Conseration of Linear Momentum (COLM) ade Bartlett, PE Presented at the 005 Pennsylania State Police Annual Reconstruction Conference, updated 9SEP05 INTRODUCTION Conseration of Linear Momentum is one of the two most powerful tools aailable to a collision reconstructionist (the other being Conseration of Energy). But why does it work and how is it applied? This article will attempt to explain the scientific basis and present numeric and graphic techniques to apply conseration of linear momentum. SCIENTIFIC BASIS As with many of the tools aailable to a reconstructionist, COLM is grounded in Newtonian mechanics as described 00 years ago by Sir Isaac, and which hae proen to be ery accurate predictors of how bodies behae in the world where we lie. Newton s Second Law can be stated as: here F m a = Eq. F = force acting on a body, lb (N) m = mass of the body, lb f *sec /ft a = acceleration experienced by the body, ft/s (m/s ) g = graity,. ft/s (9.8m/s ) Some of the other terms we will need to define are:. = weight, lbf (kg) M = Momentum, lbf sec lb m = pound mass lb f = pound force (often written as simply lb) eight is the force a mass exerts on the ground when acted on by the acceleration of graity ( = mg), so we can diide each side of this equation by graity to get mass in terms of weight and graity: m= Eq. g The most common units for mass (in the US) are: lbf which may also be written as (lbm) or ft sec sometimes (slugs). e ll see this relationship again. Acceleration is defined as the change in elocity oer a change in time, so we can rewrite Eq. as: V F = m t If we rearrange terms a little bit, we find that: F t = m V Eq. Eq. 4 So the force multiplied by the time it is acting equals the product of the mass and its change in elocity. The left-hand term in Eq.4 is called the impulse, and the right hand term is the change in momentum, since we define momentum as the product of mass and elocity: Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
M m V = Eq. 5 Keeping consistent units, with elocity in feet per second, the true units for momentum are: lbf ( ft sec ft ) ( ) = ( lbf sec sec ft ft ) ( ) = ( lbf sec sec ft ft ) ( ) = ( lbf sec sec) APPLYING COLM TO TO PARTICLES Consider two bodies of masses m and m traeling at elocities and which collide, then moe apart at elocities of and 4. FIGURE a: Two balls coming together FIGURE b: Two balls collide FIGURE c: Two balls moing apart Newton s third law says that the force acting between the two during the collision (in Figure b) are equal and opposite: F = -F But Newton s second law says So: F = m a and F = m a m a = -m a By the definition of acceleration as the change in elocity diided by the change in time we can write that as: m V t V t = m The change-in-time term cancels out since it is the same for both bodies, and the change in elocity for each body is simply the starting elocity subtracted from the final elocity, so we can write that as: m ( V V ) m ( V4 V ) mv mv = m V = 4 + m V Rearranging, we get the basic conseration of momentum equation: Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
mv mv = mv + mv4 + Eq. 6a The left side of the equation is the total momentum carried IN by the two cars just prior to the collision, while the right side of the equation is the total momentum carried OUT by the two cars just after the collision. APPLYING COLM TO A CRASH So momentum changes as a result of a force acting on a body or system oer time. In the case of a crash, the system is the two ehicles. Let s think about the forces acting on two cars as they collide: they act on each other (internal), the roadway exerts force on the tire contact patches (external), and there are some ery ery small aerodynamic forces (external). The internal forces act only inside the system - these are the forces of the two ehicles acting on one another. During a typical crash, the internal forces are in the 50,000 lb range. The external forces are those that act on the system from outside these include the forces the roadway applies to the tire contact patch, and aerodynamic drag forces acting on the bodies of the ehicles. Typical tire forces are less than 500 lb, and the aerodynamic forces are an order of magnitude smaller still. This means we can usually neglect the tire and aero forces because they are so much smaller than the internal forces. How about when a car strikes a tree? If we could quantify the forces applied to the tree, we could, in fact, apply COLM, howeer, the force exerted by the tree on the car can t be measured, calculated, or inferred with any accuracy, so COLM doesn t help us in this case. If we can t quantify the external forces acting on the system, COLM is not much help. So, to recap, we know seeral things about crashes which impact our use of COLM:. The duration ( t) is ery short (typically 0.080 to 0.0 seconds, or 80 to 0 milliseconds).. The duration of a collision is the same for both ehicles inoled.. The force exerted on Vehicle by Vehicle, is equal and opposite the force exerted on Vehicle by Vehicle (From Newton s Third Law). 