Leecommotion, the RightSide Swing
Part 2  Analysis
Dave Tutelman
 December 20, 2010
The first article
on the rightsided swing refers to a mathematical analysis of clubhead
speed. This article details that analysis. The mathematical analysis
here is not
essential to understanding the swing, though it does give insight into
where the power comes from. We first derive a model to
analyze the
rightside swing, then apply it to a particular golfer's driving
distance to validate the model.
Deriving the Model
The
rationale for the model is described in my article
on the doublependulum model of the swing. Here we will just recap the
details necessary to do the mathematical model.
The
figure is based on the notion that the hands are moving along a curved
path driven by shoulder torque. It is assumed that the shoulder torque
is transmitted to the hands through a lever arm, whose length is the
radius from the hands to the center of rotation.
This is a useful model of both the conventional swing and the
Leecommotion swing.
 In the conventional swing, the radius
is a constant
equal to the length of the left arm.
 In
the Leecommotion swing, the left arm defines the "track" along which
the hands move, but the right arm transmits the torque from the
shoulders. The folded right arm is a shorter lever arm, so the force on
the hands is greater.
The distance r
is the radius from the center of rotation to the hands moving along the
path. For the Leecommotion swing, this radius changes during the
downswing. Initially, the arms are folded and r
is relatively small. As the downswing evolves and the right arm
extends, r
assumes a larger value, eventually the same as it would for a
conventional swing. So we are working with a time function r(t).
Because r(t) varies,
the moment of inertia of the arms, hands, and club (as a unit) also
varies. So our model wants to:
 Start with the reality of a varying moment of inertia
exposed to the actual shoulder torque created by the golfer.
 Derive a timevarying shoulder torque T(t)
that gives the same acceleration to the hands as #1 would, but assuming
the moment of inertia were constant.
Why do we want to do this? Because the most convenient tool for
simulating the swing, the SwingPerfect
computer program, will accept a varying shoulder torque as input, so we
will be able to simulate the Leecommotion swing by computer.
Let us proceed to derive T(t).
The green boxes are the detailed math. If you don't want to follow it
in that much detail, you can safely skip them. The important results
will be summarized.

NonCircular
Model: Upper Bound
This note in green tells you how to calculate the torque profile for a
given noncircular swing. It depends only on high school math and
college freshman physics. But if you're not comfortable with algebra
and never expect to run the computational model yourself, you won't
miss anything if you skip the note entirely.
Let's start with notation:
 T_{o} =
The actual shoulder torque that the golfer is capable of exerting for a
golf swing.
 T(t) = The
torque profile used as input to the model, as it varies over time.
 R_{o} =
The circular radius in the latter (release) portion of the swing. If
the swing is not circular at this point, the model is inaccurate.
 r(t) = The
distance from the path to the center of rotation, as it varies over
time. Obviously, it becomes R_{o} late in
the downswing.
 I_{o} =
The moment of inertia of the arms, hands, and club when fully extended
and still with the initial wrist cock. Basically, the moment of inertia
as it would have been for the early phases of a normal, circularpath
downswing.
 I(t) = The
moment of inertia of the arms, hands, and club, when folded to the
lever arm of r(t),
also as a function of time.
 α(t) =
Angular acceleration of the arms, hands, and club, as a function of
time.
 M =
The mass of the arms, hands, and club.
 K = An
arbitrary constant used in computing the moment of inertia. Its value
depends on the shape of the body. For most shapes, the formula takes
the form I=KMr^{2} as a
quick visit to any table of moments of inertia will
attest. Luckily for us, we don't need to compute its value, because it
will cancel out of our calculations.
The goal is to compute T(t),
the torque profile to apply to a standard doublependulum model.
We depend heavily on the elementary piece of physics, T=Iα, the rotational
analogy of F=Ma.
But we will see it in the form:
First of all, the angular acceleration over time will be given by the
equation (1), where the torque is what the body produces T_{o} and the
moment of inertia is the actual moment of inertia due to the changed
lever arm I(t).
