# Should you really lean into a punch?

There's a conventional wisdom that the best way to minimize the force impact of a punch to the head is to lean into it, rather than away from it.

Is it true? If so, why?

EDIT: Hard to search for where I got this CW, but heres one, and another. The reason it seems counter-intuitive is that I'd think if you move in the direction that a force is going to collide into you with, the collision would theoretically be softer. You see that when you catch a baseball barehanded; it hurts much more when you move towards the ball, rather than away from the ball, as it hits your hand.

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youtube.com/watch?v=w26-BwJ9QcQ&NR=1 this old man takes some punches. He pushes forward to meet the fist early. This reduces the speed of the fist during the impact. youtube.com/watch?v=wMJ_b9uV1Lo here you can see the combination of both. Fast blocks and continuous retreat. –  whoplisp Jul 22 '11 at 15:13
Your links provide the answer to your question: 1. "When you can't duck/lean away from a punch, step in towards it with your head angled toward the attacking arm." 2. "...**cannot be blocked or dodged**, the best thing to do would be to duck and take the punch with your forehead" If you lean away, you will be presenting your more vulnerable chin rather than your more protected skull. –  Jacob Eggers Sep 9 '11 at 19:09

Without having heard this argument before I would guess that the plan it to reduce the degree to which the head rattles around.

Most of the brain damage (short term and long term) associate with a punch comes from the brain bouncing off the skull a few times as the head whips back and forth. Minimize the motion, minimize the damage. By leaning in you get a whole complex of muscles pre-tensioned to absorb the impact smoothly and give yourself some room to rock back without losing your balance. Your facial bones will heal.

That said, trying to analyze anything related to fighting in terms of physics is hard because bio-mechanics is always a complicated subject with a lot of subtle inter-relationships.

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+1 I think this is the correct answer (if there is one). It has to do with being stable and absorbing the impact energy in a more controlled way. Nothing to do with punch being softer if met with earlier (as other answers try to imply). –  Marek Feb 16 '11 at 12:36
Actually I think a punch is softer earlier, b/c you disrupt the opponents kinetic chain. The key point which is missing from the other answers is that the effective mass behind a punch has time dependance. Its maximized at the end of the trajectory. This is why heavyweights have knockout power and feather weights (who often have quick hands) do not. This is also why you are taught not to move back away from a blow. –  Columbia Feb 17 '11 at 19:33
@Columbia: The comment about disrupting the other guys mechanics is a good thought (at least if the other guy knows enough of what he's about to have any kind of deliberate mechanics). If you make that an answer, I'd vote for it. –  dmckee Feb 17 '11 at 22:10

Possibly because the punch has less force behind it, if you meet it earlier. Can't say I ever heard of this conventional wisdom, but then again, I haven't been into many fights or rings.

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And what if the punch is waning? (You're out of reach if you move backwards, other person looses balance.) –  muntoo Feb 17 '11 at 19:49

My guess would be that the objective is to reduce the amount of energy dissipated into a localised area, rather than reducing the force of the strike itself.

Consider that if you lean into a strike, your muscles will flex and become more rigid. The shockwave from the impact will then travel through the rigid muscle instead of being dissipated. The energy will then be spread out over a larger area (and if your stance is good enough, maybe into the ground).

The get an idea of what I mean, compare a newton's cradle made out of steel and one made from flesh like substance. Also similar to boxers wrapping their hands and wrists.

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The only possible reason I can see is that the earlier, one faces the punch the less amount of momentum he would face. A punch builds its momentum during its journey towards destination and it reaches its maximum at the end of it. It the target comes close then he would naturally encounter less momentum and hence less force for the change of momentum.

Force $F =\frac{(mv - 0)}{\delta t}$ =$\frac{mv}{\delta t}$ if the target lean towards the punch when its seed is v

Force $F\prime = \frac{mV - 0}{\delta t}$ = $\frac{mV}{\delta t}$ if the target does not move and the punch reach its maximum speed V.

Since $V > v$, $F\prime > F$

In both cases most of the transferred momentum will be absorbed and carried by the target's muscles and he will feel the pain.

It explains why the punch will be less if the target moves towards the punch.

