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The Georgia Police Academy Traffic Accident Reconstruction Level V Background and Purpose The National Highway Traffic Safety Administration (NHTSA), requires that automobiles sold in this country meet crash worthiness standards. Each automobile manufacturer must provide new automobiles for the purpose of crash testing so NHTSA can verify that the standards are being met. When crash tests are done all the data is collected and maintained. Since the tests are done in a controlled environment the vehicle’s speed and resulting crush damage are known. This testing has allowed engineers to develop a mathematical model that can be used to measure a vehicle’s speed based on the amount of crush damage it receives in a collision. Although its application is limited, this information does provide traffic accident investigators with another tool that may be used to calculate speed. Traffic Accident Reconstruction Level V, introduces students to the proper techniques for making accurate speed estimates from crush damage. Critical issues such as work, energy, the conservation of energy, equation application and the energy equation derivation are addressed. Students who successfully complete this course should have the knowledge and confidence to accurately estimate speed from vehicle crush in some instances. Terminal Performance Objective Given a simulated traffic accident, students will evaluate vehicle crush data to accurately calculate vehicle impact speeds. Enabling Objectives • Demonstrate the ability to use work, force and distance to calculate energy. • Demonstrate one method for calculating elastic potential energy when force constant and distance are known. • Define the Law of Conservation of Energy. • Demonstrate one method for calculating energy of a vehicle that skids over numerous surfaces, when weight, drag factor and distance are known. Enabling Objectives (con’t) • Demonstrate two methods for calculating vehicle speed when the amount of crush and after impact vehicle dynamics are known. • Define the terms dynamic collapse and restitution as they apply to vehicle damage. Work Technically, work is done when a force acts on an object through a distance. Work is also a measure of the effect force has on changing an object. The amount of change or work done is reflected in changes in the object's velocity, position, size, and shape. Work can be described mathematically as the product of the force (F) and the distance (d) through which it acts, provided the force and distance covered are in the same direction. This may be written in equation form: W = Fd Remember from past courses that the unit for force is pounds (lb) and the unit for distance is feet (ft). W = Fd F = force (lb) d = distance (ft) W = work (ft-lb) Example Assume a force of 300 lbs is required to push a box 20 feet. Calculate the amount of work done. W = Fd W = (300 lbs) (20 ft) W = 6000 ft-lbs If the force and distance are not in the same direction, then the equation is rewritten. W = Fd Cos 1 Example Assume the force is equal to 300 lbs, the angle is 45 degrees, and the distance is 20 feet. Calculate the amount of work done. W = Fd Cos 1 W = (300) (20) (Cos 45E) W = 4242 ft-lbs Work is the product of the magnitudes of force and distance. Therefore, work is a scalar quantity. Being a scalar quantity, work has only magnitude and sign (positive or negative), but has no direction associated with it. Work is positive if 1 is greater than or equal to 0 degrees but less than 90 degrees. Work is negative if 1 is greater than 90 degrees but less than or equal to 180 degrees. An accelerating vehicle is an example of positive work. The force accelerating the vehicle is in the direction the vehicle is traveling. In that case 1(angle) is 0 degrees. d Direction of Travel Example Assume the force acting on the vehicle in the previous slide is 1000 lbs, the distance the force acts through is 100 ft, and the angle is 0 degrees. Calculate the amount of work done. W = Fd Cos 1 W = (1000) (100) (Cos OE) W = 100,000 ft-lbs Now suppose the vehicle is decelerating. In this case the work has a negative sign since the