Crush Analysis with Under-rides and the Coefficient of Restitution

Transcription

1 Institute of Police Technology and Management s 24th Annual Special Problems in Traffic Crash Reconstruction Crush Analysis with Under-rides and the Coefficient of Restitution Jeremy Daily Russell Strickland John Daily 27 April 2006 Abstract In this paper, a detailed discussion of the damage momentum technique is presented which involves an interpretation of the damage profiles, understanding the origin of the stiffness coefficients, and derivation of v. Also, a technique for estimating the energy dissipated during deformation based on residual crush measurements is explained in the context of the damage momentum solution. A staged crash test from the Special Problems 2005 conference is used as an example to validate the technique. Also, a discussion of the coefficient of restitution is given with the derivation of its relationship with crush energy. Finally, a discussion of the misapplication of damage energy techniques is outlined for trailer underride collisions. There were four crash tests were conducted at Special Problems 2005 in which passenger vehicles were run into a stationary tractor-trailer unit. Two of the impacts were collinear into the back of the trailer, while the other two were at right angles to the trailer tandems. Analysis for the collinear impacts was limited to standard COLM techniques, while the side impacts were analyzed by means of rotational mechanics. Damage crush profiles were recorded for later use with a damage-energy technique. In this paper, we will examine the previous impacts using a damage-energy technique. Furthermore, comparison of the damage-energy solution to recorded pre-crash speed measurements will validate the technique for the side impacts. Finally, the inability to apply a damage momentum solution to trailer under-ride collisions will be explained. Jackson Hole Scientific Investigations, Inc., 7845 Timber Hill Dr, Huber Heights, OH 45424, (937) , jeremy@jhscientific.com Fairfield City Police Department, 5320 Pleasant Ave., Fairfield, OH 45014, (513) , russell-gina@fuse.net Jackson Hole Scientific Investigations, Inc., P.O. Box 2206, Jackson, WY 83001, (307) , john@jhscientific.com 1

2 Contents Contents 1. Introduction Recap of Crash Tests from Review of Conservation of Linear Momentum Damage Momentum Analysis Crush Energy and Equivalent Barrier Speed (EBS) A Planar (Two Dimensional) Impact Model Relationship to v Determining Crush Energy The General Energy Model CRASH III Deformation Model Determining Stiffness Coefficients Crash Test Data Determining v test Determine Campbell Model Coefficients Determining A, B, and G Values Determining Damage Area Determining the Location of the Damage Centroid Longitudinal Location of the Centroid in the Damage Area Lateral Location of the Centroid in the Damage Area Locating the Damage Centroid with Respect to the Local Axis of the Vehicle Crush Energy Equations Relationship between Crush Energy and the Coefficient of Restitution Physics of an Impact Taxonomy of Impacts Nature of Elasticity Relative Velocities Orientation Understanding the Coefficient of Restitution Kinematic Definition of Restitution Kinetic Definition of Restitution Energetic Definition of Restitution Computing Restitution based on Damage Energy Computing Coefficient of Restitution Based on Crash Test Data Concluding Remarks on Restitution c 2006 Jeremy Daily, Russell Strickland, and John Daily

3 Contents 5. Analysis of Underride Collisions Plymouth Voyager Van into the rear of the tractor-trailer Jeep Cherokee into the rear of the tractor-trailer Under-ride Analysis Conclusions Summary and Conclusions 47 A. Analysis of the Nissan Crash 50 B. Analysis of the Plymouth Van Crash 51 C. Analysis of the Jeep Crash 52 Copyright Information The following material contains excerpts from Fundamentals of Traffic Crash Reconstruction, an IPTM publication. It is copyrighted and has been reprinted with permission of the authors for use in IPTM training programs. c 2006 Jeremy Daily, Russell Strickland, and John Daily 3

4 1. Introduction Figure 1: Scene photo of the mini van into the rear of the trailer. 1. Introduction 1.1. Recap of Crash Tests from 2005 There were four tests performed in two days. Engineers from MacInnis Engineering used a tow cable system to pull the bullet vehicles into the trailer. The tow cable was fed through the center of the rear duals for two crashes and along the right and left side for the other two crashes. The tractor trailer was common to all crashes. The trailer was a Pine 48 ft box van with sliding tandems. The tractor was a 2004 Mack single axle day cab (VIN: 1M1AE02YX4N00138). Rear Under-ride Crash #1 A 1989 Plymouth Voyager SE (VIN: 2P4FH4531KR174080) was pulled into a stationary tractor trailer, Figure 1. It was pulled into the left rear of the box van trailer and penetrated to the rear tires of the trailer. The tractor-trailer combination had its spring brakes applied and was pushed forward a small amount due to the impact. Rear Under-ride Crash #2 A 1994 Jeep Cherokee (VIN: 1J4FJ28S6RL225912) was run into the right rear of the box van trailer in the same manner as crash #1. Rear Dual Axle Crash #3 A 1992 Nissan Sentra (VIN: 1N4EB32A7NC734928) was pulled by cable into the rear duals of the box van trailer. The trailer rotated and the tractor remained 4 c 2006 Jeremy Daily, Russell Strickland, and John Daily

5 1.1. Recap of Crash Tests from 2005 Figure 2: Scene photo of the Jeep into the rear of the trailer. Table 1: Comparison of speed estimates using reconstruction techniques to measured speeds in miles per hour. Bullet Test Vehicle Analysis Type Lower Bound Upper Bound Actual Speed Jeep In-line Momentum Voyager In-line Momentum Nissan Rotational Mechanics Honda Rotational Mechanics stationary. The impulse of the collision rocked the tractor trailer but did not tip it over. Rear Dual Axle Crash #4 A 1993 Honda Accord (VIN:1HGCB7693PA090661) was pulled into the right rear duals of the box van trailer. The trailer rocked and the tractor drive axles moved about 1 inch. Table 1 shows a comparison of the reconstructed speeds with the actual speeds measured by a RADAR system. The two under-ride collisions were problematic when trying to predict impact speeds using momentum due to the high mass ratio. The impact analysis using rotational mechanics concepts proved to be accurate. Moreover, using the middle values of the ranges yielded answers within a couple miles per hour. The analysis presented herein is valid and the evidence can be easily gathered at the scene, provided c 2006 Jeremy Daily, Russell Strickland, and John Daily 5