4. The external tire and aerodynamic forces acting during the impact are much much smaller than the internal forces the two ehicles exert on each other, meaning the tire and aero forces can usually be neglected. These lead us to the conclusion that during a typical ehicle collision, the total momentum of the ehicles remains essentially the same: Law of Conseration of Linear Momentum: In a system with no external forces acting on it, linear momentum is neither created nor destroyed, it stays the same, or is consered. In other words, neglecting the external forces at work during the ery short time span of an impact, we can say that the total momentum of the system going into the impact is the same as the total momentum of the system leaing the impact, or Momentum-IN equals Momentum-OUT. The deriation of this was shown as Equation 6, which can also be written as: M in = M out Eq. 6b Additionally, as long as there are no significant external forces at work (the bodies under consideration are acting only on each other) they are each exposed to the same force oer the same time, so the change of momentum for Vehicle is equal and opposite the change of momentum for Vehicle. This can be written as: Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
M = M Eq. 7 This is important, and we will see it again. The direction of action of the change in momentum is the principal direction of force (PDOF). Now, we hae to note that elocity is a ector quantity: it has magnitude (65 feet per second, for instance) and direction (East, for instance). This is commonly indicated in equations by an arrow or a line oer the top of the ariable representing the ector quantity. Momentum is a ector quantity, too. Because of this ector-nature, when we add momentums together, we can not simply add their magnitudes, we hae to take their directions into account. Examples of how to do this will be presented later in this article. In order to keep track of pre- and post- impact trael directions, we will use a 60-degree left-handed coordinate system, measuring the angle counter-clockwise from zero, as shown in Figure. As long as one is consistent in application throughout an analysis, one can orient the coordinate system in any way, but for simplicity s sake, all cases in this article will orient the coordinates such that ehicle is traeling at 0 degrees immediately pre-impact, along the X-axis. ith this coordinate arrangement, ehicle s approach angle, α, is always zero, so SIN(α)=0 and COS(α)=, which simplifies the numeric analysis. In crash analysis, the system is comprised of the ehicles impacting each other. If we hae two cars, then the momentum-in can be written as: M in = M + M Eq. 8 where M = pre-impact momentum of Vehicle M = pre-impact momentum of Vehicle And the momentum-out can be written as: M out = M + M 4 Eq. 9 where M = post-impact momentum of Vehicle M = post-impact momentum of Vehicle 4 Substituting Eq.8 and Eq.9 into Eq.6b, we can write the momentum equation for a two-ehicle impact as: M+ M = M + M 4 Eq. 0 Using equation 5, we can rewrite equation 0 as: mv + mv = mv + mv4 Eq. Substituting Eq. into Eq. for each mass, we get: V + V = V + V4 Eq. g g g g And we can then multiply both sides of the equation by g, (canceling it out of eery term) to get an equation in units of (lbf ft/sec): V + V = V + Eq. V4 Sometimes it s easier to work with speed in miles per hour. To do this, we can conert all the elocity terms to speed (V ft/sec =.467 S mph), to get: Conseration of Linear Momentum Page 4 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
.467 S ) + (.467 S ) = (.467 S ) + (.467 ) Eq. 4 ( S4 The speed term still has magnitude and direction, so it is a ector, and gets the little oer-head arrow to remind us of that fact. Now we can diide eery term by.467, and rewrite the original momentum equation for two ehicles in units of (lb-mph): S + S = S + Eq. 5 S4 It s worth noting that it doesn t matter which set of units we use for the motion term (feet per second, miles per hour, or een furlongs per fortnight), or the inertia term (mass, weight, or stones), as long as we use the same units for eery term in the equation. This is why Eqs. -5 are all equally alid expressions of Eq.0. For simplicity, units of lb-mph will be used for most of this article. If you eer want to use different units, ft/sec instead of mph for instance, simply replace ALL the speed terms in the momentum equations you are using with a elocity term. e ll use the following ariable conentions: α (alpha) = Vehicle approach angle S = Vehicle Pre-impact speed ψ (psi) = Vehicle approach angle S = Vehicle Pre-impact speed θ (theta) = Vehicle departure angle S = Vehicle departure speed φ (phi) = Vehicle departure angle S 4 = Vehicle departure sp S = Vehicle pre-impact speed 90 degrees, +Y axis 80 degrees Vehicle φ (psi) Direction of angle measurement +X axis S = Vehicle pre-impact speed Veh. 70 degrees FIGURE : The Standard 60 Left-Hand Coordinate System, pre-impact Conseration of Linear Momentum Page 5 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
