# FISHER SHOTOKAN KARATE

## PHYSICS

By Charles Rhodes, P.Eng., Ph.D.

INTRODUCTION:
Fisher Shotokan is a karate movement methodology that is optimized for delivering the maximum possible energy to the target in less than 0.5 seconds. Fisher Shotokan involves explosive linear movements and extraordinarily fast turns during which kinetic energy is accumulated. At the completion of each turn all of the stored kinetic energy is available for delivery to the target.

The physics involved in Fisher Shotokan turns is similar to the physics involved in rapid spins by elite figure skaters.

Fisher Shotokan is easier grasp if the principles can be communicated from instructor to student in the exact language of the branch of physics known as mechanics. Persons who fully understand mechanics can readily comprehend the superiority of Fisher Shotokan technique. This web page is intended to be a review for persons who have successfully completed an introductory university level course in mechanics together with the course's calculus and vector algebra prerequisites. If the reader has not previously been exposed to mechanics he/she should simply skip to the summary at the end of this web page. Such readers will have to personally experience Fisher Shotokan turns to grasp their superiority.

From a physics perspective a Fisher Shotokan turn involves three sequential phases: development of kinetic energy (angular momentum) about a vertical axis of rotation, moment of inertia reduction during free rotation and energy discharge. The first two phases both accumulate kinetic energy. The third phase delivers this kinetic energy to the target.

The second phase is the most difficult to understand. In the language of physics, if a freely rotating body converts internal chemical potential energy into work that reduces the body's moment of inertia about its axis of rotation, then the rotational kinetic energy increases. That is the principle behind the increases in angular velocity and kinetic energy achieved during the second phase of Fisher Shotokan turns. However, practical implementation of that principle involves movements that are not envisaged in the JKA kata manuals written by M. Nakayama and published in English during the period 1970 to 1989.

This web page reviews the physics of Fisher Shotokan turns. It is assumed herein that the reader has a some familiarity with basic physics concepts such as:
linear distance, linear velocity, linear momentum, linear acceleration, gravitational acceleration, force vectors, kinetic energy, potential energy, conservation of energy, center of momentum (CM), angular position, angular velocity, angular acceleration, moment of inertia, and torque. To assist the reader these concepts are briefly reviewed herein.

About 1905 Albert Einstein discovered that the total energy Ei of a particle is related to its rest energy Eio via the formula:
Ei = Eio / [1 - (Vi / C)^2]^0.5
where:
C = speed of light
Vi = velocity vector of the particle with respect to the observer.

In karate all velocities are very small compared to the speed of light. Hence to an excellent approximation this formula simplifies to:
Ei = Eio [1 + (1 / 2)(Vi / C)^2]

The kinetic energy Eik of a particle is defined by:
Eik = Ei - Eio
Hence for karate to an excellent approximation:
Eik = (Eio / 2)(Vi / C)^2]

The rest mass Mio of an object is given by:
Mio = Eio / C^2
Hence to an excellent approximation:
Eik = (Mio / 2)(Vi)^2
which is the Newtonian expression for kinetic energy.

GRAVITATIONAL CONSTRAINT:
Deliverable kinetic energy can be accumulated via both linear and angular motion. However, there is a practical constraint that the horizontal component of this motion is provided by forces that act through the legs. The legs also support the body weight. Each leg has associated with it a force vector from the point of foot contact with the floor to the hip. This force vector has both a horizontal and a vertical component. In order for the body center of momentum (CM) to remain at a constant level the sum of the legs' vertical force components must equal the body weight. For any particular leg angles this constraint limits the horizontal force components and hence the change in horizontal momentum per unit time. The net horizontal force together with the distance over which it can act limits the amount of kinetic energy that a karate-ka can accumulate via horizontal movement.

Sensei Fisher recognized that useful kinetic energy can only be efficiently developed over a limited range of leg angles and that no useful kinetic energy is developed at the Mid-Point Fully Loaded Position (MPFLP) when the body weight is directly over one foot and the other foot is not touching the ground. Much of the Fisher Shotokan footwork is related to placing the feet such that immediate and large horizontal acceleration in the desired direction is possible. This horizontal acceleration accumulates deliverable kinetic energy via both linear motion and angular motion.

