Introduction

What is Judo? Judo is an MMA style sport

that involves physical and mental discipline, it involves various techniques to

throw your opponent or to pin them to the ground. Judo helps an individual

develop strength, balance, speed, agility, and flexibility. In judo, one of the main techniques present

are the throwing techniques and grappling techniques. Although there are many

techniques for throwing, they all are encompassed under 3 distinct stages,

“Kuzushi” which is the breaking of your opponent’s balance, “Tsukuri” which is

when you prepare you opponent for the throw, and then the “Kake” which is the

execution of the throw. Three main throws that most judo competitors use are

the hip throws, leg throws, and arm throws since they are the easiest and most

basic to learn. Judo’s underlying concept involves enormous amounts of physics

and math, especially when the grappling and throwing aspects of the sport are

involved.

As

a former MMA fighter, I have always learned my techniques in fighting visually

from my coach rather than on paper. Fighting against an opponent came mostly

instinctively, I never gave a second thought to the math and physics behind

each punch or how to carry out the most effective throw on an individual. I was

curious to understand more about the theoretical aspects of Martial Arts and

how gathering knowledge about it could possibly make me a better and more

accurate fighter. After learning the art for almost 6 years, I still had no

clue how the techniques I use work theoretically, how effectively I can perform

them for maximum efficiency, and the various aspects of physics and math that

play behind each move.

I am going to investigate 1 main throw

within judo called hip-throw

(harai-goshi) and 2 main physics principles involved in these throws that relate

to the weight of an individual and their force. The center of mass and torque are

most prevalent in executing throws efficiently, understanding them will help

the attacker gain an advantage against the opponent. These variables vary

within the person and they affect how the throw can be carried out.

This

throw is known to be a greater advantage to smaller and faster since it basic

principle is to knock the opponents balance before can do anything. In this

throw the opponents face each other, the attacker steps forward with their

right foot to the middle between the feet of their opponent. The attacker pulls

the opponent downward and toward their right. The opponent would be stable against

a pull directly toward the attacker, but because of the position of the feet,

an instability is created. Then, holding the opponent’s arm, the attacker

rotates their hip and throws the opponent to the ground. Although this doesn’t look

like it’s possible visually, it works theoretically using physics and basic

math. The opponent’s center of mass is stable as long as it is over the support

area of his feet. A normal person’s center of mass is between their spine and

navel, because of this, many throws target that area to manipulate a person’s

center of mass to execute a throw with minimal effort. The reason why this

throw is most beneficial to lighter individuals is because if a light-weighted

person were to fight a heavier person, the center of mass would be closer to

the heavier person, giving the lighter individual an advantage since the torque

becomes greater making up for the amount of strength of the lighter individual.

To figure out where the center of mass between two individuals is going to be,

you have to use an equation where m = the weight of the individual

in kilograms, x= position of the individual, and X=position of center of mass.

For

example, if a 25 kg individual is going to throw a 75 kg individual and they

are 60 cm apart, you would find the center of mass by plugging in the numbers

accordingly to the variables.

Once

you get the answer, the center of mass will be 45cm from the starting point

which was 0. In this case, the center of mass will be closer to the 75 kg

individual creating a smaller torque giving the light-weighted person an

advantage in the throw. In this specific throw, the two opponents are around 1

foot apart from each other, and then depending on the weights of the

individuals, the center of mass will change. In the problem above, I

substituted x1 for 0 and the other for 70.

25cm 45cm 75cm

(0) (60)

If

we were to switch around the values for the x’s so that x1 is 70 and

the other x is 0, we would get another number. Nonetheless, the center of mass

would still be in the same place as it was in the above example.

+ (Positive)

25cm 52.5cm 75cm

(0) – (Neg) (70)

I

modeled a basic diagram to visualize the equation in the previous problems,

following the same properties as a number line, going right is positive and

going left is negative. If we were to do the same equation but switching the

values of the x’s, we would get the same center of mass but a different number.

+ (Positive)

25cm 15cm 75cm

(60) – (Neg) (0)

In

the equation above, x1 became -70 since the initial position is

coming from the right heading to the left end of the number line, going in the

direction of negative. Although we got a different number, it is still in the

same place since now we are working from the right to the left ends of the line

rather than the regular positive route, left to right ends. To double check

that answer and its position, just subtract 45 from 60 and you will get 15 cm,

thus proving this method of solving the equation.

During a throw, if an individual’s

center of mass moves more towards their feet, gravity will create a torque from

its pull on the individual’s center of mass. The equation to calculate torque

is

T=

torque

F=

linear force

r=distance

from axis of rotation to where force is

applied

In

this case, the force bringing a rotation and the lever arm between the pivot

point and the force would be multiplied to find the torque on the unstable

individual since the weight vector is perpendicular to the lever arm. However,

when you don’t know the vector that is perpendicular “r” in the diagram above,

you would use the sine of the angle given as an easier way to find the torque.

But in this scenario since we already know the weight vector, the equation can

be simplified to

where

Tperp.= the vector perpendicular to r.

To put this into a scenario that

corresponds with this specific throw, if an individual were to put 60N of force

onto the hip getting ready to throw their opponent, what would be the torque if

the force is 11 cm away and the direction of the force created a 45 degree

angle?

To

solve this equation, you would use,, and in the diagram, F2 being

a force parallel to r, and F3 being the force perpendicular to r.

The hole in the diagram represents the axis of rotation and the rectangle being

the hip of the opponent. Fperp. encompasses the force in newton

times the sine of the angle. Although this doesn’t accurately display the shape

of the hip and its axis of rotation, this diagram was the closest I could think

of for a model to this sort of equation since it conceptually would work the

same.

In

this problem, since we have to find the force that is exerted perpendicular, we

would have to draw a triangle to find out the leg of the triangle that we need,

F3. Based on alternate interior angles, the left angle in the

triangle is also 45 degrees, and since the identical angle is directly opposite

from F3, it becomes 60 N.

)

Now

we plug this answer into the equation to find torque and multiply it by r in

meters.

So,

after solving this equation, the torque of the opponent would be 4.667 Nm when

60 N of force is applied at a 45 degree angle to the opponent’s hip.

When the opponent is unstable, gravity

will pull the opponent down based on the opponent’s weight, this can be

conceptually thought as a vector going down from the individual’s center of

mass, the weight vector.

In

this picture, the lever arm is the horizontal line running from the individual’s

pivot point and the end of the weight vector. The weight of the opponent and

the distance of the level arm multiplied gives the torque. However, when the opponent

is stable standing up the lever arm for his weight vector is zero which makes

the torque is zero. If the person’s center of mass moves forward to where their

feet are, the lever arm won’t be zero and the torque will cause the person’s

rotation. If the person is leaning more, the lever arm will increase resulting

in a greater torque. For example, if the weight vector of two individuals are

both 20 N and one individual has a lever arm of .5 m and the other .7m, the

individual with a lever arm of .7m would have a greater torque than the other. This

particular throw is fairly simple mathematically when trying to find the center

of mass and the torque, but it gets a little more complicated with other throws

that don’t target near the center of mass.

Doing the research for this topic not

only has helped me gain more insight into the mathematical works of the art but

it also helped inspire me to learn Judo and MMA once again. Researching in

depth on how physics affects the throw and its quality affected by center of

mass and torque will make me more aware when practicing and learning new

techniques. My research wasn’t just limited to judo, I gained an interest in

what other forms of MMA were about and their underlying concepts. Even though I

won’t be calculating equations in my head whenever I practice, learning more in

depth theoretically will aid me when I need to perform a technique accurately,

using my newly found knowledge of center of mass and torque.