I am trying to design a simple drop / shock testing machine.

We want to rate our products to be shock rated to 5000g.

In order to test our products to 5000g I plan on dropping a weight onto the product imparting a shock. (See drawing)

Drop Test Rig

My question is; How high would I need to drop my 72 kg weight from the test piece in order to rate by product to survive a 5000g shock.

I understand that the impulse time and distance is quite critical in the final numbers regarding the calucations. Lets say for calucation sake, that both materials were to be steel.


1 Answer 1


it depends on how far your steal drop table travels after hitting the sample, the less travel the greater g.

The kinetic energy of the drop table when it hits the sample is $$KE=1/2mv^2 \quad and\ v=\sqrt{2gh} $$

But depending on the softness or hardness of your sample and the apparatus and your weight the force of collision changes.

$$F_{impact}X_{distance\ traveled}= 1/2mv^2 $$ For example if your steel drop weight fall 20cm and travels 2mm

$$v=1.979m/s\ and\ F=(1/2)72*1.979*1000/1mm/m = 71276.3kg\ force $$ And if your sample is one gram it will feel$ \ 71276.3*1000gr-kg/9.8 \ g\ $ acceleration.

And if the 72kg jerks up by the impact you need to add that impulse too.


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