How do they calculate hammer force?

**Daniel Vernall** · 25th September 2016, 19:42

I've seen people mentioning the force that their hammer weapon produces but an struggling to calculate my design: A spring hammer design.

As an example: 1000N tension spring. With 10kg hammer head and 50cm arm with fulcrum at 10cm.

Force applied to hammer is:
1000N x 0.1m / 0.4m = 0.25Nm

How can I calculate how much this accelerates my hammer assuming the hammer starts at 0 degrees and pivots to 180degrees.

I assume this takes one calculation to get force to 90degrees and then a different one to add in gravity as it falls.

Would be really grateful for some help.

Ps. these motor driven hammer "400kg down force" claims seem a little off. Are they just get the kgf/m and adding the force of the hammer head dropping under its own weight?

Thanks,
Dan

**maddox10** · 25th September 2016, 19:47

A hammer is a disk that doesn't make a full circle, so the calculations are the same.
But with the caveat that acceleration to a way lower rpm will make the available hitting power a lot less.

**Daniel Vernall** · 25th September 2016, 19:49

Thanks for your reply Maddox. But doesn't the weight of the head first need lifting as opposed to a disc which has momentum and is symmetrically weighted?

**maddox10** · 25th September 2016, 19:53

For power balance isn't an issue.
Balance is an issue for mechanical reliability.

On momentum. There is exactly no difference between the impact of a tooth from a disk or the impact of a simular sized hammerhead with the same speed and mass.