6 1. Introduction Figure 3: Scene photo of the Nissan into the rear of the trailer. Notice the two units did not stick together. Figure 4: Scene photo of the Honda into the rear of the trailer. Notice the lack of damage to the rear duals of the trailer. 6 c 2006 Jeremy Daily, Russell Strickland, and John Daily

7 1.2. Review of Conservation of Linear Momentum the on-scene investigator is trained to look for evidence under the trailer. In this paper, we are going to analyze in detail the damage-momentum solution for the impact speed of the Honda. The other three crashes are analyzed using WinCRASH, a commercial implementation of the CRASH3 computer program Review of Conservation of Linear Momentum Conservation of Linear Momentum (COLM) techniques are a proven tool in any reconstructionists toolbox. There are times, however, where using COLM is not useful (e.g., high mass ratios, insufficient evidence and so forth). A derivation of the momentum equation is provided in Section A collision is considered to be a closed system. That is, the only forces acting on the vehicles during the collision phase are the collision forces themselves. Ground frictional forces are neglected. This may not be valid for low speed collisions or for collisions with high mass ratios. Collisions are inelastic, that is, kinetic energy is not conserved. Linear momentum is conserved in the collision system. In other words, system momentum before equals momentum after. Collisions may be collinear or angled. Collisions are also either central or non-central. Approach and departure angles are based upon the impact circle. See Section?? for further review. Departure speeds are based upon calculations using the Work Energy Theorem. The analysis chosen is based upon the type of post-impact trajectory (spin, etc.) Change of velocity vectors ( v) may be calculated in a straight-forward manner using the vector geometry with the law of cosines. The direction of the v vectors may be calculated using the Law of Sines. This gives us the Principal Direction of Force (PDOF). If both v vectors are calculated using the Law of Cosines, then Newton s Third Law may be used to check the results of the calculation. A simple geometry check may be used to see if the v vectors are actually opposite in direction. c 2006 Jeremy Daily, Russell Strickland, and John Daily 7

8 2. Damage Momentum Analysis We see this analysis is not dependent on the amount of vehicle damage and the energy causing the damage. Neither is it dependent on pre- or post-impact vehicle rotations, except in choosing correct trajectory type. There are some impact configurations or conditions for which conservation of linear momentum is not the sole analysis. These include, but are not limited to, barrier impacts, moving barrier impacts, fixed object collisions, and in-line collisions. The in-line collisions can be either head-on or in the same direction. We will examine some of these impact types in this chapter. 2. Damage Momentum Analysis 2.1. Crush Energy and Equivalent Barrier Speed (EBS) The equivalent barrier speed (EBS), sometimes referred to as the barrier equivalent velocity (BEV), is determined by calculating the amount of energy that the vehicle dissipates during a crash. The speed is determined from v = 2KE m or S = 30KE w (1) where m is the mass (slugs or kilograms) of the vehicle and KE is the kinetic energy (ft-lb or joules) of the vehicle immediately before it was used to crush the vehicle. Therefore, there is no post-impact velocity and the vehicle comes to a complete stop. The reason it is called an equivalent speed is because it may not be an actual speed. If there is any post-impact velocity, then the EBS and the actual speed will be different. There are some instances in which the EBS will be different than the v as outlined in reference [1]: 1. The EBS will be higher than the v of a vehicle if the vehicle strikes an object that is rigid and movable. 2. The EBS will be lower than the v of a vehicle if the vehicle strikes an object that is soft and massive. An example of this would be a vehicle running into a snow bank. 3. The EBS and v will be the same whenever a vehicle strikes an object whose stiffness is proportional to the weight ratio of the object to the vehicle. In this section, we will know the equivalent barrier speed and the amount of crush energy. The coefficient of restitution will be assumed away for this section. 8 c 2006 Jeremy Daily, Russell Strickland, and John Daily

9 2.2. A Planar (Two Dimensional) Impact Model 2.2. A Planar (Two Dimensional) Impact Model Often, the collision force between two vehicles passes through one or both vehicles offset from the center of mass. When this situation occurs, we will see the vehicle upon which this force acts to both move in the direction of the force and also to rotate about its center of mass. A schematic is illustrated in Figure 5. Let us now relate this general collision model to our system of two eccentric, in-line vehicles shown in Figure 6. When looking at Figure 6, we would intuitively expect the acceleration of the center of mass of each vehicle to be different from each other, simply because the vehicles will tend to rotate away from the collision. In the same way, we would expect the acceleration at the centroid of the damage areas to be larger than the acceleration of the center of mass of each vehicle, respectively. Let us look at the governing equations for vehicle #1. Vehicle #2 may be analyzed in a similar way. F 1 = M 1 a 1 (Newton s Second Law) (2) τ 1 = I 1 α 1 (Newton s Second Law for rotation) (3) Recall I = moment of inertia = mk 2, where k = radius of gyration. So, τ 1 = m 1 k 2 1 α 1 (4) Torque is also defined as the product of a force and a lever arm: τ 1 = F 1 h 1 (5) Now consider the following relationship that equates the accelerations: a c = a 1 + h 1 α 1 (6) From rotational mechanics a = rα and r = h, so: a c a 1 = h 1 α 1 (7) Solving for α 1 : α 1 = a c a 1 h 1 (8) c 2006 Jeremy Daily, Russell Strickland, and John Daily 9

