In the Ride pace thread we veered dramatically off topic with:
@ColinJ saying:
"2 of my most painful crashes were when my bike was stationary.
Fall #1 happened when I decided to do a u-turn on a narrow road that wasn't u-turnable! I turned my front wheel too much, stopped dead, and toppled over onto my right knee. The pain was so bad that I almost threw up!
Fall #2 was just a few days later when I was distracted by 2 attractive women in a convertible sports car next to me at an uphill to a t-junction. I wasn't paying attention to clipping in when I tried to set off, my foot slipped off the pedal, and I toppled over onto my other knee. That was ALSO so painful that I nearly revisited my breakfast!
One thing you don't get in low speed crashes is masses of road rash all over your body but the vertical impact is the same no matter what speed you were going at.
There is a school of thought that suggests that you get less of a wallop when falling at speed, but I can't quite see how that would be."
#####
I replied: To the bolded point:
Straightforward physics.
Force = rate of change of momentum
Vertical stationary fall = momentum absorbed virtually instantly
Fall at speed = glancing impact + skid/bounce/roll = longer for momentum to be absorbed
######
to which @Ajax Bay replied:
Now resolve the momentum into vertical and horizontal components. How does the loss of the vertical element of momentum differ in the two scenarios (vertical drop and trajectory drop (from same height)? If the thrown item 'bounces' isn't the momentum change even greater?
#####
I replied:
thinking out loud....
We are a long way from the idealised elastic and inelastic collisions of mechanics problems...
In a vertical fall, the body stops almost instantly and all the kinetic energy is transferred to the body as crush and deformity
in the oblique fall the impact takes longer and the kinetic energy reappears as crush and deformity and in linear motion and rotational motion.
try slapping the desk infront of you now. vertically down and then at an oblique angle - does which hurts more?
#####
@ColinJ came back with:
In order for you to go from falling down to bouncing up then you must have momentarily gone through (vertically) stationary!
The skidding and rolling is definitely scrubbing off the horizontal momentum
#################################################################################
My suggestion now ia:
In the earlier posts we were thinking too much in terms of the idealised model of elastic collisions with strict conservation of energy and conservation of momentum, when in fact we are dealing with a deformable body of indeterminate shape and an inelastic impact on a rigid surface.
To simplify the thinking, let’s take a rough ball of soft plasticine or playdough.
###########################################################################
Now, in scenario 2, the internal forces allow the vertical momentum to reappear as horizontal and rotational motion as that is easier than internal deformation and @ColinJ ‘s wallop is less even though the total energy involved is higher.
#########################################################################
QED?
Tin hat in place!
@ColinJ saying:
"2 of my most painful crashes were when my bike was stationary.
Fall #1 happened when I decided to do a u-turn on a narrow road that wasn't u-turnable! I turned my front wheel too much, stopped dead, and toppled over onto my right knee. The pain was so bad that I almost threw up!
Fall #2 was just a few days later when I was distracted by 2 attractive women in a convertible sports car next to me at an uphill to a t-junction. I wasn't paying attention to clipping in when I tried to set off, my foot slipped off the pedal, and I toppled over onto my other knee. That was ALSO so painful that I nearly revisited my breakfast!
One thing you don't get in low speed crashes is masses of road rash all over your body but the vertical impact is the same no matter what speed you were going at.
There is a school of thought that suggests that you get less of a wallop when falling at speed, but I can't quite see how that would be."
#####
I replied: To the bolded point:
Straightforward physics.
Force = rate of change of momentum
Vertical stationary fall = momentum absorbed virtually instantly
Fall at speed = glancing impact + skid/bounce/roll = longer for momentum to be absorbed
######
to which @Ajax Bay replied:
Now resolve the momentum into vertical and horizontal components. How does the loss of the vertical element of momentum differ in the two scenarios (vertical drop and trajectory drop (from same height)? If the thrown item 'bounces' isn't the momentum change even greater?
#####
I replied:
thinking out loud....
We are a long way from the idealised elastic and inelastic collisions of mechanics problems...
In a vertical fall, the body stops almost instantly and all the kinetic energy is transferred to the body as crush and deformity
in the oblique fall the impact takes longer and the kinetic energy reappears as crush and deformity and in linear motion and rotational motion.
try slapping the desk infront of you now. vertically down and then at an oblique angle - does which hurts more?
#####
@ColinJ came back with:
In order for you to go from falling down to bouncing up then you must have momentarily gone through (vertically) stationary!
The skidding and rolling is definitely scrubbing off the horizontal momentum
#################################################################################
My suggestion now ia:
In the earlier posts we were thinking too much in terms of the idealised model of elastic collisions with strict conservation of energy and conservation of momentum, when in fact we are dealing with a deformable body of indeterminate shape and an inelastic impact on a rigid surface.
To simplify the thinking, let’s take a rough ball of soft plasticine or playdough.
Scenario 1.
Drop it vertically, it falls under gravity and the PE at the top is transformed to KE at the bottom, and it has vertical momentum.
It lands on a hard immovable surface.
It does not bounce.
KE and Momentum are not conserved through the collision with the ground as internal forces come into play as the body deforms
All of the kinetic energy at impact is transformed into deformation energy and the rough ball becomes a flatter lump. All momentum is lost through the action of the forces internal and reaction
Drop it vertically, it falls under gravity and the PE at the top is transformed to KE at the bottom, and it has vertical momentum.
It lands on a hard immovable surface.
It does not bounce.
KE and Momentum are not conserved through the collision with the ground as internal forces come into play as the body deforms
All of the kinetic energy at impact is transformed into deformation energy and the rough ball becomes a flatter lump. All momentum is lost through the action of the forces internal and reaction
Scenario 2
Fire the same ball horizontally.
Ball traces arc and strikes ground at an angle, it has the initial KE plus more KE from the fall under gravity.
It hits the surface and does not bounce significantly, instead it rolls, and slides along the ground leaving a smear of plasticine from the impact point, changing shape through the process
The KE at impact is then transformed into the sum of (deformation energy + rotational kinetic energy* + linear kinetic energy* + friction losses)
(* loose terminology for simplicity)
Fire the same ball horizontally.
Ball traces arc and strikes ground at an angle, it has the initial KE plus more KE from the fall under gravity.
It hits the surface and does not bounce significantly, instead it rolls, and slides along the ground leaving a smear of plasticine from the impact point, changing shape through the process
The KE at impact is then transformed into the sum of (deformation energy + rotational kinetic energy* + linear kinetic energy* + friction losses)
(* loose terminology for simplicity)
###########################################################################
Now, in scenario 2, the internal forces allow the vertical momentum to reappear as horizontal and rotational motion as that is easier than internal deformation and @ColinJ ‘s wallop is less even though the total energy involved is higher.
#########################################################################
QED?
Tin hat in place!