Physics of a punch












2














So I have odd question revolved around karate and would love to know if my physics assumptions are correct.



Basically I would like to know what equation to use to calculate the force of a punch when two things happen: when one arm punches out and the other pulls in to the hip while pivoting around the torso, this creates a circular type motion when you punch. The second when you just throw a punch and do more of a linear punch with the second hand staying out. A bonus would be if I can also calculate a radius difference, for example if the second hand pulling back actually comes in closer to the body and increases the force since the radius decreases going from a position of the punch being fully extended to a position where it comes back to the hip.



Someone uploaded this image physics of a punchand I'm no physics major but he talks a lot about linear equations, which seems wrong. I see linear all over it, but also see a note mentioning pivot, so I'm a little confused what its actually showing. I thought this would actually be more of a parabolic (might be the wrong word) or centrifugal equation like so:




F= mv2/r




So it's clear from this one that if you have more mass spinning like the second hand, the force would also increase. Seems simple but I wanted to see if my assumption are right it if they are actually using the correct equations.



Just as a background I'm a third degree black belt in karate and I'm trying to have a physics based approach to training, so this would greatly help in dispelling some common myths.










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  • Could you add some diagrams of the martial art maneuvers you are talking about? I’m having a hard time visualizing the motion involved in the described punch.
    – cms
    1 hour ago


















2














So I have odd question revolved around karate and would love to know if my physics assumptions are correct.



Basically I would like to know what equation to use to calculate the force of a punch when two things happen: when one arm punches out and the other pulls in to the hip while pivoting around the torso, this creates a circular type motion when you punch. The second when you just throw a punch and do more of a linear punch with the second hand staying out. A bonus would be if I can also calculate a radius difference, for example if the second hand pulling back actually comes in closer to the body and increases the force since the radius decreases going from a position of the punch being fully extended to a position where it comes back to the hip.



Someone uploaded this image physics of a punchand I'm no physics major but he talks a lot about linear equations, which seems wrong. I see linear all over it, but also see a note mentioning pivot, so I'm a little confused what its actually showing. I thought this would actually be more of a parabolic (might be the wrong word) or centrifugal equation like so:




F= mv2/r




So it's clear from this one that if you have more mass spinning like the second hand, the force would also increase. Seems simple but I wanted to see if my assumption are right it if they are actually using the correct equations.



Just as a background I'm a third degree black belt in karate and I'm trying to have a physics based approach to training, so this would greatly help in dispelling some common myths.










share|cite|improve this question







New contributor




lukasz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Could you add some diagrams of the martial art maneuvers you are talking about? I’m having a hard time visualizing the motion involved in the described punch.
    – cms
    1 hour ago
















2












2








2







So I have odd question revolved around karate and would love to know if my physics assumptions are correct.



Basically I would like to know what equation to use to calculate the force of a punch when two things happen: when one arm punches out and the other pulls in to the hip while pivoting around the torso, this creates a circular type motion when you punch. The second when you just throw a punch and do more of a linear punch with the second hand staying out. A bonus would be if I can also calculate a radius difference, for example if the second hand pulling back actually comes in closer to the body and increases the force since the radius decreases going from a position of the punch being fully extended to a position where it comes back to the hip.



Someone uploaded this image physics of a punchand I'm no physics major but he talks a lot about linear equations, which seems wrong. I see linear all over it, but also see a note mentioning pivot, so I'm a little confused what its actually showing. I thought this would actually be more of a parabolic (might be the wrong word) or centrifugal equation like so:




F= mv2/r




So it's clear from this one that if you have more mass spinning like the second hand, the force would also increase. Seems simple but I wanted to see if my assumption are right it if they are actually using the correct equations.



Just as a background I'm a third degree black belt in karate and I'm trying to have a physics based approach to training, so this would greatly help in dispelling some common myths.










share|cite|improve this question







New contributor




lukasz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











So I have odd question revolved around karate and would love to know if my physics assumptions are correct.



Basically I would like to know what equation to use to calculate the force of a punch when two things happen: when one arm punches out and the other pulls in to the hip while pivoting around the torso, this creates a circular type motion when you punch. The second when you just throw a punch and do more of a linear punch with the second hand staying out. A bonus would be if I can also calculate a radius difference, for example if the second hand pulling back actually comes in closer to the body and increases the force since the radius decreases going from a position of the punch being fully extended to a position where it comes back to the hip.



Someone uploaded this image physics of a punchand I'm no physics major but he talks a lot about linear equations, which seems wrong. I see linear all over it, but also see a note mentioning pivot, so I'm a little confused what its actually showing. I thought this would actually be more of a parabolic (might be the wrong word) or centrifugal equation like so:




F= mv2/r




So it's clear from this one that if you have more mass spinning like the second hand, the force would also increase. Seems simple but I wanted to see if my assumption are right it if they are actually using the correct equations.