In other words:
Our goal is to compute a new torque function T(t) that,
when applied to the simple circular double pendulum model, gives the
same angular acceleration over time. So:
Since we are constraining both cases to the same acceleration profile α(t),
we can equate (2a) and (2b) to get:
α(t) = 
T_{o}
I(t) 
= 
T(t)
I_{o} 
T(t)
= T_{o} 
I_{o
}
I(t) 
(3) 
Now we need to see how the moment of inertia varies during the early
part of the downswing. The only thing that is varying, fortunately, is
the radius from the center of the torque to the hands. If the club were
flying out from the center, releasing the wrist cock angle, we would
have major complications here. But, since it is not, we can compute the
moment of inertia to a reasonable approximation by the simple,
wellknown:
I
= K M r^{2}
Plugging this into the moments of inertia in equation (3):
T(t)
= T_{o} 
K M R_{o}^{2}_{
}
K M r^{2}(t) 
T(t)
= T_{o} 
R_{o}^{2}_{
}
r^{2}(t) 
(4) 
Equation (4) is our answer, and it is really simple. The details of the
mass and shape of the arms, hands, and club have canceled out; we don't
have to worry about them. The formula may not be trivial to use,
because we have to find r(t) for the
actual swing. That may require a frame by frame analysis of video of
the
swing. But at least we know what we need from the video, and how to use
it when we have it.
As a "sanity test" of equation (4), we would expect T(t) to
revert to T_{o} in the
latter phases of the downswing. And it will, as long as r(t) reaches R_{o} before
the wrist cock angle releases much. And, if you remember,
that was one
of the constraints we knew up front we would need to impose on the
model.
So equation (4) is the way to find T(t)
from r(t),
and it is remarkably simple. Simply scale T(t)
to the inverse square of r(t).

But
wait!
This gives distances that are too optimistic. We are not accounting for
conservation
of angular momentum.
Think about what happens when a spinning figure skater extends or
pulls
in her arms. With the arms extended, the spin slows down. With
the arms tucked in close to the body, the spin speeds up. That is
because angular momentum must be constant. That is:
ω_{1}I_{1}
= ω_{2}I_{2}
Where ω is
the angular velocity of the skater, and the subscripts 1 and 2 represent
extended and tucked arms respectively.
That is very analogous to the case we have here. We have based the
analysis on
angular acceleration. But bear in mind that acceleration accumulated
over time is velocity. And some of that velocity does not remain
accumulated as r(t) increases and thus
raises the moment of inertia. As the arms extend, conservation of
angular momentum demands that we shed some of the angular velocity that
previous acceleration had given us.
That is why I labeled this model "Upper Bound". It is a toooptimistic
estimate for clubhead speed.
I tried to quantify how optimistic it would be. The math turns out to
be messy. Not as much work as a complete rewrite of the simulation, but
a major step in that direction. It would have to account for the actual
moment of inertia and actual angular velocity, which our formula has so
far managed to make cancel out. By explicitly including them, we bring
in other complications.
So let us take a slightly different approach  and a much easier one.
We know the model we already have will overestimate our clubhead speed,
because we didn't allow for the slowing effect of angular momentum. So
let's see if we can come up with another model that will underestimate
the clubhead speed. Then we will have a lower bound to go with our
upper bound. We won't know clubhead speed exactly, but we will have a
range that it has to lie within.
The model
we will
use is shown in the diagram. Instead of spinning a moment of inertia,
we'll accelerate a point mass. So we will assume all the mass of the
arms and hands to be traveling along the curved track along with the
hands. This means that the shoulder torque will have to accelerate the
mass of the
biceps as much as it accelerates the wrists. This clearly does not give
as much
clubhead speed as we would get from the real swing, because
rotation does not require as much acceleration of mass that is closer
to the center of rotation... But we are treating it as if it did.
The things to notice about this diagram are:
 The black dashed line is the actual path of the
hands. The important thing to notice here is that it is not a circle
around the shoulder pivot; if it were, then it would be perpendicular
to the dottedline radius. So there is some angle between the actual
path and an ideal circular path.