force is opposite the direction of travel. In this case the 1(angle) is 180 degrees. Assume the force acting on this vehicle is 2000 lbs through a distance of 50 feet. d Direction of Travel Friction W = Fd Cos 1 W = (2000) (50) (Cos 180E) W = -100,000 ft-lbs Energy When work is done, energy is transferred between different objects. The amount of energy possessed by an object is an indication of its ability to do work. Work and energy are proportional. The more energy an object has, the more work it can perform. Equations can be derived to calculate the amount of energy transferred between objects when work is done under a variety of conditions. These derivations use the positive sign if the object's energy is increased and negative if it is decreased when work is done. This agrees with the previous examples given of the accelerating and decelerating vehicles. The accelerating vehicle is gaining energy and increasing its ability to do work. The decelerating vehicle is losing energy and decreasing its ability to do work. Energy can be grouped into three general categories: • Rest energy • Kinetic energy • Potential energy 1. Rest energy is energy an object possesses due to its mass. This form of energy is related to Einstein's theory of relativity and will not be discussed. 2. Kinetic energy (KE) is energy an object possesses due to its motion. A vehicle in motion has more energy than a vehicle at rest. When a vehicle accelerates, it gains velocity and also increases its energy. The energy of a moving body is equal to: • KE = ½mv² • KE = Kinetic energy (ft-lbs) • M = Mass (weight divided by the acceleration of gravity) • v = Velocity (fps) The amount of work required to accelerate the vehicle from rest to a given velocity can be calculated using this equation. The energy calculated is the energy gained by the vehicle and it is kinetic energy. In the equation mass and weight are related by the acceleration due to gravity. Considering this relationship the equation could also be written as follows: KE = wv²/2g KE = Kinetic energy(ft-lbs) w = Weight (lbs) v = Velocity (fps) g = Gravity (32.2 ft/sec²) Example Assume a 3,000 lb vehicle is accelerated from rest to a velocity of 40 fps. Calculate the amount of work done and energy gained. KE = wv²/2g KE = (3000) (40)²/(2)(32.2) KE = 4,800,000/64.4 KE = 74,534 ft-lbs The same amount of work would be required to bring the vehicle to rest. The same amount of energy that was originally gained, would have to be dissipated. Energy is dissipated when it is converted into thermal (heat) energy. This happens as a result of the friction between the tires and road surface, tires and brake system, or through rolling resistance in coasting to a stop. In the derivation of the speed from skid mark equation we learned that the following equation can be used to calculate the amount of kinetic energy that must be converted to another form in order to bring a vehicle to a stop: W = wfd W = Work (ft-lbs) w = weight (lbs) f = drag factor d = distance (ft) Example Assume a 3000 lb vehicle from the previous example skidded to a stop with a drag factor of .71 from 40 fps, it would slide approximately 35 ft. Calculate the amount of work done. W = wfd W = (3000)(.71)(35) W = 74,550 ft-lbs The following two equations can be used to calculate speed estimates from skid marks. • W = wfd • KE = wv²/2g The first equation is used to find the amount of kinetic energy the vehicle dissipated while skidding. The second equation, rearranged to solve for velocity, uses the energy calculated to find the velocity of the vehicle. • KE = wv²/2g • Multiply both sides by 2g • 2g * KE = wv²/2g * 2g/1 • • • • Cancel leaving: 2g * Ke = wv² Divide weight from both sides: 2g * KE/w = wv²/w • • • • Cancel leaving v² = 2g * KE/w Take the square root of both sides leaving v = %(2g)(KE)/w Since the quantity of one form of energy is the same as the quantity of any other form of energy, the equation can be written: v = %(2g)(E)/w Example From the previous example, the amount of kinetic energy that was dissipated while skidding was approximately 74,534 ft-lbs and the vehicle weighed 3000 lbs. Calculate the velocity. v = %(2g)(E)/w v = %(2)(32.2)(74,534)/3000 v = %4,800,000/3000 v = %1600 v = 40 fps Kinetic energy is a scalar quantity. The kinetic energy of a given mass depends on the magnitude of its velocity and not on the direction of travel. Any change in kinetic energy depends on the sign of the work done. A positive sign indicates the kinetic energy is increased. A negative sign indicates the kinetic energy is decreased. Acceleration rate is not a factor in kinetic energy. The mass and velocity of a body govern the amount of kinetic energy the body has. The rate at which the body reaches that velocity does not affect the kinetic energy. 3. Potential energy (PE) is energy an object possesses due to its position. Two forms of potential energy will be covered in this course. • Gravitational potential energy (PEh) • Elastic potential energy (PEk) • Gravitational potential energy is the energy an object possesses due to its position above some reference plane, usually the surface of the earth. Position 1 Position 2 Horizontal Motion W h1 W h2 Reference Line When the box moves from position 1 to position 2 there is no change in the vertical distance. PEh1 = PEh2 The boxes in the overhead have a downward gravitational force which is their weight (w). The distance moved is the height (h) of the center of mass above its original position. These variables can be substituted into the equation for work: W = Fd W = wh This is the equation for gravitational potential energy: PEh = wh PEh = gravitational potential energy (ft-lbs) w = weight (lbs) h = vertical distance (ft) Example Assume a vehicle weighing 3000 lbs has its center of mass raised 10 ft above the ground. Calculate the amount of gravitational potential energy the vehicle has. PEh = wh PEh= (3000) (10) PEh = 30,000 ft-lbs Calculate how many feet per second a 3000 lb vehicle would be traveling to have 30,000 ft-lbs of v = %(2g) (E)/w v = %(2) (32.2) (30,000)/3000 v = %1,932,000/3000 v = %644 v = 25.37 fps The change in gravitational potential energy depends only on the difference between the initial and final heights of the object, not on the path taken to reach the final height. The vehicle moves up and down while going from position 1 to position 2. The change in gravitational potential energy is only dependent upon the difference between h1 and h2. Peh = w(h2-h1) Position 2 Position 1 h2-h1 h2 Reference Plane h1 • Elastic potential energy is energy an object possesses because of its shape. A spring which is compressed or stretched has elastic potential energy because of its shape. The amount of elastic potential energy it has depends upon the amount of work done to compress or stretch the spring. A force is required to compress a spring from it free length. The force required is not constant. It must increase as the spring is compressed more and more. The amount of force is directly proportional to the amount the spring is compressed. This relationship is known as Hooke's Law. The proportionality constant for Hooke's Law depends on the stiffness of the spring and is called the force constant (k). A stiff spring will have a larger force constant than a spring that can be easily stretched. Free Length X Compressed A spring is compressed a certain distance, X from its free length. The force required can be calculated by the equation F = kx F = kx F = force (lbs) k = force constant (lbs/ft) x = distance compressed (ft) In the overhead the distance, x, is zero before the spring is compressed. When the spring is compressed a certain distance, x, the force is F = kx The average force would be calculated by using the following equation. Favg = (0 + kx)/2 or Favg = ½kx The work equation is W = Fd Therefore, the work done to compress the spring is W = Favg x Remember F = ½kx, so ½kx can be substituted for F, leaving W = (½kx) (x) or W = ½kx² W = work (ft-lbs) k = force constant (lb/ft) x = distance compressed/stretched This is also equal to the amount of elastic potential energy possessed by the spring. PEk = ½kx² Example Assume a spring with a force constant of 10,000 lbs/ft is compressed 2 ft from its free length. Calculate the elastic potential energy possessed by the spring. PEk = ½kx² PEk = ½ (10,000) (2)² PEk = 20,000 ft-lbs Remember this equation applies equally for a spring that is stretched and to one that is compressed. The elastic potential energy is equal to one-half the product of the force constant times the square of the distance compressed/stretched from its free length. If the distance is doubled, the energy increases by a factor of 4. Conservation of Energy When work is done energy is converted from one form into another. We are able to apply energy to accident situations because it is a conserved quantity. The total energy remains the same, but it may change forms. The law of conservation of energy states that when work is done and energy is converted from one form into another, no energy is created and no energy is destroyed. As accident investigators, this concept can be another tool which may enable us to determine accident vehicle speed. First we must account for the work done during the accident sequence. Once the total energy dissipated has been calculated it can then be converted into velocity. The total energy must be calculated by using an energy balance equation. ET = E1 + E2 + E3.....Ei ET = initial energy possessed by the vehicle E1 = first amount of energy dissipated E2 = second amount of energy dissipated E3 = third amount of energy dissipated Ei = last amount of energy dissipated or left unchanged Example Assume a vehicle weighing 3000 lbs, was initially traveling at 60 fps and then skidded 73.75 ft with a drag factor of .50, and its velocity after skidding was 35 fps. Substitute these values into the energy balance equation and calculate. ET = E1 + E2 ET = wv²/2g + wfd ET = (3000) (35²)/64.4 + (3000) (.50) (73.75) ET = 57,065.217 + 110,625 ET = 167,690.22 ft-lbs Now considering our energy balance equation this total amount of energy should equal our initial amount of energy (ET). The initial velocity was 60 fps and the weight is 3,000 lbs. KE = wv²/2g KE = (3000) (60²)/64.4 KE = 167,701.86 ft-lbs As you can see, the two are close enough to be considered equal. Example Assume a vehicle that weighs 4,000 lbs skids for 100 ft on a surface having a .80 drag factor, and then continued to slide an additional 60 ft on a second surface having a .40 drag factor. At the end of the 60 ft skid on the second surface the vehicle's brakes were released and it was still going 30 fps. Calculate the total amount of energy possessed by the vehicle. ET = E1 + E2 + E3 ET = 4,000*.80*100+4,000*.40*60+4,000*(30²)/ 64.4 ET = 320,000 + 96,000 + 55,900 ET = 471,900 ft-lbs This total represents the initial amount of energy the vehicle possessed when it first began to slide. Since this is kinetic energy, the velocity of the vehicle can be found as follows: v = %2gE/w v = %(2) (32.2) (471,900)/4000 v = %7597.59 v = 87.16 fps Example A vehicle weighing 3,500 lbs skids to a stop in 100 feet while traveling up a 4% grade. The drag factor is .60. Calculate the vehicle's initial kinetic energy. W = E1 = wfd E1 = (3500) (.60) (100) E1 = 210,000 ft-lbs A vehicle traveling 100 feet up a 4% grade will rise approximately 4 ft. The gravitational potential energy gained by the vehicle due to this height change can be found by using the gravitational potential energy equation.PEh = E2 = wh E2 = (3,500) (4) E2 = 14,000 ft-lbs The energy balance equation can be written to calculate the amount of kinetic energy the vehicle initially possessed. KE = ET = E1 + E2 ET = (210,000) + (14,000) ET = 224,000 ft-lbs The initial velocity can now be found. v = %2gET/w v = %(2) (32.2) (224,000)/3500 v = %4121.6 v = 64.19 fps If the same vehicle was sliding down the grade instead of up, the initial energy would be the sum of the initial kinetic energy and gravitational potential energy. If the vehicle stopped in 100 ft, the energy dissipated would have to equal the sum of the initial kinetic energy and gravitational potential energy: KEi + PEh = E1 Rearrange the equation to solve for KEi, leaving KEi = E1 PEh KEi = (210,000) - (14,000) KEi = 196,000 ft-lbs The velocity of the vehicle when it first began to slide would be calculated as follows: v = %2gET/w v = %(2) (32.2) (196,000)/(3500) v = %3606.4 v = 60.05 