POST-IMPACT 90 degrees, +Y axis S = S 4= Vehicle post impact speed Vehicle post impact speed 80 degrees Ф θ θ (theta) = Vehicle post impact direction Ф (phi) =Veh. post impact direction (angle) +X axis 70 degrees FIGURE : The Standard 60 LHCS, post impact ORKING ITH TO PERPENDICULAR AXES here conseration of linear momentum applies, it can be independently applied to two mutually perpendicular axes, which will be termed here the X and Y axes. By our system definition aboe, Vehicle will always be traeling along the X-axis at the point of impact. The entire reference frame will simply be rotated to accommodate its trael path orientation at the time of impact. This means we can write two separate and independent equations for linear momentum in the form of Eq.5, one for each axis: S + S = S + S Eq. 6a X X X 4 X + Eq. 6b S Y S Y = S Y + S4Y Usually, we can estimate or measure the ehicle weights and calculate the post impact speeds, leaing us with only the two pre-impact speeds as unknown. ith two equations, we can sole for any two unknown alues in the equations, and these will usually be the two pre-impact speed terms. Before starting in on the two axes, a quick reiew of geometry and resoling a ector into its perpendicular components is in order. The sum of the angles inside any triangle always equals 80 degrees. In a right triangle, then, the sum of the two non-right angles must be 90. If we can determine one of the angles in a right triangle, we know the other included angle. For instance, using the triangle in Figure 4, we know that γ + β = 90 degrees. FIGURE 4: A right triangle Conseration of Linear Momentum Page 6 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
If we hae parallel lines with a third line crossing through them, as shown in Figure 5, the two indicated angles must be equal. Now, back to ectors. Any ector ( M, for instance) can be broken down into two mutually perpendicular ectors, M X and M Y, as shown in Figure 6, by drawing a rectangle using the two perpendicular ectors as two sides. The original ector will cut diagonally across the rectangle. The angles shown are gamma (γ) and beta (β). FIGURE 5: Parallel Lines equal angles FIGURE 6: Vector components If we add our two perpendicular ectors head-to-tail we get the original ector, so we can say that the ector M is equal to the ector sum of the two components ( M = M X + M ): Y FIGURE 7: Adding ectors If we hae the constituent perpendicular ectors, the magnitude of the ector M can be found using the Pythagorean Theorem for right triangles (the length of the hypotenuse is the square root of the sum of the squares of the sides): M = M X + M Y If we hae the length of ector M, and one angle, the lengths of M x and M y can be found using the trigonometry relationships shown in Figure 7. (Notice that when working with just the magnitude of the ector, M in this case, the oer-bar is not used. The oer-bar is only used to indicate a ector.) Using the terminology conentions denoted earlier, and the relationships shown in Figure 6, we can rewrite Equations 6a and 6b without the ector notation (using just the speed magnitudes). These will be our two most heaily used equations: S+ S cosψ = S cosθ+ S4 cosφ Eq. 7a S sinψ = S sinθ + S4 sinφ Eq. 7b Notice that since we hae defined the incoming direction of Vehicle as being along the X-axis, it has no momentum in the Y-direction, and its momentum-in does NOT enter into the Y-axis momentum Conseration of Linear Momentum Page 7 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
equation. e can sole all two-ehicle the linear momentum problems for two ehicles with this pair of equations. How much simpler can it get? In addition to the speeds, we can ealuate the change in speed (or change in elocity, V) for each ehicle, using these two equations: θ S = S + S S S cos( ) Eq. 8a 4 4 φ S = S + S S S cos( ψ ) Eq. 8b e can also ealuate the Principal Direction of Force (PDOF) for each ehicle, using Equations 9a and 9b for most cases: The sin - is called the inerse sin and is usually accessed on calculators by the <INV><SIN> or < nd ><SIN> keys. PDOF = sin S sinθ ( ) S Eq. 9a PDOF = sin S ( 4 sin( ψ φ) ) S Eq. 9b Remember that the PDOF is measured relatie to the direction each ehicle was facing at impact, not its original path of trael. If the car was facing the direction it was going, then 0-degrees is pointed straight back toward the front of the car, 90-degrees indicates a passenger (right) side hit, 80-degrees indicates a rear-end hit, and 70 (or -90) being a drier s side (left) side hit.. The PDOF is NOT anchored to our 60 degree scale. Also, for co-linear crashes, these equations can t differentiate between 0 and 80, since sin(0) = sin(80) = 0. Occasionally, the angle calculated will be off by 90 or 80 degrees. A little common sense and a ector diagram will both go a long way in interpreting PDOF, and we ll see this in action in seeral examples. FIGURE 8: PDOF conention (from CRASH User s Manual) Conseration of Linear Momentum Page 8 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