 Shoulder torque produces a force perpendicular to r.
That is what torque does. The force is shown in blue, and is of a
magnitude equal to the torque divided by r.
 The force can be resolved into components, shown in
aqua. One component (the bold one labeled Fa)
is tangent to the path
of the
hands, and so accelerates the hands along their path. The other (the
very pale one) does nothing to aid or hinder the progress of the hands,
and is rather small as well; it serves to curve the path. But the left
arm pull does that, and allows for forces like this, so we can ignore
it for most analysis
purposes.

Let's
use this new model to see what form the lower bound will take.
NonCircular Model: Lower Bound
Again, the derivation depends only on high school math and college
freshman physics. And again, you won't miss much
if you skip the note entirely.
The notation is similar, but includes linear instead of angular motion:
 T_{o} = The
actual shoulder torque that the golfer is capable of exerting for a
normal, circular swing.
 T(t) = The
torque profile used as input to the model, as it varies over time.
 R_{o} =
The circular radius in the latter (release) portion of the swing. If
the swing is not circular at this point, the model is inaccurate.
 r(t) = the
distance from the path to the center of rotation, as it varies over
time. Obviously, it becomes R_{o} late in
the downswing.
 θ(t) = the
angle between the actual path and the ideal circular path, also as a
function of time.
 F =
the force produced by shoulder torque at radius r(t).
 F_{a} = the
accelerating component of F,
tangent to the path of the hands.
The goal is to compute T(t),
the torque profile to apply to a standard doublependulum model. As
before, we want to match the acceleration for the standard case to the
acceleration for the
variableradius case. Since we know F=ma,
and the mass is the same, we can do it by matching the accelerating
force F_{a}.
The derivation goes very similarly to the upper bound.
We know from the diagram above that
F =
T_{o}
/ r(t)
(5)
Also from the diagram, we can resolve F into
components to get
F_{a}
= F cos θ(t)
= T_{o} cos θ(t) /
r(t)
(6a)
In order to obtain the same velocities along the curved "track", we
need to apply the same F_{a} when
we use the standard double pendulum model. So our torque T(t) for the
standard double pendulum has to satisfy
F_{a}
= T(t) / R_{o
} (6b)
So we have two equations for F_{a},
(6a) for the curved track and (6b) for the standard double pendulum.
Since F_{a} has
to be the same for both, let's set them equal.
F_{a}
= 
T_{o} cos θ(t)
r(t) 
= 
T(t)
R_{o} 
T(t)
= 
T_{o} 
R_{o}
r(t) 
cos θ(t) 
(7) 
This is pretty simple, and applying it is similar to applying the upper
bound. But wait, it usually gets even simpler.
Remember that the angle θ is not
supposed to get too big. Well, if it stays small enough (say, under
10º) we can safely ignore the factor cos θ(t).
At 8º the error from ignoring the cosine is only 1%, and is still under
2% at 11º. That sort of error is smaller than most sources of
measurement error when it comes to modeling a golf swing. If we assume θ is
rather acute and drop
out the cosine, we are left with
T(t)
≈ 
T_{o} 
R_{o}
r(t) 
(8) 
Very simple and manageable!

Thus
our model consists of two torque profiles vs time, T_{upper}(t)
and T_{lower}(t).
One gives an upper bound to clubhead speed and the other a lower bound.
Each can be applied to the doublependulum simulator program, which
will give us a clubhead speed. The clubhead speeds (upper and lower
bounds) can be applied to a trajectory program to give us a range of
distances. Summarizing the model:
T_{upper}(t)
= T_{o} 
R_{o}^{2}_{
}
r^{2}(t) 
equation (4) 
T_{lower}(t)
= 
T_{o} 
R_{o}
r(t) 
cos θ(t)
≈ T_{o} 
R_{o}
r(t) 
equation (8) 
Rewritten as an inequality to reflect the bounding nature of the model:
T_{o} 
R_{o}
r(t) 
< T(t)
<
T_{o} 
R_{o}^{2}_{
}
r^{2}(t) 
equation (9) 

The key parameter in all this is the ratio of R_{o} to
r(t).
