Measuring punching G-Force on a Pendulum

[lownlazy]

I am building a machine for my martial arts club so that I can have a bit of fun and measure the speed and reaction times of my club mates.

The design that I feel would yield the most reliable and accurate results would be to mount an upside down 'pendulum' on a frame. At the head of the pendulum would be a padded punching surface, the other end would be steel rod axis mounted on bearings connected to a optical rotary encoder. (rotation/angle measuring sensor)

The theory being that you hit the pad and a computer measures the nanoseconds between a given angle. Do a bit of high-school geometry and you can find the m/s acceleration and multiply it by the weight of the 'pendulum' to find the g-force. (F = MA) But...

Q1) What is the mass of punched bag?

I hope its as simple as this: if I removed the 'pendulum' from the machine and it weights 1kg. Then lay it down, with a fulcrum at the axis and scales under the head end, the head end weights 750grams. Would I be correct in guessing that the head end weight (750grams) would be use to calculate g-force? Or do I some how need to include the whole 1kg mass of the whole 'pendulum'?

2) Do you think I would need to bother dialing-out the effects of gravity? Or would the be somewhat insignificant give that the measures will be taken in the first 45 degrees (start from vertical position)

I hope that makes sence! Many thanks in advance :)

 

[mfb]

If you want to do it right, you need the inertial moment of the pendulum and the distance between punch and rotation axis. If the punched bag weights much more than the rod itself, the assumption that the whole mass is at the location of the punch (and neglect the weight of the detector) could give some reasonable approximation.

To be more specific, it would help to see a sketch of the rotating object.

 

[lownlazy]

Yes I would like to calculate it properly, can you please expand on that? Or perhaps Can you point me at a website that has a formula?

 

[lownlazy]

I googled "moment of inertia" and (correct me if I am wrong) it looks like I should to calculated each part (axis, rod, head/pag) of the swing arm construction and add them all up to arrive and mass figure?

 

[Danger]

I'm a bit curious as to whether or not all of that mechanism is necessary. Can't you just hard-mount a target pad with an accelerometer chip right under the surface?

 

[lownlazy]

I had consider an accelerometer but I decided against it because they get increasingly noisy the higher the G figure you want to read from them. Obviously the g reading is relative to the weight of the punching bag but by my approximate calculations you would need a 5-10g+ bag to bring the G-Forces down to a less noisy level but then you run the increased risk of injury or breaking bones in your hand.

With the swing arm rig using an optical sensor the force figures will be. hopefully accurate but, if nothing else, consistent and comparable within a sample group.

 

[mfb]

I googled "moment of inertia" and (correct me if I am wrong) it looks like I should to calculated each part (axis, rod, head/pag) of the swing arm construction and add them all up to arrive and mass figure?

Right, but the result is not a mass, it is a mass multiplied by a distance squared.
Together with the measured acceleration, you can relate this to a torque. The torque is equal to the force multiplied by its distance to the center of rotation (assuming the direction of force is orthogonal to the rod).

 

[Danger]

I had consider an accelerometer but I decided against it because

Cool. You obviously already know far more about the subject than I do. (That, however, will not prevent me from popping in with opinions once in a while. )

 

[lownlazy]

Cool. You obviously already know far more about the subject than I do. (That, however, will not prevent me from popping in with opinions once in a while. )

thats cool :)

 

[lownlazy]

it is a mass multiplied by a distance squared.

Ahh see that makes sence. I was see "kg m2" and I was thinking what does surface area have to do with it? :)

Thanks for you help mfb, that has me pointed in the right direction.