10 2. Damage Momentum Analysis F τ α h a F is the collision force in lbs (N). a is the translational acceleration of the center of mass, in-line with the direction of force. The units are ft/sec 2 or m/sec 2. h is the lever arm upon which the force acts (ft or m). τ is the torque about the center of mass. Torque is the cross product of F and h. Units are lb-ft or N-m. α is the angular acceleration caused by τ about the center of mass. Basic units are rad/sec 2 for both systems of measure. Figure 5: The general case of a non-central collision. This figure shows the overall collision force acting on a vehicle. 10 c 2006 Jeremy Daily, Russell Strickland, and John Daily

11 2.2. A Planar (Two Dimensional) Impact Model a 1 a c a 2 h 1 F h 2 M 2 M 1 τ 2, α 2 τ 1, α 1 M 1 and M 2 are the respective vehicle masses. F is the total collision force. a 1 and a 2 are the respective accelerations of the CM of each vehicle. In an offset collision, these probably will not be the same for each vehicle. a c is the common acceleration of the crush zone of the respective vehicles. Therefore, the centroids of the damage areas must reach a common velocity since collision times are identical. h 1 and h 2 are the lever arms upon which force F acts. τ 1 and τ 2 are the respective torques acting about the mass centers. α 1 and α 2 are the respective angular accelerations of each vehicle about their centers of mass. Figure 6: A schematic of a general non-central, in-line collision. c 2006 Jeremy Daily, Russell Strickland, and John Daily 11

12 2. Damage Momentum Analysis Now, the torques in Eqs. (4) and (5) are the same, so: F 1 h 1 = m 1 k 2 1 α 1 (9) Substitute for α 1 : [ ] F 1 h 1 = m 1 k 2 ac a 1 1 h 1 Divide both sides by m 1 and multiply by h 1 : (10) F 1 m 1 h 2 1 = k2 1 (a c a 1 ) (11) Recall from Newton s Second Law that F 1 = m 1 a 1 or a 1 = F 1 m 1. Substitute this value into Eq. (11): a 1 h 2 1 = k 2 1 (a c a 1 ) a 1 h 2 1 = k 2 1 a c k 2 1 a 1 a 1 h a 1 k 2 1 = k 2 1a c We will define the effective (dynamic) mass ratio as: a 1 (k h2 1 ) = k2 1 a c ( k 2 ) a 1 = 1 k a c (12) h2 1 γ = k2 k 2 + h 2 (13) Substituting the definition of an effective mass ratio into Eq. (12) gives the simple proportion: a 1 = γ 1 a c (14) Equation (14) shows us the acceleration of the CM of vehicle #1 is a proportion, γ 1, of the acceleration of the centroid of the damage area. We may also see that γ will always be less than or equal to 1 because the denominator of Eq. (13) will never be less than the numerator. If γ = 1, then we have a central collision without rotation. This is a proof of our intuition Relationship to v Consider the following general relationship: a = v t (15) 12 c 2006 Jeremy Daily, Russell Strickland, and John Daily

13 2.3. Relationship to v For vehicle #1: a 1 = v 1 t (16) a c = v c t (17) So substituting Eqs. (16) and (17) into Eq. (14) yields: v 1 t = γ 1 v c t (18) The t terms are common in the denominators on both sides and will cancel: v 1 = γ 1 v c (19) In offset collisions, the acceleration a c and velocity change v c of the damage centroid will always be larger than the velocity change and acceleration of the center of mass. Using an energy and momentum based approach [2, 3, 4], formulas for the change in velocity for an eccentric impact can be determined as: v 1 = 2γ 1E crush ) or v 1 = m 1 (1 + γ 1m 1 γ 2 m 2 2gγ 1E crush ) (20) w 1 (1 + γ 1w 1 γ 2 w 2 We calculate v 2 knowing the change in momentum from one vehicle has to be equal and opposite the change of momentum in the other. m 1 v 1 + m 2 v 2 = 0 v 2 = m 1 m 2 v 1 (21) Newton s Third Law is satisfied because the impulse vectors, and thus the v vectors, are opposite in direction. Example 1 Recall from 2005 that the 1993 Honda Accord crashed into the rear dual axles of a stationary semi-trailer. There was no permanent damage done to the trailer and the Honda absorbed 81,200 ft-lb (110,100 J) of crush energy. The trailer rotated around its kingpin and had a moment of inertia of 421,470 lb-ft-sec 2 (571,513 kg-m 2 ). The distance from the kingpin to the point of impact is 36.4 ft (11.09 m). The Honda weighs 2900 lb (1315 kg) and the trailer weighs 13,425 lb (6090 kg). Determine the impact speed of the Honda. c 2006 Jeremy Daily, Russell Strickland, and John Daily 13

14 2. Damage Momentum Analysis Solution Since there is no permanent damage to the trailer wheels, there is no contribution to the total crush energy from the trailer. Therefore: E crush = E Honda + E trailer = E Honda + 0 We also know that the trailer has no initial lateral acceleration. As such, the v of the trailer is its actual post-impact velocity. To determine the change in velocity of both vehicles, we will use Eq. (20). Equation (20) requires the effective mass ratios of each of the vehicles. Since the Honda hit the center of the rear duals, the collision force acts through the Honda s center of mass, so its effective mass ratio is 1. On the other hand, the effective mass ratio of the trailer must be computed from Eq. (13). The square of the radius of gyration (k 2 ) of the trailer was determined to be ft 2 (93.83 m 2 ). Therefore, the effective mass ratio of the trailer is: US SI γ 2 = = k 2 2 k h = γ 2 = = k 2 2 k h = Notice that both ratios are the same because a ratio has no dimension. Let us use subscript 1 for the Honda and subscript 2 for the trailer. Thus, for the Honda: v 1 = = 2gγ 1E crush ) w 1 (1 + γ 1w 1 γ 2 w 2 2(32.2)(1)(81, 200) ( ) (2900) 0.433(13,425) v 1 = = 2γ 1E crush ) m 1 (1 + γ 1m 1 γ 2 m 2 2(1)(110, 100) ( ) (1315) 0.433(6090) = ft/s = m/s In a similar fashion, we can determine the v for the center of mass of the trailer: 14 c 2006 Jeremy Daily, Russell Strickland, and John Daily