fps The previous example demonstrates that a vehicle will require less initial velocity (kinetic energy) to slide down a grade than to slide the same distance up the grade. In some instances energy might be dissipated in subtle ways that we are unable to measure. In these cases it may be difficult to account for all the energy in an energy balance equation. Speed Estimates From Vehicle Damage Energy is dissipated in ways other than sliding. Suppose a vehicle weighing 3,000 lbs is traveling 40 fps when it collides with a solid (nondeforming) immovable barrier and comes to rest. E1 = wv²/2g E1 = (3,000) (40²)/64.4 E1 = 4,800,000/64.4 E1 = 74,534 ft-lbs This vehicle would possess 74,534 ft-lbs of kinetic energy due to its weight and velocity. This energy is dissipated in the collision with the barrier, and the result is damage to the vehicle. The quantity of energy used to do damage to a vehicle must be determined in order to estimate speed based on that damage. After a collision the vehicle is permanently damaged. The crush measurements taken on a damaged vehicle show the resulting damage to the vehicle. In the collision itself the vehicle will have slightly more damage than is revealed after the collision. This damage is referred to as dynamic collapse. A vehicle has some restitution in a collision. Restitution means the vehicle tries to return to its original shape. The effect of restitution is insignificant in nearly all severe collisions. In low-velocity impacts, restitution may have to be considered. The definition of work indicates that the amount of energy dissipated is related to the force and the distance through which it acts. The distance is determined by making crush measurements. Once this has been done, the speed can be calculated provided information exists regarding the amount of force required to produce the damage. This quantity of force is related to the dynamic force deflection characteristics, or crush resistance (stiffness), of the vehicle structure being crushed. The dynamic force-deflection characteristics of a vehicle are determined by tests, primarily barrier impact tests. In barrier impact tests it is assumed that all the energy is used in doing damage to the vehicle. The velocity at which the vehicle strikes the barrier is an indication of the amount of kinetic energy (crush energy) available to damage the vehicle. The assumed relationship between a barrier test vehicle and a similar accident vehicle is that an equal quantity of energy must be used to damage the latter the same as the barrier test vehicle. For example, suppose a 1990 Chevrolet Impala is used in a barrier impact test and reveals an average of 15 inches of crush from a 30 mph impact. You investigate and accident in which a 1990 Chevrolet Impala strikes a bridge abutment. You measure the crush damage and find that it averages 15 inches. In this scenario it would be safe to assume the accident vehicle struck the bridge at a speed of 30 mph. The barrier impact speeds are known as the equivalent barrier speed (EBS) or barrier equivalent velocity (BEV). Once the EBS for the accident vehicle has been determined, the energy dissipated in doing damage can be calculated. This energy along with energy dissipated in other ways, can be used in an energy balance equation to calculate the vehicle's initial velocity. Example Assume a 3,000 lb vehicle (vehicle 1) skids with a drag factor of .70 for 60 ft, then strikes a parked vehicle (vehicle 2) weighing 3,500 lbs in the front. Barrier impact tests run at 50 fps using the same type of vehicle as the striking vehicle (1) and tests run at 30 fps using the same type of vehicle as the struck vehicle (2) resulted in similar damage patterns as reflected by the accident vehicles. After impact, the two vehicles remained locked together and slide as a unit for 60 ft, with a drag factor of .50. In this twovehicle system, the striking vehicle has all the initial energy. The amount of energy initially possessed by the striking vehicle is calculated by quantifying the amount of work done in sliding and doing damage. Energy (work) dissipated by vehicle 1 while skidding, before impact: E1 = wfd E1 = (3,000) (.70) (60) E1 = 126,000 ft-lbs Energy (work) dissipated in doing damage to vehicle 1: E2 = w1v1²/2g E2 = (3,000) (50²)/64.4 E2 = 7,500,000/64.4 E2 = 116,459 ft-lbs Energy (work) dissipated in doing damage to vehicle 2: E3 = w2v2²/2g E3 = (3,500) (30²)/64.4 E3 = 3,150,000/64.4 E3 = 48,913 ft-lbs Energy (work) dissipated by vehicles 1 and 2 while sliding after impact: E4 = (w1 + w2)fd E4 = (3,000 + 3,500) (.50) (60) E4 = (6,500) (.50) (60) E4 = 195,000 ft-lbs Total amount of energy possessed by the striking vehicle (1) when it first began to skid: ET = E1 + E2 + E3 + E4 ET = 126,000 + 116,459 + 48,913 + 195,000 ET = 486,372 ft-lbs This amount of energy can be used to determine the velocity of the striking vehicle when it first began to skid: v = %2g ET/w1 v = %(2) (32.2) (486,372)/3,000 v = %31,322,357/3,000 v = %10,440.786 v = 102.18 fps It is obvious, from this example, that the key to this method of estimating speed is assigning an appropriate EBS value for the amount of energy required for damage. To accomplish this we could go out and find like vehicles and conduct crash tests until the actual accident scenario is reproduced. This however, is impractical. On the other hand, we can compare our accident data to crash tests which have already been performed. Yet, even with this method the possibilities are infinite due to the number of different damage profile configurations. The best way would be to have a general mathematical model. This model would relate energy to damage (crush) for as many damage profile configurations as possible. For years crash tests have been conducted which examine the dynamics of automobile collisions. The results of these tests indicate that the deceleration force on an automobile in a head-on collision was directly proportional to the deformation of the vehicle. The studies also proved that vehicles behave like linear plastic springs, in other words they dissipate energy rather than store it (the damage remains after the force of the collision is removed). Earlier in the course you were given the equation for the elastic potential energy of a compressed spring. PEK = ½kx² This equation also applies to a linear plastic spring, i.e., one that does not return to its original shape. Vehicles behave like linear plastic springs when they are crushed, dissipating energy. This is modeled by a spring at its free length attached to the vehicle. The vehicle makes contact with a barrier and the spring begins to compress, dissipating kinetic energy and slowing the vehicle. The vehicle stops when all the Ke has been dissipated by compressing the spring. If the vehicle in the previous slide was traveling 30 fps and weighed 3,000 lbs, it would initially possess approximately 41,900 ft-lbs of kinetic energy. To bring the vehicle to a stop, the spring would have to deform enough to dissipate 41,900 ft-lbs of kinetic energy. In equation form this is written: ½mv² = ½kx² The actual amount of deformation would depend on the spring's stiffness. This concept can be used to estimate the amount of energy dissipated by damage if the permanent deformation and the vehicle's stiffness are known. The process of calculating an accident vehicle's speed from damage begins by understanding the variables involved. First of all you have to determine the A and B stiffness coefficients for the accident vehicle. These stiffness coefficients reflect the force deflection characteristics (stiffness) of the vehicle structure. The A and B coefficients can be estimated from fixed barrier impact tests by plotting permanent crush against the impact speed. Remember, the deceleration force of an automobile is directly proportional to the crush. One of the variables for the A and B coefficients, b1, must be solved first. The following equation will solve for b1. b1 = (v - bo)/C b1 = change in impact speed to change in crush (unknown) v = impact velocity (mph not fps) (crash test vehicle) bo = 3 mph to 8 mph (the maximum barrier impact velocity resulting in no permanent damage. The lower number will give the lower speed in the energy dissipated doing damage equation) C = average crush (inches) (crash test vehicle) NOTE: Both (v) and bo must be multiplied by 17.6 (12)(22)/15 = 17.6 (converts to in/sec.). The A coefficient can be determined from the following equation. A = (w)(bo)(b1) / (g)(W) A = A coefficient (unknown) w = vehicle weight (lbs) (crash test vehicle) bo = substituted from the b1 equation b1 = solved from the b1 equation g = 386.4 in/sec² W = crush width (inches) (crash test vehicle) The B coefficient can be determined from the following equation. B = (w)(b1²) / (g)(W) B = B coefficient (unknown) w = vehicle weight (lbs) (crash test vehicle) b1 = change in impact speed to change in crush, substituted from b1 equation g = 386.4 (in/sec²) W = crush width (inches) (crash test vehicle) Your handouts will give you the vehicle weight of the test vehicle. Crush width is given in some of the crash test results. When it is not given the Crash III chart will be used to determine A and B coefficients. Remember that all the variables given so far reflect information from the crash test vehicle. Once the A and B coefficients have been determined, they can be substituted into the equation for energy dissipated in doing damage. There are three equations. They are based on the number of crush measurements ("C" measurements) taken from the accident vehicle. If only two crush measurements are taken the following equation is used. E=W[G+A/2(C1+C2)+B/6(C1²+C2²+ C1C2)](1+TAN²1) If four crush measurements are taken the following equation is used. E=W/6[6G+A(C1+2C2+2C3+C4)+B/3(C1² +2C2²+2C3²+C4²+C1C2+C2C3+C3C4)](1 +TAN²1) If six crush measurements are taken the following equation is used. E= W/5[5G+A/2(C1+2C2+2C3+2C4+2C5+C6) + B/6(C1²+2C2²+2C3²+2C4²+2C5²+C6²+C1C 2 +C2C3+C3C4+C4C5+C5C6)](1+TAN²1) E = energy dissipated (in-lbs) (divide end result by 12 to convert to ft-lbs) G = A²/2B (lbs) W = width of crush region (inches) (accident vehicle) A = the maximum force per inch of damage width which will not cause permanent damage (lb/inch) B = the spring stiffness per inch of damage width (lb/inch²) 1 = the angle between a line perpendicular to the vehicle's damaged surface and the force direction It is recommended that this factor not exceed 2. The tangent squared of any angle above a 45E angle added to 1 will exceed 2. C = crush measurements of damage Assume the PDOF angle is 50 degrees. Force Direction 1 d1 d2 C’s d1<d2 Tan 50 = 1.19 1+ Tan² 50 1+ 1.19² 1+1.42 2.42 Since this exceeds 2 it would be an indication that the force came from the side and not the front. Therefore, the angle should be measured from the side. Example Assume we are investigating an accident involving a 1982 Chevrolet Camero, 2 door which strikes a bridge head-on. The following crush measurements are made of the accident vehicle: C1 = 29.4 inches; C2 = 30.4 inches; C3 = 30.6 inches; C4 = 30.2 inches; C5 = 29.3 inches; C6 = 27.5 inches. The width of the crush is 71 inches. Example (continued) Calculate the impact speed of the accident vehicle. From our handouts we find that impact tests done on a 1982 Camero were done from a speed of 35.4 mph, weight 3428 lbs, crush width 62.4 inches, and average crush 24.57 inches, use 5 mph for bo. Solve for b1 first. b1 = (v - bo) / C b1 = (35.4)(17.6) - (5)(17.6) / 24.57 b1 = 623.04 - 88 / 24.57 b1 = 535.04 / 24.57 b1 = 21.77 Solve for the A coefficient. A = (w)(bo)(b1) / (g)(W) A = (3428)(5)(17.6)(21.77) / (386.4)(62.4) A = 6567225.3 / 24111.36 A = 272.24 lb/in. Solve for the B coefficient B = (w)(b1²) / (g)(W) B = (3428)(21.77²) / (386.4)(62.4) B = (3428)(473.93) / 24111.36 B = 1624642 / 24111.36 B = 67.38 lb/in. Solve for G: G = A² / 2B G = 272.24² / (2)(67.38) G = 74114.618 / 134.76 G = 549.97 lbs Now we can solve the energy dissipated by damage equation. Since we have six crush measurements taken from the accident vehicle we will use the following equation. E=W/5[5G+A/2(C1+2C2+2C3+2C4+2C5+ C6)+B/6(C1²+2C2²+2C3²+2C4²+2C5²+C6² +C1C2+ C2C3+C3C4+C4C5+C5C6)](1 + TAN²1) E=71.0/5{(5)(549)+272/2[29.4+(2)(30.4)+( 2)(30.6)+(2)(30.2)+(2)(29.3)+27.5]+67/6[2 9.4²+(2)(30.4²)+(2)(30.6²)+(2)(30.2²)+(2)(2 9.3²)+27.5²+(29.4)(30.4)+(30.4)(30.6)+(30. 6)(30.2)+(30.2)(29.3)+(29.3)(27.5)]}{1 + 0} E=14.2{[(2745)+(136)(29.4+60.8+61.2+60. 