How much g's creates a 100 lbs punch to the head?

  • Thread starter Thread starter freexd
  • Start date Start date
  • Tags Tags
    Head
AI Thread Summary
A punch delivering 100 lbs of force to the head can create an acceleration of 10 to 20 g, depending on the head's mass. The discussion highlights the complexity of calculating g-forces, especially when considering factors like head movement and resistance from the neck or wall. If the head moves only 1 or 2 mm before stopping, the deceleration could theoretically lead to infinite g-forces, complicating the calculation. The conversation emphasizes the need for precise parameters, such as deformation of the head and wall, to accurately determine the resulting g-forces. Overall, the biomechanics of a punch and its effects on the head involve intricate dynamics that go beyond simple force calculations.
freexd
Messages
24
Reaction score
0
Hi guys,

I heard in the news today that a punch of a pro boxer can create about 50 g. I was just wondering how much g's a rather weak punch(100 lbs) can create?
 
Physics news on Phys.org
The human head weighs about 5 to 10 lbs. So a 100 lbs force applied to the head could cause an acceleration of 10 to 20 g, assuming no resistance from the neck.
 
  • Like
Likes freexd
Thanks for the answer! One more question - how much Newtons creates the punch itself?
 
So the result is 444N. Thanks for your help.
 
You are welcome. And welcome to PF!
 
I wanted to ask a similar question. I hope I am allowed continue this thread.

So the punch to the head stays the same, but what would be the g force if the head only moves 1 or 2 mm and then suddenly stops?

would be great if you could post the formulas for this calculation.

Thanks
 
The acceleration (in units of g if you like) is the result of dividing force (100 pounds-force) by mass (5 to 10 pounds-mass). It does not matter whether that force is applied for a long time or short, a long distance or a short. The resulting acceleration is the same.

If the head then suddenly stops, the g force for the deceleration would be infinite.

One relevant formula is ##f=ma## -- force (f) is equal to the product of mass (m) and acceleration (a) in a coherent system of units. If force is measured in pounds-force, mass is measured in pounds-mass and acceleration is measured in g's then the units are "coherent" enough for this purpose.

Another formula is ##p=ft## -- the change in momentum (p) from a constant force (f) exerted for a time (t) is equal to the product of force and time.
 
  • Like
Likes paget
Basically i undersand what you said,

but what means 'the deceleration would be infinite'?

Does it mean that the g forces can't be calculated in my example? My case referred to an actual situation, namely if a person stands very near to a wall(space 2 mm) and gets punched to the head.
 
  • #10
paget said:
My case referred to an actual situation, namely if a person stands very near to a wall(space 2 mm) and gets punched to the head.
Neither the head, nor the wall are perfectly rigid. You have to now how much they deform to compute the acceleration.
 
  • Like
Likes paget
  • #11
If the question really is about a boxer then why has it been mostly assumed that the 'force' of the punch will last for as long as you want? I think this assumption has been made without qualification and it is just possible that people may get the wrong idea about the reality of the situation. It's obviously right to quote F=ma but that is not really enough to deal with a punching situation. Imo, the proper answer to the OP is much more complicated and must include the momentum changes of the head mass, coupled to the body and the fist mass, coupled to the other body.
I don't suggest that things should necessarily be taken further than they have been, here but that there is a massive caveat required, lest anyone jump to unjustified conclusions.
 
  • Like
Likes billy_joule, paget and jbriggs444
  • #12
I couldn't reply earlier, as I didn't have much time

Ok I see now that this is indeed a complicated example. The only thing that really interested me here was, how much the stop of the head highers the g forces, compared to the case when there is no wall behind. To calculate this, we could simply take reasonable numbers. The question is, what parameters do we need?

Some were mentioned in the previous post. Let's assume the wall moves 1 mm due to the impact, and the head caves in 1 mm as well. The fist weighs 1.5 lbs, the mass of the body is 165 lbs (both persons). The head weighs 10 lbs.

As I said, the only thing that would interest me is the difference of the g forces, when the head is suddenly stopped.

I hope it can be calculated now.
 
  • #13
If I understand what you want, you want to calculate the acceleration (g-force) of the head in the following two scenarios:

A 10 lb head accelerated by a 100 lb punch with no neck or head restraint forces.

A 10 lb head accelerated by a 100 lb punch for 2 mm and then decelerated by a wall for 2 mm (1 mm of give from the wall and 1 mm of give from the head). Do you want the force of the punch to continue during the phase where the wall is also decelerating the head (I would assume so).

I agree with sophiecentaur that the real biomechanics of a punch are far more complicated than this model of a punch, but with that caveat what you ask can be calculated.
 
  • #14
DaleSpam, you understood completely right what my question was.

DaleSpam said:
Do you want the force of the punch to continue during the phase where the wall is also decelerating the head (I would assume so).

It would be great if you could show me both ways.

Thanks in advance.
 
Back
Top