Impact force of a punch

I've tried to model a punch and the potential force that it delivers to a target.

*in order to model a punch, this calculation assumes that your body is not absorbing any of the force of impact, that all kinetic energy is transferred into the target, and that the target itself does not undergo elastic or plastic deformation upon impact*

momentum = mass x velocity

More important than momentum, is the CHANGE in momentum - this is called the IMPULSE

impulse = mass x change in velocity

Velocity provides a measure of both speed and direction, so we will assume that you can punch in a straight line! Impulse = force x time. So we get:

Force x time = mass x change in velocity

and this can be rearranged to provide:

Force = (mass x change in velocity)/time

This is the same as the F=ma equation, since accerlation = (change in velocity)/time. Observe that the greater the time over which the collision occurs, the smaller the force acting upon the object. To maximize the effect of the force on an object involved in a collision, the time must be decreased - this increases the rate of velocity change (acceleration) which is the major cause of damage. Think about hitting the rigid steering wheel in a car crash as opposed to a deformable airbag...short vs long collision time to reduce the rate of momentum change).

If you look at boxers, then they apply the reverse logic to minimize the effect of force by extending collision time. Relaxation is of central importance, and that injuries accompany tension. Hence why boxers relax their necks and allow the head to move backwards upon impact ("riding the punch") to extend the impact time of the glove to their head.

Now we can calculate some estimates. Let's assume the following:

Punch speed = 10m/s
Contact time = 0.05 seconds
Mass behind punch = 30kg
Target is stationary

In reality, you want your punch to be connected to your body weight, not simply to be the hand flying into someone's face (which would reduce the mass behind a punch by about 30 times). So assuming that you punch with your elbow aligned with your hip, and rotate the hip FORWARD toward the opponent - then I've estimated that a little under half your body mass will be transfered forward behind the punch. I have no empirical way of judging what percentage of body weight is really thrown behind the punch, so if someone has more accurate information, please do share it.

Therefore F = (30x10)/0.05 = 6000 Newtons (or kgm/s2)

So you are going to deliver 6000 Newtons of force (600kg) to the opponent. But that is not the whole story. The area over which this force is transfered is also important in determining the damage a punch will do (compare someone standing on you with stilletos on their feet rather than flat shoes). For this you can use the pressure equation:

Pressure = Force/Area

Again, let's assume that you are hitting your opponent with the bottom three nuckles of your fist, and that this represents a surface area of roughly 5 square centimetres (this is equal to 0.0005 square metres - we must use the same units in all calculations)

Pressure = 6000/0.0005 = 12 000 000 Pa (or N/m2)

This is roughly 1200 tonnes per square metre or 120kg per square centimetre!

[This message has been edited by someotherguy (edited 04-21-2005).]

My method and calculations of this are somewhat simpler.

If I hit the guy and he lies on the floor twitching, it was hard enough.

If he doesn't, it wasn't.

Just saves me calculation time in coming to my conclusions.

JohnL
[IMG]http://www.fightingarts.com/forums/ubb/cool.gif[/IMG]

Too many crazy numbers...couldent read it... got bored half way through and dident read it...

My tiny brain couldent prossess that stuff....


Narf..

someotherguy: Your calculations look correct, but the numbers are meaningless until you compare it to a real life experiment. When you do that, you can work the numbers backwards to figure a more accurate Mass. The speed that you used 10m/s is probably fair.

I'd be interested in what you find out.

Some...

I got it (& I'm not even a math geek). I'm glad somebody took the time to lay it out.

I think that you'll find a large number of MA-ists use their 1st 2 knuckles rather than their last 3 (wing tsun?).

What if you increased the percentage of body weight (mass) behind the punch?

Just wondering.

Well let's say you only punch with the hand and don't put your body into it. Hand mass probably around 1kg

Therefore Force = (1x10)/0.05 = 200 Newtons (or kgm/s2)

Therefore you are delivering 4kg per square centimetre, which not surprisingly is a 30 fold reduction. So if you take 1kg of mass travelling at 10m/s to equate to 20kg of "force", then simply multiple 20kg by the mass behind the punch.

For example, if you could shift 40kg of body weight behind the punch, then you'd do 20x40=800kg.

