# Creating a vert. drop impact that is equivalent to a faster, but lighter hor. impact

1. Jan 5, 2010

### Steve4446

There is a standard impact test that is used in the fenestration industry. An air cannon is used to fire a piece of lumber at the unit being tested. The mass of the 'missile' is 4.09 kg. The velocity of the missile is 15.26 m/s.

I am wondering if there is a way to approximate this test without an air cannon by using a free-falling object in a drop tube. I currently have a drop tube that would allow for a 2 meter freefall. I calculated that this should give the missile a velocity of 6.21 m/s in freefall. Can I compensate for this slower velocity by adding mass to the projectile to give an equivalent impact?

How much mass would be required at 6.21 m/s to equal the impact of a 4.09 kg projectile at 15.26 m/s? Or is there no way to equate the impacts?

Thank you for your help.

2. Jan 5, 2010

### mgb_phys

Re: Creating a vert. drop impact that is equivalent to a faster, but lighter hor. imp

The kinetic energy of an object is 1/2mV^2
So to give the same impact energy with only 40% of the speed you would need about 6x ( 1/ (0.4)^2 ) as much mass.

That won't necessarily give the same effect on the object - it depends on the speed that the object deforms under impact. It won't be a million miles out - how accurate do you need?

3. Jan 5, 2010

### Steve4446

Re: Creating a vert. drop impact that is equivalent to a faster, but lighter hor. imp

That is what we first tried. We based our initial calculations on the kinetic energy of the object, and therefore we increased the mass to 25 kg. However, the results of those tests seemed to suggest to us that the mass might be too large as the failures were much more catastrophic than we expected based on previous air cannon tests.

If the same object is being impacted in both the vertical and horizontal tests, is there any way to calculate what the relationship would be between the deformations in the two impact tests? [ie. Does (deformation upon impact of 4kg missile @ 15 m/s) = k * (deformation upon impact of 25kg missile @ 6 m/s) if all other factors in the test are the same?]

4. Jan 5, 2010

### mgb_phys

Re: Creating a vert. drop impact that is equivalent to a faster, but lighter hor. imp

It depends on the details of the material. If it's a plastic it migth be able to deform to absorb the impact but only upto a certain rate.

Did your two missiles have the same shape/area hitting the target?
The damage will depend on the pressure applied. A rifle bullet has a ke of around 150J, the same as an apple dropping from say 5feet - but the effect isn't quite the same !

In this case I would have thought it would be close if the area was the same.

5. Jan 5, 2010

### Staff: Mentor

Re: Creating a vert. drop impact that is equivalent to a faster, but lighter hor. imp

If you are limited to a 2m drop, could you use compressed springs to add energy to the standard projectile in that short distance? The energy of a compressed spring is 1/2 kx^2, where k is the spring constant, and x is the compression distance. You could look up the k values for some standard (large) springs, and consider adding that to your vertical "launch" apparatus...

6. Jan 6, 2010

### Steve4446

Re: Creating a vert. drop impact that is equivalent to a faster, but lighter hor. imp

mgb,
The missiles have exactly the same shape/area on the contact surface. They are both 2x4's, but the vertical one has an iron weight on top of it. In this case, the damage I am talking about is not penetration into the wood, etc, but the actual breaking of structural components in the window. For example, a rail is struck on the unit, and the force transfers through the frame causing the jamb to break apart, etc.

berkemann,
I had considered using a linear spring, but the practicalities of the current setup make it difficult to implement without significant changes.

Considering the data I have...
Hor. setup: 4.09 kg, 15.24 m/s, ?unknown deceleration
Ver. setup: 24.5 kg, 6.21 m/s, deceleration 497.4 m/s^2

Working with F= ma, I know the deceleration of the vertical setup because I was able to obtain the deceleration time (.0125 s) from high speed video footage that we shot during the tests. Since I lack the ability to create the horizontal setup because I do not have an air cannon or a drop tube high enough to attain the required velocity, I do not know its deceleration time. However, is there a way to calculate it based on the deceleration time of the vertical setup and the constant change in variables? Something like this...

If x mass @ y velocity in given collision setup results in z deceleration
then 6x mass @ .4y velocity in same collision setup will result in k*z deceleration

As I see the situation, the fact that the setups are vertical or horizontal is almost immaterial to the calculation, and doesn't really affect the answer. The only changing variables are velocity and mass. However, I don't know whether it is possible to calculate k, with the information I have.

Thanks again for your help with this.