NOTATION:
Mi = the particle mass designated by the subscript i
M = sum of all Mi
T = time
Eki = kinetic energy of particle mass Mi
Ekti = kinetic energy of particle mass Mi due to motion tangential to Center of Momentum (CM)
Ekt = kinetic energy of mass M due to rotation about the Center of Momentum (CM)
Ekri = kinetic energy of particle mass Mi due to motion radial from Center of Momentum (CM)
Ekr = kinetic energy of mass M due to radial expansion about the Center of Momentum (CM)
Ekc = kinetic energy of mass M due to linear motion of Center of Momentum (CM)
Ek = sum of all Eki
Ii = moment of inertia of particle mass Mi with respect to rotation axis
I = sum of all Ii
Bold Face = vector quantity with both direction and magnitude
Xi = position of particle mass Mi
X = position of Center of Momentum (CM)
Vi = velocity of particle mass Mi
V = velocity of Center of Momentum (CM)
Pi = momentum of particle mass Mi
P = total momentum = CM momentum
Fi = force acting on particle mass Mi
F = tangential force on a rotating rigid body exerted at radius Rt
Ri = radius from the rotation axis to particle mass Mi
R = effective radius of mass M from rotation axis
Rt = radius from pivot axis to location of linear force F
Wi = angular velocity of particle mass Mi about the rotation axis
W = angular velocity of a rigid body about the rotation axis
N = torque applied to a real body about the rotation axis
Subscript a = initial state
Subscript b = final state
Subscript r = radial motion with respect to CM
Subscript t = tangential motion with respect to CM
* = vector dot product
X = vector cross product
Note that by definition of vector dot product:
Vi*Vi = |Vi|^2 = Vi^2
Note that by definition of vector cross product:
(Vi - V)t = (Wi X Ri)

POSITION:
Particle i with mass Mi at time T has a position vector Xi with respect to a reference point at the origin.

VELOCITY:
Differentiating the position vector Xi with respect to time gives the velocity vector Vi of particle i at Xi as:
Vi = dXi / dT

LINEAR MOMENTUM OF A PARTICLE:
The linear momentum Pi of particle i is defined as:
Pi = d(Mi Xi) / dT

For |Vi| << C to a good approximation:
Mi = Mio
Hence:
Pi = Mio d Xi / dT
= Mio Vi

CENTER OF MOMENTUM POSITION:
Consider a multiplicity of particles that are at different spacial locations Xi. The vector position X of the center of momentum (CM) is defined by:
M X = Sum of all Mi Xi
where:
M = Sum of all Mi
Hence:
X = Sum of all [(Mi / M) Xi]

CENTER OF MOMENTUM VELOCITY:
The center of momentum velocity V is given by:
V = dX / dT
= d[Sum of all (Mi / M) Xi / dT]
= Sum of all [Mi Vi] / M
= Sum of all Pi / M

= P / M
Hence for an isolated system the center of momentum velocity given by:
V = P / M
= constant

VECTORS WITH RESPECT TO CENTER OF MOMENTUM (CM):
Each position vector Xi can be resolved into two vectors of the form:
Xi = X + (Xi - X)
where X indicates the position of the CM and (Xi - X) indicates the position of mass Mi with respect to the CM.

Each mass Mi has a velocity vector Vi that can be resolved into two vectors of the form:
Vi = V + (Vi - V)
where V indicates the velocity of the CM and (Vi - V) indicates the velocity of mass Mi with respect to the CM.

The velocity vector (Vi - V) can be resolved into two components. The component (Vi - V)r which is parallel to (Xi - X) is referred to as the radial velocity component with respect to the center of momentum. The other component (Vi - V)t is perpendicular to (Xi - X) and is referred to as the tangential velocity component with respect to the center of momentum. Note that these two velocity components are orthogonal to each other. Hence:
|(Vi - V)|^2 = |(Vi - V)r|^2 + |(Vi - V)t|^2

TOTAL MOMENTUM:
Define total momentum P by:
P = Sum of all Pi

dP = d(Sum of all Pi)
= Sum of all dPi
= 0
This equation is one of the most important conservation laws of the universe and is known as the law of conservation of momentum. This conservation law states that for any isolated system not subject to external influences:
P = constant

TOTAL MOMENTUM AND CENTER OF MOMENTUM MOMENTUM:
The center of momentum momentum is defined as:
M V
= M d[Sum of all (Mi / M) Xi] / dT
= d[Sum of all Mi Xi] / dT
= Sum of all d[Mi Xi] / dT
= Sum of all Pi
= P
Thus total momentum and center of momentum momentum are equal.

Note that adding total momentum components involves separate addition of radial and angular momentum components.

FORCE:
By definition application of force Fi to a free point mass Mi results in a change in linear momentum with respect to time T of:
Fi = dPi / dT
= d(MiVi) / dT
= Mi (dVi / dT)
where:
(dVi / dT) is the acceleration of mass Mi.