USING COLM FOR CO-LINEAR IMPACTS, ONE VEHICLE STATIONARY The simplest form of collision to ealuate 90 degrees, with momentum is a co-linear rear-ender +Y axis with Vehicle striking a stationary Vehicle, as shown in Figure 9. Post impact direction of trael is easy to isualize: the two cars leae the impact in the direction they were facing and traeling, but at speeds different than their original speeds. How much different? That depends on the Veh. weights (or masses) of the cars and the impact speed. Veh. FIGURE 9: Colinear Collision Example +X axis 0 degrees, Veh. and Veh. pre-impact trael direction NUMERIC SOLUTION: Starting with Equation 7a and plugging in our known alues (S =0, θ =zero,φ = zero) we get: S+ S cosψ = S cos(0) + S4 cos(0) =0 = = e get the X-axis momentum equation for an inline-impact, with one ehicle stopped before impact: = + Eq.0 S S S4 Looking at Equation 7b, though, we find that sin(ψ) = sin(θ) = sin(0) = 0, and all the terms drop out, leaing us with nothing equals nothing: S sinψ = S sinθ+ S4 sinφ =0 =0 =0 Intuitiely, this makes sense, though: all the ehicle motion was in one direction (the X-direction) we shouldn t hae anything in the other direction (the Y-direction). GRAPHIC SOLUTION: Graphically, we can show the same thing, using arrows to denote the momentum carried by each ehicle. The arrow length is proportional to the ehicle s momentum, and its direction indicates the speed ectors (and momentum ectors ) direction. Using units of lb-mph, momentum-in is shown by two arrows of length S and S added head-to-tail. In this case, though, S is zero, so it doesn t enter into the picture. The length of the two momentum-out ectors when so placed must be the same as the length of the single momentum-in ector and in the same direction: Momentum-IN S Momentum-OUT S S 4 Vehicle Change in Momentum M -IN = S M -OUT = S M Conseration of Linear Momentum Page 9 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
Example : An eastbound pickup truck weighing 60 pounds (Vehicle ) strikes a 50-pound car (Vehicle ) which is stationary at a stoplight facing east. The pickup s front bumper oerrides the sedan s rear bumper and the two ehicles trael together in a straight line to final rest. Using a skid-tostop analysis, their post-impact speed is determined to hae been approximately 8 miles per hour. Find the pickup s pre-impact speed. Numeric Solution: Since the pickup is the only one in motion pre-impact, we ll call that one Vehicle, and orient our X-axis to match its pre-impact trael direction. Now we can make a table with all our information: VEHICLE VEHICLE eight =,60 lb =,50 lb Approach Speed S = mph S = 0 mph Approach Angle α = 0 ψ = 0 cos α = cos ψ = sin α = 0 sin ψ =0 Departure Speed S = 8 mph S 4 = 8 mph Departure Angle θ = 0 degrees φ = 0 degrees Starting with equation 7a S+ S cosψ = S cosθ+ S4 cosφ Eq. 7a =0 = = Since S 4 =S, we can rewrite this as: S= ( + ) S ( + ) and diiding by, we can sole for S : S = S Substituting the known alues gies: 60+ 50 S = 8mph 60 ith weights measured to three significant figures, this reduces to: S = (.59) 8mph ith the post-impact speed calculated to two significant figures, the final solution can only carry two significant figures: S = 44mph NUMERIC ANSER Graphic Solution: An alternatie approach would be to sole the problem graphically, as shown below. For this example, I will select a scale of inch = 50,000 lbf-mph. Start with what we know: the post-impact momentums. The post impact momentum of ehicle in units of (lb-mph) is (60 lb * 8 mph) = 0,60 lb-mph =.0 inches in our newly defined scale. The post impact momentum of ehicle is found to be (50 lb * 8 mph) = 60,00 lb-mph =.0 inches. Post Impact Momentums: cos θ = cos φ = sin θ = 0 sin φ =0 S =.0in S 4 =.0in So, adding them gies us a Momentum-OUT ector that s. inches long: Conseration of Linear Momentum Page 0 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
M OUT = M IN =. in Since S =0, ALL the momentum comes from the pickup: M -IN = S =.in At 50,000 lb-mph per inch (the scale we selected earlier), that conerts to 6,500 lbf-mph, so: S = 6,500 lb-mph (60 lb) S =6,500 lb-mph S =6,500 / 60 = 45mph GRAPHIC ANSER Rounding numbers during the graphic conersions produces a result slightly different from the numerical solution. If the two ehicles left the aboe impact at different speeds, we would start out the same as with the example just worked, but insert both post-impact speeds into Eq.7a, instead of combining them. The rest of the solution would be unchanged. USING COLM FOR CO-LINEAR IMPACTS, BOTH VEHICLES IN MOTION PREIMPACT If both ehicles are in motion before the impact, collinear momentum alone is not sufficient to sole for both pre-impact speeds. Momentum will only allow you to calculate the relationship between the two pre-impact speeds. In other words, in order to calculate one ehicle s pre-impact speed, een if both departure speeds can be calculated, the other s speed must be assumed, known (ia witnesses or Eent Data Recorders, for instance), or calculated by some other means (such as energy analysis). Example : A southbound Cherolet Blazer (Vehicle, =,604 pounds) crosses the centerline and strikes a northbound Chrysler Lebaron (Vehicle, =,040 pounds) essentially head on. The Lebaron is pushed south approximately 6 feet, while the Blazer traels approximately feet. Using a skid-to-stop analysis it is determined that the Lebaron s post impact speed was approximately 7 mph, while the Blazer s post impact speed was 5 mph. How fast was the Blazer going before the impact, and what was the Lebaron s speed change? 90 degrees, +Y axis 80 degrees, Veh. pre-impact direction Veh. Veh. +X axis 0 degrees, Veh. pre-impact direction, and post impact direction for both ehicles Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
Numeric Solution e know that there s no motion in the Y-axis, so Equation 7b won t do us any good (all the sin terms still equal zero), so starting with equation 7a : S+ S cosψ = S cosθ + S4 cosφ Plugging in the alues for the cosine terms gies us: S+ S cosψ = S cosθ + S4 cosφ + S S = S S4 = - = = e can rearrange terms to sole for S in terms of S : Add S to each side: + S S = S S4 + S S + S = S + S4 S + S = S + S4 S Then diide by to sole for S in terms of S : Or we can rewrite this as S S = Plugging in the alues we do hae: S VEHICLE VEHICLE eight =,604 lb =040 lb Approach Speed S = mph S = mph Approach Angle α = 0 ψ = 80 cos α = cos ψ = - sin α = 0 sin ψ =0 Departure Speed S = 5 mph S 4 = 7 mph Departure Angle θ = 0 degrees φ = 0 degrees + S 4 S = S+ + * S 4 + S * S 040lb*7mph 040lb* S = 5mph+ 604lb 604lb + S 9 + S =.mph 0.84* cos θ = cos φ = sin θ = 0 sin φ =0 The problem now, of course, is that we hae only one equation but two unknowns. e need more information about either S or S to sole for the other one. If a witness comes forward and Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
says they were pacing Vehicle at 45mph, we can rearrange equation 7a as done earlier, but this time sole for S in terms of S : * S * S S = S4 + S 9mph = NUMERIC ANSER And then the change in speed experienced by the Lebaron can be calculated with 4 4 φ S = S + S S S cos( ψ ) Eq. 8b S = 9 + 7 (9)(7) cos(80 0) S = 6mph NUMERIC ANSER Graphic Solution Because of the differences in direction, the graphic representation of this problem is a little more complicated than for Example. Using the same 50,000 lb-mph per inch scale as before, we can draw the POST-IMPACT momentum in the same fashion: M -OUT = S = 54,060 lb-mph =.08in M -OUT = S 4 = 5,680 lb-mph =.0in Post Impact Momentums: M -OUT =.08in M -OUT =.0in So, adding them (HEAD-TO-TAIL) gies us a Momentum-OUT ector that s. inches: M OUT =.in = M IN e know that the head-to-tail addition of the two momentum-in ectors will equal this momentum-out ector, and we know that the two momentum-in ectors are in opposite directions. hat we don t know from this data is how large either of the two ectors is, all we know is their directions. But een this is instructie: e see that Vehicle HAD to be carrying much higher momentum (not necessarily speed) than Vehicle for the combination to hae the net result noted earlier. This makes sense intuitiely. Using the witness statement that S was 45mph, we find that S = 604lb * 45mph = 6,80 lb*mph, which is.4 inches in our scale, so we can measure the length of S : M IN =.in M -IN M -IN =.4 in So M -IN = S must then be.inches = 56,500 lb*mph, to the left, so S =56,500 / 040 = 8 mph GRAPHIC ANSER Again, the rounding during graphing changes the answer slightly, een with a simple diagram and fairly accurate measurements. This highlights why the graphic method may be good for rough estimates, and for isualizing a wreck, as we ll see later, but a numerical solution can be relied upon for greater accuracy. Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
hat else can we learn from the diagram here? e know that one ector added head-to-tail to another yields a resulting ector, so how about the change in momentum? For ehicle, it started out going north at 9mph (with a momentum length of. inches in our scale), and ended going south at 7mph (with a momentum ector length of.0 inches in our scale). S =. in M =.6 inches S 4 ==.0in e see that the ector we hae to add to M to get M 4 is.6 inches long, or 08,000 lb-mph. Remembering that M = S, so S = ( M / ), and that this car weighed 040 lb, we can determine that the change in elocity experienced by the Lebaron (Vehicle ) was: S = M / = (08,000 lb-mph / 040 lb) = 5 mph. GRAPHIC ANSER Again, rounding has produced a graphic result slightly different from the numeric result. USING COLM FOR NON-CO-LINEAR IMPACTS hat if the two ehicles are not traeling in the same direction (along the same axis)? This is where the 60-degree coordinate system (defined earlier) really shines. By resoling each momentum ector into its X and Y components, we can write the two separate momentum equations described earlier: One for the X-axis and one for the Y-axis. In this way, we can get two equations, which will allow us to sole for two unknowns. One of the most important aspects of this type of crash is determining the departure angles. This is one place where some people get into trouble if you simply take the direction from impact to final rest you might be right in some cases, but quite often you will be wrong because of rollout, or other postimpact phenomena. The angles required for momentum analysis are the direction of the CGs trael just as the cars separate. Oer the years, as the art and science of crash reconstruction has progressed, the understanding of how best to measure the departure angle has changed. As with many analysis techniques, the improed models come with increasing leels of complexity. In the early days, the direction from impact to final rest was used. e know now that that can lead to gross errors due to rollout. The next commonly used technique was to take the CG s path from first touch to last touch, which really doesn t get us what we seek, which is the post-impact trael (not the duringimpact trael). One more recent technique has been to use the line of action from maximum engagement to separation. But in considering the definition of what we re trying to measure (Post-Impact Trael) we see that this is closer than the earlier techniques, but it still isn t quite