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Calculating Impact of an object (small rock) on the front of a moving car

  1. May 3, 2012 #1
    Hey Guys,

    I have a problem I am working on and was wondering if someone could point me in the right direction.

    The problem:
    I need to calculate a force to apply to a given area of the front of a vehicle in relaticely small collision / impact with both a stationary and possibly a small moving object the size roughly of a softball for volume.

    The scenario:
    A car is driving at any given speed and a rock is either a) stationary and hits the front bumper or under neath of the car. or b) is propelled from a car in front and accelerates backwards at a given speed and impacts the vehicle.
    I basicallly need to calculate the force that object will have at the impact point.

    vehicle speed = 80km/hr, 50miles / hr
    vehicle weight = 3000lbs, 1360 kgs

    object size (small rock size of a softball)
    circumference of a regulation softball = 12" / 0.3048

    Now given that the radius would be:

    C = 2 pi * r

    r = C / ( 2 *pi )
    r = (0.3048 m) / (2 * pi)
    r = (0.3048 m) / (6.28318)
    r = 0.0485 m

    Volume of the softball-sized rock

    V = (4/3) * pi* (r^3)
    V = (4/3) * pi * (0.0485 m)^3
    V = (4/3) * pi* (1.142E-4 m^3)
    V = 4.782E-4 m^3

    Find the mass of the rock using density

    m = (2,750 kg / m^3) * (4.782E-4 m^3)
    m = 1.315 kg

    Now two given scenarios:

    1) the rock is not moving and the car is traveling at 50mph. What the impact firce of the rock be?
    - here there would obviously be a few factors at play. a) impact upwards b) force of car pushing down c) side angle forces
    - for simplicity sake lets assume the rock hits something underneath the car and there is an equal distribution of forces updward and downward acting here and no acceleration of the vehicle at the point of the impact over a given time....

    2) second scenario the rock is moving towards the car we can assume at a speed of 50mph also.

    Any thoughts on simplifying the above problem?

    much appreciated!
  2. jcsd
  3. May 3, 2012 #2


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    The actual Force will depend upon how long the impact lasts (i.e. the force would be a lot less and last for longer if it hits a soft part and higher for a short time if it hits a hard part). What you can work out will be the Impulse, which is equal to the change in Momentum. Momentum is Mass times velocity and you can work out the initial and final momentum if you know the initial and final velocities. Velocities are all relative so if one car is following another at the same speed then it is only the velocity of the 'rock' relative to the cars that counts. For the rock to be propelled backwards from the car, it would need to be actively thrown, I should imagine. A rock, picked up by a tyre (pretty enormous tread pattern needed) would be thrown Upwards but not backwards, so the maximum impact speed would be no more than the actual speed of the target car.

    Finding or estimating the actual force involved, even for a single impact, would be very difficult and would probably need experiments on similar steel panels with different projectiles. Thing is, you just can't tell how long the actual impact took.

    Are you trying to reconstruct some actual incident with this calculation and to get some meaningful evidence / ammunition?
    Your more complex scenario has too many unknown variables to give you anything useful, I think.
  4. May 3, 2012 #3
    I am looking for a Force to start with in an impact scenario for penetration / damage results to a panel on the front of a car... like a bumper for instance... or an oil pan or something...

    I just need a force that I can substantiate to a certain extent.... i.e. not just a number I picked.... for arguments sake I have done teh analysis a dozen times over using various values of force - I would like to pin point a collision 'type' so to speak in order to justify the analysis done of course.

    I would argue that a stone driven over will have a number of angular velocities if caught on a tire tread and pitched out the back and will of course always have an upwards acceleration component to it and will vary in magnitude of course. however the same scenario here that would send a stone into your windshield and crack it for instance.

    The argument of driving over a point of an object such as a rock is an impact that I would like to look at.
  5. May 3, 2012 #4
    As sophiecentaur stated the rock would not be pitched out the back by a tire of a moving car in front of you. It may appear that way to you in the second car, but what is actually happening is that the rock would have a vertical compenent upwards and a horizontal component in the direction of the moving car. Your car has a greater horizontal velocity than the rock and catches up to it and that is what causes the impact.

    The reason being is that the tire tread and surface of the ground are in contact and have no relative velocity - ie the patch of tire contacting the ground has zero velocity at that point. All other points of the tire move froward.

    The one way a rcok can be flung backwards is if the front car has the wheels spinning.
  6. May 3, 2012 #5
    Fair enough. I had always though that there would still be some component in the opposite direction. As when you pinch an object at a stand still on the ground it flips up but also has a direction towards the opposite side of where you were pinching it?

    that being said I am concerned with the impact. and defining a possibly force that could act in such an impact. if a moving object such as the rock scenario above has too many unknowns then what about the stationary rock impact on the underside of a car?
  7. May 3, 2012 #6


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    You may think that's what you want but the concept has no real meaning where a projectile is concerned. If you pushed against a steel plate with a static force then you could produce a displacement - true. But that wouldn't relate directly to the effect of a projectile impact. The only quantities that are relevant with a projectile would be the Kinetic Energy and the Momentum (related but not the same).
    If you want to use this work in litigation of some sort then you need to be dealing with the appropriate quantities, to deal with the 'experts'.
    Note: There will be no "upward acceleration" once released from the car. The only acceleration would be g, (downwards), as the stone is in free fall. Angular velocity would not be relevant. Would a spinning stone be worse than a non-spinning one?

    AS for the range of velocities that you are likely to get from stones (small pebbles) picked up and thrown by tyres, the maximum velocity would be in the forward direction, from the top of the wheel, which is travelling forward, always at twice the road speed. This would assume that the stone was released cleanly right at the top of the tyre. Any higher speeds are unlikely to be encountered, I think.

    Afaik, the majority of stone chip damage is due to the forward motion of the car behind as it collides with stones which have been flung more or less vertically upwards by a couple of metres.

    But you are talking in terms of substantial rocks - where would they come from - apart from total vandalism?
  8. May 4, 2012 #7
    Usually, when talking about impacts, the impact is quantified in terms of energy and/or impulse, not force. Force is just too hard to calculate and not that useful. The energy and impulse are easy to calculate. Assuming that the car is much heavier than the rock and it doesn't slow down when colliding, just take E=1/2 m v^2, where m is the mass of the rock and v is the relative velocity (mainly the velocity of the car). If you assume the rock is embedded into the car, the impulse is something like I=m*v. If the rock bounces of the car, it's something like I=2*m*v.
  9. May 6, 2012 #8


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    The initial response to impact is elastic deformation. In this stage, force is proportional to deformation, i.e. like a spring. If the energy of the impact exceeds what can be stored elastically (the elastic limit) the resistive force plateaus. The depth of damage is therefore proportional to the remaining energy.
    See e.g. http://www.pdhcenter.com/courses/s164/s164content.pdf.
    For high speed impacts, the inertial density of the impacted surface becomes important. Instead of the stress being transferred right across the surface, local stresses are enough to exceed the elastic limit and the damage is therefore more localised. E.g. push hard on a metal sheet and the whole thing will buckle, but fire a bullet at it and you'll punch a neat hole in it.
    See e.g. http://www.deas.harvard.edu/hutchinson/papers/chen.pdf
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