Impact force Calculations

by david robinson
Tags: calculations, force, impact
david robinson
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Jun23-05, 01:28 PM
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Could someone please help with this rudimentry calculation. I need to know the difference in the energy at impact between the following two scenarios.

1) 2 lbs falling under gravity from 8 feet with an impact area of 12 mm.

2) 4 lbs falling under gravity from 4 feet with an impact area of 4 mm.

Many Thanks
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brewnog is offline
Jun23-05, 01:33 PM
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Energy is not measured in lbs force.
Doc Al
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Jun23-05, 01:52 PM
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Realize that the kinetic energy (measured in ft-lbs) at impact is very different from the force of the impact (measured in lbs).

The KE is easy to figure: it's equal to the initial gravitational potential energy = mgh. So both (1) and (2) have the same energy at impact = 2*8 = 4*4 = 16 ft-lbs. (Note that they don't hit with the same momentum (mv); the 4 lb object has greater momentum at impact.)

The impact force is not easy to figure. The average force of impact is a measure of the rate of change of the momentum. (The quicker an object is brought to rest, the greater the average force.) The duration of the impact depends on many things, such as the type of materials involved (steel against steel, etc.), shape, etc. You have not provided sufficient information to calculate the force. (But perhaps an engineer could take a guess as to the relative forces involved based on the behavior of typical materials.)

david robinson
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Jun23-05, 02:05 PM
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Impact force Calculations

Thanks. If all surfaces were steel and of the same cylindrical shape would that help.
Q_Goest is offline
Jun23-05, 07:17 PM
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To calculate impact force, you need to model the two objects that are colliding. Often this isn't easy. They act as springy/spongy masses, so that some of the mass is stopped almost instantly upon collision and some of the mass farther back stops more slowly. Often the objects are modeled similar to springs with a 'spring constant' in force per unit length. Once you calculate the kinetic energy at impact you can then equate that energy to the energy of the 'spring' and find the force over time and peak force.
Crosson is offline
Jun24-05, 02:20 AM
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As stated above both cylinders would have the same energy, and exert the same force upon collision.

But for one of them that force is concentrated into a smaller area, meaning that the smaller has 3 times higher pressure!

The answer is the 4 mm cylinder hits 3 times harder..
Doc Al
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Jun24-05, 08:15 AM
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Quote Quote by Crosson
As stated above both cylinders would have the same energy, and exert the same force upon collision.
It's not obvious to me that both collisions generate the same impact force.
minger is offline
Jun24-05, 10:35 AM
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I have a post in here where I was trying to find the same problem. An older engineer eventually came to me with an *cough ancient cough* handwritten textbook. It had approximations for falling objects based on velocity at impact and static deflection.

I will search for that thread and post what I ended up using, because as said, impact force is quite difficult to actually find. Back in school they would give you a nice impulse - time graph...which obviously doesn't happen in real life.

I found the equations, and here they are:

Load Factors
Static Load: 1.0
Suddenly Applied: 2.0
Suddenly Applied and Reversed: 3.0
Dropped from a height h: k
where k = 1 + ((dē + dvē)^.5)/d
and v = velocity
and d = static deflection

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