right. The angle we seek is the direction of the CG s path of trael at the instant the cars stop working on each other. This will be the ehicle s direction of trael prior to any external forces (tire forces) haing an effect on it. The best way I know to determine this angle is to plot the ehicle s CG position through the post impact trael using scale cars in a good scale diagram, lining the cars up with scuffs, skids, and gouges. Be careful about simply taking the direction of the gougemarks, as they may hae been caused by some part of the car far remoed from the CG, and thus not follow the CG s path as the car rotates and translates. There should be an essentially straight line oer at least a short distance leaing the impact, before those external tire/roadway forces can redirect the ehicle. Conseration of Linear Momentum Page 4 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
Example : Perpendicular approach angles A Dodge Neon (Vehicle, C=,5 pounds) carrying one young male occupant (weight~45lb) collides with an eastbound Ford Mustang (Vehicle, C=,758 pounds) with four male adult occupants (total weight ~ 785lb) at a fully controlled four-way intersection. Two witnesses were in a car following the Mustang, one says the Neon was traeling South, one says it was traeling est. Examining the damage patterns on the two ehicles indicates that the Neon was perpendicular to the Mustang at impact, with the Neon s passenger side heaily impacted. The basic scene diagram is shown below, with fresh gouges in the paement near the middle of the intersection. hich way was the Neon traeling and how fast were the two cars going? V, Neon V, Mustang Directions ith the Mustang headed east, it s only proiding momentum in one direction, therefore, we see that the Neon had to supply the momentum headed south to get the cars into that corner of the intersection. So the Neon was NOT going west, it was headed south. The damage, combined with the location of the gouge marks, it is concluded that the Neon was traeling straight through the intersection going south, not turning south from a west-bound lane. Haing selected the Neon as Vehicle, we set our X-axis in the Neon s pre-trael direction, or South. Setting scale cars in a scene diagram and plotting the paths of the CGs, it is found that the departure angle for Vehicle is 60 degrees (on our 60-scale) and for Vehicle it s 56 degrees. (Note that the impact to final rest direction would be wrong). Since all the people stayed in their respectie cars, their weight should be included, as well (Usually this isn t a big deal, but in this case, the people are more than one fifth of the Mustang s total weight, so they should specifically not be neglected!). Now to the speeds: e e determined the basic direction of trael for both ehicles immediately before impact but we still need the post-impact directions. Conseration of Linear Momentum Page 5 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
Approximate postimpact paths of trael. Post impact trael for the Neon with one crush-locked front wheel gies a post-impact speed of 9mph. Because the Ford rolled out to final rest, its post-impact speed cannot be conclusiely determined from skid-to-stop, but gien the solid nature of this hit, it s not unreasonable to assume that the Ford s post impact speed was essentially the same. Numeric Solution VEHICLE VEHICLE eight =,658 lb =,54 lb Approach Speed S = mph S = mph Approach Angle α = 0 ψ = 90 cos α = cos ψ = 0 sin α = 0 sin ψ = Departure Speed S = 9 mph S 4 = 9 mph Departure Angle θ = 60 degrees φ = 55 degrees cos θ = 0.5000 cos φ = 0.575 sin θ = 0.8660 sin φ =0.89 As always, we start with equations 7a and 7b, plug in all our known ariables. Because the Neon and Mustang were thoughtful enough to take perpendicular paths to the crash, we can directly sole for S and S One term on the left drops out: S S cosψ = S cosθ+ S4 cosφ + Eq. 7a S + S cosψ = S cosθ + S 4 cosφ Leaing us with this: =0 ( 658lb * S) = (658lb*9mph *0.50) + (54lb*9mph *0.559) ( 658lb * S ) = (55+ 760) Conseration of Linear Momentum Page 6 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
S = (55 760) lb* mph / 658lb + S 4mph = NUMERIC ANSER And taking Eq. 7b we see that here one trig term equals, so it disappears S sinψ = S sinθ + S4 sinφ Eq. 7b = 54lb * S = (658lb*9mph *0.8660) + (54lb*9mph *0.89) S = (4,74+ 55,9) / 54 S 8mph = NUMERIC ANSER The Neon s change in speed is calculated with Eq.8a: S = 4 + 9 (4)(9) cos(60 0) S=. 9mph NUMERIC ANSER The Mustang s change is speed is calculated to be 6.5mph The Neon s PDOF is calculated with Eq.9a: PDOF = sin S sinθ ( ) = sin S 9 *0.8660 ( ) = 48.7.9 S4 sin( φ ψ ) 9*sin( 5) PDOF = sin ( ) = sin ( ) = 4. S 6.5 Graphic Solution e e got our post impact ector directions from the earlier analysis, we just need their lengths: For V, the output momentum is (,658 lb)*9mph = 50,50 lb-mph, and for Vehicle, the output momentum is (,54)*9 = 67,7 lb-mph. FIND A SCALE: If we add up the total numerical momentum (7,89) and diide by the aailable width of our paper (5 inches) we get,500 lb-mph per inch. For simplicity, I ll round to the nearest nice number, and I ll use a scale of inch=5,000 lb-mph. So our M, will be (50,50/5,00) =.0 inches long headed 60 degrees, while M 4. will be (67,7/5,000) =.69 inches headed 55 degrees from the origin. Conseration of Linear Momentum Page 7 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
