drossall wrote:I'm not sure what you mean by "the vertical component of the impact", because impact is not a defined physical quantity as such, and certainly not a vector one, so I'm not sure what you are calculating. There are impact forces, which are vector, and proportional to the vertical deceleration (or the horizonal deceleration if you are unlucky enough to hit a tree or a wall). There are impact energies - these do depend on the square of velocity.
However, as far as I can see, the cyclist's kinetic energy is dissipated partly by a vertical impact with the ground, and partly by frictional forces acting horizontally. The amount dissipated by the vertical impact does not depend on the cyclist's actual velocity, because it is determined by the vertical velocity, which is always 12mph according to pjclinch's description.
The forces and accelerations from impact surely break down into vectors. If the head velocity before impact is forward 40mph, vertical 12mph, at nadir is forward 40mph vertical 0mph, and for simplicity's sake we'll argue that whatever "bounce" is probably lower in acceleration magnitude and maximum rebound velocity than the deceleration, due to hysteresis, then I would argue that peak deceleration vector is the same as for forward velocity 0mph, vertical velocity 12mph. The rate of helmet deformation is governed by the perpendicular velocity.
If there is also a change in longitudinal velocity then yes the two decelerations should be merged into the resultant vector, but I assume the argument is that in practice this hypotenuse is barely longer. I'm guessing that the brain is roughly of radial symmetry so the direction of the vector is not hugely important.
Fall at 0mph = 12mph velocity at 90deg to surface
Fall at 12mph = 17mph velocity at 45deg to surface
Fall at 38mph = 40mph velocity at 17.5deg to surface
All three have the same vertical velocity.