4+58.6+27.5)+11.16(864.36+1848.32+1872 .72+1824.08+1716.98+756.25+893.76+930. 24+924.12+884.86+805.75)]}(1) E=14.2[(2745)+(136)(297.9)+(11.16)(1332 1.44)] (1) E=14.2(2745+40514.4+148667.2)(1) E=14.2(191926.6)(1) E=2,725,358.7 in-lbs/12 = 227,113.22 ft-lbs EBS = v = %2gE/w v = %(2)(32.2)(227,113.22)/3428 v = %146,626,091.79 v = %4266.65 v = 65.31 fps (15/22) = 44.53 mph Equation limitations Remember that A and B coefficients are established from barrier tests of many vehicles. If the accident vehicle is not listed in the barrier test information, then it must be assigned to a group. That group's A and B coefficients must then be used. This may or may not be accurate. A different uniform stiffness is assumed for the entire front and rear widths and the entire lengths of the sides. Frame or uni-body structure involvement is an important factor in front and rear collisions. If the uni-body structure is involved, the stiffness coefficients may be accurate. If only sheet metal is involved the stiffness coefficient may be too high. This applies to the sides of the vehicle also. There are "hard" spots at the wheel areas and "soft" spots between the wheels in the passenger area. The frame and frame-type structure provide the major stiffness to the vehicle. It is likely to be more accurate to weigh the crush measurement in favor of the deformation in the frame or uni-body structure area. Be careful when there is under-ride or override. Damage can be difficult to measure. Refer to the S.A.E. handout on measuring protocol. Do not use the equation for rollover or sideswipe accidents. There is no stiffness data for large trucks. Alternative Method For Calculating Speed From Barrier Test Data. Some of the NHTSA crush data you have been provided with does not list the width of the damage sustained by the test vehicle. Without knowing the width you would have to refer to the Crash III chart for the A & B coefficients. There is an equation which will give the desired results and is much simpler to use. From our earlier example, a 1982 Chevrolet Camero test vehicle sustained 24.57 inches of average crush from a speed of 35.4 mph. The 1982 Chevrolet Camero accident vehicle sustained an average crush of 29.56 inches. We computed a speed of 44.56 mph using the long equation. Now we will use the same information in the short formula. Ratio = Test speed - speed without sustaining damage/average crush from test From our example: Ratio = (35.4 - 5)/24.57 Ratio = 30.4/24.57 Ratio = 1.23 Once the ratio has been established it can be substituted into the following equation which will give the equivalent barrier speed. Speed = Ratio (Cavg) + speed without sustaining damage From our example: Speed = 1.23 (29.56) + 5 Speed = 41.35 mph (speed of the accident vehicle) – From the long formula a speed of 44.56 mph was computed. This is a difference of just over 3 miles an hour. Remember, the ratio for each case needs to be determined based on crash test results. In this case the ratio was 1.23. 1.23 is not a constant for every situation. Equation limitations This equation assumes a straight linear relationship between the amount of vehicle crush and the velocity of that vehicle required to create the designated amount of crush. A limitation of the linear function is its lack of input for exact crush width. This hampers ability to compensate for collisions involving deformation to only a portion of the vehicle end. The linear function assumes damage across the entire end width of the vehicle. This can be compensated for by assigning some of your C1 through C6 measurements the value of zero. These zeros would be used for calculation of Cavg. C4 C5 C6 Note: C4, C5 and C6 equal zero inches. Vehicle Crush Profile Cavg = C1+C2+C3+0+0+0 6 Derivation of the Kinetic Energy Equation The work done on an object by a force is defined as the product of the magnitude of the force and the distance that the object moves in the direction of the force: Work = Fd (1) The concept of work is that work represents the amount of energy transformed from one form to another. Therefore, work and energy have different appearances for the same thing and are both measured in foot pounds.