# Force delivered by a punch: Question

1. Dec 31, 2007

### BriK

I am trying to figure out the force delivered behind a right rear hand punch as you twist your hips and shoulders into the punch. I have looked at several sources to find most of the information but I am stuck on one part and would appreciate any help.

I found an article that helped me determine the approximate moment of inertia (I) for that of a thin wide rectangle (kind of shaped like a glass) with the top being the torso shoulders and the narrower bottom being the torso waist:

I = 1/12 M (d1^2 + d2^2)

d1 = the distance from center of torso to the edge of the shoulder
d2 = the distance from the center of the torso to the outer edge of the back

Assume: M = 35 kg, d1 = 0.22 m, d2= 0.07 m

Therefore: I = 0.1554 kg*m^2

Find total energy (Et) by adding the potential (Ep), kinetic (Ek) and rotational (Er) components:

Et = Ep + Ek + Er = m*g*h + 0.5*m*v^2 + 0.5*I*w^2

m = M = mass, assume arm is 8 kg
g = gravitational acceleration = 9.8 m/sec^2
h = height, assume slightly downward punch of 0.16 meters
v = velocity of the punch, assume 7.3 m/sec
w = angular velocity

The article I found then has this equation:

Et = (8 kg)(9.8 m/s^2)(0.16 m) + 0.5(8 kg)(7.3 m/s)^2 + 0.5(0.1554 kg m^2)(1.9^2/0.22^2) = 207.03 Joules

I can see where all parts come from but the 1.9 valve.
The 1.9 is in the term for angular velocity, which is: w = velocity/radius

Can anyone tell me where 1.9 comes from, or is information missing?

Thanks in advance for any help!

2. Dec 31, 2007

### mjsd

firstly, is m =M = 8 Kg or is M=35 Kg?
also the final value for Et (when you put numbers in calculator) is not 207.03J

3. Dec 31, 2007

### BriK

35kg is the torso mass but that is not thrown at the opponent...just the 8kg mass of the arm is thrown at the opponents with the torso twisting into it...

Yes...d1 and d2 are radius values...

4. Dec 31, 2007

### BriK

Yes the total doesnt add up for me either, its off by a good bit from the 207 J value given in the article I am looking at...

Et = (8 kg)(9.8 m/s^2)(0.16 m) + 0.5(8 kg)(7.3 m/s)^2 + 0.5(0.1554 kg m^2)(1.9^2/0.22^2)
Et = 12.544 + 213.16 + 0.013576 = 225.72 Is what I get...

5. Dec 31, 2007

### mjsd

ok why is 35 used for moment of inertia calculation? isn't the arm is moving only? which I assume is 8Kg right?

6. Dec 31, 2007

### mjsd

how did you get 0.013576?

7. Dec 31, 2007

### BriK

I misspoke, kind of...

The torso (35kg) is twisting, not being thrown like the fist/arm, thereby lengthening the punch reach and adding in a rotational energy...The 35kg is used to find the moment of inertia as a component to find the rotational energy...

Yes, the arm is 8kg...

You are correct also 0.5(0.1554 kg m^2)(1.9^2/0.22^2) = 5.8

I have been wondering if the mgh may be negative as I am not sure if the punch is going upward or downward...

8. Dec 31, 2007

### mjsd

i 've found it difficult to imagine what the system actually looks like.
anyway, i think you are on right track with this problem. perhaps don't worry too much about reproducing what other has but instead do the calculation yourself with your own diagrams and assumptions

9. Dec 31, 2007

### mjsd

mmm.... then perhaps the 1.9 comes from how quickly the torso is twisted?

and yeah perhpas the punch would go downward...

10. Dec 31, 2007

### BriK

Here is a link to the paper I am looking at for a reference...it may help to see the setup...

http://www.phys.ttu.edu/~cmyles/Phys5306/Papers/2004/Physics of Martial Arts.doc

Seems to make sense to me, but I cannot explain where the 1.9 value comes from...but yes it could be velocity of twisting...has to be I think, because angular velocity, w=v/r...but becuae he does not explain it I am shying away from guessing at it I guess...

11. Dec 31, 2007

### BriK

Here is another link that uses the same formula to define a straight punch (no twisting torso)...so the rotational energy is real low...

http://www.pims.math.ca/pi/issue6/page09-11.pdf [Broken]

But he uses "5 pi" for angular velocity...with the fist rotating 180degrees as it extends to completion...

Last edited by a moderator: May 3, 2017
12. Dec 31, 2007

### mjsd

looking at the first document, it appears that they did use -mgh for their first example, and I don't see why they would change it to +mgh in the rotational case, so I assume that one of them must have been a typo. I guess is that the latter is wrong. now, as far as 1.9 goes, they did not specify how they get that, but judging by the way they have been presenting the article, I guess that 1.9 is just some typical values one would use. Like the 8Kg, 7cm etc... no big deal. the ideas they were trying to get across were simply the concepts of how to account for all variables in the setup.

13. Dec 31, 2007

### mjsd

if you look at the numbers:
12.544, 213.16 and 5.8
you can see also that the rot energy 5.8 is fairly small by comparsion, so I guess not a surprise

14. Jan 1, 2008

### BriK

One other thought about the torso twisting into the punch...

As you twist not all portions of the torso turn the same amount...plus, if I lift my elbow I increase the diameter at my shoulders out to the length to reach my elbows...so I was thinking, if I segment the torso into several pieces and assume each section twists a portion of the full rotation with each portion above the last going further until the shoulders out to the elbows swing the full distance...that increases the radisu if the uppermost portion of the torso as it adds the length of the arms thru the shoulders to the elbows increasing the rotational portion of the energy calculation...

So I segment the torso and calculate the moment of inertia and rotational energy for each section with the final section being the shoulders out to the elbows...then add all of these for the total rotational portion...

I am thinking I need to do this becuase I know from experience throwing punches that a hooking rear hand roundhouse type of puch feels to have a larger impact...