15 2.3. Relationship to v v c v ft (11.09 m) 27.3 ft (8.32 m) Figure 7: The v of the damage centroid is different than the v of the center of mass of the trailer. v 2 = = 2gγ 2E crush ) w 2 (1 + γ 2w 2 γ 1 w 1 2(32.2)(0.433)(81, 200) ( ) 13, (13,425) 1(2900) v 2 = = 2γ 2E crush ) m 2 (1 + γ 2m 2 γ 1 m 1 2(0.433)(110, 100) ( ) (6090) 1(1315) = 7.49 ft/s = 2.28 m/s Since Eq. (20) was developed by assuming the vehicles remain in contact after the collision, the coefficient of restitution is zero. In order for the Honda to experience the v computed in this example, the impact speed must be equal to the change in velocity of the Honda plus the post-impact velocity. This post-impact velocity is the v of the damage centroid of the trailer, v c, because it was initially at rest. Computing the change in velocity of the damage centroid of the trailer involves using the concept of similar triangles. Consider Figure 7, which shows the geometric relationship of the damage centroid, the center of mass, and the point of rotation. We can consider the trailer to be a rigid body rotating about its kingpin, so the relative velocities of any point on the trailer is proportional to its distance from the kingpin. This fact allows us to use the property of equal ratios to determine a relationship between v 2 and v c. c 2006 Jeremy Daily, Russell Strickland, and John Daily 15

16 3. Determining Crush Energy US v c v 2 = = SI v c = v = v c = v 2 = 1.333(7.49) = 9.98 ft/s v c = v 2 = 1.333(2.28) = 3.04 m/s US SI v 1 = v 1 + v c v 1 = v 1 + v c = = = ft/s = m/s S = mph S = kph The actual speed measured of the Honda at impact for this staged crash was close to 31 mph (50 kph). The accuracy of a damage momentum analysis is not always guaranteed, because the energy absorbed in the crash by crushing the vehicle is empirically based. In other words, the techniques used to determine the energy in a crash are not based completely on physics, but rather a curve fit to crash test data. The curve-fitting technique is the only tractable way to obtain energy values for vehicle crashes and the details will be discussed in the next section. 3. Determining Crush Energy The acronym CRASH stands for Computer Reconstruction of Automobile Speeds on the Highway. The CRASH3 algorithm is implemented in many commercial software packages such as EDCRASH, WinCRASH, and m-crash. The computer programs provide great tools to perform analysis; however, comprehension of the physics and models implemented in the computer code enable the user to make a more sound interpretation of the evidence and results of the computer analysis. CRASH has two parts: the trajectory calculator and the damage algorithm. This chapter explains some of the aspects of the damage algorithm. 16 c 2006 Jeremy Daily, Russell Strickland, and John Daily

17 3.1. The General Energy Model L x A D = Area of Damage Narrow width rectangles Figure 8: A large rectangle that represents the damage area can be broken into a series of small rectangles. Here, L is the damage width, A D is the area of the damage projection, and x is the depth of crush The General Energy Model Consider Figure 8, in which we have taken a rectangular damage profile and have segmented it into smaller rectangles that are 1 unit wide. Each of these rectangles is x units in depth. We may now define A and B stiffness coefficients for each unit of width where: The units of A are lb/inch (N/m), The units of B are lb/in 2 (N/m 2 ), The units of G are lb (N), and the value of G is equal to A2 2B. Now consider our energy equation: E = Ax + Bx2 2 + A2 2B (22) This equation was used to determine the energy per unit width for a rectangular damage profile. There, the unit width was the entire front of the vehicle. The A and B values were also for the entire width of the vehicle. Here, since we divided the entire rectangular damage area into smaller rectangles one unit wide (e.g., 1 inch), Eq. (22) can be used to calculate the damage energy for each of the narrow rectangles. Correspondingly, the A and B values are for one of the narrow rectangular strips. If we want to calculate the total damage energy, E T, we will have to multiply E by L. E T = EL = AxL+ Bx2 L 2 + A2 L 2B The product of L and x is the area of the large rectangle, which is the area of the damage. If we use the variable A D to denote this area of damage, then Eq. (23) becomes: (23) E T = A(A D ) + Bx 2 (A D) + A2 L 2B (24) c 2006 Jeremy Daily, Russell Strickland, and John Daily 17

18 3. Determining Crush Energy Factoring out the area term gives: E T = ( A + Bx ) A D + A2 L 2 2B (25) The centroid of a rectangle, denoted as x, in relationship to its length, x, is x. Thus, we can 2 write Eq. (25) in terms of the centroid of the damage area: E T = (A + B x) A D + A2 L 2B (26) 3.2. CRASH III Deformation Model To calculate the total energy it took to damage (crush) a vehicle, three things are needed: Stiffness coefficients. These are the A, B, and G values and come from crash test data. Ways to calculate these values are discussed in Section Area of damage. This is the A D value. In essence, we integrate over the damage profile to get the damage area in square inches (or square meters). There are several ways to do this and we will explore the trapezoid rule in Section 3.3. Depth of the centroid of the damage area. This is the x value. Calculating this value is discussed in Section The following discussions and the equations developed are for a collision in which the PDOF angle is 90 to the damage face. The situation in which the PDOF is not 90 is beyond the scope of this paper Determining Stiffness Coefficients The A stiffness coefficient represents the amount of force that the vehicle can sustain before it begins to permanently deform. The B stiffness coefficient represents the amount of force needed to permanently deform the vehicle structure. The G value is the area of the force-deflection triangle to the left of the abscissa (y-axis) and is related to A and B by the relationship G = A2 2B (27) Crash Test Data Most vehicle crashes are conducted under the New Car Assessment Program (NCAP), which requires a 35-mph (56 kph) full-frontal barrier crash test. Other crash tests include: angled 18 c 2006 Jeremy Daily, Russell Strickland, and John Daily

