Leecommotion, the Right-Side Swing

Part 2 - Analysis

Dave Tutelman  --  December 20, 2010

The first article on the right-sided 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 right-side 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 double-pendulum 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:
  1. Start with the reality of a varying moment of inertia exposed to the actual shoulder torque created by the golfer.
  2. Derive a time-varying 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.


Non-Circular Model: Upper Bound

This note in green tells you how to calculate the torque profile for a given non-circular 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:
  • To  = 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.
  • Ro  = 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 Ro late in the downswing.
  • Io  = 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, circular-path 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=KMr2 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 double-pendulum 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:

α  =  T
I		             (1)

First of all, the angular acceleration over time will be given by the equation (1), where the torque is what the body produces To and the moment of inertia is the actual moment of inertia due to the changed lever arm I(t). In other words:

α(t)  =  To
I(t)		             (2a)

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:

α(t)  =  T(t)
Io		             (2b)

Since we are constraining both cases to the same acceleration profile α(t), we can equate (2a) and (2b) to get:

α(t)  =  To
I(t)		  = T(t)
Io

T(t)   =   To Io
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, well-known:

I   =   K M r2

Plugging this into the moments of inertia in equation (3):

T(t)   =   To K M Ro2
K M r2(t)

T(t)   =   To Ro2
r2(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 To in the latter phases of the downswing. And it will, as long as r(t) reaches Ro 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:

ω1I1  =  ω2I2

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 too-optimistic 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 dotted-line 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.


Non-Circular 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:
  • To = 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.
  • Ro  = 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 Ro 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).
  • Fa = 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 double-pendulum model. As before, we want to match the acceleration for the standard case to the acceleration for the variable-radius case. Since we know F=ma, and the mass is the same, we can do it by matching the accelerating force Fa. The derivation goes very similarly to the upper bound.

We know from the diagram above that

F  =  To  /  r(t)                   (5)

Also from the diagram, we can resolve F into components to get

Fa   =   F cos θ(t)   =   To cos θ(t) /  r(t)             (6a)

In order to obtain the same velocities along the curved "track", we need to apply the same Fa when we use the standard double pendulum model. So our torque T(t) for the standard double pendulum has to satisfy

Fa   =   T(t) / Ro                             (6b)

So we have two equations for Fa, (6a) for the curved track and (6b) for the standard double pendulum. Since Fa has to be the same for both, let's set them equal.

Fa   =   To cos θ(t)
r(t)		  = T(t)
Ro

T(t)   =   To   Ro
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)   ≈   To   Ro
r(t)		                        (8)

Very simple and manageable!
Thus our model consists of two torque profiles vs time, Tupper(t) and Tlower(t). One gives an upper bound to clubhead speed and the other a lower bound. Each can be applied to the double-pendulum 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:

Tupper(t)   =   To Ro2
r2(t)		            equation (4)

Tlower(t)   =   To  Ro
r(t)		  cos θ(t)   ≈   To  Ro
r(t)		                    equation (8)

Rewritten as an inequality to reflect the bounding nature of the model:

To Ro
r(t)		    <     T(t)    <    To Ro2
r2(t)		                    equation (9)

The key parameter in all this is the ratio of Ro 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:
  1. Find the shoulder torque To that gives the golfer's carry distance with a conventional swing, 250yd.
    1. Apply the club's physical characteristics to the TrajectoWare Drive computer program to determine the required clubhead speed.
    2. 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).
  2. From detailed videos of the golfer, determine r(t) and from it the ratio Ro/r(t).
  3.  Find the range of carry distances that the Leecommotion swing would give.
    1. Compute the upper and lower bounds on shoulder torque T(t) using the model (equation 9).
    2. Apply the upper and lower bounded shoulder torque T(t) to SwingPerfect and find the range of clubhead speeds.
    3. 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 foot-pounds. So we have:

To  =  53 foot-pounds

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 face-on 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 down-the-line 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 face-on frames with down-the-line 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:

r2  =  dX2 + dY2 + dZ2

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 face-on and down-the-line? 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, low-handicap 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 face-on frame paired best with the .12sec down-the-line 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 Ro, 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:
Here is a table with the results.[1]
Time
(sec)		 r(t)
(pixels)		 Lower Bound Upper Bound
Ro/r(t) T(s) [Ro/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 foot-pounds. The upper bound includes a torque value of 135 foot-pounds. 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