Cycling Equipment · Public discussion

Re: Physics - biking question

Started by Zog The Undeniable · · Last activity · 2 posts · 409 views

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Cycling Equipment
Published
5 October 2004
Last activity
5 October 2004
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Zog The Undeniable
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  1. Retro Bob said:

    OK, all you amateur physicists:

    If a 155 pound mass (me) is traveling at 10 mph (on bike) and I
    strike an object dead on with my shoulder (roughly 2 sq inches) -
    what the load presented by my shoulder to the object.

    A rough calculation of 150lbs concentrated into 2 sq inches tells
    me that the static load would be 75 lbs/inch... but obviously
    the 10 mph would be the key here... and I forgot more physics than
    I ever really understood.


    What you're looking for is "impulse", which is a force, calculated as
    the change in momentum per second.

    If your bike weighs 25lb, then total mass is 180lb. Let's do this in SI
    units as it's a lot easier, so call it 82kg.

    Momentum is mass x velocity, so 82kg x 4.5m/s = 369 kg m/s.

    To get the impulse you need to know how long it takes for the bike to
    come to a stop. Let's assume it's something rigid and guess 1/100
    second to allow for the deformation of your flesh and bone and some
    movement of the rest of your body to absorb the impact. That gives an
    impulse of

    369 x 100 = 37kN (or kg m/s^2, if you prefer - it's the same thing).
    Divided by the 2 square inches, this gives 14.4 MPa or about 2080psi,
    which is going to hurt but shouldn't actually break your shoulder.

    This is the *average* force over the deceleration - the peak force is
    likely to be quite a lot greater. Don't try it at home.

  2. Zog The Undeniable said:
    Retro Bob said:

    OK, all you amateur physicists:
    If a 155 pound mass (me) is traveling at 10 mph (on bike) and I
    strike an object dead on with my shoulder (roughly 2 sq inches) -
    what the load presented by my shoulder to the object. A rough
    calculation of 150lbs concentrated into 2 sq inches tells
    me that the static load would be 75 lbs/inch... but obviously the 10
    mph would be the key here... and I forgot more physics than
    I ever really understood.


    What you're looking for is "impulse", which is a force, calculated as
    the change in momentum per second.

    If your bike weighs 25lb, then total mass is 180lb. Let's do this in
    SI units as it's a lot easier, so call it 82kg.

    Momentum is mass x velocity, so 82kg x 4.5m/s = 369 kg m/s.

    To get the impulse you need to know how long it takes for the bike to
    come to a stop. Let's assume it's something rigid and guess 1/100
    second to allow for the deformation of your flesh and bone and some
    movement of the rest of your body to absorb the impact. That gives an
    impulse of

    369 x 100 = 37kN (or kg m/s^2, if you prefer - it's the same
    thing). Divided by the 2 square inches, this gives 14.4 MPa or about
    2080psi, which is going to hurt but shouldn't actually break your
    shoulder.

    This is the *average* force over the deceleration - the peak force is
    likely to be quite a lot greater. Don't try it at home.

    Basicaly, what you say is true but your terminology is off. Impulse is
    defined as force integrated over time: $J=\int F(t) dt$. Impulse is closely
    related to momentum and has the same units: kg*m/s. The time rate of change
    of momentum is force, not impulse.

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