IMPULSE:
An sudden application of force over a short time is known as an impulse.
(Impulse on particle i) = Fi dT
= dPi
= a step change in linear momentum of particle i

Similarly the sudden application of an impulse force through the Centre of Momentum causes a step change in total linear momentum.

KINETIC ENERGY OF A PARTICLE:
By definition:
dEki = Fi*dXi
where dEki is an elemental change in the kinetic energy Eki of point mass Mi.

The total change in kinetic energy (Ekib - Ekia) of mass Mi resulting from application of a force Fi between position Xia and position Xib is given by:
(Ekib - Ekia) = Integral from Xi = Xia to Xi = Xib of {Fi*dXi}
= Integral from Xi = Xia to Xi = Xib of {(dPi / dT)*dXi}
= Integral from Vi = Via to Vi = Vib of {dPi*Vi}
= Integral from Vi = Via to Vi = Vib of {Mi dVi*Vi}
= (Mi / 2)(Vib^2 � Via^2)

Kinetic energy Eki of mass Mi is energy related to motion. Hence it is customary to define:
Eki = 0 at Via = 0,
Eki = Eki at Vib = Vi
giving:
Eki = (Mi / 2) |Vi|^2
This is the usual expression for the kinetic energy of a point mass Mi.

KINETIC ENERGY OF A REAL BODY:
A real body consists of a multiplicity of point masses Mi.
The mass of a real body is M.
The location vector of the CM of a real body is X.
The velocity vector of the CM of a real body is V.
The linear momentum of a real body is M V.

One of the important concepts of Newtonian mechanics is total energy. The total kinetic energy Ek of a real body is given by:
Ek = [Sum of all (Mi / 2) Vi^2]
= Sum of all (Mi / 2)[(Vi - V) + V]*[(Vi - V) + V]
= Sum of all (Mi / 2)[(Vi - V)*(Vi - V) + 2 V*(Vi - V) + V*V]
= Sum of all (Mi / 2)[(|Vi - V|^2) + 2 V*(Vi - V) + |V|^2]

However, the term:
Sum of all (Mi / 2)[2 V*(Vi - V)]
= 2 V*[Sum of all (Mi / 2)(Vi - V)]
= 2 V*[(Sum of all (Mi / 2)(Vi)) - (M /2)V)]
= 2 V*(1 / 2)[(Sum of all Pi) - P]
= 0

Hence the total energy expression simplifies to:
Ek = Sum of all (Mi / 2)[(|Vi - V|^2) + |V|^2]
= Sum of all [(Mi / 2)(|Vi - V|^2)] + Sum of all [(Mi / 2)|V|^2]
= Sum of all [(Mi / 2)|(Vi - V)r|^2]
+ Sum of all (Mi / 2)|(Vi - V)t|^2]
+ Sum of all [(Mi / 2)|V|^2]

Thus in Newtonian mechanics the total kinetic energy can be expressed as:
Ek = Ekr + Ekt + Ec
where:
Ekr = Sum of all [(Mi / 2)|(Vi - V)r|^2] = kinetic energy due to radial motion with respect to the CM
Ekt = Sum of all (Mi / 2)|(Vi - V)t|^2] = kinetic energy due to tangential motion with respect to the CM
Ekc = Sum of all [(Mi / 2)|V|^2] = kinetic energy due to linear CM motion

TOTAL KINETIC ENERGY IN FISHER SHOTOKAN:
In modelling Fisher Shotokan we are concerned about the total kinetic energy of a human being in motion. In the time frame under consideration that person will neither expand nor shrink. Hence, to a good approximation:
Ekr = 0

When the human body turns while performing Fisher Shotokan, to a good approximation every part of the body turns at the same angular velocity about the pivot axis through the CM. Hence, to a good approximation:
|(Vi - V)t|^2 = |W X Ri|^2
= |W|^2 |Ri|^2
= (W Ri)^2
where:
W = angular velocity vector along the pivot axis
X = vector cross product
Ri = radius vector from the pivot axis to Mi
W = |W|
Ri = |Ri|

Thus in Fisher Shotokan the total kinetic energy simplifies to:
Ek = Sum of all [(Mi / 2)|V|^2] + Sum of all (Mi / 2)|(Vi - V)t|^2]
= (M / 2) V^2 + Sum of all (Mi / 2)(W Ri)^2
= (M / 2) V^2 + (I / 2) W^2
where:
I = Sum of all Mi Ri^2
= Moment of Inertia

Thus in Fisher Shotokan, to a good approximation the total kinetic energy is the linear kinetic energy of the CM plus the rotational kinetic energy about the pivot axis through the CM.

