Think Fencing Impact Resistance Test.

Let:

m be the mass of the Striker,

D be the diameter of the Striker,

h1 be the initial height of the Striker,

h2 be the rebound height of the Striker,

h3 be the height of the test specimen above the ground,

g be the acceleration due to gravity,

t be the time it takes the Striker to reach the specimen, i.e. the time it takes to traverse h1,

v(t) be the velocity of the Striker at time t,

x(t) be the position of the Striker at time t,

p1 be the momentum of the Striker on its downwards journey,

p2 be the momentum of the Striker on its upwards, rebound journey,

p be the change in momentum of the Striker,

F be the impact force exerted on the specimen by the Striker during the initial collision and

be the total stress the test specimen experiences during the initial collision.

Newtons Second Law of Motion gives

= () [Minus sign denotes downwards motion] which, upon integration gives the velocity as

= (2) Furthermore we find the position of the Striker at time, t, is

= (3)

By setting = and solving for t we can find the fall time of the Striker. Doing this yields = / ()

So the velocity at the moment of impact is given by substituting (4) into (2), = () Hence, the momentum of the Striker at impact is ! = () Now, to find the rebound momentum of the Striker let us consider the kinetic and potential energy of the Striker. At impact the potential energy is zero while at the peak of the Strikers rebound journey, the kinetic energy is zero. Since energy is conserved here, we can equate the potential energy at the peak of the rebound journey with the kinetic energy at the start of the rebound journey, and solve for the rebound velocity, and thus momentum. We get = Thus, the change in momentum is (7)-(6) i.e.

= + Now, the force experienced by the test specimen during a collision is equal and opposite to that

experienced by the Striker. Using the definition of a force, = !"!" . Since the test specimen can deform at most h3 during impact and h3h1 we will assume that the duration of the impact is approximately twice the time it takes the Striker to travel h3 at the velocity at time of impact, that is,

= Therefore, dividing (8) by (9) we get the force exerted on the test specimen during a Think Fencing

Impact Test;

= + () Now, the area of impact is

. So the stress loading, endured by the test specimen during impact is given by

= = + ()

Table 1.1

This table shows the values of force and stress applied to different Think Fencing products during a specific formulation trial.

A 5kg Striker raised to 3.5m has the same energy as needed to kick an AFL football 77m!

Profile Mass of Striker (kg)

Part Weight/m (gm)

Release Height (m)

Impact Force (N)

Stress Endured (MPa)

127x127mm post 5 2300 3.5 4850 1.715 150x50mm rail 5 1250 3.0 4150 1.464 140x40mm rail 5 1041 2.5 3490 1.254