POST-IMPACT MOMENTUM VECTORS 90 degrees, +Y axis Veh. Veh. +X axis 0 degrees, Veh. pre-impact trael direction Now we add the two momentum-out ectors head to tail, to find the total outgoing momentum: ADDING MOMENTUM-OUT VECTORS M OUT Vehicle Momentum out M -OUT = V 90 degrees, +Y axis Vehicle Momentum out = M -OUT = V 4 +X axis In this case, we know that ALL the X-momentum comes from the Neon, while ALL the Y- Momentum comes from the Mustang, so we draw the parallelogram around the total momentum Conseration of Linear Momentum Page 8 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
ector using the X and Y axes as two sides in order to find the two ectors along those axes which add up to the total momentum ector: BREAKING TOTAL MOMENTUM-OUT INTO TO PERPENDICULAR VECTORS M -IN and M -IN M M OUT-total = M IN-total Vehicle Momentum IN M -IN = S M -OUT 90 degrees, +Y axis M -OUT M Vehicle Momentum IN = M -IN = S +X axis The Vehicle (Neon) M -IN ector is about.5 inches long, so: S =.5in*5,000lb-mph/in = 6,500 lb-mph S =6,500 lb-mph / 658lb = 4 mph GRAPHIC ANSER The Vehicle (Mustang) M -IN ector is about 4 inches long, and running through the same analysis yields: V =4 in * 5,000 lb-mph/in = 00,000 lb-mph V = 00,000 lb-mph / 54 lb = 8 mph GRAPHIC ANSER And the dashed ectors connecting M to M and M to M 4 represent the change in momentum experienced by each ehicle, and they are about.8 inches long, or 57,000 lb-mph. e noted earlier that they should be equal and opposite (parallel), and they are oriented at the angle of the principal directions of force. The Neon s speed change from the graphic method is then found to be 57,000/658 = mph, and pointed to the Neon s left and rearward. This time, with slightly better scale resolution, the graphic answer is essentially the same as the numeric answer. The Neon s PDOF can be measured right off the diagram, and should be a little oer 45 degrees. Vector analysis, as shown by REC-TEC: Conseration of Linear Momentum Page 9 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
Vector Analysis as Displayed in ARPro 7: Conseration of Linear Momentum Page 0 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
Example 4: Non-Perpendicular approach angles A Toyota T00 pickup carrying no load (C=40 lb, Vehicle ) and two male occupants (weight~50 lb) collides with a Subaru Justy (C=805 lb, Vehicle ) with one female occupant (weight ~ 45 lb) at an oblique intersection on rural roads with speed limits of 40 mph. The impact is non-central, spinning the Justy around 80 degrees, and the Toyota approximately 90 degrees. The pickup has a flashing red light. He claims he stopped at the light, looked both ways, saw noone, proceeded through the intersection, and was struck by the Justy on the left rear. She claims he ran the light, and she didn t hae time to stop. The Justy left 4 feet of solid pre-impact skidmarks leading to the impact site. Find the impact speeds for both ehicles and the Justy s change in speed. Vehicle Toyota T00 Vehicle Justy Solution Using a skid-to-stop analysis, the Justy s post impact speed was calculated to be about 8 miles per hour, gien the 80 degree rotation and 40 foot trael distance, while the Toyota s post impact speed was determined to be 7 miles per hour. Numeric Solution VEHICLE E on Main VEHICLE N on Ash eight =,950 lb =,780 lb Approach Speed S = mph S = mph Approach Angle α = 0 ψ = 7 cos α = cos ψ = 0.7 sin α = 0 sin ψ = -0.680 Departure Speed S = 8 mph S 4 = 7 mph Departure Angle θ = 55 degrees φ = 5 degrees cos θ = 0.996 cos φ = 0.906 sin θ = -0.087 sin φ = -0.46 Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
S sinψ S sinθ + S4 sinφ 000S = Eq. 7b (.0) = 000(0)(0.648) + 000(0)(0.46) 000* S = 57780+ 6904 S = 74684 / 000= 7. 578S = ( 4,759) + ( 7,8) S = (,94) /( 578) S.4mph mph = NUMERIC ANSER = And plugging our known alues, and the newly found S into 6a: S+ S cosψ = S cosθ+ S4 cosφ Eq. 7a 000S = 000(0)(0.7660) + 000(0)(0.906) 000S = (68940) + (6,40) S= 5. 06mph NUMERIC ANSER Regarding the Justy s change in speed, we break out equations 8a and 9a: θ S = S + S S S cos( ) Eq. 8a S = (40mph) + (8mph) (40mph)(8mph)(0.996) S= mph PDOF = sin S sinθ ( ) = sin S 8mph *( 0.087) ( ) =.5 And the T00 s PDOF calculates to be 55 degrees: PDOF = sin S ( 4 sin( φ ψ ) ) = sin S 7*sin(8) ( ) = 55 * 6.4 deg. *atch out now check out the ector diagram! The true PDOF is (80-55) = 5 degrees or so. Most especially when we cross quadrants, we hae to be careful of how the calculated PDOF is really associated with our wreck. Graphic Solution: The Vehicle (Justy) M out ector is (950 lb * 8 mph) = 54,600 lb-mph, which will plot out at.84 inches (using a scale of inch = 5,000 lb-mph), at an angle of 55 degrees (or -5degrees). The Vehicle (T00) M out ector is (780 lb * 7 mph) = 64,60 lb-mph, which will make it.57 inches at 5 degrees. Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