ANGULAR MOMENTUM:
One of the most important concepts for understanding Fisher Shotokan turns is conservation of rotational kinetic energy, also known in Newtonian mechanics as conservation of angular momentum. In Fisher Shotokan conservation of angular momentum is used for increasing rotation rate and for accumulating rotational kinetic energy. The process of doing work (force X distance) to reduce the moment of inertia of a freely rotating body increases both the angular velocity (rotation rate) and the deliverable kinetic energy.

The angular momentum Li of particle i is defined by the vector cross product:
Li = Ri x Pi
= Mi Ri x Vi
= Mi Ri x (Wi X Ri)
= Mi |Ri|^2 Wi
= Mi Ri^2 W
= Ii W

CONSERVATION OF ANGULAR MOMENTUM FOR A BODY IN FREE ROTATION:
Consider particle i with mass Mi confined by a string of length Ri to revolve in the horizontal plane about a vertical axis. The initial state is:
Ri = Ria,
|(Via - V)t| = |W X Ria|
= |W| |Ria|
= W Ria
where |(Via - V)t| is the initial tangential velocity of particle i with respect to the CM.

The initial tangential kinetic energy Ektia of particle i with respect to the CM is given by:
Ektia = (Mi / 2) (W Ria)^2
= (Mi / 2) Ria^2 W^2
= (Iia / 2) W^2
where:
Iia = (Mi / 2) Ria^2
= initial moment of inertia of particle i

In general:
Ekti = (Mi Ri^2 / 2) W^2
= (Ii / 2) W^2

Recall that the angular momentum Li due to particle i is:
Li = Ii W
or
|Li| = Ii |W|
or
Li = Ii W

Then the initial state kinetic energy is:
Ektia = (Iia / 2) Wa^2
and the initial state angular momentum is:
Lia = Iia Wa

Now pull on the string from the rotation axis to gradually reduce Ria to Rib (Rib < Ria). The centrifugal force Fi on the string is given by:
Fi = Mi W^2 Ri
and an element of work done on the revolving mass is:
dEkti = - (Fi*dRi)
= -Mi W^2 Ri*dRi
= -Mi W^2 Ri^2 (dRi / Ri)

Recall that:
Ekti = = (Mi Ri^2 / 2) W^2
Hence:
dEkti = - 2 Ekti (dRi / Ri)

Rearrange this equation to get:
dEkti / Ekti = - 2 dRi / Ri

Integrate from state a to state b to get:
Ln(Ektib / Ektia) = -2 Ln(Rib / Ria)
= Ln[(Ria / Rib)^2]
or
Ektib / Ektia = (Ria/ Rib)^2
This formula shows how the kinetic energy of a freely revolving point mass Mi is affected by its radial position Ri.

Substitute for Ektia and Ektib to get:
Mi (Rib Wb)^2 / 2) / ((Mi (Ria Wa)^2) / 2) = (Ria / Rib)^2
or
Wb^2 / Wa^2 = (Ria / Rib)^4
or
Wb / Wa = (Ria / Rib)^2
or
Rib^2 Wb = Ria^2 Wa
or
Mi Rib^2 Wb = Mi Ria^2 Wa
or
Ib Wb = Ia Wa
or
Lib = Lia
This expression indicates conservation of angular momentum for a particle. Note that this expression is only valid for |V| << C. The parameter that is actually conserved is total energy, not angular momentum.

For a real body:
La = sum of all Lia
Lb = sum of all Lib
For all Mi:
Lia = Lib
Hence:
La = Lb
or
Ia Wa = Ib Wb
This expression indicates conservation of angular momentum for a real body in free rotation. Note that this expression is only valid for |V| << C. The parameter that is actually conserved is total energy, not angular momentum.

If during free rotation with |V| << C the moment of inertia I is reduced from Ia to Ib where:
Ib = Ia / 2
then:
Wb = 2 Wa
In practical application in Fisher Shotokan conservation of angular momentum is used to convert radial work done within a freely rotating body into increased angular velocity.