Just as a background I'm a third degree black belt in karate and I'm trying to have a physics based approach to training, so this would greatly help in dispelling some common myths.







forces






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asked 3 hours ago









lukaszlukasz

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Check out our Code of Conduct.












  • Could you add some diagrams of the martial art maneuvers you are talking about? I’m having a hard time visualizing the motion involved in the described punch.
    – cms
    1 hour ago




















  • Could you add some diagrams of the martial art maneuvers you are talking about? I’m having a hard time visualizing the motion involved in the described punch.
    – cms
    1 hour ago


















Could you add some diagrams of the martial art maneuvers you are talking about? I’m having a hard time visualizing the motion involved in the described punch.
– cms
1 hour ago






Could you add some diagrams of the martial art maneuvers you are talking about? I’m having a hard time visualizing the motion involved in the described punch.
– cms
1 hour ago












1 Answer
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I applaud your desire to find an equation to explain the physics of a karate punch, but to be frank I must say that an equation has limited relevance if you don't have numbers to plug into it. The lengths of the upper arm and forearm, their masses, etc., all enter into the equation.



It's probably more to the point to explain that momentum is always conserved. If you are standing upright with your feet placed on a line parallel to your shoulders, you're basically an inverted pendulum. If you throw a punch forward with your right hand, your body will move backward. Because your arm is attached to your body, your hand will be moving at less than the shoulder-to-hand velocity. But if you throw a punch forward with your right hand while your left hand is yanked backwards, the two motions have opposite momenta. In that case, your body does not need to absorb the momentum, so your punching hand can be moving at the full shoulder-to-hand velocity.



If you deliver your punch by rotating your body so that the punching arm's shoulder moves forward while the other shoulder moves backward, the forward velocity of the punching shoulder adds to the hand-shoulder velocity of the punch. The other shoulder's backward momentum balances that of the punching shoulder, just as pulling the left hand back balances the momentum of the punching hand.



If your feet are not parallel to your body but are placed one forward and one back, your back foot can press backward against the floor while you punch. In that case, you're no longer an inverted pyramid and the above argument does not apply. The entire Earth absorbs the backward momentum so your punch will not lose velocity.



Finally, if you lunge forward while delivering the punch, and keep your rear foot in contact with the ground while punching, and meanwhile turn your body while you punch, the punch will have maximum velocity.



The energy delivered by a punch -- e.g., its ability to cause damage where it lands -- increases as the square of the velocity, so every little bit of extra velocity can make a big difference if your intention is to break bones or cause internal injuries.






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    1 Answer
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    1 Answer
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    active

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    active

    oldest

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    3














    I applaud your desire to find an equation to explain the physics of a karate punch, but to be frank I must say that an equation has limited relevance if you don't have numbers to plug into it. The lengths of the upper arm and forearm, their masses, etc., all enter into the equation.



    It's probably more to the point to explain that momentum is always conserved. If you are standing upright with your feet placed on a line parallel to your shoulders, you're basically an inverted pendulum. If you throw a punch forward with your right hand, your body will move backward. Because your arm is attached to your body, your hand will be moving at less than the shoulder-to-hand velocity. But if you throw a punch forward with your right hand while your left hand is yanked backwards, the two motions have opposite momenta. In that case, your body does not need to absorb the momentum, so your punching hand can be moving at the full shoulder-to-hand velocity.



    If you deliver your punch by rotating your body so that the punching arm's shoulder moves forward while the other shoulder moves backward, the forward velocity of the punching shoulder adds to the hand-shoulder velocity of the punch. The other shoulder's backward momentum balances that of the punching shoulder, just as pulling the left hand back balances the momentum of the punching hand.



    If your feet are not parallel to your body but are placed one forward and one back, your back foot can press backward against the floor while you punch. In that case, you're no longer an inverted pyramid and the above argument does not apply. The entire Earth absorbs the backward momentum so your punch will not lose velocity.



    Finally, if you lunge forward while delivering the punch, and keep your rear foot in contact with the ground while punching, and meanwhile turn your body while you punch, the punch will have maximum velocity.



    The energy delivered by a punch -- e.g., its ability to cause damage where it lands -- increases as the square of the velocity, so every little bit of extra velocity can make a big difference if your intention is to break bones or cause internal injuries.






    share|cite|improve this answer


























      3














      I applaud your desire to find an equation to explain the physics of a karate punch, but to be frank I must say that an equation has limited relevance if you don't have numbers to plug into it. The lengths of the upper arm and forearm, their masses, etc., all enter into the equation.