19 3.2. CRASH III Deformation Model frontal-fixed barrier collisions, movable barrier collisions (both deformable and non-deformable), narrow object impact tests, and side-movable deformable barrier tests. The remaining discussion on crash test data will deal only with full-frontal barrier tests. When we obtain crash test data, we will get the average crush of the vehicle. We will call this C avg. This assumes a uniform crush profile. Procedures for dealing with a non-uniform crush profile are discussed in Ref. [5]. Test reports are available to the public through the National Highway Traffic Safety Administration web site, These reports may have a data sheet entitled Accident Investigation Division Data that will reveal: The approach speed of the vehicle in kph: v test The test mass of the vehicle in kg: m test Six vehicle crush depths in mm: C 1, C 2, C 3, C 4, C 5, and C 6 Width of the damage profile in mm: L test Unit conversions will be required to work in US units. Dimensions must agree in all equations Determining v test There are different ways in which people have defined v test when using crush energy formulas. One method is to use the total v, which is the actual change in velocity experienced by the center of mass of the vehicle. It is defined as: v = v 1 v 3 where v 1 is the approach velocity and v 3 is the rebound velocity. Using the definition of the coefficient of restitution gives: v = v 1 (1 + ǫ) The equations for v are based on the energy, E crush, dissipated during the collision. These equations also assume no restitution. As such, the simplest way of computing stiffness values is to ignore the effects of restitution. This means that v test = v approach. The CRASH III Manual (Ref. [4]) presents the idea of using an effective energy that will account for the restitution of the crash test in the computation of the v values by introducing the effective energy term E eff = E A (1 + ǫ) 2 c 2006 Jeremy Daily, Russell Strickland, and John Daily 19

20 3. Determining Crush Energy where E A is the kinetic energy of the approaching vehicle. This is misleading because the actual energy dissipated in crushing the vehicle is: E crush = 1 2 mv2 A (1 ǫ2 ) = E A (1 ǫ 2 ) Therefore, it is the recommendation of the authors to either ignore restitution when computing crush energy or include the effects of restitution on total v calculation separate from the damage energy calculation. Some authors may advocate determining stiffness coefficient based on the total v. Again, this is ill-advised because using the total v will give crush energy values that are erroneously high unless restitution is near zero. For example, if ǫ = 0.1, then E eff will be 21% higher than E A, whereas E crush is only 1% lower than E A. Therefore, ignoring the restitution for barrier impact tests at 30 or 35 mph will result in small errors. It will also result in a consistent definition of v test, namely, the change in speed before rebound. Therefore, for a fixed barrier test, use the actual approach speed to determine the stiffness values Determine Campbell Model Coefficients Campbell noted that, for early 1970 s full-sized General Motors vehicles, the impact speed and the depth of crush followed a straight line similar to the line shown in Figure 9. The model for relating frontal barrier impact speed to crush damage takes a linear form: v = b 0 + b 1 C (28) where v is the impact speed, C is the residual crush, b 0 is the zero crush speed (intercept) in units of speed, and b 1 is the slope in units of speed per length (i.e. mph/in). When a vehicle crashes into an object, energy is expended. The majority of the energy lost in a collision is due to the plastic (permanent) deformation of the vehicle(s) and/or the objects involved in the crash. Quantifying this energy is difficult since there are many different mechanisms of dissipating the energy. However, a model based on crash tests can approximate the amount of energy dissipated. This is called an empirical model because it is based on observations rather than physical principles. The linear model shown in Figure 9 is an empirical model. Since energy is dependent on mass as well as velocity, a standard weight was used for diagrams similar to Figure 9. This technique can be used if only a single crash test is available. We will outline a simple procedure to extract the empirical crush coefficients. These coefficients are used to determine 20 c 2006 Jeremy Daily, Russell Strickland, and John Daily

21 3.2. CRASH III Deformation Model 60 Impact Speed, miles per hour b 1 0 b Residual Crush, inches Figure 9: Impact speed plotted against the measured residual crush for a frontal-fixed barrier impact. The slope of the line is b 1 and the intercept is b 0. This plot resembles Fig. 1 from Campbell s 1974 paper [6]. the energy dissipated for an irregular crush profile. The first step is to define a no-damage speed for our test vehicle, b o. Typical values for the speed at which no residual crush exists are 5 mph (8 kph) for front and rear impacts in which the vehicle is protected by bumpers. On the other hand, side impact no-damage speeds are typically 2 mph (3.2 kph). When using the US system of measurement, our A and B values will be in units of lb/in and lb/in 2, respectively, so we will need to express all of our no-damage speeds and v values in inches/second. In the same way, acceleration will be expressed in inches/second 2. Thus, 5 mph = 88 in/sec, 2 mph = 35 in/sec, and 32.2 ft/sec 2 = 386 in/sec 2. Similarly, when using the SI system of measurement, our A and B values will be in units of newton/meter and newton/meter 2, respectively, so we will need to express all of our nodamage speeds and v values in meters/second. In the same way, acceleration will be expressed in meters/second 2. Thus, 8 kph = 2.22 m/sec, and 3.2 kph = 0.89 m/sec. Let us now determine the slope of the line that would intersect the y-axis (speed) at b o and the crash data point (C avg, v test ). This slope is b 1 : b 1 = v test b o C avg (29) This equation assumes uniform crush depth. Vehicles with significant taper on the front or c 2006 Jeremy Daily, Russell Strickland, and John Daily 21