Multiply the body's shoulder torque by this ratio for the lower bound,
and by the square of the ratio for the upper bound. Beautiful!
Using the Model on the Leecommotion Swing
Now we have the tools we need to see if Rock's distance gains are
really due to the physics of Leecommotion, specifically the shorter
lever arm due to pushing the hands around with a folded right arm. The
steps are:
 Find the shoulder torque T_{o}
that gives the golfer's carry distance with a conventional swing, 250yd.
 Apply the
club's physical characteristics to the TrajectoWare Drive computer program to determine the required clubhead
speed.
 Apply the golfer's and the club's physical
characteristics
to the SwingPerfect computer program to find the shoulder
torque that gives the clubhead speed we just found in step (1a).
 From detailed videos of the golfer, determine r(t) and
from it the ratio R_{o/}r(t).
 Find
the range of carry distances that the Leecommotion swing would give.
 Compute the upper and lower bounds on shoulder torque T(t) using the model (equation 9).
 Apply
the upper and lower bounded shoulder torque T(t)
to SwingPerfect and find the range of clubhead speeds.
 Apply
these speeds to TrajectoWare Drive to find the range of carry distances.
Let's begin by cataloguing the data we know about Rock, his swing, and
his club:
Personal characteristics 
Weight 
205
pounds 
Arm
extension 
24.5" 
Swing 
Shoulder
turn 
180º 
Wrist
cock 
96º 
Angle
of attack 
Assumed
level 
Impact
position 
Assumed
center of clubface 
Driver 
Conventional 10.5º OEM Driver,
so we assume: 
Head
weight 
200
grams 
Shaft
weight 
65
grams 
Length 
45" 
COR 
0.83 
1. Find shoulder torque for conventional swing
Plugging Rock's data into TrajectoWare Drive, we find that a carry
distance of 250yd requires a clubhead speed of 106.0mph.
We then turn to SwingPerfect, and play with the shoulder torque until
we find one that gives a 106mph clubhead speed. That turns out to be 53
footpounds. So we have:
T_{o} = 53
footpounds
2. Find the radius and the ratio over the course of the
downswing.
This turned out to
be
considerable work. I had a lot of pretty good
video footage of Rock's swing, and my intent was to find frames at the
appropriate moments and measure r(t)
on those frames. The "appropriate moments" would be defined by the
intervals that SwingPerfect allows the input of shoulder torque: every
.05 seconds up to .20 seconds.
The plane of hand motion is not the
plane of the videos. That is not surprising; the swing plane
is
typically slanted at about 60º. Unless the camera is not only faceon
but also slanted down at 30º from above, the plane of the picture is
not the plane of the swing. So I would have to apply some sort of math
to the measurements of the pictures. My first thought was to find the
plane of hand motion and use simple trigonometry to find the actual
distances.
Finding the plane of hand motion involves superimposing a series of
frames
from the downtheline video, as shown at the left. The picture shows
the fallacy of my initial plan. The dotted lines are from the center of
rotation  the spine at the back of the neck between the shoulders 
to a couple of hand positions. The problem is that the center of
rotation is not even close to being in the plane of hand motion. The
angle of r(t) changed,
position to position, from about 30º to about 80º. I obviously had to
change my strategy. 
The strategy I used is exemplified in this pair of frames. I paired
faceon frames with downtheline frames at the same position in the
swing. Then I measured the distances in the X, Y, and Z directions,
from the pivot at the base of the neck to the point where the hands
joined on the grip. These distances are labeled dX, dY, and dZ in the frames.
The actual distance r
can be computed by the Pythagorean Theorem extended to three dimensions:
r^{2}
= dX^{2} + dY^{2}
+ dZ^{2}
By picking pairs of frames at 0, .05, .10, .15, and .20 seconds, we can
find r(0),
r(.10),
r(.15),
and r(.20),
and use them as r(t)
for the simulation.