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With a well thrown punch, you deliver at least as much impulse with muscular action after contact as with momentum in the hand and arm. Well, maybe not with a light little distraction punch, but when you're trying to hurt the other guy. –  dmckee Feb 16 '11 at 4:33
@dmckee: But in that case, after the collision, a considerable momentum will be carried by the internal muscles of the hand/ shoulder (because of the internal elasticity of the muscle structure) and less transfer of momentum takes place in order to keep the momentum conserved. So the impulse will be less. –  user1355 Feb 16 '11 at 6:25

This movement causes the temper of the neck muscles, and strengthens the spine. The punching power is also distributed by the muscles. The soft flesh does not absorb any impact energy and is simply overwhelmed.

How much more energy is absorbed by the neck less energy will be transmitted to the brain.

Similarly a handball goalkeeper hit the ball. Thus the arm is much more tense. From personal experience he is liable to break the arm if he is distracted.

Best to do: avoid the punch ;)

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Never had my arm broken(!) blocking a kick, but I think this is the correct mechanism, and a good analogy. +1 –  wsc Feb 18 '11 at 0:56

The second link you provide clearly indicates that one should try to roll with the punch and should never lean into it. Paul Hewitt, author of the popular textbook Conceptual Physics and former Golden Gloves champ, also teaches that one should roll with a punch. He justifies it in terms of the impulse-momentum theorem: He assumes that the force between the fist and head is what brings the fist to rest whether one rolls with the punch or not, so prolonging the contact time one reduces the average force by the fist on the head.

Strangely enough, he also mentions catching a baseball barehanded as another example of the impulse-momentum theorem, claiming it will hurt much less if you give with the ball as you catch it, the opposite of what you state. My experience squares with Hewitt's on this point. Try holding your hand inches in front of a brick wall and catching a baseball. Without being able to give during the catch, it hurts.

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+1 for not being misled by the question and using basic physics to set things straight. –  Pete Jul 20 '11 at 3:18
Oops. I wrote the opposite of what I meant to write with the baseball example. Sorry. Corrected. –  Yahel Jul 20 '11 at 4:13

And don't forget; skull is the hardest bone on your body. When the fist crashes into your skull, some bones are likely to be broken. Hopefully, not the skull. So you can avoid a second punch.

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$m_1,u_1, v_1$ be mass, initial velocity and final velocity of your head.

$m_2,u_2, v_2$ be mass, initial velocity and final velocity of opponents hand.

Now, there are four cases,

• $m_1 < m_2$ AND [ $u_1 < u_2$ OR $u_1 > u_2$ ]
• $m_1 > m_2$ AND { $u_1 < u_2$ OR $u_1 > u_2$ ]

$v_1 = \frac{u_1(m_1 - m_2) + 2m_2 (-u_2) }{m_1 + m_2}$

$u_2$ is negative because direction of $u_1$ and $u_2$ opposite. Also, we are not worried about $v_2$ because its opponents hands velocity after collision. (Who cares what happens to his hand?)

Simplifying and not worrying about denominator, $v_1 = u_1 ( m_1 - m_2)- 2m_2 u_2$

So, if $m_1 < m_2$ then $v_1 > u_1$ if $u_1$ and $u_2$ have opposite directions and $v_1 < u_1$ if $u_1$ and $u_2$ have the same direction.

and if $m_1 > m_2$ then $v_1 < u_1$ if $u_1$ and $u_1$ have opposite directions and $v_1 < u_1$ if $u_1$ and $u_2$ have the same direction

Hence, if mass of your head is less than mass of your opponents hand then move away from the punch otherwise move into the punch.

So, reason for the recommendation might be that, generally, mass of head is larger than that of hand.

Assumptions:

1. Collision is elastic.
2. Your head and opponents hand are intact after collision $m_1 , m_2$ not changed.

Source: Wikipedia Elastic Collision

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The collision isn't elastic, and neither the hand nor the fist are free: they're stuck onto their respective with big sticks of muscle, tendons and bone. –  dmckee Feb 17 '11 at 22:13
@dmckee Thanks man! I took a look at inelastic collision formulas (which I should have done earlier). $v_1 = ( m_1 - C_R m_2 ) u_1 + ( C_R + 1 ) m_2 u_2$. I guess if we plot $v_1$ as a function of $C_R$ then similar analysis should give some answer(insight:). What do you think? Do you mean $C_R = 0 (approx.)$? –  Pratik Deoghare Feb 18 '11 at 7:31