22 3. Determining Crush Energy Impact Speed v (C avg, v) b o C avg Crush Figure 10: Impact speed-crush graph. Point (C avg, v) is on the graph and can be used to calculate the slope, b 1, of the graph. some otherwise irregular crush profile need different crush averaging. Neptune provides a formula to determine stiffness coefficients based on irregular crush profiles in Ref. [5] Determining A, B, and G Values Once we have b o and b 1, we can calculate the A, B, and G stiffness coefficients. These equations take the Campbell impact speed-crush data (Figure 9) and convert it to force-crush data (Figure 11). The A stiffness coefficient can be determined with the following equation: A = m Tb o b 1 L test (30) where m T is the mass of the test vehicle with instrumentation. This is usually given in kilograms and needs to be in slugs for the US system (1 kg = slugs). The B stiffness coefficient can be determined with the following equation: B = m Tb 2 1 L test (31) The G value is a straight-forward calculation based upon the triangle geometry. G = A2 2B (32) 3.3. Determining Damage Area The initial models for the CRASH III program approximated irregular damage profiles by dividing the damage area into equally spaced trapezoids. Each trapezoid was bounded on the 22 c 2006 Jeremy Daily, Russell Strickland, and John Daily

23 3.3. Determining Damage Area F B F = A + Bx A G O x = residual crush Figure 11: The geometric relationship between A, B, and G. ends by the damage face and the undamaged collision face profile. The trapezoids are bounded on the sides by the crush measurements, C n. Two, four, or six crush measurements are taken and must be equally spaced. This results in one, three, or five crush zones, as seen in Figure 12. The damage area is computed by summing the areas of all the zones, so for Figure 12: A D = A 1 + A 2 + A 3 + A 4 + A 5 After some geometry and algebra, we get the damage area for six equally spaced crush measurements: A D = L 10 (C 1 + 2C 2 + 2C 3 + 2C 4 + 2C 5 + C 6 ) (33) Example 2 Determine the stiffness coefficients for a 1993 Honda Accord in SI units based on the NHTSA crash test 1875 performed by Calspan Corp. This crash test was at 56.3 kph (assume no rebound velocity) and the test weight was 1579 kg. The overall length of the damage region is L test = 1460mm. The crush depth dimensions reported are: Location C 1 C 2 C 3 C 4 C 5 C 6 Depth (mm) The mean crush depth is calculated using the area determined in Eq. (33): C avg = A D L = L( (481) + 2(522) + 2(523)+ 2(483) + 376) 10 L = 482 mm = m c 2006 Jeremy Daily, Russell Strickland, and John Daily 23

24 3. Determining Crush Energy L zone 1 zone 2 zone 3 zone 4 zone 5 C 6 C C 5 4 C 1 C 3 C 2 x + deformed vehicle + y Figure 12: An irregular damage profile of width L pictured here is broken up into five trapezoidal zones by taking six equally spaced crush measurements. 24 c 2006 Jeremy Daily, Russell Strickland, and John Daily

25 3.4. Determining the Location of the Damage Centroid Employing Eq. (29), converting 56.3 kph to m/s, and assuming that b 0 = 2.22 m/s gives the value for b 1 : b 1 = v test b o C avg = = This value is used in Eq. (30) to determine the A coefficient: A = m Tb o b 1 L test = 1579 kg(2.22 m s )( s ) 1.46 m = 66, 842 kg s 2 If we multiply the units for A by m/m, we would get units of kg-m ( 1 sec 2 m ), which are units of newtons/meter. The value for the B coefficient is: B = m Tb 2 1 L test = 1579 kg( s ) m = 838, 239 N m 2 The value for the G coefficient is: G = A2 2B ( ) 66, 842 N 2 m = 2 ( 838, 239 N ) m 2 = 2665 N 3.4. Determining the Location of the Damage Centroid The damage centroid is the point within the damage profile at which the collision force acts. It can be thought of as the center of mass of the damage area. If the CRASH III model is to c 2006 Jeremy Daily, Russell Strickland, and John Daily 25

26 3. Determining Crush Energy IMPORTANT! work for two-vehicle collisions, the damage centroids must reach a common velocity. Therefore, sideswipe collisions cannot be modeled with CRASH III. This constraint is not placed on a COLM solution! For a COLM solution, there is no requirement that the damage centroids reach a common velocity. In the interest of brevity, the results only are presented in this section. A detailed derivation can be found in Ref. [2] Longitudinal Location of the Centroid in the Damage Area x = C C C C C2 5 + C2 6 + C 1C 2 + C 2 C 3 + C 3 C 4 + C 4 C 5 + C 5 C 6 3(C 1 + 2C 2 + 2C 3 + 2C 4 + 2C 5 + C 6 ) (34) Lateral Location of the Centroid in the Damage Area Having determined the longitudinal location of the centroid, x, we will determine the lateral location of the centroid, ȳ, which pinpoints the centroid of the damage area. We will need this location to determine where the PDOF is acting. ȳ = L ( ) 13C1 18C 2 6C 3 + 6C C C 6 (35) 30 C 1 + 2C 2 + 2C 3 + 2C 4 + 2C 5 + C Locating the Damage Centroid with Respect to the Local Axis of the Vehicle Once the centroid of the damage area has been located within the damage area, we can locate the centroid with respect to the local x-y axis of the vehicle. The local axis has its origin at the center of mass (CM) of the vehicle. Positive x is forward of the CM. Positive y is to the passenger side of the CM. The x-y location of the centroid is an ordered pair with + or signs used depending on where it is with respect to the local axis. In the x-direction The value for x locates the depth of the centroid from the damage face of the vehicle. From vehicle specification databases (such as Expert Autostats R ) we can determine various measurements such as, front overhang, wheelbase, front bumper to front axle, center of mass to front axle, etc. Using these measurements, as necessary, along with x, the location of the centroid can be located with respect to the local vehicle axis. It may be helpful to sketch a picture showing these measurements to assist in locating the centroid and to help remember the sign of the location. 26 c 2006 Jeremy Daily, Russell Strickland, and John Daily

