Math note

Here
is the basis of the mathematical model I am using. The diagram
shows how I’ve defined the various angles, shaft lengths, etc.
We are only considering the instantaneous motion of the system at the
time of two snapshots:
- The moment just before the string is released. At
this moment, everything is rotating; nothing is flying out from the
centre of rotation -- though it will the instant after the string is
removed.
- The moment of impact, calculated to correspond to
"complete release" -- a straight line from the golfer's center of
rotation down the arms and the club to the ball.
By picking these two snapshots, and having the
system
move under its own inertia, the system is very much simpler to analyze.
Mainly,
there are no radial terms to
consider; only tangential terms involving angular velocities.
Only two things you really need to note here, both of them at the
moment the string is released:
- The downswing angle ß, which measures
how far the arms move, and
- The wrist cock angle θ, which
measures the angle between the arms and the shaft of the club.
These are the simplified equations for the swing. (We
still have not included the collision with the ball at this
point.) They might still look horrific, but they are much
simpler than the full equations, and they can be solved algebraically
to give a simple expression for the condition for 100% efficiency.
To
read the equations you need to know that terms of the form
mass
x length2
measure a quantity called the moment of inertia. Moment of inertia
measures the resistance to rotational forces (torques) in the same way
that mass is a measure of resistance to a translational force.
Everybody knows "
F=ma".
Well the rotational analogue is "
torque =
moment of inertia x angular
acceleration".
The Greek symbols with dots over them are angular velocities.
| Kinetic energies
have the form: |
moment of inertia x (angular velocity)2 |
| Momenta have the
form: |
moment of inertia x angular velocity |
The subscripts on the angles,
i
and
f,
indicate initial and final velocity.
First, we require the kinetic energy and angular momentum to be the
same for the two
snapshots of the swing.


Note that the angular velocity of the two masses, beta-dot and
alpha-dot, is the same in the first snapshot.
As we did with the equivalent equations for the
collision, we can solve these two equations to calculate the final
angular
velocities for the two masses. This is how the swing efficiency curve
was
calculated. However, it is more interesting to find out the condition
for a
100% efficient swing. Ideally,
we want the blue mass to
come to a dead stop at impact, and this leads to a further
simplification, as given by the second set of equations.

This pair of equations is much simpler and has
one
non-trivial solution:
m1L12
+ m2R2 =
m2 (L1 + L2)2
The ‘bottom line’ of this analysis is that the
swing is 100% efficient when the
moment of inertia of the system in the first snap shot equals the
moment of
inertia
of the club in the second snap shot.
In words, this last equation reads, the moment
of inertia of the
initial system (with the string in place) equals the moment of inertia
of the
club when fully extended. This is analogous to the condition that the M1 = M2 in
the two-ball collision.
Most importantly, this observation tells us
that the energy transfer
occurs because of the changing moment of inertia of the club as it
swing out.
We can now calculate the optimum clubhead mass
for this passive model
by substituting the value for R,
which is given by the cosine law (from diagram a
above):
R2
= L12
+ L22 - 2L1L2
cos θ
Which then gives us the required condition for optimum clubhead mass:
| m2
= |
m1
2[1
+ cos(θ)] |
L1
L2 |