Putting these two head-to-tail we add them together we see that our total Momentum OUT ector is about 4.5 inches long: 90 degrees, +Y axis ADDING MOMENTUM-OUT VECTORS Scale: in = 5,000 lb*mph +X axis Vehicle Momentum out = M V 4 M out Vehicle Momentum out=m V Now, we know the Justy s Momentum-IN ector is along the X-axis (that s the way we set the system up), we just don t know its magnitude yet. e also know that the T00 s Momentum-IN ector is along a line of action at 7 degrees. So we can draw a ector at 7 degrees which crosses the end of our M out ector, and we know that where it crosses the X-axis is the end of the Justy s M in ector. Remember that the combination should add up (head-to-tail) to our known total momentum. FINDING MOMENTUM-IN VECTORS M -IN = S +X axis 7 degrees Vehicle Momentum IN acting along this direction. M -IN = S M OUT-total = M IN-total Looking at the diagram aboe, we see that the Justy s momentum-in ector is about inches long, which with our scale (in=5,000 lb-mph) means it had 75,000 lb-mph of momentum, and a weight of =950 lb, so S =8 mph GRAPHIC ANSER And the length of the T00 s ector is about.9 inches, which equals 47,500 lb-mph, so: S =47,500 lb-mph S = 47,500 lb-mph / 780lb S =mph GRAPHIC ANSER These numbers compare fairly well with the numeric solution, but the ector diagram proides a more informatie isual aid to help understand the eents of the crash. Referring to the ector diagram on the next page, the momentum change ectors are about 0.88 inches long, or,4 lb-mph, which yields a speed change of about miles per hour for the Conseration of Linear Momentum Page Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
Justy, pointed rearward and slightly to the right. These numbers compare pretty well with the 40 and found numerically aboe. FINDING CHANGE IN MOMENTUM & PDOF M -IN = S M -OUT M +X axis M -IN = S M -OUT M M OUT-total = M IN-total SENSITIVITY An analysis is said to be sensitie to changes in a ariable when making small changes in the ariable causes big changes in the calculated result. If an analysis is highly sensitie to one (or more) ariables, then one may hae to allow a wide range in the final answer to be sure that the truth is in there somewhere. To check an analysis sensitiity, we simply switch one alue by a small amount and see how dramatic the effect is. This can often be most easily done with a spreadsheet or an A/R program. From example for instance, to check the sensitiity of the analysis to the T00 s departure angle, we just change that alue and redo the analysis with a slightly higher and lower alue. The same can be done with the Justy s departure angles. If we hae three alues for each one, we can make a 9-cell table showing the calculated alue with each combination of alues: Calculated Impact Velocity Justy/ T00 for nine departure angle combinations: T00 DEPARTURE ANGLE 5 7 JUSTY 5 7.45 /.9 9.06 /. 40.66 /. DEPARTURE 55 8.6 /.6 40. /.8 4.8 /.59 ANGLE 57 9.7 /.4 4. /.64 4.9 / 0.85 Net Difference from Nominal Values (Justy/ T00): T00 DEPARTURE ANGLE 5 7 JUSTY 5 -.75 /.5 -.4 / 0.74 0.46 / -0.06 DEPARTURE 55 -.6 / 0.78 0 / 0.6 / -0.79 ANGLE 57-0.49 / 0.04. / -..7 / -.5 So we see that een if the departure angles we used are off by a couple degrees either way, the changes are not huge in this case, but the Justy s result is about twice as sensitie to our angle selections (primarily because it is so much lighter than the T00). The same sort of sensitiity analysis can be performed for each ariable used in the analysis to determine which (if any) hae an extraordinary affect on the results. hen we hae high weight (or momentum) ratios, slight changes in the approach or departure angle of one ehicle (typically the heaier one) can make huge changes in the calculated speeds for the other one. Conseration of Linear Momentum Page 4 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005
A ORD ABOUT MOTORCYCLES Motorcycles present special problems with respect to momentum because (a) they are much lighter than the car, and (b) the rider often gets ejected after some interaction with the car, and goes his own way. The first item means the small changes in the angles selected for the car s pre-and post-impact trael can hae an inordinate affect on the calculated motorcycle speed, and the second item means we hae to determine the departure angle and departure speed for a third item, namely the rider s body. This extra complication often means momentum is not a iable solution method due to lack of scene data. If there are three bodies inoled in an accident, and we can determine post-impact trael directions and speeds for each of them, we can only sole for two unknowns with momentum. ith motorcycles, though, we can usually assume that the rider and motorcycle were traeling at the same speed before the impact, so we only hae two unknowns: V (car) and V (rider/motorcycle). CONCLUSION: Conseration of linear momentum is one of the most powerful tools aailable to the reconstructionist. Numeric solutions can be used to sole for speeds, and graphic methods can be ery helpful in not only ealuating the crash speeds, but also for understanding the bigger picture of a wreck. If you can create the ector diagram for a crash, you e got the analysis nailed down, and can discuss essentially all the dynamic aspects of the eent. This article presented both the numeric and graphic methods, as well as a number of soled examples. CONTACT The author can be reached ia his website, http://www.mfes.com Conseration of Linear Momentum Page 5 Presented by ade Bartlett at the Annual PSP Conference, Sept. 7-9, 005