27 3.5. Crush Energy Equations In the y-direction The value for ȳ laterally locates the centroid from the center of the damage area (half of the measured damage width). From vehicle specification databases, we can determine various measurements such as vehicle width, front overhang, wheelbase, front bumper to front axle, center of mass to front axle, etc. The center of mass of the vehicle is generally located at physically half the vehicle width. When measuring the damage area of a vehicle, the center of the damage area (half the damage width) is located with respect to the center of mass of the vehicle and is called D. Since ȳ is referenced to the center of the damage width, knowing D allows us to locate the centroid with respect to the local vehicle axis. It may be helpful to sketch a picture showing these measurements to assist in locating the centroid and to help remember the sign of the location Crush Energy Equations In the previous sections, we have developed equations for damage area and for damage centroid depth. Now we substitute these values into Eq. (26) on page 18: E T = (A + B x) A D + A2 L 2B The result for six evenly spaced crush measurements (five crush zones) is: E T = L 5 ( A 2 (C 1 + 2C 2 + 2C 3 + 2C 4 + 2C 5 + C 6 ) + B 6 [ C C C C C2 5 + C2 6 + C 1C 2 + C 2 C 3 + C 3 C 4 + C 4 C 5 + C 5 C 6 ] ) + 5A2 2B (36) Example 3 Determine the crush energy of a 1993 Honda Accord given the following equally spaced crush measurements across a front width of L = in: Location C 1 C 2 C 3 C 4 C 5 C 6 Depth (in) The A stiffness coefficient was determined from the data in Example 2 to be 396 lb/in and the B coefficient is 129 lb/in 2. This is a six-crush measurement that can be cumbersome by hand. c 2006 Jeremy Daily, Russell Strickland, and John Daily 27

28 4. Relationship between Crush Energy and the Coefficient of Restitution Therefore, let us introduce the temporary variables: x 1 = C 1 + 2C 2 + 2C 3 + 2C 4 + 2C 5 + C 6 x 2 = C C C C C2 5 + C 2 6 x 3 = C 1 C 2 + C 2 C 3 + C 3 C 4 + C 4 C 5 + C 5 C 6 So from Eq. (36) we get: Solving for the x values gives: E T = L 5 ( Ax1 2 + B(x 2 + x 3 ) 6 ) + 5A2 2B (37) x 1 = (11.0) + 2(14.78)+ 2(15.0) + 2(14.0) = in x 2 = ( ) + 2( ) + 2( ) + 2( ) = in 2 x 3 = 3.3(11.0) (14.78) (15.0)+ 15.0(14.0) (7.0) = in 2 Substituting these values into Eq. (37) yields: E T = ( 396(119.86) ( ) 6 = 12.75(23, , ) = 974, 390 in-lb = 81, 200 ft-lb ) + 5(396)2 2(129) 4. Relationship between Crush Energy and the Coefficient of Restitution 4.1. Physics of an Impact An impact happens when two objects interact with large forces over a short period of time. We refer to this interaction as a collision. This causes large impulsive forces which typically deform the objects. The moment two objects come together is called incidence. Immediately after incidence the compression phase begins where the colliding objects deform and absorb kinetic energy. This compression phase has a finite duration and is terminated when the dynamic deformation reaches a maximum. Following the maximum deformation, a period of restitution occurs where the object may rebound. During the rebound phase some (not all) of the stored energy is turned back into kinetic energy as the objects depart with some relative velocity. We also 28 c 2006 Jeremy Daily, Russell Strickland, and John Daily

29 4.1. Physics of an Impact v = v in Fixed Barrier Compression Rebound (a) Location of the vehicle at incidence, t = 0. v = 0 (b) Location of the vehicle at maximum crush, t = t c. v = v out Fixed Barrier Fixed Barrier (c) Location of the vehicle at separation, t = t s. Figure 13: The different phases of an impact between a vehicle and a solid fixed barrier. define the collision as taking place with no displacement with respect to an inertial (ground) reference frame. Hence, we do not have to consider any gain or loss of potential energy during the collision phase. Figure 13 shows the two phases of a collision of a vehicle into a fixed barrier. In Fig. 13a, the vehicle is just touching the barrier and has some kinetic energy. The contact generates a force that acts over some distance to the point of maximum crush. The collision force acting through the distance is the work done in the crash. All of the energy used to do the work associated with deformation comes from the initial kinetic energy. c 2006 Jeremy Daily, Russell Strickland, and John Daily 29