      It's probably more to the point to explain that momentum is always conserved. If you are standing upright with your feet placed on a line parallel to your shoulders, you're basically an inverted pendulum. If you throw a punch forward with your right hand, your body will move backward. Because your arm is attached to your body, your hand will be moving at less than the shoulder-to-hand velocity. But if you throw a punch forward with your right hand while your left hand is yanked backwards, the two motions have opposite momenta. In that case, your body does not need to absorb the momentum, so your punching hand can be moving at the full shoulder-to-hand velocity.



      If you deliver your punch by rotating your body so that the punching arm's shoulder moves forward while the other shoulder moves backward, the forward velocity of the punching shoulder adds to the hand-shoulder velocity of the punch. The other shoulder's backward momentum balances that of the punching shoulder, just as pulling the left hand back balances the momentum of the punching hand.



      If your feet are not parallel to your body but are placed one forward and one back, your back foot can press backward against the floor while you punch. In that case, you're no longer an inverted pyramid and the above argument does not apply. The entire Earth absorbs the backward momentum so your punch will not lose velocity.



      Finally, if you lunge forward while delivering the punch, and keep your rear foot in contact with the ground while punching, and meanwhile turn your body while you punch, the punch will have maximum velocity.



      The energy delivered by a punch -- e.g., its ability to cause damage where it lands -- increases as the square of the velocity, so every little bit of extra velocity can make a big difference if your intention is to break bones or cause internal injuries.






      share|cite|improve this answer
























        3












        3








        3






        I applaud your desire to find an equation to explain the physics of a karate punch, but to be frank I must say that an equation has limited relevance if you don't have numbers to plug into it. The lengths of the upper arm and forearm, their masses, etc., all enter into the equation.



        It's probably more to the point to explain that momentum is always conserved. If you are standing upright with your feet placed on a line parallel to your shoulders, you're basically an inverted pendulum. If you throw a punch forward with your right hand, your body will move backward. Because your arm is attached to your body, your hand will be moving at less than the shoulder-to-hand velocity. But if you throw a punch forward with your right hand while your left hand is yanked backwards, the two motions have opposite momenta. In that case, your body does not need to absorb the momentum, so your punching hand can be moving at the full shoulder-to-hand velocity.



        If you deliver your punch by rotating your body so that the punching arm's shoulder moves forward while the other shoulder moves backward, the forward velocity of the punching shoulder adds to the hand-shoulder velocity of the punch. The other shoulder's backward momentum balances that of the punching shoulder, just as pulling the left hand back balances the momentum of the punching hand.



        If your feet are not parallel to your body but are placed one forward and one back, your back foot can press backward against the floor while you punch. In that case, you're no longer an inverted pyramid and the above argument does not apply. The entire Earth absorbs the backward momentum so your punch will not lose velocity.



        Finally, if you lunge forward while delivering the punch, and keep your rear foot in contact with the ground while punching, and meanwhile turn your body while you punch, the punch will have maximum velocity.



        The energy delivered by a punch -- e.g., its ability to cause damage where it lands -- increases as the square of the velocity, so every little bit of extra velocity can make a big difference if your intention is to break bones or cause internal injuries.






        share|cite|improve this answer












        I applaud your desire to find an equation to explain the physics of a karate punch, but to be frank I must say that an equation has limited relevance if you don't have numbers to plug into it. The lengths of the upper arm and forearm, their masses, etc., all enter into the equation.



        It's probably more to the point to explain that momentum is always conserved. If you are standing upright with your feet placed on a line parallel to your shoulders, you're basically an inverted pendulum. If you throw a punch forward with your right hand, your body will move backward. Because your arm is attached to your body, your hand will be moving at less than the shoulder-to-hand velocity. But if you throw a punch forward with your right hand while your left hand is yanked backwards, the two motions have opposite momenta. In that case, your body does not need to absorb the momentum, so your punching hand can be moving at the full shoulder-to-hand velocity.



        If you deliver your punch by rotating your body so that the punching arm's shoulder moves forward while the other shoulder moves backward, the forward velocity of the punching shoulder adds to the hand-shoulder velocity of the punch. The other shoulder's backward momentum balances that of the punching shoulder, just as pulling the left hand back balances the momentum of the punching hand.



        If your feet are not parallel to your body but are placed one forward and one back, your back foot can press backward against the floor while you punch. In that case, you're no longer an inverted pyramid and the above argument does not apply. The entire Earth absorbs the backward momentum so your punch will not lose velocity.



        Finally, if you lunge forward while delivering the punch, and keep your rear foot in contact with the ground while punching, and meanwhile turn your body while you punch, the punch will have maximum velocity.



        The energy delivered by a punch -- e.g., its ability to cause damage where it lands -- increases as the square of the velocity, so every little bit of extra velocity can make a big difference if your intention is to break bones or cause internal injuries.







        share|cite|improve this answer












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        answered 3 hours ago









        S. McGrewS. McGrew

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