This is the strategy. Of course, the devil is in the details. Here are
some of the imps I had to deal with:
 How
should we measure dX, dY, and dZ? I
identified the shoulder pivot and the hands, in each frame. (They are
the yellow crosses.) Then I measured the horizontal and vertical
positions of the yellow crosses, and entered each into an Excel
spreadsheet. The positions were easy to measure, because I have a
graphics viewer that gives the [x,y]
position of the cursor, in pixels. So I placed the cursor on the center
of each yellow cross, and copied the position from the viewer's status
bar into the spreadsheet.
 What
about the details of the calculations? Since I was
entering the coordinates into a spreadsheet, it was easy to let Excel
do the Pythagorean calculations for me. The results were distances in
pixels, not inches. But that is OK, since we are really interested in a
ratio
of distances. As long as the units are the same for numerator and
denominator, it does
not matter what the units are. Pixels are good enough.
 Is
the scale the same for faceon and downtheline? We have
a way of telling; we measure dZ
in both frames. It turns out I didn't have to do any scaling. dZ was the same to
within 10% (typically closer than 5%), and neither consistently bigger
nor smaller. So I wrote off the difference to measurement error and
made no corrections. (Bear in mind that I visually estimated
the pivot point at the spine between the tops of the shoulders. That
process is probably responsible for most of the error.)
 Which
frames match up? The videos were not made with two
synchronized cameras photographing the same swing. It was the same
camera in different positions for two different swings. They were taken
on the same day and not too many swings apart. Moreover, Rock is a
consistent, lowhandicap golfer, so the swings are probably pretty
similar. But it is worth asking whether it was the same swing, to which
the answer is obviously "no". That makes it fair to ask how close were the swings, and which
frames do we pair together? I compared the arm positions on the frames,
and paired the ones that matched. All the pairs except for the last
(just before impact) were no more than .01 second off. (E.g. the
.11sec faceon frame paired best with the .12sec downtheline frame.)
The last pairing was .02 second off. So I conclude that the swings were
very similar. Certainly similar enough to use for r(t).
Here is the table of r(t)
that came out of the spreadsheet.
Time
(sec) 
r(t)
(pixels) 
.00 
100 
.05 
100 
.10 
115 
.15 
155 
.20 
160 
.25 
165 
I am going to use 160 as R_{o},
because the last value SwingPerfect accepts is at .20 seconds, and
because the error in the measurement is probably about 5 pixels anyway. 
3. Find the range of clubhead speeds and carry distances.
At this point, we just have to do some rote calculation, most with the
aid of computer programs:
 Find the lower and upper bounds of T(s)
by plugging r(t)
into equation (9). We use 53 footpounds for T_{o}, as computed in step 1 above.
 Plug the upper and lower bounds into SwingPerfect to get
the range of clubhead speeds.
 Plug the clubhead speeds into TrajectoWare Drive to get the
range of carry distance.
Here is a table with the results.^{[1]}
Time
(sec) 
r(t)
(pixels) 
Lower
Bound 
Upper
Bound 
R_{o}/r(t) 
T(s) 
[R_{o}/r(t)]^{2} 
T(s) 
.00 
100 
1.60 
85 
2.56 
135 
.05 
100 
1.60 
85 
2.56 
135 
.10 
115 
1.39 
74 
1.93 
102 
.15 
155 
1.03 
55 
1.06 
56 
.20 
160 
1.00 
53 
1.00 
53 
Clubhead speed 
112
mph 
123
mph 
Carry distance 
268
yd 
296
yd 
... And a graphical representation on a TrajectoWare Drive output.
The important thing to note is that Rock's actual Leecommotion swing
falls between the upper and lower bounds of the model, so the model
does reflect reality.
Notes:
[1] The SwingPerfect
program does not accept shoulder torque more than
120 footpounds. The upper bound includes a torque value of 135
footpounds. I plotted the upper bound with 135 replaced by values
between 110 and 120. They fell on a straight line, so it was easy to
extrapolate the result out to 135.
Last modified  Jan 20, 2011