TOTAL ROTATIONAL KINETIC ENERGY:
For element of mass Mi the rotational kinetic energy Ekti is given by:
Ekti = Mi Vi^2 / 2
= Mi (Ri X Wi)^2 / 2
= Mi |Ri|^2 |Wi|^2 / 2
= Ii |Wi|^2 / 2

For a body with sufficient rigidity that all point masses Mi revolve at the same rate W about a common pivot axis:
Wi = W
Hence:
Ekt = sum of all Ekti
= sum of all (Ii |Wi|^2 / 2)
= sum of all (Ii |W|^2 / 2)
= (|W|^2 / 2) (sum of all Ii)
= I |W|^2 / 2

Conservation of angular momentum (rotational kinetic energy) gives:
Ia Wa = Ib Wb

Since Wa and Wb are in the same direction:
Ia |Wa| = Ib |Wb|

Recall that:
Eka = Ia |Wa|^2 / 2
Ekb = Ib |Wb|^2 / 2
Hence:
Ekb / Eka = Ib |Wb|^2 / Ia |Wa|^2
= Ib |Wb|^2 / ((Ib |Wb|) |Wa|)
= |Wb| / |Wa|

Thus if Ib < Ia, then |Wb| > |Wa| and Ekb > Eka. These inequalities show the physical principle of energy accumulation during Fisher Shotokan turns.

QUANTIFICATION OF THE ADVANTAGE OF FISHER SHOTOKAN TURNS:
A key part of Fisher Shotokan is for a karate-ka to rapidly turn about a pivot axis while accumulating kinetic energy for delivery to the target. The centers of mass of the pelvis, trunk and head are kept in a straight vertical line known as the body axis which is initially positioned close to the pivot axis. A turn starts with the arms and the unloaded leg rigid and horizontally extended to maximize initial moment of inertia Ia. Using the legs, hips and shoulders torque is applied to accumulate angular momentum about the vertical pivot axis.

Immediately after free rotation has commenced inward radial force is applied to the extended limbs and to the vertical head-trunk-pelvis mass to pull these masses as close as possible to the axis of rotation, thus minimizing moment of inertia Ib. This process of doing work against centrifugal force while angular momentum is conserved increases both the angular velocity Wb and the rotational kinetic energy Ektb given by:
Ektb = (Ib |Wb|^2 / 2)

For example, if at commencement of free rotation R = Ra, then:
Ia = M Ra^2
Ekta = (Ia / 2) |Wa|^2
Now retract the extended limbs to reduce R to:
R = Rb = Ra / 2
Then:
Ib = M Rb^2 = M (Ra^2 / 4) = Ia / 4
From conservation of angular momentum:
Wb = (Ia Wa / Ib)
= (Ia Wa) / (Ia / 4)
= 4 Wa
and:
Ektb = (Ib / 2) |Wb|^2
= ((Ia / 4) / 2) (4 |Wa|)^2
= 2 Ia |Wa|^2
= 4 Ekta

Thus, through the use of suitable combined head, pelvis, trunk, leg, hip and arm motion that all move distributed body mass toward the rotation axis while the body is in free rotation the Fisher method can theoretically increase the body angular velocity and the rotational kinetic energy by as much as a factor of four.

TORQUE:
Torque N is defined as the rate of change of angular momentum L. Torque is a twisting force that is used to transfer angular momentum from one body to another. Application of torque can add or remove angular momentum from an otherwise freely rotating body.

Recall that angular momentum L is defined by:
L = I W
If the body is rigid at the time of application of torque then I = constant, giving:
N = dL / dT
= I (dW /dT)

Recall that rotational kinetic energy Ekt is given by:
Ekt = (I / 2) |W|^2
Hence if I is constant during an application of torque, then during the torque application period:
dEkt / dT = I W*(dW / dT)
= W*N

Recall that if the change in energy originates with a linear force F tangent to the rotation (and hence perpendicular to the axis of rotation) at radius Rt from the rotation axis then:
dEk / dT = F*dX / dT
= F*(W X Rt)

Thus equating the dEk values for energy transfer between a linear system and a rotating system gives:
W*N = F*(W X Rt)

Since the torque vector N is aligned to the axis of rotation vector W this equation simplifies to:
|N| = |F||Rt|
or
N = F Rt
This is the common expression for mechanical torque N.

SUMMARY:
To maximize torque N, which is the rate of change of angular momentum L, as is necessary at the start of a Fisher Shotokan turn, it is necessary to have the feet spaced and in firm contact with the ground. The hip and shoulder muscles exert a large torque N about the pivot axis on the moment of inertia Ia of the upper body. While this torque is being applied the arms should be fully extended to maximize Ia.

After angular momentum is acquired and the body is in free rotation it is necessary to immediately retract the extended limbs against the centrifugal force in order to minimize the final moment of inertia Ib and hence maximize both the final angular velocity Wb and the final rotational kinetic energy Ektb.

This web page last updated April 9, 2012.