change in coefficient of friction is not linear, and the outer tire loses coefficient of friction more than 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 power power can severely affect the handling when it is at one end of the car. Tire Dynamics Another 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 high compared to the suspension spring rates that they are generally ignored. However, they can become harmonic with the chassis under certain conditions, and create a destructive chatter or wheel hop. A tire also has a self-aligning torque, 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 would have 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 the geometric center, upsetting static geometry. It should also be noted that tires can be made (intentionally or unintentionally) asymmetrical in cornering capabilities. This might be an advantage if the car continually had to turn in one direction (race track), or a definite disadvantage if it happens accidentally on a car intended to comer in both directions. Finally, most of these factors are related to steady-state conditions, whether in high-speed straight away driving or low-speed cornering. In practice, everything is always changing, or transient, and can't be isolated. When the suspension allows the tire to oscillate vertically and move laterally over bumps, the tire coefficient of friction always drops. Tire diameter also plays a part in tire coefficient of friction determination. As the diameter goes down, the contact patch gets shorter and wider causing the car to become less stable. As section height is reduced there is less distance for the sidewalls to deflect. Determination Determination of the coefficient of tire tire and road surface friction Tire 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-friction equation:

where: d = stopping distance, feet m = mass of vehicle, pounds g = acceleration due to gravity, 32.173 feet per second squared u = coefficient of friction, dimensionless number v = velocity of vehicle prior to braking, feet per second rearrangement of equation-14 (mass cancels out):



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Example-1 Calculation of coefficient of friction by field determination: Assume a Nissan pickup that takes 148 feet to come to a stop from 60 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-2 Calculation 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 to center 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 to overturning, allowing ample warning to the driver that an unstable condition exists, if the driver enters the curve below critical rollover speed (equation-11b). Example-3 Assume 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, it will not first go into a skid before it reaches it’s critical overturning speed. Critical skid speed is much lower than critical overturning speed. When going into a skid the vehicle alerts the driver of a problem prior to reaching the point of overturning. In example-3 the bus will not give any warning, as the overturn condition will occur without any skidding. Example-4 Assume 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 tire coefficient of friction of 0.7: It’s critical skid speed will be (by equation-12a):

65 miles per hour



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Example-4 (continued) The same 1992 Voyager will have a critical overturning speed of : a = ½ x 5 = 2.5 feet h = 2.08 feet r = 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 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 on a car traveling at the design velocity equals the required centripetal force.

Dividing Fx by Fy gives:



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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 horizontal component Fx provides the centripetal force. If a car goes around a banked turn at less than the design speed, friction will keep it from skidding outward. If the car’s speed is above the design velocity with respect to the bank angle the car will enter into a skid. The drawback with the banking of curves is, a given angle is only correct for one speed of vehicle, an average speed is assumed and the road is angled to that speed. Example-5: A 1,000 foot banked curve is being designed to handle vehicles traveling at 50 mph. The proper angle 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 equilibrium at 50 mph on this banked curve.



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Forces on two wheeled Vehicles

Figure-4 shows a simplified two-wheeled vehicle. Assume that the two wheeled vehicle is coming toward you with a velocity v, while following a curved path whose center O is to the right. To make it travel in the curved path, the front wheel is turned so that there is a force F exerted by the ground on the tires towards O. The inertia of a body may be considered as concentrated at it’s center of gravity. To accelerate the whole vehicle towards O, a force directed towards O is required; but this force 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.



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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 field experimentation 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 the weight at each wheel is recorded. Next elevate the rear wheels by placing blocks of wood, concrete blocks or jacks under the rear wheels on top of the wheel scales. Elevation must be a minimum of 20 inches above the ground. When rear wheels are elevated, front wheel scales will show that there is now more weight on 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 wheel weight when the car is angled, the answer is referred to as “D” in the example shown below. Example-7: (refer to to figure-6) A determination 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 pounds D = 1,795 – 1,668 = 127 pounds Taking 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.



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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 prior to 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 7 Dskid = 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 7 Dskid = 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 increased friction 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 and the tire coefficient of friction with the road surface is 0.75, what was the vehicles initial speed prior to 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.



