# Re: Physics - biking question

Discussion in 'Cycling Equipment' started by Zog The Undeniable, Oct 5, 2004.

1. ### Zog The Undeniable Guest

Retro Bob wrote:

> 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.

Tags:

2. ### Jim Smith Guest

Zog The Undeniable <[email protected]> writes:

> Retro Bob wrote:
>
> > 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.