30 4. Relationship between Crush Energy and the Coefficient of Restitution 4.2. Taxonomy of Impacts Nature of Elasticity The discussion in this section is based upon classical mechanics, which have a foundation in Newton s three laws of motion as well as the work done by Huygens and others just before Newton s time. We use classical mechanics for crash reconstruction because, in most cases, the results we obtain are reasonable and may be done with relatively simple math. The variables involved are simple to define and are measurable by means of testing. The core of classic impact analysis is the impulse-momentum model. There are other impact models that are based upon elastic wave propagation, contact mechanics, and plastic deformation or hydrodynamic modeling which are far beyond the scope of this paper. We may place collisions into two general categories: an elastic or inelastic collision: An elastic collision is one in which kinetic energy is conserved. As such, an elastic collision is a conservative system, meaning a system in which kinetic energy is conserved. Real world systems are never completely elastic, as kinetic energy is never completely saved. The work done to deform the objects is therefore called reversible work. In elastic collisions, the colliding objects bounce off of each other. An inelastic collision is one in which kinetic energy is not conserved. This does not mean that all of the kinetic energy of the system goes away in the collision. In fact, usually it does not, as the colliding objects will move away with some velocity after the impact. The amount of kinetic energy that may be lost in a collision is consistent within the bounds set by conservation of linear momentum. In a completely inelastic collision, the kinetic energy required to do the work to deform the objects is transformed into other forms of energy, such as heat. As such, the work done to deform the colliding objects is called irreversible work, because we cannot get it back. A characteristic of inelastic collisions is the two objects tend to stick together. Real world collisions usually fall somewhere between being elastic and inelastic. A measure of the elasticity of a collision is the coefficient of restitution, which, by Newton s definition, is a ratio of the relative velocity of approach to the relative velocity of recession. The coefficient of restitution may vary between zero and one. A value of one indicates a completely elastic collision, while a value of zero indicates a completely inelastic collision. For example, if we drop a steel ball bearing on a hard steel plate, it will bounce quite a bit. We would say that collision has a relatively high coefficient of restitution. If we take a soft lead ball and drop it on the same plate, it will not bounce quite as much. Hence, its coefficient of restitution is lower than that of the steel ball, but still has some value. If we replace the lead ball with a soft lump of clay, then we will see an inelastic collision, as the clay will hit and stick to the surface. As 30 c 2006 Jeremy Daily, Russell Strickland, and John Daily

31 4.2. Taxonomy of Impacts the coefficient of restitution becomes closer to zero, more of the system kinetic energy is being dissipated into other forms of energy, primarily heat. We also consider that the only forces acting on the bodies during the collision are the impulsive (collision) forces themselves. As such, for either one or two dimensional collisions, we ignore the effects of any ground frictional forces. This is an example of an unconstrained impact. In the real world, this assumption may not always be valid and depends on several things, such as relative vehicle masses, presence of significant ground forces, or other constraints that may affect vehicle motion Relative Velocities Consider for a moment the steel plate and ball mentioned above. How might the relative velocity at impact affect the impact behavior? Let us think about what happens if we drop the steel ball on the plate from a height of 14 feet. This will result in an impact velocity of about 30 fps. If we examine this ball after the impact, we probably will not be able to discern and permanent deformation in the ball. If we do the same thing with the lead ball, then we may see a small flat spot on it, but still no great deformation. Now, let us fire the balls in turn out of an air gun at 300 fps. The steel ball will still bounce off the plate, but there may well be some measurable deformation in it. It s coefficient of restitution, in the sense of classical mechanics, will likely be less because of this permanent deformation. The lead ball, being softer (less internal strength), will probably be flattened by the impact and may have little or no bounce at all. As a final example, we will replace the air gun with a high velocity rifle that is capable of launching the balls at 3000 fps. In this case, the steel ball will likely penetrate through the plate, resulting in large plastic flows of both the ball and the plate. This is the beginning of hydrodynamic behavior where the extreme stresses make the solid behave like a fluid. In a similar way, the lead ball may also penetrate the plate, even though it is much softer than the plate. These high velocity impacts are called ballistic impacts, and we may not use classical impact mechanics to determine impact behavior. As we may see with our thought experiment, the coefficient of restitution is a function not only of the material properties of the impacting bodies, but also on the relative velocity at impact. We will discuss this further in Section Orientation There are generally two different types of impacts: collinear and oblique: A collinear impact occurs when the direction of travel coincides with the direction of force. c 2006 Jeremy Daily, Russell Strickland, and John Daily 31

32 4. Relationship between Crush Energy and the Coefficient of Restitution Head-on or rear-end collisions are examples of collinear impacts. Also, a collinear impact occurs whenever one vehicle or object is stationary. An oblique impact occurs when the line of force is not coincident with the direction of travel. The oblique impact is the general case of any planar (two dimensional) impact which means a central impact is a special case of an oblique impact. In traffic crash reconstruction, we may also categorize collisions as being either central or noncentral. A central collision is one in which the impulse force (PDOF) passes through the center of mass of the vehicle (object). A non-central collision means the impulse force does not go through the center of mass. In a central collision for both vehicles, we would expect the postimpact velocity (both magnitude and direction) to be similar and for the vehicles to move away from the collision with little or no rotation. These constraints are not placed on a non-central collision Understanding the Coefficient of Restitution The coefficient of restitution provides the analyst with a technique of dealing with energy losses in a collision. It is fairly simple to use with in-line collisions but it fails to provide all the missing pieces to a planar impact problem. There are three basis for the definition of the coefficient of restitution: kinematic, kinetic, and energy. These definitions will be presented and the relationship between them will be explained Kinematic Definition of Restitution Sir Issac Newton provided the first definition of restitution with a formula based on the relative velocities of each object: ǫ = v 2,out v 1,out v 1,in v 2,in (38) where v refers to the magnitude of the velocity normal to the impact plane. For in-line collisions, these velocities are the actual velocity magnitudes. If the collision is oblique, then the velocity used in Eq. (38) is the component of the velocity vector normal (perpendicular) to the plane of impact. While Newton was correct with most everything he wrote, he mistakenly presumed the coefficient of restitution was only a material property. It has been since shown that restitution also depends on the relative velocities themselves. Also, application of Eq. 38 for eccentric impacts may lead to an apparent increase in energy which is physically inadmissible. 32 c 2006 Jeremy Daily, Russell Strickland, and John Daily