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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, feet u = coefficient of friction (dimensionless number) see pages 4 through 7 g = 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 driver increases 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.



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Acceleration, Acceleration, Deceleration, G Forces and Collisions

Acceleration The 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 force the acceleration would be roughly uniform, but engine power and frictional forces vary with speed. Various factors factors contribute to a non-uniform acceleration. This article article will concentrate on situations where acceleration is considered to be uniform. Deceleration When 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 a negative acceleration. The Description of Motion A body is in motion if, as time elapses, it’s position changes. If the body is restricted to motion in a straight line--linear line--linear motion--then as time elapses, its distance from some fixed point on the line changes. If the distance moved in a given time is noted, the average velocity can be observed. More detailed observations of the distance traveled at intervals of a second or less will show how the velocity varies from time to time. If the body is 17 in, from a starting point after 43 seconds, and 20 in, from the starting point 1 second later, its velocity between the 43rd and 44th second was 3 in./second. If the velocity velocity is known from second to second in this way, way, its rate of change--the acceleration--can be calculated, and at the end of any interval an average acceleration can be worked out. Four quantities used to describe the linear motion of a vehicle are, time, distance, velocity velocity and acceleration. The relationship between these four quantities allows many diverse computations to be completed. Equations of Motion Consider 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, feet Vi = initial velocity, feet per second Vf = final velocity, feet per second a = acceleration, feet per second squared t = time, seconds The distance traveled by the vehicle is the product of average velocity 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 Acceleration a = rate of change of velocity



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Acceleration, Deceleration, Deceleration, G Forces and Collisions 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 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 on a dry road surface, the drivers weight is 180 pounds, a. What is the the distanc distancee require required d to come come to a full stop stop ? b. How much force force will will the driver driver have to to resist during during the panic stop stop ? 60 mph x 1.46 = 87.6 feet per second From equation 18:

10.a (answer)= 170 feet by 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 pounds For 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).



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

Example-10: (continued) In the vector diagram above above (Figure-8) 126 pounds pounds of horizontal braking force is exerted forward, the drivers weight (180 pounds) is being acted upon by gravity which points straight down toward the center 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 of 35 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 travel during the stop and how much negative force will the 180 pound driver driver experience ? by equation-19: 40 mph x 1.46 feet per second per mph = 58.4 feet per second by 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 pounds and have a rotational pull of 12 and and ½ degrees While feeling 184.3-180 = 4.3 pounds of force on his body.



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Acceleration, Deceleration, Deceleration, G Forces and Collisions 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 from where 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 second by 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. force.

The occupant would weigh: 7,328 pounds for 0.08 seconds, assuming 2 feet of seat belt across the mid section And 2 feet across the chest with a 2 inch wide belt 48 x 2=96 square inches of contact area, the occupant’s seat belt would absorb 76 pounds per square inch. Conservation of Energy during a Collision Sometimes Sometimes 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 elastic impact 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) = (m1)(V1)+(m2)(V2) (m1)(V1)+(m2)(V2) ( equation-25 ) Where: v1 and v2 are velocity velocity before impact V1 and V2 are velocity velocity after impact A 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 coefficient of elasticity’s (e), the equation becomes: (m1)(v1)(e1)+(m2)(v2)(e2) (m1)(v1)(e1)+(m2)(v2)(e2) = (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 2,75 2,750 0 pounds and is moving at a velocity velocity of 35 mph prior to impact. Vehicle A is slowed down to 40 mph by the impact, assuming an elasticity coefficient of 0.9 for vehicle A and 0.75 for vehicle B, what is the velocity of vehicle B immediately after impact ?



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Acceleration, Deceleration, G Forces and Collisions (continued)

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 energy has been transformed into other kinds of energy. When two vehicles collide, there is always some distortion to both of them. They become either temporarily or permanently dented. Energy is given off by one vehicle to deform the other, energy is transferred and stored in the compressed region for either a short time or if the material is inelastic it will remain deformed and the stored energy is lost as heat. If the deformed material is elastic, it will decompress itself and return to it’s original shape. Thus stored energy once again becomes 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-collis (pre-collision) ion) kinetic energy is lost and absorbed by the vehicles ? Note: The term slug is a gravitational system unit of mass that when subjected to 1 pound of force accelerates at 1 foot per second squared where weight in pounds is 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 second Total 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 the collision ? 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 the collision ?

