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I Calculating the velocity of a vehicle necessary to overturn

  1. Aug 8, 2016 #1
    The situation in a recent motor vehicle accident :
    An 850-lb motorcycle-plus-driver strikes a 2850-pound car-plus-driver on the passenger side with enough force to completely overturn it (in other words lift it completely off the ground and onto its roof), that car's momentum then enough to effectuate the complete flip-over of the vehicle onto the driver's side. What is the minimum velocity of the motorcycle-plus-driver necessary to accomplish this. The car is traveling at 10 mph in a perpendicular direction (in other words, its velocity is zero in the direction of impact).
     
  2. jcsd
  3. Aug 8, 2016 #2

    BvU

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    Hi doc, :welcome:

    Can't say: depends on several unknowns: heights of centers of mass, height of point of impact, moment of inertia of car, etc.
    It's not as easy to turn over a Ferrari as it is to turn over a Golf
     
  4. Aug 8, 2016 #3
    Hi, Understood, and there are, indeed other variables connected with friction, since the motorcycle wasn't on its wheelswhen it struck the car. ---But, let's forget the overturn. No matter what the value of those other variables, the momentum of the cycle must be sufficient to lift that 2850 lbs off the ground to begin with, so there must be a minimum velocity required for that to occur. That's what I'm seeking.
    Thanks for replying.
     
  5. Aug 8, 2016 #4

    sophiecentaur

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    I don't think so. For a start, if the car is not going dead straight there will be a centripetal contribution to the sideways forces. The exact height where the impact could be said to have occurred is also relevant.
    The point has been made on almost every thread that turns up on PF, that vehicle impacts are just too complex and the necessary data is never available to make calculations that could give reliable results. Even knowing the dimensions of the two colliding vehicles is just not enough. There are 'expert witnesses' who's opinions are respected by courts but the outcomes of court cases probably hang more on the quality of the arguments and results of previous cases. Data about impacts over many historical incidents is used. Length of tyre marks and deformation of the vehicles can give a good indication of the speeds involved. An experienced lawyer will know where to get hold of such stuff but Joe Public (PhD in Physics or whatever) cannot present a case.
     
  6. Aug 8, 2016 #5
    Thanks for responding, and I do "get" your red text. However, I'm not looking for exact calculations, but rather seeking a way to find a lower limit to necessary velocity --- and let's say we idealize the situation. In other words, neglecting the other factors mentioned, let's look at a rest mass of 2850 lbs, of 5.5-foot height, 7-foot width and 15-foot length, and assume that its center of mass is located at its exact geometric center, at which impact occurs, and likewise assume that the motorcycle is an 850-pound point mass that strikes the center of mass of the car. My problem in seeking a ballpark lower-limit velocity (I know that anything like a precise number is untenable in this situation) is that the insurance company does not seem interested in gathering such data. For example, indeed, the site of impact is known as is the degree of vehicle deformation, but no one seems to be factoring that into their evidence-gathering. Hence I'm seeking to show that, ignoring these other energy-loss factors (most of which would argue for a higher velocity than the minimum required in the ideal situation) can one calculate a minimum velocity necessary to lift that weight off the ground in the ideal situation. Thanks for any help with the idealized problem. And, by the way, my PhD is in cellular and molecular biology, where precision is far less achievable. Would be happy to reciprocally offer my services as an expert in human pathophysiology to anyone who can help me. Health/wellness questions or simply curiosity, I'm available --- and grateful for any help.
     
  7. Aug 8, 2016 #6

    CWatters

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    Here is how I would ballpark a figure... Draw the car tipped so the the centre of mass is above the wheels on one side. The centre of mass will be higher than it was before. Calculate the change in potential energy = mass of car*g*change in height . Assume that energy came from the kinetic energy of the bike = 0.5*mass of bike*v^2. Solve for V.

    The bike may well have been going faster than calculated as this ignores energy used to deform the car and bike.

    In the past threads on this subject have been closed by the moderators as it's nearly impossible to give accurate answers and we are not expert witnesses.
     
  8. Aug 8, 2016 #7

    BvU

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    The ballpark figure comes out 16 km/h, bicycle speed. (in funny units 10 mph). Correct me if I am wrong...

    [edit] why don't I do so myself: 21 km/h (13 mph) (-- I swapped width and height.)
     
  9. Aug 9, 2016 #8

    davenn

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    and you need to also understand that on the forum here, we shy away from giving info in cases like these
    as it will never be able to be used as evidence. It's just a really bad idea and the forum owner or members
    don't want to leave themselves open to litigation as a result of info given

    most threads like this get locked


    Dave
     
  10. Aug 9, 2016 #9

    Orodruin

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    Thread closed.

    Crash reconstruction is a complex matter and cannot be accomplished by giving a few input variables to a bunch of strangers on an anonymous internet forum.
     
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