Then take the surface area of impact, something between 3-5 square centimetres, and divide the number by the area. So this example would be 800/5=160kg/cm2

That way you can work it out for any variation in mass and area that you want to explore. Remember that this is just "a model" and that the reality of a strike is much more complicated. The impact time would be different depending on whether you hit tissue or bone, and what kind of tissue - and so this will affect momentum change. Certain types of bone can absorb energy more than others, for example, the ribs can deform more than cranial bone and do not break clean but rather crumple. But this provides some rough indication

[This message has been edited by someotherguy (edited 04-22-2005).]

F=ma is true its one of the laws of motion,
But you need to clarify two things.
1. It does not apply to what you are talking about.

Because what you are saying is that there is no force if there is no acceleration.
So if something is travelling at a constant speed and hits you the force will be zero since F= m x 0. Which is wrong since a train travelling at a constant speed will kill you if it hits you.
You are looking at the wrong equation,
go to your physics teacher and ask him to explain it to you.

2. Its the energy transfered that causes damage,

Acceleration makes a differnce but you need to look into a different equation and then substitute it into this one.

He put it in terms of momentum (or impulse, which is the change of momentum)

(mass x velocity at impact) / (time of impact)

If you increase mass or velocity you get a larger force.

If you decrease the impact time, you get a larger force.

[QUOTE]Originally posted by MAGr:
It does not apply to what you are talking about.

Because what you are saying is that there is no force if there is no acceleration.[/QUOTE]

Correct, if you are not accelerating then you are not experiencing a force, that is what F=ma makes perfectly clear. If you are moving in an atmosphere, then air resistance comes into play and it must be balanced by an equal and opposite force to maintain constant velocity (which is why you don't just turn the engine off once you reach your desired speed, if you were in space, you could do this).

[QUOTE]Originally posted by MAGr:
So if something is travelling at a constant speed and hits you the force will be zero since F= m x 0.[/QUOTE]

No, because you experience acceleration when that object hits you, hence you experience a force.

[QUOTE]Originally posted by MAGr:
Its the energy transfered that causes damage[/QUOTE]

No, it is the rate of energy transfer that causes the damage, hence the rate of momentum change. Rate of momentum change is related to rate of velocity change, which is acceleration. Again, I refer to the car-crash example. If you hit the windscreen rather than the airbag, then the rate of energy transfer (for the same amount of total energy loss/gain) is greater on hitting a windscreen with your skull than hitting an airbag. In a crash, you experience decceleration from a constant velocity

[QUOTE]Originally posted by MAGr:
you need to look into a different equation and then substitute it into this one[/QUOTE]

You could use the kinetic energy equation, but telling people how many joules they deliver to a target is not going to be particularly understandable

[QUOTE]Originally posted by MAGr:
You are looking at the wrong equation,
go to your physics teacher and ask him to explain it to you.[/QUOTE]

Perhaps you should follow your own advice

[This message has been edited by someotherguy (edited 04-22-2005).]

[QUOTE]Originally posted by someotherguy:
Perhaps you should follow your own advice

[This message has been edited by someotherguy (edited 04-22-2005).][/QUOTE]

Perhaps i should. [IMG]http://www.fightingarts.com/forums/ubb/smile.gif[/IMG]
Although you should have said deceleration so as to avoid confusion.

Sorry, deceleration is really just a negative acceleration, my maths conditioned me out of referring to deceleration or centrifugal force etc! But i realised most people separate the two in their mind.

I didn't mean any offense in that last comment either, I was just being cheeky!

The way I see it, the train example doesn't apply because a train w/ no acceleration is sitting still. It had to accelerate from 0 to "X" mph. The fact it reached "X" mph & remained constant doesn't disqualify the formula. The impact on your body would be the same @ the instant the train hit "X" mph or an hour after it hit "X" mph (remaining constant).

Isn't this correct?

[QUOTE]Originally posted by hedkikr:
The way I see it, the train example doesn't apply because a train w/ no acceleration is sitting still. It had to accelerate from 0 to "X" mph. The fact it reached "X" mph & remained constant doesn't disqualify the formula. The impact on your body would be the same @ the instant the train hit "X" mph or an hour after it hit "X" mph (remaining constant).[/QUOTE]

A train at any constant speed, be it 0mph or 100mph, will not be accelerating (even though it needs a force to equally counter various friction forces).

You are right about the impact being the same if you're hit while it is at constant velocity, or if it reaches X velocity during a period of acceleration. At the moment of impact it's kinetic energy would be the same, or it's momentum would be the same. If the impact conditions were the same in both examples, then the acceleration you experience would be the same too.