# Am I doing this right? The Boxers punch (not for school)

1. Apr 23, 2015

### SpanishOmelette

Salutations, all.

I have recently been pondering a rather basic, initiate question, just to test my classical physics.

So, if a boxer of 40kg throws a metre-long punch that hits his opponent in 0.25s, assuming he can channel his entire body weight into the punch, is the force generated 160N?

This is also operating on ignoring air resistance and assuming it is a straight punch.

Now, here is my work. This is to see if I am doing things right.

Going by linear acceleration, his fist travelled 1M in 0.25s. Therefore, in 1s he could travel 4m. So I now assume that his acceleration is 4m/s/s.

Using the formula F=MA, I multiply that acceleration by the weight of 40kg. Therefore, 160N.

Have I been working my questions correctly?

Thankyou and much gratitude,

Mahmoud.

2. Apr 23, 2015

### jbriggs444

Where did you come up with the 1 meter again? You were not given the acceleration and you are trying to calculate the force.

3. Apr 23, 2015

### SpanishOmelette

What I was trying to do is work out the acceleration by using the distance travelled in 0.25s and the mass of the boxer...

I assumed that if the fist travelled 1 Metre in 0.25s, then it would travel 4 in 1. That is linear acceleratioon, no?

Mahmoud.

4. Apr 23, 2015

### Merlin3189

"Using the formula F=MA, I multiply that acceleration by the weight of 40kg. Therefore, 160N."
But it is the fist that travelled 1m in 0.25sec, not the 40kg mass of his body.

5. Apr 23, 2015

### SpanishOmelette

But I am merely working in approximations. A skilled boxer can propel almost his entire body into his punch. As I am not an expert in biomechanics, I simply decided to opt that he does, in fact, use his entire body weight.

As A) this is merely to practice the use of the acceleration rule,

and B) this is not a real punch, and am not working on the set of "Fight Science!" by the National Geographic!

Mahmoud.

6. Apr 23, 2015

### Merlin3189

""I assumed that if the fist travelled 1 Metre in 0.25s, then it would travel 4 in 1.""
Only at constant speed, not accelerating.
The good news is that, starting from rest and travelling 1m in 0.25sec, the acceleration needs to be 32 m per sec per sec.

7. Apr 23, 2015

### Merlin3189

As indeed am I - and more power to your elbow for doing so!
But his fist must move relative to his body, else his arm will not be extended. However much his body accelerates, his fist must accelerate more.
If by "a metre long punch" you mean the target is 1m from the boxer's shoulder at the start, perhaps the fist moves 0.7m and the rest of the body 0.3m?
The fist must move faster than the body during the "throw" (else the arm would not extend) but I think it will still be moving faster when it hits the target (and gets slowed) otherwise would imply the fist accelerated to a high speed and then slowed to let the body catch up.
I think your comment (quoted) means that at the point of impact, the body is moving, though slower than the fist, and the arm "locks out" just at that instant, so that the momentum of the body is transmitted to the target. I would think the notion of putting your body into a punch is more to do with this accurate timing, than the idea that you can launch your whole body at speed towards your target with your arm outstretched like a spear! Anyone could do that, but I think would not fare too well in a boxing ring.

But back to the plot. Say the body managed to move 0.25m while the arm extended by its reach of 0.75m in 0.25sec. Then at the moment of impact (using our approximations again),
the body will have accelerated at $\frac { 2*0.25m}{(0.25sec)^2} = 8\frac {m}{sec^2}$ to $8*0.25 =2 \frac {m}{sec}$
and the fist will have accelerated at $\frac { 2*0.75m}{(0.25sec)^2} = 24\frac {m}{sec^2}$ to $24*0.25 = 6 \frac {m}{sec}$

"Using the formula F=MA, I multiply that acceleration by the weight of 40kg. " now gives $F=40*8=320N$ that he has to apply to the floor to accelerate his body.
Plus some force to accelerate the fist, say another 24N for 1kg * 24m/sec.
But this is the force to accelerate him, not that to slow him and his fist down when he reaches the target. That is a whole new can of approximate worms for you to open.

This Scientific American comment suggests that a good boxer can do more than double this speed (for the fist), though they don't comment on the amount of body motion involved. They also give a measurement of the force of impact, which might help check your own calculations.

Personally, I'd say a boxer would be staking all (or most) of his chips on a knockout punch, if he put much body movement into a punch. Were the punch to miss or be deflected, he'd be very unbalanced and vulnerable. What, apart from his opponent or the floor, is going to provide the force to decelerate his 40kg after he has launched himself forward by pushing on his feet behind him? I'm no boxer, but I'd guess the technique is maybe a slight forward motion, but keeping your mass between your front and back foot, so that you can accelerate forwards and backwards at any time. Maybe 0.25m would be possible within a stable stance allowing a similar overrun to stop if needed?

8. Apr 23, 2015

### jbriggs444

How did you figure 1 meter in 0.25 seconds if you have neither velocity nor acceleration to work from?

9. Apr 23, 2015

### SpanishOmelette

I didn't figure it. I just imagined a boxer lobbing a punch at 1m per Second, and then worked from there.

10. Apr 23, 2015

### jbriggs444

You started with an assumption of 1 meter in 0.25 seconds to arrive at 4 meters per second. Now you tell us that the 1 meter came from an assumed velocity of one meter per second. So that raises two questions:

1. Where did the one second come from?
2. How can 1 meter per second be equal to 4 meters per second?

11. Apr 25, 2015

### SpanishOmelette

Well... obviously I misunderstood the concept of linear acceleration. So I assume...

Hang on!

So what would actually happen, the?