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 Force involved in this accident:



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Static Friction Force It 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 pedal throughout 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 be traveling 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 second Mass 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 a considerably considerably heavier vehicle than the vehicle in example-14 which is traveling at a much lower rate of speed. Application of the brakes during a collision will cut down on the G force experienced by the impacted vehicle, however, driver’s that are rear ended often have their foot slip 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, distance , feet Vi = initial velocity, feet per second Vf = final velocity, feet per second a = 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 average acceleration. Acceleration 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 travel the 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 incorrect---see appendix for actual answer)



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Acceleration, Deceleration, G Forces and Collisions (continued) Distance as a Function of Acceleration

Example-17: A 1965 Mustang with 210 hp, 289 cuin. engine and 4 barrel carburetor could reach 60 mph in 8.5 seconds from a standing start, how far did the Mustang travel upon reaching 60 mph ? 60 mph x 1.46 = 87.6 feet per second from 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 the speed 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 to gravity 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 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.


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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 feet at 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) (63.41_degr ees)

Solving for v2 =

The tangent of 63.41 degrees is 1.998

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


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Acceleration, 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 speed once 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 developed against 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 it is 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), psi L = Length of stroke, feet A = area of piston in square inches N = 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.


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William A. Greco 2008 [email protected] page-23 of 33 Vehicle 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 the lower 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 maximum torque 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 friction values 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 particular engine, shifts for maximum acceleration using a manual transmission should be executed at 2,400 RPM, with run up to maximum speed in high gear to 4,400 RPM. Maximum Speed, Road and Aerodynamic Resistance Aerodynamic forces 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 is another factor to be considered when calculating the maximum speed of a vehicle. The power consumption of road 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 feet V = vehicle velocity, feet per minute CD = Coefficient of Drag BHP = 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 drag what is the calculated aerodynamic drag force developed and horsepower that this vehicle will use to reach 110 mph (160.6 feet feet per second) ? By equation-34: By equation-35: The maximum calculated speed for this vehicle is approximately 110 mph.


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

Gearing Torque 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 the driving wheels exceeds the torque necessary to overcome the tractive resistance caused by the combined rolling and air resistance. Axle ratios and transmission ratios are combined below final 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 Acceleration The 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 is 282 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 shift point should be made at 38 mph at 2,400 rpm and will be accelerating at 0.42g x 32.173 = 13.5 feet per second squared. 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 differential this same vehicle in second gear would have a maximum of 0.502 g’s of acceleration.


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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 the end 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 four gears varied in acceleration values, unfortunately the last gear always consumes the longest amount of time and has the lowest acceleration value. The ’65 Mustang owners manual indicates that the transmission gearbox gearbox had the following ratio’s: st nd rd th 1 gear = 2.78 2 gear = 1.93 3 gear = 1.36 and 4 gear 1:1 The rear axle ratio was 3 : 1, tire radius = 15 inches (1.25 feet); the car weighed 2,955 pounds The 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 second A = acceleration, feet per second squared Optimum time spent in first gear = (26 mph x 1.46 = 37.96 feet per second)

st

Gear st

1 nd 2 rd 3 th 4

gear gear gear gear

nd

rd

th

Calculated Values for 1 ,2 ,3 ,4 Gears at Optimum Shift Points Pounds Force Acceleration Acceleration Speed MPH Available g’s Ft sec squared Shift Point 1,390 0.47 15.12 26 929 0.31 10.11 37.3 581 0.20 6.33 53 Average 263 0.11 3.65 --------

Time in gear Seconds 2.5 1.20 2.64 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 feet in 8.81 seconds and finished the quarter mile at a speed of 84.6 mph.



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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 seconds Hp = 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 the elapsed 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 are inaccurate. For more exact answers the method shown in example-26 should be used.



