pash wrote:
From equations of motion:
v^2=u^2+2as
where:
v=final velocity
u=initial velocity
a=acceleration (or deceleration)
s=distance travelled
v=0 cos we are stopping
u=60mph or 26.822 metres per second
a=?
s=4 metres
a=-u^2/2s = 26.822^2 / (2 * 4) = -89.927 (got to be accurate in this game) m/s^2
WTF does that mean to Joe Public? Not much, so lets put it in terms of "g"
a = -9.2g
So the forces you exert stopping from 60mph in the length of a car are more than Eurofighter can pull in it's combat envelope...
The limit to deceleration on a straight run, on a good road in the dry will be looping the bike forwards, where the torque applied by the front brake against is greater than the product of the mass of the bike and rider and the distance of the centre of gravity from the front wheel spindle. The torque needed to be applied by the disc can be calculated as half the average diameter of the disc multiplied by the force needed to brake at 9.2g. The force needed can be calculated from Newton's Second Law: force = mass x acceleration... blah blah blah...
I am sure Performance Bikes did some tests recently on wavey discs etc. and I think they will have recorded g data at the limit.
Blimey, 'A' Level maths at 6:16 am, I can't even figure out how to make a brew at that time in the morning
Some years ago, Mrs Adam stuffed in an aircraft whist doing a 'touch and go' routine for her PPL. Hers was more touch than go

and after measuring the start of the skid marks (on the airfield

) and estimating the speed of the aircraft in its landing configuration (about 45kts) someone worked out she must've pulled about -10g. I forget the actual figures but remember the weals the harness left on her body for a few days afterwards. Her Log Book has an entry by her instructor - 'Very heavy landing, aircraft destroyed'
