4

Intuitively I know faster punches = more speed is better. But also mass is important for generating force as well as energy. In high school physics we learn E = 1/2 mv^2 so energy is proportional to speed of punch squared. But also Force = mass x acceleration and acceleration is (speed - 0) / time of punch so there's that speed term again (this time it isn't squared).

So I'm wondering is it true that if I double the speed of my punch, itd have 2^2 = 4 times the knockout power? (assuming energy is what does the damage). Or would it only have twice the knock out power (twice the acceleration from 0).

5

Generally it is the force that is important - not the energy but.

There's a simple misunderstanding here, in that looking at E = 1/2 mv2

Let's carefully look at the units - mass * velocity * velocity

Now velocity is distance / time

So we end up with (mass * distance2)/time2

So we can see that by halving the time it takes to deliver a technique - we get 4 times the energy.

Now lets look at F = MA

That is mass * acceleration >> acceleration being distance / time2

So (mass * distance)/time2

Meaning once again if we half the time taken to deliver the technique we get 4 times the force.

The main difference between the two is that for Energy we want as much mass as possible travelling as fast as possible, for Force we want as much mass as possible travelling as fast as possible to stop as quickly as possible.

If we look at modern cars - they travel the same speed as old cars - but have crumple zones at the front so that when they hit something they slow down more gently. The energy involved is the same - but the force is lower because the acceleration is more spread out - making it more survivable. So in general it is the force that matters. Humans can travel very quickly indeed (kinetic energy) its the sudden stop that kills you (force).

Now when it comes to knocking someone out - there are so many more variables than energy and/or force - (such as the location and direction that that force/energy is applied). The main thing to realise is that the defender wants to minimise the force and energy that they receive - either by dodging the attacks entirely - or reducing the speed with which a technique hits (thus reducing energy equation directly and the acceleration/deceleration part of F=MA) - or lengthening the time over which the technique hits (again similar effect to both equations - a lesser deceleration).

| improve this answer | |
  • Maybe it would be worth adding that it is the inertia (acceleration + deceleration) which reaches the brain itself floating in your head which knocks you out. This, in turn, is, for the reasons you stated, only correlated to the force exerted by the punch itself, but moderated by many other factors (neck muscles play a huge role). – Philip Klöcking Jul 12 '19 at 15:53
0

The answer from @Collett89 is a good one. However, there are some other aspects which should also be considered for a KO.

First, delivering a punch with constant speed would give you the momentum p = mv, which represents the start energy given to the punch. And it is difficult to keep this energy through a long distance, as there are other forces working on it like gravity, the damping effect of your muscles, friction, etc. Therefore, it would be better to start the punch with a velocity which is comfortable for you, and accelerate it to full speed shortly before the contact. This is the idea of the exploding energy Fajin in tai chi chuan.

And the second thing is that all these forces are vectors. Which means they have a direction. If you don't hit the punch with the correct angle to the surface, you will deliver only a part of your energy.

And the last thing I would mention is the pressure. Pressure is the P=Force/Area, which plays a big role in this regard. If you minimize the contact area of your punch, it will be much more effective. But don't forget, all these forces affect on both sides. Maybe hitting with the tip of a finger with an enormous force would be very effective, but it would also break your finger.

There can also be some other physical parameters, which I may not know. But I am only an engineer, not a physician.

| improve this answer | |
  • 3
    The acceleration of the punch (from an inch away) is meaningless to the target - its the deceleration within the target that counts.. (if i'm hit by a baseball travelling 100mph - it doesn't matter whether the pitcher is stood next to me - or 20 metres away - its the deceleration of the ball during its impact with me - and subsequent acceleration of the part of me that got hit that is important - if it hit me somewhere soft it will slow down over a longer period of time than if it hit me somewhere solid and the force I receive will be lower) – Collett89 Jul 11 '19 at 9:50
  • @Collett89 Of course the damage is caused through the deceleration within the target, but acceleration is necessary for increasing the force applied from the punch, f=ma. And it is difficult to keep a high accelaration through a long distance because of the hindrances. That was my point. Or am I missing something? – Endery Jul 11 '19 at 11:13
  • 1
    I agree it is difficult to keep a high acceleration, but the overall speed of the "tool" at impact is what matters. So a lower acceleration for a greater time can give the same energy and force to the attack (the benefit of a short technique being a reduced reaction time for defender to block/evade). Intuitively - being hit by a car travelling at 30mph will hurt the same regardless of how long it took for the car the car to get to 30. – Collett89 Jul 11 '19 at 13:18
  • @Collett89 OK I will try to edit and correct my answer, I hope I understood it. Thanks... – Endery Jul 11 '19 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.