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Appendix II

Drunk Driving

Although not directly related to Motor Vehicle Dynamics, drunk driving is one of the most frequent causes 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 two killer 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 have consumed is known as the BAC (Blood Alcohol Concentration). In most of the United States the BAC is considered to be 0.08, in Pennsylvannia the BAC is 0.08.

where: BAC = Blood alcohol Concentration Ounces = Ounces of Alcoholic Beverage percent_alcohol = percent by volume of Alcoholic Beverage weight = body weight of individual in pounds hours = 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 containing 4.0 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: st

Gear st

1 nd 2 rd 3 th 4

gear gear gear gear

nd

rd

th

Calculated Values for 1 ,2 ,3 ,4 Gears at Optimum Shift Points Pounds Force Acceleration Acceleration Speed MPH Available g’s Ft sec squared Shift Point 1,390 0.47 15.12 26 929 0.31 10.11 37.3 581 0.20 6.33 53 Average 263 0.11 3.65 --------

Time in gear Seconds 2.5 1.20 2.64 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.



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Please keep in mind that equation-43 is used only as a special case for example 26 and is not to be used for general 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) a u du/dx

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

Simplifying and collecting powers:

2



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Appendix III.

Differentiation of equation 43 – Continued:

Power rule:

-n

d/dx xn = n x(n-1)

1-0.296 = 0.704

n

a = 1/a

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

(-) and (-) = + The derivative is :

Substituting t (time) in seconds for x:

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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.



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As graph-1 on page 30 indicates that as time from initial start and speed increase, the rate of acceleration decreases. There are many reasons for the decrease in the rate of acceleration. 1. As speed speed increases, increases, the the vehicle vehicle must must overcome overcome greater greater air resistance resistance force. force. 2. Less force force is availabl availablee for acceleration acceleration because because of of the lower gear ratios ratios in higher higher gears. gears. Although the calculations shown here are strictly newtonian in nature and do not support the number 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 still require a mechanism (scientific explanation) to generate an inertial reaction force upon acceleration. The contention here is that acceleration is acted upon by inertia, in short acceleration must overcome inertia, the higher the inertia the harder it is to accelerate. Available horsepower used for acceleration for example 26 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 manufacturers rated horsepower at the flywheel, most of the brake horsepower is used to drive accessories, overcome mechanical friction, overcome tire slipage and 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 being shared by air resistance plus the amount of horsepower to maintain speed.

Gear

1 2 3 4

Force versus distance from example example 26 Total Time Total Distance Force available Seconds (feet) For acceleration (pounds) 2.5 47 1390 3.7 76 928 6.34 238 581 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.

y ax b

y  8113 x

0.4797



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Integrating using the indefinite integral:

 8113

0.4797

x



To find the integral of x basic monomial is:



cdt

ct c constant

 8113 

0.4797

x

-0.4797

, find the anti-derivative. The formula for the anti-derivative of a

1 – 0.4797 = 0.5203

 x 0.5203  8113     0.5203  The indefinite integral also has some unknown constant. This can be proven by completing the opposite operation (derivative) in which C would go to 0. The value of C can be found in cases when an initial condition of the functions is given, however C in this case goes to zero since no value for C was required.

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



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

 8113 x

0.4797

t 550

 hp 

8113 x 0.5203



0.5203 t 550

(equation  45)



Where: hp = horsepower at the drive wheels t = time from the start of the ¼ mile run 550 = 550 foot pounds per second per horsepower x = 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 reached 47 feet to attain the 300 Pound feet of torque at 2400 rpm prior to the next shift point. 8113  47 0.5203 0.5203 2.5  550



 84hp @ drive  wheels

At the end of the run: 8113 81131320 13200.5203



0.5203 16.56 6.56  550

 72hp @ drive  wheels

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

 x v

x2  x1 (distance_difference)

x2  x1 t2  t1





x v t a

dv dt



x (average_velocity) t

x  t 0 t

v  lim

dv

dx

dx

dt

v

dv dx

dx dt

x 2  x1

a

dv dt

v



t2  t1 

x  x0

d dx d 2 ( ) 2 dt dt dt

v a  a1  a2 ( t ) t

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

v



t

Automotive Dynamics of Vehicle Motion 4 2008

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