Re:Bontrager’s WaveCel material more effective at preventing...
by lambda-11235 on Thursday August 29, 2019 @01:14AM
(#20826864)
You're right about the velocity, but the impact force really depends on the properties of the material one lands on.
First consider that our kinetic energy is
K = mv^2/2
Additionally is the material has a spring constant k, then the force that results when the material deforms by x amount is
F(x) = kx
If we let d be the max distance the material, then we can integrate over the force to get the energy absorbed, which is equal to the kinetic energy.
int_0^d F(x) dx = K
kd^2/2 = mv^2/2
Now there are two ways the impact could work. Imagine we have a really thick crash pad, then we consider it to deform without limit and have a constant k, giving
d = sqrt(mv^2/k)
F_max = F(d) = v sqrt(mk)
But what about a helmet? It can only over it's thickness, at which point your head is basically in contact with the concrete. Thus, we have a constant deformation distance d, and get
k = mv^2/d^2
F(d) = mv^2/d
As we can see, depending on how we consider it we get either a force of sqrt(10) or 10 times the original amount. I probably failed to take into account something that someone who knew more about materials could point out.
tl;dr it's complicated because impact force does not necessarily scale linearly with velocity.
update: so after some conversation with others. It seems that head on concrete is better modeled by a constant d. So a drop of 10 times the height results in 10 times the force.
First consider that our kinetic energy is
K = mv^2/2
Additionally is the material has a spring constant k, then the force that results when the material deforms by x amount is
F(x) = kx
If we let d be the max distance the material, then we can integrate over the force to get the energy absorbed, which is equal to the kinetic energy.
int_0^d F(x) dx = K
kd^2/2 = mv^2/2
Now there are two ways the impact could work. Imagine we have a really thick crash pad, then we consider it to deform without limit and have a constant k, giving
d = sqrt(mv^2/k)
F_max = F(d) = v sqrt(mk)
But what about a helmet? It can only over it's thickness, at which point your head is basically in contact with the concrete. Thus, we have a constant deformation distance d, and get
k = mv^2/d^2
F(d) = mv^2/d
As we can see, depending on how we consider it we get either a force of sqrt(10) or 10 times the original amount. I probably failed to take into account something that someone who knew more about materials could point out.
tl;dr it's complicated because impact force does not necessarily scale linearly with velocity.
update: so after some conversation with others. It seems that head on concrete is better modeled by a constant d. So a drop of 10 times the height results in 10 times the force.
See also http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html