1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Punch speed and exerted g's

  1. Oct 17, 2012 #1
    Hey guys I am a high school senior and this is my first year taking physics. I am really interested in the subject and like to find out real world problems for fun. One of the things I am trying to do is find my punching speed. I downloaded an acceleration app for my phone that displays the amount of felt g's using the accelerometer. I then tried to change this into m/s^2 and find out the speed of my punch and the force being exerted upon it. It came out to be around 110 m/s and about 748 N of force. I don't think that this is correct so I was wondering what I am doing wrong? Is my phone not accurate (more than likely) or do I need a different equation other than average acceleration and f=ma? If you want the equations I am using just ask and I will upload some photos.
     
  2. jcsd
  3. Oct 17, 2012 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF,

    How many "g's" did the phone say it felt? If you assume constant acceleration, you can figure out how fast your fist should be moving at the end of the displacement using the kinematics equations. You know a, you know d, you assume vi = 0, and then you can find vf.
     
  4. Oct 17, 2012 #3
    It said that it experienced 8.5 g's. I assumed that vi was 0 and that my arm travels about 26 inches. I converted everything into m and calculated it and got 110 m/s as vf. Then did f=ma and got 74 n or ~168 ft. lbs of force.

    Oh and I said that my arm was about 15 lbs or 6.8 kg, which is about right.
     
  5. Oct 17, 2012 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, for one thing you decelerate over a much smaller time interval and distance interval than you accelerate. So the sudden stop when your arm reaches full extension is going to cause the phone to feel a huge acceleration.

    Secondly, you forgot to take a square root:

    $$v_f^2 = 2ad$$

    2*a*d = 110 m2/s2

    You need to take the square root to get the final speed. 36 km/h sounds plausible I guess. So maybe I was wrong about the deceleration. Maybe the phone doesn't measure that, or it takes an average, or something.
     
  6. Oct 17, 2012 #5
    Well I am trying to figure the speed at which somebody would be hit, so I am thinking that the deceleration from my arm would be almost equal to if not less than from hitting a person. I am kind of missing you on the square root thing, do you mean I need to take the square root of 2ad? Because vf is squared?
     
  7. Oct 18, 2012 #6

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah, of course.
     
  8. Oct 18, 2012 #7
    The impulse delivered in a punch has very little to do with the speed or acceleration of the fist, and much more to do with what the puncher does with the rest of his/her body. A trained person can do much more damage with an elbow strike than with a punch, yet the speed and acceleration of the elbow is far less than he can achieve with his fist. Why is that?

    You can think of this in terms of impulse, energy, or momentum. Any way you look at it, you have to apply a force for a period of time or over a distance of travel. Any force imparted has to be transmitted to an equal and opposite force reacted by the ground. Nearly all the training to deliver an effective strike is aimed at maximizing the reaction force transmitted to the ground.

    To get to the ground, the force has to follow an efficient load path, or it will be absorbed by your body, which is the connection between the force delivered to the victim and the reaction to the ground. So you learn how to shape that load path to keep it as short and as simple as possible, and how to make it pass through the strongest parts of your body.

    The other important thing is delivering an effective strike is to maximize duration of force. That is why in Jujitsu I was always train to never deliver a strike to a person’s face. Aim for the back of the head. The face is just in the way. Punch through the face and head.

    Nothing can demonstrate this better than the “one inch punch” made famous by Bruce Lee. It is possible to deliver nearly full punching power with only one inch of fist acceleration distance if you do everything correctly in the rest of your body. This was never meant to be a credible fighting technique, but rather a training method to remove the speed and acceleration of the fist from the equation all together and focusing on training the rest of the body to effectively develop the reaction force to the ground.

    So if you want to understand the physics of delivering an effective punch, learn how to draw a free body diagram and understand the load paths through your body. Speed and acceleration are important, but more so for what the rest of your body is doing than for your fist.
     
  9. Oct 18, 2012 #8
    I completely understand what you are saying, I am a black belt in karate, I just wanted to do this for its and giggles. See if the accelerometer measurement was any accurate.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook